Thumb test

In article <[email hidden]>,

Tim McNamara said:

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.

Tim McNamara said:

Quoted message said:

Quoted message said:

>>> 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

Quoted message said:

Quoted message said:

Quoted message said:

>>> "[...] unit casing tension is equivalent to inflation pressure
>>> times the radius of curvature divided by pi [...]".

Quoted message said:

Quoted message said:

Quoted message said:

>>> 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.

Quoted message said:

Quoted message said:

Quoted message said:

>> 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.

Quoted message said:

Quoted message said:

Quoted message said:

> 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.

Quoted message said:

Quoted message said:

Quoted message said:

> 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.

Quoted message said:

Quoted message said:

Quoted message said:

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:

Quoted message said:

Quoted message said:

Quoted message said:

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?

Quoted message said:

Quoted message said:

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.

Quoted message said:

Quoted message said:

Quoted message said:

2. When I am on my bike, is it casing tension or inflation
pressure that holds my rims off the ground?

Quoted message said:

Quoted message said:

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

Quoted message said:

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

On 05 Oct 2006 00:47:35 GMT, [email hidden] wrote:

[snip]

Quoted message said:

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

Michael Press said:

In article <[email hidden]>,

Ben C said:

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

I nominate this for Post of the Month.

--
Tom Sherman - Here, not there.

Tim McNamara said:

In article <[email hidden]>,

Tim McNamara said:

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.

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

In article <[email hidden]>,

Quoted message said:

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.

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.

In article
<[email hidden]>,

Quoted message said:

Michael Press said:

In article
<[email hidden]>,

Tim McNamara said:

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.

The "Avocet Duro" *is* a Fasgrip, one that is 28mm. Different widths
have different names.

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

In article <[email hidden]>,

Quoted message said:

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

In article <[email hidden]>,

(Luns Tee) said:

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
<[email hidden]>

Quoted message said:

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.
<[email hidden]>

--
Michael Press

Tim McNamara said:

Quoted message said:

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.

Quoted message said:

Quoted message said:

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.

Quoted message said:

Quoted message said:

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.

Quoted message said:

Quoted message said:

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.

Quoted message said:

Quoted message said:

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]

Quoted message said:

Quoted message said:

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.

Quoted message said:

Quoted message said:

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.

Quoted message said:

Quoted message said:

[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.]

Quoted message said:

Quoted message said:

The two effects, lower casing tension and broader angle reduces
downward pull on the rim. The rim stands on the tire!

Quoted message said:

Quoted message said:

Remember? That's how this subject got introduced a while ago.

Quoted message said:

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.

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

In article <[email hidden]>,

Quoted message said:

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

In article <[email hidden]>,

Quoted message said:

Tim McNamara said:

Quoted message said:

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.

Quoted message said:

Quoted message said:

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.

Quoted message said:

Quoted message said:

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.

Quoted message said:

Quoted message said:

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.

Quoted message said:

Quoted message said:

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]

Quoted message said:

Quoted message said:

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.

Quoted message said:

Quoted message said:

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.

Quoted message said:

Quoted message said:

[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.]

Quoted message said:

Quoted message said:

The two effects, lower casing tension and broader angle reduces
downward pull on the rim. The rim stands on the tire!

Quoted message said:

Quoted message said:

Remember? That's how this subject got introduced a while ago.

Quoted message said:

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.

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.

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.

Michael Press said:

Quoted message said:

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.

Quoted message said:

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

"Michael Press" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

In article <[email hidden]>,

Quoted message said:

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, some great writing indeed!

JL

"A foolish consistency is the hobgoblin of little minds." Emerson

In article <[email hidden]>,

Michael Press said:

Quoted message said:

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

In article <[email hidden]>,

Quoted message said:

Michael Press said:

Quoted message said:

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.

Quoted message said:

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.

Quoted message said:

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

Michael Press said:

Quoted message said:

Quoted message said:

> 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.

Quoted message said:

Quoted message said:

Quoted message said:

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.

Quoted message said:

Quoted message said:

What do you mean by "a greater force of air pressure on a wider
tire cross-section at the contact patch". Greater than what?

Quoted message said:

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.

Quoted message said:

Quoted message said:

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.

Quoted message said:

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

In article <[email hidden]>,

(Luns Tee) said:

In article <[email hidden]>,

Michael Press said:

Quoted message said:

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!

I won't.

Quoted message said:

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

Quoted message said:

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

So, as I was out riding this evening and pondering a number of things,
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?

Tim McNamara said:

So, as I was out riding this evening and pondering a number of things,
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?

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:

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