Tire Thread Count - What difference does it make?

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]

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.

tomschmitz.orgContact%20PatchFrame1Source1.htmOpen ↗

Regards,

Tom

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:

draco.acs.uci.edu8b.25.htmlOpen ↗

Jobst Brandt [email hidden]

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:

tomschmitz.orgContact%20PatchFrame1Source1.htmOpen ↗

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]

Quoted post said:

Originally posted by Ted Bennett

BBaCardi 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 [/B]

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

<[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:

tomschmitz.orgContact%20PatchFrame1SourcOpen ↗
e1.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

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.

tomschmitz.orgContact%20PatchFrame1Source1.htmOpen ↗

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.

tomschmitz.orgContact%20PatchFrame1Source1.htmOpen ↗

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.

tomschmitz.orgContact%20PatchFrame1SourOpen ↗
ce1.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