Thumb test
Ben C
measure tyre pressure or casing tension?
Tim McNamara
Ben C <spam...@spam.eggs> wrote:
> Does the "thumb test" (squeezing a tyre to see if it's hard enough)
> measure tyre pressure or casing tension?
Directly the latter and indirectly the former, I think.
Ben C
This is what I was starting to think too. What this means of course is
that if you have a fat tyre and a thin tyre that feel the same, the fat
tyre will actually be at a lower pressure (as it typically should be),
and have higher rolling resistance.
ecmcd...@gmail.com
Hm. I'm not sure, and I'm just a fly on the wall in this forum, but
this makes little sense to me. Doesn't the larger tire/tyre require
more pressure to get to the same hardness as a small tire? The large
tire has more rubber, so it's got more "stretchiness" or elasticity
across it's cross section. Further, wouldn't a large tire with the same
"thumb feel" as a thin tire (ie, really hard), have less rolling
resistance than a large tire with less pressure? I must be
misunderstanding your statement, sorry....
Ben C
> Ben C wrote:
>> On 2006-10-02, Tim McNamara <tim...@bitstream.net> wrote:
>> > In article <slrnei1jti....@bowser.marioworld>,
>> > Ben C <spam...@spam.eggs> wrote:
>> >
>> >> Does the "thumb test" (squeezing a tyre to see if it's hard enough)
>> >> measure tyre pressure or casing tension?
>> >
>> > Directly the latter and indirectly the former, I think.
>>
>> This is what I was starting to think too. What this means of course is
>> that if you have a fat tyre and a thin tyre that feel the same, the fat
>> tyre will actually be at a lower pressure (as it typically should be),
>> and have higher rolling resistance.
>
> Hm. I'm not sure, and I'm just a fly on the wall in this forum, but
> this makes little sense to me. Doesn't the larger tire/tyre require
> more pressure to get to the same hardness as a small tire?
Well, the larger tyre requires less pressure for a given casing tension.
I read that here recently and am still getting my head around it. It's
also mentioned here by Jobst Brandt:
http://www.sheldonbrown.com/brandt/rim-support.html
"[...] unit casing tension is equivalent to inflation pressure times the
radius of curvature divided by pi [...]".
I was a bit surprised by this at first, but then if you think pressure
is force per unit area, if you increase the area of the inside of the
casing, you need more force for a given pressure. Not sure if this
reasoning is bogus or not though.
So if the thumb test measures casing tension, then the larger tyre does
require less pressure to get to the same thumb-hardness.
> The large tire has more rubber, so it's got more "stretchiness" or
> elasticity across it's cross section. Further, wouldn't a large tire
> with the same "thumb feel" as a thin tire (ie, really hard), have less
> rolling resistance than a large tire with less pressure?
A hard tyre basically has less rolling resistance than if it's soft. But
rolling resistance depends on pressure rather than on casing tension, or
at least, most of the graphs you see are of pressure against RR.
I think this also starts to explain the counter-intuitive result that
fatter tyres have lower RR than thinner tyres at the same pressure.
There are some explanations given in terms of fatter tyres needing to
deform less which I don't fully understand. But if at equal pressures
the fat tyre actually feels harder, it's easier to see intuitively that
it would roll better.
I might make a graph of RR against casing tension (as opposed to
pressure) for different tyre diameters...
jobst....@stanfordalumni.org
>>>> Does the "thumb test" (squeezing a tyre to see if it's hard
>>>> enough) measure tyre pressure or casing tension?
>>> Directly the latter and indirectly the former, I think.
>> This is what I was starting to think too. What this means of
>> course is that if you have a fat tyre and a thin tyre that feel the
>> same, the fat tyre will actually be at a lower pressure (as it
>> typically should be), and have higher rolling resistance.
> Hm. I'm not sure, and I'm just a fly on the wall in this forum, but
> this makes little sense to me. Doesn't the larger tire/tyre require
> more pressure to get to the same hardness as a small tire? The large
> tire has more rubber, so it's got more "stretchiness" or elasticity
> across it's cross section. Further, wouldn't a large tire with the
> same "thumb feel" as a thin tire (ie, really hard), have less
> rolling resistance than a large tire with less pressure? I must be
> misunderstanding your statement, sorry...
The thumb test accurately exposes the thumb to tire pressure. How
that gets to the rim is another problem, but to flatten a portion of
the tire by pressing on it is directly related to pressure (assuming
the tire is a >23mm cross section). I'm assuming the thumb is not
that of a giant and has a width of 10-15mm contact.
The size of the tire does not affect that sensation, just as it
doesn't affect the contact patch area. What is different is the
compliance of the tire when loaded. A fat tire can absorb more
deflection than a narrow one.
Jobst Brandt
carl...@comcast.net
>Does the "thumb test" (squeezing a tyre to see if it's hard enough)
>measure tyre pressure or casing tension?
Dear Ben,
Mostly, it measures the operator's belief in his sensitivity to
pressing one squashy pad against a less squashy pad as the contact
surface broadens.
(It's good enough for riding around.)
But you were asking about casing tension versus air pressure.
Thumb pressure measures casing tension.
Yes, there's 100 psi pushing the casing outward against your thumb.
But when you push the tire a quarter of an inch inward locally,
there's still 100 psi pushing against your thumb.
The 100 psi didn't go away. It didn't increase significantly due to
the tiny change in tire volume. But something stopped you from pushing
any further.
Think of a trampoline.
Air pressure does not hold the trampoline up, any more than it holds
the trampoline down. What holds the tramoline taut is the springs
pulling it tight at the edges. Push down, and the trampoline dents
easily. Push a little further, and more force is needed. The
resistance is nicely progressive . . .
Just like a tire.
A tire is a doughnut-shaped trampoline. What stretches the tire tight
in all directions is the expanding spring of the air pressure.
To dent a tire, you must overcome tension in every direction around
the dent. The deeper you try to push the casing in, the more tension
you must overcome.
_________________________ _ = tire casing
||||||||||||||||||||||||| | = 100 psi upward force
-> ______ ______ <- exaggerated shortening for dent
\/ pulling casing in against tension
Since 100 psi of air pressure is trying to keep the rest of the tire
casing tight everywhere else, you won't get very far.
Here's a simple demonstration that everyone knows--push on a tire
valve. There's no side tension holding the metal valve stem up against
your finger because the metal valve is a special case in the otherwise
uniform doughnut.
Notice that the valve doesn't move (dent) as you slowly apply more and
more pressure. It just sits there until your thumb pressure matches
the air pressure resisting it.
At that point, the valve moves dramatically because almost all the
resisting pressure is no longer being applied--the end of the valve is
just waving around inside the 100 psi air chamber.
But if you were simply pushing a tiny metal piston down into a deep
metal tube set into the tire, you'd push harder and harder without any
movement--and then the piston would begin to move steadily inward and
keep moving steadily because the air pressure resisting your thumb had
been overcome.
Here's another simple demonstration with a bicycle.
Toss a metal tool with a rounded handle less than an inch thick on the
garage floor--a slim socket wrench handle will work well.
Now roll your roughly 1-inch wide 700c front tire at 100 psi onto the
thin round handle.
Lean on the handlebars.
Presumably you weigh well over 100 pounds, but it's unlikely that you
can make the tire touch the garage floor on both sides of the round
metal handle.
A local pressure of 100 psi is not enough to overcome the casing
tension of a tire inflated all the way around to 100 psi.
Cheers,
Carl Fogel
jobst....@stanfordalumni.org
> Here's another simple demonstration with a bicycle.
> Toss a metal tool with a rounded handle less than an inch thick on
> the garage floor--a slim socket wrench handle will work well.
> Now roll your roughly 1-inch wide 700c front tire at 100 psi onto
> the thin round handle.
> Lean on the handlebars.
> Presumably you weigh well over 100 pounds, but it's unlikely that
> you can make the tire touch the garage floor on both sides of the
> round metal handle.
> A local pressure of 100 psi is not enough to overcome the casing
> tension of a tire inflated all the way around to 100 psi.
Invalid experiment!
When a tire is pressed against a flat surface, the tire flattens until
the flat contact area times inflation pressure equal the load.
Remember, how tight must a wire be pulled so that it doesn't sag at
midspan? Casing tension does not come into play at tire-to-ground
contact where it has no curvature, only inflation pressure. In
contrast your complex experiment, involves pressure, casing curvature
in two directions with cord angle and more.
Jobst Brandt
carl...@comcast.net
Dear Jobst,
The point is that the casing tension must be overcome in all
directions to make a dent in the doughnut-like surface.
The dent does not raise the air pressure significantly.
But the dent does require dragging material inward in all directions,
material that's being held in tension over the rest of the doughnut by
100 psi.
Consider the reverse case, an imaginary thumb pushing outward from
inside the tire. What would stop it from pulling the tire wildly out
of shape?
Stick a nail inside a tubeless tire, a nail large enough to make a
bulge in the uninfalted tire.
When you inflate the tire, the air pressure will pushe the tire
outward in all directions.
The increasing tension, however, will pull the tire onto the nail and
puncture the tire.
Cheers,
Carl Fogel
jobst....@stanfordalumni.org
>>> Here's another simple demonstration with a bicycle.
>>> Toss a metal tool with a rounded handle less than an inch thick on
>>> the garage floor--a slim socket wrench handle will work well.
>>> Now roll your roughly 1-inch wide 700c front tire at 100 psi onto
>>> the thin round handle.
>>> Lean on the handlebars.
>>> Presumably you weigh well over 100 pounds, but it's unlikely that
>>> you can make the tire touch the garage floor on both sides of the
>>> round metal handle.
>>> A local pressure of 100 psi is not enough to overcome the casing
>>> tension of a tire inflated all the way around to 100 psi.
>> Invalid experiment!
>> When a tire is pressed against a flat surface, the tire flattens
>> until the flat contact area times inflation pressure equal the
>> load.
>> Remember, how tight must a wire be pulled so that it doesn't sag at
>> midspan? Casing tension does not come into play at tire-to-ground
>> contact where it has no curvature, only inflation pressure. In
>> contrast your complex experiment, involves pressure, casing
>> curvature in two directions with cord angle and more.
> The point is that the casing tension must be overcome in all
> directions to make a dent in the donut-like surface.
It may do that but for bicycle tires that have little curvature with
respect to the minor diameter even these distortions are small. This
has no effect on pressing against a flat surface as I pointed out.
> The dent does not raise the air pressure significantly.
...and who said it did?
> But the dent does require dragging material inward in all
> directions, material that's being held in tension over the rest of
> the donut by 100 psi.
That may be a dynamic effect, the one that causes rolling resistance,
but it has no effect on contact pressure between tire and road.
> Consider the reverse case, an imaginary thumb pushing outward from
> inside the tire. What would stop it from pulling the tire wildly
> out of shape?
The reverse case does not apply. There is no curvature at the ground
contact and thumb pushing is not a dynamic effect.
> Stick a nail inside a tubeless tire, a nail large enough to make a
> bulge in the uninfalted tire.
You're grasping at straws. All this does not apply to the contact
pressure with the road or a thumb flattening the tire locally.
> When you inflate the tire, the air pressure will push the tire
> outward in all directions.
> The increasing tension, however, will pull the tire onto the nail
> and puncture the tire.
And how does this apply to contact pressure with an essentially flat
surface?
Jobst Brandt
carl...@comcast.net
Dear Jobst,
You seem determined to miss every point--including the nail. :)
Cheers,
Carl Fogel
Tim McNamara
Ben C <spam...@spam.eggs> wrote:
> On 2006-10-02, Tim McNamara <tim...@bitstream.net> wrote:
> > In article <slrnei1jti....@bowser.marioworld>,
> > Ben C <spam...@spam.eggs> wrote:
> >
> >> Does the "thumb test" (squeezing a tyre to see if it's hard
> >> enough) measure tyre pressure or casing tension?
> >
> > Directly the latter and indirectly the former, I think.
>
> This is what I was starting to think too. What this means of course
> is that if you have a fat tyre and a thin tyre that feel the same,
> the fat tyre will actually be at a lower pressure (as it typically
> should be), and have higher rolling resistance.
After I typed this, I thought that the thumb test would also be
influenced by the stiffness of the tire wall: the stiffness of the
fabric casing, the thickness and durometer of the rubber, etc.
jobst....@stanfordalumni.org
>>>> Does the "thumb test" (squeezing a tyre to see if it's hard
>>>> enough) measure tyre pressure or casing tension?
>>> Directly the latter and indirectly the former, I think.
>> This is what I was starting to think too. What this means of
>> course is that if you have a fat tyre and a thin tyre that feel the
>> same, the fat tyre will actually be at a lower pressure (as it
>> typically should be), and have higher rolling resistance.
> After I typed this, I thought that the thumb test would also be
> influenced by the stiffness of the tire wall: the stiffness of the
> fabric casing, the thickness and durometer of the rubber, etc.
... not more than the un-inflated tire. Obviously some tires have a
stiff tread that doesn't deflect even when flat, an example is a
motorcycle tire. I assume those who use their thumbs are also smart
enough not to do that to a knobby tread or a thick road tread. The
bare side wall on older tires could give an accurate feel, but even
that has been taken from us by the "I get too many flat tires" folks.
We may revert to the non-pneumatic tire era soon.
Jobst Brandt
Tim McNamara
ecmcd...@gmail.com wrote:
> Ben C wrote:
> > On 2006-10-02, Tim McNamara <tim...@bitstream.net> wrote:
> > > In article <slrnei1jti....@bowser.marioworld>,
> > > Ben C <spam...@spam.eggs> wrote:
> > >
> > >> Does the "thumb test" (squeezing a tyre to see if it's hard
> > >> enough) measure tyre pressure or casing tension?
> > >
> > > Directly the latter and indirectly the former, I think.
> >
> > This is what I was starting to think too. What this means of course
> > is that if you have a fat tyre and a thin tyre that feel the same,
> > the fat tyre will actually be at a lower pressure (as it typically
> > should be), and have higher rolling resistance.
>
> Hm. I'm not sure, and I'm just a fly on the wall in this forum, but
> this makes little sense to me. Doesn't the larger tire/tyre require
> more pressure to get to the same hardness as a small tire?
All other things being equal, the wider tire will have higher casing
tension (called "hoop stress," IIRC) at the same PSI. But it will take
more air to achieve that PSI because of course the volume of the tire is
larger.
> The large tire has more rubber, so it's got more "stretchiness" or
> elasticity across it's cross section. Further, wouldn't a large tire
> with the same "thumb feel" as a thin tire (ie, really hard), have
> less rolling resistance than a large tire with less pressure? I must
> be misunderstanding your statement, sorry....
Because of the fabric casing, a tire doesn't stretch much. It's not
like a balloon. The casing bears the load of the tension, not the
rubber.
Interestingly enough, there is a lot of data showing that- all other
things being equal- a wider tire has lower rolling resistance than a
narrower tire. But of course, all things are not equal. Wider tires,
for example, need a heavier casing because the tension is higher. There
is a fair amount of debate whether rolling resistance tests are
consistent with the behavior of tires in the real world. The problem is
that real world testing introduces all kinds of weird confounds that
make accurate measurement difficult.
The most recent attempt I know of is in the latest Bicycle Quarterly
(formerly "Vintage Bicycle Quarterly"), which compared 16 different
tires in a roll-down test on a soapbox derby course. The results
suggested that at moderate speeds (10.6 to 16.9 mph) some tires were 20%
faster than others (the Deda Tre Giro d'Italia 700 x 24 being the
fastest and the Rivendell Nifty Swifty 650B x 32 being the slowest);
tires with cotton casings rather than nylon appeared to perform better
when other factors were similar; tires with soft rubber compounds fared
better than those with harder compounds (the reverse of the findings
when steel rollers are used, apparently); subjective impression of
"fastness" do not correlate well with actual performance; and air
pressure made less difference that tire construction factors (size,
tread, rubber thickness, etc). The results also showed that tires roll
faster in hot weather than cold.
It was an interesting study but I am inclined to take it with a grain of
salt as yet. Some of the results appear confounding to me, which may be
due to there having been significant errors of measurement or that there
are real differences of which I don't understand the causes. Some
things were obvious- large tread features were generally problematic;
thinner rubber was faster than thicker rubber for example, wider tires
were faster (three widths of the Michelin Pro2 Race were tested- about a
2% difference). These things are known. What seemed to make an
interesting difference was the tires' ability to absorb surface
irregularities. The shocker was the Avocet Cross, with an inverted
tread like a car tire and very thick rubber, came in 8th!
The ordinal ranking of the top ten tires was (widths are actual): Deda
Tre Giro d'Italia (700 x 24); Clement del Mondo (700 x 28 tubular);
Michelin Pro2 Race (700 x 25); Continental Ultra Gator (700 x 23),
Mistuboshi Trimline (650B x 37), Panaracer Pasela (700 x 35); Clement
Criterium (700 x 21); Avocet Cross (700 x 35); Avocet Duro (700 x 28).
There is not a Web version of this article. You'll have to get a copy
from Vintage Bicycle Press, with which I have no connection other than
being a subscriber. I'm intrigued as the results support some things
that I have previously dismissed as myth and lore. Being a rather
skeptical sort, I have to question whether these results are accurate,
being that they seem to fly in the face of previous tests, or whether
the results point out some things that haven't been taken into account.
Tim McNamara
Ben C <spam...@spam.eggs> wrote:
> Well, the larger tyre requires less pressure for a given casing
> tension. I read that here recently and am still getting my head
> around it. It's also mentioned here by Jobst Brandt:
>
> http://www.sheldonbrown.com/brandt/rim-support.html
>
> "[...] unit casing tension is equivalent to inflation pressure times
> the radius of curvature divided by pi [...]".
>
> I was a bit surprised by this at first, but then if you think
> pressure is force per unit area, if you increase the area of the
> inside of the casing, you need more force for a given pressure. Not
> sure if this reasoning is bogus or not though.
If you have an inflation pressure of 100 psi, a tire with more inside
surface area will have a casing under greater tension because there are
more square inches. If I grok correctly.
jobst....@stanfordalumni.org
http://www.sheldonbrown.com/brandt/rim-support.html
Casing stress is arrived upon by cutting across the circular minor
diameter of the tire (the tire is a circular cross section having no
structural belt as radial tires do to change that) and take the two
halves as solid sections being pressed apart by inflation pressure.
That gives the lineal separation force which is the casing tension.
The above mentioned formula reduces to just that. For cord stress,
adjusting for 45 degree bias ply SQR(2) gets involved but this is
about casing tension which is the same regardless of fabric structure.
Jobst Brandt
carl...@comcast.net
Dear Jobst,
Here's an explanation of how tire tension resists thumbs (and larger
thing).
The tire is inflated to 100 psi.
The force of the air pressure pushes the tire outward, expanding the
tire.
Two forces oppose what would otherwise be an endless expansion (given
enough soap-bubble matterial).
First one atmosphere of air pressure pushes back.
When your pumping raises the pressure inside the tire to match the
outside atmosphere, the second force takes over--casing tension.
The doughnut trying to expand in all directions puts the tire casing
in tremendous tension.
In the case of our tire, the almost sideways tension is enough to hold
back the force of 100 pounds per square inch that's not canceled out
by 1 atmosphere on each side of the tire.
Once the forces match, the tire stops expanding and becomes motionless
and achieves a Buddah-like contentment--no net force, no net movement.
But you can disturb its repose by pushing on it with your thumb. The
instant that you push on it with 10 pounds of force (or 20 or 50), the
tire flees from your touch because the forces are no longer balanced.
In a moment, you'd sprain your thumb as it banged into the rim,
except--
Well, it's not the air pressure stopping your thumb. The air pressure
doesn't rise to fight your thumb. It remains 100 psi everywhere inside
the tire.
Yet obviously the forces have balanced, since your thumb stops moving
inward, even though you maintain your hopeful pressure against the
tire.
What's changed is the casing tension, which has risen until it plus
the 100 psi match the atmosphere and the pressure of your thumb.
Yes, the casing tension does depend on the air pressure, which is what
turns the doughnut shape into a taut 3-d trampoline.
But the motion that begins when you push on the tire stops only
because of the increasing local tension.
A dramatic example of this in reverse is when the bead of an inflated
tire slips off the rim for a few inches. Suddenly, there's a local
area of exposed inner tube that's too stretchy to supply the necessary
tension.
The inner tube swells wildly, as if a monster thumb had pushed it
outward from inside the tire.
(It doesn't matter whether you push in or out.)
If the inner tube were strong enough and elastic enough, you might end
up with a huge aneurysm. More likely, the inner tube splits and
explodes with a bang.
You can also work through the physics with a trampoline, a rope, a
pulley, and two weights, one large and one small.
Hang a 100 pound weight from the pulley and let it sit on the ground.
Its downward force represents the outward air pressure in the mistaken
view of what happens.
Dangle a 50 pound weight from the rope on the other end of the pulley.
Drat.
The 100 pound weight doesn't move. Neither does the 50 pound weight.
Doesn't look like pressing on a tire with your thumb.
Now let's see how a tire really works.
Heave the 100 pound weight onto the trampoline. It accelerates
downward until the tension on the trampoline skin matches the
force of gravity.
Then the weight stops and just sits there, with the trampoline bulging
but motionless--no net force, no movement.
Just like a tire expanding outward with 100 psi of air pressure and
enormous tension.
Now attach the rope to the weight, run it over the pulley, and hang
the 50 pound weight from it.
Instead of sitting still, the 100 pound weight rises. It still weighs
100 pounds, so that hasn't changed, any more than the air pressure
changes inside a tire when you push on it with your thumb.
All that's changed is the tension of the trampoline, which no longer
has to be tight enough to hold up 100 pounds--it rises to the point
where a 50 pound weight would cause it to sag.
Then the 100 pound weight stops moving because the forces have
balanced again.
All that changes is the tension.
Cheers,
Carl Fogel
Blair P. Houghton
>The point is that the casing tension must be overcome in all
>directions to make a dent in the doughnut-like surface.
Jobst just told you that it's not a dent.
In fact, flattening the tire at the contact patch relieves
the casing tension.
It doesn't overcome it.
>The dent does not raise the air pressure significantly.
Doesn't need to. Just needs to increase the
contact surface.
Same thing happens when you press with your thumb.
Try it sometime.
>But the dent does require dragging material inward in all directions,
>material that's being held in tension over the rest of the doughnut by
>100 psi.
>
>Consider the reverse case, an imaginary thumb pushing outward from
>inside the tire. What would stop it from pulling the tire wildly out
>of shape?
See, now you should be getting a clue. That side is causing
the casing tension; increasing the pressure from that side
increases the casing tension; pressing from the outside
relieves the casing tension. The casing is under tension at
all stretch other than zero stretch. Tubes stretch a lot,
tires stretch a little (spokes stretch a teeny bit).
>Stick a nail inside a tubeless tire, a nail large enough to make a
>bulge in the uninfalted tire.
>
>When you inflate the tire, the air pressure will pushe the tire
>outward in all directions.
>
>The increasing tension, however, will pull the tire onto the nail and
>puncture the tire.
Where do you get this stuff?
Have you ever done anything even remotely like this?
How do you know whether the nail would be more, less,
or equally likely to pierce the surface when the tube
is pulling backwards against the point when uninflated
or thinned-out by the expansion when inflated or has a
tension tending to pull its intermolecular bonds apart
when inflated? I think depending on the type of rubber,
the relative sizes of the tire and nail, and the sharpness
of the point, it could be any of those answers.
--Blair
Blair P. Houghton
>You seem determined to miss every point--including the nail. :)
I'm going to sit here and tell you flat-out that you just
projected.
Here, you're fond of balloons, do this:
Take a balloon, preferably the kind that are shaped
like a baloney rather than a basketball. Inflate it.
Now press the nose of the balloon against your hand.
Notice what happens to the tension of the rubber against
your palm. It decreases. The rubber relaxes there.
It contracts towards its uninflated size.
That isn't increasing the tension in the rubber,
it's relieving it.
It's still under inflated tension, because its uninflated
size is like a centimeter in diameter and you've inflated
it to a diameter of several inches.
But it's less.
You can make that end wrinkle and slew, which you can't
do with the fully-tensioned sides.
I'd say you shouldn't go in so much for the physics
discussions until you've had a few more birthday parties.
You need more of a grounding in real data to understand
a lot of this stuff.
--Blair
David L. Johnson
> Well, the larger tyre requires less pressure for a given casing tension.
> I read that here recently and am still getting my head around it. It's
> also mentioned here by Jobst Brandt:
>
> http://www.sheldonbrown.com/brandt/rim-support.html
>
> "[...] unit casing tension is equivalent to inflation pressure times the
> radius of curvature divided by pi [...]".
>
> I was a bit surprised by this at first, but then if you think pressure
> is force per unit area, if you increase the area of the inside of the
> casing, you need more force for a given pressure. Not sure if this
> reasoning is bogus or not though.
It is right, of course. And, the area increase is roughly proportional to
the tire width.
>
> So if the thumb test measures casing tension, then the larger tyre does
> require less pressure to get to the same thumb-hardness.
But a thumb-press is not measuring casing tension. It is moving the
casing perpendicular to the direction of the tension, so feels nothing
from that directly, only the internal pressure.
--
David L. Johnson
__o | Arguing with an engineer is like mud wrestling with a pig... You
_`\(,_ | soon find out the pig likes it!
(_)/ (_) |
David L. Johnson
> The ordinal ranking of the top ten tires was (widths are actual): Deda
> Tre Giro d'Italia (700 x 24); Clement del Mondo (700 x 28 tubular);
> Michelin Pro2 Race (700 x 25); Continental Ultra Gator (700 x 23),
> Mistuboshi Trimline (650B x 37), Panaracer Pasela (700 x 35); Clement
> Criterium (700 x 21); Avocet Cross (700 x 35); Avocet Duro (700 x 28).
I wonder when this was done. The comparisons with most of these tires
just no longer matter. Certainly neither the Clement del Mondo nor the
Criterium are available any longer, and if they were, they would cost a
bleeding fortune. They were beautiful, handmade silk tires, but their
kind has gone extinct.
Or, it may be that someone has resurrected the names, though not the tires.
--
David L. Johnson
__o | When you are up to your ass in alligators, it's hard to remember
_`\(,_ | that your initial objective was to drain the swamp. -- LBJ
(_)/ (_) |
jim beam
> <carl...@comcast.net> wrote:
>> You seem determined to miss every point--including the nail. :)
>
> I'm going to sit here and tell you flat-out that you just
> projected.
>
> Here, you're fond of balloons, do this:
>
> Take a balloon, preferably the kind that are shaped
> like a baloney rather than a basketball. Inflate it.
>
> Now press the nose of the balloon against your hand.
>
> Notice what happens to the tension of the rubber against
> your palm. It decreases. The rubber relaxes there.
> It contracts towards its uninflated size.
>
> That isn't increasing the tension in the rubber,
> it's relieving it.
isn't that because because it's in contact with your hand? as i
understand it, if your hand was entirely frictionless, tension would
remain as the skin of the balloon has to be in equilibrium.
jim beam
> On Mon, 02 Oct 2006 13:39:44 -0500, Ben C wrote:
>
>> Well, the larger tyre requires less pressure for a given casing tension.
>> I read that here recently and am still getting my head around it. It's
>> also mentioned here by Jobst Brandt:
>>
>> http://www.sheldonbrown.com/brandt/rim-support.html
>>
>> "[...] unit casing tension is equivalent to inflation pressure times the
>> radius of curvature divided by pi [...]".
>>
>> I was a bit surprised by this at first, but then if you think pressure
>> is force per unit area, if you increase the area of the inside of the
>> casing, you need more force for a given pressure. Not sure if this
>> reasoning is bogus or not though.
>
> It is right, of course. And, the area increase is roughly proportional to
> the tire width.
>
>> So if the thumb test measures casing tension, then the larger tyre does
>> require less pressure to get to the same thumb-hardness.
>
> But a thumb-press is not measuring casing tension. It is moving the
> casing perpendicular to the direction of the tension, so feels nothing
> from that directly, only the internal pressure.
>
Blair P. Houghton
>The ordinal ranking of the top ten tires was (widths are actual): Deda
>Tre Giro d'Italia (700 x 24); Clement del Mondo (700 x 28 tubular);
>Michelin Pro2 Race (700 x 25); Continental Ultra Gator (700 x 23),
>Mistuboshi Trimline (650B x 37), Panaracer Pasela (700 x 35); Clement
>Criterium (700 x 21); Avocet Cross (700 x 35); Avocet Duro (700 x 28).
I just went looking for the Dedas (of course I want them! for I
must go faster than any other 43-year-old solo huffer on Pecos
Road!) and someone has included a partial table of the rolling-
resistance coefficients here:
========
http://www.woodlandscycling.org/index.php?option=com_joomlaboard&func=view&catid=5&id=5335
The rule of thumb is that .0001 of Crr, equates to about 1 watt at 25-30 mph. So, for example, the Pro 2 Race takes 4 more watts than the Deda Tre Giro
Deda Tre Giro d'Italia......... 0.0038
Michelin Pro 2 Race...........0.0042
Vittoria Diamante Pro Rain....... 0.0044
Continental GP Force (rear specific).. 0.0057
Contintal Podium (Tubular).......0.0060
Specialized S-Works Mondo........0.0061
Continental GP 3000...........0.0067
Tufo Hi-Composite Carbon (Tubular)...0.0077
==========
>There is not a Web version of this article. You'll have to get a copy
>from Vintage Bicycle Press, with which I have no connection other than
>being a subscriber. I'm intrigued as the results support some things
>that I have previously dismissed as myth and lore. Being a rather
>skeptical sort, I have to question whether these results are accurate,
>being that they seem to fly in the face of previous tests, or whether
>the results point out some things that haven't been taken into account.
I think the 2:1 ratio of Crr's in that table and the
obvious mixing of sizes in the list indicates that
unless you buy exactly those items that were tested
you will not be able to tell by looking at the tire
specs whether it will have a high Crr or a low one.
I.e., get the ones that look cool, and go improve the
engine.
--Blair
carl...@comcast.net
<david....@lehigh.edu> wrote:
>On Mon, 02 Oct 2006 13:39:44 -0500, Ben C wrote:
>
>> Well, the larger tyre requires less pressure for a given casing tension.
>> I read that here recently and am still getting my head around it. It's
>> also mentioned here by Jobst Brandt:
>>
>> http://www.sheldonbrown.com/brandt/rim-support.html
>>
>> "[...] unit casing tension is equivalent to inflation pressure times the
>> radius of curvature divided by pi [...]".
>>
>> I was a bit surprised by this at first, but then if you think pressure
>> is force per unit area, if you increase the area of the inside of the
>> casing, you need more force for a given pressure. Not sure if this
>> reasoning is bogus or not though.
>
>It is right, of course. And, the area increase is roughly proportional to
>the tire width.
>
>>
>> So if the thumb test measures casing tension, then the larger tyre does
>> require less pressure to get to the same thumb-hardness.
>
>But a thumb-press is not measuring casing tension. It is moving the
>casing perpendicular to the direction of the tension, so feels nothing
>from that directly, only the internal pressure.
Dear Dave,
A thumb press measures the casing tension.
At 100 psi, the inflated tire is motionless.
But the air pressure on each side of the tire is unequal.
What stops the tire from moving is the not-truly-perpendicular tension
of the casing. The casing can't be truly flat. It must always be at at
slight angle if there's any pressure differential, just as a rope
cannot be stretched truly horizontal against the force of gravity.
When you apply a 30-pound pressure against the taut casing, your thumb
moves the casing inward.
(Or outward, if you're inside the tire--it doesn't matter.)
Then your thumb stops moving.
The tire pressure remains at 100 psi.
What stops your thumb from moving is the local increase in tire casing
tension on the doughnut-shaped trampoline kept taut by internal air
pressure.
To repeat, the tire pressure does not change.
It's the tension of the casing that changes until it's enough to
balance the forces again and stop your thumb from moving. Otherwise,
your thumb would move endlessly because of the net force.
Hang a 100-pound weight from a pulley and let it sit on the ground.
That's the mistaken raw air-pressure idea.
Put a 50-pound weight on the other end of the rope. That's your thumb
pressure in this mistaken theory.
The 100-pound weight just sits there. The 50-pound weight just hangs
there. There's no motion, unlike your thumb against the tire casing.
The 50-pound weight (or a 50-pound push from the ground) can't raise a
100-pound weight.
Here's how the tire actually works.
Put the 100-pound weight on a trampoline. The weight accelerates
toward the ground. The trampoline surface sags, going into tension, no
longer at that perpendicular angle.
When the forces balance, the weight stops moving, just as a tire stops
expanding when the forces of its inflation and its casing tension
match. No net force, no motion.
Hang the 50-pound weight from the pulley again and attach the rope to
the motionless 100-pound weight.
Now the 100-pound weight rises, just as the tire casing moves under
your thumb. Then it stops, just as the tire casing stops moving under
your thumb. The weight still weighs 100 pounds, just as the tire
pressure remained 100 psi.
The weight simply moves to where all the forces balance again. Since
the weight (and air pressure) never changed, the motion just tells you
where the tension balances the thumb/pulley force. The 100-pound
weight simply moves upward to where a 50-pound weight would push the
trampoline.
Indirectly, the thumb press does measure air pressure, since air
pressure will determine casing tension. And that's good enough for
bicycles. But the air pressure never changes. Only the tension and the
small angles at which it resists your thumb change when you press
against a tire, a trampoline, a drum head, or a stretched rubber band.
Cheers,
Carl Fogel
Michael Press
Ben C <spam...@spam.eggs> wrote:
> Does the "thumb test" (squeezing a tyre to see if it's hard enough)
> measure tyre pressure or casing tension?
I get the tire pressure by measuring the loaded roll-out
of the tire, then reading the pressure from a chart:
Roll out, 90 psi: 2099 mm
...
Roll out, 120 psi: 2107 mm
--
Michael Press
Michael Press
<timmcn-50E4C3....@news.iphouse.com>,
Tim McNamara <tim...@bitstream.net> wrote:
> The ordinal ranking of the top ten tires was (widths are actual): Deda
> Tre Giro d'Italia (700 x 24); Clement del Mondo (700 x 28 tubular);
> Michelin Pro2 Race (700 x 25); Continental Ultra Gator (700 x 23),
> Mistuboshi Trimline (650B x 37), Panaracer Pasela (700 x 35); Clement
> Criterium (700 x 21); Avocet Cross (700 x 35); Avocet Duro (700 x 28).
Conspicuously absent is the Avocet Fasgrip.
--
Michael Press
carl...@comcast.net
Dear Jim,
You're probably right about the friction. It's the equivalent of
twisting part of the bulging rubber surface and pulling it into a knot.
In any case, it doesn't really matter. Total tension must incease
The only way to reduce the tension in a local spot on an expanded
balloon surface is to contract it.
Obviously, its contraction must raise the tension in the elastic
surface all around it, so the the edges are pulling all around the hand
and trying to force it outward.
Consider a trampoline--if the section under your foot contracts, then
the rest of the trampoline must tighten. The total tension must be
greater, not the same or less.
Consider a taut string--if you pull a section together in the middle so
that it hangs slack, then tension must increase in the rest of the
string. The total tension must be greater, not the same or less.
Back to the balloon.
Since the same total force (all the outward air pressure) is now
contained by a smaller total skin (the whole original balloon surface
less amount of local contraction), the tension must rise all around the
edge of your hand--a slightly smaller bag is constraining the same
volume of gas. The rest of the ballon must bulge out as much as your
hands pushed it inward.
The only way that total tension can drop is if you press inward on the
entire surface (raise the exterior air pressure).
Cheers,
Carl Fogel
Tim McNamara
Michael Press <ja...@abc.net> wrote:
Avocet Duro is the FasGrip design. It was formerly maked 700 x 32 and
is now marked 700 x 28 to reflect reality. I have the old ones which
said "FasGrip" on the label. Now they say "Carbon 12." The testers
only used the one size of Avocet slick road tires.
Tim McNamara
"David L. Johnson" <david....@lehigh.edu> wrote:
> On Mon, 02 Oct 2006 17:11:40 -0500, Tim McNamara wrote:
>
> > The ordinal ranking of the top ten tires was (widths are actual):
> > Deda Tre Giro d'Italia (700 x 24); Clement del Mondo (700 x 28
> > tubular); Michelin Pro2 Race (700 x 25); Continental Ultra Gator
> > (700 x 23), Mistuboshi Trimline (650B x 37), Panaracer Pasela (700
> > x 35); Clement Criterium (700 x 21); Avocet Cross (700 x 35);
> > Avocet Duro (700 x 28).
>
> I wonder when this was done. The comparisons with most of these
> tires just no longer matter. Certainly neither the Clement del Mondo
> nor the Criterium are available any longer, and if they were, they
> would cost a bleeding fortune. They were beautiful, handmade silk
> tires, but their kind has gone extinct.
>
> Or, it may be that someone has resurrected the names, though not the
> tires.
The tests were done over three days in 2006. Some of the tires used as
reference were old stock, such as the del Mundos. I don't know about
the Clement Criterium- didn't somebody buy the Clement name and try to
reissue tires made in Thailand under that label a few years back? The
Mitsuboshi Trimline is also recently out of production.
As I said, I am skeptical about accepting the results at face value. I
really don't know how good the methodology was and how well confounds
were controlled for.
Tim McNamara
Those numbers match the findings of TOUR Magazine's measurements in
10/05, according to the table published in Bicycle Quarterly as a
comparison to their study.
> >There is not a Web version of this article. You'll have to get a
> >copy from Vintage Bicycle Press, with which I have no connection
> >other than being a subscriber. I'm intrigued as the results support
> >some things that I have previously dismissed as myth and lore.
> >Being a rather skeptical sort, I have to question whether these
> >results are accurate, being that they seem to fly in the face of
> >previous tests, or whether the results point out some things that
> >haven't been taken into account.
>
> I think the 2:1 ratio of Crr's in that table and the obvious mixing
> of sizes in the list indicates that unless you buy exactly those
> items that were tested you will not be able to tell by looking at the
> tire specs whether it will have a high Crr or a low one.
>
> I.e., get the ones that look cool, and go improve the engine.
The conclusion of the authors was that they would choose tires based not
only on rolling resistance but also other issues, such as comfort,
durability, etc. The authors are randonneurs and one author did find
that he rode significantly faster times on a 600K and 1000K brevet
(setting personal bests and course records) while his times on his
regular tires in other brevets were about typical. He felt that was an
indication that there was something to the results of their testing.
Bill Sornson
> Does the "thumb test" (squeezing a tyre to see if it's hard enough)
> measure tyre pressure or casing tension?
Yes.
Michael Press
<timmcn-C2C14F....@news.iphouse.com>,
Tim McNamara <tim...@bitstream.net> wrote:
> In article <jack-19B646.2...@newsclstr02.news.prodigy.com>,
> Michael Press <ja...@abc.net> wrote:
>
> > In article
> > <timmcn-50E4C3....@news.iphouse.com>,
> > Tim McNamara <tim...@bitstream.net> wrote:
> >
> > > The ordinal ranking of the top ten tires was (widths are actual): Deda
> > > Tre Giro d'Italia (700 x 24); Clement del Mondo (700 x 28 tubular);
> > > Michelin Pro2 Race (700 x 25); Continental Ultra Gator (700 x 23),
> > > Mistuboshi Trimline (650B x 37), Panaracer Pasela (700 x 35); Clement
> > > Criterium (700 x 21); Avocet Cross (700 x 35); Avocet Duro (700 x 28).
> >
> > Conspicuously absent is the Avocet Fasgrip.
>
> Avocet Duro is the FasGrip design. It was formerly maked 700 x 32 and
> is now marked 700 x 28 to reflect reality. I have the old ones which
> said "FasGrip" on the label. Now they say "Carbon 12." The testers
> only used the one size of Avocet slick road tires.
OK, but the 28 mm Duro is 66 thread per inch side wall
while the 25 mm Road is 127 tpi. Big difference.
--
Michael Press
Ben C
> On Mon, 02 Oct 2006 03:38:16 -0500, Ben C <spam...@spam.eggs> wrote:
>
>>Does the "thumb test" (squeezing a tyre to see if it's hard enough)
>>measure tyre pressure or casing tension?
[...]
Thank you to everyone for the explanations.
carlfogel> Thumb pressure measures casing tension.
carlfogel> [...]
carlfogel> Think of a trampoline.
carlfogel> Air pressure does not hold the trampoline up, any more than
carlfogel> it holds the trampoline down. What holds the tramoline taut
carlfogel> is the springs pulling it tight at the edges. Push down, and
carlfogel> the trampoline dents easily. Push a little further, and more
carlfogel> force is needed. The resistance is nicely progressive . . .
carlfogel> Just like a tire.
carlfogel> A tire is a doughnut-shaped trampoline. What stretches the
carlfogel> tire tight in all directions is the expanding spring of the
carlfogel> air pressure.
carlfogel> [...]
I am leaning towards this interpretation. Here is the explanation of
casing tension:
jobst.brandt> Casing stress is arrived upon by cutting across the
jobst.brandt> circular minor diameter of the tire (the tire is a
jobst.brandt> circular cross section having no structural belt as radial
jobst.brandt> tires do to change that) and take the two halves as solid
jobst.brandt> sections being pressed apart by inflation pressure. That
jobst.brandt> gives the lineal separation force which is the casing
jobst.brandt> tension.
Now, suppose I literally do cut across the minor diameter of my tyre, in
two places. I now have a piece of basically fabric hosepipe with no air
in it. It's a bit curved, but we'll pretend it's straight.
I now insert a piece of dowel into either end (to keep the hose
cylindrical). I attach clamps securely around the ends and attach the
clamps to strong springs such that they pull the hose out until it's in
the same tension as it was when it was on the bike and pumped up to
100psi. There's just atmospheric pressure in the hose (let's say I make
a small puncture somewhere, or the dowel is air-permeable, or it's a
clincher tyre anyway and therefore not a hose but more like a section of
guttering).
Now I squeeze it, against the force of the springs. We assume for the
sake of simplicity that the casing itself doesn't extend when stressed
(otherwise the length of the section I cut out has to be taken into
account).
Does it feel basically the same as it did when it was on the bike? If
so, it does seem that I'm measuring casing tension. The trampoline
springs are doing exactly the same job that the air was.
On the other hand I also believe this:
jobst.brandt> When a tire is pressed against a flat surface, the tire
jobst.brandt> flattens until the flat contact area times inflation
jobst.brandt> pressure equal the load.
If you think of the end of my thumb as a disk, with an area of half a
square inch, then to press that disk completely flat against either the
thin or the fat tyre will require the same force-- a force equivalent to
50lb.
But, at that point, the penetration of the disk along its normal into
the narrower tyre will be deeper, because of its narrower radius of
curvature.
If instead of asking, how hard do to I need to press to get the disk
flat against the tyre, I ask, how hard do I need to squeeze the tyre for
my thumb and forefinger to move 1/8 inch closer together, then for a fat
tyre and a thin tyre at the same pressure, I will have to squeeze the
fat tyre harder. This squares (at least qualitatively so far) with the
reasoning about trampoline springs.
David L. Johnson
>>But a thumb-press is not measuring casing tension. It is moving the
>>casing perpendicular to the direction of the tension, so feels nothing
>>from that directly, only the internal pressure.
>
> Dear Dave,
>
> A thumb press measures the casing tension.
No. By that logic, the road also "measures" casing tension -- since there
is no significant difference between the thumb pushing with a given force
on the tire, and the road pushing up. But that would mean that the size
of the contact patch would decrease with tire width (at a fixed pressure),
since the casing tension increases with tire width.
--
David L. Johnson
__o | If all economists were laid end to end, they would not reach a
_`\(,_ | conclusion. -- George Bernard Shaw
(_)/ (_) |
jobst....@stanfordalumni.org
>> The ordinal ranking of the top ten tires was (widths are actual):
>> Deda Tre Giro d'Italia (700 x 24); Clement del Mondo (700 x 28
>> tubular); Michelin Pro2 Race (700 x 25); Continental Ultra Gator
>> (700 x 23), Mistuboshi Trimline (650B x 37), Panaracer Pasela (700
>> x 35); Clement Criterium (700 x 21); Avocet Cross (700 x 35);
>> Avocet Duro (700 x 28).
> I wonder when this was done. The comparisons with most of these
> tires just no longer matter. Certainly neither the Clement del
> Mondo nor the Criterium are available any longer, and if they were,
> they would cost a bleeding fortune. They were beautiful, handmade
> silk tires, but their kind has gone extinct.
> Or, it may be that someone has resurrected the names, though not the
> tires.
It's not the tires themselves that are interesting but the relation of
inflation pressure to RR for tires with essentially smooth tread.
http://www.sheldonbrown.com/brandt/rolling-resistance-tubular.html
The RR curves for smooth tires are nearly identical but with a
multiplier. One could take one of the curves and multiply it to
generate the others. Exceptions are those that have significant tread
profile that causes tread squirm losses. Conspicuous are the two
Specialized tires with raised center ridge.
Tubular tires, that have less RR than all the clinchers, are out of
place by the losses caused by elastomeric rim glue while their slopes
shows that they have inherently low RR due to their thin casings and
tread, and the thin latex inner tubes. The effect of inflation
pressure (slope) for them diminishes with less lossy material.
All the smooth tires have a characteristic that at infinite pressure
(to the left) approaches zero RR. As is apparent, the tubulars have a
constant offset from that characteristic. When I first saw these
curves, I thought they would be apparent to observers and resolve age
old questions. Instead they gave rise to endless speculation and
interpretation.
Track glue was used for a purpose that remained unknown after WWII
when those who understood the problem had retired and left us with two
kinds of glue, much like john Starley left us with tied and soldered
spokes. No one seemed to understand the purpose.
Track tires users in events like the 1000m and 4000m pursuit know that
these events are won on 1/100 of seconds and that RR in that realm is
important. Track tires were made with bare cloth base tapes to be
coated with track glue and mounted on rims with track glue. Track
glue is partially dried shellac that hardens as non elastic cement.
Rubberized tubular base tapes do not lend themselves to use with track
glue.
Jobst Brandt
carl...@comcast.net
Dear Ben,
Basically, an inflated elastic skin automatically occupies the lowest
possible tension shape. If you force a deformation anywhere locally,
the tension rises.
It might be easier to work through the geometry than devise spring
devices inside short sections. Air and rubber are sufficient.
The simplest tension figure that we can describe is a straight line,
say a rubber string stretched tight between two fixed points.
If we push the rubber string in any direction, the force increases the
tension.
Push the tight rubber string sideways at any point and the distance
obviously increases because a straight line is the shortest distance
between two points.
Instead of pushing the rubber string sideways, grab a point on the
rubber string and pull it toward either fixed point. Tension drops on
one side, but must rise to more than the original tension on the other
side--your pull has introduced a third fixed point and stretched a
section of the original rubber string further and tauter.
Let's get rid of those pesky fixed points.
An endless rubber string is just a rubber band. Imagine a rubber band
laid flat to form an air-tight seal between two plates.
Pump air into the sealed space and the rubber band will automatically
expand to form a circle with even pressure all around.
Push inward on either side of the rubber band circle, and the circle
will start to flatten.
The air pushed out of that section doesn't vanish. It's pushed into
the unflattened areas--which must bulge outward and cause the tension
to rise in the rubber band.
If there's just the right friction on the two points where the rubber
band is flattened, you could get a local tension lowering in the
chord, but that would just raise the tension even higher in the
unflattened sections of the rubber band.
A thumb press requires flattening two spots, one where our thumb is,
and one (or more) to oppose the thumb press. Locally, we can try to
lower tension by flattening a curve into a chord, but that relies on
friction to keep the flattened area from adjusting its tension as the
rest of the rubber band expands.
We could simplify to a single chord instead of two by grabbing two
spots on the rubber band and pulling them together (or apart). Either
way, the tension must rise in the rubber band.
If we pull the two points apart, the tension is obvious between our
fingers.
If we push the two points together, the rest of the rubber band has to
bulge as the more air is squeezed into what's left of the circle.
So much for one-dimensional rubber bands. Let's do a two-dimensional
surface, such as a trampoline.
Let's skip the ordinary flat trampoline and move straight to an
endless trampline--the two-dimensional surface of a sphere.
If we inflate a round balloon, every part of its surface automatcally
goes to the lowest possible even tension.
Mark a circle on the surface and label it "trampoline"--the tension
from inflation serves as the tension springs for an ordinary flat
trampoline.
We can push two (or more) spots inward on our balloon. The force will
cause of the balloon to bulge outward and its skin elsewhere to
increase in tension. If friction where we push our hands together
reduces the tension locally (a chord is shorter than a curve), then
there's even less balloon skin left elsewhere to constrain the same
amount of air and tension and pressure must rise even further.
Transform a sphere into a toroid and we have a tire that's still an
endless trampoline. The only way to reduce total tension is to press
inward everywhere by raising atmospheric pressure. Otherwise, the tire
must bulge out somewhere else whenever we push it inward (or outward)
locally with a thumb.
The most intuitive way to see what happens may be to imagine an inner
tube or tire force-filled with water. The skin of the toroid is
obviously tight. Push inward on it locally anywhere, and you know that
the water is bulging it outward elsewhere and raising the tension.
Less intuitively, pulling the water-filled tube outward anywhere must
also raise the tension. You have to pull at two points and apply force
to distort the tube. Pulling on the tube must increase its tension. As
soon as you let go and release the tension, the tube snaps back into
its original form, which is a slightly lower pressure.
Just like an air-filled tire--push it inward or outward locally, and
you raise the tension and pressure. As soon as you stop pushing, the
deforming tension is lost and the tire snaps back into its original
shape.
Cheers,
Carl Fogel
Tim McNamara
jobst....@stanfordalumni.org wrote:
> Tim McNamara writes:
>
> >>>> Does the "thumb test" (squeezing a tyre to see if it's hard
> >>>> enough) measure tyre pressure or casing tension?
>
> >>> Directly the latter and indirectly the former, I think.
>
> >> This is what I was starting to think too. What this means of
> >> course is that if you have a fat tyre and a thin tyre that feel
> >> the same, the fat tyre will actually be at a lower pressure (as it
> >> typically should be), and have higher rolling resistance.
>
> > After I typed this, I thought that the thumb test would also be
> > influenced by the stiffness of the tire wall: the stiffness of the
> > fabric casing, the thickness and durometer of the rubber, etc.
>
> ... not more than the un-inflated tire.
That makes sense, although what I was thinking about was flexing
uninflated tires which have a noticeable resistance.
> Obviously some tires have a stiff tread that doesn't deflect even
> when flat, an example is a motorcycle tire. I assume those who use
> their thumbs are also smart enough not to do that to a knobby tread
> or a thick road tread. The bare side wall on older tires could give
> an accurate feel, but even that has been taken from us by the "I get
> too many flat tires" folks. We may revert to the non-pneumatic tire
> era soon.
Judging by the number of cyclists I see every day with no pump or repair
tools, that might not be a bad idea for some.
Tim McNamara
jobst....@stanfordalumni.org wrote:
> David L. Johnson writes:
>
> >> The ordinal ranking of the top ten tires was (widths are actual):
> >> Deda Tre Giro d'Italia (700 x 24); Clement del Mondo (700 x 28
> >> tubular); Michelin Pro2 Race (700 x 25); Continental Ultra Gator
> >> (700 x 23), Mistuboshi Trimline (650B x 37), Panaracer Pasela (700
> >> x 35); Clement Criterium (700 x 21); Avocet Cross (700 x 35);
> >> Avocet Duro (700 x 28).
>
> > I wonder when this was done. The comparisons with most of these
> > tires just no longer matter. Certainly neither the Clement del
> > Mondo nor the Criterium are available any longer, and if they were,
> > they would cost a bleeding fortune. They were beautiful, handmade
> > silk tires, but their kind has gone extinct.
>
> > Or, it may be that someone has resurrected the names, though not
> > the tires.
>
> It's not the tires themselves that are interesting but the relation
> of inflation pressure to RR for tires with essentially smooth tread.
>
> http://www.sheldonbrown.com/brandt/rolling-resistance-tubular.html
>
> The RR curves for smooth tires are nearly identical but with a
> multiplier. One could take one of the curves and multiply it to
> generate the others. Exceptions are those that have significant
> tread profile that causes tread squirm losses. Conspicuous are the
> two Specialized tires with raised center ridge.
Interestingly, the BQ study found that tire pressure had less impact on
rolling speed than other factors (assuming the results are accurate).
> Tubular tires, that have less RR than all the clinchers, are out of
> place by the losses caused by elastomeric rim glue while their slopes
> shows that they have inherently low RR due to their thin casings and
> tread, and the thin latex inner tubes. The effect of inflation
> pressure (slope) for them diminishes with less lossy material.
The BQ test tried some of the tires with butyl and latex tubes. The
tires rolled slower with latex tubes. They did note that these latex
tubes were not as thin as those in tubular tires.
As I've stated though, the results of the test are different enough that
I am somewhat skeptical, and wonder about the contribution of confounds.
Someone with a better understanding of this stuff than I would have to
read the test and its methodology.
Tim McNamara
Michael Press <ja...@abc.net> wrote:
> In article <timmcn-C2C14F....@news.iphouse.com>,
> Tim McNamara <tim...@bitstream.net> wrote:
>
> > In article
> > <jack-19B646.2...@newsclstr02.news.prodigy.com>,
> > Michael Press <ja...@abc.net> wrote:
> >
> > > In article <timmcn-50E4C3....@news.iphouse.com>,
> > > Tim McNamara <tim...@bitstream.net> wrote:
> > >
> > > > The ordinal ranking of the top ten tires was (widths are
> > > > actual): Deda Tre Giro d'Italia (700 x 24); Clement del Mondo
> > > > (700 x 28 tubular); Michelin Pro2 Race (700 x 25); Continental
> > > > Ultra Gator (700 x 23), Mistuboshi Trimline (650B x 37),
> > > > Panaracer Pasela (700 x 35); Clement Criterium (700 x 21);
> > > > Avocet Cross (700 x 35); Avocet Duro (700 x 28).
> > >
> > > Conspicuously absent is the Avocet Fasgrip.
> >
> > Avocet Duro is the FasGrip design. It was formerly maked 700 x 32
> > and is now marked 700 x 28 to reflect reality. I have the old ones
> > which said "FasGrip" on the label. Now they say "Carbon 12." The
> > testers only used the one size of Avocet slick road tires.
>
> OK, but the 28 mm Duro is 66 thread per inch side wall while the 25
> mm Road is 127 tpi. Big difference.
It is, and I thought about that too. Interestingly I found that some of
my own riding supported some of the things they wrote. For example, I
tried the Rivendell Rolly Polys and found them slower than the
Continental Ultra 2000 (both 700 x 28) that I usually use. The Avocet
700 x 25 is simply unavailable in shops locally. Only one shop bothered
to stock Avocet road tires and they are long gone. They couldn't keep
the 700 x 25 (marked 7000 x 28 in the old days) in stock- I was never
able to buy a set.
Someday I'll mail order a couple of them. None of the local bike shops
want to deal with Avocet, they are too much hassle I guess.
Tim McNamara
jobst....@stanfordalumni.org wrote:
From the responses of people more knowledgeable in the topic than
myself, it's clear that I don't grok this correctly. So, let me ask:
1. In terms of riding a bike, what is the significance of casing
tension? Does it affect how the tire feels, rolling resistance,
traction, etc?
2. When I am on my bike, is it casing tension or inflation pressure
that holds my rims off the ground?
carl...@comcast.net
Dear Tim,
Those coasting tests are well-meant, but unlikely to be accurate.
Real-world coasting tests have far too many variables to detect subtle
differences.
One run may roll over a slightly rougher section of the apparently
uniform road surface.
Or it may miss the undetectable half-inch high, ten-foot-long hump in
the apparently flat road.
Or it may shorten the course with slightly tighter turns.
Meanwhile, one rider cannot possibly duplicate his position and wind
drag second by second for two runs, much less two different riders.
And consider the effect of an imperceptible 0.5 mph headwind changing
to an imperceptible 0.5 mph tailwind on a "perfectly still day"
between runs for one rider:
http://www.kreuzotter.de/english/espeed.htm
Hands-on tops, no rpm, no watts, -2 for a 2% downhill slope:
wind bike
speed speed
mph mph
headwind 0.5 15.1
no wind 0.0 15.6
tailwind -0.5 16.1
Hell, even the temperature, barometric pressure, and humidity changes
between repeated runs can change the speed noticeably.
Sadly, calculators or controlled spin-down tests are usually needed to
detect small differences reliably. We can probably detect gross
differences (MTB knobbies versus narrow racing tires) by just coasting
next to friends or timing repeated runs, but smaller differences are
too easily swamped by the variables.
My maximum speed coasting down the steepest part of my daily ride has
varied from 36.8 to 43.0 mph so far this week.
Repeated runs minutes apart with great care on the approach and tuck
might narrow that figure.
But given normal wind variation and the slightly different line and
tuck each time for each run, I expect that there would still be a
surprising range of speeds for a dozen runs down my hill.
Spread that out over a long morning, with time spent changing to
different tires, tubes, and inflations, and small differences are just
not likely to show up.
Cheers,
Carl Fogel
JL
news:slrnei1jti....@bowser.marioworld...
> Does the "thumb test" (squeezing a tyre to see if it's hard enough)
> measure tyre pressure or casing tension?
The answer has already been given by several posters here but seems to have
become lost in a sea of confusion. I'll offer the following in an attempt
at clarification.
The casing tension per unit length of casing can be calculated by
multiplying the internal air pressure times the cylindrical diameter divided
by 2x the wall thickness. This is commonly known as tangential stress or
hoop stress and can be verified by any modern text on thin-walled pressure
vessels.
This means that the casing tension is 'proportional' to the cylindrical
diameter. For example if two different size tires are inflated to 100psi,
say the cylindrical diameter of one tire is 12 inches and the diameter of
another tire is 1.20 inches, (assuming equal casing thickness) the tension
in the larger diameter tire will be 10x greater than the small tire.
Yet both tires will resist thumb indentation roughly the same because the
internal air pressure is equal in both cases. Therefore it would be more
accurate to say that the thumb test measures internal air pressure directly
and measures casing tension indirectly.
JL
Blair P. Houghton
Surely.
But how many brands of tires label the packaging with
the Crr number?
It'd be neato-keen if we could have those at hand when
Froogling.
All we have are the ones in this one list comprising
a fraction of a percent of the market.
Anyone up for picketing at ISO headquarters Saturday
morning to demand a couple of extra digits on the molded-in
markings? 23-622-0038, perhaps?
--Blair
"Go right ahead. I'll be riding."
Ben C
[snip]
> The most intuitive way to see what happens may be to imagine an inner
> tube or tire force-filled with water. The skin of the toroid is
> obviously tight. Push inward on it locally anywhere, and you know that
> the water is bulging it outward elsewhere and raising the tension.
Yes, although here we've gone to a stretchy skin containing an
incompressible fluid. An inner tube is a stretchy skin, but an outer
tyre is better imagined as a non-stretchy fabric casing containing a
compressible fluid (air).
In the water-balloon, it's the skin that provides the springiness; in
the tyre it's the air itself.
On the subject of the thumb test, I did some "back of the envelope"
calculations.
If an object full of compressed air has a radius of curvature R, and I
press part of it flat to form a circular contact patch, how does force
relate to penetration distance? By penetration distance, I mean the
distance in the direction of the pressing force between the point at the
centre of the contact patch and where that point was before I pressed it
in.
The force required is given by P*pi*r**2 where P is the air pressure
inside the object, and r is the radius of the contact patch.
I use ** to mean "raised to the power of".
r**2, the radius of the contact patch squared I make 2Rd - d**2 where d
is the penetration distance.
So F = P*pi*(2Rd - d**2)
It is sufficient to see here that F is proportional to R.
Now I have read it elsewhere in r.b.t that unit casing tension is given
by PR/pi.
So R cancels out-- whatever the radius of curvature, the force required
for a given penetration distance is proportional to unit casing tension.
If you want the same penetration for a given force at lower R, you will
require a higher pressure to keep the casing tension the same.
This "model" is an approximation, but I think the thumb test is close to
"squeeze the tyre with a given test force and see how far it depresses".
In that case, the amount it depresses depends on casing tension, and you
should expect the same depression for a fatter tyre at a lower pressure
as you'd get for a narrower tyre at a higher pressure.
Blair P. Houghton
>Blair P. Houghton wrote:
>> <carl...@comcast.net> wrote:
>>> You seem determined to miss every point--including the nail. :)
>>
>> Notice what happens to the tension of the rubber against
>> your palm. It decreases. The rubber relaxes there.
>> It contracts towards its uninflated size.
>>
>> That isn't increasing the tension in the rubber,
>> it's relieving it.
>
>isn't that because because it's in contact with your hand? as i
>understand it, if your hand was entirely frictionless, tension would
>remain as the skin of the balloon has to be in equilibrium.
No it doesn't*.
If it were a spherical** balloon and the only forces on
it were the air pressures on either side, it might have
a constant relation between tension and pressure over
its whole surface. But once it contacts something else
outside, that no longer applies, and a similar slacking
on the flattened side should occur.
Try it on a slippery, flat, incompressible surface, if your
hand is too compliant and sticky.
--Blair
"I'll wait while you butter
your balloons..."
* - the opposite, in fact; if the skin of your hand sticks
to the balloon, the slackening in the balloon tries to
pull your skin towards the middle, so your skin is pulling
back outward, so as to stretch the balloon, not shrink it.
With slippery hands, the balloon can slacken more.
** - The fact that it's a baloney-shaped balloon was used
only to evoke your sense memory. I assume all of us have
squeezed a few in our younger days. The differing curvatures
in the balloon imply that the pressure/tension ratios vary
around the surface.
carl...@comcast.net
wrote:
Dear JL,
The casing is motionless, forces balanced, internal air pressure
against outer air pressure and casing tension.
To push the casing in, you push against the tensioned casing.
Even a tiny pressure causes a tiny movement and an increase in the
casing tension of flexible, indented, deformed body.
But the internal pressure does not change.
Consider a flimsy sidewall 4.00 x 18 motorcycle trials tire, rolling
down the road at 20 mph in a wheelstand contest for ten minutes. (I've
trailed along behind on two wheels, hoping that the showoffs would
break their necks.)
The rider and motorcycle total 400 pounds, all loaded on the rear
axle. Even without wheelie contests, the front end is popped up
routinely in trials.
The tire pressure is only 4 psi, measured with low-pressure gauges
that read in half-pound increments.
The tire's contact patch starts out at 4 inches wide across these 5
rows of closely spaced knobs and never widens appreciably:
http://12.158.74.10/product_images/368_0620L.jpg
Mounted on an 18-inch diameter rim, the tire stands just under 26
inches high. The sidewalls are flimsy--a flat 1-ply trials tire
squashes flat under the parked machine with no rider.
According to the raw air pressure theory, we have 400 pounds to
distribute on a 4-inch wide contact patch at 4 psi.
That would mean a 25-inch long contact patch on a tire whose diameter
is just under 26 inches.
Pushing on a tense casing measures the casing tension that holds the
rim up. The thumb moves inward because the casing tension changes
locally. The air pressure stays the same.
Or so it seems to me. I do appreciate your nice exposition of casing
tension in an undeformed section.
Cheers,
Carl Fogel
Tim McNamara
"JL" <jlulmail...@comcast.net> wrote:
OK. Here's where I may be confunded. If I take a piece of fabric and
stretch it, the tighter I stretch it for harder it feel to my thumb.
There's no inflation pressure to feel, since there's ambient pressure on
both sides.
Tim McNamara
carl...@comcast.net wrote:
> Those coasting tests are well-meant, but unlikely to be accurate.
>
> Real-world coasting tests have far too many variables to detect
> subtle differences.
>
> One run may roll over a slightly rougher section of the apparently
> uniform road surface.
>
> Or it may miss the undetectable half-inch high, ten-foot-long hump in
> the apparently flat road.
>
> Or it may shorten the course with slightly tighter turns.
At risk of sounding like I'm defending the report, I'll point out what
they wrote about their methodology.
The roll down test was done on a soapbox derby track near WoodlandPark
in Seattle. The course was 245 meters long starting at a 6% grade,
decreasing to a 4.5% grade and then reducing to a 0.5% grade over the
final 184 meters. The initial 16 meters were very smooth, the remainder
was uniform, moderately rough asphalt. There were no holes, ridged or
overlays. The course was swept prior to the testing. The rider coasted
from a standing start with no pedaling, holding the same position and
wearing the same clothing for all of the runs. The bike was timed over
184 meters. Speeds were between 10.6 mph and 16.9 mph. There was a set
of reference tires used to calibrate the test to try to compensate for
changing meteorological conditions. Two independent timers were used
and the times averaged. Multiple runs were made with each tire and the
measurements averaged.
All that being said, I remain concerned about the possibility of
unidentified confounds affecting the outcome. The testers made a
laudable effort at minimizing these. I see two possible major
compounds: using stopwatches to time the rider rather than a mechanical
trigger system, and the short timed coasting distance (184 m) which was
covered in 25.3 seconds (Deda Tre) to 30.6 seconds (Nifty Swifty). The
short distance may magnify the apparent differences caused by confounds
such as human error in timing and other conditions mentioned by Carl.
Michael Press
<1159852714.9...@k70g2000cwa.googlegroups.com>,
carl...@comcast.net wrote:
> The only way to reduce the tension in a local spot on an expanded
> balloon surface is to contract it.
When you push in on an inflated balloon you decrease the
area of the local patch; hence you decrease the local
tension.
Another way to see this is that the local strain energy in
an elastic structure is proportional to the local
curvature. Decreasing the curvature decreases the strain
energy.
--
Michael Press
Michael Press
<timmcn-439DAB....@news.iphouse.com>,
Tim McNamara <tim...@bitstream.net> wrote:
Both views are useful.
Analyzing casing stress and strain is fun, and informative.
It is essential for designing a good tire.
We can ignore the casing and think about the wheel
being supported by unequal forces from air pressure
acting on unequal areas, if we want.
Here is a crossection of a wheel seen edge on with the
axle in the plane of the diagram and horizontal.
The top cross section of the tire is not deformed.
The bottom cross section is deformed to a wider shape.
The air pressure pushes up against the rim at the lower
cross-section, and pushes down against the rim in the
upper cross-section; but the area is greater at the lower
than the upper cross-section. The unbalanced force
is equal to the load at the hub.
#############
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########## ^^^^^^^^^^^^^^^ ##########
##### ^^^^ ||||||||||||||| ^^^^^######
##### |||| ||||| ####
#### ###
### ###
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#### ####
#### ####
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##### ||| |||| ######
#######vvv|||||||||||||||||vvvv#######
##########vvvvvvvvvvvvvvvvv##########
#################################
##############################
###########################
#####################
#############
Axle goes here
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
###################
############################
###################################
##########################################
####################################################
#########################################################
############################################################
######################^^^^^^^^^^^^^^^^^^^^^^#####################
#################^^^^^||||||||||||||||||||||^^^^##################
#############^^^^||||| ||||^^^^^#############
####### ^^^^ |||||^^^^ #######
#### |||| |||| ####
# #
# #
#### ####
####### |||||||| ||||||||| #######
##########vvvvvvvv||||| ||||vvvvvvvvv#########
#################vvvvv||||||||||||||||||||||vvvv##################
######################vvvvvvvvvvvvvvvvvvvvvv#####################
###############################################################
#########################################################
####################################################
################################################
###################################
############################
Road surface. ###################
==============================================================================
/////////////////////////////////////////////////////////////////////////////
--
Michael Press
Phil Holman
"Tim McNamara" <tim...@bitstream.net> wrote in message
news:timmcn-439DAB....@news.iphouse.com...
Can't resist it.....the rim is standing on the casing cords :-)
Phil H
jobst....@stanfordalumni.org
> Basically, an inflated elastic skin automatically occupies the
> lowest possible tension shape. If you force a deformation anywhere
> locally, the tension rises.
Hold it. That is incorrect. Casing tension is inversely related to
curvature so that when you load a bicycle wheel, casing tension is
reduced in the area where the cross section bellies out, the free
standing radius of curvature being smaller in that area.
It is for this reason that a small cross section tire has less casing
stress than a larger one at the same pressure. aka small tires can be
inflated to higher pressures than large ones of similar quality. I
think that is or should be common knowledge.
> It might be easier to work through the geometry than devise spring
> devices inside short sections. Air and rubber are sufficient.
Tires are not rubber although their tread and "air chamber" is rubber.
I use that term because there are tubeless bicycle tires. The casings
are made of essentially inelastic fiber so any model constructed of
rubber is incorrect for demonstrating stress in tires.
> The simplest tension figure that we can describe is a straight line,
> say a rubber string stretched tight between two fixed points.
As I said, an elastic model has little if anything to do with a
pneumatic bicycle tire.
> If we push the rubber string in any direction, the force increases
> the tension.
> Push the tight rubber string sideways at any point and the distance
> obviously increases because a straight line is the shortest distance
> between two points.
> Instead of pushing the rubber string sideways, grab a point on the
> rubber string and pull it toward either fixed point. Tension drops
> on one side, but must rise to more than the original tension on the
> other side--your pull has introduced a third fixed point and
> stretched a section of the original rubber string further and
> tauter.
> Let's get rid of those pesky fixed points.
And lets get a real tire, not a rubber band.
> An endless rubber string is just a rubber band. Imagine a rubber
> band laid flat to form an air-tight seal between two plates.
Imagine a tire casing made of Kevlar cord.
> Pump air into the sealed space and the rubber band will
> automatically expand to form a circle with even pressure all around.
>... and a raft of other inappropriate examples.
Jobst Brandt
jobst....@stanfordalumni.org
http://www.sheldonbrown.com/brandt/rim-support.html
I think this puts the cart before the horse. Casing tension depends
on two parameters, inflation pressure and tire size. I think you know
the answer to that already. The harder the tire is inflated the
harsher the ride for any cross section and the larger the cross
section the softer the ride for any inflation pressure. Let the
manufacturer decide what casing stress is acceptable. It's not the
user's problem.
> 2. When I am on my bike, is it casing tension or inflation pressure
> that holds my rims off the ground?
Inflation pressure does that but does so by casing tension. As I
said, it's only a problem for the engineer who wonders how loads get
from the rim to the ground. The item in the FAQ explains that.
http://www.sheldonbrown.com/brandt/rim-support.html
Jobst Brandt
jobst....@stanfordalumni.org
>> From the responses of people more knowledgeable in the topic than
>> myself, it's clear that I don't grok this correctly. So, let me
>> ask:
>> 1. In terms of riding a bike, what is the significance of casing
>> tension? Does it affect how the tire feels, rolling resistance,
>> traction, etc?
>> 2. When I am on my bike, is it casing tension or inflation
>> pressure that holds my rims off the ground?
> Both views are useful.
> Analyzing casing stress and strain is fun, and informative. It is
> essential for designing a good tire.
> We can ignore the casing and think about the wheel being supported
> by unequal forces from air pressure acting on unequal areas, if we
> want.
Whaich unequal areas? The rim is the same width around it's
circumference and is exposed to uniform air pressure around that area.
> Here is a crossection of a wheel seen edge on with the axle in the
> plane of the diagram and horizontal. The top cross section of the
> tire is not deformed. The bottom cross section is deformed to a
> wider shape. The air pressure pushes up against the rim at the
> lower cross-section, and pushes down against the rim in the upper
> cross-section; but the area is greater at the lower than the upper
> cross-section. The unbalanced force is equal to the load at the hub.
I think you have that incorrectly. The rim is the same width top and
bottom. I suggest you read the FAQ item and start from there:
http://www.sheldonbrown.com/brandt/rim-support.html
> ////////////////////////////////////////////////////////////////////////////
Jobst Brandt
carl...@comcast.net
Dear Tim,
I agree that the testers made good-faith efforts. I hope that I'd be
just as willing as you are to point them out.
But I'm still awfully skeptical. I can't, for example, see how anyone
can adjust for random 0.5 mph wind variations during numerous runs over
an open 245 meter course.
I expected the speeds "between 10.6 and 16.9 mph" to correspond to the
"30.6 seconds (Nifty Swifty)" and the "25.3 seconds (Deda Tre)"--but
they don't seem to work out when I chuck them into a spreadsheet:
184 meters = 603.67 feet
603.67 feet / 10.6 mph = 38.8 seconds (10.6 mph -- minimum?)
603.67 feet / 13.45 mph = 30.6 seconds (30 .6 seconds Nifty Swifty)
603.67 feet / 16.27 mph = 25.3 seconds (25.3 seconds Deda Tre)
603.67 feet / 16.9 mph = 24.4 seconds (16.9 mph -- maximum?)
Were those two bikes not the fastest and slowest? Or am I just
misunderstanding you?
Cheers,
Carl Fogel
JL
<carl...@comcast.net> wrote in message
news:p0n5i21uukruas8ef...@4ax.com...
Carl, I'm not certain how to interpret your wheelie example but regarding
whether
"Pushing on a tense casing measures the casing tension that holds the rim
up",
let's look at your previous example of a rubber string attached between two
fixed points A and B. When pushed at midpoint C by a finger the rubber
string stretches under the load, increasing in tension as it forms triangle
A_C_B. Actually the string can also be inelastic, starting out slack, and
will still increase in tension and form triangle A_B_C when a load is
applied at the midpoint.
The string (elastic or inelastic) held in tension by the finger can be
considered to be in a state of equilibrium. Let's say the pressure applied
by the push of the finger is analogous to the air pressure pushing against
the inside of a tire giving it its curved taut shape and casing tension.
If someone now pushes with their thumb on the string at point C from the
other side, i.e., opposing the finger pressure, so as to flatten the
triangle or even reverse the triangle, what is this person's thumb pushing
against, the tension in the string or the finger pressing on the string
(e.g. tire pressure)?
JL
carl...@comcast.net
wrote:
Dear JL,
Let me see if I'm following you.
First, the string.
A .________________.B
Push mid-point C downward, analogous to air pressure pushing outward
A. .B
' . '
C
A-C-B stretches as it goes into tension.
(Sorry, but this seems like the spot for a quick digression. Like a
spoke, a tire casing stretches very little, but under the tremendous
total pressure it does stretch. Other posters have mentioned an 8mm
increase in about 2100 mm for a rise from 90 to 120 psi in rollout. A
quick measurement of a flat-rolled folding nominal 700 x26 tire shows
it to be 75mm wide, so its reasonable to expect the tire ACB to
stretch 8/2100 * 75 mm between 90 and 120 psi, or 0.285 mm. At 120
psi, my presumably average width and power thumb tip and fingernail
dent that tire about 1mm.)
Anyway, your question is what does the thumb press against at point C?
The air pressure is pushing outward at 120 psi everywhere.
It's held motionless by the stretched casing's tension.
When you push, the tire moves inward, then stops as you maintain a
steady pressure.
The air pressure is still pushing outward at 120 psi.
But the casing tension must have changed. We just put a tiny dent in
its curved shape.
(At least the way that I do it, I can see an indent, not a mere
flattening. I gave up trying to take a picture.)
The air pressure does push outward, which I think is where you're
headed.
But what changes, if you see the distinction, is the casing tension.
The original poster was asking whether the thumb press measures casing
tension or air pressure.
I see a dent in the tire, like the dent in a trampoline, not a mere
flattening, so I think casing tension.
To argue against my position and for air pressure, I'd try to argue
that the initial thumb contact is so small that it's greater than 120
psi--only 15 to 25 pounds, but concentrated in a tenth of a square
square inch, so it greater than 120 psi and moves the tire inward,
slowing down and finally stopping as the roughly cone-like
depression's growing area and steady force arrive at 120 psi.
But that tire is stretched awfully tight, like a trampoline. Pushing a
dent into it takes force to overcome that tension, again like a
trampoline.
Possibly I'm just wrong, or both forces combine. But my efforts to
measure thumb pressure and area don't encourage me to think that I'm
getting 120 psi--squeeze a bathroom scale as if you're testing a tire
to see what kind of force you get, and then take a close look at the
area of thumb that you apply. Maybe I'm just pressing wrong.
Anyway, here's an example to ponder. Imagine a tiny round manhole
cover sitting inside the tire, pasted over the opening to the valve
stem by 120 psi, a zillionth of an inch wider than the opening.
Now put a perfectly fitting rod down the valve stem and push on it.
The rod won't move until the push provides just over 120 psi. Then the
rod will move steadily until it hits the rim, since the pressure won't
rise appreciably inside the tire.
There's no casing tension involved in that example.
If air pressure is the primary resistance, then it should tend toward
an abrupt motion when 120 psi is reached.
(Again, I realize that a small point can provide 120 psi with only a
small total thumb pressure, so this is debatable--but until pressure
reaches 120 psi, there should be no motion at all.)
A trampoline theory, in contrast, provides for movement at any
pressure, which is what it looks like to me when I push with my thumb.
Push up on a bowling ball sitting at rest in a trampoline, and the
bowling ball rises higher to whatever position a bowling ball less the
amount of the push would occupy.
With the trampoline/tension theory, there's no large range of initial
pressure that fails to produce any movement.
To use very rough numbers, the air-pressure theory means (I think)
that a thumb area of one-fifth of an inch should not make the
slightest impression on a 100 psi sidewall as you apply from 0 to 19
pounds of force. You might as well press on a steel plate.
But a trampoline/tension theory seems to allow for movement as soon as
you start pushing.
I think that Tim McNamara put my point better than I can, if he'll
pardon a little light editing:
"OK. Here's where I may be confounded. If I take a piece of fabric and
stretch it, the tighter I stretch it, the harder it feels to my thumb.
There's no inflation pressure . . ."
Tim gives a pretty good description of the trampoline question, and
his post also has a much nicer attitude than some of my posts of late.
Much of my chattering here is due to my impression that many posters
are talking about a much broader flattening rather than a dent and are
also insisting dogmatically that the tension cannot be involved.
I do appreciate how nicely you're addressing my dogmat--er, clumsy
efforts.
Cheers,
Carl Fogel
JL
<carl...@comcast.net> wrote in message
news:7ta6i292p7kuvql95...@4ax.com...
Carl, it seems to me the stretched fabric and trampoline examples both have
an element missing, namely a uniform pressure pressing against their
surfaces so as to the replicate the air pressure inside an inflated tire.
Perhaps if a large water balloon were resting on the stretched fabric or
trampoline, pushing down over a reasonable area, this might complete the
picture. Now apply the thumb test.
JL
carl...@comcast.net
wrote:
Dear JL,
Unfortunately, the only thing that I can think of is to stretch an air
tight trampoline over a very deep chamber and raise the air pressure.
Somehow, I don't think that this is what you're after, but maybe I'm
wrong. It's really just the tire writ large.
The trampoline will bulge upward just like a tire bulging outward.
Place the weight representing the thumb press onto the bulging
trampoline representing the inflated tire.
Let's say that the air pressure bulging the trampoline upward is 100
psi.
The stretched trampoline's tension must provide 100 psi of force to
contain the air pressure--the forces must balance.
Place a 1 pound weight on the top of the bulging trampoline, a weight
on a flat, inch-square post.
I think that the trampoline will start to indent and that the local
tension will change. Even a tiny force will be acting at nearly right
angles to the tension surface, which is so faintly curved as to be
almost flat. The air pressure will remain 100 psi at all points.
In contrast, cover the opening to a valve stem from the inside with a
1-square-inch metal plate. The air pressure will force the plate up
against the underside of the trampoline with 100 pounds of force.
There's no side tension because the plate is loose. It won't budge
until we push down on it with a 101 pound force on a thin rod stuck
down through the valve hole.
r
| o |
| d |valve stem
| | |
| | |
| | |
___| | |__________trampoline/tire inflated to 100 psi
XXXXXXXXXX
loose metal 1-square-inch plate held up by 100 psi
shouldn't move until 101 pound weight rests on rod
no side tension on loose metal plate (?)
I think that this is just repeats what I've said before. I'm trying to
see flaws in the examples, but I'm going to need help. Maybe the
diagram above will let you spot a problem, or suggest something more
along the lines of what you had in mind with the large water balloon.
Cheers,
Carl Fogel
JL
<carl...@comcast.net> wrote in message
news:bqi6i2t6q8lhjqla5...@4ax.com...
Carl, your valve stem example with a 1 sq in plate is a very close
approximation of a track pump. If the inside barrel diameter of the pump =
1.128 in. and plunger disc surface area is exactly 1 sq in, how much
downward force on the pump handle is required on the final stroke to bring a
tire up to 100 psi? What is causing the resistance?
JL
jobst....@stanfordalumni.org
> I agree that the testers made good-faith efforts. I hope that I'd be
> just as willing as you are to point them out.
> But I'm still awfully skeptical. I can't, for example, see how
> anyone can adjust for random 0.5 mph wind variations during numerous
> runs over an open 245 meter course.
> I expected the speeds "between 10.6 and 16.9 mph" to correspond to
> the "30.6 seconds (Nifty Swifty)" and the "25.3 seconds (Deda
> Tre)"--but they don't seem to work out when I chuck them into a
> spreadsheet:
> 184 meters = 603.67 feet
> 603.67 feet / 10.6 mph = 38.8 seconds (10.6 mph -- minimum?)
> 603.67 feet / 13.45 mph = 30.6 seconds (30 .6 seconds Nifty Swifty)
> 603.67 feet / 16.27 mph = 25.3 seconds (25.3 seconds Deda Tre)
> 603.67 feet / 16.9 mph = 24.4 seconds (16.9 mph -- maximum?)
> Were those two bikes not the fastest and slowest? Or am I just
> misunderstanding you?
The test shown on the web site:
http://www.sheldonbrown.com/brandt/rolling-resistance-tubular.html
These were made by a company that took the effort to build an
instrumented test stand for bicycle tires using a large diameter steel
drum normally used for automobile tires. By strain gauge
measurements, RR was directly measured at a controlled speed for a
series of tires with other parameters such as inner tube used, (except
tubulars) held constant. By measuring over a range of inflation
pressures, anomalies are visible and inflation effects separated from
tread pattern effects.
Coast down tests, on the road and on small drums (as Michelin showed
at InterBike last year) are flawed from the start as you point out,
the wind effects being larger than the differences to be measured.
Jobst Brandt
Michael Press
jobst....@stanfordalumni.org wrote:
> Michael Press writes:
>
> >> From the responses of people more knowledgeable in the topic than
> >> myself, it's clear that I don't grok this correctly. So, let me
> >> ask:
>
> >> 1. In terms of riding a bike, what is the significance of casing
> >> tension? Does it affect how the tire feels, rolling resistance,
> >> traction, etc?
>
> >> 2. When I am on my bike, is it casing tension or inflation
> >> pressure that holds my rims off the ground?
>
> > Both views are useful.
>
> > Analyzing casing stress and strain is fun, and informative. It is
> > essential for designing a good tire.
>
> > We can ignore the casing and think about the wheel being supported
> > by unequal forces from air pressure acting on unequal areas, if we
> > want.
>
> Whaich unequal areas? The rim is the same width around it's
> circumference and is exposed to uniform air pressure around that area.
I drew the picture. The cross-section of the tire at the
contact patch is wider than the cross-section at the top
of the wheel. Equal pressure, larger area --> greater
force.
--
Michael Press
jobst....@stanfordalumni.org
>>>> From the responses of people more knowledgeable in the topic than
>>>> myself, it's clear that I don't grok this correctly. So, let me
>>>> ask:
>>>> 1. In terms of riding a bike, what is the significance of casing
>>>> tension? Does it affect how the tire feels, rolling resistance,
>>>> traction, etc?
>>>> 2. When I am on my bike, is it casing tension or inflation
>>>> pressure that holds my rims off the ground?
>>> Both views are useful.
>>> Analyzing casing stress and strain is fun, and informative. It is
>>> essential for designing a good tire.
>>> We can ignore the casing and think about the wheel being supported
>>> by unequal forces from air pressure acting on unequal areas, if we
>>> want.
>> Which unequal areas? The rim is the same width around it's
>> circumference and is exposed to uniform air pressure around that
>> area.
> I drew the picture. The cross-section of the tire at the contact
> patch is wider than the cross-section at the top of the wheel.
> Equal pressure, larger area --> greater force.
That the tire bellies out under load is not contested, however, the
contact area with the rim remains unchanged and the pressure around
the rim is uniform. I can only deduce that you believe the bending
strength of the casing is supporting the rim. It is not. To make
that more evident, think of a paper thin silk track tire whose side
walls require no perceptible force to bend.
I think the description of how the tire supports the rim is clearly
stated in the FAQ item:
http://www.sheldonbrown.com/brandt/rim-support.html
Your diagram raises all the old problems that keep recurring with that
description.
Jobst Brandt
carl...@comcast.net
wrote:
Dear JL,
Air pressure--no casing tension.
No movement from 0 to 100 psi, then a foot of movement at just over
100 psi (a little friction and the slight increase in air pressure).
And the pressure remains 100 psi during that foot of travel.
If we had a ten-story pump and a pressure-relief valve to keep the
tire pressure from rising over 100 psi, the handle would move a
hundred feet under the 101 pound weight.
Of course, in a thumb press, a widening contact patch will flatten out
some to provide more resistance as area increases.
But even pressing with something flat, circular, and non-expanding
like a nail head seems to provide a trampoline-like dent, not a
contact-patch-with-the-ground flattening.
That seems to suggest that the casing is being pulled into tension
with thumb-print dent.
Here's a picture that seems to show denting, not flattening:
http://server5.theimagehosting.com/image.php?img=225a%20dent.jpg
I put the details and two pictures in a new thread, since this one is
getting tangled. I thought that an actual picture might help, but
please don't take it as being presented as proof. The idea is for
people to have something concrete and visual to refer to in their
explanations.
Does it look as if there's a lot of local stretching and tension?
The question isn't facetious--the side of a toroid is trickier than my
feeble geometry can handle. It took me a while to notice that the
sidewall where most of us do our thumb pressing is much flatter than
the curved contact patch that most of us think of.
I'm also wondering what it would look like the other way, with a thumb
press from the inside denting the sidewall outward.
I think that the difference in air pressure would produce different
curves, but that both cases would require raising the local tension.
That is, I think that the inward dent would be sharper, with the air
pressure forcing the sides toward the denter, while an outward dent
would be more gradual:
narrow inward dent? wider outward dent?
.
______ ____________________ ' ' ________
. .
.. 100 psi interior
Cheers,
Carl Fogel
o...@ozarkbicycleservice.com
Michael Press wrote:
> In article
> <timmcn-50E4C3....@news.iphouse.com>,
> Tim McNamara <tim...@bitstream.net> wrote:
>
> > The ordinal ranking of the top ten tires was (widths are actual): Deda
> > Tre Giro d'Italia (700 x 24); Clement del Mondo (700 x 28 tubular);
> > Michelin Pro2 Race (700 x 25); Continental Ultra Gator (700 x 23),
> > Mistuboshi Trimline (650B x 37), Panaracer Pasela (700 x 35); Clement
> > Criterium (700 x 21); Avocet Cross (700 x 35); Avocet Duro (700 x 28).
>
>
The "Avocet Duro" *is* a Fasgrip, one that is 28mm. Different widths
have different names.
Tim McNamara
carl...@comcast.net wrote:
I think that those were just the highest and lowest speeds seen during
the course of the testing.
Tim McNamara
jobst....@stanfordalumni.org wrote:
OK, that's where I'm baffled. On the one hand, casing tension is not
the rider's problem. Other the other hand, inflation pressure holds the
rim off the ground through casing tension. But on the gripping hand,
when I squeeze the tire with my thumb I am not feeling casing tension, I
am feeling inflation pressure. I must be missing something because 2
and 2 are not equaling 4. I dunno why my brain doesn't want to wrap
around this.
Tim McNamara
Tim McNamara <tim...@bitstream.net> wrote:
> If I take a piece of fabric and stretch it, the tighter I stretch it
> for harder it feel to my thumb.
Wow. That was spectacularly bad writing and incompetent proofreading.
I'm embarrassed. I realize I speak American and not English, but
reading that you'd swear Japanese was my native language.
Luns Tee
<jobst....@stanfordalumni.org> wrote:
>> I drew the picture. The cross-section of the tire at the contact
>> patch is wider than the cross-section at the top of the wheel.
>> Equal pressure, larger area --> greater force.
>
>That the tire bellies out under load is not contested, however, the
>contact area with the rim remains unchanged and the pressure around
>the rim is uniform. I can only deduce that you believe the bending
>strength of the casing is supporting the rim. It is not. To make
>that more evident, think of a paper thin silk track tire whose side
>walls require no perceptible force to bend.
What Michael describes isn't inaccurate, although it's
incomplete. The net force of air pressure acting on the full width of
the tire at the bulge is greater than the force of air pressure on the
rim bed, and the difference of this force must be transferred the rim
somewhere.
How it gets transferred depends somewhat on the rim/tire
interface. For a clincher rim, this imbalance is supported entirely
where the tire wraps around the lip of the rim. The pressure the tire
exerts on the rim is the cord tension divided by the radius of the lip
(plus air pressure, but it's small relative to this), and the net
force exerted is a function of how much of an arc this wraps around:
this arc ending with the ~45 degree exit angle of the casing, which
varies with tire deflection.
Things here are a little tricky to analyze, and I'm actually not
entirely sure how to properly model the interface of a hook bead to the
rim. A frictionless straight wall rim combined with a rigid inelastic
tire bead is tractable though, and the differences between that an an
actual interface are internal differences.
Michael's description is wrong in saying that the area times the
pressure accounts for the entire load at the axle though. There's a
component that comes from the reduction in cord tension in the loaded
area. Draw a line through the widest part of the tire, and tally up
all the forces through it - cord tension contributes just as air
pressure does.
For tubulars, an interesting thing to note is that unlike with
clinchers, the pressure put on the bed of the rim isn't simply the air
pressure. Air pressure is contained by the curve of the casing times
the cord tension. The curve of the rim bed is flatter than the
tire's natural radius, so as the tire conforms to this, the casing
supports less of the air pressure than it does where unsupported. What's
left is supported by the rim.
Where a tire is loaded, the cord tension is less, so the casing
is able to support even less of the air pressure, leaving more pressure
to be conveyed to the rim. Reduction in the casing tension pushes on the
tubular rim bed. The effect of the tire wrapping around the edges of
the rim still applies too.
We can even get analytical expressions for what the pressure on
the rim would be, except it would have to assume a frictionless
interface between the casing and the rim, which of course is not the
case.
Friction/glue and casing shear will allow for gradients in the
cord tension. The compression put onto the casing gets transmitted by
friction/shear where the casing first meets the rim, rather than being
carried all the way across the bed (where thanks to the gluing, the
cords instead remain at full tension) to be transmitted as pressure.
These differences are all internal to the contact area though (much like
unrelieved residual stress), and are net zero on the external picture.
-Luns
jobst....@stanfordalumni.org
>>>>>> Well, the larger tyre requires less pressure for a given casing
>>>>>> tension. I read that here recently and am still getting my
>>>>>> head around it. It's also mentioned here by Jobst Brandt:
http://www.sheldonbrown.com/brandt/rim-support.html
>>>>>> "[...] unit casing tension is equivalent to inflation pressure
>>>>>> times the radius of curvature divided by pi [...]".
>>>>>> I was a bit surprised by this at first, but then if you think
>>>>>> pressure is force per unit area, if you increase the area of
>>>>>> the inside of the casing, you need more force for a given
>>>>>> pressure. Not sure if this reasoning is bogus or not though.
>>>>> If you have an inflation pressure of 100 psi, a tire with more
>>>>> inside surface area will have a casing under greater tension
>>>>> because there are more square inches. If I grok correctly.
Don't worry about the surface area, just look at the casing as two
semicircles being pushed apart by air pressure for which only the
length of the diameter is important. The above calculation reduces
the tire cross section into a diameter.
>>>> Casing stress is arrived upon by cutting across the circular
>>>> minor diameter of the tire (the tire is a circular cross section
>>>> having no structural belt as radial tires do to change that) and
>>>> take the two halves as solid sections being pressed apart by
>>>> inflation pressure. That gives the lineal separation force which
>>>> is the casing tension.
>>>> The above mentioned formula reduces to just that. For cord
>>>> stress, adjusting for 45 degree bias ply SQR(2) gets involved but
>>>> this is about casing tension which is the same regardless of
>>>> fabric structure.
>>> From the responses of people more knowledgeable in the topic than
>>> I, it's clear that I don't grok this correctly. So, let me ask:
>>> 1. In terms of riding a bike, what is the significance of casing
>>> tension? Does it affect how the tire feels, rolling resistance,
>>> traction, etc?
>> I think this puts the cart before the horse. Casing tension
>> depends on two parameters, inflation pressure and tire size. I
>> think you know the answer to that already. The harder the tire is
>> inflated the harsher the ride for any cross section and the larger
>> the cross section the softer the ride for any inflation pressure.
>> Let the manufacturer decide what casing stress is acceptable. It's
>> not the user's problem.
>>> 2. When I am on my bike, is it casing tension or inflation
>>> pressure that holds my rims off the ground?
>> Inflation pressure does that, but does so by casing tension. As I
>> said, it's only a problem for the engineer who wonders how loads
>> get from the rim to the ground. The item in the FAQ explains that.
http://www.sheldonbrown.com/brandt/rim-support.html
> OK, that's where I'm baffled. On the one hand, casing tension is
> not the rider's problem. Other the other hand, inflation pressure
> holds the rim off the ground through casing tension. But on the
> gripping hand, when I squeeze the tire with my thumb I am not
> feeling casing tension, I am feeling inflation pressure. I must be
> missing something because 2 and 2 are not equaling 4. I dunno why
> my brain doesn't want to wrap around this.
Before getting started let me say that atmospheric pressure does not
play a role in this although it is often mentioned. Inflation
pressure is the relative pressure between the air in the tube and the
outside environment, which could also be a vacuum. All that counts is
what the gauge measures, the rest is extraneous.
Let's separate the variables. The area of the road contact patch is
determined by the wheel load divided by the inflation pressure, a flat
area for practical purposes and has no relationship to casing tension.
Tire pressure can be assessed by pressing ones thumb against the tire
until the thumb contact patch is flat.
Granted this takes some skill and experience but is what the skilled
thumb senses. That force is independent of tire cross section,
assuming we have a flexible thin walled tire as most of those who
worry about these things use. Tire casing tension has no part in this
test, only inflation pressure as was explained by geometry. Casing
tension cannot enter into it because the casing is normal to the
applied force.
Inflation pressure acts uniformly around the rim circumference so it
cannot exert any net force. The rim does not change shape so its area
remains constant. The tire having no significant casing rigidity also
cannot transmit force between ground and rim other than by tension,
that being the only other force present.
That casing tension is not uniform around the rim is apparent from the
belly that a tire has when loaded. Wheel load changes casing tension
both in angle and magnitude. The effect of angular change is given by
the sine of the angle times tension, and tension magnitude is given by
the circular cross section of the bulge, which is smaller than it was
before loading. [the bulge is circular]
The angle of a typical (25mm) tire casing, tubular or clincher, lies
at a 45 degree angle where it departs from the rim. Under load, it
bulges when vertically compressed, so that the angle at which it pulls
from the rim is more horizontal, giving it a smaller vertical
component. Thus it does not pull down as much as when unloaded.
Meanwhile, the circular cross section being smaller, has its tension
reduced for the same pressure. Its tension is defined by the diameter
of the circular section divided by 2*pi as we know from smaller cross
section tires permitting higher inflation than fat tires with similar
casings.
[Also consider that spare tires in automobiles are inflated to their
operating pressure in the unloaded condition and don't change pressure
measurably when the car is let off the jack.]
The two effects, lower casing tension and broader angle reduces
downward pull on the rim. The rim stands on the tire!
Remember? That's how this subject got introduced a while ago.
Jobst Brandt
jobst....@stanfordalumni.org
>>> I drew the picture. The cross-section of the tire at the contact
>>> patch is wider than the cross-section at the top of the wheel.
>>> Equal pressure, larger area --> greater force.
>> That the tire bellies out under load is not contested, however, the
>> contact area with the rim remains unchanged and the pressure around
>> the rim is uniform. I can only deduce that you believe the bending
>> strength of the casing is supporting the rim. It is not. To make
>> that more evident, think of a paper thin silk track tire whose side
>> walls require no perceptible force to bend.
> What Michael describes isn't inaccurate, although it's incomplete.
> The net force of air pressure acting on the full width of the tire
> at the bulge is greater than the force of air pressure on the rim
> bed, and the difference of this force must be transferred the rim
> somewhere.
I think if you review what I just responded to Tim McNamara you'll see
a more detailed analysis that applies to this subject.
> How it gets transferred depends somewhat on the rim/tire interface.
> For a clincher rim, this imbalance is supported entirely where the
> tire wraps around the lip of the rim. The pressure the tire exerts
> on the rim is the cord tension divided by the radius of the lip
> (plus air pressure, but it's small relative to this), and the net
> force exerted is a function of how much of an arc this wraps around:
> this arc ending with the ~45 degree exit angle of the casing, which
> varies with tire deflection.
There is no difference between a clincher and tubular or a clincher
with or without hooked bead. The analysis of support for the rim is
the same and fits the simple model described. There is a place where
tire casing and rim diverge and this is the same for all clinchers and
tubulars alike. After that the problem is identical. The pressure on
the rim is bases entirely on air pressure and is invariant.
The difference for tubulars is that the rim curvature is slightly
smaller than the tire so it rests on the edges of the contact with the
rim, the rest of the area filled with rim glue. This contact is
plastic and squirms fore and aft as the tire rolls, gradually wearing
through the base tape by fretting. This does not affect how the tire
supports the rim. For that purpose the rim sees only inflation
pressure.
> Things here are a little tricky to analyze, and I'm actually not
> entirely sure how to properly model the interface of a hook bead to
> the rim. A frictionless straight wall rim combined with a rigid
> inelastic tire bead is tractable though, and the differences between
> that an an actual interface are internal differences.
I don't believe so. The analysis is entirely within the realm of tire
minor diameter shape and bulging under load, there being only two
effects; that of reduced tension and angle at the departure from the
rim.
> Michael's description is wrong in saying that the area times the
> pressure accounts for the entire load at the axle though. There's a
> component that comes from the reduction in cord tension in the
> loaded area. Draw a line through the widest part of the tire, and
> tally up all the forces through it - cord tension contributes just
> as air pressure does.
That's a gross misstatement because it does not address how the load
between rim and road is transmitted by the tire. It glosses over the
mechanism which is dirt simple but obscure.
> For tubulars, an interesting thing to note is that unlike with
> clinchers, the pressure put on the bed of the rim isn't simply the
> air pressure. Air pressure is contained by the curve of the casing
> times the cord tension. The curve of the rim bed is flatter than
> the tire's natural radius, so as the tire conforms to this, the
> casing supports less of the air pressure than it does where
> unsupported. What's left is supported by the rim.
There is no difference between clincher and tubular. Besides the
radius of curvature of the rim is slightly smaller than that of a tire
designed for hat rim. That makes way for the bump of the seam and
assurance that the side of the tire doesn't lift off when inflated.
That will cause early tire failure as it sticks and unsticks with each
revolution.
> Where a tire is loaded, the cord tension is less, so the casing is
> able to support even less of the air pressure, leaving more pressure
> to be conveyed to the rim. Reduction in the casing tension pushes
> on the tubular rim bed. The effect of the tire wrapping around the
> edges of the rim still applies too.
I don't understand how you determined that cord tension is less,
although it is so for geometric reasons. What do you mean by
conveying pressure to the rim. Inflation pressure against the rim is
uniform around the rim circumference. Casing tension is not.
> We can even get analytical expressions for what the pressure on the
> rim would be, except it would have to assume a frictionless
> interface between the casing and the rim, which of course is not the
> case.
Pleas review my response to Tim McNamara. I think you'll find the
model for that expression there.
> Friction/glue and casing shear will allow for gradients in the cord
> tension. The compression put onto the casing gets transmitted by
> friction/shear where the casing first meets the rim, rather than
> being carried all the way across the bed (where thanks to the
> gluing, the cords instead remain at full tension) to be transmitted
> as pressure. These differences are all internal to the contact area
> though (much like unrelieved residual stress), and are net zero on
> the external picture.
I think your model makes invalid assumptions about friction and the
balance of forces on a tubular. Glue has nothing to do with this. If
you have ridden tubulars you will recall that a new track tire with a
bare cloth base tape mounted on a new rim with no glue can be ridden
and it supports the rim the same as any clincher by the means that I
described at length.
However, in brief:
http://www.sheldonbrown.com/brandt/rim-support.html
Jobst Brandt
carl...@comcast.net
[snip]
>Let's separate the variables. The area of the road contact patch is
>determined by the wheel load divided by the inflation pressure, a flat
>area for practical purposes and has no relationship to casing tension.
>Tire pressure can be assessed by pressing one's thumb against the tire
>until the thumb contact patch is flat.
>
>Granted this takes some skill and experience but is what the skilled
>thumb senses.
[snip]
Dear Jobst,
I'm fascinated, having been brought up as a sidewall denter and
utterly unaware that tread flatteners existed.
I'm guessing that you press down on the middle of the tread on the top
of the tire, pushing it against the ground, instead of gripping the
tire and pushing against the sidewall--is this correct? Or do you grip
the rim with your fingers?
Could you take a moment and press on a bathroom scale and let us know
what kind of force you usually apply?
Is the flattened patch covered by your thumb and just felt instead of
seen? Or do you push down until you see the tire flattening around
your thumb?
That is, can you do it in the dark, or do you have to look at it?
I'm not arguing about sensitivity or how well the method works, just
absurdly curious because your method is so different from what I've
always seen. Maybe it's because I spent too much time around dirt
motorcycles, where pushing on a knob would seem wrong when the bare
sidewall is handy.
It's a little like the first time that I saw an actor in a foreign
film holding a cigarette sticking the wrong way out from between his
thumb and finger. It looked strange and silly, but after a moment I
realized that there must be a whole world out there of people who
thought that U.S. actors looked weird with the cigarette sticking out
the other way.
From experience, I know that it's beastly hard to photograph, but this
might be worth a picture or two.
Cheers,
Carl Fogel
Johnny Sunset aka Tom Sherman
Michael Press wrote:
> In article <slrnei1jti....@bowser.marioworld>,
> Ben C <spam...@spam.eggs> wrote:
>
> > Does the "thumb test" (squeezing a tyre to see if it's hard enough)
> > measure tyre pressure or casing tension?
>
> of the tire, then reading the pressure from a chart:
>
> Roll out, 90 psi: 2099 mm
> ...
> Roll out, 120 psi: 2107 mm
I nominate this for Post of the Month.
--
Tom Sherman - Here, not there.
carl...@comcast.net
Dear Tim,
Nonsense.
You probably just did the kind of editing that computers encourage.
Replace "for" with "the" and stuff a comma in, if you like.
Then slap an "s" on "feel" and declare victory.
If you had a record of your editing, I'll bet that you'd find the
original missing phrases that were changed to improve things.
I only mentioned light editing when I quoted you because English
majors feel compelled to confess to violating the sanctity of the
written word as we scribble with noses "as sharp as a pen, and a table
of green fields."
(Google it if you want to see what's arguably the most mystifying,
what-the-hell passage in English literature. Or browse around for
comments on the discovery that bored French type-setters made over
2,000 mistakes in "Ulysses," many of which had been explained as Joyce
being brilliant.)
A few minutes ago, I silently inserted an apostrophe in another
poster's reply. I felt as daring and guilty as if I'd taken an inner
tube from a bike shop trash can and stuffed it into my pocket.
What you wrote was clear enough to be followed.
Here's an internet example of really bad editing:
In many online texts of "The Conduct of Life," Emerson's first
sentence mysteriously reads:
"It chanced during one winter, a few years ago, that our cities
wsing the theory of the Age."
Wsing?
What the hell? Ralph Waldo may not be the clearest writer, but "wsing"
has gotta be a typo.
Let's see . . . Using? Easing? Swing? Sang?
Nope.
It was written before typewriters, so maybe we need to look for nearby
letters in the etaoinshrdlu linotype instead of the modern qwerty
keyboard?
Nope.
But it must be a verb of some kind: "our cities something-ed the
theory of the Age."
You'd go mad trying to guess this riddle, so I'll reveal the answer,
which could not possibly be deduced. It wasn't a typo. It was a whole
block of missing words and letters. If you look, you can find the
correct text:
"It chanced during one winter, a few years ago, that our cities w
***ere bent on discuss***
ing the theory of the Age."
Like a lot of Ralph Waldo's writing, it still doesn't say much, but it
really was a coherent sentence until something bad happened to it,
either a terrible typesetter or an awful scanner.
Cheers,
Carl Fogel
Tim McNamara
jobst....@stanfordalumni.org wrote:
OK, I can follow that. Where I get stuck is that the road pressing
against the tire and the thumb pressing against the tire amount to the
same thing. I don't see how the thumb is resisted by inflation pressure
but the road is resisted by casing tension. What's the difference?
Both forces are normal to the casing.
Michael Press
I am not discussing the bending stress of the casing. I
take the casing to be a flexible, inextensible sausage
casing with greater air pressure inside than outside. The
cross-section at the contact patch is wider than the
cross-section at the antipodal point. At the contact patch
cross-section the greater width means that the force
upward from air pressure is greater than the force
downward from air pressure at the antipodal cross-section.
The force difference is equal to the load at the hub.
--
Michael Press
Michael Press
<1159998184.9...@m73g2000cwd.googlegroups.com>,
o...@ozarkbicycleservice.com wrote:
Check. Still, the Duro has 66 thread per inch side walls,
while the 25 mm version has 127 tpi side walls, making for
a measurable difference in rolling resistance.
--
Michael Press
Michael Press
carl...@comcast.net wrote:
> Like a lot of Ralph Waldo's writing, it still doesn't say much, but it
> really was a coherent sentence until something bad happened to it,
> either a terrible typesetter or an awful scanner.
While some of it says a great deal. `Self Reliance' is
brilliant.
--
Michael Press
Michael Press
lu...@mochi.EECS.Berkeley.EDU (Luns Tee) wrote:
> What Michael describes isn't inaccurate, although it's
> incomplete. The net force of air pressure acting on the full width of
> the tire at the bulge is greater than the force of air pressure on the
> rim bed, and the difference of this force must be transferred the rim
> somewhere.
It is complete. I explicitly announced that I intended to
neglect stress in the casing. I entered into this thread
responding to Tim McNamara in
<timmcn-439DAB....@news.iphouse.com>
> 2. When I am on my bike, is it casing tension or inflation pressure
> that holds my rims off the ground?
saying that both approaches are worth pursuing.
<jack-906D16.1...@newsclstr02.news.prodigy.com>
--
Michael Press
jobst....@stanfordalumni.org
Nowhere did I say "the road is resisted by casing tension". That is
why you have difficulty understanding the effects. When the tire is
flattened against a contact surface, only inflation pressure is
pushing against that surface, assuming a flexible tire casing and
tread.
Jobst Brandt
jobst....@stanfordalumni.org
>>>>> Both views are useful.
http://www.sheldonbrown.com/brandt/rim-support.html
That is the definition of inflation pressure or gauge pressure.
> The cross-section at the contact patch is wider than the
> cross-section at the antipodal point. At the contact patch
> cross-section the greater width means that the force upward from air
> pressure is greater than the force downward from air pressure at the
> antipodal cross-section. The force difference is equal to the load
> at the hub.
So how does the force reach the ground from the rim. I see no
reference to the rim in your description. You say the tire casing is
flexible, and I concur, but you don't explain how the load is
transmitted between rim and road, and it is not inflation pressure
that supports the rim because it presses against the rim uniformly
around its circumference. Beyond that, the contact area of the tire
with the rim (clincher or tubular) also remains constant.
Jobst Brandt
Michael Press
jobst....@stanfordalumni.org wrote:
> So how does the force reach the ground from the rim. I see no
> reference to the rim in your description. You say the tire casing is
> flexible, and I concur, but you don't explain how the load is
> transmitted between rim and road, and it is not inflation pressure
> that supports the rim because it presses against the rim uniformly
> around its circumference. Beyond that, the contact area of the tire
> with the rim (clincher or tubular) also remains constant.
I have not attempted to describe a force path through the
casing. I have only described a simple view of how the hub
load is balanced by a greater force of air pressure on a
wider tire cross-section at the contact patch.
--
Michael Press
Tim McNamara
jobst....@stanfordalumni.org wrote:
OK, thanks, I think I finally get it. When the tire is flattened
against the road (or against one's thumb) the casing is basically
sandwiched between the road and the air inside the tube, and thus there
is a balance between the inflation pressure and the load. I was cooking
up red herrings for myself with casing tension.
jobst....@stanfordalumni.org
>> So how does the force reach the ground from the rim. I see no
>> reference to the rim in your description. You say the tire casing
>> is flexible, and I concur, but you don't explain how the load is
>> transmitted between rim and road, and it is not inflation pressure
>> that supports the rim because it presses against the rim uniformly
>> around its circumference. Beyond that, the contact area of the
>> tire with the rim (clincher or tubular) also remains constant.
> I have not attempted to describe a force path through the casing. I
> have only described a simple view of how the hub load is balanced by
> a greater force of air pressure on a wider tire cross-section at the
> contact patch.
What do you mean by "a greater force of air pressure on a wider tire
cross-section at the contact patch". Greater than what? The area of
the contact patch is given by the load divided by inflation pressure,
nothing more. I think that has been stated here often enough. What
is " balanced by a greater force of air pressure on a wider tire
cross-section at the contact patch." Where is the balance and what
does this have to do with the hub?
I think the matter is getting more obscure by the minute. Let's not
get the hub and spokes into this or we will be worse off than at the
beginning of this thread.
Jobst Brandt
JL
"Michael Press" <ja...@abc.net> wrote in message
news:jack-C6FCF6.2...@newsclstr02.news.prodigy.com...
Michael, some great writing indeed!
JL
"A foolish consistency is the hobgoblin of little minds." Emerson
Luns Tee
Michael Press <ja...@abc.net> wrote:
>> What Michael describes isn't inaccurate, although it's
>> incomplete. The net force of air pressure acting on the full width of
>> the tire at the bulge is greater than the force of air pressure on the
>> rim bed, and the difference of this force must be transferred the rim
>> somewhere.
>
>It is complete. I explicitly announced that I intended to
>neglect stress in the casing. I entered into this thread
>responding to Tim McNamara in
You can't neglect stress in the casing. Casing tension is
determined by the air pressure (constant) and the radius the casing
bends around. This radius is smaller in the bulge than it is in the
undisturbed tire, and thus the tension is lower at the bulge. This
change in the tension contributes to the force at the axle.
Air pressure on the broader width of the tire's bulge does
contribute to supporting the load at the axle, but is only one component
of that load - it is NOT the complete picture.
The only way you could neglect tension is if the cords were
contained frictionlessly in the casing and allowed to slide around. What
would happen is that the casing in the loaded area, while it wants to
reduce in tension, gets cord pulled out of that area by the rest of
the tire where then tension wants to stay the same - imagine twisting a
slinky to reduce its diameter.
However, this is not a stable structure! The uniform donut is an
unstable equilibrium. If the tire has a larger minor diameter in one
spot than elsewhere, the air pressure in the larger section will steal
cord away from the rest of the casing, which as it reduces in diameter
will have even less area for the air pressure to support tension. This
effect would feed on itself, with the anyeurism growing until the tire
were a sphere with an empty loop dangling off of it.
The picture is reminiscent of the long thin balloons clowns use
to make balloon sculptures. When you start inflating the balloon, the tail
of the balloon stays in pretty much its uninflated form, while the
rubber by where you're blowing puffs up to a much larger diameter. The
balloon doesn't grow with a uniform diameter until all of the balloon
has left the small-diameter state.
But this is wandering off into the absurd. What makes it absurd
is the assumption that tension is constant - don't do that!
You can't nelgect tension!
-Luns
Luns Tee
<jobst....@stanfordalumni.org> wrote:
>> What Michael describes isn't inaccurate, although it's incomplete.
>> The net force of air pressure acting on the full width of the tire
>> at the bulge is greater than the force of air pressure on the rim
>> bed, and the difference of this force must be transferred the rim
>> somewhere.
>
>I think if you review what I just responded to Tim McNamara you'll see
>a more detailed analysis that applies to this subject.
We're talking across purposes here, but let it be said that I
don't disagree with anything written in that post or in the FAQ. Support
at the rim consists of two components: the reduction in tension, and the
change in the casing's exit angle where it leaves the rim. What I was
intending to convey was that Michael's expanation of air pressure on the
extra width of the bulge, is actually identically equivalent to the
effect of the casing exit angle (but neglects the change in Tension).
The exit angle flattening, changes the arc on which air pressure
acts, and the air pressure on the extra width of that arc is identical
to the change in the vertical component of the casing tension at the
end of the arc.
>> How it gets transferred depends somewhat on the rim/tire interface.
>> For a clincher rim, this imbalance is supported entirely where the
>> tire wraps around the lip of the rim. The pressure the tire exerts
>> on the rim is the cord tension divided by the radius of the lip
>> (plus air pressure, but it's small relative to this), and the net
>> force exerted is a function of how much of an arc this wraps around:
>> this arc ending with the ~45 degree exit angle of the casing, which
>> varies with tire deflection.
>
>There is no difference between a clincher and tubular or a clincher
>with or without hooked bead. The analysis of support for the rim is
>the same and fits the simple model described. There is a place where
>tire casing and rim diverge and this is the same for all clinchers and
>tubulars alike. After that the problem is identical. The pressure on
>the rim is bases entirely on air pressure and is invariant.
We're talking across somewhat different problems here. The
problem at the point of divergence is identical, but the point of
divergence is the boundary between unsupported tire casing, and a
combined unit of rim and the rest of the tire. The forces you've
described are correct, but they're applied to the tire portion of this
combined unit, not the rim. The remainder of my discussion was about how
it gets from this portion of the tire into the rim, and this is quite
different between tubular, and clincher, with or without hooked bead.
>The difference for tubulars is that the rim curvature is slightly
>smaller than the tire so it rests on the edges of the contact with the
>rim, the rest of the area filled with rim glue. This contact is
>plastic and squirms fore and aft as the tire rolls, gradually wearing
>through the base tape by fretting. This does not affect how the tire
>supports the rim. For that purpose the rim sees only inflation
>pressure.
If the curvature the tire meets is indeed a much smaller radius
than the tire itself, then the load would be entirely on the edges of
contact, and the bed of the reduced curvature area actually wouldn't see
much of the air pressure, if any at all! The air pressure bears on the
tire casing, but the casing is suspended between those edges like a
hammock - picture a tufo tire - rather than being supported by the rim
bed.
>> Where a tire is loaded, the cord tension is less, so the casing is
>> able to support even less of the air pressure, leaving more pressure
>> to be conveyed to the rim. Reduction in the casing tension pushes
>> on the tubular rim bed. The effect of the tire wrapping around the
>> edges of the rim still applies too.
>
>I don't understand how you determined that cord tension is less,
>although it is so for geometric reasons. What do you mean by
>conveying pressure to the rim. Inflation pressure against the rim is
>uniform around the rim circumference. Casing tension is not.
Tension is the pressure times the radius of the curve, and the
bulged of the compressed tire is at a smaller radius. This tension is
what supports the 'hammock' I mention above, and if the tension is
reduced, the hammock will want to sag against the rim. It either sags
and bears on the rim with more pressure, or of the ends of the hammock
are well anchored (be it by glue or friction), then this reduced
tension gets supported by those anchors.
What's confusing things is again the distinction between the rim
itself, and the combined unit of rim and the section of tire that's
directly wedded to it. Air pressure acts uniformly on the casing, but
that isn't transferred directly to the rim unless the interface is flat
or the casing is under zero tension. If the interface has a radius,
then the casing tension on that radius supports some portion of the
air pressure.
>> Friction/glue and casing shear will allow for gradients in the cord
>> tension. The compression put onto the casing gets transmitted by
>> friction/shear where the casing first meets the rim, rather than
>> being carried all the way across the bed (where thanks to the
>> gluing, the cords instead remain at full tension) to be transmitted
>> as pressure. These differences are all internal to the contact area
>> though (much like unrelieved residual stress), and are net zero on
>> the external picture.
>
>I think your model makes invalid assumptions about friction and the
>balance of forces on a tubular. Glue has nothing to do with this. If
>you have ridden tubulars you will recall that a new track tire with a
>bare cloth base tape mounted on a new rim with no glue can be ridden
>and it supports the rim the same as any clincher by the means that I
>described at length.
It does matter, but for the internal picture of the rim/tire
section unit. The differences are of the distribution of internal
forces within this unit, not of the unsupported tire's relationship to
this unit.
-Luns
jobst....@stanfordalumni.org
>>> What Michael describes isn't inaccurate, although it's incomplete.
>>> The net force of air pressure acting on the full width of the tire
>>> at the bulge is greater than the force of air pressure on the rim
>>> bed, and the difference of this force must be transferred the rim
>>> somewhere.
>> It is complete. I explicitly announced that I intended to neglect
>> stress in the casing. I entered into this thread responding to Tim
>> McNamara in:
> You can't neglect stress in the casing. Casing tension is
> determined by the air pressure (constant) and the radius the casing
> bends around. This radius is smaller in the bulge than it is in the
> undisturbed tire, and thus the tension is lower at the bulge. This
> change in the tension contributes to the force at the axle.
> Air pressure on the broader width of the tire's bulge does
> contribute to supporting the load at the axle, but is only one
> component of that load - it is NOT the complete picture.
Please explain how the width of the tire (the bulging under load)
changes the net force on the rim (let's leave axle, hub, bearings and
spokes out of this). As is apparent, the rim does not get wider from
loading so its contact with the tire does not change, at least not more
than a few thousandths of an inch, the bead having a radius of about
0.05" and the angular change of the tire less than 10°.
> The only way you could neglect tension is if the cords were
> contained frictionlessly in the casing and allowed to slide around.
> What would happen is that the casing in the loaded area, while it
> wants to reduce in tension, gets cord pulled out of that area by the
> rest of the tire where then tension wants to stay the same - imagine
> twisting a slinky to reduce its diameter.
I am not neglecting tension in the casing as you must have seen in the
piece to Tim McNamara, which I chose not to repeat for brevity. Let's
not get trampolines, nails, rubber bands, and slinky into this. The
tire is a simple enough model. It is valid to assume the casing is a
thin tubular flexible inelastic membrane.
Why do you want to dig into the casing structure that has practically
nothing to do with how the tire supports the rim. A tire made of
mylar tubing gets rid of the TPI, bead or no bead of a clincher, or
tubular tire, although it supports the rim by the same mechanism as
an actual tire.
> However, this is not a stable structure! The uniform donut is an
> unstable equilibrium. If the tire has a larger minor diameter in
> one spot than elsewhere, the air pressure in the larger section will
> steal cord away from the rest of the casing, which as it reduces in
> diameter will have even less area for the air pressure to support
> tension. This effect would feed on itself, with the aneurysm
> growing until the tire were a sphere with an empty loop dangling off
> of it.
Are you sure you are talking about how the tire supports the rim or
are you addressing tire manufacturing problems. This has nothing to
do with the mechanism that supports the rim from the road. I don't
see any connection between this and your introducing instability.
Your model is in my estimation inappropriate and does not affect how
the tire transmits load from the rim to the road. There is no
aneurysm and no effect measurable in the rest of the tire when the rim
presses the tire on the road. Even if there were effects at other
parts of the tire, how do you propose they work to separate the rim
from the road at the load point?
Can we focus on the article to Tim McNamara for a moment? It is
appended below in the event that it is not available on your
newsreader.
What do you find missing in the force analysis given in that item?
> The picture is reminiscent of the long thin balloons clowns use to
> make balloon sculptures. When you start inflating the balloon, the
> tail of the balloon stays in pretty much its uninflated form, while
> the rubber by where you're blowing puffs up to a much larger
> diameter. The balloon doesn't grow with a uniform diameter until
> all of the balloon has left the small-diameter state.
Please leave the trampolines and slinkys... and elastic balloons out
of this. The don't clarify the issue.
> But this is wandering off into the absurd. What makes it absurd
> is the assumption that tension is constant - don't do that!
Who made the assumption tension is constant? I'm beginning to think
you are putting me on, or trying my patience with absurd models. In
any event, you are not talking about how the tire supports the rim and
seem to choose not to do so.
Jobst Brandt
=========================================================================
======================================================================
jobst....@stanfordalumni.org
>>> What Michael describes isn't inaccurate, although it's incomplete.
>>> The net force of air pressure acting on the full width of the tire
>>> at the bulge is greater than the force of air pressure on the rim
>>> bed, and the difference of this force must be transferred the rim
>>> somewhere.
>> I think if you review what I just responded to Tim McNamara you'll
>> see a more detailed analysis that applies to this subject.
I guess this posting was another perspective on the issue so let's
look at this new proposal.
> We're talking across purposes here, but let it be said that I don't
> disagree with anything written in that post or in the FAQ. Support
> at the rim consists of two components: the reduction in tension, and
> the change in the casing's exit angle where it leaves the rim. What
> I was intending to convey was that Michael's explanation of air
> pressure on the extra width of the bulge, is actually identically
> equivalent to the effect of the casing exit angle (but neglects the
> change in Tension).
From this I see your free body diagram is amiss. The width of the
tire when compressed does not change the pressure on the rim. The
pressure in the tire and that seen by the rim remains unchanged as
does the width of the rim exposed to that pneumatic force.
> The exit angle flattening, changes the arc on which air pressure
> acts, and the air pressure on the extra width of that arc is
> identical to the change in the vertical component of the casing
> tension at the end of the arc.
That is not true! There are two effects. One is that casing tension
is slightly reduced and the other is that the tension is acting at a
greater angle from the vertical than that of the rest of the tire.
The change in the angle affects the net downward pull of the casing by
the sine of the angle and because the tire has symmetry the horizontal
component has no effect on supporting the rim.
The casing can transmit only tension and thereby cannot deliver an
upward force from the bulging shape of the tire.
>>> How it gets transferred depends somewhat on the rim/tire interface.
>>> For a clincher rim, this imbalance is supported entirely where the
>>> tire wraps around the lip of the rim. The pressure the tire exerts
>>> on the rim is the cord tension divided by the radius of the lip
>>> (plus air pressure, but it's small relative to this), and the net
>>> force exerted is a function of how much of an arc this wraps around:
>>> this arc ending with the ~45 degree exit angle of the casing, which
>>> varies with tire deflection.
So if that is so, how come tubular tires under the same inflation
pressure react identically to a clincher. In fact you cannot tell
whether the tire is a tubular or not without letting the air out and
trying to expose a clincher bead or a base tape. The anchoring of
the clincher bead serves only to span the gap across the open rim,
something the base tape does identically. The recess (bed) in the
clincher rim plays no part in this.
I detect willful misinterpretation. Nowhere did I say the bed of a
tubular rim is MUCH smaller, and it isn't. I explained how much
smaller and why that is so. Effectively the tubular diverges from the
rim identically to a clincher of the same size. What the tires
(clincher or tubular) do in the span between the rim edges has no
effect on the way they support the rim. I suggest you inspect some
tubulars and clinchers and see if you can see any difference. At
InterBike the two types of tires were displayed intermixed and only bu
pressing the sidewall could one type be distinguished from the other.
>>> Where a tire is loaded, the cord tension is less, so the casing is
>>> able to support even less of the air pressure, leaving more
>>> pressure to be conveyed to the rim. Reduction in the casing
>>> tension pushes on the tubular rim bed. The effect of the tire
>>> wrapping around the edges of the rim still applies too.
>> I don't understand how you determined that cord tension is less,
>> although it is so for geometric reasons. What do you mean by
>> conveying pressure to the rim. Inflation pressure against the rim
>> is uniform around the rim circumference. Casing tension is not.
> Tension is the pressure times the radius of the curve, and the
> bulged of the compressed tire is at a smaller radius. This tension
> is what supports the 'hammock' I mention above, and if the tension
> is reduced, the hammock will want to sag against the rim. It either
> sags and bears on the rim with more pressure, or of the ends of the
> hammock are well anchored (be it by glue or friction), then this
> reduced tension gets supported by those anchors.
Let's leave the trampolines, nails, slinkys and toy balloons out of
this. I think the explanation of how the casing angle transmits its
forces to the rim are tersely and unambiguously outlined in:
http://www.sheldonbrown.com/brandt/rim-support.html
and expanded upon in the previous article. I sense that you are
trying to slip in under the circus tent and say that's where you were
all along although you lost your ticket. If you agree with what I
wrote why are you trying to rephrase that in terms that obscure the
effects by a misleading model?
> What's confusing things is again the distinction between the rim
> itself, and the combined unit of rim and the section of tire that's
> directly wedded to it. Air pressure acts uniformly on the casing, but
> that isn't transferred directly to the rim unless the interface is flat
> or the casing is under zero tension. If the interface has a radius,
> then the casing tension on that radius supports some portion of the
> air pressure.
STOP that! I can't take much more of this obfuscation. You are
testing my patience to the utmost.
>>> Friction/glue and casing shear will allow for gradients in the
>>> cord tension. The compression put onto the casing gets
>>> transmitted by friction/shear where the casing first meets the
>>> rim, rather than being carried all the way across the bed (where
>>> thanks to the gluing, the cords instead remain at full tension) to
>>> be transmitted as pressure. These differences are all internal to
>>> the contact area though (much like unrelieved residual stress),
>>> and are net zero on the external picture.
>> I think your model makes invalid assumptions about friction and the
>> balance of forces on a tubular. Glue has nothing to do with this.
>> If you have ridden tubulars you will recall that a new track tire
>> with a bare cloth base tape mounted on a new rim with no glue can
>> be ridden and it supports the rim the same as any clincher by the
>> means that I described at length.
As I mentioned, an unglued track tubular (one with a bare cloth base
tape) mounted on a clean rim has no friction, stiction or other
retention other than casing constriction from inflation (see equations
in "the Bicycle Wheel"):
----------------------------------------------------------------------
8. Constriction Force of Inflated Tire
T = C - E Tire tension from inflation
P = 0.8 MPa Tire pressure
d = 0.025 m (25 mm) Diameter of tire cross section
r = d/2 = 0.0125 m Radius of tire cross section
a = 45 deg Cord angle of casing (45 typical)
A = pi * r^2 Area of tire cross section
C = P * A * 2 * tan^2(a) Constricting component
E = P * A Expanding component
A = pi * 0.0125 = 4.909e-4 m^2
C = 0.8e6 * 4.909e-4 * 2 * 1 = 785.4 N
E = 0.8e6 * 4.909e-4 = 392.7 N
T = 785.4 - 392.7 = 392.7 N
For T = 0 the cord angle must be 35.27 degrees as it is in most hoses.
----------------------------------------------------------------------
> It does matter, but for the internal picture of the rim/tire section
> unit. The differences are of the distribution of internal forces
> within this unit, not of the unsupported tire's relationship to this
> unit.
What means this? I must ask "internal picture", " rim/tire section
unit", "distribution of internal forces", "unsupported tire's
relationship", I'm sure there is enough jargon there to support a
lengthy confusion.
Jobst Brandt
Michael Press
jobst....@stanfordalumni.org wrote:
> Michael Press writes:
>
> >> So how does the force reach the ground from the rim. I see no
> >> reference to the rim in your description. You say the tire casing
> >> is flexible, and I concur, but you don't explain how the load is
> >> transmitted between rim and road, and it is not inflation pressure
> >> that supports the rim because it presses against the rim uniformly
> >> around its circumference. Beyond that, the contact area of the
> >> tire with the rim (clincher or tubular) also remains constant.
>
> > I have not attempted to describe a force path through the casing. I
> > have only described a simple view of how the hub load is balanced by
> > a greater force of air pressure on a wider tire cross-section at the
> > contact patch.
>
> What do you mean by "a greater force of air pressure on a wider tire
> cross-section at the contact patch". Greater than what?
Greater than the force of air pressure at the antipodal
cross-section where the casing is less wide than at the
contact patch.
> The area of
> the contact patch is given by the load divided by inflation pressure,
> nothing more. I think that has been stated here often enough. What
> is " balanced by a greater force of air pressure on a wider tire
> cross-section at the contact patch." Where is the balance and what
> does this have to do with the hub?
>
> I think the matter is getting more obscure by the minute. Let's not
> get the hub and spokes into this or we will be worse off than at the
> beginning of this thread.
No danger of that here, though you have now mentioned it.
--
Michael Press
jobst....@stanfordalumni.org
>>>> So how does the force reach the ground from the rim. I see no
>>>> reference to the rim in your description. You say the tire
>>>> casing is flexible, and I concur, but you don't explain how the
>>>> load is transmitted between rim and road, and it is not inflation
>>>> pressure that supports the rim because it presses against the rim
>>>> uniformly around its circumference. Beyond that, the contact
>>>> area of the tire with the rim (clincher or tubular) also remains
>>>> constant.
>>> I have not attempted to describe a force path through the casing.
>>> I have only described a simple view of how the hub load is
>>> balanced by a greater force of air pressure on a wider tire
>>> cross-section at the contact patch.
>> What do you mean by "a greater force of air pressure on a wider
>> tire cross-section at the contact patch". Greater than what?
> Greater than the force of air pressure at the antipodal
> cross-section where the casing is less wide than at the contact
> patch.
The casing cannot transmit pressure this apparent greater force,
having only tension and that doesn't change appreciably. What changes
is the angle at which the casing departs form the rim. How does this
"greater force" gets to the rim in your perception? The rim is the
same width around its circumference so it cannot be inflation
pressure. All that is left is casing tension and you don't mention
how it affects lift on the rim.
>> The area of the contact patch is given by the load divided by
>> inflation pressure, nothing more. I think that has been stated
>> here often enough. What is " balanced by a greater force of air
>> pressure on a wider tire cross-section at the contact patch."
>> Where is the balance and what does this have to do with the hub? I
>> think the matter is getting more obscure by the minute. Let's not
>> get the hub and spokes into this or we will be worse off than at
>> the beginning of this thread.
> No danger of that here, though you have now mentioned it.
So? How does the tire support the rim... and its not through a larger
area for inflation pressure to lift it.
Jobst Brandt
Michael Press
lu...@mochi.EECS.Berkeley.EDU (Luns Tee) wrote:
I won't.
> You can't nelgect tension!
Thank you, Luns, I am considering tension. The problem
with considering tension is to think about which direction
is to be considered. You wrote
> Michael's description is wrong in saying that the area times the
> pressure accounts for the entire load at the axle though. There's a
> component that comes from the reduction in cord tension in the loaded
> area. Draw a line through the widest part of the tire, and tally up
> all the forces through it - cord tension contributes just as air
> pressure does.
Less tension taken in which direction results in a force
opposing the load at the hub? (compared with the force
resultant at the antipodal portion of the casing)
--
Michael Press
Luns Tee
<jobst....@stanfordalumni.org> wrote:
>I guess this posting was another perspective on the issue so let's
>look at this new proposal.
>
>> We're talking across purposes here, but let it be said that I don't
>> disagree with anything written in that post or in the FAQ. Support
>> at the rim consists of two components: the reduction in tension, and
>> the change in the casing's exit angle where it leaves the rim. What
>> I was intending to convey was that Michael's explanation of air
>> pressure on the extra width of the bulge, is actually identically
>> equivalent to the effect of the casing exit angle (but neglects the
>> change in Tension).
>
>From this I see your free body diagram is amiss. The width of the
>tire when compressed does not change the pressure on the rim. The
>pressure in the tire and that seen by the rim remains unchanged as
>does the width of the rim exposed to that pneumatic force.
My FBD is complete, but what is amiss is your following of
what's being said, as we now have three different problems that we're
talking across.
Draw a cross section of a rim, tire, and the road. Draw a line
across through where the tire just departs the rim. Draw another line
through the widest part of the tire. Everything you've written up to
this point has been about forces through the first line - problem 1.
What Michael has said about pressure on a changing width is about
forces through the second line: the increased width is the width of
the tire's bulge, not the width of the rim - problem 2.
These are different sets of forces, but the point I was trying
to make here is that both sets of forces are the same, since the casing
between these dividing lines is under no net force.
The third problem, which is what I'd been delving into, is what
happens to the forces _above_ the top line. You've been referring to
everything above the point of departure as the rim, but there's still
tire up there too. The forces of the casing - tension times the sine of
the angle - are pulling on that last bit of casing, which has to in turn
transmit it to the rim. It all gets to the rim eventually, where it gets
transferred is less obvious, and so far I seem to be very much alone in
considering it.
>> The exit angle flattening, changes the arc on which air pressure
>> acts, and the air pressure on the extra width of that arc is
>> identical to the change in the vertical component of the casing
>> tension at the end of the arc.
>
>That is not true! There are two effects. One is that casing tension
>is slightly reduced and the other is that the tension is acting at a
>greater angle from the vertical than that of the rest of the tire.
>The change in the angle affects the net downward pull of the casing by
>the sine of the angle and because the tire has symmetry the horizontal
>component has no effect on supporting the rim.
Again, what you've said is not inconsistent with the paragraph
immediately preceding it. The extra width referred to is at the tire's
bulge, not at the rim.
>The casing can transmit only tension and thereby cannot deliver an
>upward force from the bulging shape of the tire.
There IS an upward force! As the tire casing wraps around the
lip of the rim and goes to a more vertical trajectory, there is a
pressure between the casing and that lip which pushes upwards on the
rim. In a clincher, this is countered by a downward force of the
clinch, so the total force is down.
For a tubular, consider resting the rim on a pressurized length
of tygon hose. I hope you'll agree that aside from constriction issues,
the hose is little different from the tire. The hose is still entirely
under tension - how does it support the rim if tensioned casing can't
deliver an upward force? You might think it's relates to air pressure
on the bed of the rim, but consider doing the same experiment with a
clincher rim: the hose can support that just as well even though
contact is only on the lips of the rim.
>>>> How it gets transferred depends somewhat on the rim/tire interface.
>
>So if that is so, how come tubular tires under the same inflation
>pressure react identically to a clincher. In fact you cannot tell
>whether the tire is a tubular or not without letting the air out and
>trying to expose a clincher bead or a base tape.
The differences are all internal to what happens above the point
of departure. Below that, where it's casing-only and no rim, everything
is identical.
>>>The difference for tubulars is that the rim curvature is slightly
>>>smaller than the tire so it rests on the edges of the contact with the
>>>rim, the rest of the area filled with rim glue. This contact is
>>>plastic and squirms fore and aft as the tire rolls, gradually wearing
>>>through the base tape by fretting. This does not affect how the tire
>>>supports the rim. For that purpose the rim sees only inflation
>>>pressure.
>
>> If the curvature the tire meets is indeed a much smaller radius than
>> the tire itself, then the load would be entirely on the edges of
>> contact, and the bed of the reduced curvature area actually wouldn't
>> see much of the air pressure, if any at all! The air pressure bears
>> on the tire casing, but the casing is suspended between those edges
>> like a hammock - picture a tufo tire - rather than being supported
>> by the rim bed.
>
>I detect willful misinterpretation. Nowhere did I say the bed of a
>tubular rim is MUCH smaller, and it isn't. I explained how much
>smaller and why that is so. Effectively the tubular diverges from the
>rim identically to a clincher of the same size. What the tires
>(clincher or tubular) do in the span between the rim edges has no
>effect on the way they support the rim. I suggest you inspect some
>tubulars and clinchers and see if you can see any difference. At
>InterBike the two types of tires were displayed intermixed and only bu
>pressing the sidewall could one type be distinguished from the other.
Okay, perhaps I shouldn't have used the word 'indeed', but
there's a subtle point I was trying to make, which I was trying to
emphasize. Air pressure does not bear directly on the rim bed for a
tubular - it bears on the casing, and the pressure of the casing
pressing on the rim bed is less than the air pressure. How much less is
a function of the geometry - in the extreme case (Tufo), there's no
pressure at all. If the thickness of the base tape, and volume of the
glue present a a radius that's the same as the tire's natural radius,
there's still no pressure. It's not until the contact is flatter than
that (and the thickness of the base tape may be enough to accomplish
this) that pressure is felt by the bed.
>> Tension is the pressure times the radius of the curve, and the
>> bulged of the compressed tire is at a smaller radius. This tension
>> is what supports the 'hammock' I mention above, and if the tension
>> is reduced, the hammock will want to sag against the rim. It either
>> sags and bears on the rim with more pressure, or of the ends of the
>> hammock are well anchored (be it by glue or friction), then this
>> reduced tension gets supported by those anchors.
>
>Let's leave the trampolines, nails, slinkys and toy balloons out of
>this. I think the explanation of how the casing angle transmits its
>forces to the rim are tersely and unambiguously outlined in:
>
>http://www.sheldonbrown.com/brandt/rim-support.html
>
>and expanded upon in the previous article. I sense that you are
>trying to slip in under the circus tent and say that's where you were
>all along although you lost your ticket. If you agree with what I
>wrote why are you trying to rephrase that in terms that obscure the
>effects by a misleading model?
What you've written describes how forces get to the casing at
its point of departure from the rim, but the casing there is still
casing and not the rim itself. What I've been writing about is how
that last bit of casing transmits these forces to the rim. So far, as
far as I can see, not a word you've written has touched on this.
>> What's confusing things is again the distinction between the rim
>> itself, and the combined unit of rim and the section of tire that's
>> directly wedded to it. Air pressure acts uniformly on the casing, but
>> that isn't transferred directly to the rim unless the interface is flat
>> or the casing is under zero tension. If the interface has a radius,
>> then the casing tension on that radius supports some portion of the
>> air pressure.
>
>STOP that! I can't take much more of this obfuscation. You are
>testing my patience to the utmost.
Do you object to any of what I actually said in that sentance?
My point in saying that is that air pressure in a tubular isn't
necessarily the pressure felt on the rim bed. What pressure does a spoke
hole feel?
>> It does matter, but for the internal picture of the rim/tire section
>> unit. The differences are of the distribution of internal forces
>> within this unit, not of the unsupported tire's relationship to this
>> unit.
>
>What means this? I must ask "internal picture", " rim/tire section
>unit", "distribution of internal forces", "unsupported tire's
>relationship", I'm sure there is enough jargon there to support a
>lengthy confusion.
It means that what you call the rim in all your analysis is not
just the rim, but the rim together with the portion of tire that's in
contact with it. The forces from the point of departure aren't transferred
along the infinitessimal strip of departure, but are carried through to
the rim across a somewhat broader area. In the case of the tubular, if
there's no friction or glue (I did say IF), if the tire actually makes
contact to the rim bed, some of it even gets into there.
-Luns
Luns Tee
<jobst....@stanfordalumni.org> wrote:
>Luns Tee writes:
>
>>>> What Michael describes isn't inaccurate, although it's incomplete.
>>>> The net force of air pressure acting on the full width of the tire
>>>> at the bulge is greater than the force of air pressure on the rim
>>>> bed, and the difference of this force must be transferred the rim
>>>> somewhere.
>
>>> It is complete. I explicitly announced that I intended to neglect
>>> stress in the casing. I entered into this thread responding to Tim
>>> McNamara in:
>
>> You can't neglect stress in the casing. Casing tension is
>> determined by the air pressure (constant) and the radius the casing
>> bends around. This radius is smaller in the bulge than it is in the
>> undisturbed tire, and thus the tension is lower at the bulge. This
>> change in the tension contributes to the force at the axle.
>
>> Air pressure on the broader width of the tire's bulge does
>> contribute to supporting the load at the axle, but is only one
>> component of that load - it is NOT the complete picture.
>
>Please explain how the width of the tire (the bulging under load)
>changes the net force on the rim (let's leave axle, hub, bearings and
>spokes out of this). As is apparent, the rim does not get wider from
>loading so its contact with the tire does not change, at least not more
>than a few thousandths of an inch, the bead having a radius of about
The extra width of tire means more air pressure is pressing
upwards on the arc casing between the widest part of the bulge, and
where the tire meets the rim - the segment between the two lines in my
post from a moment ago. The casing has to then transmit this force to
the rim, but that's what Michael was decidedly NOT writing about.
Jobst,
Not all my writing is addressed to you. If you'll follow the
attributions, this posting that you're responding to was my reply to
Michael Press' article, not yours, in which he stated himself that he is
neglecting tension changes. The venture to the absurd structures was
exploring the consequences of assuming constant tension, and was only
intended to highlight the folly of that assumption. It was not meant to
describe how a tire supports the rim, but rather to show that assuming
constant tension casing indeed _cannot_ support the rim.
-Luns
Luns Tee
Less tension in the casing in the loaded area behaves the same
as less tension in the spokes in that area - I believe you're familiar
with the latter concept.
The difference is that the load path at the spokes is the spokes
alone. At the bulge of the tire, there's two parallel load paths: the
(pretensioned, vertical) casing - analogous to how spokes work - and
the varying width of the air chamber - which you've described. The
load path where the tire meets the rim is the tire alone, but there,
both the tension and angle contribute.
-Luns
Tim McNamara
as one often does during a bike ride, my thoughts turned to this
discussion. As I am currently understanding it, the loading of the tire
against the road deforms it until the area of contact patch reaches an
equilibrium with the tire pressure. A 100 pound load on a tire inflated
to 10o psi would result in a contact patch of one square inch. At the
contact patch, the casing tension is locally reduced due to the change
in curvature caused by bulging out under the load.
OK. So this may be a useless tangent, but I was reminded of drag racing
slicks- big, low pressure tires- which wrinkle along the sidewall
between the rim and the ground (until the light turns green, anyway).
That seems to me to me a pretty graphic demonstration of the decrease in
tension in the casing that's easy to see. Correct?
jobst....@stanfordalumni.org
>> I guess this posting was another perspective on the issue so let's
>> look at this new proposal.
>>> We're talking across purposes here, but let it be said that I
>>> don't disagree with anything written in that post or in the FAQ.
>>> Support at the rim consists of two components: the reduction in
>>> tension, and the change in the casing's exit angle where it leaves
>>> the rim. What I was intending to convey was that Michael's
>>> explanation of air pressure on the extra width of the bulge, is
>>> actually identically equivalent to the effect of the casing exit
>>> angle (but neglects the change in Tension).
>> From this I see your free body diagram is amiss. The width of the
>> tire when compressed does not change the pressure on the rim. The
>> pressure in the tire and that seen by the rim remains unchanged as
>> does the width of the rim exposed to that pneumatic force.
> My FBD is complete, but what is amiss is your following of what's
> being said, as we now have three different problems that we're
> talking across.
> Draw a cross section of a rim, tire, and the road. Draw a line
> across through where the tire just departs the rim. Draw another line
> through the widest part of the tire. Everything you've written up to
> this point has been about forces through the first line - problem 1.
Let's do that and at the same time realize that this could just as
well be solid aluminum rim from that line to its inner circumference
for the purposes of analyzing what supports the rim. Whether there is
an air chamber or an empty deep-V rim in there, or even a tubular
tire, has no bearing on this analysis.
> What Michael has said about pressure on a changing width is about
> forces through the second line: the increased width is the width of
> the tire's bulge, not the width of the rim - problem 2.
The second line is only remotely connected to the problem. The width
of the bulge does not affect the upward force on the rim other than
change the angle at which the casing separates from the rim, be that a
tire with a steel bead wire, a tubular or the clincher. It is the
effect of that angle that determines how much vertical force the tire
exerts on the rim. The area of the belly of the tire is a secondary
effect of the problem.
> These are different sets of forces, but the point I was trying to
> make here is that both sets of forces are the same, since the casing
> between these dividing lines is under no net force.
I don't understand what you mean by "no net force". The casing is
under tension defined by inflation pressure and casing curvature.
> The third problem, which is what I'd been delving into, is what
> happens to the forces _above_ the top line. You've been referring
> to everything above the point of departure as the rim, but there's
> still tire up there too. The forces of the casing - tension times
> the sine of the angle - are pulling on that last bit of casing,
> which has to in turn transmit it to the rim. It all gets to the rim
> eventually, where it gets transferred is less obvious, and so far I
> seem to be very much alone in considering it.
Nothing occurs above the interface between the tire casing and the
rim. That is entirely like leaving foreign objects reside inside the
air chamber, basically what the bead of the tire is with respect to
the pneumatic suspension the tire furnishes. That is why I suggest an
unglued tubular tire in my example. It operates the same way and
cannot be detected as a tubular except for the label.
>>> The exit angle flattening, changes the arc on which air pressure
>>> acts, and the air pressure on the extra width of that arc is
>>> identical to the change in the vertical component of the casing
>>> tension at the end of the arc.
It's the reverse. Flattening of the tire changes the angle and that
change is dependent on inflation pressure which describes casing
tension.
>> That is not true! There are two effects. One is that casing
>> tension is slightly reduced and the other is that the tension is
>> acting at a greater angle from the vertical than that of the rest
>> of the tire. The change in the angle affects the net downward pull
>> of the casing by the sine of the angle and because the tire has
>> symmetry the horizontal component has no effect on supporting the
>> rim.
> Again, what you've said is not inconsistent with the paragraph
> immediately preceding it. The extra width referred to is at the
> tire's bulge, not at the rim.
I don't see the similarity but if that is your perception then why are
you disagreeing with the analysis I gave?
>> The casing can transmit only tension and thereby cannot deliver an
>> upward force from the bulging shape of the tire.
> There IS an upward force! As the tire casing wraps around the
> lip of the rim and goes to a more vertical trajectory, there is a
> pressure between the casing and that lip which pushes upward on the
> rim. In a clincher, this is countered by a downward force of the
> clinch, so the total force is down.
There is only tension at that place, and contact pressure is the same
as it is around the entire rim other than the slight decrease in
casing tension from reduced curvature. There is no more net force
there than anywhere else around the rim. The difference is in the
angle at which the tire casing is pulling away from the contact line.
If you separate the vertical and horizontal components of that force
you will find a reduced downward force in the rim section over the
contact patch with the road.
> For a tubular, consider resting the rim on a pressurized length of
> Tygon hose. I hope you'll agree that aside from constriction
> issues, the hose is little different from the tire. The hose is
> still entirely under tension - how does it support the rim if
> tensioned casing can't deliver an upward force? You might think it's
> relates to air pressure on the bed of the rim, but consider doing
> the same experiment with a clincher rim: the hose can support that
> just as well even though contact is only on the lips of the rim.
Stop explaining to me what I just told you. Are you trying to put me
in your position. I explained how it does that and that it is
identical to an actual bicycle tire of the same cross section. I've
seen this line of argumentation where the dissenter starts explaining
how things work using the other's words. We are getting close to that
position. As I said, it reminds me of someone sneaking under the edge
of a circus tent and claiming he had been there all the time.
>>>>> How it gets transferred depends somewhat on the rim/tire interface.
>> So if that is so, how come tubular tires under the same inflation
>> pressure react identically to a clincher. In fact you cannot tell
>> whether the tire is a tubular or not without letting the air out
>> and trying to expose a clincher bead or a base tape.
> The differences are all internal to what happens above the point of
> departure. Below that, where it's casing-only and no rim,
> everything is identical.
Well???? What is it you are getting at? What is different and what
has it got to do with how the tire supports the rim. What do you mean
"the differences are all internal"? What are these "differences" with
respect to load transfer?
>>>> The difference for tubulars is that the rim curvature is slightly
>>>> smaller than the tire so it rests on the edges of the contact
>>>> with the rim, the rest of the area filled with rim glue. This
>>>> contact is plastic and squirms fore and aft as the tire rolls,
>>>> gradually wearing through the base tape by fretting. This does
>>>> not affect how the tire supports the rim. For that purpose the
>>>> rim sees only inflation pressure.
>>> If the curvature the tire meets is indeed a much smaller radius than
>>> the tire itself, then the load would be entirely on the edges of
>>> contact, and the bed of the reduced curvature area actually wouldn't
>>> see much of the air pressure, if any at all! The air pressure bears
>>> on the tire casing, but the casing is suspended between those edges
>>> like a hammock - picture a Tufo tire - rather than being supported
>>> by the rim bed.
>> I detect willful misinterpretation. Nowhere did I say the bed of a
>> tubular rim is MUCH smaller, and it isn't. I explained how much
>> smaller and why that is so. Effectively the tubular diverges from
>> the rim identically to a clincher of the same size. What the tires
>> (clincher or tubular) do in the span between the rim edges has no
>> effect on the way they support the rim. I suggest you inspect some
>> tubulars and clinchers and see if you can see any difference. At
>> InterBike the two types of tires were displayed intermixed and only
>> bu pressing the sidewall could one type be distinguished from the
>> other.
> OK, perhaps I shouldn't have used the word 'indeed', but
> there's a subtle point I was trying to make, which I was trying to
> emphasize.
You shouldn't have used the words MUCH SMALLER. That's a gross
exaggeration that tries to discredit what I said... but doesn't.
> Air pressure does not bear directly on the rim bed for a
> tubular - it bears on the casing, and the pressure of the casing
> pressing on the rim bed is less than the air pressure.
That is why I said the FBD line across the rim-to-tire contact could
just as well be the line of a solid rim with no recess. What occurs
between those two casing to rim contact interfaces has no bearing on
the problem. As I said, there could be a large void in the rim and
have no effect on the problem.
> How much less is a function of the geometry - in the extreme case
> (Tufo), there's no pressure at all. If the thickness of the base
> tape, and volume of the glue present a a radius that's the same as
> the tire's natural radius, there's still no pressure. It's not
> until the contact is flatter than that (and the thickness of the
> base tape may be enough to accomplish this) that pressure is felt by
> the bed.
STOP!
As I said, an unglued track tire on a clean rim responds the same as
the clincher. Or even a mylar torus resting on a clincher rim. All
this glue and rim shape between the edges where the tire rests, is a
smoke screen and has no effect on the problem.
>>> Tension is the pressure times the radius of the curve, and the
>>> bulged of the compressed tire is at a smaller radius. This
>>> tension is what supports the 'hammock' I mention above, and if the
>>> tension is reduced, the hammock will want to sag against the rim.
>>> It either sags and bears on the rim with more pressure, or of the
>>> ends of the hammock are well anchored (be it by glue or friction),
>>> then this reduced tension gets supported by those anchors.
>> Let's leave the trampolines, nails, slinkys and toy balloons out of
>> this. I think the explanation of how the casing angle transmits its
>> forces to the rim are tersely and unambiguously outlined in:
http://www.sheldonbrown.com/brandt/rim-support.html
>> and expanded upon in the previous article. I sense that you are
>> trying to slip in under the circus tent and say that's where you
>> were all along although you lost your ticket. If you agree with
>> what I wrote why are you trying to rephrase that in terms that
>> obscure the effects by a misleading model?
> What you've written describes how forces get to the casing at its
> point of departure from the rim, but the casing there is still
> casing and not the rim itself. What I've been writing about is how
> that last bit of casing transmits these forces to the rim. So far,
> as far as I can see, not a word you've written has touched on this.
Well??? Get with it and explain what you find lacking in the analysis
I presented. You keep alluding to mysterious "The differences are all
internal to what happens above the point of departure." What is it
that occurs down there that affects supporting the rim from the road?
>>> What's confusing things is again the distinction between the rim
>>> itself, and the combined unit of rim and the section of tire that's
>>> directly wedded to it. Air pressure acts uniformly on the casing, but
>>> that isn't transferred directly to the rim unless the interface is flat
>>> or the casing is under zero tension. If the interface has a radius,
>>> then the casing tension on that radius supports some portion of the
>>> air pressure.
>> STOP that! I can't take much more of this obfuscation. You are
>> testing my patience to the utmost.
> Do you object to any of what I actually said in that sentence?
> My point in saying that is that air pressure in a tubular isn't
> necessarily the pressure felt on the rim bed. What pressure does a spoke
> hole feel?
Yes. It has nothing to do with supporting the rim. Its as if I were
to say that the grease in the bearings had an effect on tire
deflection but that it is inside the hub... or the like. What is this
"distinction between rim and tire section that's directly wedded to
it." that sounds as much like newspeak as I can imagine. You invent
these terms and don't explain what their function is other than to
obfuscate.
>>> It does matter, but for the internal picture of the rim/tire
>>> section unit. The differences are of the distribution of internal
>>> forces within this unit, not of the unsupported tire's
>>> relationship to this unit.
>> What means this? I must ask "internal picture", " rim/tire section
>> unit", "distribution of internal forces", "unsupported tire's
>> relationship", I'm sure there is enough jargon there to support a
>> lengthy confusion.
> It means that what you call the rim in all your analysis is not just
> the rim, but the rim together with the portion of tire that's in
> contact with it. The forces from the point of departure aren't
> transferred along the infinitesimal strip of departure, but are
> carried through to the rim across a somewhat broader area. In the
> case of the tubular, if there's no friction or glue (I did say IF),
> if the tire actually makes contact to the rim bed, some of it even
> gets into there.
You are incorrect there. The load transfer occurs at the departure
point of the tire from the rim to such a small degree that it is less
than the tire wall thickness. It's like a string coming off a pulley.
The departure point defines the effective point of loading, the
remainder of the contact with the pulley being immaterial because it
could be a drum with many wraps of the string or just a turning
point. What is behind the contact point is immaterial to the span of
the string.
Where the casing is terminated is also immaterial to rim support. It
could be a clincher with a hooked bead, one with a straight bead and a
wire in its edge or a tubular sewn shut to make a hose. That is why a
toroidal mylar tube can substitute for the tire and display the same
characteristics as it supports the rim. It also makes no difference
whether that mylar tube is sitting on a clincher rim or a tubular rim
with a bed that exactly matches its minor diameter.
Jobst Brandt
carl...@comcast.net
Dear Tim,
I'm not sure, but I think that you're mistaken about when low-pressure
wrinkle-wall tires wrinkle--as I understand it, the wrinkling comes
after the green light, not before it.
In drag racing, the driving rim turns so hard against the traction of
the tire that the rim starts to wrap the tire around around itself.
This is actually considered a good thing--wrinkle wall tires have
unusually thin sidewalls (too fragile for street use) run with low
pressure.
When the tires twist up like this (also called wadding), they act as a
shock absorber between the engine and the ground and help to prevent
the tire from breaking loose at the drag strip starting line.
Here's a typical description:
"Slicks for drag racing have a super sticky surface, but the sidewalls
are made thin, to "wind up" as the vehicle accelerates, absorbing some
of the shock of launching. Because of this, they get the name
"wrinkle-walls" because the sidewalls actually wrinkle. The sidewalls
are so thin, in fact, that they should not be driven on the street at
all. Brush a curb, and you may need a new tire."
http://www.geocities.com/g_wellwood/automotive/acceleration.html
Here's a still picture that shows the spiral wrinkling:
http://www.vetteweb.com/features/vemp_0611w_corvettes_nostalgia_drag_racing/photo_08.html
or http://tinyurl.com/pwg6g
And here's a super-slow motion video of spiral wrinkles, called
wadding in the audio:
http://www.youtube.com/watch?v=9O0b90G8Yhg
Again, these wrinkles are after the green light and show a torsional
increase in tension.
But you seem to be talking about wrinkles before the green light,
which wouldn't be spiral, so maybe I'm misunderstanding you.
Cheers,
Carl Fogel
Michael Press
jobst....@stanfordalumni.org wrote:
I know that the rim does not change width. I described the
picture of the air pressure effects, and stand by it. I
was incorrect thinking that the static picture could be
interpreted solely in terms of air pressure.
> >> The area of the contact patch is given by the load divided by
> >> inflation pressure, nothing more. I think that has been stated
> >> here often enough. What is " balanced by a greater force of air
> >> pressure on a wider tire cross-section at the contact patch."
> >> Where is the balance and what does this have to do with the hub? I
> >> think the matter is getting more obscure by the minute. Let's not
> >> get the hub and spokes into this or we will be worse off than at
> >> the beginning of this thread.
>
> > No danger of that here, though you have now mentioned it.
>
> So? How does the tire support the rim... and its not through a larger
> area for inflation pressure to lift it.
Luns knows what I was describing and has helped me correct
my misapprehension. I will not attempt to paraphrase what
he is saying.
--
Michael Press