Progressive vs straight rate springs - Progressive's perspective
Erik Astrup
Everything I've ever read, and everything I've heard from suspension guru's
like Theede and Lindemann always indicate that straight rate springs are
superior to progressive rate springs. Here's what "Progressive Suspension"
has to say.
Comments?
========
Fork Springs:
Our famous progressive rate fork springs are available for most
motorcycles. Why progressive rate? A straight rate fork spring is a compromise.
If it is set up to deliver a "cushy" ride during the first part of the travel it
will not be stiff enough at the end of the travel to prevent bottoming.
Conversely, if the straight rate spring is set up stiff enough to prevent
bottoming, then the initial travel will be too stiff causing a harsh ride. A
progressive rate cures these problems allowing a soft initial rate for a
comfortable ride yet progressing to a stiffer rate to prevent the forks from
slamming on the big bump. The best of two worlds. If progressive rate is not the
way to go then why do the manufacturers spend huge amounts of time and money on
engineering linkage rear suspension? Linkage provides a rising rate suspension
which is another name for progressive rate. Our springs are made from the
highest quality materials, in fact the springs are warranted for life in case of
breakage or sacking.
------------------------------------------------------------------------
Erik Astrup - Team Iguana Racing (Ret) *FOR SALE* '89 Honda Transalp
http://www.mother.com/~eastrup/home.htm
Need to know when a roadrace is going to be on TV??
http://www.mother.com/~eastrup/racing/tv.htm
------------------------------------------------------------------------
Benjamin Justin Cain
Steve Zwinger (s...@Eng.Sun.COM) wrote:
:
: It was my understanding that the progressive rate for the forks was
: supplied by the air trapped between the oil level and the fork cap.
: The resistance offered by the air is non-linear and provides the most
[snip]
Why wouldn't it be linear? Doesn't P=nRT/V to first order? And V
varies linearly with the compression of the fork. I don't get it.
--
Ben Cain : Biker Scum - Yellowshirt Brigade #514
AMA HRCA TMGP CMRA/WERA DoD # 1/137 KoK3
1990 VFR750F "Deal's Gimp"
1989 YSR50/3
http://www.users.cts.com/king/d/drlubell/bscum.html
"Being shot out of a cannon will always be better than being squeezed out
of a tube. That is why God made fast motorcycles, Bubba..."
-Hunter S. Thompson-
Steve Zwinger
In article t...@news.tamu.edu, bjc...@mulder.tamu.edu (Benjamin Justin Cain) writes:
>Steve Zwinger (s...@Eng.Sun.COM) wrote:
>:
>: It was my understanding that the progressive rate for the forks was
>: supplied by the air trapped between the oil level and the fork cap.
>: The resistance offered by the air is non-linear and provides the most
>
>[snip]
>
>Why wouldn't it be linear? Doesn't P=nRT/V to first order? And V
>varies linearly with the compression of the fork. I don't get it.
>
>--
You are right. I wish I could blame this on today being 4/1, but
I don't think anyone would buy that.
I have seen graphs of fork resistance vs suspension travel with
different oil levels. This is in the Honda CR owners manual. It
shows, and I've read in other places that oil height only effects
the (approx) last 1/3 of the fork travel. For some reason I associated
that with the pressure not being linear as the forks compress (wrong?).
Could it be that the compressing fork springs and damper rod may cause
this non-linear volume in the fork tube. (just a guess). As they
compress they go from being in air to being in oil. One other
explaination may be that although the air is linear, it may be at
a higher rate than the fork springs and therefore be negligable
on the first part of the travel (compared to the spring), and very
large on the last part of the travel (again compared to the spring).
Bottom line is I believe that the fork oil level provides a non-linear
spring effect to the front fork. I don't know the physics behind it.
I guess I can't directly answer your question.
---
Steve Zwinger
steven....@west.sun.com
David Wilson
Air in enclosed chambers is not linear, it increases exponentially.
That is why adding pressure to a fork stiffens the last part of the
total travel. Also, adding air pressure acts just like adding preload
on a spring, that is why you add air pressure to bring up the front of a
machine if it is riding too low. An example: compare a spring with a
rate of 20 lbs with, say 0mm preload and 20 lbs of air without a
spring. At 1mm of travel the spring is exerting 2 Lbs (I believe) and
the air is at least 20+ Lbs. OK which is smother in initial travel.
The Spring. This is why adding air increases initial stiffness (usually
not desired) and final (last 1/4 travel) of a fork, but not much in the
middle. It is also why air is usually only a quick fix, not the best
solution. Most forks work best with no added initial air presure.
Regarding rear shocks, if you graph out the rising rate used by most off
road machine, you will see the graph is linear through most of the
travel. The linkage only makes the initial travel softer and bottoming
stiffer (first 20 and last 20 percent). Mostly it is Damper Unit that
the rising rate is trying to change, not the spring. ATK motorcycles
first used a single shock with no linkage and a good shock bumper to do
the same thing. ATK and Horst Leightner just did it in a simple manner,
Japan just likes to make it more complex. KTM motorcycles are again
designing a no linkage shock system, with the aid of a new damper
design, to eliminate weight and complexity. Maybe in 98.
Dave Wilson
Off road racer since 1968
Steve Zwinger
In article 796...@news.internetmci.com, eas...@mother.com (Erik Astrup) writes:
>
>
>Everything I've ever read, and everything I've heard from suspension guru's
>like Theede and Lindemann always indicate that straight rate springs are
>superior to progressive rate springs. Here's what "Progressive Suspension"
>has to say.
>
>Comments?
>
>========
>
> If progressive rate is not the
>way to go then why do the manufacturers spend huge amounts of time and money on
>engineering linkage rear suspension?
It was my understanding that the progressive rate for the forks was
supplied by the air trapped between the oil level and the fork cap.
The resistance offered by the air is non-linear and provides the most
effect in the last portion of the suspension travel. This is what
is varied when the oil height is raised/lowered. Also the bottoming
cones in the suspension offers additional resistance. Just because
the front spring is linear, that doesn't necessarily imply that
fork resistance is also linear. There are more pieces to the puzzle
than just the spring.
Rear suspensions (shocks) do not have an air chamber incorporated it
their design. (at least the one's I've rebuilt)
I'm not a mechanical engineer, but I've installed gold valves, delta
valves, and even revalved forks on my own. I don't profess to be an
expert, so my opinion is worth what you paid for it.
---
Steve Zwinger
Jamie Worthington
Steve Zwinger <s...@Eng.Sun.COM> wrote in article
<5hs3u7$q...@engnews2.Eng.Sun.COM>...
> In article t...@news.tamu.edu, bjc...@mulder.tamu.edu (Benjamin Justin
Cain) writes:
> >Steve Zwinger (s...@Eng.Sun.COM) wrote:
> >:
> >: It was my understanding that the progressive rate for the forks was
> >: supplied by the air trapped between the oil level and the fork cap.
> >: The resistance offered by the air is non-linear and provides the most
> >
> >[snip]
> >
> >Why wouldn't it be linear? Doesn't P=nRT/V to first order? And V
> >varies linearly with the compression of the fork. I don't get it.
> >
> >--
>
> You are right. I wish I could blame this on today being 4/1, but
> I don't think anyone would buy that.
>
> I have seen graphs of fork resistance vs suspension travel with
> different oil levels. This is in the Honda CR owners manual. It
> shows, and I've read in other places that oil height only effects
> the (approx) last 1/3 of the fork travel. For some reason I associated
> that with the pressure not being linear as the forks compress (wrong?).
> Could it be that the compressing fork springs and damper rod may cause
> this non-linear volume in the fork tube. (just a guess). As they
> compress they go from being in air to being in oil. One other
> explaination may be that although the air is linear, it may be at
> a higher rate than the fork springs and therefore be negligable
> on the first part of the travel (compared to the spring), and very
> large on the last part of the travel (again compared to the spring).
>
> Bottom line is I believe that the fork oil level provides a non-linear
> spring effect to the front fork. I don't know the physics behind it.
> I guess I can't directly answer your question.
>
>
> ---
> Steve Zwinger
> steven....@west.sun.com
>
>
>
The answer the the question (if I have the question right) is that an air
spring is an exponentially progressive rate spring. It is NOT a linear
spring.
For example, say you have a fork with 16 inches of travel and its volume is
16 cubic inches and 1sq in of area.and 1psi of preload at full extension At
0" of travel, there is 1 pound of force. Now we move the fork 8 inches. The
volume is halved, therefore the pressure is doubled. If we continue moving
the fork by halving the remaining available travel, the pressure doubles
for each halving of the distance. See chart
travel volume pressure
0 16 1
8 8 2
12 4 4
14 2 8
15 1 16
As can be seen with this chart, the last inch of travel caused the pressure
to rise by 8 psi(to 16), while it took the first 14 inches of travel to
only raise the pressure (and force) to 8.
rate=change in force/change in distance
for first 8 inches of travel, the effective rate(it always is changing) is
rate = (1)/(8) =1/8 pound per inch
for the last inch of travel
rate = (8)/(1) = 8 pounds per inch
So as you can see, the rate at which the force is applied changes with
position, therefore an airspring is a rising rate spring. Varying the
volume of oil in a fork definitley can impact how your forks work during
hard bumps or heavy braking. Not much effect is noticed over small bumps.
Jamie Worthington
Ian Drysdale
Yes this is a good explanation - you can further complicate the problem by
going for a non circular container - I.E - the Citroen hydraulic / pnuematic
suspension system that was copied by Rolls Royce and Mecedes ( when the
patents had run out ).
This is a brilliant piece of design with an engine driven hydraulic pump for
variable ride height and conical nitrogen chambers with rubber bladders that
act as the the 'sring'.
Some models use purely spherical chambers - lets see a quick calculation on
that one Jamie.
Boge made a similar unit for BMW bikes that used the bumps on the road to pump
the shocks up - the only time I know of it being used on a bike. As far as I
know the variable ride height Gold Wings were purely pnuematic - anyone know
better ??
If you get a chance to have a look at a Citroen 'Height Control Valve' in
pieces do so - they are a marvil of engineering. Of course you could just go
out and buy one to see how they work as I did. Wonder how this works ...BIONG..
I was a bastard of a kid.
IAN
Bert Kellerman
I though this thread was about springs, in which case the force wouldn't
be linear. F=1/2KX^2 where k is the spring constant diff depending
on #of coils and spring thickness and x is displacement, would be the
equation needed. With Gas and liquid however PV=nRT would be right.
Just my two pennies.
B
Matt Ho
> Everything I've ever read, and everything I've heard from suspension guru's
> like Theede and Lindemann always indicate that straight rate springs are
> superior to progressive rate springs. Here's what "Progressive Suspension"
> has to say.
>
> Comments?
>
> ========
>
> Fork Springs:
> Our famous progressive rate fork springs are available for most
> motorcycles. Why progressive rate? A straight rate fork spring is a
compromise.
> If it is set up to deliver a "cushy" ride during the first part of the
travel it
> will not be stiff enough at the end of the travel to prevent bottoming.
> Conversely, if the straight rate spring is set up stiff enough to prevent
> bottoming, then the initial travel will be too stiff causing a harsh ride. A
> progressive rate cures these problems allowing a soft initial rate for a
> comfortable ride yet progressing to a stiffer rate to prevent the forks from
> slamming on the big bump. The best of two worlds. If progressive rate is
not the
> way to go then why do the manufacturers spend huge amounts of time and
money on
> engineering linkage rear suspension? Linkage provides a rising rate suspension
> which is another name for progressive rate. Our springs are made from the
> highest quality materials, in fact the springs are warranted for life in
case of
> breakage or sacking.
This progressive claim makes sense. I think the conflict here araise from
the different applications for the bike was intended for. The
Theede/Lindemann claim is correct if the bike is a race bike where the
road surface is very smooth and the asphalt smoothness varies very little
from section to section, of course there are always exceptions but mosts
good tracks are very smooth. Because of this, the straight rate springs
can accomodate all of the bumps at one particular track for that rider.
It is different on the streets. The progressive rate springs are better
for the streets since the road smoothness varies a lot and it would be
nearly impossible to cover all conditions with one fork rate no matter how
well you tune it. If you have infinite travel then it is a different
discusion subject all together. So the progressive rate fork springs make
compromises by dividing up the travel and give each portion its own spring
rate. What this does is the travel is shorten but it can cover more
terrains.
What I found out through talking with the Fox and Ohlins tech guys is that
there is no one setting for all conditions and all riders. This is why top
racers spend a lot of time chasing their suspension settings and some even
have on-the-fly preload adjusters so they can change the fork preloads in
the middle of the race.
--
Matt Ho
mh...@lbl.gov
Bill Guiffre
eas...@mother.com (Erik Astrup) writes: >
>
> Everything I've ever read, and everything I've heard from suspension guru's
> like Theede and Lindemann always indicate that straight rate springs are
> superior to progressive rate springs. Here's what "Progressive Suspension"
> has to say.
>
> Comments?
>
> ========
>
(you've got to play your hand... sorry, getting carried away),
that if you take a linear rate spring, as you compress it
that linear rate changes so that you are in effect getting
a non-linerar rate spring or progressive rate spring.
I think it makes sense. Take a simple spring in your hand,
It appears harder to compress the last half than the first
half.
But, again, I'm not an engineer.
Bill Guiffre
Greenville, SC
CCS SE Expert #181
Cagiva Alazzurra 650
Graham Byrnes
In article <01bc3f00$37212440$7d50c7cd@megabuck>, "Jamie Worthington"
<mega...@rust.net> wrote:
> Steve Zwinger <s...@Eng.Sun.COM> wrote in article
> <5hs3u7$q...@engnews2.Eng.Sun.COM>...
> > In article t...@news.tamu.edu, bjc...@mulder.tamu.edu (Benjamin Justin
> Cain) writes:
> > >Steve Zwinger (s...@Eng.Sun.COM) wrote:
> > >
> > >Why wouldn't it be linear? Doesn't P=nRT/V to first order? And V
> > >varies linearly with the compression of the fork. I don't get it.
You do, actually. The V is in the denominator, isn't it? So it's hyperbolic
(not exponential as someone else wrote).
Jamie gave a good exposition... what you need to know is that raising
the oil level is a good way of preventing bottoming without much effect
in the initial travel. So if your front wheel patters you need to back
off the damping or use softer springs. If it then starts to bottom you
can up the fork oil to prevent the bottoming without bringing back the
patter.
What I want to know is why thousands of shop manuals trot out the crap that
increasing preload makes the suspension stiffer. It doesn't, it increases
ride height. Together with prgressive springing/linkages/air, that means
it actually gets softer, although *slightly* harder to bottom.
Graham
Jamie Worthington
Please disregard my prior post, as it is now april 2 and I'm not allowed to
be helpful anymore. I must go back to my sardonic and argumentative ways.
Jamie Worthington
Jamie Worthington
Graham Byrnes <G.By...@latrobe.edu.au> wrote in article
<G.Byrnes-030...@graham-byrnes.maths.latrobe.edu.au>...
> What I want to know is why thousands of shop manuals trot out the crap
that
> increasing preload makes the suspension stiffer. It doesn't, it increases
> ride height. Together with prgressive springing/linkages/air, that means
> it actually gets softer, although *slightly* harder to bottom.
>
> Graham
>
In general, when cranking on the preload, you are increasing the force
required to make the suspension move. "Technically" this is not stiffening
the suspension as the spring rate will still be the same(assuming linear
rate spring). But what this does is allow the bike to go over bigger bumps
or harder braking without bottoming the suspenion, as it requires more
force to overcome the higher preload. This may or may not solve your
suspension woes, though. If the preload is too high, the front end can
effectively become solid, and the bike will still act like its
bottoming.(and the tire then becomes the suspension). rig for a rough ride.
As for changing the ride height, yes it will raise the front end a little,
but the reason for adjusting preload is to make the front work better over
bumps or under braking. If you want to change the steering angle it's
better to raise or lower the forks. If you want to get picky and isolate
the preload change from the geometry change, you really should adjust the
fork height to compensate for the extra preload.(assuming there is some sag
in the suspension)
If there is no sag, the geometry doesn't change.
But these changes all seem to affect one another, and while a "simple"
preload adjustment may stop the suspension from bottoming under braking, if
the change is major, it could also affect turn-in at the corner. It also
jacks weight to the back wheel.
Tuning a suspension is not so simple anymore, is it folks. :-)
Jamie
Frank Ball
On 1 Apr 1997 21:35:20 GMT, Benjamin Justin Cain @ bjc...@mulder.tamu.edu wrote:
& Steve Zwinger (s...@Eng.Sun.COM) wrote:
& :
& : It was my understanding that the progressive rate for the forks was
& : supplied by the air trapped between the oil level and the fork cap.
& : The resistance offered by the air is non-linear and provides the most
& [snip]
& Why wouldn't it be linear? Doesn't P=nRT/V to first order? And V
& varies linearly with the compression of the fork. I don't get it.
Forks are very non-linear due to the trapped air. P=K/V is not linear
as V changes. Graph it if you don't believe me, as V gets small P gets
really big really fast.
Forks are non-linear, but this non-linearity is gradual and constantly
changing. I have a dual rate spring (stock) on the back of my XT350 and
I think it sucks. The soft rate is way soft and the bike wallows on it,
until you suddenly hit the firm rate. I think progressive may have
progressively wound springs that also vary gradually, but most bikes
have rear shock linkages that do this already, so are they really
needed?
Frank Ball 1UR-M fra...@sr.hp.com
Hewlett-Packard (707) 794-4168 work
1212 Valley House Drive (707) 794-3038 fax
Rohnert Park CA 94928-4999 (707) 538-3693 home
Bert Kellerman
> Forks are very non-linear due to the trapped air. P=K/V is not linear
Hello??? It IS linear. No exponents people... PV=nRT IS linear.
B
Chuck Zwicky
Does anyone make a pnuematic fitting for a standard fork tube? It
would be nice to be able to control the air pressure in the fork...
Chuck Zwicky
Does anyone make a pnuematic fitting for a standard fork tube? It
would be nice to be able to control the air pressure in the fork...
"It doesn't matter what you ride, as long as you ride"
1986 YAMAHA SRX-6 :
White Brothers pipe with Supertrapp alloy can,
Dynojet stage 3 with K&N filters,
Progressive Suspension, Stainless Brake Lines.
1982 SUZUKI GS 450Txz Stock. Sold it.
1974 HONDA CB200. 1324 actual miles. Stock. For Sale
Fred Farzanegan
In article <G.Byrnes-030...@graham-byrnes.maths.latrobe.edu.au>, G.By...@latrobe.edu.au (Graham Byrnes) writes:
|> What I want to know is why thousands of shop manuals trot out the crap that
|> increasing preload makes the suspension stiffer. It doesn't, it increases
Because they're right? Good thing we've got the internet to ferret out
all that bad information.
What does stiffer mean to you? Not bottoming out is one thing. I guess you're
referring to there's more _travel_ and therefore over the course of the travel
the effects of the spring are reduced- but in best case you confuse the issue.
The discussion centered around not bottoming out and increasing preload is
one way to accomplish it.
For a straight rate spring, the spring rate remains the same. But for a
progressive or dual rate spring (common on many stock bikes), the spring rate
is _not_ constant and increasing preload does change the rate.
|> ride height. Together with prgressive springing/linkages/air, that means
|> it actually gets softer, although *slightly* harder to bottom.
Huh? Once you've topped out the suspension (easy to do on rear shocks),
you are absolutely making it stiffer. How do you get _softer_ out of that?
-fred
DoD #835
Andrew J. Hutton
Bill Guiffre (Greenville, SC) (Bill_G...@ChampInt.com) wrote:
: I think it makes sense. Take a simple spring in your hand,
: It appears harder to compress the last half than the first
: half.
All springs compress like that, progressive just get MORE solid the last bit
than even a regular one does. They also compress from the TOP DOWN, rather
than uniformly which may have an effect as well.
99 out of 100 getting progressive springs helps, and the reason is they're
generally stiffer and MOST North Americans are heavier than the individual
the bike was designed for and it's the fact they're also generally heavier
springs that makes the difference.
If you replace with a good quality non-progressive spring that is suited to
your weight/driving you'll also see a drastic improvement.
--
High-Speed Reliable Internet Access Services, Since December 20th, 1993
613-723-6624 * Access Services * Leased Line Access * Web Hosting * MORE
Honda CBR900RR, Tecnomagnesio Rims, Metzler Tyres and waiting for spring..
Bert Kellerman
> Some models use purely spherical chambers - lets see a quick calculation on
> that one Jamie.
Spheres?? That sou8nd pretty smart! Actually I would think a half
sphere would be better because once you get lower than the middle of the
sphere, It would bottom out easily. HMMM
b
Bert Kellerman
> The answer the the question (if I have the question right) is that an air
> spring is an exponentially progressive rate spring. It is NOT a linear
> spring
All springs are exponentially progressive. If you wanted a linear spring
you would have to make it progressive in the reverse direction.
B
race...@aol.com
In article <5hu0ft$lko$2...@zippy.intcom.net>, Bill Guiffre (Greenville, SC) <Bill_G...@ChampInt.com> writes:
>
>I think it makes sense. Take a simple spring in your hand,
>It appears harder to compress the last half than the first
>half.
>
>
Thats because it does take more force to compress it. A straight rate spring means that 1" of compression always takes the same amount of force. If you want to compress a 10lb/in spring it will
take 10 lbs of force to compress it 1in, The second inch will take 20 lbs or 10 additional lbs for 1 inch. A progressive spring would act differently, ie 5lbs might be 1 inch to start with but at
the end of the travel it might only be 1/2 an inch. I have used race tech springs for 2 seasons and they are much better than the old progressive springs that came with the bike.
John
Racer 577
John
Racer 577
CCS EX #577
90 Duc 750
89 EX 627
Victor Ortega
Bert Kellerman <SPAMb...@iglou.com> writes:
> > Forks are very non-linear due to the trapped air. P=K/V is not linear
>
> Hello??? It IS linear. No exponents people... PV=nRT IS linear.
Pressure and volume are not linearly related. Here's your exponent:
-1
P = KV
Pressure is related to the inverse of volume. When the volume
doubles, the pressure drops by half.
Victor
Dennis McGuire
On Thu, 3 Apr 1997, Bert Kellerman wrote:
> > Forks are very non-linear due to the trapped air. P=K/V is not linear
>
> Hello??? It IS linear. No exponents people... PV=nRT IS linear.
>
Yes, but adiabatic compression of an air spring is NOT!
Graham Byrnes
> In general, when cranking on the preload, you are increasing the force
> required to make the suspension move.
Well, no, assuming you have some sag in the suspension to start with. If
you don't can I watch? :-)
Graham
Jamie Worthington
Sure, why not. I always wondered what a hardley rode like.
Jamie
Graham Byrnes <G.By...@latrobe.edu.au> wrote in article
<G.Byrnes-040...@graham-byrnes.maths.latrobe.edu.au>...
WRJ JRS
I want to mention to the group talking about raising oil levels in forks.
DON'T RAISE THE OIL LEVEL WITHOUT FIRST CHECKING IF YOUR FORKS ARE CLOSE
TO HYDROLIC LOCK. (sorrry about spelling). Run your fork to bottoming
without the spring in it and measure the oil level from the top. Then put
the spring back in carefully without compressing it. (On most bikes this
will leave about 4 inches/100mm sticking out. If your oil level is close
to the top of the compressed fork DON'T RAISE THE OIL LEVEL. If you get
the oil close to the top of the fork in the compressed state, when it
heats up it could create a solid oil column under braking. This is a very
dangerous condition. The least of your problems is that the suspension
will act like a solid front end, with large amounts of shudder, wheel hop,
and possible loss of control. Worse, especially on a "upside down" front
end you can blow the fork seals with such force that the oil will be all
over the front tire/tyre, with attendant loss of control. This type of
failure can get real ugly. I am not saying that you cannot raise the oil
level at all, just that you should proceed with caution and not put in too
much at once. BE CAREFUL! I hope all your experiments in this area are
successful. Good luck.
Bill J. WRJ...@aol.com
AFM 61
FZR 1000 racebike (gone but not forgotten)
GSXR 750 (the new kid on the block)
WRJ JRS
One thing many posters to this thread havn't mentioned is that the
Progressive (the company) springs are usually wound with a heavy guage
wire with about 2-3 inches already coil bound! These springs are HEAVY.
They also displace damping OIL which can cause some forks to bottom. A
interesting alternative is the system which uses two straight rate springs
with a adjustable cross-over. I bought my set through ATK. This set-up
worked MUCH better than the Progressive springs. The crossover adjuster,
(a bolt and washer /w nut), does displace some oil also, but the control
is worth it. My $.02 Good luck
Fred Farzanegan
In article <19970404074...@ladder01.news.aol.com>, wrj...@aol.com (WRJ JRS) writes:
|> One thing many posters to this thread havn't mentioned is that the
|> Progressive (the company) springs are usually wound with a heavy guage
|> wire with about 2-3 inches already coil bound! These springs are HEAVY.
|> They also displace damping OIL which can cause some forks to bottom. A
You should install progressively wound spring with the tight coils UP.
The reduces the effect of oil displacement compared to stock springs
and _slightly_ reduces unsprung weight (i.e. the lower spring supports
the heavier upper portion).
For _racing_, everyone recommends STRAIGHT rate springs. The best reason
is that its easier to control your suspension adjustments without having
to deal with a variable (k) spring rate and you can do _damping_ adjustments
to deal with track conditions. Set your race bike for the correct sag
and then start with damping adjustments. _If_ you don't have the ability
to control damping, then you might get an improvement from progressive
springs to make up for the lack of adjustment.
Progressive springs are _great_ for the street where you have to deal with
a variety of conditions from table-smooth to potholes. The variable
spring rate gives a soft ride in the initial travel and helps prevent
bottoming out in severe conditions.
-fred
WERA #9
CCS #191
DoD #835
Matt Ho
> I though this thread was about springs, in which case the force wouldn't
> be linear. F=1/2KX^2 where k is the spring constant diff depending
> on #of coils and spring thickness and x is displacement, would be the
> equation needed.
I thought that the force of the spring is F=KX, at least according to my
physics book.
--
Matt Ho
mh...@lbl.gov
Matt Ho
In article <G.Byrnes-030...@graham-byrnes.maths.latrobe.edu.au>,
G.By...@latrobe.edu.au (Graham Byrnes) wrote:
> What I want to know is why thousands of shop manuals trot out the crap that
> increasing preload makes the suspension stiffer. It doesn't, it increases
> ride height.
It depends on what you mean by stiffer. Adding preload does make the shock
initally stiffer but the spring constant is still the same so subsequent
travel still have the same force. The equation for the spring force is
F=KX but when you increase the preload, the new equation is F=K(X+X0)
where X0 is the amount of the preload (initial compression). For example,
in the old set-up, 5" into the shock travel, the force is 100 lbs. With a
preload of 2", the new set-up will yield a force of 100 lbs. at 3" because
of the 2" preload plus the 3" travel making 5" total compression.
If you want to graph this, it is the similar y=mX+b we learned in algebra
years and years ago. Y is the force exerted by the spring. The slope m is
the spring constant. The X is the total compression of the spring and B is
the spring preload. In the previous example, B = K*X0.
So to answer your question, adding preload does make the shock stiffer at
the same travel distance (X) compared to the old set-up but it does not
make it stiffer if we take the preload into consideration.
Alright, that's enough. Coffee break time :-).
--
Matt Ho
mh...@lbl.gov
Dennis McGuire
Well, your two cents worth is overpriced! :) The force due to
deflection of a linear spring is F=Kx where K is the spring rate and x is
the deflection. The ENERGY stored in a linear spring is E=1/2Kx^2.
See my previous post on the non-linearity of gas springs.
On Wed, 2 Apr 1997, Bert Kellerman wrote:
> I though this thread was about springs, in which case the force wouldn't
> be linear. F=1/2KX^2 where k is the spring constant diff depending
> on #of coils and spring thickness and x is displacement, would be the
Graham Byrnes
> |> What I want to know is why thousands of shop manuals trot out the crap that
> |> increasing preload makes the suspension stiffer. It doesn't, it increases
>
> Because they're right? Good thing we've got the internet to ferret out
> all that bad information.
>
> What does stiffer mean to you? Not bottoming out is one thing.
No, stiffer means more force to achieve the same deflection.
I guess you're
> referring to there's more _travel_ and therefore over the course of the travel
> the effects of the spring are reduced- but in best case you confuse the issue.
Huh? The spring rate is the same, yes? So if there was only the spring, the
stiffness would be exactly the same with extra pre-load, no? OTOH, the same
reaction force occurs after a smaller displacement, so the bike sits a a little
higher (by exactly the amount of the preload change).
Now, you also have the trapped air, where P=K/V. The rate is the derivative,
so the effective spring rate is KV^(-2), ie it decreases as the volume increases
(which is why it's progressive). Now, you've increased your spring pre-load
and the bike is sitting a little higher all the time, so the volume is, uh,
bigger, yes? So the spring rate due to the trapped air has just *decreased*.
If you increase the preload by 5mm, you would need to increase the oil
height by 5mm to retain the same stiffness. You would also by 5mm further
from bottoming.
The same arguments apply to the rear if you allow for the progressive linkage.
> The discussion centered around not bottoming out and increasing preload is
> one way to accomplish it.
I'd say a fairly poor way.
Cheers,
Graham
Matt Ho
In article <5hrr1e$o...@engnews2.Eng.Sun.COM>, s...@Eng.Sun.COM wrote:
> Rear suspensions (shocks) do not have an air chamber incorporated it
> their design. (at least the one's I've rebuilt)
Some are filled with Nitrogen under pressure which could be thought of as
"air filled".
--
Matt Ho
mh...@lbl.gov
Matt Ho
> > Forks are very non-linear due to the trapped air. P=K/V is not linear
>
> Hello??? It IS linear. No exponents people... PV=nRT IS linear.
Hello,
I suggest you sit down and graph the equation P=K/V and you will see that
it is not linear. K is a constant so if you replace the above equation
with
y=K/x
you will see a curved line with a steep negative slope near zero and a
near flat but still negative slope near infinity.
--
Matt Ho
mh...@lbl.gov
Rich Sturges
> The answer the the question (if I have the question right) is that an air
> spring is an exponentially progressive rate spring. It is NOT a linear
(Jamie Worthington's excellent explaination deleted)
> So as you can see, the rate at which the force is applied changes with
> position, therefore an airspring is a rising rate spring. Varying the
> volume of oil in a fork definitley can impact how your forks work during
> hard bumps or heavy braking. Not much effect is noticed over small bumps.
Don't forget that the dynamic forces that the oil provides does NOT
contribute to holding up the front end. The compression and rebound
damping
that results from the oil in the forks must be considered separately
from mech-
anical and air spring rates.
The effect of changing fork oil height is derived primarily from the
change in trapped air volume. Most forks have damping holes that are
always
submerged in oil.
Progressive springs were (are) indespensible on older bikes that do
not
have air tight forks.
And one final point: comparing race bike setup (reference to using
linear
springs) does not take into account the LARGE bumps often encountered on
the
street, nor the desire for a 'nice' ride.
rich
Michael Krone
Fred:
Is the question rather -- I'm going to make some changes to my front suspension,
should I consider progressive springs or adjust the straight rate springs with
washers/oil to improve and adjust static sag and rebound????
The wisdom I listen to most carefully suggests that the latter is the way to go
(which I guess you could call the Lindemann solution) rather than progressives or
cartridge emulators (but I'm still interested in emulators).
Michael Krone
Fred Farzanegan wrote:
>
> In article <G.Byrnes-030...@graham-byrnes.maths.latrobe.edu.au>, G.By...@latrobe.edu.au (Graham Byrnes) writes:
>
> |> What I want to know is why thousands of shop manuals trot out the crap that
> |> increasing preload makes the suspension stiffer. It doesn't, it increases
>
> Because they're right? Good thing we've got the internet to ferret out
> all that bad information.
>
> What does stiffer mean to you? Not bottoming out is one thing. I guess you're
> referring to there's more _travel_ and therefore over the course of the travel
> the effects of the spring are reduced- but in best case you confuse the issue.
> The discussion centered around not bottoming out and increasing preload is
> one way to accomplish it.
>
Erik Astrup
On Thu, 10 Apr 1997 15:01:07 -0700, Michael Krone <ma...@pge.com> wrote:
>
>Fred:
>
>Is the question rather -- I'm going to make some changes to my front suspension,
>should I consider progressive springs or adjust the straight rate springs with
>washers/oil to improve and adjust static sag and rebound????
>
>The wisdom I listen to most carefully suggests that the latter is the way to go
>(which I guess you could call the Lindemann solution) rather than progressives or
>cartridge emulators (but I'm still interested in emulators).
>
I am pretty certain that if you get springs from RT they set you up with
straight rate springs. That's what they sent my buddy for his 400 Bandit.
------------------------------------------------------------------------
Erik Astrup - Team Iguana Racing (Ret) *Sold* '89 Honda Transalp
Home of the Transalp Owners Mailing List!
http://www.mother.com/~eastrup/
Need to know when a roadrace is going to be on TV??
http://www.mother.com/~eastrup/racing/tv.htm
------------------------------------------------------------------------
Brian McLaughlin
I find this thread quite humorous. It is quite obvious that many of
you did not pay very close attention in your physics and math courses.
Let me put this to rest once and for all. I will start with the
trapped gas.
PV=nRT is the ideal *gas* law. It is not applicable to liquids as one
person stated.
P = pressure
V = volume
n = the number of moles (essentially the number of gas molecules, ther
are 6.02x10^23 molecules per mole)
R = the ideal gas law constant
T = the absolute (Kelvin) temperature.
For our purposes we will assume no temperature rise and no gas
escaping from the fork. Therefore our equation becomes
PV=K where K is a constant equal to nRT.
Or P=K/V.
A linear equation can be written as y = mx + b, where m and b are
constants, m being the slope of the line. In this equation the
variables y and x are linearly related. The graph of y versus x will
give a straight line. Graphing P=K/V (P on y axis, V on x axis) will
not yield a straight line. Therefore P and V *are not liearly
related*. That is why the effect of air is not apparent until the
final third of travel. It is a progressive spring. In the last third
of travel the 'spring constsnt' of the air becomes large enough to be
noticed. That is why raising fork oil level helps resist bottoming
without making the fork harsh over small bumps.
For those who know calculus the proof of non linearity is much more
straight forward. The derivative of a linear function is a constant.
For example y=mx+b, dy/dx = m which is a constant. Lokking at
P = K/V, dP/dV = -K/V^2, which is not a constant. Therefore P and V
are not linearly related. And you thought that all that calculus was a
waste of time.
Some one else stated that all springs are not linear, someone even
backed this up with F = 1/2 kx^2. Well both of these are incorrect.
Straight wound springs are linear (as long as the elastic limit of the
material is not exceeded, and the spring does not coil bind). The
correct expression for the for force is F = -kx (the minus sign
because the direction of the force is opposite to the displacement).
This is known as Hooke's law. k is the spring constant and it is
determined from the metal used, the diameter of the wire used to form
the spring, and the number of live coils. The spring constant of a
progressive spring increases as it is compressed because the number of
live coils is reduced (the coils close togather will touch before the
coils spread further apart will touch, when coils touch they are no
longer live coils, that is the spring will start to coil bind at one
end and work its way to the other. Straight rate springs do not do
that , they coil bind all at once.)
I think that is enough for now, I won't even begin to bring up the
topic of stacking different rate springs (Are those kits still
available?) Now maybe you will pay closer attention in class.
--
Brian
TZ250E (1993-95) 2 strokes smoke,
FZR600 (1990) 4 strokes choke!
Brian McLaughlin
G.By...@latrobe.edu.au (Graham Byrnes) wrote:
>No, stiffer means more force to achieve the same deflection.
You need to understand preload. According to your definition adding
preload will make the suspension stiffer because it will take more
force to get the wheel to move the same amount. See my previous reply
to you.
Brian McLaughlin
chuck....@wavefront.com (Chuck Zwicky) wrote:
>Does anyone make a pnuematic fitting for a standard fork tube? It
>would be nice to be able to control the air pressure in the fork...
Adding air causes more problems than it solves. Sticktion is increased
because the additional pressure pushes the seal against the fork tube
(look at the back side of a fork seal). Many people tended to over
fill them and blew seals. Better off spending the time and get the
suspension right without the use of air.
Brian McLaughlin
a...@sparta.achilles.net (Andrew J. Hutton) wrote:
>Bill Guiffre (Greenville, SC) (Bill_G...@ChampInt.com) wrote:
>: I think it makes sense. Take a simple spring in your hand,
>: It appears harder to compress the last half than the first
>: half.
>All springs compress like that, progressive just get MORE solid the last bit
>than even a regular one does. They also compress from the TOP DOWN, rather
>than uniformly which may have an effect as well.
No. Progressive springs compress uniformly (review Newton's third
law). The tighter coils being closer togather will coil bind there by
increasing the spring rate.
Fred Farzanegan
In article <334D63...@pge.com>, Michael Krone <ma...@pge.com> writes:
|> Fred:
|>
|> Is the question rather -- I'm going to make some changes to my front suspension,
|> should I consider progressive springs or adjust the straight rate springs with
|> washers/oil to improve and adjust static sag and rebound????
|>
|> The wisdom I listen to most carefully suggests that the latter is the way to go
|> (which I guess you could call the Lindemann solution) rather than progressives or
|> cartridge emulators (but I'm still interested in emulators).
It really depends on the bike. I assume you're asking about a race bike,
because (I think) we're all in agreement that racebikes should use straight
rate springs.
If your bike does _not_ have a cartridge-type fork. (fzr, older cbrs, etc),
then the RaceTech emulators are _excellent_. The provide you with a variable
damping rate which can also be adjusted. I used them on my FZR and they made
a _huge_ difference. For this type of fork, I recommend racetech over lindemann.
LE only welds-up, redrills damping rod forks, while Racetech allows you to get the
benefits of variable damping.
Racetech also has their own washer system to update cartridge forks with a
new shim stack for newer bikes. LE does the same thing at their shop. I haven't
done either (the F3 handles so much better than the FZR, I can't imagine a huge
improvement).
So, if your question is:
streetbike?: use progressive springs (if you need them)
racebike with damping rod?: use racetech cartridge emulators
racebike with cartridge: you're on your own.
-fred
WERA #9
CCS #|9|
DoD #835
Brian McLaughlin
G.By...@latrobe.edu.au (Graham Byrnes) wrote:
>> In general, when cranking on the preload, you are increasing the force
>> required to make the suspension move.
>Well, no, assuming you have some sag in the suspension to start with. If
>you don't can I watch? :-)
Well, yes, Jamie is correct. Adding preload means that a higher force
is necessary to move the spring even if you have sag. Why do you think
that inceasing preload reduces sag?
Try this.
Suppose our spring has a constant of 100 lbs/inch (that means it will
take 100 lbs to compress 1 inch, 200 lbs to compress 2 inches, etc).
With no preload the spring will compress regardless of the weight set
on it. If I put 1 inch of preload on the spring, a wieght of less than
100 lbs will not compress the spring, only when the force on the
spring exceeds 100 lbs will it begin to move. If I put 2 inches of
preload on the spring, it will require a force of 200 lbs before the
spring will compress.
The term usually used with too much preload is not stiff but harsh.
There is a difference. Your wheel will move less over a given bump
when the preload is higher. If the wheel is resticted from moving
enough the jolt is sent into the frame and is felt as harshness.
Stiffer preload is not the answer for poor suspension. Increasing
preload reduces compliance. Often people crank up preload when they
should really be buying a stiffer spring or having their shock damping
adjusted (the best money spent after rider training is suspension).
Brian McLaughlin
Bert Kellerman <SPAMb...@iglou.com> wrote:
>> Forks are very non-linear due to the trapped air. P=K/V is not linear
>Hello??? It IS linear. No exponents people... PV=nRT IS linear.
The relationship between P and V IS NOT LINEAR. Do you know any
calculus? dP/dV = -nRTV^-2. Doesn't look like a constant with V on the
right side, therefore the relationship between P and V is not linear.
Brian McLaughlin
wrj...@aol.com (WRJ JRS) wrote:
>One thing many posters to this thread havn't mentioned is that the
>Progressive (the company) springs are usually wound with a heavy guage
>wire with about 2-3 inches already coil bound! These springs are HEAVY.
>They also displace damping OIL which can cause some forks to bottom. A
>interesting alternative is the system which uses two straight rate springs
>with a adjustable cross-over. I bought my set through ATK. This set-up
>worked MUCH better than the Progressive springs. The crossover adjuster,
>(a bolt and washer /w nut), does displace some oil also, but the control
>is worth it. My $.02 Good luck
>Bill J. WRJ...@aol.com
>AFM 61
>FZR 1000 racebike (gone but not forgotten)
>GSXR 750 (the new kid on the block)
This was (are they still available?) a good system. The problem
compared to a progessive spring is the time to fine tune the
suspension. The multiple spring setup has far more potential.
race...@aol.com
In article <5ilskb$j...@wiscnews.wiscnet.net>, bmcla...@waukesha.tec.wi.us (Brian McLaughlin) writes:
>> In general, when cranking on the preload, you are increasing the force
>>> required to make the suspension move.
>
>>Well, no, assuming you have some sag in the suspension to start with. If
>>you don't can I watch? :-)
>
>Well, yes, Jamie is correct. Adding preload means that a higher force
>is necessary to move the spring even if you have sag. Why do you think
>that inceasing preload reduces sag?
>
>
If the spring is straight rate changing preload with sag will not make it any harder to start the suspension moving. Although at a given position is the travel the spring force will be greater. The
front end of a bike weighs 250#, This means the spring is compressed until 250# of force is produced. If you add more preload then the sag is reduced, but the spring is only compressed enough to
create 250# of force, but now when the fork is topped out there is more force in the spring than when there was less preload.
Brian McLaughlin
race...@aol.com wrote:
Have you ever had physics? Increasing preload on a spring will
increase the force necessary for the spring to move. As an example
suppose we have a 100 lb/inch spring. Without preload any force will
compress the spring, if 1 inch of preload is put in, it will take 100
lbs of force to make the spring move, any less force and the spring
will not compress. If another inch is added, for a total of 2 inches,
it will take a force of at least 200 lbs before the spring will begin
to compress. So for a given amount of wheel movement, it will require
more force with higher preload (since the spring will be compressed
more). It will also require more force to start the suspension moving
due to the preload. Small bumps are much easier to feel with the
preload at maximum as compared to having the preload at minimum.
Benjamin Justin Cain
Brian McLaughlin (bmcla...@waukesha.tec.wi.us) wrote:
: Have you ever had physics? Increasing preload on a spring will
: increase the force necessary for the spring to move. As an example
: suppose we have a 100 lb/inch spring. Without preload any force will
: compress the spring, if 1 inch of preload is put in, it will take 100
: lbs of force to make the spring move, any less force and the spring
: will not compress. If another inch is added, for a total of 2 inches,
: it will take a force of at least 200 lbs before the spring will begin
: to compress. So for a given amount of wheel movement, it will require
: more force with higher preload (since the spring will be compressed
: more). It will also require more force to start the suspension moving
: due to the preload. Small bumps are much easier to feel with the
: preload at maximum as compared to having the preload at minimum.
The problem with yelling so loudly at everyone, Brian, is that
everyone can hear you when you're wrong.
I took a physics class once, and Graham has an excellent point which
I have yet to see addressed at all here. And his point stands in
direct contradiction to what you say here, as does basic physics
applied to the question of fork suspension.
--
Ben Cain : Biker Scum - Yellowshirt Brigade #514
AMA HRCA TMGP CMRA/WERA DoD # 1/137 KoK3
1990 VFR750F "Deal's Gimp"
1989 YSR50/3
http://www.users.cts.com/king/d/drlubell/bscum.html
"Being shot out of a cannon will always be better than being squeezed out
of a tube. That is why God made fast motorcycles, Bubba..."
-Hunter S. Thompson-
Benjamin Justin Cain
Brian McLaughlin (bmcla...@waukesha.tec.wi.us) wrote:
:
: >: increase the force necessary for the spring to move. As an example
: >: suppose we have a 100 lb/inch spring. Without preload any force will
: >: compress the spring, if 1 inch of preload is put in, it will take 100
: >: lbs of force to make the spring move, any less force and the spring
: >: will not compress. If another inch is added, for a total of 2 inches,
: >: it will take a force of at least 200 lbs before the spring will begin
: >: to compress. So for a given amount of wheel movement, it will require
: >: more force with higher preload (since the spring will be compressed
: >: more). It will also require more force to start the suspension moving
: >: due to the preload. Small bumps are much easier to feel with the
: >: preload at maximum as compared to having the preload at minimum.
:
: >The problem with yelling so loudly at everyone, Brian, is that
: >everyone can hear you when you're wrong.
: What did I say that is wrong?
Your spring analogy is off. A fork spring is modelled by a vertical
spring, sitting on the floor with a (say) 200 lb weight on it. When
you preload, you effectively put a shim under the weight. If the
weight didn't move, your analogy would be correct. But the weight
will move. Assuming the shim is massless, of course, the weight will
move up a distance equal to the thickness of the shim. The spring
doesn't compress - the ride hight increases. It makes sense. If you're
the spring, you don't care about anything that's going on except the
fact that you see a 200lb weight compressing you. Preload, shmeeload.
You only see the weight of the bike compressing you. So, no; for a
given amount of wheel movement you do NOT have to exert a greater
force on the spring that's preloaded. To reach a certain given relative
position (of the wheel with respect to the steering head), you
DO have to exert greater force. So all preloading does is increase your
ride height (unless the bottoming in your suspension occurs before
the spring is fully compressed for some reason, in which case all
you've done is incresed the force required to bottom it, but I'm
not sure I've ever met a suspension that worked like that). And you
still haven't increased the force required to produce a given
amount of wheel travel, nor is there any more force required to
"start the suspension moving."
: The bottom line is that increasing the preload increases the force
: necessary to bottom the suspension.
This is true only if something other than the spring compressing
fully is responsible for your suspension bottoming (like the slider
banging into a stop or something at the end of the tube maybe?).
But if your suspension, like mine, bottoms as a result of the spring
being fully compressed, this is absolutely untrue.
Brian McLaughlin
bjc...@scully.tamu.edu (Benjamin Justin Cain) wrote:
>Brian McLaughlin (bmcla...@waukesha.tec.wi.us) wrote:
>: Have you ever had physics? Increasing preload on a spring will
>: increase the force necessary for the spring to move. As an example
>: suppose we have a 100 lb/inch spring. Without preload any force will
>: compress the spring, if 1 inch of preload is put in, it will take 100
>: lbs of force to make the spring move, any less force and the spring
>: will not compress. If another inch is added, for a total of 2 inches,
>: it will take a force of at least 200 lbs before the spring will begin
>: to compress. So for a given amount of wheel movement, it will require
>: more force with higher preload (since the spring will be compressed
>: more). It will also require more force to start the suspension moving
>: due to the preload. Small bumps are much easier to feel with the
>: preload at maximum as compared to having the preload at minimum.
>The problem with yelling so loudly at everyone, Brian, is that
>everyone can hear you when you're wrong.
What did I say that is wrong?
>I took a physics class once, and Graham has an excellent point which
>I have yet to see addressed at all here. And his point stands in
>direct contradiction to what you say here, as does basic physics
>applied to the question of fork suspension.
>--
The bottom line is that increasing the preload increases the force
necessary to bottom the suspension. In racing the suspension should be
set to use the full travel. Increasing preload makes your fork stiffer
over its full travel.
Graham Byrnes
In article <5j3ir7$6...@news.tamu.edu>, bjc...@scully.tamu.edu (Benjamin
Justin Cain) wrote:
> : >: more). It will also require more force to start the suspension moving
> : >: due to the preload.
Yeah, but if there is any sag it has *already* started to move. If you
set up your bike so that it has no sag with you sitting on it, I still
want to watch :-)
> fact that you see a 200lb weight compressing you. Preload, shmeeload.
> You only see the weight of the bike compressing you. So, no; for a
> given amount of wheel movement you do NOT have to exert a greater
> force on the spring that's preloaded.
Exactly.
>
> : The bottom line is that increasing the preload increases the force
> : necessary to bottom the suspension.
>
> This is true only if something other than the spring compressing
> fully is responsible for your suspension bottoming (like the slider
> banging into a stop or something at the end of the tube maybe?).
That's how every fork I've ever met bottoms, actually. Preloading
does actually increase the force required to bottom.
> But if your suspension, like mine, bottoms as a result of the spring
> being fully compressed, this is absolutely untrue.
I've never seen any fork (or shock) that worked this way. Are you sure?
Let's get serious. I think we agree that a spring satisfies F=-kx, where
x is the displacement from its natural length.
If you don't, go read some physics.
Now, use the spring
to hold up a mass m. The downward force due to gravity is mg, so the
spring will compress until the upward force due to the spring is equal
to the downward force due to gravity:
mg-kx=0.
So regardless of preload, the spring will end up mg/k shorter than its natural
length, once in equilibrium.
Now suppose the spring is preloaded by a distance p. The force required
to compress it further is now pk, but if mg>pk it will still compress
further.
So the additional force is mg-pk and the spring will be further compressed
by a distance (mg-pk)/k, which is less than if the spring had no preload
(when it compressed by mg/k).
Suppose you now hit a bump (or increase the fork load by tipping into a
corner). Suppose the force applied to the spring is Fb.
The unpreloaded spring is now supporting mg+Fb, so it compresses a total
of (mg+Fb)/k. This is a *change* in compression of Fb/k, ie the fork
strokes through this distance in response to the bump.
The preloaded spring is also supporting mg+Fb. pk of that is used to overcome
the preload, leaving mg+Fb-pk to further compress the spring, so it
compresses by (mg+Fb-pk)/k. Before the bump it was compressed by (mg-pk)/k,
so the change in compression in response to the bump is
(mg+Fb-pk)/k-(mg-pk)/k=Fb/k
Exactly the same as the unpreloaded spring. This is what I would call
the stiffness: how much the fork deflects in response to a given bump
*without* bottoming. (And assuming the preload was not so great that
there was no initial sag with the rider on the bike).
This suggests to me that if the fork relied simply on a spring, preload
would *not* affect stiffness. It will slightly increase bottoming resistance,
because there is slightly more travel available.
Now in fact all modern forks rely on the trapped air too. Roughly, the
force exerted by the air is given by
F=pAx/(T+L-x),
where A is the cross-sectional area of the narrower part of the forks,
p is atmospheric pressure, L is the bottomed out oil height and T is the total
travel of the forks. The displacement is x, again. Because this is non-linear
(hyperbolic in fact, the x is in the denominator), the pressure increases
more quickly as the fork compresses.
Consider a TZ250. L=95mm (on mine), T=110mm (according to Yamaha),
A=(2.15)^2*pi=14.5 cm^2, p=1kg/cm^2.
So at mid-stroke (x=55mm), F=5.3kg
At full stroke, F=16.8kg
That is for each fork. Compare it with the standard springs, with k=7kg/cm.
So the air can make more than 2cm worth of difference in travel.
The effective spring constant you get by differentiating:
k(x)=pA(T+L)/(T+L-x)^2.
For the TZ, that is about 1.3kg/cm at mid-stroke and 3.3kg/cm at bottoming.
But as you increase preload, the initial point is higher up and the effect
of the air is slightly less (because of its progressivity).
So overall, the effective spring rate at the equilibrium point will be
(slightly) less.
Hence More deflection for a given bump.
So there you go...
Graham
Brian McLaughlin
bjc...@scully.tamu.edu (Benjamin Justin Cain) wrote:
>Brian McLaughlin (bmcla...@waukesha.tec.wi.us) wrote:
>: What did I say that is wrong?
>Your spring analogy is off. A fork spring is modelled by a vertical
>spring, sitting on the floor with a (say) 200 lb weight on it. When
>you preload, you effectively put a shim under the weight. If the
>weight didn't move, your analogy would be correct. But the weight
>will move. Assuming the shim is massless, of course, the weight will
>move up a distance equal to the thickness of the shim. The spring
>doesn't compress - the ride hight increases. It makes sense. If you're
>the spring, you don't care about anything that's going on except the
>fact that you see a 200lb weight compressing you. Preload, shmeeload.
>You only see the weight of the bike compressing you. So, no; for a
>given amount of wheel movement you do NOT have to exert a greater
You are correct. I reread the articles and had misinterpreted what had
been said. I will have to read more carefully in the future.
>force on the spring that's preloaded. To reach a certain given relative
>position (of the wheel with respect to the steering head), you
>DO have to exert greater force. So all preloading does is increase your
This is what I meant. I should have been more careful. I measure my
suspension movement from the bottom of the travel (confuses people who
try to peek at my setup notes - and me sometimes).
>ride height (unless the bottoming in your suspension occurs before
>the spring is fully compressed for some reason, in which case all
>you've done is incresed the force required to bottom it, but I'm
>not sure I've ever met a suspension that worked like that). And you
>still haven't increased the force required to produce a given
>amount of wheel travel, nor is there any more force required to
>"start the suspension moving."
>: The bottom line is that increasing the preload increases the force
>: necessary to bottom the suspension.
>This is true only if something other than the spring compressing
>fully is responsible for your suspension bottoming (like the slider
>banging into a stop or something at the end of the tube maybe?).
>But if your suspension, like mine, bottoms as a result of the spring
>being fully compressed, this is absolutely untrue.
I know of no fork that uses the coil binding of the spring to limit
full travel. This would not be a good thing. Fork travel is limited by
the amount of movement of the slider and tube.
Brian McLaughlin
G.By...@latrobe.edu.au (Graham Byrnes) wrote:
>Exactly the same as the unpreloaded spring. This is what I would call
>the stiffness: how much the fork deflects in response to a given bump
>*without* bottoming. (And assuming the preload was not so great that
>there was no initial sag with the rider on the bike).
>This suggests to me that if the fork relied simply on a spring, preload
>would *not* affect stiffness. It will slightly increase bottoming resistance,
>because there is slightly more travel available.
Everything you have written is correct. Where we differ is on the
definition of stiffer. I have always used the term stiffer to refer to
the suspension being more difficult to bottom. For my defition adding
preload makes the suspension stiffer. Putting in a spring of higher
spring rate makes the suspension firmer (takes more force per inch of
compression is how I define firm rather than stiff).
>Now in fact all modern forks rely on the trapped air too. Roughly, the
>force exerted by the air is given by
>F=pAx/(T+L-x),
>where A is the cross-sectional area of the narrower part of the forks,
>p is atmospheric pressure, L is the bottomed out oil height and T is the total
>travel of the forks. The displacement is x, again. Because this is non-linear
>(hyperbolic in fact, the x is in the denominator), the pressure increases
>more quickly as the fork compresses.
Your formula will work with standard forks. For USD forks the problem
is slightly more complicated. The trapped volume has the cross
sectional area of the larger tube(A), where as the decrease in volume
is due to the smaller tube being pushed up, so the reduction in volume
uses the cross sectional area of the smaller tube(a).
I came up with
F=pa^2x/A(T+L)-ax
>Consider a TZ250. L=95mm (on mine), T=110mm (according to Yamaha),
>A=(2.15)^2*pi=14.5 cm^2, p=1kg/cm^2.
Should really use p=1.01x10^5 Pa
>So at mid-stroke (x=55mm), F=5.3kg
>At full stroke, F=16.8kg
Then your forces would be in Newtons. Kilograms are not force units (I
am sure that you know this. The kgf is a completely idiotic unit).
>That is for each fork. Compare it with the standard springs, with k=7kg/cm.
>So the air can make more than 2cm worth of difference in travel.
>The effective spring constant you get by differentiating:
>k(x)=pA(T+L)/(T+L-x)^2.
>For the TZ, that is about 1.3kg/cm at mid-stroke and 3.3kg/cm at bottoming.
>But as you increase preload, the initial point is higher up and the effect
>of the air is slightly less (because of its progressivity).
>So overall, the effective spring rate at the equilibrium point will be
>(slightly) less.
It will then be more (slightly) at full stroke.
>Hence More deflection for a given bump.
So adding preload will make the suspension easier to bottom (hence
softer)? :)
Erik Astrup
On 15 Apr 1997 19:47:33 GMT, bjc...@scully.tamu.edu (Benjamin Justin Cain)
wrote:
>Brian McLaughlin (bmcla...@waukesha.tec.wi.us) wrote:
>
>: Have you ever had physics? Increasing preload on a spring will
>: increase the force necessary for the spring to move. As an example
>: suppose we have a 100 lb/inch spring. Without preload any force will
>: compress the spring, if 1 inch of preload is put in, it will take 100
>: lbs of force to make the spring move, any less force and the spring
>: will not compress. If another inch is added, for a total of 2 inches,
>: it will take a force of at least 200 lbs before the spring will begin
>: to compress. So for a given amount of wheel movement, it will require
>: more force with higher preload (since the spring will be compressed
>: more). It will also require more force to start the suspension moving
>: due to the preload. Small bumps are much easier to feel with the
>: preload at maximum as compared to having the preload at minimum.
>
>The problem with yelling so loudly at everyone, Brian, is that
>everyone can hear you when you're wrong.
>
>I took a physics class once, and Graham has an excellent point which
>I have yet to see addressed at all here. And his point stands in
>direct contradiction to what you say here, as does basic physics
>applied to the question of fork suspension.
Ok, so what about us physics free induhviduals who just want to
know what the REAL answer is!!??!!
Who's right?
------------------------------------------------------------------------
Erik Astrup - Team Iguana Racing (Ret)
1995 Triumph Tiger
http://www.mother.com/~eastrup/home.htm
Paul Belkus
They are both correct. In classic mechanical engineering terms when you
preload a spring you compress it. In the motorcycle world when you
adjust the preload on the front or rear suspension you are increasing or
decreasing the distance from the top of the spring to the fork or rear
of the motorcycle which will raise or lower it. The conflict is in the
definition of preload.
Paul
Matt Ho
> Erik Astrup wrote:
>
> Ok, so what about us physics free induhviduals who just want to
> know what the REAL answer is!!??!!
>
> Who's right?>
In article <335725...@tiac.net>, Paul Belkus <pa...@tiac.net> wrote:
> They are both correct. In classic mechanical engineering terms when you
> preload a spring you compress it. In the motorcycle world when you
> adjust the preload on the front or rear suspension you are increasing or
> decreasing the distance from the top of the spring to the fork or rear
> of the motorcycle which will raise or lower it. The conflict is in the
> definition of preload.
Thank God this issue is finally resolved. Now I can sleep at night ;-).
--
Matt Ho
mh...@lbl.gov
Graham Byrnes
In article <5j5e2s$19...@wiscnews.wiscnet.net>,
bmcla...@waukesha.tec.wi.us (Brian McLaughlin) wrote:
> Everything you have written is correct. Where we differ is on the
> definition of stiffer. I have always used the term stiffer to refer to
> the suspension being more difficult to bottom. For my defition adding
> preload makes the suspension stiffer. Putting in a spring of higher
> spring rate makes the suspension firmer (takes more force per inch of
> compression is how I define firm rather than stiff).
:-) Ok, I'm happy to say that more preload doesn't make them more firm :-)
>
> Your formula will work with standard forks. For USD forks the problem
> is slightly more complicated. The trapped volume has the cross
> sectional area of the larger tube(A), where as the decrease in volume
> is due to the smaller tube being pushed up, so the reduction in volume
> uses the cross sectional area of the smaller tube(a).
> I came up with
> F=pa^2x/A(T+L)-ax
Yeah, I was simplifying. Actually you need to take account of the volume
occupied by the spring, the wall thickness of the bottom part and how
far the inner tube comes up into the outer... The crude approximation
I did is a lower bound, however. Which is also why I used 1kg/cm^2 rather
than 101kPa.
>
> Then your forces would be in Newtons. Kilograms are not force units (I
> am sure that you know this. The kgf is a completely idiotic unit).
Yes, but more people can relate to the weight of a 1kg mass than to a N.
All the suspension people I know talk in kg/cm rather than N/cm
(and certainly not in N/m!).
If any of my students used kgf I'd deduct marks, of course :-)
>
> So adding preload will make the suspension easier to bottom (hence
> softer)? :)
No, the total force to bottom will still increase, just the local
spring-rate will decrease. So my claim with your definitions is that
more preload makes forks stiffer but less firm. OTOH I tend to think
of firmness as related to compression damping, but that would be another
can of worms...
Cheers,
Graham