Re: Tire Thread Count - What difference does it make?

Douglas Landau said:

How much footprint there is depends upon the PSI to which the tire is
inflated, not how stiff it is. If the tires are inflated to 20PSI,
and the bike+rider weigh 200 lbs, then the tires are going to lay
down ten square inches of rubber, no matter what.

So one of my car tires, which are about 10" wide, will leave the same
footprint as one of my 700c x 23mm bike tires if they're both inflated
to the same psi & have the same weight pressing down on them?

Douglas Landau said:

Quoted message said:

How much footprint there is depends upon the PSI to which the tire is
inflated, not how stiff it is. If the tires are inflated to 20PSI,
and the bike+rider weigh 200 lbs, then the tires are going to lay
down ten square inches of rubber, no matter what.

Jay Hill said:

So one of my car tires, which are about 10" wide, will leave the same=20
footprint as one of my 700c x 23mm bike tires if they're both inflated =

Quoted message said:

to the same psi & have the same weight pressing down on them?

That's basically correct. It's a bit slippery because the appropriate=20
pressures are so different. If you actually were to run a 23 mm tire at =

20 psi, it would bottom out on the rim, so pneumatic effects would=20
disappear.

However, if you were to pump up your 10" wide car tires to 100 psi, and=20
to load each tire with a 100 pounds, you'd get a contact patch 10 inches =

wide and 1 inch front to back.

This is true in theory. In practice, a tire with a thicker, stiffer=20
tread may have a slightly smaller contact patch (and slightly higher=20
average pressure against the pavement) due to the stiffness of the tire=20
not permitting it to completely conform. Thus, your car tires, in=20
practice, would actually have a slightly _smaller_ contact patch than=20
your bike tires, for the same pressure and loading.

Sheldon "Counterintuitive" Brown
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Phone 617-244-9772 FAX 617-244-1041
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Hard-to-find parts shipped Worldwide
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ZeeExSixAre said:

What I don't get is whether the thick-sidewall tire has more or less
rolling resistance compared to the thin-walled tire.

The thick-sidewall tire has more rolling resistance, because more
viscoelastic material is deformed as it is loaded and unloaded.
--
terry morse Palo Alto, CA http://bike.terrymorse.com/

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

Quoted message said:

Douglas Landau said:

How much footprint there is depends upon the PSI to which the tire is
inflated, not how stiff it is. If the tires are inflated to 20PSI,
and the bike+rider weigh 200 lbs, then the tires are going to lay
down ten square inches of rubber, no matter what.

So one of my car tires, which are about 10" wide, will leave the same
footprint as one of my 700c x 23mm bike tires if they're both inflated
to the same psi & have the same weight pressing down on them?

Sure, at least in the clean case. I admit that not all cases are clean.
In fact, the case in which I frst read what I was quoting is itself not
clean.

What I wrote, I quoted straight from "the Boonie Book", a book about
dirtbiking from the 70s which I had when I was a kid. The author said
that about dirtbike tires, addressing the issue of whether a rider riding
sideways across a slope should weight the uphill peg or the downhill peg
for best traction. The author claimed that it does not matter, and that
was his reasoning.

However, I say that in fact, if you park your dirtbike on clean rock, you
will see that in fact only a few knobs of each tire are touching the rock,
nowhere near the number of square inches of rubber which should be according
to the theory.

So, the contactPatch=Load/PSI formula is too simplistic to be absolutely
correct. Bacardi is right, to some extent. I should have truncated the
"no matter what". That said, the case of roadbike tires is a pretty clean
case.

Doug

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

Quoted message said:

Douglas Landau said:

How much footprint there is depends upon the PSI to which the tire is
inflated, not how stiff it is. If the tires are inflated to 20PSI,
and the bike+rider weigh 200 lbs, then the tires are going to lay
down ten square inches of rubber, no matter what.

So one of my car tires, which are about 10" wide, will leave the same
footprint as one of my 700c x 23mm bike tires if they're both inflated
to the same psi & have the same weight pressing down on them?

Sure, at least in the clean case. I admit that not all cases are clean.
In fact, the case in which I frst read what I was quoting is itself not
clean.

What I wrote, I quoted straight from "the Boonie Book", a book about
dirtbiking from the 70s which I had when I was a kid. The author said
that about dirtbike tires, addressing the issue of whether a rider riding
sideways across a slope should weight the uphill peg or the downhill peg
for best traction. The author claimed that it does not matter, and that
was his reasoning.

However, I say that in fact, if you park your dirtbike on clean rock, you
will see that in fact only a few knobs of each tire are touching the rock,
nowhere near the number of square inches of rubber which should be according
to the theory.

So, the contactPatch=Load/PSI formula is too simplistic to be absolutely
correct. Bacardi is right, to some extent. I should have truncated the
"no matter what". That said, the case of roadbike tires is a pretty clean
case.

Doug

Jay Hill said:

Quoted message said:

How much footprint there is depends upon the PSI to which the tire
is inflated, not how stiff it is. If the tires are inflated to
20PSI, and the bike+rider weigh 200 lbs, then the tires are going
to lay down ten square inches of rubber, no matter what.

Quoted message said:

So one of my car tires, which are about 10" wide, will leave the
same footprint as one of my 700c x 23mm bike tires if they're both
inflated to the same psi & have the same weight pressing down on
them?

Well that's a bad example because bare tire, no rim, no inflation, can
support a person's full weight a because it is so rigid. Most of us
at one time or another sat on such a tire at an auto shop or played on
a swing made of an old tire.

In theory the idea is correct but practically automotive tires don't
have enough compliance to behave that way with small loads.

Jobst Brandt
[email hidden]

Douglas Landau said:

Quoted message said:

Quoted message said:

How much footprint there is depends upon the PSI to which the tire
is inflated, not how stiff it is. If the tires are inflated to
20PSI, and the bike+rider weigh 200 lbs, then the tires are going
to lay down ten square inches of rubber, no matter what.

Quoted message said:

Quoted message said:

So one of my car tires, which are about 10" wide, will leave the
same footprint as one of my 700c x 23mm bike tires if they're both
inflated to the same psi & have the same weight pressing down on
them?

Quoted message said:

Sure, at least in the clean case. I admit that not all cases are
clean. In fact, the case in which I first read what I was quoting is
itself not clean.

Quoted message said:

What I wrote, I quoted straight from "the Boonie Book", a book about
dirtbiking from the 70s which I had when I was a kid. The author
said that about dirtbike tires, addressing the issue of whether a
rider riding sideways across a slope should weight the uphill peg or
the downhill peg for best traction. The author claimed that it does
not matter, and that was his reasoning.

Clean or dirty, on what foot you stand has something to do with side
slope pedal clearance but nothing to do with traction. Let's not get
all crossed up here.

Quoted message said:

However, I say that in fact, if you park your dirtbike on clean
rock, you will see that in fact only a few knobs of each tire are
touching the rock, nowhere near the number of square inches of
rubber which should be according to the theory.

Oh [censored]! That's like saying when you ride over a metal mesh the
contact patch is only 1/10 or the like. You can also balance a
bicycle on the head of a nail with less than 1/10 square inch area.
The same goes for a contact patch on a chip-seal road.

Contact patch in this context means what the flattened area of the tire
that presses against the road is, not what the actual intimate material
contact is. If you want to go into that, true contact, as is used in
tribology, is a difficult area to define.

Quoted message said:

So, the contactPatch=Load/PSI formula is too simplistic to be
absolutely correct. Bacardi is right, to some extent. I should
have truncated the "no matter what". That said, the case of
roadbike tires is a pretty clean case.

Don't be such a sophist, contact patch is a valid term for bicycle
tires on smooth surfaces. The area of that patch is governed by
inflation pressure. Odd surfaces of tire or road don't fit into that
description. That smooth tread has finally arrived with bicycles took
so long because the most recalcitrant technical conservatives are
found among bicyclists.

Jobst Brandt
[email hidden]

Douglas Landau said:

However, I say that in fact, if you park your dirtbike on clean
rock, you will see that in fact only a few knobs of each tire are
touching the rock, nowhere near the number of square inches of
rubber which should be according to the theory.

Quoted message said:

So, the contactPatch=Load/PSI formula is too simplistic to be
absolutely correct. Bacardi is right, to some extent. I should
have truncated the "no matter what". That said, the case of
roadbike tires is a pretty clean case.

You might also consider this:

http://draco.acs.uci.edu/rbfaq/FAQ/8b.25.html

Jobst Brandt
[email hidden]

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

Quoted message said:

BaCardi said:

Douglas Landau said:

How much footprint there is depends upon the PSI to which the tire is
inflated, not how stiff it is. If the tires are inflated to 20PSI, and
the bike+rider weigh 200 lbs, then the tires are going to lay down ten
square inches of rubber, no matter what.

Quoted message said:

Yes, but a thinner tire leads to greater deformation under load and thus
a larger contact area with the road. Greater contact area with road =greater
rolling resistance.

You might want to go back and read what Doug is patiently trying to
explain to you.

The area of the contact patch is determined by the pressure in the tire
and the weight carried.

I am foolish to enter into a discussion re how many angels can dance
on the head of a pin so I shall just throw this in and back quickly
towards the door....

I offer the following as nothing more than a backyard experiment, but
might be food for thought.

http://www.tomschmitz.org/Contact%20PatchFrame1Source1.htm

Regards,

Tom

BaCardi said:

Ted Bennett said:

BaCardi said:

Douglas Landau wrote:
> How much footprint there is depends upon the PSI to which the tire
> is inflated, not how stiff it is. If the tires are inflated to
> 20PSI, and the bike+rider weigh 200 lbs, then the tires are going
> to lay down ten square inches of rubber, no matter what.
Yes, but a thinner tire leads to greater deformation under load and
thus a larger contact area with the road. Greater contact area with
road =greater rolling resistance.

You might want to go back and read what Doug is patiently trying to
explain to you.
The area of the contact patch is determined by the pressure in the tire
and the weight carried.
--
Ted Bennett Portland OR

OK, and your premise is based on the area of contact patch IS determined

Quoted message said:

by the pressure in the tire and weight. Right? So, so basically what you

and

Quoted message said:

Doug are saying is not the same as what Jobst patiently explained. I

think

Quoted message said:

you ought to go back and read what Jobst is saying.

I don't see any disagreement between Jobst and me in this case.

--
Ted Bennett
Portland OR

Tom Schmitz said:

Quoted message said:

The area of the contact patch is determined by the pressure in the
tire and the weight carried.

Quoted message said:

I am foolish to enter into a discussion re how many angels can dance
on the head of a pin so I shall just throw this in and back quickly
towards the door....

Quoted message said:

I offer the following as nothing more than a backyard experiment,
but might be food for thought.

Quoted message said:

http://www.tomschmitz.org/Contact%20PatchFrame1Source1.htm

That's too bad that the test was done with a tire that had a thick and
clefted tread rubber that squirmed in a way that did not represent the
contact patch that would appear if a light weight road tire had been
used. This experiment was like stepping on a jellyfish and measuring
contact area. The contact patch of a toroidal shaped tire is a canoe
form. This one was not.

Jobst Brandt
[email hidden]

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

Quoted message said:

Tom Schmitz writes:

Quoted message said:

Quoted message said:

I offer the following as nothing more than a backyard experiment,
but might be food for thought.

Quoted message said:

http://www.tomschmitz.org/Contact%20PatchFrame1Source1.htm

That's too bad that the test was done with a tire that had a thick and
clefted tread rubber that squirmed in a way that did not represent the
contact patch that would appear if a light weight road tire had been
used. This experiment was like stepping on a jellyfish and measuring

Jobst -

The tire in question, when new, weighed in at about 220 grams. The tread was
not unusually thick.
The "clefted tread" you refer to is nothing more than typically useless
rain siping.

Quoted message said:

contact area. The contact patch of a toroidal shaped tire is a canoe
form. This one was not.

I would like to see pictures of such a shape if you can provide them. I
would agree that one might
expect a canoe shape of a new tire's imprint, but what of a tire that has
1500 miles of use? Such a tire
will not have a perfectly toroidal shape as there will be a flat wear
pattern in the center of the tread.

Even so, I should think that this would not have an effect on the area of
the contact patch, only it's shape.
I shall find it personally enlightening to be able to do the test over with
new tires, in various sizes, in a more
controlled fashion. This I hope to be able to do quite soon as I have
everything I need except a round tuit.

Regards,

Tom

Ted Bennett said:

BaCardi said:

Ted Bennett said:

BaCardi wrote:
> Douglas Landau wrote:
> > How much footprint there is depends upon the PSI to which
> > the tire is inflated, not how stiff it
> > is. If the tires are inflated to 20PSI, and the bike+rider
> > weigh 200 lbs, then the tires are going to lay down ten
> > square inches of rubber, no matter what.
>Yes, but a thinner tire leads to greater deformation under load
>and thus a larger contact area with the road. Greater contact
>area with road =greater rolling resistance.
You might want to go back and read what Doug is patiently trying
to explain to you. The area of the contact patch is determined by
the pressure in the tire and the weight carried.
--
Ted Bennett Portland OR

OK, and your premise is based on the area of contact patch IS determined

Quoted message said:

by the pressure in the tire and weight. Right? So, so basically
what you

and

Quoted message said:

Doug are saying is not the same as what Jobst patiently explained. I

think

Quoted message said:

you ought to go back and read what Jobst is saying.

I don't see any disagreement between Jobst and me in this case.
--
Ted Bennett Portland OR

Foolish! Try again! Re-read the paragraphs on the characteristics by
force that deformed the material is not returned.

--

Tom Schmitz said:

Quoted message said:

Quoted message said:

I offer the following as nothing more than a backyard experiment,
but might be food for thought.

http://www.tomschmitz.org/Contact%20PatchFrame1Source1.htm

Quoted message said:

Quoted message said:

That's too bad that the test was done with a tire that had a thick
and clefted tread rubber that squirmed in a way that did not
represent the contact patch that would appear if a light weight
road tire had been used. This experiment was like stepping on a
jellyfish and measuring...

Quoted message said:

The tire in question, when new, weighed in at about 220 grams. The
tread was not unusually thick. The "clefted tread" you refer to is
nothing more than typically useless rain siping.

Quoted message said:

Quoted message said:

... contact area. The contact patch of a toroidal shaped tire is a
canoe form. This one was not.

Quoted message said:

I would like to see pictures of such a shape if you can provide
them. I would agree that one might expect a canoe shape of a new
tire's imprint, but what of a tire that has 1500 miles of use? Such
a tire will not have a perfectly toroidal shape as there will be a
flat wear pattern in the center of the tread.

Oh! The tire was worn flat. That means its unloaded contact area is
already fairly large and need hardly deform to make a fairly large
stamp pad mark on paper. That is to be expected since that area is
somewhat independent of inflation pressure.

Quoted message said:

Even so, I should think that this would not have an effect on the
area of the contact patch, only it's shape. I shall find it
personally enlightening to be able to do the test over with new
tires, in various sizes, in a more controlled fashion. This I hope
to be able to do quite soon as I have everything I need except a
round tuit.

I did my tests with a new round cross section smooth tire. Since the
tread rubber has an obvious stiffness, the area of that patch,
although canoe shaped did not match inflation pressure better than by
eyeball. The only tire that will accurately reflect that would be a
less than 120 gram track tubular. Above that other effects change the
shape and size of the contact as you demonstrate while tread stiffness
reduces contact of a smooth tread round cross section tire. In that
case contact pressure is not uniform across the area but higher in the
center due to bending forces.

In any case, the compliance of the tire being constant, inflation
pressure has the effects described although in practice they may not
be exactly the computed ones of thin walled round cross section tires.
The trend is accurate for traction purposes.

Car tires, in contrast, are mostly radial and therefore belted. Belts
can give a broad, non-circular cross section as you can see on many
cars that allude to being race cars in disguise.

Jobst Brandt
[email hidden]

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

Quoted message said:

Douglas Landau said:

Quoted message said:

> How much footprint there is depends upon the PSI to which the tire
> is inflated, not how stiff it is. If the tires are inflated to
> 20PSI, and the bike+rider weigh 200 lbs, then the tires are going
> to lay down ten square inches of rubber, no matter what.

Quoted message said:

Quoted message said:

So one of my car tires, which are about 10" wide, will leave the
same footprint as one of my 700c x 23mm bike tires if they're both
inflated to the same psi & have the same weight pressing down on
them?

Quoted message said:

Sure, at least in the clean case. I admit that not all cases are
clean. In fact, the case in which I first read what I was quoting is
itself not clean.

Quoted message said:

What I wrote, I quoted straight from "the Boonie Book", a book about
dirtbiking from the 70s which I had when I was a kid. The author
said that about dirtbike tires, addressing the issue of whether a
rider riding sideways across a slope should weight the uphill peg or
the downhill peg for best traction. The author claimed that it does
not matter, and that was his reasoning.

Clean or dirty, on what foot you stand has something to do with side
slope pedal clearance but nothing to do with traction. Let's not get
all crossed up here.

Quoted message said:

However, I say that in fact, if you park your dirtbike on clean
rock, you will see that in fact only a few knobs of each tire are
touching the rock, nowhere near the number of square inches of
rubber which should be according to the theory.

Oh [censored]! That's like saying when you ride over a metal mesh the
contact patch is only 1/10 or the like. You can also balance a
bicycle on the head of a nail with less than 1/10 square inch area.
The same goes for a contact patch on a chip-seal road.

Contact patch in this context means what the flattened area of the tire
that presses against the road is, not what the actual intimate material
contact is. If you want to go into that, true contact, as is used in
tribology, is a difficult area to define.

Ok, fine. I used "footprint" when responding to Bacardi, when I
should have used "contact patch". Then I used "contact patch" when
responding to Jay when I should have used "footprint".

Quoted message said:

Quoted message said:

So, the contactPatch=Load/PSI formula is too simplistic to be
absolutely correct. Bacardi is right, to some extent. I should
have truncated the "no matter what". That said, the case of
roadbike tires is a pretty clean case.

Don't be such a sophist, contact patch is a valid term for bicycle
tires on smooth surfaces. The area of that patch is governed by
inflation pressure.

Now -there's- the pot calling the kettle black! In any case, you are
agreeing with me here, as you do again in your next post where you write

Quoted message said:

Well that's a bad example because ...

Doug

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

Quoted message said:

Tom Schmitz writes:>

Quoted message said:

I would like to see pictures of such a shape if you can provide
them. I would agree that one might expect a canoe shape of a new
tire's imprint, but what of a tire that has 1500 miles of use? Such
a tire will not have a perfectly toroidal shape as there will be a
flat wear pattern in the center of the tread.

Oh! The tire was worn flat. That means its unloaded contact area is
already fairly large and need hardly deform to make a fairly large
stamp pad mark on paper. That is to be expected since that area is
somewhat independent of inflation pressure.

Indeed, it is. I wish that I had possessed the presence of mind to record
the imprint of that tire with no load on it. That would have been a good
baseline. I shall remember that in the next go-round.

Quoted message said:

Quoted message said:

Even so, I should think that this would not have an effect on the
area of the contact patch, only it's shape. I shall find it
personally enlightening to be able to do the test over with new
tires, in various sizes, in a more controlled fashion. This I hope
to be able to do quite soon as I have everything I need except a
round tuit.

I did my tests with a new round cross section smooth tire. Since the
tread rubber has an obvious stiffness, the area of that patch,
although canoe shaped did not match inflation pressure better than by
eyeball. The only tire that will accurately reflect that would be a
less than 120 gram track tubular. Above that other effects change the
shape and size of the contact as you demonstrate while tread stiffness
reduces contact of a smooth tread round cross section tire. In that
case contact pressure is not uniform across the area but higher in the
center due to bending forces.

In any case, the compliance of the tire being constant, inflation
pressure has the effects described although in practice they may not
be exactly the computed ones of thin walled round cross section tires.
The trend is accurate for traction purposes.

This is the part that interests me. I would like to see what happens with
brand new tires, identical except for section width. The tires I could
afford to accumulate are not top of the line, but they are what one would
call light road tires - in the range of 220-230 grams - that one would
actually ride on daily. Unfortunately, they are siped. I have some Michelin
Hi-Lites (smooth tread) in a couple of section widths, but the construction
of that tire is unique with its woven carcass.

I no longer have a wheel to mount sew-ups on, though I could build one from
junque-at-hand. My pile of tubulars, though, is very short and doesn't
include anything in the track category.

Regards,

Tom

Tom said:

I offer the following as nothing more than a backyard experiment, but
might be food for thought.

http://www.tomschmitz.org/Contact%20PatchFrame1Source1.htm

Nice piece of work. I think the results are a little bit off, but your
basic experiment is neat.

It seems to me that the act of mounting a bike would cause the wheel to
roll fore and aft slightly as well as rocking from side to side. Any
motion would cause the contact area measurement to be exaggerated. In
addition, you may load the front tire more heavily than the nominal
value momentarily during the mounting process.

I'm not convinced that the contact patch is 2 x Load/Presure with a
"light road tire." The stiffness of the casing may make the contact
patch slightly greater than L/P (i.e. 5-10%), but I'd be very surprised
to see a factor of two.

And by the way, the ideal gas law has nothin' to do with it. You could
have an incompressible fluid in your tire and the balance of force would
still predict the same contact area for a given pressure.

--
Dave
dvt at psu dot edu

dvt said:

Tom said:

I offer the following as nothing more than a backyard experiment, but
might be food for thought.

http://www.tomschmitz.org/Contact%20PatchFrame1Source1.htm

Nice piece of work. I think the results are a little bit off, but your
basic experiment is neat.

It seems to me that the act of mounting a bike would cause the wheel
to roll fore and aft slightly as well as rocking from side to side.
Any motion would cause the contact area measurement to be exaggerated.
In addition, you may load the front tire more heavily than the nominal
value momentarily during the mounting process.

I was as careful as I could be to not have any of these things happen.
Of course, this being a backyard experiment, I didn't have 100% control
over everything. That's why I want to repeat it with an apparatus.

Quoted message said:

I'm not convinced that the contact patch is 2 x Load/Presure with a
"light road tire." The stiffness of the casing may make the contact
patch slightly greater than L/P (i.e. 5-10%), but I'd be very
surprised to see a factor of two.

Well, the nice thing about conducting my own research is that it gives
me something to think about, learn about, and I always reserve the right
to be wrong.

Quoted message said:

And by the way, the ideal gas law has nothin' to do with it. You could
have an incompressible fluid in your tire and the balance of force
would still predict the same contact area for a given pressure.

That is a tongue in cheek comment - I think I clearly stated that gas
laws have nothing to do with the experiment. Though I might rethink that
if more closely controlled experiments bear out the initial
results...... (that was a bit tongue in cheek, too)

Regards,

Tom

Quoted message said: