Jetski vs a Motorcycle
Sean
I am switching from jetsking (wave runners - sitdown models) to motorcycling
(don't ask why) and want to get opinions from people who have ridden
jetski's and motorcycles to tell me the difference between the two other
than obvious differences such as water and land. Which one did you enjoy
the most and had the most use and fun out of? I know its probably personal
preference but I would like to hear your preference. I previously owned a
jetski and loved it before I sold it (financially). I never road a motor
bike in my life but hoping that their is a simularity between the two. If
you have an option between motorcycling or jetsking, which would you choose
(honestly)?
Donn Schaefer
On Thu, 15 Jan 1998, Sean wrote:
> I never road a motor
> bike in my life but hoping that their is a simularity between the two. If
> you have an option between motorcycling or jetsking, which would you choose
> (honestly)?
Hahahaha- tough question. Nope.
On a motorbike, the terrain changes. Water on the other hand is pretty
much the same. I find jetsking very dull.
I would pick windsurfing above jetsking. When you wipe out on motorbike,
you will probably be injured.
Peter Johansson
Sean <se...@megsinet.net> wrote:
> I am switching from jetsking (wave runners - sitdown models) to motorcycling
> (don't ask why) and want to get opinions from people who have ridden
> jetski's and motorcycles to tell me the difference between the two other
> than obvious differences such as water and land.
I've only ridden a jet-ski once, but they are *very* different from
motorcycles. The biggest difference is that you don't counntersteer a
jet ski -- and it takes a while to break the habit if you are a MC
rider.
> Which one did you enjoy
> the most and had the most use and fun out of?
It's a totally different experience.
> I never road a motor
> bike in my life but hoping that their is a simularity between the two. If
> you have an option between motorcycling or jetsking, which would you choose
Do yourself a favor and take the Motorcycle Safety Course. Jet skis
handle totally different than motorcycles, and pavement is a lot
harder than water. If you try to ride a MC the way you ride a jet
ski, you *will* go down.
Peter Johansson
pe...@widgetworks.com
andy the pugh
Sean <se...@megsinet.net> wrote:
> I am switching from jetsking (wave runners - sitdown models) to motorcycling
> (don't ask why) and want to get opinions from people who have ridden
> jetski's and motorcycles to tell me the difference between the two other
> than obvious differences such as water and land.
The really obvious difference is steering, Jetskis (which steer the jet)
don't counterseer. Apparently bike riders have terrible trouble with
this.
Bikes are far more versatile than jetskis, you can't cross oceans on a
Jetski, but you can cross continents on a bike.
--
ap
Freesoft
On Thu, 15 Jan 1998 22:58:04 -0600, "Sean" <se...@megsinet.net> wrote:
>I am switching from jetsking (wave runners - sitdown models) to motorcycling
>(don't ask why) and want to get opinions from people who have ridden
>jetski's and motorcycles to tell me the difference between the two other
>than obvious differences such as water and land. Which one did you enjoy
>the most
Bike, you can simple walk out of your house and ride. You don't need
to allot time to load, drive to water, and unload before riding.
Quicker accessability and more riding time with less preparation time
equate to more enjoyment.
>and had the most use
No comparison, the bike is the most useful. A jetski is strictly a
recreational vehicle. The bike will get you to the store and work.
>and fun out of?
They're both fun but, in my opinion, different types of fun. The ski
allows you to get a little crazier, after all, when you fall, you land
in water (with the exception of the occasional log or tree stump).
The bike is going to hurt when you fall. The traffic density on the
water is quite a bit less than the road and IMHO, there seems to be a
slightly lower idiot ratio with water craft than on the road.
> I know its probably personal
>preference but I would like to hear your preference.
You're right, it is a personal preference...
For me, a bike comes slightly after the daily driver and work shop
tools. It positively comes before the water craft, computer, and
kid's education (just kidding).
>I previously owned a
>jetski and loved it before I sold it (financially).
The bike will cost a lot more if you get quality equipment. A good
used or new bike will generally exceeds the cost of a jetski. Leather
protection exceeds the cost of a ski vest and wet suit. A decent
helmet will cost $125 and up.
And since you'll ride it more often, due to accessability, you'll
spend more on maintenance (tires, fluids, cables, shocks, lights,
battery, etc).
Don't forget the insurance expense.
>I never road a motor
>bike in my life but hoping that their is a simularity between the two.
There is a vague similarity between them, very vague...
>If
>you have an option between motorcycling or jetsking, which would you choose
>(honestly)?
I've already chosen, a bike because I love cruising. And a boat for
group fun (skiing, fishing, etc...)
Seriously, check with your local motorcycle shop for an AMF course.
In this area (Texas), the course costs $100. They supply a bike and
instructors. You have to demonstrate the ability to ride a bicycle,
sit through a few hours of classroom, and then get to ride a
motorcycle through basic maneuvers.
If you decide it's not for you, you're out $100 which is a lot cheaper
than the several thousand (minimum) you'll fork out by jumping in and
buying. A cheap experiment.
If, after taking the course, you decide to buy, you would have already
passed most of your license requirements and possibly enjoy a discount
in insurance rates. You would have also learned basic skills which
might save your life.
Mark
97 XLH with way too few miles
76 Camaro, Down to frame, milage is irrelevant
92 S-10 Blazer, daily driver, 70k miles
94 S-10 Pickup, daily driver, 40k miles
Remove :-) from email to reply
Ken Vail
andy the pugh wrote in message
<1d2z2p1.6y...@ra201017.shef.ac.uk>...
>Sean <se...@megsinet.net> wrote:
>
>Bikes are far more versatile than jetskis, you can't cross oceans on a
>Jetski, but you can cross continents on a bike.
I prefer personal watercraft on water, & motorcycles on land.
Bradley V. Stone
--snip--
>
>They're both fun but, in my opinion, different types of fun. The ski
>allows you to get a little crazier, after all, when you fall, you land
>in water (with the exception of the occasional log or tree stump).
>The bike is going to hurt when you fall. The traffic density on the
>water is quite a bit less than the road and IMHO, there seems to be a
>slightly lower idiot ratio with water craft than on the road.
>
You obviously haven't been to Minnesota (land of 10000+ lakes!) on a
beautiful summer day. Idiots galore.
I prefer my motorcycle for the same reasons you stated, though. I
would rather zip past the lakes. A great change of scenery. And if
it's hot, I stop on the side of the road, and jump in...hahaha!!
Bradley V. Stone
bvs...@usa.net
http://prairie.lakes.com/~bvstone/
1992 Yamaha FJ1200
1969 Suzuki T250
"You put oatmeal in the toaster?"
Mike & Trish
Just sold our 2nd ski and bought another bike. I had ridden bikes
for 30+ years and wanted something "different" with the ski's. I had
surfed since the 50's, thought it was an "evolutionary" thing to do.
Had alot of fun, was nearly killed in open ocean however when crashing 7
miles out at 62 mph. It taught me that the ocean was not the safe haven
I had always believed it to be. I had vast experience in the water and
on bikes, decided to buy yet another ski. After awhile, I realized (for
ME!), that the ski's were far more limiting, and work to maintain, as
well as costly. The bike is more immediate and rewarding gratification
without the clean up and flushing hassles.
I find them both exceptionally dangerous, yet both an absolute
blast. I recommend you ride all you can afford, ride with as much fun as
you can, and as safely as you can at the same time. I have survived alot
in life, but I really respect the "odds" in my old age. The bike's have
been MUCH cheaper to own and maintain till now, but I just bought my 1st
Harley........so that "may" change?????? If it does, I will change again
as I need too. Just have fun, and be safe!
The only other thing I would recommend is buy what you really want
if you can. Do NOT worry about stereotypes or prejudices, follow your
heart, your soul will follow........ Good Luck. Mike
Death to Spam
Bradley V. Stone wrote:
> >The bike is going to hurt when you fall. The traffic density on the
> >water is quite a bit less than the road and IMHO, there seems to be a
> >slightly lower idiot ratio with water craft than on the road.
>
> You obviously haven't been to Minnesota (land of 10000+ lakes!) on a
> beautiful summer day. Idiots galore.
Huh? You haven't ridden/driven in Minneapolis or St Paul? There are
"idiots galore" on the roads in those cities. In the summer they crash
into each other, in the winter they crash into each other and everything
else.
> I prefer my motorcycle for the same reasons you stated, though. I
> would rather zip past the lakes. A great change of scenery. And if
> it's hot, I stop on the side of the road, and jump in...hahaha!!
Yep, obviously not from the cities. Mosta the roads on the lakes near
the cities are posted "no parking".
--
David
http://ophelia.fullcoll.edu/~davidd
900SS K100RS ST1100 GS1085 RD400
New opinions are always suspected, and usually opposed, without any
other reason but because they are not already common.
-- John Locke, 'Essay Concerning Human Understanding'
Keith McCammon
Not that I disagree with countersteering on motorcycles, but with skiing it's not
necesarry to turn using that because your skis are unweighted in the transition,
unlike a motorcycle or other craft.
Ian Frechette wrote:
> countersteer any time their center of gravity is to the inside of
> their inside ski. (e.g. Turning left, and leaned left far enough that
> their center of gravity is to the left of their left ski. You have to
> drive your skis back under your center of gravity to straighten up or
> turn the other way. That's countersteering.)
>
> For those who don't believe in countersteering, or kinetic science, I know
> of at least a half dozen different tests you can do yourself to prove it.
Ian Frechette
In article <69oeva$8vo$1...@homer.widgetworks.com>,
>I've only ridden a jet-ski once, but they are *very* different from
>motorcycles. The biggest difference is that you don't counntersteer a
>jet ski -- and it takes a while to break the habit if you are a MC
>rider.
I just had to say something here. Jet skis and boats do in fact
countersteer. The difference is that they do it all the way through
the corner. Also, it's not about the control imputs (you don't push left
to go left), it about what the rider and craft do to make the turn that
defines countersteering IMHO.
When you countersteer a bike you basically drive the front end of the
bike out from under the center of gravity and then the bike starts to
tip over to the inside of the turn and then due to the rake and tire profile
the front tire turns itself back into the turn to try to stabilize or
even right the bike.
Now steer from the back and see what happens.
On a jet ski you turn the bars in the direction of the turn and that redirects
the jet in the back to rotate toward the inside of the turn. This forces the
back end and the bottom of the craft to slide out from under the center of
gravity and you lean into the turn. The flared sides of a boat tend to make
it want to float upright again, and the jet/prop is constantly trying to
tip it over and turn the boat at the same time. All these forces balance
so that in a good boat or jet ski most of the forces of turning are felt in
and up and down direction (relative to the boat) and not side to side.
BTW, not all boats countersteer, but any that leans enough to balance the
cornering force with gravity does. If you can stand up while cornering
without getting thrown to the outside of the turn then it's countersteering.
Obviously a jet ski falls into this category.
If you think about it, any combination of human and craft that must
balance in a corner to stay upright, must countersteer. That means,
motorcycles, bicycles, unicycles, some boats, skateboards, snowboards,
and any monoski (water or snow). A regular skiier is forced to
countersteer any time their center of gravity is to the inside of
their inside ski. (e.g. Turning left, and leaned left far enough that
their center of gravity is to the left of their left ski. You have to
drive your skis back under your center of gravity to straighten up or
turn the other way. That's countersteering.)
For those who don't believe in countersteering, or kinetic science, I know
of at least a half dozen different tests you can do yourself to prove it.
ian
Stan Malyshev
In article <69u5ut$n...@peabody.colorado.edu>,
Ian Frechette <frec...@rintintin.Colorado.EDU> wrote:
[]
>When you countersteer a bike you basically drive the front end of the
>bike out from under the center of gravity and then the bike starts to
>tip over to the inside of the turn and then due to the rake and tire profile
>the front tire turns itself back into the turn to try to stabilize or
>even right the bike.
No. Countersteering is a gyroscopic thing.
>If you think about it, any combination of human and craft that must
>balance in a corner to stay upright, must countersteer.
Humans and craft will retain their properties regardless of what I think.
In any case, your notions are, as Janice would say, 100% fully inferior.
Dredhead
Stan Malyshev wrote:
>
> Humans and craft will retain their properties regardless of what I think.
> In any case, your notions are, as Janice would say, 100% fully inferior.
Stan, your tone when writing about Janice is almost reverential.
You're scaring me.
--
dredhead AT enteract DOT com
1974 CB750 /1975 CB360T/196? CA95
SOHC/4 #484 DoD #33-1/3 HRMC #146
Sparky The Penguin rides with me.
Z?
Windy
On Mon, 19 Jan 1998 05:03:01 -0600, Dredhead
<dred...@lookinmysig.forreply> wrote:
>
>You're scaring me.
>
You don't keep mirrors in the house then?
--
~*~*~*~* " W I N D Y" *~*~*~*~
To Love Life You Have To Live It!
NGG #13 - unlucky for some
Tactician & Incendiary Devices
http://www.ziplink.net/~holm/ngg/ngg.html
Dan Nitschke
Dredhead wrote:
>
> Stan Malyshev wrote:
> >
> > Humans and craft will retain their properties regardless of what I think.
> > In any case, your notions are, as Janice would say, 100% fully inferior.
>
> Stan, your tone when writing about Janice is almost reverential.
Almost as if Stan were part of Teh Janice Colletcivev (tm).
> You're scaring me.
You're only one 'r' away from correct.
/* dan: THE Anti-Ged -- Ignorant Yank (tm) #1, none-%er #7 */
Dan Nitschke { peDA...@best.com } el...@redbrick.com
..:..:..:..:..:..:..:..:..:..:..:..:..:..:..:..:..:..:.
He wears a disguise to look like human guys, but he's
not a man, he's a Chicken Boo. -- Animaniacs
Ian Frechette
In article <69v6jn$g...@scam.XCF.Berkeley.EDU>,
>No. Countersteering is a gyroscopic thing.
No.. Countersteering is a balance thing. Gyroscopic forces do nothing
but stabilize the bike (if we're talking bikes). Countersteering is about
driving the bottom of your balanced craft to the right to initate a left
turn, or driving it to the left to initiate a right turn. Getting
your center of gravity falling to the inside of the turn without having
to change your shape. Leaning off the side of a bike to turn without
countersteering, changes the combined shape of the rider and bike which
moves the center of gravity without having to drive the tires out from
under the center of gravity. The problem is, that method is SLOW.
Now if you want to argue that countersteering is the ACT of turning the
handle bars to the right to initiate a left turn, or vise versa, I
can live with that opinion, but knowing that other balancing craft
also countersteer by my expanded definition can be quite useful. Learning
to snowboard, for instance. Riding a unicycle for another.
>Humans and craft will retain their properties regardless of what I think.
>In any case, your notions are, as Janice would say, 100% fully inferior.
To each his own, I suppose. The difference is, I can prove what I say
is true, quite easily with a few examples.
Have you ever seen or ridden a ski-bob? It rides just like a
motorcycle or bicycle, including countersteering, and it has no
rotating parts. (It's a sitdown snow cycle, basically) There is actually
a water cycle (not a jet ski) that has a ski in front which allows it to
balance and turn just like a motorcycle but also with no rotating parts.
How about the original push scooter (stand up, 4 inch diameter tires, etc)
It countersteers, just like a bicycle or motorcycle. Prove to me that those
4 inch, 16 oz tires are exerting any measurable gyroscopic forces on the
bike/rider. I say there is none, yet they are quite manueverable without
having to lean off the side, to turn. I've seen scooters with wheels as
small as those on rollerblades and they operate the same.
Not to make it sound overly simplistic, but all you have to do to understand
countersteering is balance a broomstick on your finger. To force it to tip
left you *must* move the bottom of the stick to the right. The same is
true of any other balanced object with a relatively fixed shape.
What happens after the lean is initiated depends entirely on the craft. Some
will right themselves, some will continue to fall unless corrected by the
operator.
Instead of chanting the "gyroscopic forces" mantra, do some experimentation
of your own. I can prove that countersteering is possible without gyroscopic
forces being necessary. That is not to say they don't play a role, I just
don't see that they play an important role.
ian
Stan Malyshev
In article <6a30ou$n...@peabody.colorado.edu>,
Ian Frechette <frec...@rintintin.Colorado.EDU> wrote:
>In article <69v6jn$g...@scam.XCF.Berkeley.EDU>,
>>No. Countersteering is a gyroscopic thing.
>
>No.. Countersteering is a balance thing. Gyroscopic forces do nothing
>but stabilize the bike (if we're talking bikes). Countersteering is about
Gyroscopic forces are fierce. Try flicking a bike at 150 - you are unlikely
to have the strength.
Countersteering messes with the axis of rotation, which causes the plane
of rotation to tilt very forcefully. Turning the axis to the left causes
a tilt to the right, and vice versa.
>Instead of chanting the "gyroscopic forces" mantra, do some experimentation
>of your own.
I prefer to get my information from motorcycle chassis design literature.
If you care to partake of the clue repository, look at
http://www.eurospares.com. Specifically, look at the list of related
literature, and the chassis design list archives.
Trevor Dennis
Dan Nitschke <peda...@best.com> said
>Almost as if Stan were part of Teh Janice Colletcivev (tm).
If he starts using bulleted indents he's our man.
--
Trevor Dennis /`\ .(o~)-(o~). /`\ tre...@tdennnis.demon.co.uk
The Polite Brit / , \( _______ )/ , \ tden...@ford.com
OGH #1 ___/ /_\ /`"-------"`\ /_\ \___ Southern England
jgs`~//^\~_`\ <__ __> /`_~/^\\~`
`~//^\\~`~//^\\~`
Christin Swider
Riding a jetski is like doing circles in a parking lot.
Motorcycling is more like skiing. You have to have some skill and an
understanding that screwing up will kill you.
Your bike won't slow down and start riding in a circle to come get you when you
fall off either.
Dave
Team K
Sean wrote:
> I am switching from jetsking (wave runners - sitdown models) to motorcycling
> (don't ask why) and want to get opinions from people who have ridden
> jetski's and motorcycles to tell me the difference between the two other
> than obvious differences such as water and land. Which one did you enjoy
> the most and had the most use and fun out of? I know its probably personal
> preference but I would like to hear your preference. I previously owned a
> jetski and loved it before I sold it (financially). I never road a motor
> bike in my life but hoping that their is a simularity between the two. If
Keith McCammon
Stan can you define what a 'gyroscopic thing' means exactly. What is generating
this gyroscopic force that you speak of?
Stan Malyshev wrote:
> No. Countersteering is a gyroscopic thing.
>
> >If you think about it, any combination of human and craft that must
> >balance in a corner to stay upright, must countersteer.
>
Dredhead
Keith McCammon wrote:
>
> Stan can you define what a 'gyroscopic thing' means exactly. What is generating
> this gyroscopic force that you speak of?
>
Dan Nitschke
Dredhead wrote:
>
> Keith McCammon wrote:
> >
> > Stan can you define what a 'gyroscopic thing' means exactly. What is generating
> > this gyroscopic force that you speak of?
> >
> Wheels. Spinning.
So far, so good.
> Around an axel.
Axel Rose? A triple axel? Axel Foley? I doubr any of those would
be a good substitute for an axle.
But then, as you know, I'm not very mechanically declined.
/* dan: THE Anti-Ged -- Ignorant Yank (tm) #1, none-%er #7 */
Dan Nitschke : peDA...@best.com : nits...@redbrick.com
()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()()
It seems to me as though I've been upon this stage before, and
juggled away the night for the same old crowd. -- Al Stewart
Ian Frechette
In article <6a34v0$r...@scam.XCF.Berkeley.EDU>,
>Gyroscopic forces are fierce. Try flicking a bike at 150 - you are unlikely
>to have the strength.
So what you're saying is that when gyroscopic forces do come into play
they do more to forces oppose effective countersteering than anything else.
I've said as much in previuos discussions on this issue.
And I think I recently said that gyroscopic forces do *more* to stabilize
the bike, keeping it in whatever attitude it happens to be in.
Most people don't spend much time at 150 though. My bike flicks quite nicely
at lower speeds.. 20-40mph where gryoscopics forces are not so significant.
At speeds much lower than that, the bike can fall over faster than I can
drive the tires back under it, so flickabily is limited, although something
smaller like a bicycle or scooter has no problems until I'm going less than
5mph.
>Countersteering messes with the axis of rotation, which causes the plane
>of rotation to tilt very forcefully. Turning the axis to the left causes
>a tilt to the right, and vice versa.
I agree that this is true, but only at much higher speeds, and it does
not negate what I've said. The effect I described is still happening, but
the amount of head rotation you need to drive the tires out from under
the bike at high speeds is very little, and you're fighting gyroscopic
forces to move it at all, as you say.
>>Instead of chanting the "gyroscopic forces" mantra, do some experimentation
>>of your own.
>
>I prefer to get my information from motorcycle chassis design literature.
>If you care to partake of the clue repository, look at
>http://www.eurospares.com. Specifically, look at the list of related
>literature, and the chassis design list archives.
Where the first thing the guy says is..
"I am not an engineer; these articles are based on my own experience, the
experience of a friend of mine, and
whatever information I have been able to glean from various books,
magazines, etc. "
The only difference between him and myself is that I am an engineer, albeit
not a mechanical one.
Oddly, that set of articles doesn't say anything about gyroscopic
forces or countersteering and what role they play in chassis design. It
has comments like:
"wheelbase: longer = stable shorter = maneuverable"
or
"rake and trail: The trend is toward steeper rake and shorter trail"
Both true, but no explaination of why.
What amazes me is how many people are willing to sit here and waste hours
trying to convince other people what they've heard other people say is
true, when they could spend some time on their own motorcycle figuring it
out for themselves.
If you choose to believe that countersteering = gyroscopic forces alone,
then I suppose I could call my effect something different, but what would
be the point? It behaves the same for all "reasonable" speeds, and has
the added benefit of explaining how other craft with little or no gyroscopic
forces are able to countersteer.
Fundamentally I believe countersteering to be the act of initiating a turn
on a balancing vehicle where the goal is to lean in a specific direction
faster than gravity can pull you over. Gyroscopic forces can help do that
but driving the contact patch back and forth under the center of gravity
to roll a 500 pound vehicle like a motorcycle is much more effective, and
works at speeds where gyroscopic forces are insignficant.
ian
Dredhead
Dan Nitschke wrote:
>
> Dredhead wrote:
> >
> > Keith McCammon wrote:
> > >
> > > Stan can you define what a 'gyroscopic thing' means exactly. What is generating
> > > this gyroscopic force that you speak of?
> > >
> > Wheels. Spinning.
>
> So far, so good.
>
> > Around an axel.
>
> Axel Rose? A triple axel? Axel Foley? I doubr any of those would
> be a good substitute for an axle.
>
one of the '57 Spagthorpes?
Keith McCammon
I think even if you had zero gyroscopic force, like no wheels, countersteering
would work the same way.
Imagine if you had a vehicle like a bicycle without wheels, but with skis instead.
Zero gyroscopic force, yet it would countersteer the same way.
Gyroscopic force is one force you are overcoming when you use countersteering to
steer, it is not a contributor to countersteering.
Rather, I think centrifugal force acts on the entire vehicle, with the contact
patch as the axis of this rotation. And of course the centrifugal force gets
proportionately higher as you go further from the axis of rotation.
Think about what would happen if you located most of the mass of the vehicle below
the contact patch (underground!)
Countersteering would not work!
Keith McCammon
Actually, unless your name is Bono or Kennedy, skiing is a pretty safe sport in
regard to fatalities compared to others like bicycling.
Trevor Dennis
> >Gyroscopic forces are fierce. Try flicking a bike at 150 - you are unlikely
> >to have the strength.
>
> So what you're saying is that when gyroscopic forces do come into play
> they do more to forces oppose effective countersteering than anything else.
> I've said as much in previuos discussions on this issue.
He is getting confused between gyroscopic "inertia" and
gyroscopic "precession". Two quite separate forces.
Dan Nitschke
Trevor Dennis wrote:
>
> Ian Frechette wrote:
>
> > In article <6a34v0$r...@scam.XCF.Berkeley.EDU>,
>
> > >Gyroscopic forces are fierce. Try flicking a bike at 150 - you are unlikely
> > >to have the strength.
> >
> > So what you're saying is that when gyroscopic forces do come into play
> > they do more to forces oppose effective countersteering than anything else.
> > I've said as much in previuos discussions on this issue.
>
> He is getting confused between gyroscopic "inertia" and
> gyroscopic "precession". Two quite separate forces.
And you're _completely_ overlooking the effects of peristalsis.
/* dan: THE Anti-Ged -- Ignorant Yank (tm) #1, none-%er #7 */
Dan Nitschke ! peDA...@best.com ! nits...@redbrick.com
--------------------------------------------------------
Dressing up in costumes; playing silly games. Hiding out
in treetops, and shouting out rude names. -Peter Gabriel
Brian McLaughlin
frec...@rintintin.Colorado.EDU (Ian Frechette) wrote:
>In article <69v6jn$g...@scam.XCF.Berkeley.EDU>,
>>No. Countersteering is a gyroscopic thing.
>No.. Countersteering is a balance thing. Gyroscopic forces do nothing
>but stabilize the bike (if we're talking bikes). Countersteering is about
Stability of a 2 wheeled single track vehicle comes from trail, not
the rotating wheels. Bicycles have been made with counter rotating
wheels to cancel the angular momentum of the wheels, stability was not
affected.
--
Brian
TZ250E (1993-95) 2 strokes smoke,
FZR600 (1990) 4 strokes choke!
R1100RTA (1997)
Brian McLaughlin
Dredhead <dred...@lookinmysig.forreply> wrote:
>Keith McCammon wrote:
>>
>> Stan can you define what a 'gyroscopic thing' means exactly. What is generating
>> this gyroscopic force that you speak of?
>>
>Wheels. Spinning. Around an axel.
Rotating wheels have angular momentum. Objects cannot have or possess
force. I have as of yet to even find the term 'gyroscopic force' in a
physics text, it is not in any of mine. These forces are the result of
torques and conservation of angular momentum.
Brian McLaughlin
Keith McCammon <kei...@mothership.org> wrote:
>I think even if you had zero gyroscopic force, like no wheels, countersteering
>would work the same way.
>Imagine if you had a vehicle like a bicycle without wheels, but with skis instead.
>Zero gyroscopic force, yet it would countersteer the same way.
Rotating wheels do not have gyroscopic forces either (whatever
gyroscopic force is).
>Gyroscopic force is one force you are overcoming when you use countersteering to
>steer, it is not a contributor to countersteering.
Check out the directions of the torques (from moving the handlebars)
and the reaction of the front wheel to the applied torque. I do not
think it is one effect or the other but a combination of both.
>Rather, I think centrifugal force acts on the entire vehicle, with the contact
>patch as the axis of this rotation. And of course the centrifugal force gets
>proportionately higher as you go further from the axis of rotation.
No such thing as centrifugal force, it seems to exist to an observer
in an accelerated frame of reference. It is a ficticious force.
>Think about what would happen if you located most of the mass of the vehicle below
>the contact patch (underground!)
>Countersteering would not work!
Are you sure? Have you tried it?
Dan Nitschke
Ged Martin wrote:
>
> On Thu, 22 Jan 1998 08:19:32 -0800, Dan Nitschke scribbled:
>
> >And you're _completely_ overlooking the effects of peristalsis.
>
Well, *you're* overstepping the bounds of podiatry.
/* dan: THE Anti-Ged -- Ignorant Yank (tm) #1, none-%er #7 */
Dan Nitschke -- peDA...@best.com -- nits...@redbrick.com
o0o0o0o0o0o0o0o0o0o0o0o0o0o0o0o0o0o0o0o0o0o0o0o0o0o0o0o0o0
Upon my every statement you can have complete reliance. I
know more than you do; call me Dr. Science! -- Dr. Science
AARON WARD
Please please please! Could someone actually do the math/physics and
calculate what kind of angular momentum a typical tire would have? I
mean, those tires are damn heavy and spinning that fast? Shoot, I know
that bicycle tires can get awful ornery just by spinning them by hand.
Here's what I KNOW about gyros:
A spinning bicycle tire, when hung by a string on one axle, will not
simply fall flat. It will instead revolve around the string like a damn
oscillating fan going in a circle! It remains, for the most part,
perpendicular to the ground while it spins! Amazing!
Supporting a bicycle and spinning the front tire causes funny things to
happen. When the bike is leaned over on it's right side by pushing on
the frame, the front tire and handlebars swing to the left. When the
handlebars and front tire are swung to the left, the frame leans over on
its right side. Amazing! Must be that damn gyroscopic precession.
I think this all has something to do with countersteering...
--
*******************************************************************
Rider Aaron 1987 CBR600F "Hurricane"
"Who is the Great Cornholio, and why is he taking my toilet paper?"
Brian McLaughlin
da...@bnr.ca (Dan Kuryliak) wrote:
>In article <6a7u1k$r...@wiscnews.wiscnet.net>, bmcla...@waukesha.tec.wi.us (Brian McLaughlin) writes:
>|> Dredhead <dred...@lookinmysig.forreply> wrote:
>|>
>|> >Keith McCammon wrote:
>|> >>
>|> >> Stan can you define what a 'gyroscopic thing' means exactly. What is generating
>|> >> this gyroscopic force that you speak of?
>|> >>
>|> >Wheels. Spinning. Around an axel.
>|>
>|> Rotating wheels have angular momentum. Objects cannot have or possess
>|> force. I have as of yet to even find the term 'gyroscopic force' in a
>|> physics text, it is not in any of mine. These forces are the result of
>|> torques and conservation of angular momentum.
> The way I understand it:
> Above speeds of about 20 to 30 km/h (15-20 mph), the
> motorcycle's wheels act like gyroscopes. These
> spinning masses of wheel and tire combine to create a
> gyroscopic force. If you've ever held a spinning
> bicycle wheel in your hand and felt how difficult it
> is to "steer" it, you'll understand.
I have a wheel that I use when I do demonstrations for my physics
classes. Reread the posts. The term 'gyroscopic forces' is a poor
term, it is not used in physics.
> If you have ever played with a spinning gyroscope, you'll remember
> that pushing forward on the left end of the axle didn't result in
> the wheel turning to the right. Instead a force called gyroscopic precession,
> redirects the force 90 degrees in the direction of rotation.
> The wheel actually leaned to the left, rotating on an axis
> around the center of the wheel. A motorcycle wheel works
> the same way.
What you describe is based on the definition of torque (torque =
dl/dt). When I apply a torque to an object, it results in a *change*
in angular momentum. The direction of the torque and the direction of
the *change* in angular momentum are the same. Usually the term
precession is used if the torque is applied contiuously, as in the
precession of a spinning top. Precession is the result of torque and
the vector nature of angular momentum.
andy the pugh
Keith McCammon <kei...@mothership.org> wrote:
> Think about what would happen if you located most of the mass of the
> vehicle below the contact patch (underground!)
>
> Countersteering would not work!
>
More to the point, you wouldn't want it to, as in such a hypothetical
situation you would need to lean out of the corner.
Of course, if leaned out of the corner you would have no conventional
cornering force, so would need huge amounts of positive steering input.
The moral is clear, don't add more negative mass matter than the mass of
the vehicle to your bike.
Of course, negative mass makes a really neat propulsion system.
--
ap
Death to Spam
AARON WARD wrote:
>
>
> Here's what I KNOW about gyros:
>
> A spinning bicycle tire, when hung by a string on one axle, will not
> simply fall flat. It will instead revolve around the string like a damn
> oscillating fan going in a circle! It remains, for the most part,
> perpendicular to the ground while it spins! Amazing!
>
> Supporting a bicycle and spinning the front tire causes funny things to
> happen. When the bike is leaned over on it's right side by pushing on
> the frame, the front tire and handlebars swing to the left. When the
> handlebars and front tire are swung to the left, the frame leans over on
> its right side. Amazing! Must be that damn gyroscopic precession.
>
> I think this all has something to do with countersteering...
Aha! Another precessionist. Only true believers are allowed to join
the First Church of the Immaculate Precession. Aaron, are you ready to
follow the guiding force?
Times have been hard on The Church, most of our congregation has
precessed off the roads lately. Those doing it on mountain roads have
gone on to join that great winged gyro in the sky. We welcome your
contributions and please do have a current life insurance policy with
The Church named as beneficiary.
Your mission will be to ride your bicycle around and get the True Word
of Precession to the masses...
--
David
http://ophelia.fullcoll.edu/~davidd
900SS K100RS ST1100 GS1085 RD400
New opinions are always suspected, and usually opposed, without any
other reason but because they are not already common.
-- John Locke, 'Essay Concerning Human Understanding'
Richard H. Clark
> AARON WARD wrote:
> >
> >
> > Here's what I KNOW about gyros:
> >
> > A spinning bicycle tire, when hung by a string on one axle, will not
> > simply fall flat. It will instead revolve around the string like a damn
> > oscillating fan going in a circle! It remains, for the most part,
> > perpendicular to the ground while it spins! Amazing!
> >
> > Supporting a bicycle and spinning the front tire causes funny things to
> > happen. When the bike is leaned over on it's right side by pushing on
> > the frame, the front tire and handlebars swing to the left. When the
> > handlebars and front tire are swung to the left, the frame leans over on
> > its right side. Amazing! Must be that damn gyroscopic precession.
> >
Ah... But now what happens if you spin that front tire backwards?
Could this be the reason they don't put reverse on bikes?
> > I think this all has something to do with countersteering...
>
And, the driveshaft, don't forget... Shafties don't countersteer,
rather, they've got that extra gizmo in the gearbox that makes
them *seem* like they countersteer...
Richard
Brian McLaughlin
wa...@nevada.edu (AARON WARD) wrote:
>Please please please! Could someone actually do the math/physics and
>calculate what kind of angular momentum a typical tire would have? I
>mean, those tires are damn heavy and spinning that fast? Shoot, I know
>that bicycle tires can get awful ornery just by spinning them by hand.
Calculations are left as an exercise for the reader.
Ian Frechette
In article <6aaml8$2n9$1...@news.nevada.edu>, AARON WARD <wa...@nevada.edu> wrote:
>
>Please please please! Could someone actually do the math/physics and
>calculate what kind of angular momentum a typical tire would have? I
Ok, math and physics, and a little engineering coming up.
>mean, those tires are damn heavy and spinning that fast? Shoot, I know
>that bicycle tires can get awful ornery just by spinning them by hand.
Guess what, it's not really about how fast it spins or how heavy it is.
Ok, here are some interesting things to look at.
I went in search of information on gyroscopes and conservation of
momentum and torque and so on. A couple things came out of this that
are somewhat important to this whole thread.
As someone pointed out, precession only comes into play when a force is
applied continuously, although but that is relevant to this discussion.
Next, we all know by now that when you apply a torque to a spinning wheel,
the wheel pivots 90 degrees from where you applied the torque.
To see this pick up one of your AOL CD coasters and spin it rapidly around
a pen held horizontally with the disk in a vertical orientation. Press
gently, sideways near the top and it'll react as though you pressed 90
degrees around the disc in the direction of rotation. Easy.
Note how fast it reacts. That's the frequency of precesssion. If it
continued spinning the same speed, and you continued applying your pressure
and it was free to pivot how it wanted, it would continue precessing in
the same direction 90 degrees forward of where you applied your torque.
If you tried to stop that precession, it would apply pressure to whatever
you used to stop it. It's applying a torque around the axis of precession.
But here's the catch. It can't give back more torque than you put in.
So if you take a spinning disk and apply one foot pound of torque, it is
simply conserved and given back 90 degrees forward on the disk as one
foot pound of torque. In addition the formula
frequency of precession = (Input Torque) / ((moment of inertia) (rotational freq))
Shows that the frequency of prececession, is inversely proportional to
the moment of inertia AND the rotational frequency.
This means that as you spin the disk faster, and apply the same torque to
to try to push it off it's axis of rotation, it precesses slower.
But there's no torque multiplier. What you put in, you get back and no more.
So, what the hell does all this mean to a motorcycle?
Let's ignore the effects of the rear wheel for a moment and concentrate
on the front wheel and the handlbars as steering input. Spin the front
wheel fast, so that we know there's enough energy in the wheel
to translate 100% of our input torqee (steering input) into what I'll
call precession torque (rotation of the bike around it's long axis).
You'll quickly notice two things.
1. To rotate the bike from a still position you must overcome the
moment of intertia of the bike itself. We're talking a 500lb bike, plus
say 200 lb rider, with a center of gravity somewhere above the front axle.
This is quite a lot of inertia to overcome. Let me ellaborate with
thought experiment.
First, attach a 200lb weight to the seat to represent the rider.
Now find the center of gravity of the bike and run a rod right through
the center down the long axis of the bike. I see it going between the
front forks, through the top of the engine, and coming out the back
somewhere above the rear axle. Lift both ends of this rod so the bike
is fully suspended.
The bike should now be free to spin around its long axis and it should be
balanced so that it can remain still in in any orientation.
Now simply attach the handlebars to one end of this rod. Now orient the
bike so that it's leaning about 45 degrees to the left. Grab the handlebars
and flick the bike 90 degrees to the right and stop it so that it's leaning
45 degrees to the right. Note, gravity plays no role here because it's
balanced around it's center of mass.
You should be thinking "flick???!!!" about now? It should be obvious to
anyone that flicking 700 pounds worth of mass is not possible
through direct input.
So back to the precessional torque. Remember what I said, you only get
back as much torque as you put in. So now remove the 200lb weight, climb
up on the motorcycle, have someone spin the front wheel really fast,
and now by using the handlbars only, FLICK the bike 90 degrees from one
side to the other. What's changed? It's the same handlebars. The bike
plus rider is the same weight, it's more or less balanced, and you've
applied X ft/lbs of torque to the bars and that has been converted into X
ft/lbs precessional torque at the axle of the front wheel, which we'll
pretend is close enough to the center of gravity to mean that the axis of
precession is lined up with the long axis of the bike. You only get out
as much torqe as you put in, so the answer is, nothing. Nothing has
changed, you're still trying to manhandle 700 lbs as though it was directly
connected to your handlebars.
That's the "ideal" situation. No contact with the road, and ignoring
the frequency of precession. And 100% of input torque being translated
90 degrees into precession torque along the long axis of the bike.
so.. what was that second thing?
2. The faster the front wheel spins the slower the frequency of precession.
This means that if the wheels spins twice as fast, the frequency of
precession decreases by 1/2. About now the precessionists are saying
"I told you so.. ", but we'll find that this doesn't really help
anyone.
Ok, so put this all together. If conservation of momentum, and torque
were responsible for leaning a bike at all speeds, then all other
influences aside it would feel as though you had to overcome the entire
rotational interia of the bike everytime you tried to initate a turn, and then
STOP the rotational momentum of the bike, to stop and hold a given lean angle.
I know this is not the case, as it's quite possible to flick a bike from one
side to the other in a very short time, using the pressure from only a couple
fingers. I've tested it myself by coasting down a mountain road with my
hands braced against the tank applying only a slight pressure with my thumbs
to the bars to initiate and hold a turn, and even to quickly lean from
one side to the other.
Ok, but what about point 2? The precessionists were just about to tell me
that point 2 is totally consistent with what they feel when they're
going 150 mph. Well, I started out thinking that the precession
frequency would be quite slow at low speeds and even lower at high
speeds, but I was surprised by what I found. Let's start with the
150mph example, and plug it into the equation from above.
We'll start with a few assumptions.
The handlebars are 2 feet long front end to end.
The front tire and wheel weigh 20 lbs and for maximum effect we'll
say that the mass is concentrated at a distance of 12 inches from the axis
of rotation, in an ideal ring with a moment of inertia of MR^2
With this 2 foot diameter tire, the tire turns 35 revolutions per second.
We'll apply a force to the handlbars of X lbs (yes.. english units)
X lb * 32 ft/sec^2 * 1 ft
precession freq = -------------------------- = .045*X rps or 16X degrees/sec
20 lb * 1 ft^2 * 35/sec
So if you apply 1 lb of force the wheel will WANT to precess only 16 degrees
in one second. 2lbs.. 32 degrees.. and so on. Remember this is not
a force multiplier. Applying 10lbs of force doesnt mean that the bike
will rotate 160 degrees per second, it means that nearly 100% of the
applied torque is translated into precessional torque and you're
trying to lean 700lbs of bike directly through the handlebars. At this
point you're fighting the full mass of the bike any time you try to move
the bars at all. Yes, gyroscopic effects are felt at this speed, but
they are not helping you countersteer at all.
I was actually surprised at how high the precession frequency was at
this speed. I expected it to be lower, now watch what happens at 50mph.
X lb * 32 ft/sec^2 * 1 ft
precession freq = -------------------------- = .135*X rps or 48.6X degrees/sec
20 lb * 1 ft^2 * 11/sec
That means that 1 lb of pressure results in a precession rate of 48.6
degrees per second. 2 lbs of pressure and we're talking 97 degrees per
second. That sounds wonderful doesn't it? Truly snappy. But once again
a gyroscope is NOT a force multiplier. A free floating gyroscope of this
size and speed WANTS to precess at 97 degrees per second, but it's got
to overcome the rotational inertia of the bike and rider. Actually,
it doesn't.. You do, and you're not going to do it with 2 lbs of pressure.
So what happens to a gyroscope if it can't precess as fast as it wants?
It rotates around the axis of the original applied torque. For the
motorcycle, it means that if you apply 2 lbs of pressure and the
bike only only precesses at 30 degrees per second instead of 97, then
the handlebars will actually turn in the direction of pressure. All
that fancy math and the damn thing is still outracking.
As you go slower, the precession frequency increases even further and
the the gyroscope becomes even less effective at translating your
applied torque into precessional torque. A motorcycle can be
countersteered quite easily at 25mph, but the frequency of precession
has increased to 97X degrees/sec. But it's now taking even more force
to translate any useful torque into the lean of the bike, and now
clearly most of the applied torque is going directly into
rotating the handlbars. You can look at the handlbars when you apply
a pressure to induce countersteering and see it turning first in
the direction of torque and then turning back into the lean. An ideal
gyroscope would allow no rotation of the handbars as all the torque would
be translated 90 degrees into lean.
So matter what else you think, it's been proven that outtracking is
happening. That's convenient because otherwise the bike would be
forced to rotate around it's contact patches instead of its center
of mass.
Plus, if a bike did lean in this fashion without outtracking, it would
not feel balanced at all. If you initiated a lean to the left you'd
feel as though the top of the bike and the seat were moving horizontally
to the left as the bike pivoted about the contact patches. We all know
(at least anyone who's ridden a bike) that this is not the case. You
feel the force of rotation as you snap the bike from side to side, but
overall you always feel as though you are still directly above the bike
and "down" is toward the seat.
So whether you like it or not, outtracking is occuring. One might argue
that the lean is induced entirely by gyroscopic effects, and that
yes, the outtracking is occuring but it just "happens" to match
the angle that the handlbars are allowed to rotate because of the
slight loss of effectiveness in conserving torque. In other words, because
not 100% of the torque is translated 90 degrees, and some does turn
the wheel, that amount is just enough to allow the bike to outtrack
so that precession can rotate the bike around the precession axis, close
to the center of gravity and long axis of the bike..
Sounds complicated and just a little too coincidental doesn't it?
Occam's Razor paraphrased, states that the simpler explaination is
most often the correct one. Luckly there is a much simpler explaination
which explains the observed effects.
That explanation is that at speeds where the percentage of conserved
torque is low (which I claim is all but the highest speeds), it is possible
to turn the handlbars far enough to literally drive the contact patch
out from under the center of gravity of the bike, and induce a lean that
is as fast or faster than the rate of precession. Because the bike is balanced
and there's nothing holding the top, the bike simply rotates around its
own center of gravity (CG of bike plus rider actually).
The rotation is stopped in the same way that it was started, by driving
the tires back under the center of gravity. The nice thing about the second
part is that you don't even have to do it. The rake and trail of the fork
and the tire profile all conspire to make the front tire want to turn into
the lean. On most bikes a person has to maintain countersteering pressure
throughout a sharp turn to keep the bike from righting itself. It is possible
to steepen the rake or choose a different tire profile that'll keep the bike
stable in a turn with minimal correction by the rider but you also have to
do more work to turn back the other way.
If you look at this way, leaning the bike faster than gyroscopic precession,
with minimal input torque, then what conservation of angular momentum does
exist, actually works opposite countersteering. The bike is forced to
lean, and that lean actually causes gyroscopic precession that forces
the front tire to turn in the direction of the lean, thus driving the tire
back under the bike. This is a stabilizing influence. It's easier for
gyroscopic precession to influence the mass of the front tire/wheel/forks
and head, than for the spinning wheel to influence the whole rest of the
bike as the precessionists would have us believe.
I'm pretty sure this last effect is actually what's going on when you
try to steer the bike by leaning off alone. Arguments have been made that
because it's hard to turn when you lean off, that you're feeling the gyroscopic
forces "holding you up". I contend that what's really happening is that a
small bit of bike lean, CAN make the front wheel precess into the lean
quick enough (only a degree or so is needed) to drive the front tire
back under the CG thus keeping you upright and not turning very fast.
So if outtracking is so effective, who or what is doing all the work? Gravity
and the friction of the contact patch against the ground. With a small
deflection of the bars, a large sideways force is excerted on the contact
patch which translates directly into rotational force.
And I'd like to expand on that some. Some people seam to have this
impression of outracking which would have all us outtrackers driving
off the edge of the road at the slightest whim. But look at it at speed.
At 50mph I can generate 45 degrees of rotation along the long axis of the
bike, or just over a foot of outtracking in less than a second with only
1 degree of rotation of the handlebars. Of course as soon as you start to
lean the bike starts to carve a curved arc on the ground, so what you
see of the outracking over the 73 feet it takes to cover that second, is
not a full foot, but the horizontal distance from the center of gravity to
a point above the contact patch is. To do the same thing
in half the distance increase the bar rotation to only 1.6 degrees.
1/2 second, 1 foot of outracking, in 35 feet at 50mph.
It's fast, it takes very little effort and it even works on vehicles
that don't have wheels, of which I can name many, and have before.
The biggest problem with precessionism is that it is not a force
multiplier and that bikes can be leaned much faster and with less
effort than what can be achieved with a direct connection between
the handlbars and the long axis of the bike, which is what a perfect
gyroscope would give you.
ian
P.S. In case you don't think the bike is a handlful, here are a couple eqs
for the moment of inertia of some shapes to model the bike as.
How about a flat slab rotating around a long axis with a width of a.
I=1/2Ma^2 Turn it 90 degrees.. make a=3 feet (height of bike)
I = 1/2*700*9 ft^2*lb = 3150 ft^2*lb
and torque = d(freq)*I/dt
Let's say we want to increase rotational freq from 0 to 30 degrees / sec
or 0.08333 rps inside 1/2 sec. so dt = 1/2 and d(freq) = .0833 rps
and I = 3150 ft^2*lb..
We get 525 ft^2lb/sec^2 which in normal notation is 16 ft-lbs of
torque.
That means using that model of the bike, you have to apply 16 lbs of
pressure one foot from the center of the bars to accelerate the lean
rate up to 30 degrees per second in 1/2 second, and then reverse the
process to slow it back down to 0 degrees per second. The intent was
to cover 60 degrees in 1 second, but really it was somewhat less because
we spent most of the time accelerating and decelerating. (I think..
someone integrate that please) Plus it took
16 lbs of pressure first in one direction and then in the other to
control it all. And these are under ideal conditions.
I could be REALLY generous and model the bike as a solid cylinder with
a diameter of only 2 feet then
I = 1/2 M R^2 = 350 ft^2*lb
so we can cut the total input torque to acheive the above effect down
to 1.7 ft-lbs of torque. That's almost reasonable, but I don't
think the model of the bike is, and it's still a very slow rate of
rotation..
I claim I can do the same work in half the time and half the effort,
using good ol' outtracking.
AARON WARD
Ian Frechette (frec...@rintintin.Colorado.EDU) wrote:
: Ok, math and physics, and a little engineering coming up.
You forgot the common sense.
: Guess what, it's not really about how fast it spins or how heavy it is.
Yes, it is.
A more massive gyro will be more stable when subjected to disturbance than
a small gyro.
Or with motorcycles, a big heavy motorcycle tire will convert my "twist of
the handlebars" to a "push on the side of the tank" more completely than
an AOL CDRom.
Get that? The front tire takes a twist of handlebars and gives back a
push on the side of the tank. So I could push on the side of the tank of
the motorcycle like I'm trying to knock the bike over instead of twisting
the handlebars.
And vice-versa! When you do knock a bike over by pushing on the tank (or
rotating along the long roll axis), the handlebars will twist on their
axis provided the front wheel is spinning.
: But here's the catch. It can't give back more torque than you put in.
: So if you take a spinning disk and apply one foot pound of torque, it is
: simply conserved and given back 90 degrees forward on the disk as one
: foot pound of torque. In addition the formula
Good. See? Just pretend like your push on the handlebars was really a
push on the side of the tank.
: to translate 100% of our input torque (steering input) into what I'll
: call precession torque (rotation of the bike around it's long axis).
Exactly. That's what I just said. We agree. Now, the question is, are
you strong enough to push your bike over? You say that you aren't strong
enough.
You say this because you lack common sense.
: 1. To rotate the bike from a still position you must overcome the
: moment of inertia of the bike itself. We're talking a 500lb bike, plus
: say 200 lb rider, with a center of gravity somewhere above the front axle.
And here is your error. You assume that this is the case.
What the hell? Do you go into a low-earth orbit when you ride? Free-fall
is the only place where this free-body analysis works.
Listen.
Motorcycles are balancing! I can "overcome the moment of inertia" with my
pinky finger.
Take your bike off it's center/side stand. Balance it (like you do when
you are riding). Now see how difficult it is to overcome the inertia.
Fool.
John M. Feiereisen
>Ian Frechette (frec...@rintintin.Colorado.EDU) wrote:
>: Ok, math and physics, and a little engineering coming up.
>You forgot the common sense.
You must not have been looking very hard. I spotted it.
>: Guess what, it's not really about how fast it spins or how heavy it is.
>Yes, it is.
>A more massive gyro will be more stable when subjected to disturbance than
>a small gyro.
What does stability have to do with this? Ian was talking about
precession.
>Or with motorcycles, a big heavy motorcycle tire will convert my "twist of
>the handlebars" to a "push on the side of the tank" more completely than
>an AOL CDRom.
Wrong.
>Get that?
Yeah, I got it. It's still wrong.
>The front tire takes a twist of handlebars and gives back a
>push on the side of the tank.
Wrong again.
>So I could push on the side of the tank of
>the motorcycle like I'm trying to knock the bike over instead of twisting
>the handlebars.
But when you're sitting on your bike, there's nothing for your feet to
push off of.
Due to precession, a twist of the handlebars (applying a moment to the
front wheel about an axis parallel to the steering head) will produce
a *moment* about an axis perpendicular to both the steering head and
the front axle. Moments and forces are completely different.
Let's call the moment parallel to the steering head 'steering moment',
and the moment perpendicular to the steering head and front axle
'precession moment'. What Ian is saying is that the precession moment
will *never* exceed the steering moment -- no matter how massive the
front wheel may be.
The 'force multiplier' in the steering situation is the lateral
movement of the front wheel contact patch. A small steering moment
will turn the front wheel slightly and cause the contact patch to move
sideways. The resultant lateral force at the front wheel contact
patch, coupled with the vertical distance from the contact patch to
the bike/rider GC, will produce a relatively large roll moment about
the bike/rider CG. The result: light steering inputs produce huge
moments to roll the bike - quickly.
>And vice-versa! When you do knock a bike over by pushing on the tank (or
>rotating along the long roll axis), the handlebars will twist on their
>axis provided the front wheel is spinning.
Hey, precession happens. But we're not knocking bikes over by pushing
on the tanks. We're riding them and countersteering!
>: But here's the catch. It can't give back more torque than you put in.
>: So if you take a spinning disk and apply one foot pound of torque, it is
>: simply conserved and given back 90 degrees forward on the disk as one
>: foot pound of torque. In addition the formula
>Good. See? Just pretend like your push on the handlebars was really a
>push on the side of the tank.
You may do that if you wish. However, it appears that Ian is more
interested in doing things correctly.
>: to translate 100% of our input torque (steering input) into what I'll
>: call precession torque (rotation of the bike around it's long axis).
Minor point Ian: the precession torque is about an axis perpendicular
to both the steering head and the front axle. This is close to, but
not exactly parallel to the long axis of the bike. (Hey, if we're
going to be pedantic, why not be *really* pedantic? :-) )
>Exactly. That's what I just said. We agree. Now, the question is, are
>you strong enough to push your bike over? You say that you aren't strong
>enough.
I believe Ian is saying he's not strong enough to do it quickly enough
to make corners interesting.
>You say this because you lack common sense.
No, I believe he says it because he understands the dynamics.
>: 1. To rotate the bike from a still position you must overcome the
>: moment of inertia of the bike itself. We're talking a 500lb bike, plus
>: say 200 lb rider, with a center of gravity somewhere above the front axle.
>And here is your error. You assume that this is the case.
And he assumes correctly.
>What the hell? Do you go into a low-earth orbit when you ride? Free-fall
>is the only place where this free-body analysis works.
Wrongo. You've never taken (or at least passed) a statics or dynamics
course, have you?
>Listen.
Can we heckle, too? How about calling you names, you ignorant
edu-breath newbie puke?
>Motorcycles are balancing!
Balancing what?
>I can "overcome the moment of inertia" with my pinky finger.
Theoretically, you could push an ocean liner with your pinkie finger,
too.
>Take your bike off it's center/side stand. Balance it (like you do when
>you are riding). Now see how difficult it is to overcome the inertia.
Stand there and very quickly lean it back and forth 45 degrees to
either side over and over again. Now compare your effort to the
effort you exert doing this in the twisties.
>Fool.
Physician, heal thyself.
>*******************************************************************
>Rider Aaron 1987 CBR600F "Hurricane"
>"Who is the Great Cornholio, and why is he taking my toilet paper?"
A Hurricane pilot *and* a Beavis & Butthead fan. Oh, now I
understand.
--
John M. Feiereisen feierejm(at)utrc(dot)utc(dot)com
Ian Frechette
In article <6al68h$4ba$1...@news.nevada.edu>, AARON WARD <wa...@nevada.edu> wrote:
>Motorcycles are balancing! I can "overcome the moment of inertia" with my
>pinky finger.
That may be true, but I'll bet you can't rotate it 90 degrees in less than
a second and stop it at the end of the lean all with your pinky, balanced
the whole way or not.
>Take your bike off it's center/side stand. Balance it (like you do when
>you are riding). Now see how difficult it is to overcome the inertia.
Interia first, and then momentum. I was out this morning on my bike
flicking it back and forth at about 40mph fast enough to seriously
weight and unweight the front suspension, covering somewhere on the
order of 60 degrees of rotation per second, and that's back and forth
both ways, overcoming the momentum of a fast lean and turning it back
to initate a new lean the other way. No way can you do all that with
light direct applied pressure to the bike itself.
If I continue pushing with your model of countersteering, the bike
would simply flop over on the ground. If you don't stop the rotation
by driving the wheels back under the balance point, then you HAVE
to stop the rotation by hand. It is quite easy to initate a very sharp
turn, AND stop the lean, and maintain the turn, by only applying
pressure to one handlbar along one vector. How exactly do you explain that?
If you admit that the wheel turns into the lean by itself, to stop the lean,
then you're saying that gyroscopic precession starts the turn and outtracking
stops it. But you can't have 100% effective gyroscopic precession AND
outtracking, because 100% torque translation wouldn't let the handlbars
move at all. If you don't have 100% torque translation, then some of
it must go into turning the bars, and no matter what else you think,
that will cause outtracking, and induce a lean.
BTW, all you have to do to see how the bars move is wrap a clothes hanger
around the center of the bars and let the other end hang out about a foot
straigh back over the tank. Watch it while you ride. Throw the bike around
at 30-50mph and you'll see it move quite a bit. Several degrees in some
cases. Remember 2 degrees of head rotation at 50mph translates into
over a foot of outtracking and thus 45 degrees of rotation in only 30 feet.
Don't say it isn't so, until you try it yourself. I brought the EQs to
the game because some people are too fucking lazy to go ride their bikes
and see it for themselves, but the emperical evidence is there for anyone
who really wants to know the truth.
>Fool.
Ya know, I haven't resorted to personal attacks yet, but you're pushing it.
>*******************************************************************
>Rider Aaron 1987 CBR600F "Hurricane"
>"Who is the Great Cornholio, and why is he taking my toilet paper?"
ian
AARON WARD
John M. Feiereisen (feie...@nospam.utrc.utc.com) wrote:
: >Ian Frechette (frec...@rintintin.Colorado.EDU) wrote:
Lemme explain: I am very curious about thing I don't understand. Never
ever have I gotten a good explanation of countersteering. I have already
tried and tried. So I knew if I asked questions politely, nothing would
happen.
Sorry Ian for being so asshole-ish. Sorry John, didn't mean to ruffle
your feathers too.
Thanks to both for clearing up what was fuzzy for so long. Everyone
usually just makes fun of the countersteering thread. You guys actually
helped me understand. Seriously, no sarcasm intended.
I remember a long time ago someone joking about needing to countersteer
their office chair or something. This is usually the kind of treatment
that countersteering gets because no one ever wants to explain it.
Thanks again, guys.
Sincerely,
Aaron S. Ward
seeing the light... dimly
John M. Feiereisen
>I remember a long time ago someone joking about needing to countersteer
>their office chair or something. This is usually the kind of treatment
>that countersteering gets because no one ever wants to explain it.
Welcome to rec.moto. I hope you brought your Nomex suit. ;-)
Brian McLaughlin
feie...@nospam.utrc.utc.com (John M. Feiereisen) wrote:
>Due to precession, a twist of the handlebars (applying a moment to the
>front wheel about an axis parallel to the steering head) will produce
>a *moment* about an axis perpendicular to both the steering head and
>the front axle. Moments and forces are completely different.
>Let's call the moment parallel to the steering head 'steering moment',
>and the moment perpendicular to the steering head and front axle
>'precession moment'. What Ian is saying is that the precession moment
>will *never* exceed the steering moment -- no matter how massive the
>front wheel may be.
ARRRRGGGGHHHH, why do engineers have to make up new terms? Torque
dammit! Moment is just as bad as silly units like pound mass or
kilogram force.
>The 'force multiplier' in the steering situation is the lateral
>movement of the front wheel contact patch. A small steering moment
>will turn the front wheel slightly and cause the contact patch to move
>sideways. The resultant lateral force at the front wheel contact
>patch, coupled with the vertical distance from the contact patch to
>the bike/rider GC, will produce a relatively large roll moment about
>the bike/rider CG. The result: light steering inputs produce huge
>moments to roll the bike - quickly.
Why isn't sterring easier at high speeds? The contact patch should
squirt out faster at high speeds.
>>And vice-versa! When you do knock a bike over by pushing on the tank (or
>>rotating along the long roll axis), the handlebars will twist on their
>>axis provided the front wheel is spinning.
>Hey, precession happens. But we're not knocking bikes over by pushing
>on the tanks. We're riding them and countersteering!
Personally I think it is a mix of phenomina invloved. The directions
of applied torque and the reaction of the front wheel work out too
well to not be involved in some way.
--
Brian AP#1
Ian Frechette
In article <6alpi9$n...@newshost1.res.utc.com>,
John M. Feiereisen <feie...@nospam.utrc.utc.com> wrote:
>In <6al68h$4ba$1...@news.nevada.edu>, wa...@nevada.edu (AARON WARD) wrote:
>>Ian Frechette (frec...@rintintin.Colorado.EDU) wrote:
>>: to translate 100% of our input torque (steering input) into what I'll
>>: call precession torque (rotation of the bike around it's long axis).
>
>Minor point Ian: the precession torque is about an axis perpendicular
>to both the steering head and the front axle. This is close to, but
>not exactly parallel to the long axis of the bike. (Hey, if we're
>going to be pedantic, why not be *really* pedantic? :-) )
You should have seen my first cut of that post. It was much more pedantic
but it was getting way too long. I was going to mention that the precession
axis was not only NOT along the long axis of bike through the CG, but
also at an angle perpendicular to the forks, 90 degrees from the rake,
meaning that some of what precession torque does exist is lost to misaligned
vectors.
Makes you wonder if any precessionist rider choppers. I wonder how they
do it.
There are so many things conspiring against effective precession (in
the direction needed to lean the bike) that I don't have the room to
list them all.
However, I do believe that precession is happening the other way around
all the time. The lean induced by outtracking, produces a tremendous
torque around the long axis of the bike, and thus even with less than
100% torque translation, causes the front wheel to precess into the lean
very quickly, instantly *trying* to drive the front wheel under the CG.
It's helped or hurt by the rake, trail, and tire profile of course, but
I believe it's a real effect.
>>: moment of inertia of the bike itself. We're talking a 500lb bike, plus
>>: say 200 lb rider, with a center of gravity somewhere above the front axle.
Oh heh.. I did mention that the CG was above the axle.
>>Motorcycles are balancing!
>Balancing what?
Luckly, balancing all those people on top of them who think they're
using precession to turn.
>A Hurricane pilot *and* a Beavis & Butthead fan. Oh, now I
>understand.
Wasn't the old hurricane the "ground magnet", with its gear driven cams and
very high CG, or am I thinking of a different bike? Anyone riding a tall
bike with a high CG should be able to appreciate this discussion.
>John M. Feiereisen feierejm(at)utrc(dot)utc(dot)com
ian
John M. Feiereisen
In <6ann9m$s...@wiscnews.wiscnet.net>, bmcla...@waukesha.tec.wi.us
(Brian McLaughlin) wrote:
>feie...@nospam.utrc.utc.com (John M. Feiereisen) wrote:
>>Due to precession, a twist of the handlebars (applying a moment to the
>>front wheel about an axis parallel to the steering head) will produce
>>a *moment* about an axis perpendicular to both the steering head and
>>the front axle. Moments and forces are completely different.
>>Let's call the moment parallel to the steering head 'steering moment',
>>and the moment perpendicular to the steering head and front axle
>>'precession moment'. What Ian is saying is that the precession moment
>>will *never* exceed the steering moment -- no matter how massive the
>>front wheel may be.
>ARRRRGGGGHHHH, why do engineers have to make up new terms? Torque
>dammit! Moment is just as bad as silly units like pound mass or
>kilogram force.
Hey, I didn't make up the term 'moment', but I guess I do sort of use
it interchangeably with 'torque' just about anywhere. You piqued my
interest.
From the American Heritage College Dictionary:
Moment: a. The product of a quantity and its perpendicular distance
from a reference point.
b. The tendency to cause rotation about a point or an axis.
Torque: a. The moment of a force equal to the vector product of the
force and the radius vector from the axis about which the
force produces torsion or rotation.
b. A turning or twisting force.
Interesting. I never knew this. There is a difference. (But I do
know when and when not to treat either quantity as a vector or scalar.
But to be rigorous, according to the definitions above, ot looks like
only the torque is a vector.)
Yeah, you're probably right. Exclusive use of the term 'torque' would
probably avoid confusion with terms such as 'polar moment of inertia'.
(Besides, do you suppose the Snap-On guy would understand me if I
asked for a 'moment wrench'?)
BTW, I agree with you, users of 'kilograms force' should be shot.
>>The 'force multiplier' in the steering situation is the lateral
>>movement of the front wheel contact patch. A small steering moment
>>will turn the front wheel slightly and cause the contact patch to move
>>sideways. The resultant lateral force at the front wheel contact
>>patch, coupled with the vertical distance from the contact patch to
>>the bike/rider GC, will produce a relatively large roll moment about
>>the bike/rider CG. The result: light steering inputs produce huge
>>moments to roll the bike - quickly.
>Why isn't sterring easier at high speeds? The contact patch should
>squirt out faster at high speeds.
The torques necessary to roll the bike increase as the roll
accelerations increase. Yes, the front wheel does outtrack faster at
higher speeds. This leads to faster steering.
<snip>
>Personally I think it is a mix of phenomina invloved. The directions
>of applied torque and the reaction of the front wheel work out too
>well to not be involved in some way.
No doubt there. It most certainly is a mix of phenomena. But looking
at Ian's discussion of the ratio of precession torque to steering
torque (How'd I do? How'd I do?), he makes a pretty compelling
argument that precession is only a small player.
Brian McLaughlin
feie...@nospam.utrc.utc.com (John M. Feiereisen) wrote:
>>Why isn't sterring easier at high speeds? The contact patch should
>>squirt out faster at high speeds.
>The torques necessary to roll the bike increase as the roll
>accelerations increase. Yes, the front wheel does outtrack faster at
>higher speeds. This leads to faster steering.
But at high speeds steering takes a huge amount of effort (Brainerd
turn 1 at about 150 to 160 mph, kink at Road America at about 120
mph), these turns take quite a push on the bars - I quite literally
throw my weight on the bar to turn quick enough. The steering does not
seem any faster.
><snip>
>>Personally I think it is a mix of phenomina invloved. The directions
>>of applied torque and the reaction of the front wheel work out too
>>well to not be involved in some way.
>No doubt there. It most certainly is a mix of phenomena. But looking
>at Ian's discussion of the ratio of precession torque to steering
>torque (How'd I do? How'd I do?), he makes a pretty compelling
>argument that precession is only a small player.
The only way to truely find out would be to make a wheel with no
angular momentum (counter rotating weights) and see how the steering
is affected.
Death to Spam
John M. Feiereisen wrote:
>
> In <6al68h$4ba$1...@news.nevada.edu>, wa...@nevada.edu (AARON WARD) wrote:
>
> >Ian Frechette (frec...@rintintin.Colorado.EDU) wrote:
> >: Ok, math and physics, and a little engineering coming up.
>
> >You forgot the common sense.
Wot's common sense gotta do with posting in rec.mc, John? :)
> Hey, precession happens. But we're not knocking bikes over by pushing
> on the tanks. We're riding them and countersteering!
He's gonna think you are putting a spin on things (maybe he's a young
democrat). We could send him off to the First Church of the Immaculate
Precession.
> >Good. See? Just pretend like your push on the handlebars was really a
> >push on the side of the tank.
We've got some great pretenders here!
> Can we heckle, too? How about calling you names, you ignorant
> edu-breath newbie puke?
Them's fightin' words, son. You wouldn't be dragging me into this just
cause I've an .edu address, too, would ya?
Ian Frechette
In article <6apu1t$u...@newshost1.res.utc.com>,
John M. Feiereisen <feie...@nospam.utrc.utc.com> wrote:
>>Why isn't sterring easier at high speeds? The contact patch should
>>squirt out faster at high speeds.
>
>The torques necessary to roll the bike increase as the roll
>accelerations increase. Yes, the front wheel does outtrack faster at
>higher speeds. This leads to faster steering.
Should lead to faster steering, but at high enough speeds, precession does
come into play in several different ways.
1. If it's spinning fast enough, a much higher percentage of torque is
translated, and you are now trying to push against the full moment
of intertia of the bike.
2. If close to 100% torque translation is occuring, then you simply can't
turn the handlbars at all (nutation may come in but that would be nasty)
and that means that you *can't* outtrack. Not only does precession
cause the bike to lean, it causes it to not be able to outtrack, and
that means that the moment of inertia of the bike increases, because
it can't rotate around it's CG, but only around it's contact patches
without the contact patches being able to move sideways (in the bike's
frame of reference) I doubt that this is every really a problem.
3. The rear wheel does nothing good. The only force you an exert on it
is the lean, and that causes precession which makes the rear wheel
try to turn the bike into the lean, and that force will in turn counteract
the precession torque of the front wheel. Not a large force, but
not a useful one either if you're trying to turn quickly.
4. The frequency of precession slows with increased speed so all things
be equal, if precession is a significant force at high speeds, if you
apply the same torque as you would at a low speed, you'll lean slower.
4. Let's say outtracking is still occuring, but that the gyroscope *is*
much more effective, but not 100% effective, then I suspect that 2 and 4
are not the problem, but that 1 is, but in a direction you wouldn't
suspect. What's happening is that you still can induce a significant
lean torque using outtracking, but that is torgue down the long axis of
the bike and is translated 90 degrees at the front wheel to precession
that turns the front wheel into the lean. This torque is exactly opposite
that of countersteering. If you don't apply more force to oppose it,
it'll successfully drive the front tire back under the bike and stop
the lean, or even reverse it. You're fighting precession the whole way
now. What's that feel like? Apply pressure to the left bar to go left,
and it immediately starts to lean but stabilizes without continuing to
go over. You have to apply more and more pressure to induce more lean.
Constrast this with when you're going 40mph. A slight continuous pressure
can dump the bike right on its side.
>>Personally I think it is a mix of phenomina invloved. The directions
>No doubt there. It most certainly is a mix of phenomena. But looking
>at Ian's discussion of the ratio of precession torque to steering
>torque (How'd I do? How'd I do?), he makes a pretty compelling
>argument that precession is only a small player.
A small player at low speeds. A bigger player at very high speeds, but in
no productive fashion.
CyberWoof
How about real data?
For my summer project I'm going to mount a potentiometer on my bike's steering
head to measure steering input and a dual-axis gyroscope on the back to record
angle of lean and rate of turn. Add a couple of A/D converters and a Mac
PowerBook with custom data-recording software and I've got a flight recorder.
Maybe I'll even get clever and build an electronic speedometer and record that
data, too. I'll present the data and my interpretation on my web site. Then
we'll know.
--Timberwoof at aol dot com - http://members.aol.com/timberwoof
Spam Reading Offer: http://members.aol.com/timberwoof/spamoff.html
These people hve sent me unauthorized junk mail:
Do...@merteuil.com
sple...@earthlink.net
Brian McLaughlin
frec...@rintintin.Colorado.EDU (Ian Frechette) wrote:
> Now simply attach the handlebars to one end of this rod. Now orient the
> bike so that it's leaning about 45 degrees to the left. Grab the handlebars
> and flick the bike 90 degrees to the right and stop it so that it's leaning
> 45 degrees to the right. Note, gravity plays no role here because it's
> balanced around it's center of mass.
> You should be thinking "flick???!!!" about now? It should be obvious to
> anyone that flicking 700 pounds worth of mass is not possible
> through direct input.
Recall that torque = moment of inertia * angular acceleration.
Depending upon what the moment of inertia is 700 lbs may be easier to
rotate than you think. Also the rotation of a motorcyle when leaning
over does not take place through the center of mass (for this to
happen the tires would have to slide sideways, 2ft high cg and 45 deg
lean the tires would have to slide 1.4 feet sideways), the rotation
takes place at the contact patch. Do your same experiment with the
bike sitting on the ground. Just let your bike fall over, where does
the rotation take place? Why should it be different if the bike is
moving (esp. the rear tire)?
> So back to the precessional torque. Remember what I said, you only get
> back as much torque as you put in. So now remove the 200lb weight, climb
> up on the motorcycle, have someone spin the front wheel really fast,
> and now by using the handlbars only, FLICK the bike 90 degrees from one
> side to the other. What's changed? It's the same handlebars. The bike
> plus rider is the same weight, it's more or less balanced, and you've
> applied X ft/lbs of torque to the bars and that has been converted into X
> ft/lbs precessional torque at the axle of the front wheel, which we'll
> pretend is close enough to the center of gravity to mean that the axis of
> precession is lined up with the long axis of the bike. You only get out
> as much torqe as you put in, so the answer is, nothing. Nothing has
> changed, you're still trying to manhandle 700 lbs as though it was directly
> connected to your handlebars.
>
How much torque is actually supplied to a handle bar? Freddie Spencer
and Eddie Lawson both bent tubular steel handlebars when racing 1000cc
superbikes.
<snip>
>Ok, but what about point 2? The precessionists were just about to tell me
>that point 2 is totally consistent with what they feel when they're
>going 150 mph. Well, I started out thinking that the precession
>frequency would be quite slow at low speeds and even lower at high
>speeds, but I was surprised by what I found. Let's start with the
>150mph example, and plug it into the equation from above.
>We'll start with a few assumptions.
>The handlebars are 2 feet long front end to end.
>The front tire and wheel weigh 20 lbs and for maximum effect we'll
>say that the mass is concentrated at a distance of 12 inches from the axis
>of rotation, in an ideal ring with a moment of inertia of MR^2
>With this 2 foot diameter tire, the tire turns 35 revolutions per second.
>We'll apply a force to the handlbars of X lbs (yes.. english units)
rotational frequency (and precession frequency) is in *radians* per
second.
> X lb * 32 ft/sec^2 * 1 ft
>precession freq = -------------------------- = .045*X rps or 16X degrees/sec
> 20 lb * 1 ft^2 * 35/sec
Revolutions/seconds are not the same as per second, rework with
radians per second.
Ian Frechette
In article <19980130040...@ladder03.news.aol.com>,
CyberWoof <cybe...@aol.com> wrote:
>How about real data?
>
>For my summer project I'm going to mount a potentiometer on my bike's steering
>head to measure steering input and a dual-axis gyroscope on the back to record
>angle of lean and rate of turn. Add a couple of A/D converters and a Mac
>PowerBook with custom data-recording software and I've got a flight recorder.
>Maybe I'll even get clever and build an electronic speedometer and record that
>data, too. I'll present the data and my interpretation on my web site. Then
>we'll know.
More power to you. A few things though.
- You can already get an electronic speedometer for a bike.. It's called
a bicycle computer. Here's some info I threw together on how I mounted
mine. http://shell.rmi.net/~frechett/bikecomputer.html
The nice thing is that you can calibrate using the tire circumfrence in
milimeters so it's very accurate. Mine reads accurate to 160mph, not
that my bike goes that fast.
- You'll either need a digital device to measure the torque that you're
putting into the bars, or make a device that, when activated, applies
a fixed amount of torque and then base the rest of your measurments on
that baseline.
I think you could save yourself a lot of time if you start by calculating, or
measuring the true moment of inertia of the bike if rotated around its long
axis through the CG. Then set up your gyros, go riding and measure how fast
you can lean the bike through 60 degrees, from a steadystate 30 degree lean
on one side to the same angle on the other side. Using the moment of
inertia of the bike you can calculate the amount of torque required to
produce the measured lean rate.
If the measured input torque on the bars is less than the torque calculated
above then you know that precession can't be doing all the work. I think
you'll find that the torque required to initate, maintain and stop the lean
is very high and that the input torque at the bars is typically much lower.
As you go very very fast, you may find that the input torque matches more
closely the torque required to produce the measured lean rate.
>-- Imberwoof at aol dot com - http://members.aol.com/timberwoof
ian
andy the pugh
CyberWoof <cybe...@aol.com> wrote:
> How about real data?
>
> For my summer project I'm going to mount a potentiometer on my bike's steering
> head to measure steering input and a dual-axis gyroscope on the back to record
> angle of lean and rate of turn.
Strain gauge your bars and front axle too.
> Add a couple of A/D converters and a Mac
> PowerBook
I approve of your hardware choice.
--
ap
andy the pugh
Ian Frechette <frec...@rintintin.Colorado.EDU> wrote:
> There are so many things conspiring against effective precession (in
> the direction needed to lean the bike) that I don't have the room to
> list them all.
>
I have just thought of another issue. The precessional torque to lean
the bike has to be transferred to the bike throught the front axle and
forks. Does a fork brace affect countersteering? It should.
--
ap
CyberWoof
a.c....@shef.ac.uk (andy the pugh) wrote, after quoting me...
>> For my summer project I'm going to mount a potentiometer on my bike's
steering
>> head to measure steering input and a dual-axis gyroscope on the back to
record
>> angle of lean and rate of turn.
>
>Strain gauge your bars and front axle too.
How much do the things bend? How much does that affect steering? My test bed is
a 400cc CB-1, which has 41mm bars, which look quite chunky. Besides, I've
already allocated three of my four analog channels. I was thinking of also
measuring engine speed by tapping the appropriate trace in my tachometer and
using three bits for the shifter (up, down, and neutral) and one for the brakes
(using the handy 12V source that comes on when the brakes are applied).
>> Add a couple of A/D converters and a Mac
>> PowerBook
>
>I approve of your hardware choice.
You'll like this, then: My A/D converter is a Beehive ADB I/O
(http://www.bzzzzzz.com). It's got four 8-bit analong inputs and four 1-bit
digital input/output lines. It connects to the Mac through the ADB port. In
early experiments with Xcmds and Director, I got 10 readings per second out of
it on a Centris 610 (68040/40). With custom software on my PowerBook 100
(68000/16) I should be able to get at least that. Gawd, I'd hate to think of
what I'd have to do if I were using a PC portable. Ayieee!
CyberWoof
frec...@rintintin.Colorado.EDU (Ian Frechette) wrote...
>>>>>
In article <19980130040...@ladder03.news.aol.com>,
CyberWoof <cybe...@aol.com> wrote:
>How about real data?
>
>For my summer project I'm going to mount a potentiometer on my bike's steering
>head to measure steering input and a dual-axis gyroscope on the back to record
>angle of lean and rate of turn. Add a couple of A/D converters and a Mac
>PowerBook with custom data-recording software and I've got a flight recorder.
>Maybe I'll even get clever and build an electronic speedometer and record that
>data, too. I'll present the data and my interpretation on my web site. Then
>we'll know.
<snip>
- You can already get an electronic speedometer for a bike.. It's called
a bicycle computer. Here's some info I threw together on how I mounted
mine. http://shell.rmi.net/~frechett/bikecomputer.html
The nice thing is that you can calibrate using the tire circumfrence in
milimeters so it's very accurate. Mine reads accurate to 160mph, not
that my bike goes that fast.
- You'll either need a digital device to measure the torque that you're
putting into the bars, or make a device that, when activated, applies
a fixed amount of torque and then base the rest of your measurments on
that baseline.
>>>>>
Thanks! (And it's nice to see that CU is maintaining the tradition of naming
computers after dogs. Do you know the "correct" pronunciation for cubldr?) Now,
to what you really said...
Does that electronic speedo have a variable voltage output I can connect to my
A/D converter?
Oy, Strain gauges on the clip-ons. That's hard. (And ooh, yuck. Squishy
clip-ons.) My approach is to just look at the angles the steering head goes
through around curves and compare them to the lean angle and rate of turn. The
idea of applying a fixed force to the handlebars scares me. Here, you try it
first. }: ) (Check out
http://members.aol.com/cyberwoof/motorcycle/dynamics.html. That's where I'll be
posting my progress and results.)
andy the pugh
CyberWoof <cybe...@aol.com> wrote:
> >Strain gauge your bars and front axle too.
>
> How much do the things bend?
Not a lot, but strain gauges are very sensitive.
> You'll like this, then: My A/D converter is a Beehive ADB I/O
> (http://www.bzzzzzz.com). It's got four 8-bit analong inputs and four 1-bit
> digital input/output lines. It connects to the Mac through the ADB port. In
> early experiments with Xcmds and Director, I got 10 readings per second out of
> it on a Centris 610 (68040/40). With custom software on my PowerBook 100
> (68000/16) I should be able to get at least that.
Sounds convenient, but a a bit slow. There are PCMCIA a/d cards running
at 100kHz+ now at 12 bits (possibly 16).
--
ap
John M. Feiereisen
In <6at81t$g...@wiscnews.wiscnet.net>, bmcla...@waukesha.tec.wi.us
(Brian McLaughlin) wrote:
>Also the rotation of a motorcyle when leaning
>over does not take place through the center of mass (for this to
>happen the tires would have to slide sideways, 2ft high cg and 45 deg
>lean the tires would have to slide 1.4 feet sideways), the rotation
>takes place at the contact patch. Do your same experiment with the
>bike sitting on the ground. Just let your bike fall over, where does
>the rotation take place?
About the contact patch.
Now do this on a perfectly frictionless surface. You'll find that the
tires do indeed slide out from under the bike and the CG goes pretty
much straight down.
>Why should it be different if the bike is
>moving (esp. the rear tire)?
In actuality, the *instantaneous* center of rotation moves all over
the map. Yes, at a standstill the center of rotation during a roll
maneuver is at the contact patch. But this is only because the
contact patch isn't moving sideways. When you're moving, the contact
patch may be moving one way while the GC is moving another. You've
got to look at two points on the machine, look at their instantaneous
velocities, and find the intersection of the lines perpendicular to
these velocity vectors. (As you roll on the throttle coming out of a
corner, your instantaneous center of rotation might actually be way
over your head.)
The center of gravity is a convenient reference point for all your
bookkeeping. You can completely describe the motion of the bike by
specifying only the position, velocity, and acceleration of the GG and
the angle, angular velocity, and angular acceleration (in two planes)
of the bike about the GC. Note that the polar moment of inertia is
dependent upon the location of the axis of rotation. The moment of
inertia about the contact patch will be a lot higher than the moment
of inertia about the GC. (What was that? Did somebody say "Parallel
Axis Theorem"???)
<pedant mode off>
Trevor Dennis
Ian Frechette <frec...@rintintin.Colorado.EDU> said
>
>- You can already get an electronic speedometer for a bike.. It's called
> a bicycle computer.
But does it have a data output? It's of little use if he can't
log the speed relative to other data.
>- You'll either need a digital device to measure the torque that you're
> putting into the bars, or make a device that, when activated, applies
> a fixed amount of torque and then base the rest of your measurments on
> that baseline.
A second set of handlebars mounted to the first via a load cell?
That might be more effort than he wants to go to.
>As you go very very fast, you may find that the input torque matches more
>closely the torque required to produce the measured lean rate.
As gyroscopic _inertia_ comes into the equation.
--
Trevor Dennis /`\ .(o~)-(o~). /`\ tre...@tdennnis.demon.co.uk
The Polite Brit / , \( _______ )/ , \ tden...@ford.com
OGH #1 ___/ /_\ /`"-------"`\ /_\ \___ Southern England
jgs`~//^\~_`\ <__ __> /`_~/^\\~`
`~//^\\~`~//^\\~`
Ian Frechette
In article <19980131094...@ladder02.news.aol.com>,
CyberWoof <cybe...@aol.com> wrote:
>frec...@rintintin.Colorado.EDU (Ian Frechette) wrote...
>Does that electronic speedo have a variable voltage output I can connect to my
>A/D converter?
No, although you could probably simply do what it does yourself. Use its
wheel sensor and count the number of revolutions per second, or measure
the actual time for each revolution and calculate the speed yourself.
You'll have to make something to convert their pulse into a digital signal..
(the magnet passing by the small coil induces a current one way, then
the other.. )
>Oy, Strain gauges on the clip-ons. That's hard. (And ooh, yuck. Squishy
>clip-ons.) My approach is to just look at the angles the steering head goes
>through around curves and compare them to the lean angle and rate of turn. The
That's not good enough. If, for the sake of argument you were able to
produce a truly effective gyroscope with the front wheel, then when
you apply pressure to the handlbars it won't move at all, as 100% of
the torque is translated 90 degrees at the axle. However, the lean rate
will depend entirely on the pressure applied and not the head rotation.
Go play with a gyroscope. The old bicycle wheel hanging from one side of
the axle is a great one. It'll precess under it's own weight. Add some
more weight to the other end of the axle and it'll precess faster, given
the same rotational rate, but in neither case does the axle move significantly
from the horizonal plane (other than slight oscilations.. nutation)
Ok, granted, you might find that most of the time precession plays little
or no role and that head rotation at a given speed creates a known
amount of outtracking and thus lean angle, but in all that you can't
prove that precession isn't occuring some, unless you measure the torque
required on the bike to produce that lean, and contrast that with the
input torque. But, outtracking alone, does not explain the high speed
behavior, so measurments there would be interesting.
Anyway, I'm not really thinking squishy clipons. How about simply attaching
the equivalent of a torque wrench to the center of the handlebars, either
aligned along the bars themselves, or even back over the tank. Find someway
to continuously measure the applied torque and then give it a little shove
when you want to turn. The applied torque can change, but should be
measured continuously, so you can plot it again the angular acceleration
and lean angle. You only need to measure the deflection of the torque
wrench to get the force. You're already measuring one deflection angle,
why not another.
>idea of applying a fixed force to the handlebars scares me. Here, you try it
>first. }: ) (Check out
I know, it scares me too, although I was thinking that if it were something
you could turn on and off, and could apply only a small force, you'd be ok.
We know that even a very small force can produce a measurable lean. In fact
with a small fixed force you might start by stabilizing the bars with your
hands, turn *it* on, and then just gently let off pressure with your hands
and let it lean under the pressure of the device.
You can't use a weight because of centrifugal forces, and most springs won't
provide the fixed force you want, and they're hard to turn off. I was thinking
something like a homemade air pneumatic piston, with an air supply you can
turn on and off. Snagging the innards from a small electric car/bike airhorn
might be the trick.
>http://members.aol.com/cyberwoof/motorcycle/dynamics.html. That's where I'll be
>posting my progress and results.)
I'll keep an eye out.
>--Timberwoof at aol dot com - http://members.aol.com/timberwoof
ian
AARON WARD
Here is how to settle this.
Put a motorcycle up on a rear stand so the rear wheel can spin freely.
Jack up another bike and line it up so that the front tire touches the
rear of the first bike, but support the other bike in such a way that you
can turn the handlebars.
Start up the first bike and using the RPMs as a guide, carefully bring the
front tire of the second bike up to some known speed.
Turn the handlebars and see what happens.
If necessary, substitute a bicycle for the second bike.
Keep in mind that the front tire will be spinning backwards.
--
Ian Frechette
In article <6at81t$g...@wiscnews.wiscnet.net>,
Brian McLaughlin <bmcla...@waukesha.tec.wi.us> wrote:
>frec...@rintintin.Colorado.EDU (Ian Frechette) wrote:
>Recall that torque = moment of inertia * angular acceleration.
>Depending upon what the moment of inertia is 700 lbs may be easier to
>rotate than you think. Also the rotation of a motorcyle when leaning
I just don't see how. All other things aside, the center of mass of the
bike and the center of mass of the rider are separated by some distance.
That increases the moment of inertia. I did show two possible moments
for a 700lb bike plus rider combo but the second seems overly optimistic
to me.
>over does not take place through the center of mass (for this to
>happen the tires would have to slide sideways, 2ft high cg and 45 deg
>lean the tires would have to slide 1.4 feet sideways), the rotation
>takes place at the contact patch. Do your same experiment with the
The above statement is only true in a non-moving frame of reference.
The tires CAN move sideways from the bike's frame of reference
because the ground is moving below the bike and the tires can follow
a curved path. But that assumes SOME head rotation, and thus
outtracking. If you're having a hard time visualizing this, set the bike
up on a treadmill and attach ropes to the frame of the bike near the
engine, so that it can't move sideways. The bike is not moving anywhere,
but the tires are rolling as the treadmill moves under it. Now you can
induce a lean at any speed by simply turning the front wheel a little
and the tires will ride out from under the place where your ropes
are attached (close to the CG).
Now put it on the road. If the bike is going straight newton's laws say
the mass of the bike will want to continue going straight due to
inertia. To induce lean, you need only turn the front wheel out
from under the CG (which will try to continue on straight), while the
front tire, followed by the rear follows a curved path on the ground one
way, which translates to a force at the contact patches perpendicular to
the direction of travel. That is a purely rotational torque. The bike
rotates around it's CG, with the tires going one way, the top of the
bike with the rider going the other. It leans.. Continue that way
and it crashes to the ground, so at some point the tires are driven
back under the CG (note that "under" is a relative term here because
of centrifugal force. During a turn, "under" is at whatever point
the centrifugal force balances gravity's force trying to make the bike
fall over) This turning back under the CG is handled by a combination
of rake, trail, tire profile and sometimes the rider applying opposite
pressure.
>bike sitting on the ground. Just let your bike fall over, where does
>the rotation take place? Why should it be different if the bike is
>moving (esp. the rear tire)?
You're right that if the bike were rotating around the contact patches
then the moment of inertia could be measured by just letting it fall
over. BUT, note that the axis of rotation is now even further from the
CG and more importantly some distance from ALL the mass of the bike
and rider. This increases the moment of inertia of the whole thing
tremendously. Push the bike back and forth rapidly only 15 degrees or
so, to minimize the effects of gravity until you think you can acheive
a lean frequency of 60 degrees per second, and tell me how much force
you had to apply.
I've been trying to be as kind to the precessionist as possible by letting
the ideal case for precession be that the bike does rotate around it's
long axis through the CG and that what outtracking that does occur
is incidental, and is a result of a slight loss of torque translation
from the handlebars to precession. A perfect coincidence.
If I were being strict, I would say that for only precession to do the
work, the handlebars do not move at all. 100% of input torque is
translated into lean, AT the precession frequency. There is no
outtracking so the bike is forced to lean around an axis at the contact
patches (at a high moment of inertia). Furthermore, the axis of precession
is perpendicular to the fork angle, so for a bike like an old chopper
the precession axis could be pointing more at the ground than back through
the bike. Plus, even if the axis of precession were horizontal, it would
be somewhat lower than the CG so more loss of effective torque.
>How much torque is actually supplied to a handle bar? Freddie Spencer
>and Eddie Lawson both bent tubular steel handlebars when racing 1000cc
>superbikes.
You can choose to apply as much as you want, but I know that what is
required at normal speeds is ussually very little. And yes, at very
high speeds, you will be fighting gyroscopic precession and thus
having to apply much more torque, but the rest of us don't spend all
our times over 100mph.. (well most of us don't at least)
>rotational frequency (and precession frequency) is in *radians* per
>second.
>
>> X lb * 32 ft/sec^2 * 1 ft
>>precession freq = -------------------------- = .045*X rps or 16X degrees/sec
>> 20 lb * 1 ft^2 * 35/sec
>
>Revolutions/seconds are not the same as per second, rework with
>radians per second.
Actually, natural frequency, cycles per second *IS* 1/T
f = 1/T and radian freq is 2*pi*f so rad freq = 2*pi/T
I fed cycles/sec in, and I wanted cycles/sec out. If I wanted rads/sec
out, I would have used converted cycles/sec to 2pi*rad/sec and
stuck that in and gotten that out and divided by 2pi*rads to get cycles per
sec again.. Nothing gained.
A radian is only a measure of angle, and it disappears when you
use it for something else. For instance, distance travelled
by a wheel is circumfrence, with units of meters, multiplied
by the number of radians travelled.
meters * 2*pi radians and the units are in meters only.
What's confusing is realizing that a frequency with units like
f = X / (rad*sec) is just f = X / (2pi*sec)
I could have reworked the eq with grams and meters too, and it would
still come out the same. Actually, I did, and it did.
I suggest you go back and find your old physics book. I had to, to be
sure, when I worked this out the first time, because I got X/(rad*sec).
>Brian AP#1
ian
Bleah.. I'm going to have to cut back on this thread soon. I haven't been
in school for years and all this math and physics is driving me nuts.
CyberWoof
Andy the pugh quoted me, then said...
>> You'll like this, then: My A/D converter is a Beehive ADB I/O
>> (http://www.bzzzzzz.com). It's got four 8-bit analong inputs and four 1-bit
>> digital input/output lines. It connects to the Mac through the ADB port. In
>> early experiments with Xcmds and Director, I got 10 readings per second out
of
>> it on a Centris 610 (68040/40). With custom software on my PowerBook 100
>> (68000/16) I should be able to get at least that.
>
>Sounds convenient, but a a bit slow. There are PCMCIA a/d cards running
>at 100kHz+ now at 12 bits (possibly 16).
Let's see. A portable computer that will take a PCMCIA card will be a coupla
thousand bucks; the a/d card that will fit it will be several hundred. 16 bits
times four channels times 100k samples per second times sixty seconds times ten
minutes works out to roughly 2GB of data ... if I can cram it into the hard
drive fast enough. Do I really *need* 10 microsecond resolution for this
project? Let's get a grip on reality here ... my little PowerBook and ADB I/O
have one tremendous advantage over the system you propose: they're paid for.
--Timberwoof at aol dot com - http://members.aol.com/timberwoof
mull...@advinc.com
In article <19980131223...@ladder03.news.aol.com>,
cybe...@aol.com (CyberWoof) wrote:
>
> Andy the pugh quoted me, then said...
>
> >> You'll like this, then: My A/D converter is a Beehive ADB I/O
> >> (http://www.bzzzzzz.com). It's got four 8-bit analong inputs and four 1-bit
> >> digital input/output lines. It connects to the Mac through the ADB port. In
> >> early experiments with Xcmds and Director, I got 10 readings per second out
> of
> >> it on a Centris 610 (68040/40). With custom software on my PowerBook 100
> >> (68000/16) I should be able to get at least that.
> >
> >Sounds convenient, but a a bit slow. There are PCMCIA a/d cards running
> >at 100kHz+ now at 12 bits (possibly 16).
>
> Let's see. A portable computer that will take a PCMCIA card will be a coupla
> thousand bucks; the a/d card that will fit it will be several hundred. 16 bits
> times four channels times 100k samples per second times sixty seconds times
ten
> minutes works out to roughly 2GB of data ... if I can cram it into the hard
> drive fast enough. Do I really *need* 10 microsecond resolution for this
> project? Let's get a grip on reality here ... my little PowerBook and ADB I/O
> have one tremendous advantage over the system you propose: they're paid for.
>
> --Timberwoof
If you beanie-propeller heads are really gonna do this, you might also
consider that Burr Brown (and others) make some very inexpensive, accurate
accelerometers. 3-axis data could be added in for a very small amount of
money.
I think these chips were originally made for air-bag deployment sensors.
Jim - who think thinks that bikes should not have software.
-------------------==== Posted via Deja News ====-----------------------
http://www.dejanews.com/ Search, Read, Post to Usenet
AARON WARD
Ian Frechette (frec...@rintintin.Colorado.EDU) wrote:
: inertia. To induce lean, you need only turn the front wheel out
: from under the CG (which will try to continue on straight), while the
This is "outtracking" right? I'm not *totally* sure of what outtracking
means exactly.
: the direction of travel. That is a purely rotational torque. The bike
: rotates around it's CG, with the tires going one way, the top of the
: bike with the rider going the other. It leans.. Continue that way
Can't we just watch a motorcycle race and see this phenonemon happen? It
is possible to determine the CG of a leaning bike visually, isn't it?
: You're right that if the bike were rotating around the contact patches
<snip>
: This increases the moment of inertia of the whole thing
So what you're saying is that bikes DO NOT rotate while leaning about the
contact patches, right? If they did, that would mean that outtracking WAS
NOT taking place?
Now it seems to me that it's only a matter of visual verification to see
where the axis of lean roataion is for bikes (maybe a head-on shot of a
bike race where the bikes are coming off a straight and hitting a turn).
If the axis is the CG, then countersteering is due to outtracking and
precession is negligible. If the axis is the contact patches,
countersteering is due to precession and outtracking is negligible.
So far from what I have been reading, that is how I see it. But we newbie
edu-pukes are usually wrong; that's why we are going to school.
: translated into lean, AT the precession frequency. There is no
: outtracking so the bike is forced to lean around an axis at the contact
: patches (at a high moment of inertia). Furthermore, the axis of precession
Now whether this is true or not can be visually verified, right? I know
I'll be looking hard the next time I see a race.
Aaron S. Ward
wa...@nevada.edu
Wondering if any real-world observations even matter when it comes to
physics.
Ian Frechette
In article <6b31uq$9lc$1...@news.nevada.edu>, AARON WARD <wa...@nevada.edu> wrote:
>Ian Frechette (frec...@rintintin.Colorado.EDU) wrote:
>: inertia. To induce lean, you need only turn the front wheel out
>: from under the CG (which will try to continue on straight), while the
>
>This is "outtracking" right? I'm not *totally* sure of what outtracking
>means exactly.
Yes, this is outtracking.
>Can't we just watch a motorcycle race and see this phenonemon happen? It
>is possible to determine the CG of a leaning bike visually, isn't it?
It should be. But I'll get to this in a second.
>So what you're saying is that bikes DO NOT rotate while leaning about the
>contact patches, right? If they did, that would mean that outtracking WAS
>NOT taking place?
That's what I'm saying. I know, and I suspect everyone knows, at some
level that the bike doesn't rotate around the contact patches. It rotates
around it's CG (center of mass is probably more correct). It would feel
very very unnatural if it rotated around it's contact patches.
The actual axis of rotation changes a bit depending on how fast you're moving
because the rear tire doesn't exactly follow in the path of the front. The
axis is actually tilted up, closer to the ground at the back than
the front, but roughtly through the CG. I've explained that in some long
past post, and it explains why putting lots of weight up and behind the
rear wheel (like a tail box) is bad for handling.
>Now it seems to me that it's only a matter of visual verification to see
>where the axis of lean roataion is for bikes (maybe a head-on shot of a
>bike race where the bikes are coming off a straight and hitting a turn).
>
>If the axis is the CG, then countersteering is due to outtracking and
>precession is negligible. If the axis is the contact patches,
>countersteering is due to precession and outtracking is negligible.
I think we can all agree that outtracking IS occurring. The question
is still, is it causing the lean, or is it happening simply because
of a slight loss of efficiency between input torque and precessional torque?
So maybe when you push on the bars, 90% of the torque translates into
precession and that's enough to flick the bike over, the other 10% is lost
to head rotation, which results in outtracking, but that amount just
HAPPENS to make the front tire follow a path on the ground that will allow
the precesison torque to make the bike rotate around its CG. I think
this scheme is totally silly, but that's never stopped anyone before.
They can claim that precession is doing all the work, and the outtracking
just allows the precession to rotate the bike more freely.
It's much easier to believe, and I think to demonstrate, that 90% of the
torque goes into head rotation and that causes enough outtracking to
force the tires out from under the CG, inducing angular acceleration (lean)
around the long axis of the bike through the CG. The other 10% of the
torque may go into precession, which can't hurt, but I doubt it does
much good either.
>: translated into lean, AT the precession frequency. There is no
>: outtracking so the bike is forced to lean around an axis at the contact
>: patches (at a high moment of inertia). Furthermore, the axis of precession
>
>Now whether this is true or not can be visually verified, right? I know
>I'll be looking hard the next time I see a race.
What you'll see is the bike rotating around its CG.. What made it do that
is still a question in some people's mind. Not mine, but I'm trying
to be fair. The fact that outtracking works (and countersteering) for
balanced vehicles that have no rotating wheels, is probably the biggest
nail in the precessionist coffin. Maybe the First Church of the Immaculate
Precession can resurrect em.
>Aaron S. Ward
ian
Brian McLaughlin
feie...@nospam.utrc.utc.com (John M. Feiereisen) wrote:
>In <6at81t$g...@wiscnews.wiscnet.net>, bmcla...@waukesha.tec.wi.us
>(Brian McLaughlin) wrote:
>>Also the rotation of a motorcyle when leaning
>>over does not take place through the center of mass (for this to
>>happen the tires would have to slide sideways, 2ft high cg and 45 deg
>>lean the tires would have to slide 1.4 feet sideways), the rotation
>>takes place at the contact patch. Do your same experiment with the
>>bike sitting on the ground. Just let your bike fall over, where does
>>the rotation take place?
>About the contact patch.
>Now do this on a perfectly frictionless surface. You'll find that the
>tires do indeed slide out from under the bike and the CG goes pretty
>much straight down.
Of course, except I do not live in a world of zero friction.
>>Why should it be different if the bike is
>>moving (esp. the rear tire)?
>In actuality, the *instantaneous* center of rotation moves all over
>the map. Yes, at a standstill the center of rotation during a roll
>maneuver is at the contact patch. But this is only because the
>contact patch isn't moving sideways. When you're moving, the contact
>patch may be moving one way while the GC is moving another. You've
>got to look at two points on the machine, look at their instantaneous
>velocities, and find the intersection of the lines perpendicular to
>these velocity vectors. (As you roll on the throttle coming out of a
>corner, your instantaneous center of rotation might actually be way
>over your head.)
You still have not addressed the rear wheel. Where does the force come
from to move the rear wheel sideways? Afterall the rear wheel does not
steer into the turn. Have you ever raced? I have followed riders with
national and international experience. I have watched them lean a bike
from the very edge of the track lap after lap and they do not end up
in the dirt. I am an experimentalist, based upon my experience the
rotation is about the contact patch.
>The center of gravity is a convenient reference point for all your
>bookkeeping. You can completely describe the motion of the bike by
>specifying only the position, velocity, and acceleration of the GG and
>the angle, angular velocity, and angular acceleration (in two planes)
>of the bike about the GC. Note that the polar moment of inertia is
>dependent upon the location of the axis of rotation. The moment of
>inertia about the contact patch will be a lot higher than the moment
>of inertia about the GC. (What was that? Did somebody say "Parallel
>Axis Theorem"???)
Rotation about the contact patch will be aided by gravity, whereas
rotation about the cg will not. If rotation is about the cg, if rolled
quick enough, the suspension should extend and possibly the wheels
come off the ground. Never seen this happen. Go to a race track and
work on a corner and watch very carefully. I have seen no evidence
supporting roll about the cg. I do not see where the force comes from
to move the rear wheel as far as it should. Isn't friction at the
contact patches a constraint?
><pedant mode off>
To be proper you should be using the center of mass, not the center of
gravity. They are *not* the same.
Brian McLaughlin
frec...@rintintin.Colorado.EDU (Ian Frechette) wrote:
>>rotational frequency (and precession frequency) is in *radians* per
>>second.
>>
>>> X lb * 32 ft/sec^2 * 1 ft
>>>precession freq = -------------------------- = .045*X rps or 16X degrees/sec
>>> 20 lb * 1 ft^2 * 35/sec
>>
>>Revolutions/seconds are not the same as per second, rework with
>>radians per second.
>Actually, natural frequency, cycles per second *IS* 1/T
>f = 1/T and radian freq is 2*pi*f so rad freq = 2*pi/T
>I fed cycles/sec in, and I wanted cycles/sec out. If I wanted rads/sec
>out, I would have used converted cycles/sec to 2pi*rad/sec and
>stuck that in and gotten that out and divided by 2pi*rads to get cycles per
>sec again.. Nothing gained.
Doesn't work that way. Check your units. I got ft/(sec*rev) for the
units for your answer. The formulas do not work the way you are trying
to use them.
>A radian is only a measure of angle, and it disappears when you
>use it for something else. For instance, distance travelled
>by a wheel is circumfrence, with units of meters, multiplied
>by the number of radians travelled.
That is why frequency and rotational frequency both have units of Hz.
>meters * 2*pi radians and the units are in meters only.
>What's confusing is realizing that a frequency with units like
>f = X / (rad*sec) is just f = X / (2pi*sec)
>I could have reworked the eq with grams and meters too, and it would
>still come out the same. Actually, I did, and it did.
>I suggest you go back and find your old physics book. I had to, to be
>sure, when I worked this out the first time, because I got X/(rad*sec).
I did, they are here on my shelf at work. The equations you used
*must* use radians per second for the units of angular rotation. An
example in my book shows the solution divided by 2pi to get rot/time.
The problem started with rot/min, converted to get radians/minute,
found the precessional frequency and converted back to rot/min.
Oh yeah, I teach this stuff (college physics) for a living.
Rework with rad/sec and then convert answer to rpm and you will not
get the same answer.
Brian McLaughlin
>Ian Frechette <frec...@rintintin.Colorado.EDU> wrote:
Back in the days of flimsy forks, the addition of a fork brace gave
better feed back and made the steering feel much more positive (didn't
seem to be as much delay between input and turning).
Brian McLaughlin
frec...@rintintin.Colorado.EDU (Ian Frechette) wrote:
>>: translated into lean, AT the precession frequency. There is no
>>: outtracking so the bike is forced to lean around an axis at the contact
>>: patches (at a high moment of inertia). Furthermore, the axis of precession
>>
>>Now whether this is true or not can be visually verified, right? I know
>>I'll be looking hard the next time I see a race.
>What you'll see is the bike rotating around its CG.. What made it do that
>is still a question in some people's mind. Not mine, but I'm trying
>to be fair. The fact that outtracking works (and countersteering) for
>balanced vehicles that have no rotating wheels, is probably the biggest
>nail in the precessionist coffin. Maybe the First Church of the Immaculate
>Precession can resurrect em.
I am going to have to search through my videos to see if I still have
it. It was a straight on shot at the end of a straight. There were
several shots of bikes in a line. As each on peeled off in a turn, the
amount of wheel movement (out tracking) could be judged by the line of
wheels behind. It did not appear to be much.
How far do the wheels move (sideways) during out tracking? I don't see
why the rotation has to take place about the cm? The system has
constraints acting (friction).
My problem is not how the lean is caused, but I do not think the
rotation is around the cm. If out tracking is occuring, how far are we
talking about? When I lean hard into a corner (straight up to dragging
toes on a TZ250 ASAP) it does not feel like rotation is taking place
about the cm. What about at the edge of the track? I know I have been
within 6 inches of the edge of a track, leaned over (at least 45 deg)
and didn't end up in the dirt. If out tracking is taking place, the
wheels do not move to the side very much (could only be a couple of
inches). How does this work with rotation about the cm?
John M. Feiereisen
In <6b4hk5$18...@wiscnews.wiscnet.net>, bmcla...@waukesha.tec.wi.us
(Brian McLaughlin) wrote:
>feie...@nospam.utrc.utc.com (John M. Feiereisen) wrote:
>>In <6at81t$g...@wiscnews.wiscnet.net>, bmcla...@waukesha.tec.wi.us
>>(Brian McLaughlin) wrote:
>>>Also the rotation of a motorcyle when leaning
>>>over does not take place through the center of mass (for this to
>>>happen the tires would have to slide sideways, 2ft high cg and 45 deg
>>>lean the tires would have to slide 1.4 feet sideways), the rotation
>>>takes place at the contact patch. Do your same experiment with the
>>>bike sitting on the ground. Just let your bike fall over, where does
>>>the rotation take place?
>>About the contact patch.
>>Now do this on a perfectly frictionless surface. You'll find that the
>>tires do indeed slide out from under the bike and the CG goes pretty
>>much straight down.
>Of course, except I do not live in a world of zero friction.
Nor do I live in a world where a 2D, lumped parameter model for the
roll maneuver of a motorcycle includes all the relevent physics.
However, it can give a pretty darn good answer and still leave enough
flops available for doing important stuff, like playing Quake or
flaming people on rec.moto.
>>>Why should it be different if the bike is
>>>moving (esp. the rear tire)?
>>In actuality, the *instantaneous* center of rotation moves all over
>>the map. Yes, at a standstill the center of rotation during a roll
>>maneuver is at the contact patch. But this is only because the
>>contact patch isn't moving sideways. When you're moving, the contact
>>patch may be moving one way while the GC is moving another. You've
>>got to look at two points on the machine, look at their instantaneous
>>velocities, and find the intersection of the lines perpendicular to
>>these velocity vectors. (As you roll on the throttle coming out of a
>>corner, your instantaneous center of rotation might actually be way
>>over your head.)
>You still have not addressed the rear wheel.
Hello, Rear Wheel. How are you today? ;-)
>Where does the force come
>from to move the rear wheel sideways?
Believe it or not, it all starts at the front wheel contact patch.
And if you look very closely, you'll see that both it and the GC are
located ahead of the rear wheel contact patch. This makes a 2D
analysis a bit problematic. Include that pesky third dimension into
the mix, and you'll see that a lateral aceleration of the front wheel
contact patch will produce a lateral force at the rear wheel contact
patch.
>Afterall the rear wheel does not
>steer into the turn. Have you ever raced? I have followed riders with
>national and international experience. I have watched them lean a bike
>from the very edge of the track lap after lap and they do not end up
>in the dirt.
That's good. They probably wouldn't get much national or
international experience if they did that very often.
>I am an experimentalist, based upon my experience the
>rotation is about the contact patch.
Well, your experience is wrong. I'm an experimentalist as well. But
I make pretty heavy use of both simple and complex analytical models
to help me understand what it is I'm experiencing in the laboratory.
>>The center of gravity is a convenient reference point for all your
>>bookkeeping. You can completely describe the motion of the bike by
>>specifying only the position, velocity, and acceleration of the GG and
>>the angle, angular velocity, and angular acceleration (in two planes)
>>of the bike about the GC. Note that the polar moment of inertia is
>>dependent upon the location of the axis of rotation. The moment of
>>inertia about the contact patch will be a lot higher than the moment
>>of inertia about the GC. (What was that? Did somebody say "Parallel
>>Axis Theorem"???)
>Rotation about the contact patch will be aided by gravity, whereas
>rotation about the cg will not.
Rotation about the contact patch and rotation about the CG are the
same, except that you need to add a translation to one. (Rotation
about the contact patch plus a translation of the contact patch is
exactly equal to rotation about the GC.)
All you do, you do through your tires. All you can do through your
tires is generate forces (and, if you consider camber effects, small
yaw moments, excuse me, *torques*). To effect rotation of the bike
about the contact patch, you'd need to generate a roll torque at the
contact patch. Ain't gonna happen. To effect rotation of the bike
about the CG, all you need is a lateral force at the contact patch.
Assuming your bike isn't shod with Dai Yung tires, this is pretty easy
to do.
>If rotation is about the cg, if rolled
>quick enough, the suspension should extend and possibly the wheels
>come off the ground.
Yes, you're right.
>Never seen this happen.
You're not looking hard enough.
You can countersteer so severely that your front end becomes quite
light. I believe one instructor here (who, oddly, understands
countersteering) has even mentioned he's seen some of his better
students' front wheels leave the ground in an aggressive countersteer.
When you dive into a lean, the fastest you can ever hope accelerate
your CG toward the ground is at the rate of 1 g. You can very easily
roll your bike quickly enough that your suspension must extend to keep
the wheels on the ground before the rest of your bike falls. Somebody
here mentioned their goal of instrumenting a bike for a study of
countersteering. They might consider adding front and rear suspension
positions if they've got enough channels left.
>Go to a race track and
>work on a corner and watch very carefully. I have seen no evidence
>supporting roll about the cg.
Would it surprise you to find that you can analyse straight-line
translation as rotation about a point that is infinitely far away?
Yes, I thought so.
Would it surprise you to find out that the bike rolls about *neither*
the CG or the contact patch, and that its instantaneous roll center at
any moment in time could be anywhere between the bike's CG and the far
side of the Andromeda galaxy? (Multiply both sides of the equation by
the Hubble constant, and we might be able to extend that all the way
out to the limits of the visible universe.) Of course, the polar
moment of inertia of the bike about a point that far away would be
staggering.
>I do not see where the force comes from
>to move the rear wheel as far as it should. Isn't friction at the
>contact patches a constraint?
Of course it is. However, there's a bit of (what's it called in the
world of tire dynamics? Slip?). The direction of motion of the
contact patch isn't necessarily exactly perpendicular to the axle.
Even discounting this, and assuming that there's absolutely no motion
of the contact patch parallel to the axle, your front wheel steers
relative to your rear wheel. The front wheel contact patch can (and
does) move laterally. As a result, the line connecting the two points
describing the front and rear contact patches moves laterally relative
to the CG.
>To be proper you should be using the center of mass, not the center of
>gravity. They are *not* the same.
That's news to me. Are you sure you're not thinking of 'centroid'?
Death to Spam
Brian McLaughlin wrote:
> You still have not addressed the rear wheel. Where does the force come
> from to move the rear wheel sideways? Afterall the rear wheel does not
> steer into the turn. Have you ever raced? I have followed riders with
> national and international experience. I have watched them lean a bike
> from the very edge of the track lap after lap and they do not end up
> in the dirt. I am an experimentalist, based upon my experience the
> rotation is about the contact patch.
Ride through a puddle in a parking lot at low speed and then make a turn
afterward. You'll see that the front wheel does off-track and then turn
into the turn. The rear wheel tracks to the inside of the front.
Anyway, in slow turns you can get the front wheel to go on one side of
something and the rear on the other. I think the same thing happens at
higher speeds, but not to the same degree.
> Rotation about the contact patch will be aided by gravity, whereas
> rotation about the cg will not.
Maybe it's somewhere in between.
> If rotation is about the cg, if rolled
> quick enough, the suspension should extend and possibly the wheels
> come off the ground. Never seen this happen.
I have. Do some quick side to side swerves on a light bike and you can
sometimes get the front wheel to lift as the bike goes through
verticle. (Playing in parking lots with course bikes) This may be
mostly attributable to the undamped rebound from shitty, spongy forks,
though, but it does happen. I'm talking with just enough throttle to
maintain speed.
Ian Frechette
In article <6b4hk5$18...@wiscnews.wiscnet.net>,
Brian McLaughlin <bmcla...@waukesha.tec.wi.us> wrote:
>feie...@nospam.utrc.utc.com (John M. Feiereisen) wrote:
>Of course, except I do not live in a world of zero friction.
But you do ride in a moving frame of reference, which you don't seem
to be able to grasp.
>You still have not addressed the rear wheel. Where does the force come
Sigh.. The rear wheel simply follows the front, albiet not directly behind.
I have explained before that this does change the actual axis of rotation
a bit. It makes it not horizontal but slightly tilted up, from back to
front. It becomes less tilted with higher speeds because less head
rotation is needed to cause outtracking at higher speeds, so the rear
tire more closely follows the path of the front.
In EITHER case, the CG still wants to continue straight.
>from to move the rear wheel sideways? Afterall the rear wheel does not
>steer into the turn. Have you ever raced? I have followed riders with
>national and international experience. I have watched them lean a bike
>from the very edge of the track lap after lap and they do not end up
>in the dirt.
So do it yourself, but be sure not to throw your weight off the bike on
the inside of the turn before you initiate. Ride right down the line,
keep your body upright and then crank the bike over as hard as it needed.
See you in the dirt. Better yet, try the same trick at 35 or 40, instead
of 100. I've already stated that things may be a bit different at very
high speeds. I also think that in the circumstance you mention above
that people are getting off the line the only way they can, which is
by shifting their body weight to the inside of the turn, until they're a few
inches inside the line and then flopping it over via outtracking as normal.
>I am an experimentalist, based upon my experience the
>rotation is about the contact patch.
As am I, which is why I know it's around the CG when only torque is applied
to the bars, and I do not hang out.
>Rotation about the contact patch will be aided by gravity, whereas
You're correct.. But when the bike is balanced, it takes a long time
for it to fall from vertical to a usable angle. The only way you can
aid that process along, is to hang your body off the side, which
effectively changes the CG of the bike so that it is to one side or the
other of the contact patches. It is also true that if you do nothing
but lean off, the bike WILL rotate around it's contact patches, but
it does so VERY slowly, and if you maintain speed, you'll find that
it only leans so far. It will not continue to fall over, It'll just
ride around in a *big* circle. Try it..
>rotation about the cg will not. If rotation is about the cg, if rolled
>quick enough, the suspension should extend and possibly the wheels
>come off the ground. Never seen this happen. Go to a race track and
I, however, have. You must have missed the thread about this a while
back. I've done this very trick on small bikes. When the bike is snapped
quickly from vertical into a lean, the suspension does extend. If you snap
it from a hard lean one direction to a hard lean the other, the suspension
will compress and then extend somewhere just past center, and it is quite
possible to leap the front tire completely off the ground. I've even dumped
a scooter on the ground by doing this. It landed on it side with the tires
on one side of my track and the top of the bike on the other, exactly
where I'd expect. I have leaped the tire off the ground with my TDM
but not on purpose.
>work on a corner and watch very carefully. I have seen no evidence
>supporting roll about the cg. I do not see where the force comes from
>to move the rear wheel as far as it should. Isn't friction at the
>contact patches a constraint?
Friction at the contact patches is what makes it work. You're in a moving
frame of reference. The tires are FREE to turn under the bike. From the
bike's point of reference it appears that the front, followed by the rear tire
are moving sideways, while the bike rotates around it's CG. It has to, it
is a balanced object.
>To be proper you should be using the center of mass, not the center of
>gravity. They are *not* the same.
I've stated as much before, but everyone knows what we're talking about
here.
>Brian AP#1
ian
Brian McLaughlin
So how far do the wheels move laterally in a turn?
>>To be proper you should be using the center of mass, not the center of
>>gravity. They are *not* the same.
>That's news to me. Are you sure you're not thinking of 'centroid'?
Nope, look up the calculation of cm and cg. the formulas are not the
same. For the cg and cm to be in *exactly* the same spot, the
gravitational field has to be constant across the object. For small
objects the difference is very small, but they are not the same.
Brian McLaughlin
Death to Spam <ab...@aol.edu> wrote:
>Brian McLaughlin wrote:
>> If rotation is about the cg, if rolled
>> quick enough, the suspension should extend and possibly the wheels
>> come off the ground. Never seen this happen.
>I have. Do some quick side to side swerves on a light bike and you can
>sometimes get the front wheel to lift as the bike goes through
>verticle. (Playing in parking lots with course bikes) This may be
>mostly attributable to the undamped rebound from shitty, spongy forks,
>though, but it does happen. I'm talking with just enough throttle to
>maintain speed.
I have done and seen this also, but this is not what i was refering
to. What you are speaking about is caused by inertia. The cm of the
bike is moving upward, and wants to keep going upward. Gravity must
stop it from going up. Due to inertia the cm will 'overshoot' the
ideal turn around point. Done quickly enough the bike can come off the
ground.
I am referring to leaning into a corner, not a transition from one
side to another. If rotation is about the cm, as you lean, the contact
patch will be moving away from the cm (gravity will cause the cm to
fall, lessening this distance, but the cm can only fall as fast as
gravity allows). If the bike is rotated extremely fast from vertical
to full lean, the suspension may not be able to extend enough to
account for the difference in distance between cm and the contact
patch. If this happens the wheels will come off the ground.
Rick and/or Sara Baartman
Brian McLaughlin wrote:
>
> Nope, look up the calculation of cm and cg. the formulas are not the
> same. For the cg and cm to be in *exactly* the same spot, the
> gravitational field has to be constant across the object. For small
> objects the difference is very small, but they are not the same.
>
Yeah for objects the size of a motorcycle, the difference is about one
part in ten million. In other words, it is insignificant even when
compared with the difference between c.of g. for rider with sunglasses
on as compared with sunglasses off. C.of g. and c.of m. are not equal.
True statement. But not all true statements are relevant. What's the
point of bringing it up?
--
rick
andy the pugh
Trevor Dennis <tre...@tdennis.demon.co.uk> wrote:
> A second set of handlebars mounted to the first via a load cell?
> That might be more effort than he wants to go to.
>
Strain gauging the clip ons or bars shoud do the trick, you will be down
near the resolution limits, but you should get somthing.
--
ap
Ian Frechette
In article <6b751v$l...@wiscnews.wiscnet.net>,
Brian McLaughlin <bmcla...@waukesha.tec.wi.us> wrote:
>So how far do the wheels move laterally in a turn?
Picture this example. Make a bike that has no rake angle so the forks
point straight down. Now make it so that you can turn the front tire
a full 90 degrees, so that it is perpendicular to the direction of the
bike. Stand the bike up vertically with the front wheel turned 90
degrees. Let the bike go?
What will happen? Anyone who's ever ridden a unicycle knows.
It'll start to tip slowly.. say to the left, due to the force of gravity,
but as soon as it's even a little non-vertical the front tire will squirt
out from under the bike to the right. The whole bike will crash to the
ground with the front tire a foot and a half to the right and the handlbars
a foot and a half to the left. The rear tire will basically stay
where it was.. The axis of rotation was through the rear contact patch
up through the engine (or thereabouts). It did not rotate around the
front contact patch. The front tire tried to roll out from under the CM
and the CM basically tried to fall straight down.
Ok, now set the bike in motion.. Walking along beside it,
at 4mph.. 6 feet/sec In one quick movement, turn the front wheel only
45 degrees to the right and let go of the bike.
The path of the front tire looks like ------------\,
\
The path of the rear tire looks like -------__
The path of the CM looks like ------------X
^--- Point of impact
Once again the front tire will squirt out to the right of the CM
the CM wants to continue on straight and the bars will fall to the
left of the CM. The axis of rotation is near the middle of the bike
in the front and still very near the rear contact patch in the back.
The front tire continued to go 6 ft/sec but at a 45 degree angle therefore
with a sideways vector of 4.2 ft/sec. In less than a second it will have
attempted to move 4.2 feet to the right of it's original path.. the
path the CM is still on.
But notice that the rear tire started to follow a curved path of it's own.
From the CM, both the front and the rear tire appeared to move to the
right, the front much more than the rear.
At the same slow speed, if you had not turned the bars far enough, the sideways
component would have been too small to allow the CM to fall straight down
and thus it would have fallen further to the left and the axis of rotation
would have been lower.. closer to the contact patches.
So what happens when you go faster? If I go 60 ft/sec instead of 6..
I only have to turn the bars 4 degrees to acheive the same 4.2 ft/sec
sideways velocity component, but both the front and the rear tire travel 60 feet
in that second. Most importantly, the rear tire has plenty of time to follow
the front tire's path, by the time the bike hits the ground, both the front
and the rear tire are following the same path (see note 1 below), and are
both going 4.2ft/sec to the side and if the wheelbase is 5 feet then the
rear tire is only 4 inches closer to the original path than the front.
From the CM both the front and rear wheel are moving to the side.
note 1: I'm assuming that the 4 degree turn is made, and then the front
tire is held on a fixed line 4 degrees from the original path so the front
and rear tires line up together along this new straight path. If you
continued to hold a 4 degree rotation of the bars with respect to the bike
then the front tire would continue to follow a curved path to the right,
the bike would lean much faster and you'd hit the ground much much sooner.
Obviously the real numbers are much less dramatic. You only need to move
out a foot or so to induce serious lean, and once you've established slight
outtracking, you need only hold the line or even start correcting
as the bike leans over.
If you can find an old beater of a bicycle to play with you should be
able to demonstrate this whole thing. Just tighten down the head bearings
a bit so that the bars stick in whatever orientation you put them.
Could tie a long string to one bar, give the bike a push and when it reaches
the end of the string, let it tug just enough to turn and let go of the string
and watch what it does. Might even be more fun if you left the head
tightness alone and watched how the bike leans and then corrects itself.
[snip].. CG versus CM
>Nope, look up the calculation of cm and cg. the formulas are not the
>same. For the cg and cm to be in *exactly* the same spot, the
>gravitational field has to be constant across the object. For small
>objects the difference is very small, but they are not the same.
I suppose if we were riding around a small asteroid this might be worrying.
>Brian AP#1
ian
Ian Frechette
In article <6b4p89$17...@wiscnews.wiscnet.net>,
Brian McLaughlin <bmcla...@waukesha.tec.wi.us> wrote:
>frec...@rintintin.Colorado.EDU (Ian Frechette) wrote:
>>>
>>>> X lb * 32 ft/sec^2 * 1 ft
>>>>precession freq = -------------------------- = .045*X rps or 16X degrees/sec
>>>> 20 lb * 1 ft^2 * 35/sec
>Doesn't work that way. Check your units. I got ft/(sec*rev) for the
Well, for one the ft shouldn't be there, and I can prove that right quick.
The original EQ is
precession freq = torque/(moment of inertia * rotational freq)
torque = F * d (distance from axis of rotation)
F = mass * accel I'm using lb as a unit of mass and not as a force
F = lb * ft/sec^2
d = ft
moment of Inertia = MR^2 = lb*ft^2
So the units of the eq are
lb * ft/sec^2 * ft
pfreq = --------------------
lb*ft^2 * rfreq
So you're left with only (1/sec^2)/rfreq
All the other units are gone. I'm surprised you'd make such a mistake.
Same thing happens if you work it with grams and meters..
So on to the frequency question.
It's simple. natural frequency has units of 1/sec, period.. not
rev/sec, cycles/sec.. just 1/sec
A radian is no more a unit than a cycle or a rev. It is however a multiplier
of them. 1 radian = 1/2pi cycle and therefore JUST 1/2pi no units.
So rework the eq with rad/sec as rfreq and you get
1/sec^2 1 1 2pi
-------- = ------ = ----------- = ------ (units of 1/s)
X rad/sec X rad*sec X 1/2pi*sec X sec
That's all. To convert it to rad/sec you have to multiply by 2pi again
for (2pi)^2 rad/(X*sec)
It's not magic, it's not cheating. The fundamental conversion you're lacking
are:
1 1 1 2pi*2pi rad
--------- = ----- and ------ = ------------
cycle*sec sec rad*sec sec
I knew this was right when I started, but I just broke out my hp48 (relegated
to alarm clock duty for the last few years) and let it do some unit
conversions.
1_rad/sec has base units of 1/(2pi*sec)
1_(rad*sec) has base units of 2pi/sec
Obviously 1_rad/(rad*sec) falls in the middle with units of 1/sec
1_rad has base units of 1/2pi
>units for your answer. The formulas do not work the way you are trying
>to use them.
In fact, they work exactly the way I'm using them. Watch.
Xn = 33/sec natural frequency (150mph.. 24 inch tire)
Xr = Xn*2pi = 207 rad/sec radian frequency
pfreq = precession freq
Set every lb and ft to 1.. let them all cancel out.
Using Xn..
pfreq = (1/sec^2)/(Xn*sec) = 1/(33*sec) = .0303 cycles/sec
Using Xr
pfreq = (1/sec^2)/(Xr*sec) = 1/(207*rad*sec) .. multiply by (2pi)^2 to
convert to rad/sec.. = .19 rad/sec
divide by 2pi to get natural freq again.. guess what.. .0303 cycles/sec
>>I suggest you go back and find your old physics book. I had to, to be
>>sure, when I worked this out the first time, because I got X/(rad*sec).
>
>I did, they are here on my shelf at work. The equations you used
>*must* use radians per second for the units of angular rotation. An
>example in my book shows the solution divided by 2pi to get rot/time.
No they could use 360 degrees/sec except that every instance of degrees
would simply be substituted for 1/360
>The problem started with rot/min, converted to get radians/minute,
>found the precessional frequency and converted back to rot/min.
Try rerunning their EQ with Hz. However, I would like to
see the EQ they used first. I suspect that they've already got
a multiplier somewhere in the EQ to handle converting to rad/sec.
A (2pi)^2 would do it. Perhaps it's hidden in the moment of inertia.
Mail me the problem, offline. I want to see it from beginning to
end. The 1/sec^2 divided by rad/sec flips this EQ on it's head.
The final units are NOT in rad/sec without some conversion first.
I want to see how they handled that. It works much cleaner using
the natural frequency.
>Oh yeah, I teach this stuff (college physics) for a living.
That sorta scares me.
>Rework with rad/sec and then convert answer to rpm and you will not
>get the same answer.
I did, and I did. I also finally plugged the whole EQ into my
hp48 tonight with units of Hz and rad/sec and the result is the same. I do
find it puzzling that most physics books don't explain radians better though.
>Brian AP#1
Perhaps you could give this problem to your class and see what units
they get. ;)
ian
P.S. I noticed that you stripped out my treadmill example. Does a moving
frame of reference make sense yet?
CyberWoof
bmcla...@waukesha.tec.wi.us (Brian McLaughlin) wrote...
>feie...@nospam.utrc.utc.com (John M. Feiereisen) wrote:
<snip>
>>[someone else wrote]:
<snip>
>>>To be proper you should be using the center of mass, not the center of
>>>gravity. They are *not* the same.
>
>>That's news to me. Are you sure you're not thinking of 'centroid'?
>
>Nope, look up the calculation of cm and cg. the formulas are not the
>same. For the cg and cm to be in *exactly* the same spot, the
>gravitational field has to be constant across the object. For small
>objects the difference is very small, but they are not the same.
The center of mass is calculated using parallel lines; the center of gravity is
calculated using lines that converge at the center of mass of the other body --
in this case, the Earth. In this particular case, it's a difference that makes
no difference. The gravitational vectors coming out of the earth on the area of
a motorcycle are so close to parallel that you can't measure the error. If you
built a motorcycle the size of the Golden Gate Bridge, then you might have to
worry about this ... but financing and licensing the damn thing would be a
bigger headache. Jugglers, basketball players, and tennis players don't
generally worry about this minute difference. I suggest that we don't either.
--Timberwoof at aol dot com - http://members.aol.com/timberwoof
Spam Reading Offer: http://members.aol.com/timberwoof/spamoff.html
they sent me spam:
Do...@merteuil.com
sple...@earthlink.net
canc...@juno.com
vit...@mailexcite.com
John M Feiereisen
In article <34D7A6...@triumf.ca>,
Rick and/or Sara Baartman <kr...@triumf.ca> wrote:
>Brian McLaughlin wrote:
>> Nope, look up the calculation of cm and cg. the formulas are not the
>> same. For the cg and cm to be in *exactly* the same spot, the
>> gravitational field has to be constant across the object. For small
>> objects the difference is very small, but they are not the same.
>part in ten million. In other words, it is insignificant even when
>compared with the difference between c.of g. for rider with sunglasses
>on as compared with sunglasses off. C.of g. and c.of m. are not equal.
>True statement. But not all true statements are relevant. What's the
>point of bringing it up?
I think you're missing the point. There is no point in bringing it up.
This is rec.moto, remember.
--
John
I'm only posting from this account because the Usenet guy's truck is
broken and he hasn't come to pick up my posts since the evening before
last.
John M. Feiereisen
In <6b751v$l...@wiscnews.wiscnet.net>, bmcla...@waukesha.tec.wi.us
(Brian McLaughlin) wrote:
>So how far do the wheels move laterally in a turn?
I would guess it depends on the particular corner, your speed, and
your riding style. It could be a little, or it could be a lot. It
all depends on how much of a lean you develop before the bike starts
tracing a curved path. If you lean the bike way over before you
really start turning, the contact patches will move to the outside of
the turn (and the roll center will be near the CG). If you start the
turn immediately as you start the bike leaning, the contact patches
will move only a little toward the outside of the turn (and the roll
center will be near, but not at, the contact patches).
It's pretty well accepted that at low speeds precession is only a
minor player and that outtracking is the dominant effect. Outtracking
will tend to roll the bike about its CG. At high speeds, precession
is more pronounced. Precession 'returns' your steering torque along
roll axis. This roll torque, coupled with the constraint the tires
place on the system, would tend to rotate the bike about the contact
patch. However, there is *always* some degree of outtracking, so the
roll center will not be at the contact patch. It may be close though.
So, at low speeds outtracking dominates and the roll center is near
the CG. At high speeds precession probably dominates and the roll
center is near the contact patch.
But still, Ian Frechette's bit about the precession torque never
exceeding the steering torque makes me believe that outtracking still
has to be significant. Otherwise your arms would get tired providing
all the roll torque.
>>>To be proper you should be using the center of mass, not the center of
>>>gravity. They are *not* the same.
>>That's news to me. Are you sure you're not thinking of 'centroid'?
>Nope, look up the calculation of cm and cg. the formulas are not the
>same. For the cg and cm to be in *exactly* the same spot, the
>gravitational field has to be constant across the object. For small
>objects the difference is very small, but they are not the same.
Yeeeeeow! That one flew in under the radar. I'd like to call a
nonuniform gravity field cheating, but it really isn't. Now I've got
pedant envy. I am humbled.
Now, I think it's only fitting that we include relativistic effects.
Even at speeds as low as DoD nominal, they could be 'significant' if
you use enough decimal places...
Brian McLaughlin
Rick and/or Sara Baartman <kr...@triumf.ca> wrote:
>Brian McLaughlin wrote:
>>
>> Nope, look up the calculation of cm and cg. the formulas are not the
>> same. For the cg and cm to be in *exactly* the same spot, the
>> gravitational field has to be constant across the object. For small
>> objects the difference is very small, but they are not the same.
>>
>Yeah for objects the size of a motorcycle, the difference is about one
>part in ten million. In other words, it is insignificant even when
>compared with the difference between c.of g. for rider with sunglasses
>on as compared with sunglasses off. C.of g. and c.of m. are not equal.
>True statement. But not all true statements are relevant. What's the
>point of bringing it up?
Someone mentioned being pedantic and also most people do not know that
there is a difference.
mull...@advinc.com
In article <6b9o1q$j...@mozo.cc.purdue.edu>,
feie...@widget.ecn.purdue.edu (John M Feiereisen) wrote:
> I think you're missing the point. There is no point in bringing it up.
> This is rec.moto, remember.
> --
> John
Hey. Watch it there. Not all of us are the shaft driven, countersteered
jumpstarted gun fanatics that one would think.
> I'm only posting from this account because the Usenet guy's truck is
> broken and he hasn't come to pick up my posts since the evening before
> last.
And I'm only posting form this account 'cause my boss said "how come
you spend all your time fooling around with the damn usenet???"
Jim (formerly JR...@watson.ibm.com)
Ian Frechette
In article <34D7A6...@triumf.ca>,
>on as compared with sunglasses off. C.of g. and c.of m. are not equal.
>True statement. But not all true statements are relevant. What's the
>point of bringing it up?
He says he's a physicist.. You sound like more of an engineer. I remember
how much all my physics profs hated the engineers because they were
always rounding away the insignificant bits. Same answer to 5 decimal
places but the eq just isn't right.. ;)
>rick
ian
Rick Baartman
>>>>> "Frechette" == Ian Frechette <frec...@rintintin.Colorado.EDU>
>>>>> wrote the following on 3 Feb 1998 06:31:25 GMT
Frechette> precession freq = torque/(moment of inertia * rotational freq)
Frechette> So on to the frequency question.
Frechette> It's simple. natural frequency has units of 1/sec,
Frechette> period.. not rev/sec, cycles/sec.. just 1/sec
Frechette> A radian is no more a unit than a cycle or a rev. It is
Frechette> however a multiplier of them. 1 radian = 1/2pi cycle and
Frechette> therefore JUST 1/2pi no units.
Sorry, Ian, Brian is right on this one. The frequencies in the above
equation are ANGULAR frequencies. Ang.freq.=2pi*cycling freq. Therefore,
in terms of cycling freq., the above eqn. becomes
2pi*precession cyc.freq.=torque/(mom.of inertia*2pi*rot.cyc.freq.)
or,
prec.cyc.freq.=torque/(4*pi*pi*mom.of inertia*rot.cyc.freq.)
In other words, your calculations of prec.freq. are about FORTY times too
large.
BTW, for the same reason the formula
power=torque*frequency
becomes
power=2pi*torque*cyc.freq.
Converting power from ft*lb/sec to hp and cyc/sec to cyc/min (RPM), we get
(power in hp)=2pi*(torque in ft*lb)*RPM/60/550
yielding the well-known
(power in hp)=(torque in ft*lb)*RPM/5252,
since 60*550/(2pi)=5252.
Hope this helps.
--
rick baartman
(delete the extra 'a' when replying)
Ian Frechette
In article <j4en1fp...@alph04.triumf.ca>,
Rick Baartman <baaa...@alph04.triumf.ca> wrote:
>Sorry, Ian, Brian is right on this one. The frequencies in the above
>equation are ANGULAR frequencies. Ang.freq.=2pi*cycling freq. Therefore,
>in terms of cycling freq., the above eqn. becomes
>
>2pi*precession cyc.freq.=torque/(mom.of inertia*2pi*rot.cyc.freq.)
>prec.cyc.freq.=torque/(4*pi*pi*mom.of inertia*rot.cyc.freq.)
Uggg..
2pi radians/sec = 1 cycle/sec = 1 Hz = 1/sec
Therefore 2pi radians = 1 cycle = 1
Let me repeat that.. 2pi rad = 1
The problem is you threw away the radians/sec on both sides of
the EQ. You cannot do that. Normally you'd never notice
if the radians disappeared from the EQ, but because the final result is
not in rad/sec, but in 1/(rad*sec) it makes a big difference to this EQ.
If you want to divide through, divide though by 2pi rad instead of
just 2pi.
If Pn is natural frequency of precession. Y/sec then radian frequency
of precession is 2*pi*Y rad/sec
If Xn is natural frequency of rotation.. X/sec then radian frequency of
rotation = 2pi*X rad/sec
Then the eq must be
2pi*Y rad/sec = torque/(mom.of.inertia*2pi*X rad/sec)
divide through by 2pi rad
Y/sec = torque/(mom.of.inertia * 2pi rad * 2pi rad * X /sec)
But since 2pi rad = 1, both instances of 2pi rad go away..
If you don't like that, then prove to me that 2pi radians does not equal
1 cycle and that 1 cycle/sec does not have base units of 1/sec.
2pi rad/sec = 1Hz = 1/sec Prove it ain't so.
>In other words, your calculations of prec.freq. are about FORTY times too
>large.
If only that were true. It would really drive a stake through the heart
of the precessionism theory, because then the precession rate would be
so slow at ALL speeds as to be totally worthless. I'm pretty sure that
precession torque is mostly worthless for inducing lean, but the
precession frequencies are certainly reasonable.
[HP EQ snipped]
I don't see the relevance of the hp eq here. HP was an arbitrarily
created unit by Watt. The original 550 constant could have been 5252 to begin
with and Watt could have specified that you plug in rpm instead of
radians/sec and it would work as you know it now.
I could also specify that 1 hp = 31512 ft-lb/sec
and specify that you enter the frequency in degrees per second
into hp = torque*frequency and the results would once again, be the same.
In the precession equation, we're plugging a frequency in
and we want a frequency out. I can plug in Hz and get out rad/sec or
plug in rad/sec and get Hz. The key is that a radian is just a constant
that represents the value 1/2pi with no base units of its own.
>rick baartman
ian
andy the pugh
Ian Frechette <frec...@rintintin.Colorado.EDU> wrote:
> He says he's a physicist.. You sound like more of an engineer. I remember
> how much all my physics profs hated the engineers because they were
> always rounding away the insignificant bits.
That's strange, I did a physics degree and we were always happy to work
to the nearest order of magnitude.
--
ap
Brian McLaughlin
frec...@rintintin.Colorado.EDU (Ian Frechette) wrote:
>In article <6b4p89$17...@wiscnews.wiscnet.net>,
>Brian McLaughlin <bmcla...@waukesha.tec.wi.us> wrote:
>>frec...@rintintin.Colorado.EDU (Ian Frechette) wrote:
>>>>
>>>>> X lb * 32 ft/sec^2 * 1 ft
>>>>>precession freq = -------------------------- = .045*X rps or 16X degrees/sec
>>>>> 20 lb * 1 ft^2 * 35/sec
ok, lets see:
processional freq (since omega is used, this is angular frequency)
= X lb*32ft/s^2 * 1ft/(20lb * 1ft^2 * 220 rad/sec) = .0073X rad/sec =
0012X rev/sec = .042X deg/sec.
Doesn't give your answer.
Units of Hz applies to frequency (f or nu in physics texts) and
angular frequency (small omega). Rev/sec are not the same units. The
2pi is a conversion factor between radians and rotations and carries
units. Rev/sec is not the same as Hz.
>>Doesn't work that way. Check your units. I got ft/(sec*rev) for the
>Well, for one the ft shouldn't be there, and I can prove that right quick.
>The original EQ is
No it shouldn't. Found my error. But it still leaves 1/(sec*rev) which
does not equal rev/sec. Revolutions cannot be dropped like radians. By
definition radians are unitless (the ratio of arc length/radius) the
same cannot be said for rotations.
>precession freq = torque/(moment of inertia * rotational freq)
>torque = F * d (distance from axis of rotation)
>F = mass * accel I'm using lb as a unit of mass and not as a force
>F = lb * ft/sec^2
>d = ft
>moment of Inertia = MR^2 = lb*ft^2
>So the units of the eq are
> lb * ft/sec^2 * ft
>pfreq = --------------------
> lb*ft^2 * rfreq
>So you're left with only (1/sec^2)/rfreq
>All the other units are gone. I'm surprised you'd make such a mistake.
>Same thing happens if you work it with grams and meters..
But that leaves you with sec/rotation. Rotation does not disappear as
does radians. Look at the equation for torque. Torque = moment of
inertia (I) * angular acceleration. For the equation to work out the
units of angular accleration must match those for angular velocity.
Units of ft-lbs for torque means that rad/s^2 were used for angular
acceleration. Look at the fotmula:
I * rad/sec^2
Angular velocity of precession = ------------------------------------
(rad/sec) I * rad/sec
The 2 moment of inetia terms will cancel leaving 1/sec or rad/sec.
Recall the the unit of angular measure is the rad, angular velocity is
the rad/sec, and angular acceleration is rad/sec^2. Multiplying both
sides by 1/2pi does not give the correct units. The two angular
frequency terms are not both located correctly for this to happen, one
is in the numerator and the other is in the denominator.
>So on to the frequency question.
>It's simple. natural frequency has units of 1/sec, period.. not
>rev/sec, cycles/sec.. just 1/sec
"... the *angular velocity* of precession about the z-axis is
omega sub p = d(phi)/dt = torque/angular momentum.
This result is known as simple precession."
p170 Classical Mechanics by Barger and Olsson.
You are solving for an angular velocity, not a frequency. Angular
velocity is often referred to as angular frequency since the units of
angular velocity are rad/ sec, or Hz.
>A radian is no more a unit than a cycle or a rev. It is however a multiplier
>of them. 1 radian = 1/2pi cycle and therefore JUST 1/2pi no units.
Rev *is* a unit and cannot be ignored in the unit analysis!
>So rework the eq with rad/sec as rfreq and you get
>1/sec^2 1 1 2pi
>-------- = ------ = ----------- = ------ (units of 1/s)
>X rad/sec X rad*sec X 1/2pi*sec X sec
>That's all. To convert it to rad/sec you have to multiply by 2pi again
>for (2pi)^2 rad/(X*sec)
>It's not magic, it's not cheating. The fundamental conversion you're lacking
>are:
> 1 1 1 2pi*2pi rad
>--------- = ----- and ------ = ------------
>cycle*sec sec rad*sec sec
This is true. You are using rev/sec which is not Hz (cycles/sec)!!!!
>I knew this was right when I started, but I just broke out my hp48 (relegated
>to alarm clock duty for the last few years) and let it do some unit
>conversions.
>1_rad/sec has base units of 1/(2pi*sec)
>1_(rad*sec) has base units of 2pi/sec
>Obviously 1_rad/(rad*sec) falls in the middle with units of 1/sec
>1_rad has base units of 1/2pi
>>units for your answer. The formulas do not work the way you are trying
>>to use them.
>In fact, they work exactly the way I'm using them. Watch.
>Xn = 33/sec natural frequency (150mph.. 24 inch tire)
Xn has units rev/sec. You cannot just drop the rev unless you have
redefined torque using angular acceleration of rev/sec^2.
>Xr = Xn*2pi = 207 rad/sec radian frequency
>pfreq = precession freq
>Set every lb and ft to 1.. let them all cancel out.
>Using Xn..
>pfreq = (1/sec^2)/(Xn*sec) = 1/(33*sec) = .0303 cycles/sec
Units are .0303 1/rev*sec.
>Using Xr
>pfreq = (1/sec^2)/(Xr*sec) = 1/(207*rad*sec) .. multiply by (2pi)^2 to
>convert to rad/sec.. = .19 rad/sec
1/(207 sec) *is* in rad per sec! Multiply by 1/2pi to convert to
rev/sec, .00077 rev/sec.
>divide by 2pi to get natural freq again.. guess what.. .0303 cycles/sec
The formula is solving for angular velocity, not for natural
frequency! Look up the derivation of the formula.
>>>I suggest you go back and find your old physics book. I had to, to be
>>>sure, when I worked this out the first time, because I got X/(rad*sec).
>>
>>I did, they are here on my shelf at work. The equations you used
>>*must* use radians per second for the units of angular rotation. An
>>example in my book shows the solution divided by 2pi to get rot/time.
>No they could use 360 degrees/sec except that every instance of degrees
>would simply be substituted for 1/360
No. The definition of angular displacement is the radian. The formula
that you used is solving for angular velocity (or angular frequency),
this is not the same as the frequency that you are thinking of.
>>The problem started with rot/min, converted to get radians/minute,
>>found the precessional frequency and converted back to rot/min.
>Try rerunning their EQ with Hz. However, I would like to
>see the EQ they used first. I suspect that they've already got
>a multiplier somewhere in the EQ to handle converting to rad/sec.
>A (2pi)^2 would do it. Perhaps it's hidden in the moment of inertia.
No, see above. There is no 2pi hidden in the equation. Since the
precession is defines as d(phi)/dt where phi is the angle that the
angular momentum is sweeping out, the units of precession are those of
an angular velocity. Look up the *derivation* of the formula.
>Mail me the problem, offline. I want to see it from beginning to
>end. The 1/sec^2 divided by rad/sec flips this EQ on it's head.
>The final units are NOT in rad/sec without some conversion first.
>I want to see how they handled that. It works much cleaner using
>the natural frequency.
>>Oh yeah, I teach this stuff (college physics) for a living.
>That sorta scares me.
It would if you were in my class :)
>>Rework with rad/sec and then convert answer to rpm and you will not
>>get the same answer.
>I did, and I did. I also finally plugged the whole EQ into my
>hp48 tonight with units of Hz and rad/sec and the result is the same. I do
>find it puzzling that most physics books don't explain radians better though.
The results from Hz and rad/sec should be the same. The problem is
that rev/sec are *not* Hz. You cannot simply drop the rev. Radians are
defined by the formula for arc length and are therefore unitless. The
way rotation is setup, rev/sec are not Hz. Besides the formula is not
solving for frequency, but angular velocity.
Don't use your conversion, I got different answers than you did.
>>Brian AP#1
>Perhaps you could give this problem to your class and see what units
>they get. ;)
> ian
>P.S. I noticed that you stripped out my treadmill example. Does a moving
> frame of reference make sense yet?
I am still waiting to see the numbers for the amount the wheels move
out. Outtracking takes place, but I do not see how the rotation can
occur about the cm. The amount of outtracking would be greater than
what is seen.
Rick Baartman
>>>>> "andy" == andy the pugh <a.c....@shef.ac.uk>
>>>>> wrote the following on Mon, 9 Feb 1998 00:39:05 +0000
andy> Ian Frechette <frec...@rintintin.Colorado.EDU> wrote:
>> He says he's a physicist.. You sound like more of an engineer. I
>> remember how much all my physics profs hated the engineers because
>> they were always rounding away the insignificant bits.
andy> That's strange, I did a physics degree and we were always happy to
andy> work to the nearest order of magnitude.
I find Ian's comments extremely puzzling as well. I'm still in physics, and
I deal with lots of engineers. My experience is that when I say two devices
should be around 1 inch apart, a drawing comes back from engineering with
their locations specified to within 0.001 inch.
Brian McLaughlin
frec...@rintintin.Colorado.EDU (Ian Frechette) wrote:
>In article <6b4hk5$18...@wiscnews.wiscnet.net>,
>Brian McLaughlin <bmcla...@waukesha.tec.wi.us> wrote:
>>feie...@nospam.utrc.utc.com (John M. Feiereisen) wrote:
>>Of course, except I do not live in a world of zero friction.
>But you do ride in a moving frame of reference, which you don't seem
>to be able to grasp.
>>You still have not addressed the rear wheel. Where does the force come
>Sigh.. The rear wheel simply follows the front, albiet not directly behind.
>I have explained before that this does change the actual axis of rotation
>a bit. It makes it not horizontal but slightly tilted up, from back to
>front. It becomes less tilted with higher speeds because less head
>rotation is needed to cause outtracking at higher speeds, so the rear
>tire more closely follows the path of the front.
The problem I am having is the distance the wheels have to move
sideways (for rotation about the cm) compared to the distance traveled
forward. Full lean can be reached in a fraction of a second. The
wheels do not seem to move very far to the outside.
>In EITHER case, the CG still wants to continue straight.
>>from to move the rear wheel sideways? Afterall the rear wheel does not
>>steer into the turn. Have you ever raced? I have followed riders with
>>national and international experience. I have watched them lean a bike
>>from the very edge of the track lap after lap and they do not end up
>>in the dirt.
>So do it yourself, but be sure not to throw your weight off the bike on
>the inside of the turn before you initiate. Ride right down the line,
>keep your body upright and then crank the bike over as hard as it needed.
I have done it leaning the wrong way (one track has 2 rights connected
by a short left kink, easiest to just stay on the right side). Doesn't
seem to steer significantly different.
>See you in the dirt. Better yet, try the same trick at 35 or 40, instead
>of 100. I've already stated that things may be a bit different at very
>high speeds. I also think that in the circumstance you mention above
>that people are getting off the line the only way they can, which is
>by shifting their body weight to the inside of the turn, until they're a few
>inches inside the line and then flopping it over via outtracking as normal.
>>I am an experimentalist, based upon my experience the
>>rotation is about the contact patch.
>As am I, which is why I know it's around the CG when only torque is applied
>to the bars, and I do not hang out.
>>Rotation about the contact patch will be aided by gravity, whereas
>You're correct.. But when the bike is balanced, it takes a long time
>for it to fall from vertical to a usable angle. The only way you can
>aid that process along, is to hang your body off the side, which
>effectively changes the CG of the bike so that it is to one side or the
>other of the contact patches. It is also true that if you do nothing
>but lean off, the bike WILL rotate around it's contact patches, but
>it does so VERY slowly, and if you maintain speed, you'll find that
>it only leans so far. It will not continue to fall over, It'll just
>ride around in a *big* circle. Try it..
Why if I lean off and move cm to one side will the bike rotate about
the contact patch, yet when the bike outtracks and the cm is no longer
above the wheels the rotation suddenly takes place about the cm. Isn't
the idea of outtracking to move the cm from above the wheels? Then
gravity should lean the bike. So what is the difference between the 2
situations. For outtracking the only vertical force I see is gravity.
What other forces help to lean the bike?
>>rotation about the cg will not. If rotation is about the cg, if rolled
>>quick enough, the suspension should extend and possibly the wheels
>>come off the ground. Never seen this happen. Go to a race track and
>I, however, have. You must have missed the thread about this a while
>back. I've done this very trick on small bikes. When the bike is snapped
>quickly from vertical into a lean, the suspension does extend. If you snap
>it from a hard lean one direction to a hard lean the other, the suspension
>will compress and then extend somewhere just past center, and it is quite
>possible to leap the front tire completely off the ground. I've even dumped
>a scooter on the ground by doing this. It landed on it side with the tires
>on one side of my track and the top of the bike on the other, exactly
>where I'd expect. I have leaped the tire off the ground with my TDM
>but not on purpose.
Of course the rotation takes place about the cm when the wheels are
off the ground. I was not talking about transitions. I was talking
about leaning into a corner (straight up to full lean). If rotation is
taking place about the cm, the distance from the cm to the contact
patch will change. The cm has inertia and only gravity can make it
lower. If the rotation is very fast the suspension should extend (the
distance from cm to point on road where tires touch (since the cm can
only drop at a fixed rate). If this distance exceeds the distance that
the suspension extends, the wheels should come off the ground as max
lean is approached.
The phenomena you describe above is due only to inertia of the cm and
would occur regardless of the mechanism causing the steering.
>>work on a corner and watch very carefully. I have seen no evidence
>>supporting roll about the cg. I do not see where the force comes from
>>to move the rear wheel as far as it should. Isn't friction at the
>>contact patches a constraint?
>Friction at the contact patches is what makes it work. You're in a moving
>frame of reference. The tires are FREE to turn under the bike. From the
>bike's point of reference it appears that the front, followed by the rear tire
>are moving sideways, while the bike rotates around it's CG. It has to, it
>is a balanced object.
The bike is not free to rotate about the cm because of the sideways
friction of the tires. Just because the tires are rolling doesn't mean
that they are free to slide sideways. The angle of the front wheel
will control the amount that the wheels outtrack.
So how far do the wheels have to move for a bike with a 2 foot high
(from ground) cm at 45 deg of lean? At say 60 mph with a time of lean
of 0.25 sec from straight up to 45 deg, at what angle would the bars
have to be turned for outtracking?
The movement of the wheels cannot be much. I have taken turns
completely on the inside and I end up way over the grass. Much further
than rotation about the cm would indicate.
The bike is only free to rotate about the cm as far as the outtracking
wheels will allow. From my experiance, the distance of outtracking
cannot be more than a couple of inches. Most turns on tracks are not
taken at 100 mph, been very close (within 6 inches of edge of track
many times and turned a low speeds (less than 60 mph) and didn't end
up in the grass. On a TZ a minimal amount of hang off is needed. My
chest stays mostly over the tank and my knee does not touch until
around 45 deg of lean.
I have exited corners and came up on the inside of another rider,
Using the other moving bike as a reference (the bike was upright at
the edge of the track before I was) I rotated towards the bike until
our elbows were touching, didn't seem like rotation about the cm to
me.
I will have to borrow a video camera and do some taping, weather
permitting.
>I've stated as much before, but everyone knows what we're talking about
>here.
Are you sure;)
Brian McLaughlin
frec...@rintintin.Colorado.EDU (Ian Frechette) wrote:
>Obviously the real numbers are much less dramatic. You only need to move
>out a foot or so to induce serious lean, and once you've established slight
>outtracking, you need only hold the line or even start correcting
>as the bike leans over.
Where does the foot come from? What is serious lean? If the cm is 2 ft
from the ground, a foot would be an anglr of about 30 deg. I know my
wheels do not move to the side a foot. I have been much closer to the
edge than that on public roads (without hanging off) and did not end
up in the dirt.
John M. Feiereisen
In <6bnk35$14...@wiscnews.wiscnet.net>, bmcla...@waukesha.tec.wi.us
(Brian McLaughlin) wrote:
>frec...@rintintin.Colorado.EDU (Ian Frechette) wrote:
<snip>
>Why if I lean off and move cm to one side will the bike rotate about
>the contact patch, yet when the bike outtracks and the cm is no longer
>above the wheels the rotation suddenly takes place about the cm.
When you initially lean off, your CM is going one way while the bike's
CM is going the other way. The combined CM of the bike/rider system
doesn't move either way -- until you start developing a lean of the
bike/rider in the desired direction.
>Isn't
>the idea of outtracking to move the cm from above the wheels?
Yup.
>Then gravity should lean the bike.
Gravity is only a small player at the start of a lean. Don't forget
the lateral force at the contact patch.
>So what is the difference between the 2
>situations. For outtracking the only vertical force I see is gravity.
The dominant force in outtracking is the lateral traction force at the
front wheel contact patch.
In hanging off, all you're doing is moving the relative positions of
the bike's CM and the rider's CM. Hanging off will not move the
combined bike/rider CM either way.
>What other forces help to lean the bike?
Discounting aerodynamics forces, coriolis acceleration, etc., the
*only* 'forces' that lean the bike/rider are the lateral traction
forces at the contact patches due to outtracking, and the precession
torque arising from your steering torque.
The lateral traction force due to outtracking will tend to roll-rotate
the bike/rider about the bike/rider's CM. The precession torque will
tend to roll-rotate the bike/rider about the front axle (about an axis
perpendicular to both the axle and the steering head). The tires tend
to constrain the system by preventing lateral movement at the contact
patches, but you can generate lateral movement of the contact patches
by steering.
In reality, all the effects add up and you rotate about some place
else entirely.
<snip>
>Of course the rotation takes place about the cm when the wheels are
>off the ground. I was not talking about transitions. I was talking
>about leaning into a corner (straight up to full lean). If rotation is
>taking place about the cm, the distance from the cm to the contact
>patch will change.
The *vertical* distance from the CM to the contact patch will change.
Discounting the additional suspension compression due to cornering
forces, the overall distance between the CM and the contact patch
remains unchanged.
>The cm has inertia and only gravity can make it lower.
Yes.
>If the rotation is very fast the suspension should extend (the
>distance from cm to point on road where tires touch (since the cm can
>only drop at a fixed rate). If this distance exceeds the distance that
>the suspension extends, the wheels should come off the ground as max
>lean is approached.
Yes.
>The phenomena you describe above is due only to inertia of the cm and
>would occur regardless of the mechanism causing the steering.
But if you were able to somehow magically *force* rotation about the
contact patch, you wouldn't be constrained by gravity as you describe
above -- you'd be *forcing* the CM toward the ground. Hence, in
reality outtracking tends to rotate the bike/rider about the CM.
And I remember one cycle instructor here saying that one of his
students was able to countersteer so aggressively that his front wheel
left the ground. IIRC that was from a straight line into a corner.
>>Friction at the contact patches is what makes it work. You're in a moving
>>frame of reference. The tires are FREE to turn under the bike. From the
>>bike's point of reference it appears that the front, followed by the rear tire
>>are moving sideways, while the bike rotates around it's CG. It has to, it
>>is a balanced object.
Hey Ian, how's this for throwing a monkey wrench into the works:
In a reference frame moving with the CM of the bike/rider, rotation
will *always* and *only* be about the CM. In a reference frame moving
with the contact patch of the bike, rotation will *always* and *only*
be about the contact patch.
And in a reference frame that moves with the bike CM and rotates with
the bike about the bike's CM, the bike doesn't move at all!
>The bike is not free to rotate about the cm because of the sideways
>friction of the tires. Just because the tires are rolling doesn't mean
>that they are free to slide sideways. The angle of the front wheel
>will control the amount that the wheels outtrack.
Yes.
>So how far do the wheels have to move for a bike with a 2 foot high
>(from ground) cm at 45 deg of lean? At say 60 mph with a time of lean
>of 0.25 sec from straight up to 45 deg, at what angle would the bars
>have to be turned for outtracking?
Now you're getting into the realm where only a full 3D analysis
including tire dynamics will give you the right answer. If you assume
that the whole bike magically and instantaneously rotates (yaw) some
theta degrees, if the CM continues in a straight line both wheels will
need to move laterally 1.414 feet to effect a 45 deg lean angle. In
0.25 seconds at 60 mph you'll move 22 feet down the road. theta =
atan(1.414/22.0) = 3.7 degrees. I would assume this is reasonably
close to the correct answer (probably within +/- 100%).
>The movement of the wheels cannot be much. I have taken turns
>completely on the inside and I end up way over the grass. Much further
>than rotation about the cm would indicate.
But I doubt you develop your full lean angle before the bike starts
tracing a curved path in a planform view. If you kept your CM moving
in a straight line until you achieved your full lean angle, you'd find
that you're not so far over the grass.
>The bike is only free to rotate about the cm as far as the outtracking
>wheels will allow.
Only if you constrain the CM to move in a straight line.
mull...@advinc.com
In article <j4zpk01...@alph04.triumf.ca>,
baaa...@alph04.triumf.ca wrote:
>
> I find Ian's comments extremely puzzling as well. I'm still in physics, and
> I deal with lots of engineers. My experience is that when I say two devices
> should be around 1 inch apart, a drawing comes back from engineering with
> their locations specified to within 0.001 inch.
> --
> rick baartman
>
And then that's when *you* say "Oh. Those were supposed to be TWO
inches apart. Sorry. Could you do it again, for free?" For everybody
knows that for a physics person, two equals three. EXCEPT when
figuring lunch checks, which are done on HP calculators to 5 significant
figures.
True story: The American Physical Society held its annual meeting in
Las Vegas one year. They were told in no uncertain terms to *never*
come back, as there was a decided lack of drinking, whoring, and gambling
during their stay. Hell, I could have told them about that.
Jim (an engineer in the land-o-physics)
Trevor Dennis
mull...@advinc.com wrote:
> And then that's when *you* say "Oh. Those were supposed to be TWO
> inches apart. Sorry. Could you do it again, for free?" For everybody
> knows that for a physics person, two equals three. EXCEPT when
> figuring lunch checks, which are done on HP calculators to 5 significant
> figures.
I've had people steal back a sketch and change it when
things didn't work out. I learned a long time ago *NEVER*
to trust the information provided. I've saved a lot of
work that way.
--
Trevor Dennis /`\ .(o~)-(o~). /`\ tre...@tdennnis.demon.co.uk
The Polite Brit / , \( _______ )/ , \ tden...@ford.com
OGH #1 ___/ /_\ /`"-------"`\ /_\ \___ Southern England
jgs`~//^\~_`\ <__ __> /`_~/^\\~`
`~//^\\~`~//^\\~`
Rick Baartman
>>>>> "Ian" == Ian Frechette <frec...@rintintin.Colorado.EDU>
>>>>> wrote the following on 8 Feb 1998 11:27:55 GMT
Ian> In article <j4en1fp...@alph04.triumf.ca>, Rick Baartman
Ian> <baaa...@alph04.triumf.ca> wrote:
>> Sorry, Ian, Brian is right on this one. The frequencies in the above
>> equation are ANGULAR frequencies. Ang.freq.=2pi*cycling
>> freq. Therefore, in terms of cycling freq., the above eqn. becomes
>>
>> 2pi*precession cyc.freq.=torque/(mom.of inertia*2pi*rot.cyc.freq.)
>> prec.cyc.freq.=torque/(4*pi*pi*mom.of inertia*rot.cyc.freq.)
Ian> Uggg..
Ian> 2pi radians/sec = 1 cycle/sec = 1 Hz = 1/sec
Entirely wrong, Ian. A radian is an angle corresponding to when the length
of arc is equal to the radius. 1 radian is 1 without units since it is the
ratio of two lengths: Arclength/radius. For this reason in physical formulas
like the ones you quote, physicists don't mention radians. Omega (the usual
Greek letter for angular frequency) = time derivative of angle, is written
with units 1/sec. The cycling frequency is odd man out here. To denote the
difference, scientists usual denote its units as Hz. 1 Hz is not equal to
1/sec. 1 Hz = 2pi/sec.
Also, I could quote lots of literature on this subject, but my experience
is that this accomplishes nowt. I know how to derive the formulas you
quote. I can derive them from first principles. (Mechanics is my physics
field of specialty.) You can define Hz any way you like, but if you use a
formula form known physics and stick in your own definitions of units,
you'll be taken to task.
Is this a troll? ... I've been hooked into responding to these before...
Brian McLaughlin
frec...@rintintin.Colorado.EDU (Ian Frechette) wrote:
>In article <j4en1fp...@alph04.triumf.ca>,
>Rick Baartman <baaa...@alph04.triumf.ca> wrote:
>>Sorry, Ian, Brian is right on this one. The frequencies in the above
>>equation are ANGULAR frequencies. Ang.freq.=2pi*cycling freq. Therefore,
>>in terms of cycling freq., the above eqn. becomes
>>
>>2pi*precession cyc.freq.=torque/(mom.of inertia*2pi*rot.cyc.freq.)
>>prec.cyc.freq.=torque/(4*pi*pi*mom.of inertia*rot.cyc.freq.)
>Uggg..
>2pi radians/sec = 1 cycle/sec = 1 Hz = 1/sec
>Therefore 2pi radians = 1 cycle = 1
>Let me repeat that.. 2pi rad = 1
You have got to be kidding. Do you know the difference between
frequency and angular frequency (same as angular velocity)? They have
the same units but not the same value. An oscillation of 1 cycle/sec
is *equivalent* to 2pi rad/sec. The frequency is 1 Hz but the angular
frequency is (approx) 6.28 Hz, to avoid confusion the units are
usually left as 6.28 rad/sec. When I took math 6.28 didn't equal 1.
You seem to be confusing a unit conversion.
>The problem is you threw away the radians/sec on both sides of
>the EQ. You cannot do that. Normally you'd never notice
>if the radians disappeared from the EQ, but because the final result is
>not in rad/sec, but in 1/(rad*sec) it makes a big difference to this EQ.
Nope, the radians disappear in the equation. Radians are not a unit of
measure. By definition (arc length/radius) radians are unitless. Why
does it drop out of the calculation for torque? Recall torque is
moment of inertia(kg*m^2) * angular acceleration(rad/sec^2), but the
unit for torque is N*m not N*m*rad.
>If you want to divide through, divide though by 2pi rad instead of
>just 2pi.
You have stated that rad are not a unit, why do I need to cancel them
here?
>If Pn is natural frequency of precession. Y/sec then radian frequency
> of precession is 2*pi*Y rad/sec
The equation that you first wrote was for the angular frequency of
precession. Why? Because you are ignoring the derivation of the
equation. The derivation starts with
ang. vel. of precession = d(angle)/dt, where the angle is measured in
radians. Therefore the precession frequency given by the equation is
an angular frequency, not the natural frequency as you call it.
>If Xn is natural frequency of rotation.. X/sec then radian frequency of
> rotation = 2pi*X rad/sec
>Then the eq must be
>2pi*Y rad/sec = torque/(mom.of.inertia*2pi*X rad/sec)
No. By definition the left hand side of the equation is in rad/sec,
not in cycles/sec, so the 2pi is unnecessary on the left hand side.
>divide through by 2pi rad
>Y/sec = torque/(mom.of.inertia * 2pi rad * 2pi rad * X /sec)
>But since 2pi rad = 1, both instances of 2pi rad go away..
Since when does 6.28 = 1?
>If you don't like that, then prove to me that 2pi radians does not equal
>1 cycle and that 1 cycle/sec does not have base units of 1/sec.
>2pi rad/sec = 1Hz = 1/sec Prove it ain't so.
2pi rad/sec = 6.28 rad/sec = 6.28 1/sec which is not equal to one
(didn't even need my hp 48 to figure it out). Now 6.28 rad/sec is
*equivalent* to 1 cycle/sec. The whole reason to keep the radians even
though they are not a unit is to avoid the confusion you are having
(or pretending to have).
What you say would work if the angular frequency on the right hand
side of the equation was in the numerator, problem is the left side of
the equation is in rad/sec, the ang. freq. term on the right hand side
is also rad/sec. Both sides of the equation cannot be multiplied by
the same number to get both converted to cycles per second.
In rotational motion, angles are defined in radians, angular velocity
in radians/sec, and angular acceleration in rad/sec^2.
Let T = torque (N-m)
L = angular momentum (kg * m^2/sec)
I = moment of inertia (kg * m^2)
wp = angular velocity of precession (rad/sec)
w = angular velocity of spinning object (rad/sec)
t = time (sec)
phi = the angle that L sweeps out as a response to the applied torque
T, that is by definition, the angle of precession (measured in
radians).
<--------------------------------------L
|
|
\/ dL
T is pointing out of screen.
T = dL/dt
but, d(phi) = dL/L
(tan d(phi) = dL/L, but for small angles measured in *radians*, tan
d(phi) = d(phi) for proof look at the series expansion for tangent,
hence, d(phi) = dL/L)
dL = T dt so
d(phi) = T dt/L finally
d(phi)/dt = T/L = T/I w = wp
By definition of angle of precession, the formula that you used is for
the angular velocity of precession (which means rad/sec as units). L =
I w with w in rad/sec.
To convert wp to cycles per second would require multiplying both
sides of the equation by 1 cycle/2pi rad. Converting w to cycles/sec
would require multiplying both sides by 2pi rad/1 cycle. This however
would convert wp back to radians/sec. So both wp and w cannot both be
in cycles (or revolutions or degrees) per second. The only way to get
the formula the way you want it would be to start over with a fresh
sheet and re-do all of rotational dynamics in terms of revolutions
(not sure if this is even possible) instead of radians (remember the
present units of torque and angular momentum are based upon the
angular measure of radians).
Ian you may just be jerking me around, that is not the reason I
continue to carry this on, the problem is that there may actually be
some people who believe what you are saying in reference to this
equation.
Putting in rot/sec for w will not give wp with units of rot/sec
regardless of what you have to say.
Ian Frechette
In article <6bnfk0$8...@wiscnews.wiscnet.net>,
Brian McLaughlin <bmcla...@waukesha.tec.wi.us> wrote:
>frec...@rintintin.Colorado.EDU (Ian Frechette) wrote:
>
>>In article <6b4p89$17...@wiscnews.wiscnet.net>,
>>Brian McLaughlin <bmcla...@waukesha.tec.wi.us> wrote:
>>>frec...@rintintin.Colorado.EDU (Ian Frechette) wrote:
>>>>>
>>>>>> X lb * 32 ft/sec^2 * 1 ft
>>>>>>precession freq = -------------------------- = .045*X rps or 16X degrees/sec
>>>>>> 20 lb * 1 ft^2 * 35/sec
>
>ok, lets see:
>
>processional freq (since omega is used, this is angular frequency)
>
>= X lb*32ft/s^2 * 1ft/(20lb * 1ft^2 * 220 rad/sec) = .0073X rad/sec =
>0012X rev/sec = .042X deg/sec.
Ok. So here's the fundamental difference between your calculations and
mine.
My physics reference and my hp48 both say that
1/(rad*sec) = 4pi^2 rad/sec while you say 1/(rad*sec) = 1 rad/sec
You say this is the difference between velocity and frequency. Correct?
Explain that.
>Doesn't give your answer.
>Units of Hz applies to frequency (f or nu in physics texts) and
>angular frequency (small omega). Rev/sec are not the same units. The
What is the base unit of a revolution? If something is rotating
at 1 revolution per second, how is its frequency not 1 cycle/sec?
Read nothing into my question. I am simply asking.
>2pi is a conversion factor between radians and rotations and carries
>units. Rev/sec is not the same as Hz.
Again, my reference says that 2pi is a conversion between fn and rad/sec.
fn = X/sec rps=2pi*f in units of rad/sec = X*2pi rad/sec
You're saying that
v = X rev/sec w = 2pi*v in rad/sec ? What's the difference?
>No it shouldn't. Found my error. But it still leaves 1/(sec*rev) which
>does not equal rev/sec. Revolutions cannot be dropped like radians. By
I can accept that if you can explain why a rev is not a cycle.
>definition radians are unitless (the ratio of arc length/radius) the
>same cannot be said for rotations.
I agree that radians are unitless, as are cycles, but what's the base
unit of a rotation? Is not a rotation just 2pi rads?
>> 1/(sec^2) / rfreq
>But that leaves you with sec/rotation. Rotation does not disappear as
No.. at worst it leaves me with 1/(rot*sec) but that's an incidental error.
>does radians. Look at the equation for torque. Torque = moment of
>inertia (I) * angular acceleration. For the equation to work out the
This is pivotal, so to speak.
I defined torque as force perpendicular to the end of a lever of a
specific length. With units of kg*m^2/sec^2
That's obviously the problem. Linear torque.
> I * rad/sec^2
>Angular velocity of precession = ------------------------------------
> (rad/sec) I * rad/sec
>
>The 2 moment of inetia terms will cancel leaving 1/sec or rad/sec.
This is all I missed. I was applying a non-angular torque.
Ok, so 1 kg*m^2/sec^2 (linear) = 1 kg*m^2 rad/sec^2 (angular).
(given I = MR^2 and a=linear tangential accel/R and l.t.a. = kg*m/sec^2)
Lovely..
>Recall the the unit of angular measure is the rad, angular velocity is
>the rad/sec, and angular acceleration is rad/sec^2. Multiplying both
>sides by 1/2pi does not give the correct units. The two angular
>frequency terms are not both located correctly for this to happen, one
>is in the numerator and the other is in the denominator.
Ok, I see how 1/(rad*sec) is avoided now, but doesn't the above
eq yield 1/sec which is *still* 2pi rad/sec? You say rad is a non-unit
like I say a cycle is a non-unit, but 1/sec and 1 rad/sec are still
not equivalent are they?
>"... the *angular velocity* of precession about the z-axis is
>omega sub p = d(phi)/dt = torque/angular momentum.
>
>This result is known as simple precession."
>p170 Classical Mechanics by Barger and Olsson.
>You are solving for an angular velocity, not a frequency. Angular
>velocity is often referred to as angular frequency since the units of
>angular velocity are rad/ sec, or Hz.
>This is true. You are using rev/sec which is not Hz (cycles/sec)!!!!
Fine.. Why?
>Xn has units rev/sec. You cannot just drop the rev unless you have
>redefined torque using angular acceleration of rev/sec^2.
That makes sense, but are you saying tht 1/sec = rad/sec? and if
so how is it that you are telling me that you can convert rad/sec to
rev/sec but I can't convert it to cycles/sec?
I see the only problem as being that I have not expressed
angular torque in rev *or* cycles per second, and that without initial
conversion of the angular torque, the eq requires rad/sec.
>>pfreq = (1/sec^2)/(Xn*sec) = 1/(33*sec) = .0303 cycles/sec
>
>Units are .0303 1/rev*sec.
You're right.. That's a result of the lack of rads or revs in angular torque.
A mistake on my part. I can live with that.
>1/(207 sec) *is* in rad per sec! Multiply by 1/2pi to convert to
>rev/sec, .00077 rev/sec.
I feel we're a lot closer but if Hz *is* 1/sec and rad/sec is 2pi*f then
how can 1/(207*sec) = 1/207 rad/sec?
Side note: All that aside, if it's what you say, then it shows
how worthless this precession frequency (angular frequency) is from
a steering point of view. Go back to the original EQ with
X ft-lb of torque on the bars with a 2 foot diameter 20lb tire
rotating at 33 rev/sec (150mph)
(X*32)/(20*207) rad/sec = X*8/1036 rad/sec = X*8/(2pi*1036) rev/sec
= X*.00123 rev/sec = X*0.44 degrees/sec
Plug in 1 foot-lb of torque..
That's nice and slow, as you might expect at that speed but bring
it down to 50mph where we know the bike is quite flickable.
Multiply the above result by 3 (divide by 1/3) and we get
.83 degrees/sec
To acheive a 45 degree/sec roll rate ( assuming 0 moment of inertia of
bike/rider) we need 33 ft-lbs of torque applied to the bars. I know
I can acheive the same effect with a small fraction of the effort, so
precession must be a minor player at this speed. Factor in the
mass and distribution of mass (moment of inertia) of the bike/rider
and the fact that the precession axis is perpendicular to the forks
and therefore not inline with the long axis of the bike either through
the CG or the contact patches and precession becomes downright worthless
for initiating a turn. Note: it's still useful for stabilizing the
bike. Tell me why.
>No. The definition of angular displacement is the radian. The formula
>that you used is solving for angular velocity (or angular frequency),
>this is not the same as the frequency that you are thinking of.
And yet, I still find numerous equations converting what appears to
be the frequency I'm thinking of to the frequency you're thinking of.
I admit that my values are off, but I'm still not convinced it's by
as much as 4pi^2.
>>>Oh yeah, I teach this stuff (college physics) for a living.
>
>>That sorta scares me.
>
>It would if you were in my class :)
Only if you failed to explain why in one physics reference it says
that 1/sec = 2pi rad/sec while in another it seems to indicate
that 1/sec = rad/sec.
>The results from Hz and rad/sec should be the same. The problem is
>that rev/sec are *not* Hz. You cannot simply drop the rev. Radians are
You converted rev/sec to rad/sec with nothing more than a factor of
2pi.. Again, how is a rev not a cycle?
>Don't use your conversion, I got different answers than you did.
I won't, but I don't think that the conversion was the problem.
The problem was that I lacked units for angular acceleration in
in angular torque. But we may still disagree by a factor of 2pi
as I'm still not convinced that (rad/sec^2)/(rad/sec) = rad/sec
but is in fact 2pi rad/sec. All I need is a decent explanation
(or a pointer to one) of why sometimes 1/sec = 2pi rad/sec and sometimes
1/sec = 1 rad/sec.
>>P.S. I noticed that you stripped out my treadmill example. Does a moving
>> frame of reference make sense yet?
>
>I am still waiting to see the numbers for the amount the wheels move
>out. Outtracking takes place, but I do not see how the rotation can
>occur about the cm. The amount of outtracking would be greater than
>what is seen.
I'll get to this in another post, and I think John is doing a decent
job of explaining this, but in short, here's how I see it.
The outtracking should be measured as the distance from the curved
path drawn on the ground required to exactly balance gravity's force
trying to pull the bike over, to the path of the tires. This means that
the CG is forced to follow a curved path to keep the bike balanced while
the tires follow a different curved path to start and stop lean. Note that
it's impossible to maintain an outtracking stance as that leads
continuous angular acceleration in one direction, which can only continue
until you hit the ground.
>Brian AP#1
ian
Ian Frechette
In article <6bq9es$d...@wiscnews.wiscnet.net>,
Brian McLaughlin <bmcla...@waukesha.tec.wi.us> wrote:
>frec...@rintintin.Colorado.EDU (Ian Frechette) wrote:
>
>
>You have got to be kidding. Do you know the difference between
>frequency and angular frequency (same as angular velocity)? They have
>the same units but not the same value. An oscillation of 1 cycle/sec
That's the problem then. I saw the equation specifying "precession frequency"
and damn if I didn't go using units of frequency. So I'm looking
for angular velocity.
Fine..
Good news.. You're right..
Bad news.. I'm right. Take any precession frequency I had calculated before
and divide by 4pi^2. Now that is SLOW...
It totally defeats the theory that precession is a signifcant factor
in countertsteering at any speed. The precession
frequency is way too small to be useful, the precession torque is
no more than what you put in. The precession axis is not in line
with any useful axis of rotation of the bike, through the CG *or*
through the contact patches.
I actually started out in the beginning, to prove exactly what you've
hown, that the precession frequency was too slow to be useful. I got
tripped up on the units and came up with a result that was too big. A math
error no doubt, but now the proof is complete.
Now, as you say, the next step is to get out the video camera. I wish
I had one.
>You seem to be confusing a unit conversion.
I said that. And I asked you in my other post to explain, but I guess
you were too busy attacking this post, to notice.
>You have stated that rad are not a unit, why do I need to cancel them
>here?
Actually, what I stated was that a rad was unitless but was a multiplier.
It appears that instead it should be said that a rad/sec = 1/sec
but that a cycle is a multiplier and is 2pi. What I had was that
cycle/sec = 1/sec and that rad was 1/2pi.
I just had it backwards.
>Ian you may just be jerking me around, that is not the reason I
>continue to carry this on, the problem is that there may actually be
>some people who believe what you are saying in reference to this
>equation.
No, not jerking you around. I made an honest mistake.
But the nice part is that you've proved what I started out trying
to prove quite nicely for me. I did not go about this lightly, or to
provoke anyone, into a huge physics debate.
Basically, I had 1 cycle/sec = 1/sec and if you'd simply said
that 1 cycle/sec = 2pi/sec = 2pi rad/sec instead, it would have been
clear 10 posts ago. The fact that you made a units error yourself
in your first reply didn't exactly give me confidence.
And ultimately it boiled down to the units of angular acceleration. As
you say, if that could be expressed in cycles per second, everything
would have been hunky-dory, but I was using linear torque, with no
regard to angular acceleration. I said that yesterday.
Now can we get back to what makes a motorcycle steer.
>Brian AP#1
ian