Tire puzzler
jl...@comcast.net
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.
D'ohBoy
The bottom spokes actually support the rim ;-)
D'ohBoy
carl...@comcast.net
Joe Riel
> 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
carl...@comcast.net
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
Joe Riel
>>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
jl...@comcast.net
>
> 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
Michael Press
<1158865826....@i42g2000cwa.googlegroups.com>,
jl...@comcast.net wrote:
> 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
jobst....@stanfordalumni.org
> An oldie but goodie:
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
jobst....@stanfordalumni.org
> An oldie but goodie:
I don't believe the answer is found in the question but I am surprised
jl...@comcast.net
jobst....@stanfordalumni.org wrote:
> I don't believe the answer is found in the question...
That's what makes it a puzzler.
> 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
Tosspot
Nono, it hangs from the top spoke!
Chris Y.F.N.W.
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
jobst....@stanfordalumni.org
> 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
DaveH
>from the rim to the tire?
>
>Hint: this is more of a thought exercise; the answer can be found in
>the question.
I think a free-body analysis of a differential element of the tire
sidewall would show that in fact it is essentially the sidewall in the
vacinity of the contact patch that supports the load.
Counterintuitive since the sidewall material is flexible and thus
would not normally be expected to bear substantial compressive loads.
Still thinking about this one...good question. Long time since I
studied Statics.
Dave
DaveH
wrote:
Wrong---sidewalls are clearly not in a state of compression.
Dave
jobst....@stanfordalumni.org
>> 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.
> I think a free-body analysis of a differential element of the tire
> sidewall would show that in fact it is essentially the sidewall in
> the vicinity of the contact patch that supports the load.
> Counterintuitive since the sidewall material is flexible and thus
> would not normally be expected to bear substantial compressive
> loads.
> Still thinking about this one... good question. Long time since I
> studied Statics.
So? How does it support the rim from the ground? It can't be
standing on the tire as we have so often heard about spokes.
Jobst Brandt
wvantwiller
$742e...@news.sonic.net:
So, what happens when you let the air out of the tire?
DaveH
Jobst,
You said "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 first effect you observe -- "the casing pulls more to the sides
than downward" -- suggests that indeed, the rim is to some extent
standing or resting on the tire bead/sidewall--rim interface. This,
in the case of a clincher.
In the case of a tubular tire, the rim is obviously to some extent
sitting a portion of the inner cross-section. I've not read the
"standing on spokes" debate. Clearly--the rim can be treated as a
solid cylinder of appropriate cross-section.
I did correct my erroneous conclusion that the sidewall is in a
compressive state...clearly it is in tension, consistent with your
explanation. It is interesting to note your description of reduced
tension due to increased radius of curvature.
OP did say: "In other words, the air pressure exerted against the rim
is exactly the same 360 degrees around its circumference." All this
tells us is that the sum of all forces exerted by the compressed body
of air on the rim interior is zero. Therefore, the tire is not in a
state of acceleration wrt to the rim, and vice verse. Nothing
more...mystery inserted where it doesn't exist.
I'm trying to develop the proper conceptual model.
The most elementary model description would be that of a flexible
torus containing a body of fluid (tire) surrounding a rigid
cylindrical body -- rim.
In fact--how does any inflated device bear a load? That is the
essence of the problem, I think. The clincher scenario possibly being
a specialized case.
Well--this is an interesting problem which I've obviously yet to
solve. Lots of words here with not much said, I fear.
Regards,
Dave
DaveH
After more thought...indeed, the rim does sit on the tire. Force is
transferred exactly at the rim/tire interface basically at the bead
sidewall juncture ("...casing leaves the rim at a 45 deg angle...",
etc.). IOW, I'm completely in agreement with your explanation Jobst,
upon further thought.
If the sector of rim corresponding to the tire/ground contact patch
were suddenly removed, we would have a decidely different tire
deformation configuration.
That the tire proper on a typical bike wheel is essentially an
inflated torus bladder makes little difference. A solid or inflated
tire's transfer of forces only differ in the distribution and
magnitude of forces.
Indeed, every bit of tire could be removed save that twixt the rim and
road and the rim would still be supported. Thus, the fact that
"...air pressure exerted against the rim is exactly the same 360
degrees around its circumference.. " has no bearing on the analysis
of rim to ground force transfer.
Regarding spokes: If someone has claimed that the rim is "sitting" on
the spokes, this is obviously incorrect. Spokes clearly do not bear
non-negligible compressive loads.
Finally, a detailed stress analysis of a cross-section of a clincher
tire within the contact patch would probably show a complex mix of
compressive, tensile and shearing forces. A tubular would show a
different mapping, but in all cases the rim is most definitely sitting
on the tire.
OTOH....I could be completely wrong! (Don't think so though)
Dave
jobst....@stanfordalumni.org
>>>> 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.
>>> I think a free-body analysis of a differential element of the tire
>>> sidewall would show that in fact it is essentially the sidewall in
>>> the vicinity of the contact patch that supports the load.
>>> Counterintuitive since the sidewall material is flexible and thus
>>> would not normally be expected to bear substantial compressive
>>> loads.
>>> Still thinking about this one... good question. Long time since I
>>> studied Statics.
>> So? How does it support the rim from the ground? It can't be
>> standing on the tire as we have so often heard about spokes.
> You said "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 first effect you observe -- "the casing pulls more to the sides
> than downward" -- suggests that indeed, the rim is to some extent
> standing or resting on the tire bead/sidewall--rim interface. This,
> in the case of a clincher.
> In the case of a tubular tire, the rim is obviously to some extent
> sitting a portion of the inner cross-section. I've not read the
> "standing on spokes" debate. Clearly--the rim can be treated as a
> solid cylinder of appropriate cross-section.
The tubular tide is not different from a clincher. It supports the
rim the same way.
> I did correct my erroneous conclusion that the sidewall is in a
> compressive state... clearly it is in tension, consistent with your
> explanation. It is interesting to note your description of reduced
> tension due to increased radius of curvature.
That's the crux of the exercise. I'm sure you won't find many folks
who will argue against the rim standing on the tire. The way it does
this is that it pulls down less than when unloaded, just as the spokes
in a bicycle wheel do. We've been here before and used a large toy
balloon on which one can sit.
The point is that slicing the sausage too thin, loses its essence, so
to speak. Compression and tension are a mechanical continuum and to
try to separate them in force diagrams is fruitless. For material
strength assessment it is appropriate to analyze stress, but getting
stress confused with structural load path creates a collision.
> OP did say: "In other words, the air pressure exerted against the
> rim is exactly the same 360 degrees around its circumference." All
> this tells us is that the sum of all forces exerted by the
> compressed body of air on the rim interior is zero. Therefore, the
> tire is not in a state of acceleration with respect to to the rim,
> and vice versa. Nothing more... mystery inserted where it doesn't
> exist.
In that event, please clarify whether the rim is standing on the tire
or not. I get the feeling you are dodging to avoid Paul Kaddy who
will tell you that it cannot stand on the tire... or a car stand on
its wheels, since they have flexible tires.
> I'm trying to develop the proper conceptual model. The most
> elementary model description would be that of a flexible torus
> containing a body of fluid (tire) surrounding a rigid cylindrical
> body -- rim.
Lets leave the generic mathematical terms out of this. Just use a
tubular tire or a toy balloon.
> In fact--how does any inflated device bear a load? That is the
> essence of the problem, I think. The clincher scenario possibly
> being a specialized case.
Nothing special about it.
> Well--this is an interesting problem which I've obviously yet to
> solve. Lots of words here with not much said, I fear.
http://www.sheldonbrown.com/brandt/rim-support.html
Does this FAQ item not answer your question.
Jobst Brandt
jobst....@stanfordalumni.org
>>>> 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.
>>> I think a free-body analysis of a differential element of the tire
>>> sidewall would show that in fact it is essentially the sidewall in
>>> the vicinity of the contact patch that supports the load.
>>> Counterintuitive since the sidewall material is flexible and thus
>>> would not normally be expected to bear substantial compressive
>>> loads.
>>> Still thinking about this one... good question. Long time since I
>>> studied Statics.
>> So? How does it support the rim from the ground? It can't be
>> standing on the tire as we have so often heard about spokes.
> After more thought...indeed, the rim does sit on the tire. Force is
> transferred exactly at the rim/tire interface basically at the bead
> sidewall juncture ("...casing leaves the rim at a 45 deg angle...",
> etc.). IOW, I'm completely in agreement with your explanation
> Jobst, upon further thought.
> If the sector of rim corresponding to the tire/ground contact patch
> were suddenly removed, we would have a decidedly different tire
> deformation configuration.
> That the tire proper on a typical bike wheel is essentially an
> inflated torus bladder makes little difference. A solid or inflated
> tire's transfer of forces only differ in the distribution and
> magnitude of forces.
> Indeed, every bit of tire could be removed save that twixt the rim
> and road and the rim would still be supported. Thus, the fact that
> "...air pressure exerted against the rim is exactly the same 360
> degrees around its circumference.. " has no bearing on the analysis
> of rim to ground force transfer.
> Regarding spokes: If someone has claimed that the rim is "sitting"
> on the spokes, this is obviously incorrect. Spokes clearly do not
> bear non-negligible compressive loads.
Tire casings do not support compressive loads. As I said, I see a
collision here. Please explain.
> Finally, a detailed stress analysis of a cross-section of a clincher
> tire within the contact patch would probably show a complex mix of
> compressive, tensile and shearing forces. A tubular would show a
> different mapping, but in all cases the rim is most definitely
> sitting on the tire.
How much more detailed can you get than to resolve the forces acting
on the rim.
> OTOH....I could be completely wrong! (Don't think so though)
Jobst Brandt
DaveH
I'm not sure of where I've gone wrong here. You see a
collision--where? I made the correction in a previous post that
casings do not (substantially) support compressive loads.
I've not read any of other debates regarding this, rims and spokes,
and have not heard of Paul Kaddy.
Dave
jobst....@stanfordalumni.org
>>>> So? How does it support the rim from the ground? It can't be
>>>> standing on the tire as we have so often heard about spokes.
>>> After more thought...indeed, the rim does sit on the tire. Force
>>> is transferred exactly at the rim/tire interface basically at the
>>> bead sidewall juncture ("...casing leaves the rim at a 45 deg
>>> angle...", etc.). IOW, I'm completely in agreement with your
>>> explanation Jobst, upon further thought.
By the way, how does the Isle of Wight (IOW) come into this?
> I'm not sure of where I've gone wrong here. You see a
> collision--where? I made the correction in a previous post that
> casings do not (substantially) support compressive loads.
> I've not read any of other debates regarding this, rims and spokes,
> and have not heard of Paul Kaddy.
He likes to post a p.k. especially on how the spoked wheel supports
loads. If you believe the rim stand on the tire. then you are in
conflict in the claim that the tire supports the rim because, as you
agree, the deformation of the casing at the ground contact pint causes
a net upward force, yet all its fibers are pulling downward.
How do you resolve that conflict in terms?
Jobst Brandt
jobst....@stanfordalumni.org
>>>> So? How does it support the rim from the ground? It can't be
>>>> standing on the tire as we have so often heard about spokes.
>>> After more thought...indeed, the rim does sit on the tire. Force
>>> is transferred exactly at the rim/tire interface basically at the
>>> bead sidewall juncture ("...casing leaves the rim at a 45 deg
>>> angle...", etc.). IOW, I'm completely in agreement with your
>>> explanation Jobst, upon further thought.
By the way, how does the Isle of Wight (IOW) come into this?
>>> If the sector of rim corresponding to the tire/ground contact
> I'm not sure of where I've gone wrong here. You see a
> collision--where? I made the correction in a previous post that
> casings do not (substantially) support compressive loads.
> I've not read any of other debates regarding this, rims and spokes,
> and have not heard of Paul Kaddy.
He likes to post as p.k., especially on how the spoked wheel supports
DaveH
Are there any adults in this thread who would care to discuss the
puzzler?
Dave
JL
news:45158d36$0$34553$742e...@news.sonic.net...
>
> 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
carl...@comcast.net
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
JL
<carl...@comcast.net> wrote in message
news:0ouih2lk3cp272dd8...@4ax.com...
> >
> 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
jobst....@stanfordalumni.org
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);
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.
> 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.
> 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.
> 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
carl...@comcast.net
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
JL
<jobst....@stanfordalumni.org> wrote in message
news:451982e2$0$34500$742e...@news.sonic.net...
I was using the solo inner tube example to show that a rim can be held off
the ground without the two effects described in your article, i.e., pulling
on the rim by the tire casing and side wall tension.
Use a tubular tire instead of a solo inner tube in the example if you
prefer. Don't glue it to the rim, in fact use a layer of Vaseline between
the tire and rim. The tire will support a vertical load even though it can't
pull on the rim.
JL
carl...@comcast.net
Dear JL,
The sidewall tension is provided by the constraint of the tubular
tire, which encases the tube.
Without the inelastic tire to constrain it and raise the pressure (and
tension), a thin bicycle inner tube would just lengthen enormously
without gaining much pressure (or tension).
Think of a toy balloon. Its volume expands far more than its pressure
and surface tension increase because the thin rubber is so elastic.
There is some increase in tension in the "sidewall" of the balloon,
enough to support a very small weight.
But if you want something that can support your weight, you have to
constrain the balloon inside something far less elastic to raise the
pressure and "sidewall" tension.
Again, you can demonstrate this by pinching a bicycle inner tube
between your thumb and finger and pumping it up. The tube is so
elastic that it lengthens enormously, bulges only slightly, and
provides so little "sidewall" tension that you can pinch it together
with a thumb and finger, just as you can pinch an elastic balloon
together at any point with very little pressure because the thin,
elastic walls provide scarcely any tension.
Cheers,
Carl Fogel
Blair P. Houghton
>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.
I'll tell that to Eddie Merckx next time I see him riding sewups...
--Blair
Blair P. Houghton
><jobst....@stanfordalumni.org> wrote in message
>news:45158d36$0$34553$742e...@news.sonic.net...
>
>>
>> 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."
I'm not sure it's entirely sufficient, either.
About 15 years ago, Marvin Minsky mentioned that he
liked to ask Nobel laureates, "how does a tire work?"
then stand amused watching them try to figure it out.
But then, he used to hang around with Nobel laureates,
and I used to converse with Marvin Minsky.
The world's changed for us all.
>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 with-casing and without-casing cases are two different cases.
He redunded.
The wheel is supported by the entire bead of a tire with
a casing that has a bead-and-rim interface. In the case
of tubeless tires, this is the only thing supporting the
wheel.
Once the wheel is married to the tire through the bead
and rim interface, it becomes part of the torus. And
the freaky thing is, in automobile tires, it's entirely
a friction fit. The pressure in the tire pushes the sidewall
out which forces the bead against the metal. The static
coefficient of friction at that normal force is high enough
that almost no tires slip on their rims under even the
strongest acceleration of the vehicle. (One example, sadly,
was the original equipment tires on my Lexus, which didn't
have a good enough friction with the new alloy they chose
for the wheels, so they would slip slightly when I matted
it, which in a GS400 is a decent tug. And when it was
new, I really liked making the road go away. This led to
the need to rebalance the rear wheels every thousand miles.
Lexus has since listed other tire models that fit and
will grip those rims, but for a while my car was a lot
like Ken Griffey Jr.: A superman with glass legs.
But I digress...)
This is different from a solid rim of a certain diameter
sitting inside the donut-hole of a tube that is much larger.
That's a matter of having air pressure under the rim (in
the tube) that is higher than the air pressure over it (which
isn't in contact with the tube). Which is the same as putting
a book on top of a balloon, if you can get it to balance.
> "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.
The tire and the pressure in it convert the upward force
on the contact patch to a transverse force distributed
around the entire bead and the rim.
But it's not equal at all points on the bead interface.
The imbalance could be as simple as a net upward force
at all points which is equivalent to a differential
radial force that rotates as you go around the rim and a
differential tangential force that rotates 90 degrees out
of phase with the radial force around the rim.
The pressure on the rest of the rim surface (between the
beads) is equal around the rim, because that's how gas
pressures work. Any increase at the bottom is equal to the
increase at the top, unless the system is moving so fast
that the propagation of pressure waves becomes an issue.
--Blair
Blair P. Houghton
>Again, you can demonstrate this by pinching a bicycle inner tube
>between your thumb and finger and pumping it up. The tube is so
>elastic that it lengthens enormously, bulges only slightly, and
>provides so little "sidewall" tension that you can pinch it together
>with a thumb and finger, just as you can pinch an elastic balloon
>together at any point with very little pressure because the thin,
>elastic walls provide scarcely any tension.
Do this at 110 psi.
Go ahead.
--Blair
carl...@comcast.net
Let us know what happens when you try to inflate an ordinary
thin-walled bicycle inner-tube to 110 psi, unconstrained by a tire.
carl...@comcast.net
Eddie would probably explain that without the edges of the sew-up tire
sewed together to constrain the tube, it would behave exactly as
predicted.
That's why we ride on tires, not naked inner tubes.
You cannot raise the rim without tension, which is the result of
constraining the air pressure. If the "constraint" is as elastic as
inner tube, there won't be enough pressure or tension.
That's why you can pump up and sit on a soccer ball, whose inelastic
side go into tension and constrain the air, but you cannot sit on a
child's balloon of the same size, since its sides offer little
constraint, negligible tension, and scarcely any air pressure.
Paul Hobson
You probably couldn't get a bare tube up to 110 psi. Unconstrained, I'm
guessing it'd blow up long before then.
\\paul
carl...@comcast.net
Dear Paul,
"Long" is an understatement.
Attach an ordinary bicycle inner tube to a floor pump with an air
gauge.
It's unlikely that you'll be able to reach 10 psi, much less that
silly 110 psi figure. The inner tube will elongate enormously, while
bulging only slightly. The rubber fails long before you achieve much
pressure.
The analogy of a balloon is obvious--no matter how you blow up a toy
balloon, there's very little pressure, hence very little tension, and
thus almost no ability to support a load.
You can inflate a soccer ball to a modest pressure, putting its
inelastic sides in enough tension to let you sit on it. The toy ballon
squashes flat or pops under your weight.
The stronger and less elastic the material, the greater the
constraint, the higher the pressure, and the greater the tension
available to support a load.
Tire sidewalls are good at this.
Unconstrained bicycle inner tubes are useless.
Anyone with a floor pump, a spare tube to waste, and a pair of ear
plugs can demonstrate this behavior. Step on the tube from time to
time with your shoe, squashing it to the floor. Stop pumping and
remove the ear plugs after the loud bang when the tube fails.
Thicker-walled inner tubes that are large enough, the kind used in
large truck and tractor tires, provide enough constraint, pressure,
and tension to support your weight and are often used to float down
rivers.
Cheers,
Carl Fogel
Mark Hickey
>You probably couldn't get a bare tube up to 110 psi. Unconstrained, I'm
>guessing it'd blow up long before then.
Well, you COULD get 110psi into the tube, but you'd have to be under
about 260 feet of water. ;-)
Mark Hickey
Habanero Cycles
http://www.habcycles.com
Home of the $795 ti frame
Joe Riel
> Paul Hobson <gtg...@mail.gatech.edu> wrote:
>
>>You probably couldn't get a bare tube up to 110 psi. Unconstrained, I'm
>>guessing it'd blow up long before then.
>
> Well, you COULD get 110psi into the tube, but you'd have to be under
> about 260 feet of water. ;-)
That's 110psivg. You still couldn't do it 8-).
--
Joe Riel
Blair P. Houghton
>On Wed, 27 Sep 2006 00:05:22 GMT, Blair P. Houghton <b@p.h> wrote:
>
>> <carl...@comcast.net> wrote:
>>>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.
>>
>>I'll tell that to Eddie Merckx next time I see him riding sewups...
>
>sewed together to constrain the tube, it would behave exactly as
>predicted.
>
>That's why we ride on tires, not naked inner tubes.
>
>You cannot raise the rim without tension, which is the result of
>constraining the air pressure. If the "constraint" is as elastic as
>inner tube, there won't be enough pressure or tension.
Depends on the inner tube. A sewup is just an inner tube
with a butyl inner tube in it, if I put it in a clincher.
It's the glue on sewups that keeps the pressure from
pushing the tire away from the rim at the top when the
bottom is under compression.
>That's why you can pump up and sit on a soccer ball, whose inelastic
>side go into tension and constrain the air, but you cannot sit on a
>child's balloon of the same size, since its sides offer little
>constraint, negligible tension, and scarcely any air pressure.
You never had a childhood:
http://toys-gifts-store.onlineshoppingday.co.uk/104683.html
--Blair
Blair P. Houghton
>On Wed, 27 Sep 2006 00:50:25 GMT, Blair P. Houghton <b@p.h> wrote:
>
>> <carl...@comcast.net> wrote:
>>>Again, you can demonstrate this by pinching a bicycle inner tube
>>>between your thumb and finger and pumping it up. The tube is so
>>>elastic that it lengthens enormously, bulges only slightly, and
>>>provides so little "sidewall" tension that you can pinch it together
>>>with a thumb and finger, just as you can pinch an elastic balloon
>>>together at any point with very little pressure because the thin,
>>>elastic walls provide scarcely any tension.
>>
>>Do this at 110 psi.
>>
>>Go ahead.
>
>thin-walled bicycle inner-tube
So now you have 40 other qualifications for your
example?
>to 110 psi, unconstrained by a tire.
I bet I can put a bicycle on it and sit on it and the
rim doesn't crush the tube to the ground.
I bet you can't even hope to get its sides to touch
when you squeeze it with two fingers.
--Blair
Blair P. Houghton
>Thicker-walled inner tubes that are large enough, the kind used in
>large truck and tractor tires, provide enough constraint, pressure,
>and tension to support your weight and are often used to float down
>rivers.
The point here is that you jumped someone's perfectly
good example of a certain demonstration of the lifting
power of a pressurized rubber tube and you won't admit
that your "pinch it" retort doesn't have anything
to do with the problem.
Now you're on about the wall thickness and pretending that
the principles that apply to bicycle tubes aren't the same
that apply to tractor tubes.
Hint: they're exactly the same.
The scales are different, but the effect on a rim
is no different.
This was never about the strength of the rubber,
and you know it.
--Blair
Mark Hickey
> <carl...@comcast.net> wrote:
>>On Wed, 27 Sep 2006 00:50:25 GMT, Blair P. Houghton <b@p.h> wrote:
>>
>>> <carl...@comcast.net> wrote:
>>>>Again, you can demonstrate this by pinching a bicycle inner tube
>>>>between your thumb and finger and pumping it up. The tube is so
>>>>elastic that it lengthens enormously, bulges only slightly, and
>>>>provides so little "sidewall" tension that you can pinch it together
>>>>with a thumb and finger, just as you can pinch an elastic balloon
>>>>together at any point with very little pressure because the thin,
>>>>elastic walls provide scarcely any tension.
>>>
>>>Do this at 110 psi.
>>>
>>>Go ahead.
>>
>>Let us know what happens when you try to inflate an ordinary
>>thin-walled bicycle inner-tube
>
>So now you have 40 other qualifications for your
>example?
His "qualification" is just a clarification - precisely what he meant
originally (and what any reasonable reader would have concluded).
>>to 110 psi, unconstrained by a tire.
>
>I bet I can put a bicycle on it and sit on it and the
>rim doesn't crush the tube to the ground.
>
>I bet you can't even hope to get its sides to touch
>when you squeeze it with two fingers.
Errrrrr, one of us is living in a universe with different physics than
the other, apparently.
Mark Hickey
Naaaah, I was using a floor pump on a boat - one with a 260 foot hose.
JL
<carl...@comcast.net> wrote in message
news:hijjh29ejc8pa318k...@4ax.com...
Carl, it seems that you prefer to argue ancillary details rather than stay
on point. First of all, if you are literally going to try the inner tube
experiment, select a tube size small enough such that it stays stretched
over the rim when some air is pumped into it. Secondly, while it may make
for a humorous mental image, inflating to 110 psi isn't called for in the
stated example, nor would it be advisable. A tube inflated to 5 psi would
support a 10 lb load with a 2 sq-in contact patch under the rim.
Although the tube example in this case may be limited to 'very low pressure'
supporting a 'very modest load', the point is that the two effects cited in
Jobst's article, namely casing pull and side wall tension, don't explain how
the load is being carried. True you can't get on the bike (with only inner
tubes) and ride away but that's not the point.
JL
carl...@comcast.net
Dear JL,
Please take an ordinary bicycle inner tube, attach it to a floor pump,
and try to pump it up to even 5 psi--you can't do it.
The tube will elongate enormously. You'd need to find an inner tube
for a 6-inch tire if you hoped to fit it onto a 700c rim.
The 110 psi nonsense is from another poster.
Since no one appears to be willing to perform the simple experiment,
here's a picture of what happens if you try to pump an unconstrained
700c inner tube up:
http://home.comcast.net/~carlfogel/download/212_700c_80_pump_strokes_0_psi_4_feet_wide.jpg
or http://tinyurl.com/n3bxa
After 80 strokes with a large floor pump, the 700c tube has obviously
expanded far beyond the 700c rim. The tape measure indicates a 4-foot
wide circle.
The pump indicator registered 0 psi, but the picture showing that
proves only that the shiny glass over the indicator reflects my
camera's automatic flash amazingly well.
Something as broad and heavy as my size 12 shoe with half my 190 pound
weight on it will easily flatten the inflated tube.
Another way to show how little load can be carried is to simply twist
the tube--it's almost effortless to twist it and constrict the section
between your hands into a tiny knot because there's so little tension
in the sidewall.
Mercifully, the tube did not explode. Instead, it sprang a hissing
leak at a seam, typical behavior for such over-expanded tubes.
There is no important difference in how a load is supported by a soap
bubble, a toy balloon, a bare inner tube, or a bicycle tire.
The strength and elasticity of the "sidewall" determine how much
pressure differential can be produced, which determines the tension of
the sidewall, which determines the load that it can support.
With the soap bubble, the soap film is so elastic that scarcely any
pressure difference can be obtained--adding more air simply expands
the volume with almost no pressure increase. If the soap film were
infinitely puncture-resistant, it still could support little more than
a feather--think of a cloth pillow case collapsing under its own
weight because there's no pressure difference.
With the toy balloon, there a little more pressure because the thin
rubber isn't as elastic as the soap bubble film. But the toy balloon
can't carry much load because the slight pressure difference puts so
little tension on the sidewalls. If you try to blow it up to a higher
pressure, it expands so easily that hardly any pressure or tension
builds up.
People who haven't fooled around with bicycle inner tubes don't
understand that they're not much better than balloons. You simply
can't raise pressure enough to register on a good pump--the tube just
expands. The wall is thicker and the faint pressure is a bit higher,
so it will support more than a balloon, but it doesn't support much.
The tire works like a soccer ball. The sidewall material is so strong
and inelastic that you can easily build up a considerable pressure
difference and substantial tension, enough to support your weight.
The sewup and the clincher behave the same way. The sewup is closed
over an inner tube by the strong tire being sewn together over the
inner tube. The clincher is closed by the rim, constraining the inner
tube. If you want to squash such tires flat, you have to overcome the
enormous tension in the sidewalls.
For those curious about details, the clincher is held against the rim
by air pressure, as Jobst has demonstrated by cutting the tire bead,
which is primarily an aid to mounting the tire.
It may seem at first that the small strip of tire pressed against the
flange can't hold things in place, but a little calculation shows
otherwise.
A quarter inch strip of tire pressing against the inside of one rim
flange is about 80 inches long, so there's about 20 square inches of
tire mashed against the metal--at 100 psi, you have 2,000 lbs of force
pushing 20 square inches of material similar to a brake pad against
the rim.
Tubulars have much lower flanges, but they're glued on and benefit
from the other trick that keeps bicycle tires in place. The tire
casing cords are laid at an angle, going X instead of ||. When the
tire is inflated, it swells a bit, being slight elastic. The angled
cords cause it to constrict, instead of expanding as you'd expect, so
it tightens down on the rim.
Jobst has repeatedly pointed this out, too, giving the angles at which
pressure hoses are braided to keep them from expanding or contracting,
as opposed to the angles that cause tires to contract when inflated.
The contraction is so counter-intuitive that the example of a Chinese
finger-trap toy is often used to explain it. The toy is just a tube
made of strips loose woven in a criss-cross pattern. If you stick a
finger in each end of the tube and pull, the tube lengthens and its
diameter shrinks, squeezing your fingers and trapping them. To
release, you have to push your fingers together, which shortens the
tube while expanding its diameter--just like a tire inflating.
Here's a crude ASCII diagram:
xxxxxxxxxxxxx (uninflated, thin, long)
versus
XXXXXXX (inflated, thickens, shortens, and constricts)
Cheers,
Carl Fogel
JL
news:72elh2tvak8god44f...@4ax.com...
Yes, it's the strength of the casing that allows a higher internal pressure
without expanding wildly. I am in complete agreement with you on this. I
just don't think that Jobst's article fully explains the mechanics of what's
carrying the load.
JL
Blair P. Houghton
>Blair P. Houghton <b@p.h> wrote:
>
>> <carl...@comcast.net> wrote:
>>>On Wed, 27 Sep 2006 00:50:25 GMT, Blair P. Houghton <b@p.h> wrote:
>>>
>>>> <carl...@comcast.net> wrote:
>>>>>Again, you can demonstrate this by pinching a bicycle inner tube
>>>>>between your thumb and finger and pumping it up. The tube is so
>>>>>elastic that it lengthens enormously, bulges only slightly, and
>>>>>provides so little "sidewall" tension that you can pinch it together
>>>>>with a thumb and finger, just as you can pinch an elastic balloon
>>>>>together at any point with very little pressure because the thin,
>>>>>elastic walls provide scarcely any tension.
>>>>
>>>>Do this at 110 psi.
>>>>
>>>>Go ahead.
>>>
>>>Let us know what happens when you try to inflate an ordinary
>>>thin-walled bicycle inner-tube
>>
>>So now you have 40 other qualifications for your
>>example?
>
>His "qualification" is just a clarification - precisely what he meant
>originally (and what any reasonable reader would have concluded).
We're talking about bicycle wheels, which operate at 110
psi. Playing around with tubes at 1.5 bar is not the same
thing and is therefore not reasonable. Hence I mocked him.
>>>to 110 psi, unconstrained by a tire.
>>
>>I bet I can put a bicycle on it and sit on it and the
>>rim doesn't crush the tube to the ground.
>>
>>I bet you can't even hope to get its sides to touch
>>when you squeeze it with two fingers.
>
>Errrrrr, one of us is living in a universe with different physics than
>the other, apparently.
Air pressure is air pressure, no matter how soft the
rest of the tube is. You're not squishing the sides of
a 110 psi tube together with two fingers. Unless they're
this guy's:
Then I'd make it even-money.
--Blair
G.T.
The question is whether you can get 110psi into an unconstrained normal
bicycle inner tube, or will it stretch too thin before it gets anywhere
near 110psi?
Greg
--
"All my time I spent in heaven
Revelries of dance and wine
Waking to the sound of laughter
Up I'd rise and kiss the sky" - The Mekons
carl...@comcast.net
wrote:
Dear Greg,
Here's the answer, to quote from elsewhere in this thread:
The 110 psi nonsense is from another poster.
Since no one appears to be willing to perform the simple experiment,
here's a picture of what happens if you try to pump an unconstrained
700c inner tube up:
http://home.comcast.net/~carlfogel/download/212_700c_80_pump_strokes_0_psi_4_feet_wide.jpg
or http://tinyurl.com/n3bxa
After 80 strokes with a large floor pump, the 700c tube has obviously
expanded far beyond the 700c rim. The tape measure indicates a 4-foot
wide circle.
The pump indicator registered 0 psi, but the picture showing that
proves only that the shiny glass over the indicator reflects my
camera's automatic flash amazingly well.
Something as broad and heavy as my size 12 shoe with half my 190 pound
weight on it will easily flatten the inflated tube.
Another way to show how little load can be carried is to simply twist
the tube--it's almost effortless to twist it and constrict the section
between your hands into a tiny knot because there's so little tension
in the sidewall.
Mercifully, the tube did not explode. Instead, it sprang a hissing
leak at a seam, typical behavior for such over-expanded tubes.
Cheers,
Carl Fogel
Blair P. Houghton
>The question is whether you can get 110psi into an unconstrained normal
>bicycle inner tube, or will it stretch too thin before it gets anywhere
>near 110psi?
No, the question is how does a tire work.
A tube that has elasticity at 110 psi would be baloonish,
if you assumed ordinary butyl rubber, but it's
not unreasonable to posit a material that has linear
elasticity from some low pressure to beyond 110 psi.
It would have to start out about as thick as macaroni and
as big around as a dinner plate, and expand to the size
of a bicycle tire.
That material could be used to make a bicycle tire.
Now. Try to squish it with your fingers until they
touch. It won't be much easier than doing it to the
110 psi tire on my bike now.
--Blair
"Well...0 psi, but I'm going to
fix that before my morning ride..."
Blair P. Houghton
>Dear Greg,
>
>Here's the answer, to quote from elsewhere in this thread:
>
>The 110 psi nonsense is from another poster.
>
>Since no one appears to be willing to perform the simple experiment,
You presume it needs to be performed.
>here's a picture of what happens if you try to pump an unconstrained
>700c inner tube up:
>
>http://home.comcast.net/~carlfogel/download/212_700c_80_pump_strokes_0_psi_4_feet_wide.jpg
>or http://tinyurl.com/n3bxa
>
>After 80 strokes with a large floor pump, the 700c tube has obviously
>expanded far beyond the 700c rim. The tape measure indicates a 4-foot
>wide circle.
>
>The pump indicator registered 0 psi, but the picture showing that
>proves only that the shiny glass over the indicator reflects my
>camera's automatic flash amazingly well.
>
>Something as broad and heavy as my size 12 shoe with half my 190 pound
>weight on it will easily flatten the inflated tube.
>
>Another way to show how little load can be carried is to simply twist
>the tube--it's almost effortless to twist it and constrict the section
>between your hands into a tiny knot because there's so little tension
>in the sidewall.
>
>Mercifully, the tube did not explode. Instead, it sprang a hissing
>leak at a seam, typical behavior for such over-expanded tubes.
None of which is relevant to how a tire works.
An innertube at 1.5 bar is not a bicycle tire.
Nor does a bicycle tire necessarily have to be inelastic.
It's just easier that way, given common materials.
--Blair
Ben C
> G.T. <getn...@dslextreme.com> wrote:
>>The question is whether you can get 110psi into an unconstrained normal
>>bicycle inner tube, or will it stretch too thin before it gets anywhere
>>near 110psi?
>
> No, the question is how does a tire work.
>
> A tube that has elasticity at 110 psi would be baloonish,
> if you assumed ordinary butyl rubber, but it's
> not unreasonable to posit a material that has linear
> elasticity from some low pressure to beyond 110 psi.
If you mean it has an elasticity that's proportional to pressure, and
that at 110 psi it has about the same elasticity as an ordinary bicycle
tyre does at 110 psi, then yes, it would work.
> It would have to start out about as thick as macaroni and
> as big around as a dinner plate, and expand to the size
> of a bicycle tire.
This would be a most useful design from the point of view of carrying a
spare tyre with you on rides.
> That material could be used to make a bicycle tire.
>
> Now. Try to squish it with your fingers until they
> touch. It won't be much easier than doing it to the
> 110 psi tire on my bike now.
Indeed, but that's because at 110psi, the sidewall has the same
elasticity as the tyre on your bike now.
Joe Riel
> On 2006-09-29, Blair P Houghton <b@p.h> wrote:
>> G.T. <getn...@dslextreme.com> wrote:
>>>The question is whether you can get 110psi into an unconstrained normal
>>>bicycle inner tube, or will it stretch too thin before it gets anywhere
>>>near 110psi?
>>
>> No, the question is how does a tire work.
>>
>> A tube that has elasticity at 110 psi would be baloonish,
>> if you assumed ordinary butyl rubber, but it's
>> not unreasonable to posit a material that has linear
>> elasticity from some low pressure to beyond 110 psi.
>
> If you mean it has an elasticity that's proportional to pressure, and
> that at 110 psi it has about the same elasticity as an ordinary bicycle
> tyre does at 110 psi, then yes, it would work.
That isn't possible. If it were, then a bicycle tire would
expand as it was inflated.
--
Joe Riel
Ben C
I thought they did? Inner tubes certainly do.
But "elasticity proportional to pressure" is a slightly strange idea,
since it has been observed by others that the high elasticity of a
balloon is what makes it increase in volume rather than in pressure as
you blow it up.
But it's not too hard to imagine a material whose stretchiness is
roughly proportional to extension. More precisely, you apply 10N to a
strip of the stuff, and it gets 10cm longer, but apply another 10N, and
it only gets another 5cm longer, and so on.
If you built a balloon or torus out of this material, it would inflate a
lot initially, but would eventually reach 110psi at some volume.
Blair P. Houghton
Everyone go look up "Hooke's Law".
In a linear elastic system, tension is proportional to
stretch distance. That's all I mean. Just that it doesn't
have a big nonlinearity the way a modern bicycle tire does.
--Blair
Ben C
I see, I assumed you meant "elasticity is proportional to
extension" by linear elasticity". This is quite different from Hooke's
Law.
A tyre whose casing obeyed Hooke's Law would inflate until it got very
big and then burst. I think it would be like a balloon-- more air in and
it gets bigger, but the pressure stays the same. If it were very stiff
to start with though, it could hold 110psi, and then you'd just inflate
it to the size you needed.
You could even make a universal tyre this way that would fit rim sizes
from 20" up to 700c; although it would increasingly become a mountain
bike tyre as you got to the larger diameters.
It would also be easy to fit to the rim. You'd inflate it until its
diameter was a bit too big, slide it on easily, and then let a bit of
air out to shrink it on tightly.
jobst....@stanfordalumni.org
>> The question is whether you can get 110psi into an unconstrained
>> normal bicycle inner tube, or will it stretch too thin before it
>> gets anywhere near 110psi?
> No, the question is how does a tire work.
> A tube that has elasticity at 110 psi would be baloonish, if you
> assumed ordinary butyl rubber, but it's not unreasonable to posit a
> material that has linear elasticity from some low pressure to beyond
> 110 psi.
> It would have to start out about as thick as macaroni and as big
> around as a dinner plate, and expand to the size of a bicycle tire.
> That material could be used to make a bicycle tire.
> Now. Try to squish it with your fingers until they touch. It won't
> be much easier than doing it to the 110 psi tire on my bike now.
Just forget about the material and rim and clinch and move on to a
tubular tire of which we know only that it is sitting on a suitable
rim, preferably to which it is not glued, just to get all the
irrelevant features out of the picture.
The tubular tire inflated to some reasonable pressure that with a
reasonable vertical load makes it react like a bicycle wheel as we
know it as the model for this exercise; all the psi and bare inner
tube diversions aside. This tire, as pneumatic tires in general,
support the rim as was described. I don't see what is so difficult
with that. The questions asked about what it the description means
seem to be more smoke screens than searches for clarification.
http://www.sheldonbrown.com/brandt/rim-support.html
Had I known the clincher was a point of diversion, I would have made
it unglued tubular, but thought that might be to vague for those who
don't know what a tubular tire is or how it it holds air.
Jobst Brandt
JL
<jobst....@stanfordalumni.org> wrote in message
news:451ea7e3$0$34574$742e...@news.sonic.net...
> Blair P. Houghton writes:
>
>
> http://www.sheldonbrown.com/brandt/rim-support.html
>
> Had I known the clincher was a point of diversion, I would have made
> it unglued tubular, but thought that might be to vague for those who
> don't know what a tubular tire is or how it it holds air.
>
> Jobst Brandt
Jobst, the article seems to imply that the load is being taken up by the
casing:
"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."
If the vertical load is taken up by the tire casing, explain how the load
then gets from the casing to the ground. It seems that the ground pushes up
on the outside of the tire casing at the contact patch with a pressure equal
to the air pressure inside the tire. In other words the reaction at the
ground is taken up by the internal air pressure, not the casing. Can you
please clarify?
JL.
jobst....@stanfordalumni.org
http://www.sheldonbrown.com/brandt/rim-support.html
>> Had I known the clincher was a point of diversion, I would have
>> made it unglued tubular, but thought that might be to vague for
>> those who don't know what a tubular tire is or how it it holds air.
> Jobst, the article seems to imply that the load is being taken up by
> the casing:
> "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."
> If the vertical load is taken up by the tire casing, explain how the
> load then gets from the casing to the ground. It seems that the
> ground pushes up on the outside of the tire casing at the contact
> patch with a pressure equal to the air pressure inside the tire. In
> other words the reaction at the ground is taken up by the internal
> air pressure, not the casing. Can you please clarify?
Try visualizing a tubular and apply the basic effect of inflation
pressure and ground contact surface. You must think in both modes,
what is pressing on the ground, which seems to be apparent from the
years of discussion on this, and the way that force supports the rim.
True, the tire casing contains the force and inflation pressure and
presses on the road, but that there is no net unbalanced inflation
pressure on the rim leaves the air pressure as a non-conductor of
force to it. That is why the casing tension gets involved as
described. I see that the two modes of analysis are a hurdle to
perception of what occurs.
On the road, inflation pressure on an ideal flexible tire is uniform
over the effective ground contact patch. This is not true for the
tire to rim interface where the bulge of the tire is variable over the
length of tire deformation from a circular cross section as are the
casing forces and angles.
Tire shape is determinable from the outline of the contact patch and
distance from the road to the inner edge of the tire where it bears on
the rim, but this is a geometric problem, each slice of which is
essentially a circular arc from ground to rim. This is more complex
because these are not radial tires and the 45 degree bias makes these
arcs slightly non circular.
This is evident when the tire has a few adjacent tight cords or
several broken cords from a cut. Their effect is not perpendicular to
the wheel center line but sweeps diagonally into and out of the
contact patch. Whether this effect is balance by opposing bias plies
in a uniform tire is not clear to me and gets into more analysis that
I care to pursue. For practical purposes, the tire behaves as a
radial ply casing.
Jobst Brandt
jobst....@stanfordalumni.org
to radial tires, will recall that the notable visible feature was that
radial tires "bellied out" at the road contact zone in contrast to
their bias ply forerunners.
The reason for this in bias ply tires is that the cords affected by
the contact patch begin before the contact patch and end at beyond at
the far side of it, the most bellied out zone being before and after
the middle of the contact patch. This is more pronounced for the
fatter donut shaped tires that car tires have in which the tire cross
section is a small multiple of its major diameter while on bicycle
tires it is about 25:1. On bicycle tires the effect is hardly visible
while on a car tire the cross section at the middle of the contact
patch often looks even narrower than the rest of the tire.
Jobst Brandt
* * Chas
<jl...@comcast.net> wrote in message
news:1158865826....@i42g2000cwa.googlegroups.com...
> 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.
>
Levitation!
Chas.
Blair P. Houghton
>A tyre whose casing obeyed Hooke's Law would inflate until it got very
>big and then burst.
Bursting is a non-linear effect. Hooke's Law would have
nothing to say about it.
>I think it would be like a balloon-- more air in and
>it gets bigger, but the pressure stays the same. If it were very stiff
>to start with though, it could hold 110psi, and then you'd just inflate
>it to the size you needed.
You understand the concept. I don't know why you
don't understand the law.
>You could even make a universal tyre this way that would fit rim sizes
>from 20" up to 700c; although it would increasingly become a mountain
>bike tyre as you got to the larger diameters.
I don't think it would work that way. You'd be expanding
it in all directions as you inflated it. It would have to
be sized to both rim and tire diameter. Because if it was
1 inch wide on a 20 inch rim, it'd be 1.35 inches wide on a
27-inch rim.
>It would also be easy to fit to the rim. You'd inflate it until its
>diameter was a bit too big, slide it on easily, and then let a bit of
>air out to shrink it on tightly.
Don't forget the glue. It's the only thing keeping it from
precessing under load.
--Blair
Blair P. Houghton
Clinchers are not a diversion. They're nearly magic.
Car tire beads don't hook. They stick by friction.
Sometimes glue and sometimes bolt-on bead keepers
are used, but they're not necessary in general-purpose
tires. And there's no tube in there. The only interface
between the rim and the tire is a vertical ring of friction
that keeps the air in and keeps the entire vehicle off
the ground. And it sticks when you're matting the pedal,
and it almost always stays sealed even though you're
driving over bumps and rocks and curbs and junk.
Bike clinchers are not entirely different. The tube
is just in there to help keep the air in, because the
sidewall is much smaller so even with triple the pressure
the normal force on the bead seat is too small to make
an airtight seal. Same reason bike tires have more
aggressive bead hooks. Other than that, the tube is
irrelevant to transferring load from wheel to ground;
it could be replaced by thin strips of caulk or o-rings
at the tire beads. The rim's transverse surface has
the same air pressure on it all the way around the rim.
The bead transfers the entire working load.
With an unglued tubular, one of three things can be happening:
1. The rim rides on the tube like it is a rail of infinite
length. Whether it's full of air or foam rubber is
irrelevant. Whether it touches the rim anywhere but
a spot at the bottom the same size as the contact patch
is also irrelevant. The entire rim is supported by that
rim-tube contact patch at the bottom.
2. The rim is gripped all around its circumference by the
tube, leading to a distribution of tangential and normal
forces between the tube and rim. If the contact surface
is large it acts like an extended contact patch. Being
large and at the same pressure as the tire, it is of course
a much larger total force. Changing the size of the contact
patch (lifting the wheel, sitting on the bike, hitting a
bump, etc.) creates a small change in the rim-tube contact
force and the size of the rim-tube contact surface. But
there's no reason to believe that any particular portion of
the interface has a normal force different from any other.
The entire load can be distributed as varying tangential
forces around the wheel, holding it up by static friction.
3. A combination of the these. I suspect the reality
is a lot of #1 and a tiny bit of #2. There's a competing
action between making a tire stick to the rim and keeping
the rim off the ground. The latter wants more air pressure
and the former wants less.
With a glued tubular there will be more #2, but probably
not a lot more.
Oh. And you're not going to compress any of these until
your fingers touch. Not even if the tire is just a tube
of stretchy rubber with enough of a spring constant to
hold the size of a tire at full pressure.
--Blair
"Ask Marvin."
Ben C
> Ben C <spam...@spam.eggs> wrote:
>>A tyre whose casing obeyed Hooke's Law would inflate until it got very
>>big and then burst.
>
> Bursting is a non-linear effect. Hooke's Law would have
> nothing to say about it.
Correct. I admit I was imprecise. But the point is that most materials
only obey Hooke's Law as far as their elastic limits.
>>I think it would be like a balloon-- more air in and
>>it gets bigger, but the pressure stays the same. If it were very stiff
>>to start with though, it could hold 110psi, and then you'd just inflate
>>it to the size you needed.
>
> You understand the concept. I don't know why you
> don't understand the law.
I do understand Hooke's Law, I just would be inclined to describe a
material that obeyed Hooke's Law as having "constant elasticity" rather
than "linear elasticity". But I may be at odds with common usage here.
>
>>You could even make a universal tyre this way that would fit rim sizes
>>from 20" up to 700c; although it would increasingly become a mountain
>>bike tyre as you got to the larger diameters.
>
> I don't think it would work that way. You'd be expanding
> it in all directions as you inflated it. It would have to
> be sized to both rim and tire diameter. Because if it was
> 1 inch wide on a 20 inch rim, it'd be 1.35 inches wide on a
> 27-inch rim.
Yes exactly, that's what I meant by "become a mountain bike tyre". You
can fit quite a wide range of tyre widths though on a given rim.
>>It would also be easy to fit to the rim. You'd inflate it until its
>>diameter was a bit too big, slide it on easily, and then let a bit of
>>air out to shrink it on tightly.
>
> Don't forget the glue. It's the only thing keeping it from
> precessing under load.
What do you mean by "precessing" in this context?
Blair P. Houghton
>I do understand Hooke's Law, I just would be inclined to describe a
>material that obeyed Hooke's Law as having "constant elasticity" rather
>than "linear elasticity". But I may be at odds with common usage here.
I think "constant" is more correct, but I didn't want
to confuse the hammerheads.
>>>You could even make a universal tyre this way that would fit rim sizes
>>>from 20" up to 700c; although it would increasingly become a mountain
>>>bike tyre as you got to the larger diameters.
>>
>> I don't think it would work that way. You'd be expanding
>> it in all directions as you inflated it. It would have to
>> be sized to both rim and tire diameter. Because if it was
>> 1 inch wide on a 20 inch rim, it'd be 1.35 inches wide on a
>> 27-inch rim.
>
>Yes exactly, that's what I meant by "become a mountain bike tyre". You
>can fit quite a wide range of tyre widths though on a given rim.
But a mountain bike tire is fatter and smaller than a road tire.
There'd have to be variant forms for it.
>> Don't forget the glue. It's the only thing keeping it from
>> precessing under load.
>
>What do you mean by "precessing" in this context?
I mean that the tube effectively rotates backwards on
the rim, because the rim rolls forward on the tube.
The rim's diameter is a constant, but the tube's isn't.
The tube is in fact longer than the rotating diameter of
the wheel is, due to compression at the contact patch.
So if it's allowed to the tube will relieve this imbalance
by acting like it's larger than the rim.
Without the glue, or sufficient static friction, it will
have to stretch away from the valve body until the tension
is enough to keep it in place.
--Blair