Tire Pressure and Sidewalls

On 17/03/16 08:40, Frank Krygowski wrote:
On 3/16/2016 5:50 PM, James wrote:
On 17/03/16 01:02, Frank Krygowski wrote:
On 3/16/2016 2:25 AM, James wrote:
On 16/03/16 13:17, Frank Krygowski wrote:
On 3/15/2016 10:47 PM, James wrote:
Dear Frank (and Joe),

Did you notice my reply (see below)? Am I barking mad?

I noticed it, and I don't think it's wrong. But I'm not sure it's the
answer, because I'm no longer sure what our precise question is!
That's
a consequence of normal thread drift.

So: Let's re-state the problem (so we all agree on what we're
trying to
solve).



The question was of how tyre width and inflation pressure affect the
spring rate.

I suspect that it is proportional to the rate of change of area
flattened as the tyre is deformed under load.

OK. Spring rate, as usually defined, is change in force divided by
change in distance - and in this instance, we're talking about a
vertical distance between the rim and the ground. So let's call that
"y".

K = dF/dy

As you said, dA = dF/P (and it is entirely reasonable to say P is
constant for a given tire on a given ride).

That gives
K = dA/dy * P
based on zero sidewall support. (And I note that the units work out
fine.)

I think that's reasonable for thinwall tires. Super-thick Joerg-duty
sidewalls would add some vertical stiffness, I think. As a first
approximation, we could add that as a parallel constant, maybe Kw.

K = dA/dy * P + Kw

I think Kw is more complex, more akin to the buckling of a tall column;
i.e. with higher stiffness initially as portions of the thick wall
remain closer to vertical, and lower stiffness as the bulge increases.
But I don't want to try to guess at that function.

Again, I think that for performance-oriented road tires ("supple tires")
Kw can be neglected.

So if K = dA/dy * P

obviously more pressure gives a stiffer tire. Duh.

dA/dy depends on tire geometry, i.e. diameter and width. And I note
that dA/dy is not dimensionless; it's got units of length (inches or
mm). It's not a percentage-wise change. That seems to say that dA/dy
is larger for wider tires or larger diameter wheels. That seems strange
to me, so double-check me on that.

Of course, wider tires typically go with lower pressures, and I suppose
the pressure is the dominant term. But for a given pressure, my
equation seems to say that a larger wheel diameter gives a stiffer ride.
I'd have thought that was wrong.


I don't see how the diameter of the wheel makes a difference, until
perhaps the diameter is very small, like less than 16" for example.

Before the wheel has a load, the perimeter of the tire is a true circle.
When load is applied the rim sinks downward. The portion of the tire
in contact with the road forms a chord of the circle.

For each millimeter of downward motion a larger diameter wheel generates
a longer chord compared to a smaller diameter wheel.

Larger wheels have other advantages, of course. But we're working on
the spring constant of a pneumatic tire right now.


I see. My point was that I doubt there would be an appreciable
difference caused by that effect, between the wheels commonly used on
adult size bicycles - e.g. 26", 27" 700c, 650c.

However there there is an appreciable difference in tyre width available
for that general size of wheel, and that certainly affects the spring
rate (which I guess is not constant in this case).

--
JS

Am 16.03.2016 um 16:02 schrieb Frank Krygowski:
On 3/16/2016 2:25 AM, James wrote:

The question was of how tyre width and inflation pressure affect the
spring rate.

I suspect that it is proportional to the rate of change of area
flattened as the tyre is deformed under load.

OK. Spring rate, as usually defined, is change in force divided by
change in distance - and in this instance, we're talking about a
vertical distance between the rim and the ground. So let's call that "y".

K = dF/dy

As you said, dA = dF/P (and it is entirely reasonable to say P is
constant for a given tire on a given ride).

That gives
K = dA/dy * P
based on zero sidewall support. (And I note that the units work out fine.)

Now let's apply the further simplifying assumption that a bike tire
would be like a car tire (it's not quite correct but almost).

Then A = chord (c) * width (w), with the width being constant.

K = dc/dy * P * w, so ignoring the higher-order effects, the spring rate
would be approximately proportional to width * pressu a narrow
high-pressure tire has similar stiffness as a wide medium-pressure tire.
A 2-inch tire can only be ridden at 100 psi if you have other
suspension - not really a surprise.

On 3/16/2016 9:54 PM, James wrote:
On 17/03/16 08:40, Frank Krygowski wrote:
On 3/16/2016 5:50 PM, James wrote:
On 17/03/16 01:02, Frank Krygowski wrote:
On 3/16/2016 2:25 AM, James wrote:
On 16/03/16 13:17, Frank Krygowski wrote:
On 3/15/2016 10:47 PM, James wrote:
Dear Frank (and Joe),

Did you notice my reply (see below)? Am I barking mad?

I noticed it, and I don't think it's wrong. But I'm not sure it's
the
answer, because I'm no longer sure what our precise question is!
That's
a consequence of normal thread drift.

So: Let's re-state the problem (so we all agree on what we're
trying to
solve).



The question was of how tyre width and inflation pressure affect the
spring rate.

I suspect that it is proportional to the rate of change of area
flattened as the tyre is deformed under load.

OK. Spring rate, as usually defined, is change in force divided by
change in distance - and in this instance, we're talking about a
vertical distance between the rim and the ground. So let's call that
"y".

K = dF/dy

As you said, dA = dF/P (and it is entirely reasonable to say P is
constant for a given tire on a given ride).

That gives
K = dA/dy * P
based on zero sidewall support. (And I note that the units work out
fine.)

I think that's reasonable for thinwall tires. Super-thick Joerg-duty
sidewalls would add some vertical stiffness, I think. As a first
approximation, we could add that as a parallel constant, maybe Kw.

K = dA/dy * P + Kw

I think Kw is more complex, more akin to the buckling of a tall column;
i.e. with higher stiffness initially as portions of the thick wall
remain closer to vertical, and lower stiffness as the bulge increases.
But I don't want to try to guess at that function.

Again, I think that for performance-oriented road tires ("supple
tires")
Kw can be neglected.

So if K = dA/dy * P

obviously more pressure gives a stiffer tire. Duh.

dA/dy depends on tire geometry, i.e. diameter and width. And I note
that dA/dy is not dimensionless; it's got units of length (inches or
mm). It's not a percentage-wise change. That seems to say that dA/dy
is larger for wider tires or larger diameter wheels. That seems
strange
to me, so double-check me on that.

Of course, wider tires typically go with lower pressures, and I suppose
the pressure is the dominant term. But for a given pressure, my
equation seems to say that a larger wheel diameter gives a stiffer
ride.
I'd have thought that was wrong.


I don't see how the diameter of the wheel makes a difference, until
perhaps the diameter is very small, like less than 16" for example.

Before the wheel has a load, the perimeter of the tire is a true circle.
When load is applied the rim sinks downward. The portion of the tire
in contact with the road forms a chord of the circle.

For each millimeter of downward motion a larger diameter wheel generates
a longer chord compared to a smaller diameter wheel.

Larger wheels have other advantages, of course. But we're working on
the spring constant of a pneumatic tire right now.


I see. My point was that I doubt there would be an appreciable
difference caused by that effect, between the wheels commonly used on
adult size bicycles - e.g. 26", 27" 700c, 650c.

Yes, those sizes are all fairly close in outside diameter. I do have a
bike with 20" wheels and another with 16" wheels, IIRC. (That one's on
loan to a friend.)


However there there is an appreciable difference in tyre width available
for that general size of wheel, and that certainly affects the spring
rate (which I guess is not constant in this case).

Yes, it's interesting that a tire seems to be a naturally progressive
spring.

BTW, in looking for something else, I just came across this:
http://www.sheldonbrown.com/tires.html
Note the section titled "How a Tire Supports its Load"
and the link to Encyclopedia Brittanica.

--
- Frank Krygowski

Frank Krygowski writes:

On 3/16/2016 9:54 PM, James wrote:
On 17/03/16 08:40, Frank Krygowski wrote:
On 3/16/2016 5:50 PM, James wrote:
On 17/03/16 01:02, Frank Krygowski wrote:
On 3/16/2016 2:25 AM, James wrote:
On 16/03/16 13:17, Frank Krygowski wrote:
On 3/15/2016 10:47 PM, James wrote:
Dear Frank (and Joe),

Did you notice my reply (see below)? Am I barking mad?

I noticed it, and I don't think it's wrong. But I'm not sure it's
the
answer, because I'm no longer sure what our precise question is!
That's
a consequence of normal thread drift.

So: Let's re-state the problem (so we all agree on what we're
trying to
solve).



The question was of how tyre width and inflation pressure affect the
spring rate.

I suspect that it is proportional to the rate of change of area
flattened as the tyre is deformed under load.

OK. Spring rate, as usually defined, is change in force divided by
change in distance - and in this instance, we're talking about a
vertical distance between the rim and the ground. So let's call that
"y".

K = dF/dy

As you said, dA = dF/P (and it is entirely reasonable to say P is
constant for a given tire on a given ride).

That gives
K = dA/dy * P
based on zero sidewall support. (And I note that the units work out
fine.)

I think that's reasonable for thinwall tires. Super-thick Joerg-duty
sidewalls would add some vertical stiffness, I think. As a first
approximation, we could add that as a parallel constant, maybe Kw.

K = dA/dy * P + Kw

I think Kw is more complex, more akin to the buckling of a tall column;
i.e. with higher stiffness initially as portions of the thick wall
remain closer to vertical, and lower stiffness as the bulge increases.
But I don't want to try to guess at that function.

Again, I think that for performance-oriented road tires ("supple
tires")
Kw can be neglected.

So if K = dA/dy * P

obviously more pressure gives a stiffer tire. Duh.

dA/dy depends on tire geometry, i.e. diameter and width. And I note
that dA/dy is not dimensionless; it's got units of length (inches or
mm). It's not a percentage-wise change. That seems to say that dA/dy
is larger for wider tires or larger diameter wheels. That seems
strange
to me, so double-check me on that.

Of course, wider tires typically go with lower pressures, and I suppose
the pressure is the dominant term. But for a given pressure, my
equation seems to say that a larger wheel diameter gives a stiffer
ride.
I'd have thought that was wrong.


I don't see how the diameter of the wheel makes a difference, until
perhaps the diameter is very small, like less than 16" for example.

Before the wheel has a load, the perimeter of the tire is a true circle.
When load is applied the rim sinks downward. The portion of the tire
in contact with the road forms a chord of the circle.

For each millimeter of downward motion a larger diameter wheel generates
a longer chord compared to a smaller diameter wheel.

Larger wheels have other advantages, of course. But we're working on
the spring constant of a pneumatic tire right now.


I see. My point was that I doubt there would be an appreciable
difference caused by that effect, between the wheels commonly used on
adult size bicycles - e.g. 26", 27" 700c, 650c.

Yes, those sizes are all fairly close in outside diameter. I do have
a bike with 20" wheels and another with 16" wheels, IIRC. (That one's
on loan to a friend.)


However there there is an appreciable difference in tyre width available
for that general size of wheel, and that certainly affects the spring
rate (which I guess is not constant in this case).

Yes, it's interesting that a tire seems to be a naturally progressive
spring.

BTW, in looking for something else, I just came across this:
http://www.sheldonbrown.com/tires.html
Note the section titled "How a Tire Supports its Load"
and the link to Encyclopedia Brittanica.

I believe that Jobst wrote that. Parts of it are inaccurate. To
first-order, the sidewall arc is a constant radius whose center is
directly above the edge of the contact patch. That is, the tangent at
the ground is parallel to the ground. As such, changing the tension in
the sidewalls has no direct force effect at the ground contact (the
force has no vertical component). It does effect the rim contact.

--
Joe Riel

On Thursday, March 17, 2016 at 12:07:10 PM UTC-4, JoeRiel wrote:
Frank Krygowski writes:

On 3/16/2016 9:54 PM, James wrote:
On 17/03/16 08:40, Frank Krygowski wrote:
On 3/16/2016 5:50 PM, James wrote:
On 17/03/16 01:02, Frank Krygowski wrote:
On 3/16/2016 2:25 AM, James wrote:
On 16/03/16 13:17, Frank Krygowski wrote:
On 3/15/2016 10:47 PM, James wrote:
Dear Frank (and Joe),

Did you notice my reply (see below)? Am I barking mad?

I noticed it, and I don't think it's wrong. But I'm not sure it's
the
answer, because I'm no longer sure what our precise question is!
That's
a consequence of normal thread drift.

So: Let's re-state the problem (so we all agree on what we're
trying to
solve).



The question was of how tyre width and inflation pressure affect the
spring rate.

I suspect that it is proportional to the rate of change of area
flattened as the tyre is deformed under load.

OK. Spring rate, as usually defined, is change in force divided by
change in distance - and in this instance, we're talking about a
vertical distance between the rim and the ground. So let's call that
"y".

K = dF/dy

As you said, dA = dF/P (and it is entirely reasonable to say P is
constant for a given tire on a given ride).

That gives
K = dA/dy * P
based on zero sidewall support. (And I note that the units work out
fine.)

I think that's reasonable for thinwall tires. Super-thick Joerg-duty
sidewalls would add some vertical stiffness, I think. As a first
approximation, we could add that as a parallel constant, maybe Kw.

K = dA/dy * P + Kw

I think Kw is more complex, more akin to the buckling of a tall column;
i.e. with higher stiffness initially as portions of the thick wall
remain closer to vertical, and lower stiffness as the bulge increases.
But I don't want to try to guess at that function.

Again, I think that for performance-oriented road tires ("supple
tires")
Kw can be neglected.

So if K = dA/dy * P

obviously more pressure gives a stiffer tire. Duh.

dA/dy depends on tire geometry, i.e. diameter and width. And I note
that dA/dy is not dimensionless; it's got units of length (inches or
mm). It's not a percentage-wise change. That seems to say that dA/dy
is larger for wider tires or larger diameter wheels. That seems
strange
to me, so double-check me on that.

Of course, wider tires typically go with lower pressures, and I suppose
the pressure is the dominant term. But for a given pressure, my
equation seems to say that a larger wheel diameter gives a stiffer
ride.
I'd have thought that was wrong.


I don't see how the diameter of the wheel makes a difference, until
perhaps the diameter is very small, like less than 16" for example.

Before the wheel has a load, the perimeter of the tire is a true circle.
When load is applied the rim sinks downward. The portion of the tire
in contact with the road forms a chord of the circle.

For each millimeter of downward motion a larger diameter wheel generates
a longer chord compared to a smaller diameter wheel.

Larger wheels have other advantages, of course. But we're working on
the spring constant of a pneumatic tire right now.


I see. My point was that I doubt there would be an appreciable
difference caused by that effect, between the wheels commonly used on
adult size bicycles - e.g. 26", 27" 700c, 650c.

Yes, those sizes are all fairly close in outside diameter. I do have
a bike with 20" wheels and another with 16" wheels, IIRC. (That one's
on loan to a friend.)


However there there is an appreciable difference in tyre width available
for that general size of wheel, and that certainly affects the spring
rate (which I guess is not constant in this case).

Yes, it's interesting that a tire seems to be a naturally progressive
spring.

BTW, in looking for something else, I just came across this:
http://www.sheldonbrown.com/tires.html
Note the section titled "How a Tire Supports its Load"
and the link to Encyclopedia Brittanica.

I believe that Jobst wrote that. Parts of it are inaccurate. To
first-order, the sidewall arc is a constant radius whose center is
directly above the edge of the contact patch. That is, the tangent at
the ground is parallel to the ground. As such, changing the tension in
the sidewalls has no direct force effect at the ground contact (the
force has no vertical component). It does effect the rim contact.

--
Joe Riel

Just curious.

If one tire 's sidewalls are a lot more flexible (less tension?) than another would that not increase the contact patch?

Sir Ridesalot writes:

On Thursday, March 17, 2016 at 12:07:10 PM UTC-4, JoeRiel wrote:
Frank Krygowski writes:

On 3/16/2016 9:54 PM, James wrote:
On 17/03/16 08:40, Frank Krygowski wrote:
On 3/16/2016 5:50 PM, James wrote:
On 17/03/16 01:02, Frank Krygowski wrote:
On 3/16/2016 2:25 AM, James wrote:
On 16/03/16 13:17, Frank Krygowski wrote:
On 3/15/2016 10:47 PM, James wrote:
Dear Frank (and Joe),

Did you notice my reply (see below)? Am I barking mad?

I noticed it, and I don't think it's wrong. But I'm not sure it's
the
answer, because I'm no longer sure what our precise question is!
That's
a consequence of normal thread drift.

So: Let's re-state the problem (so we all agree on what we're
trying to
solve).



The question was of how tyre width and inflation pressure affect the
spring rate.

I suspect that it is proportional to the rate of change of area
flattened as the tyre is deformed under load.

OK. Spring rate, as usually defined, is change in force divided by
change in distance - and in this instance, we're talking about a
vertical distance between the rim and the ground. So let's call that
"y".

K = dF/dy

As you said, dA = dF/P (and it is entirely reasonable to say P is
constant for a given tire on a given ride).

That gives
K = dA/dy * P
based on zero sidewall support. (And I note that the units work out
fine.)

I think that's reasonable for thinwall tires. Super-thick Joerg-duty
sidewalls would add some vertical stiffness, I think. As a first
approximation, we could add that as a parallel constant, maybe Kw.

K = dA/dy * P + Kw

I think Kw is more complex, more akin to the buckling of a tall column;
i.e. with higher stiffness initially as portions of the thick wall
remain closer to vertical, and lower stiffness as the bulge increases.
But I don't want to try to guess at that function.

Again, I think that for performance-oriented road tires ("supple
tires")
Kw can be neglected.

So if K = dA/dy * P

obviously more pressure gives a stiffer tire. Duh.

dA/dy depends on tire geometry, i.e. diameter and width. And I note
that dA/dy is not dimensionless; it's got units of length (inches or
mm). It's not a percentage-wise change. That seems to say that dA/dy
is larger for wider tires or larger diameter wheels. That seems
strange
to me, so double-check me on that.

Of course, wider tires typically go with lower pressures, and I suppose
the pressure is the dominant term. But for a given pressure, my
equation seems to say that a larger wheel diameter gives a stiffer
ride.
I'd have thought that was wrong.


I don't see how the diameter of the wheel makes a difference, until
perhaps the diameter is very small, like less than 16" for example.

Before the wheel has a load, the perimeter of the tire is a true circle.
When load is applied the rim sinks downward. The portion of the tire
in contact with the road forms a chord of the circle.

For each millimeter of downward motion a larger diameter wheel generates
a longer chord compared to a smaller diameter wheel.

Larger wheels have other advantages, of course. But we're working on
the spring constant of a pneumatic tire right now.


I see. My point was that I doubt there would be an appreciable
difference caused by that effect, between the wheels commonly used on
adult size bicycles - e.g. 26", 27" 700c, 650c.

Yes, those sizes are all fairly close in outside diameter. I do have
a bike with 20" wheels and another with 16" wheels, IIRC. (That one's
on loan to a friend.)


However there there is an appreciable difference in tyre width available
for that general size of wheel, and that certainly affects the spring
rate (which I guess is not constant in this case).

Yes, it's interesting that a tire seems to be a naturally progressive
spring.

BTW, in looking for something else, I just came across this:
http://www.sheldonbrown.com/tires.html
Note the section titled "How a Tire Supports its Load"
and the link to Encyclopedia Brittanica.

I believe that Jobst wrote that. Parts of it are inaccurate. To
first-order, the sidewall arc is a constant radius whose center is
directly above the edge of the contact patch. That is, the tangent at
the ground is parallel to the ground. As such, changing the tension in
the sidewalls has no direct force effect at the ground contact (the
force has no vertical component). It does effect the rim contact.

--
Joe Riel

Just curious.

If one tire 's sidewalls are a lot more flexible (less tension?) than another would that not increase the contact patch?

The question would be better asked the other way---if the sidewalls were
sufficiently rigid could that decrease the contact patch? Yes, in so
far as the sidewalls are no longer tangent to the contact.

--
Joe Riel

On 18/03/16 00:57, Frank Krygowski wrote:
On 3/16/2016 9:54 PM, James wrote:
On 17/03/16 08:40, Frank Krygowski wrote:
On 3/16/2016 5:50 PM, James wrote:
On 17/03/16 01:02, Frank Krygowski wrote:
On 3/16/2016 2:25 AM, James wrote:
On 16/03/16 13:17, Frank Krygowski wrote:
On 3/15/2016 10:47 PM, James wrote:
Dear Frank (and Joe),

Did you notice my reply (see below)? Am I barking mad?

I noticed it, and I don't think it's wrong. But I'm not sure it's
the
answer, because I'm no longer sure what our precise question is!
That's
a consequence of normal thread drift.

So: Let's re-state the problem (so we all agree on what we're
trying to
solve).



The question was of how tyre width and inflation pressure affect the
spring rate.

I suspect that it is proportional to the rate of change of area
flattened as the tyre is deformed under load.

OK. Spring rate, as usually defined, is change in force divided by
change in distance - and in this instance, we're talking about a
vertical distance between the rim and the ground. So let's call that
"y".

K = dF/dy

As you said, dA = dF/P (and it is entirely reasonable to say P is
constant for a given tire on a given ride).

That gives
K = dA/dy * P
based on zero sidewall support. (And I note that the units work out
fine.)

I think that's reasonable for thinwall tires. Super-thick Joerg-duty
sidewalls would add some vertical stiffness, I think. As a first
approximation, we could add that as a parallel constant, maybe Kw.

K = dA/dy * P + Kw

I think Kw is more complex, more akin to the buckling of a tall
column;
i.e. with higher stiffness initially as portions of the thick wall
remain closer to vertical, and lower stiffness as the bulge increases.
But I don't want to try to guess at that function.

Again, I think that for performance-oriented road tires ("supple
tires")
Kw can be neglected.

So if K = dA/dy * P

obviously more pressure gives a stiffer tire. Duh.

dA/dy depends on tire geometry, i.e. diameter and width. And I note
that dA/dy is not dimensionless; it's got units of length (inches or
mm). It's not a percentage-wise change. That seems to say that dA/dy
is larger for wider tires or larger diameter wheels. That seems
strange
to me, so double-check me on that.

Of course, wider tires typically go with lower pressures, and I
suppose
the pressure is the dominant term. But for a given pressure, my
equation seems to say that a larger wheel diameter gives a stiffer
ride.
I'd have thought that was wrong.


I don't see how the diameter of the wheel makes a difference, until
perhaps the diameter is very small, like less than 16" for example.

Before the wheel has a load, the perimeter of the tire is a true circle.
When load is applied the rim sinks downward. The portion of the tire
in contact with the road forms a chord of the circle.

For each millimeter of downward motion a larger diameter wheel generates
a longer chord compared to a smaller diameter wheel.

Larger wheels have other advantages, of course. But we're working on
the spring constant of a pneumatic tire right now.


I see. My point was that I doubt there would be an appreciable
difference caused by that effect, between the wheels commonly used on
adult size bicycles - e.g. 26", 27" 700c, 650c.

Yes, those sizes are all fairly close in outside diameter. I do have a
bike with 20" wheels and another with 16" wheels, IIRC. (That one's on
loan to a friend.)


However there there is an appreciable difference in tyre width available
for that general size of wheel, and that certainly affects the spring
rate (which I guess is not constant in this case).

Yes, it's interesting that a tire seems to be a naturally progressive
spring.

BTW, in looking for something else, I just came across this:
http://www.sheldonbrown.com/tires.html
Note the section titled "How a Tire Supports its Load"
and the link to Encyclopedia Brittanica.


I understand how a rim and spokes work together as a pre tensioned
structure. As you load up the axle, the force tries to flatten the
bottom of the rim, which reduces the tension in those spokes right at
the bottom, and as the rim bulges slightly either side of where the rim
touches the ground, so the tension in those spokes increases. The
change in tension in the spokes effectively trying the keep the rim
circular.

I guess the fibres in the side walls are doing something, but I'm not
sure how to explain what or how. However, I did stumble across an
article that seems very close to what we're discussing, with formulas
that could potentially be adapted to a tyre shape.

http://scholarworks.rit.edu/cgi/view...ontext=article

--
JS

On 18/03/16 02:07, Joe Riel wrote:
Frank Krygowski writes:

On 3/16/2016 9:54 PM, James wrote:
On 17/03/16 08:40, Frank Krygowski wrote:
On 3/16/2016 5:50 PM, James wrote:
On 17/03/16 01:02, Frank Krygowski wrote:
On 3/16/2016 2:25 AM, James wrote:
On 16/03/16 13:17, Frank Krygowski wrote:
On 3/15/2016 10:47 PM, James wrote:
Dear Frank (and Joe),

Did you notice my reply (see below)? Am I barking mad?

I noticed it, and I don't think it's wrong. But I'm not sure it's
the
answer, because I'm no longer sure what our precise question is!
That's
a consequence of normal thread drift.

So: Let's re-state the problem (so we all agree on what we're
trying to
solve).



The question was of how tyre width and inflation pressure affect the
spring rate.

I suspect that it is proportional to the rate of change of area
flattened as the tyre is deformed under load.

OK. Spring rate, as usually defined, is change in force divided by
change in distance - and in this instance, we're talking about a
vertical distance between the rim and the ground. So let's call that
"y".

K = dF/dy

As you said, dA = dF/P (and it is entirely reasonable to say P is
constant for a given tire on a given ride).

That gives
K = dA/dy * P
based on zero sidewall support. (And I note that the units work out
fine.)

I think that's reasonable for thinwall tires. Super-thick Joerg-duty
sidewalls would add some vertical stiffness, I think. As a first
approximation, we could add that as a parallel constant, maybe Kw.

K = dA/dy * P + Kw

I think Kw is more complex, more akin to the buckling of a tall column;
i.e. with higher stiffness initially as portions of the thick wall
remain closer to vertical, and lower stiffness as the bulge increases.
But I don't want to try to guess at that function.

Again, I think that for performance-oriented road tires ("supple
tires")
Kw can be neglected.

So if K = dA/dy * P

obviously more pressure gives a stiffer tire. Duh.

dA/dy depends on tire geometry, i.e. diameter and width. And I note
that dA/dy is not dimensionless; it's got units of length (inches or
mm). It's not a percentage-wise change. That seems to say that dA/dy
is larger for wider tires or larger diameter wheels. That seems
strange
to me, so double-check me on that.

Of course, wider tires typically go with lower pressures, and I suppose
the pressure is the dominant term. But for a given pressure, my
equation seems to say that a larger wheel diameter gives a stiffer
ride.
I'd have thought that was wrong.


I don't see how the diameter of the wheel makes a difference, until
perhaps the diameter is very small, like less than 16" for example.

Before the wheel has a load, the perimeter of the tire is a true circle.
When load is applied the rim sinks downward. The portion of the tire
in contact with the road forms a chord of the circle.

For each millimeter of downward motion a larger diameter wheel generates
a longer chord compared to a smaller diameter wheel.

Larger wheels have other advantages, of course. But we're working on
the spring constant of a pneumatic tire right now.


I see. My point was that I doubt there would be an appreciable
difference caused by that effect, between the wheels commonly used on
adult size bicycles - e.g. 26", 27" 700c, 650c.

Yes, those sizes are all fairly close in outside diameter. I do have
a bike with 20" wheels and another with 16" wheels, IIRC. (That one's
on loan to a friend.)


However there there is an appreciable difference in tyre width available
for that general size of wheel, and that certainly affects the spring
rate (which I guess is not constant in this case).

Yes, it's interesting that a tire seems to be a naturally progressive
spring.

BTW, in looking for something else, I just came across this:
http://www.sheldonbrown.com/tires.html
Note the section titled "How a Tire Supports its Load"
and the link to Encyclopedia Brittanica.

I believe that Jobst wrote that. Parts of it are inaccurate. To
first-order, the sidewall arc is a constant radius whose center is
directly above the edge of the contact patch. That is, the tangent at
the ground is parallel to the ground. As such, changing the tension in
the sidewalls has no direct force effect at the ground contact (the
force has no vertical component). It does effect the rim contact.


Yep. As in my reply to Frank, this is a good reference.

http://scholarworks.rit.edu/cgi/view...ontext=article

--
JS

On 18/03/16 02:37, Joe Riel wrote:
Sir Ridesalot writes:



Just curious.

If one tire 's sidewalls are a lot more flexible (less tension?) than another would that not increase the contact patch?

The question would be better asked the other way---if the sidewalls were
sufficiently rigid could that decrease the contact patch? Yes, in so
far as the sidewalls are no longer tangent to the contact.


And the natural followup question, how much?

Compared to 100psi in a 23mm tyre, or 40psi in a 40mm tyre, I wouldn't
imagine the normal range of sidewall thicknesses available would
contribute significantly to the static spring rate.

However, it is obvious that a stiff sidewall will dissipate more energy
than a thin and supple sidewall of a similar tyre at similar inflation
pressure. So the sidewall stiffness, IMO, contributes to dynamic
damping in effect, which manifests as rolling resistance.

--
JS

James writes:

On 18/03/16 02:37, Joe Riel wrote:
Sir Ridesalot writes:



Just curious.

If one tire 's sidewalls are a lot more flexible (less tension?) than another would that not increase the contact patch?

The question would be better asked the other way---if the sidewalls were
sufficiently rigid could that decrease the contact patch? Yes, in so
far as the sidewalls are no longer tangent to the contact.


And the natural followup question, how much?

Compared to 100psi in a 23mm tyre, or 40psi in a 40mm tyre, I wouldn't
imagine the normal range of sidewall thicknesses available would
contribute significantly to the static spring rate.

However, it is obvious that a stiff sidewall will dissipate more
energy than a thin and supple sidewall of a similar tyre at similar
inflation pressure. So the sidewall stiffness, IMO, contributes to
dynamic damping in effect, which manifests as rolling resistance.

That is my working assumption. The model I've created assumes a
flexible, lossless, sidewall. I've been thinking of simple ways to
measure, approximately, sidewall losses. One thought is to cut a
section of the sidewall out and use it as the hinge in a pendulum with
sufficient weight to bring it to typical tension. However, if losses
are due to the belts rubbing on one another, the pressure in a tire
should increase those losses, and that setup has no such pressure.

A more indirect method is to measure the coefficient of restitution of
an inflated tire by bouncing it. With the spring constant known (or
assumed) that can be fitted to an appropriate lossy model. I like
it because it is extremely simple and doable.

--
Joe Riel