Tire puzzler

Quoted message said:

An oldie but goodie:

A bicycle literally rides on a cushion of air; the air being contained
in pressurized rubber compartments. When the tire contacts the ground,
the load is distributed evenly over a small area, sometimes referred to
as the contact patch. The air pressure inside the tire increases
slightly from the applied load but it increases everywhere equally
inside the tire, not just at the contact patch. In other words, the
air pressure exerted against the rim is exactly the same 360 degrees
around its circumference. How does the vertical load get transferred
from the rim to the tire?

Hint: this is more of a thought exercise; the answer can be found in
the question.

The bottom spokes actually support the rim ;-)

D'ohBoy

Quoted message said:

An oldie but goodie:

A bicycle literally rides on a cushion of air; the air being contained
in pressurized rubber compartments. When the tire contacts the ground,
the load is distributed evenly over a small area, sometimes referred to
as the contact patch. The air pressure inside the tire increases
slightly from the applied load but it increases everywhere equally
inside the tire, not just at the contact patch. In other words, the
air pressure exerted against the rim is exactly the same 360 degrees
around its circumference. How does the vertical load get transferred
from the rim to the tire?

Hint: this is more of a thought exercise; the answer can be found in
the question.

Dear J.,

http://www.sheldonbrown.com/brandt/rim-support.html

Cheers,

Carl Fogel

Quoted message said:

A bicycle literally rides on a cushion of air; the air being contained
in pressurized rubber compartments. When the tire contacts the ground,
the load is distributed evenly over a small area, sometimes referred to
as the contact patch. The air pressure inside the tire increases
slightly from the applied load but it increases everywhere equally
inside the tire, not just at the contact patch. In other words, the
air pressure exerted against the rim is exactly the same 360 degrees
around its circumference. How does the vertical load get transferred
from the rim to the tire?

Because, as others have clearly explained, a tensioned element cannot
support a compressive load, it cannot be through the sidewalls of the
lower half of the tire. Obviously, then, the rim must be supported by
the tension of the sidewalls in the upper of the tire. Just draw the
force diagram.

--
Joe [snicker] Riel

Joe Riel said:

Quoted message said:

A bicycle literally rides on a cushion of air; the air being contained
in pressurized rubber compartments. When the tire contacts the ground,
the load is distributed evenly over a small area, sometimes referred to
as the contact patch. The air pressure inside the tire increases
slightly from the applied load but it increases everywhere equally
inside the tire, not just at the contact patch. In other words, the
air pressure exerted against the rim is exactly the same 360 degrees
around its circumference. How does the vertical load get transferred
from the rim to the tire?

Because, as others have clearly explained, a tensioned element cannot
support a compressive load, it cannot be through the sidewalls of the
lower half of the tire. Obviously, then, the rim must be supported by
the tension of the sidewalls in the upper of the tire. Just draw the
force diagram.

--
Joe [snicker] Riel

Dear Joe,

At last I've caught you out!

Your theory obviously neglects the crucial contribution of the inner
tube, whose 2 mm of stout--nay, incompressible!--rubber braces the
sidewalls.

Cheers,

Carl Fogel

Quoted message said:

Quoted message said:

Because, as others have clearly explained, a tensioned element cannot
support a compressive load, it cannot be through the sidewalls of the
lower half of the tire. Obviously, then, the rim must be supported by
the tension of the sidewalls in the upper of the tire. Just draw the
force diagram.

--
Joe [snicker] Riel

Dear Joe,

At last I've caught you out!

Your theory obviously neglects the crucial contribution of the inner
tube, whose 2 mm of stout--nay, incompressible!--rubber braces the
sidewalls.

I just performed an experiment in the lab that appears to confirm your
hypothesis. Removing the inner tube caused the hub to sink towards
the floor. It's all in the tube.

--
Joe Riel

Quoted message said:

Dear J.,

http://www.sheldonbrown.com/brandt/rim-support.html

Cheers,

Carl Fogel

I take it that since the above link was posted absent any qualifying
remarks that you support the position presented therein, or at least
are satisfied with the explanation.

Anyway, thanks for the link. It does provide an interesting point of
view, albeit in a somewhat familiar kind of way.

JL

In article
<[email hidden]>,

Quoted message said:

An oldie but goodie:

A bicycle literally rides on a cushion of air; the air being contained
in pressurized rubber compartments. When the tire contacts the ground,
the load is distributed evenly over a small area, sometimes referred to
as the contact patch. The air pressure inside the tire increases
slightly from the applied load but it increases everywhere equally
inside the tire, not just at the contact patch. In other words, the
air pressure exerted against the rim is exactly the same 360 degrees
around its circumference. How does the vertical load get transferred
from the rim to the tire?

The force along the normal vector to an area patch is
equal to the area times the pressure.
The contact patch has more area normal to the rim than any
other comparable section of the tire.

--
Michael Press

jlulm post anonymously:

Quoted message said:

An oldie but goodie:

Quoted message said:

A bicycle literally rides on a cushion of air; the air being
contained in pressurized rubber compartments. When the tire contacts
the ground, the load is distributed evenly over a small area,
sometimes referred to as the contact patch. The air pressure inside
the tire increases slightly from the applied load but it increases
everywhere equally inside the tire, not just at the contact patch.
In other words, the air pressure exerted against the rim is exactly
the same 360 degrees around its circumference. How does the
vertical load get transferred from the rim to the tire?

Quoted message said:

Hint: this is more of a thought exercise; the answer can be found in
the question.

I don't believe the answer is not found in the question but I am
surprised that this was brought up at the same time that I mentioned
it in the spoke load thread. However, it is the same concept.

Jobst Brandt

jlulm post anonymously:

Quoted message said:

An oldie but goodie:

Quoted message said:

A bicycle literally rides on a cushion of air; the air being
contained in pressurized rubber compartments. When the tire contacts
the ground, the load is distributed evenly over a small area,
sometimes referred to as the contact patch. The air pressure inside
the tire increases slightly from the applied load but it increases
everywhere equally inside the tire, not just at the contact patch.
In other words, the air pressure exerted against the rim is exactly
the same 360 degrees around its circumference. How does the
vertical load get transferred from the rim to the tire?

Quoted message said:

Hint: this is more of a thought exercise; the answer can be found in
the question.

I don't believe the answer is found in the question but I am surprised
that this was brought up at the same time that I mentioned it in the
spoke load thread. However, it is the same concept.

Jobst Brandt

Quoted message said:

I don't believe the answer is found in the question...

That's what makes it a puzzler.

Quoted message said:

but I am surprised that this was brought up at the same time
that I mentioned it in the spoke load thread.

Actually this topic was posted several hours before you mentioned it in
the spoke thread.

JL

D'ohBoy said:

Quoted message said:

An oldie but goodie:

A bicycle literally rides on a cushion of air; the air being contained
in pressurized rubber compartments. When the tire contacts the ground,
the load is distributed evenly over a small area, sometimes referred to
as the contact patch. The air pressure inside the tire increases
slightly from the applied load but it increases everywhere equally
inside the tire, not just at the contact patch. In other words, the
air pressure exerted against the rim is exactly the same 360 degrees
around its circumference. How does the vertical load get transferred
from the rim to the tire?

Hint: this is more of a thought exercise; the answer can be found in
the question.

The bottom spokes actually support the rim ;-)

Nono, it hangs from the top spoke!

It is transferred along the entire circumference of the rim, I would
think. Therefore the weight would be supported by the entire rim.

- -
Comments and opinions compliments of,
"Your Friendly Neighborhood Wheelman"

My web Site:
http://geocities.com/czcorner

To E-mail me:
ChrisZCorner "at" webtv "dot" net

Chris who? said:

It is transferred along the entire circumference of the rim, I would
think. Therefore the weight would be supported by the entire rim.

What is "it" and how does "it" accomplish this... where does the reach
the ground from the rim? I ask because your response is an entirely
vague one that I haven't heard before.

Jobst Brandt

Quoted message said:

An oldie but goodie:

A bicycle literally rides on a cushion of air; the air being contained
in pressurized rubber compartments. When the tire contacts the ground,
the load is distributed evenly over a small area, sometimes referred to
as the contact patch. The air pressure inside the tire increases
slightly from the applied load but it increases everywhere equally
inside the tire, not just at the contact patch. In other words, the
air pressure exerted against the rim is exactly the same 360 degrees
around its circumference. How does the vertical load get transferred
from the rim to the tire?

Hint: this is more of a thought exercise; the answer can be found in
the question.

Are there any adults in this thread who would care to discuss the
puzzler?
Dave

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

Quoted message said:

http://www.sheldonbrown.com/brandt/rim-support.html

Does this FAQ item not answer your question.

Jobst Brandt

I don't see how the referenced article explains the mechanics of how a load
is actually carried by a pneumatic tire.

The article states:

"Under load, in the ground contact zone, the tire bulges so that two effects
reduce the downward pull (increase the net upward force) of the casing.
First, the most obvious one is that the casing pulls more to the sides than
downward (than it did in its unloaded condition); the second is that the
side wall tension is reduced. The reduction arises from the relationship
that unit casing tension is equivalent to inflation pressure times the
radius of curvature divided by pi. As the curvature reduces when the tire
bulges out, the casing tension decreases correspondingly. The inflated tire
supports the rim primarily by these two effects."

Here's why the explanation doesn't satisfy the question:

"First, the most obvious one is that the casing pulls more to the sides than
downward (than it did in its unloaded condition)"

The tire casing (and its pull on the rim) can be taken completely out of the
picture. Consider a rim with just the inner tube installed on it. Slowly
pump the tube to say 10 psi and observe how the rim lifts off the ground as
the weight of the bicycle is taken up by the increased air pressure inside
the tube. The tube is not able to pull on the rim.

"the second is that the side wall tension is reduced. The reduction arises
from the relationship that unit casing tension is equivalent to inflation
pressure times the radius of curvature divided by pi. As the curvature
reduces when the tire bulges out, the casing tension decreases
correspondingly."

So what? The side wall tension can also be taken completely out of the
picture using the solo inner tube example from above; i.e., there is no pull
on the rim from inner tube side wall tension. If you need to convince
yourself of this, envision a layer of Vaseline between the tube and rim.

"The inflated tire supports the rim primarily by these two effects."

While perhaps describing some physical attributes of tire casings under
load, these don't explain the phenomenon of what's actually supporting the
load. In the above example something besides tire casing pull and side wall
tension is holding up the rim. The same mechanics that are at work holding
up the rim in the solo inner tube example come into play in all pneumatic
tires.

JL

JL said:

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

Quoted message said:

http://www.sheldonbrown.com/brandt/rim-support.html

Does this FAQ item not answer your question.

Jobst Brandt

I don't see how the referenced article explains the mechanics of how a load
is actually carried by a pneumatic tire.

The article states:

"Under load, in the ground contact zone, the tire bulges so that two effects
reduce the downward pull (increase the net upward force) of the casing.
First, the most obvious one is that the casing pulls more to the sides than
downward (than it did in its unloaded condition); the second is that the
side wall tension is reduced. The reduction arises from the relationship
that unit casing tension is equivalent to inflation pressure times the
radius of curvature divided by pi. As the curvature reduces when the tire
bulges out, the casing tension decreases correspondingly. The inflated tire
supports the rim primarily by these two effects."

Here's why the explanation doesn't satisfy the question:

"First, the most obvious one is that the casing pulls more to the sides than
downward (than it did in its unloaded condition)"

The tire casing (and its pull on the rim) can be taken completely out of the
picture. Consider a rim with just the inner tube installed on it. Slowly
pump the tube to say 10 psi and observe how the rim lifts off the ground as
the weight of the bicycle is taken up by the increased air pressure inside
the tube. The tube is not able to pull on the rim.

"the second is that the side wall tension is reduced. The reduction arises
from the relationship that unit casing tension is equivalent to inflation
pressure times the radius of curvature divided by pi. As the curvature
reduces when the tire bulges out, the casing tension decreases
correspondingly."

So what? The side wall tension can also be taken completely out of the
picture using the solo inner tube example from above; i.e., there is no pull
on the rim from inner tube side wall tension. If you need to convince
yourself of this, envision a layer of Vaseline between the tube and rim.

"The inflated tire supports the rim primarily by these two effects."

While perhaps describing some physical attributes of tire casings under
load, these don't explain the phenomenon of what's actually supporting the
load. In the above example something besides tire casing pull and side wall
tension is holding up the rim. The same mechanics that are at work holding
up the rim in the solo inner tube example come into play in all pneumatic
tires.

JL

Dear JL,

I think that if you actually try to support a normally loaded bicycle
with just an inner tube under the rim, you'll find that it doesn't
work--the elastic tube simply expands wildly elsewhere while remaining
squashed flat under the rim.

Unless the tube is constrained by a tire casing (or is so huge that it
begins to function like a sidewall--think of truck-tire inner-tube),
the example doesn't seem to work.

Cheers,

Carl Fogel

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

Quoted message said:

Quoted message said:

Dear JL,

I think that if you actually try to support a normally loaded bicycle
with just an inner tube under the rim, you'll find that it doesn't
work--the elastic tube simply expands wildly elsewhere while remaining
squashed flat under the rim.

Unless the tube is constrained by a tire casing (or is so huge that it
begins to function like a sidewall--think of truck-tire inner-tube),
the example doesn't seem to work.

Cheers,

Carl Fogel

Are you saying you could not slowly pump up an unconstrained tube such that
the rim lifts off the ground?

JL

someone writes:

http://www.sheldonbrown.com/brandt/rim-support.html

Quoted message said:

Quoted message said:

Does this FAQ item not answer your question.

Quoted message said:

Quoted message said:

Jobst Brandt

Quoted message said:

I don't see how the referenced article explains the mechanics of how
a load is actually carried by a pneumatic tire.

Quoted message said:

The article states:

# Under load, in the ground contact zone, the tire bulges so that two
# effects reduce the downward pull (increase the net upward force) of
# the casing. First, the most obvious one is that the casing pulls
# more to the sides than downward (than it did in its unloaded
# condition); the second is that the side wall tension is reduced. The
# reduction arises from the relationship that unit casing tension is
# equivalent to inflation pressure times the radius of curvature
# divided by pi. As the curvature reduces when the tire bulges out,
# the casing tension decreases correspondingly. The inflated tire
# supports the rim primarily by these two effects.

Quoted message said:

Here's why the explanation doesn't satisfy the question:

# First, the most obvious one is that the casing pulls more to the
# sides than downward (than it did in its unloaded condition);

Less downward pull is equivalent to upward push. That is the
significance of this subject and why it applies to the spoked wheel
and was first introduce in that regard.

Quoted message said:

The tire casing (and its pull on the rim) can be taken completely
out of the picture. Consider a rim with just the inner tube
installed on it. Slowly pump the tube to say 10 psi and observe how
the rim lifts off the ground as the weight of the bicycle is taken
up by the increased air pressure inside the tube. The tube is not
able to pull on the rim.

You are building an invalid free body diagram. That is why a tubular
tire is a better model for this. A bare inner tube will not have
enough pressure to support a rider but it will work the same way for
lighter loads, say just the bicycle.

The attachment to the rim does not need to be a tire bead. It can be
the point at which the tubular tire base tape rests on the rim and
does not flex while the rest of the tire casing deforms with load. It
is that part of the tire that is important.

# The second is that the side wall tension is reduced. The reduction
# arises from the relationship that unit casing tension is equivalent
# to inflation pressure times the radius of curvature divided by pi.
# As the curvature reduces when the tire bulges out, the casing
# tension decreases correspondingly.

Quoted message said:

So what? The side wall tension can also be taken completely out of
the picture using the solo inner tube example from above; i.e.,
there is no pull on the rim from inner tube side wall tension. If
you need to convince yourself of this, envision a layer of Vaseline
between the tube and rim.

It can't or there would be nothing between the rim and the ground with
any forces in it. The inner tube is only an air seal and does not
contribute.

# The inflated tire supports the rim primarily by these two effects.

Quoted message said:

While perhaps describing some physical attributes of tire casings
under load, these don't explain the phenomenon of what's actually
supporting the load. In the above example something besides tire
casing pull and side wall tension is holding up the rim. The same
mechanics that are at work holding up the rim in the solo inner tube
example come into play in all pneumatic tires.

I think you should give this some more thought.

By the way, who are you?

Jobst Brandt

JL said:

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

Quoted message said:

Quoted message said:

Dear JL,

I think that if you actually try to support a normally loaded bicycle
with just an inner tube under the rim, you'll find that it doesn't
work--the elastic tube simply expands wildly elsewhere while remaining
squashed flat under the rim.

Unless the tube is constrained by a tire casing (or is so huge that it
begins to function like a sidewall--think of truck-tire inner-tube),
the example doesn't seem to work.

Cheers,

Carl Fogel

Are you saying you could not slowly pump up an unconstrained tube such that
the rim lifts off the ground?

JL

Dear JL,

Take an ordinary bicycle tube, pinch it between your thumb and finger,
and start pumping.

The thin tube will lengthen enormously and bulge a bit everywhere
else, but you'll have little trouble keeping the tube pinched together
with just the pressure of your thumb and finger.

You'll never get much pressure because the rubber is so elastic that
it simply expands when not constrained by an essentially non-elastic
tire. The elasticity prevents you from getting much pressure because
the volume simply increases.

In contrast, a large truck or tractor inner tube has much thicker and
less elastic walls. They act more like a tire's sidewalls, so such
large inner tubes can easily support your weight.

To support a normally loaded bike wheel, you'd need much thicker
rubber walls on an inner tube, thick enough to imitate the inelastic
sidewalls of a tire.

Cheers,

Carl Fogel