Tire Thread Count - What difference does it make?

Is that a genuine 290tpi? For example, Continental routinely
misrepresent their thread counts because they add up the
TPIs of all the carcass layers of one tyre.

Yes, I believe so. However, the 290 TPI was not for clinchers. The ones I got were 220 TPI. I remembered wrong!

To do as you say and add up plys is very irresponsible. Threads-per-inch should be just what it says.

Check these tires out. They are on sale for $25, they are lighter than Michelin Pro Race, have 220 tpi and you will know the difference over low tpi tires, if your frame doesn't absorb the shock. It was night and day on an aluminum bike. I have 2,000 miles on these with not a single flat (and I do occasionally run over glass.)

http://www.performancebike.com/shop/Profile.cfm?SKU=17216

To others, imagine your forward momentum as a horizontal vector. With stiff tires (low tpi), when you hit a bump, the front end of your bike lifts up, converting the horizontal momentum to an upward force which never is reconverted to forward motion. The upward motion transmits throughout the frame into your hands and muscles. This is rolling resistance.

A supple high tpi tire will deform and although the flex of the tires absorb some energy and equals heat loss, the energy wasted is minor compared to what is transmitted into your body from hard tires.

What goes on over a bump goes on constantly over asphalt. High tpi tires give the best of both worlds: comfort and low rolling resistance.

Quoted post said:

Originally posted by carlfogel

Basically, the less material that you flex into an
endless bulge as the tire meets the road, the
less energy you waste distorting it. This is the
real advantage of light tires and light tubes, the
latter becoming part of the tire under normal
pressures.

Some materials absorb more energy as they
flex (rubber versus latex, for example), but
the big problem with your tires is likely just
that there's so much of them. Big soggy chunks
of rubber, kevlar, and cotton with block treads
squirming uselessly to the sides don't roll as
well as thin, smooth wafers of highly pressurized
latex and silk.

Carl Fogel

I like this line of reasoning. However, if you are saying that stiffer tires = less rolling resistance due to less footprint where the tire rubber meets the road, then why would lighter thinner casing tires such as those with higher TPI count = less rolling resistance? Wouldn't you get more rolling resistance due to the thinner tire deforming more under the rider's weight and hence a bigger footprint where the tire rubber meets the road?

Quoted message said:

Oh, I've forgot, I used from latex to heavy duty tubes,
tubes with slime, and tubes surrounded by a second tube. I
can't tell the difference either. Right now, I stick to
the chepest butyl tube that I can find. I usually use
tires until there is nothing left. If a tire gets a big
tear, I cut a plastic cup and use the plastic to boot it.
I get rid of tires when I have two or three boots.

On that note, my commuter bikes have next to always been
old 27" ten-speeds. I use those thick, thorn-resistant
tubes because, when they do flat, they allow me to keep
riding to my destination before I do something about it.
Run-flat, if you will.

Quoted message said:

never is reconverted to forward motion. The upward motion
transmits throughout the frame into your hands and
muscles. This is rolling resistance.

?????

--
Phil, Squid-in-Training

Quoted post said:

Originally posted by some alcoholic
I like this line of reasoning. However, if you are saying that stiffer tires = less rolling resistance due to less footprint where the tire rubber meets the road, then why would lighter thinner casing tires such as those with higher TPI count = less rolling resistance? Wouldn't you get more rolling resistance due to the thinner tire deforming more under the rider's weight and hence a bigger footprint where the tire rubber meets the road?

You miss two of Carl’s points. The first has to do with the nature of the materials. In the railway wheel analogy the steel is behaving elastically at the contact patch – it is being loaded and unloaded along the same (linear) path in the load versus elongation plot. As work is force times the distance the force moves, no net work is being done.

In the case of the bicycle tyre, you are dealing with an anelastic material. Because its force vs elongation characteristic exhibits hysteresis, work must be done on it during loading and unloading. The rate of doing work is power and this power comes from the rider’s legs.

The area of the contact patch of a tyre is almost solely a function of force on the tyre (the weight) and its pressure. This just comes from the definition of pressure (force/area). The force supported by the sidewalls is negligible for a feasible bicycle tyre. Try pressing a fully deflated tyre with your thumb. So the thick walled tyre will have essentially the same contact patch area as its thin walled counterpart and will absorb more of the rider’s power in hysteresis loss as its walls deform as it rolls along. Just like Carl said.

John Retchford

Quoted post said:

Originally posted by John Retchford
You miss two of Carl’s points. The first has to do with the nature of the materials. In the railway wheel analogy the steel is behaving elastically at the contact patch – it is being loaded and unloaded along the same (linear) path in the load versus elongation plot. As work is force times the distance the force moves, no net work is being done.

In the case of the bicycle tyre, you are dealing with an anelastic material. Because its force vs elongation characteristic exhibits hysteresis, work must be done on it during loading and unloading. The rate of doing work is power and this power comes from the rider’s legs.

The area of the contact patch of a tyre is almost solely a function of force on the tyre (the weight) and its pressure. This just comes from the definition of pressure (force/area). The force supported by the sidewalls is negligible for a feasible bicycle tyre. Try pressing a fully deflated tyre with your thumb. So the thick walled tyre will have essentially the same contact patch area as its thin walled counterpart and will absorb more of the rider’s power in hysteresis loss as its walls deform as it rolls along. Just like Carl said.

John Retchford

Alright, then I must be still missing the two points you say, because after a review I still don't understand it. If a tire's contact is alsmost "solely a function of force on the tyre (the weight) and its pressure" like you say, then what difference does TPI have to do with it? That's what we've all been discussing to date.

Quoted post said:

Originally posted by BaCardi
Alright, then I must be still missing the two points you say, because after a review I still don't understand it. If a tire's contact is alsmost "solely a function of force on the tyre (the weight) and its pressure" like you say, then what difference does TPI have to do with it? That's what we've all been discussing to date.

To fit more threads in a given distance (“TPI”) means the threads must be thinner. Using the same number of thinner plies makes a thinner, more flexible tyre wall. Deforming this thinner, more flexible wall takes less work. The thinner wall deforms the same amount as a thicker wall on an otherwise equivalent tyre because the contact patch area is the same.

John Retchford

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

Quoted message said:

.... However, if you are saying that stiffer tires = less
rolling resistance due to less footprint where the tire
rubber meets the road, then why would lighter thinner
casing tires such as those with higher TPI count = less
rolling resistance? Wouldn't you get more rolling
resistance due to the thinner tire deforming more under
the rider's weight and hence a bigger footprint where the
tire rubber meets the road?

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.

dl

Quoted message said:

The area of the contact patch of a tyre is almost solely a
function of force on the tyre (the weight) and its
pressure. This just comes from the definition of pressure
(force/area). The force supported by the sidewalls is
negligible for a feasible bicycle tyre. Try pressing a
fully deflated tyre with your thumb. So the thick walled
tyre will have essentially the same contact patch area as
its thin walled counterpart and will absorb more of the
rider's power in hysteresis loss as its walls deform as it
rolls along. Just like Carl said.

Wow, this is cool - I know what you're talking about,
hysteresis and such. (I'm a college student studying
mechanical engineering) What I don't get is whether the thick-
sidewall tire has more or less rolling resistance compared
to the thin-walled tire.

--
Phil, Squid-in-Training

Quoted post said:

Originally posted by Douglas Landau
BaCardi <[email hidden]> wrote in message news:<[email hidden]>...

Quoted message said:

.... However, if you are saying that stiffer tires = less
rolling resistance due to less footprint where the tire
rubber meets the road, then why would lighter thinner
casing tires such as those with higher TPI count = less
rolling resistance? Wouldn't you get more rolling
resistance due to the thinner tire deforming more under
the rider's weight and hence a bigger footprint where the
tire rubber meets the road?

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.

dl

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.

Quoted post said:

Originally posted by Zeeexsixare
> The area of the contact patch of a tyre is almost solely a

Quoted message said:

function of force on the tyre (the weight) and its
pressure. This just comes from the definition of pressure
(force/area). The force supported by the sidewalls is
negligible for a feasible bicycle tyre. Try pressing a
fully deflated tyre with your thumb. So the thick walled
tyre will have essentially the same contact patch area as
its thin walled counterpart and will absorb more of the
rider's power in hysteresis loss as its walls deform as it
rolls along. Just like Carl said.

Wow, this is cool - I know what you're talking about,
hysteresis and such. (I'm a college student studying
mechanical engineering) 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 walled tyre will have more rolling resistance than the thin walled one, other things being equal. The rider must do more work on the walls in the region of the rolling contact patch to deform them.

John Retchford

Quoted post said:

Originally posted by BaCardi
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 keep saying this, but that does not necessarily make it correct. Douglas Landau and I have suggested to you that the contact patch area is set by the pressure in the tyre and the downward force on it. Is pressure not equal to force/area any more? Where does the larger area come from? And why do you think that "greater contact area with road = greater rolling resistance?

John Retchford

John Retchford said:

Quoted message said:

I like this line of reasoning. However, if you are saying
that stiffer tires = less rolling resistance due to less
footprint where the tire rubber meets the road, then why
would lighter thinner casing tires such as those with
higher TPI count = less rolling resistance? Wouldn't you
get more rolling resistance due to the thinner tire
deforming more under the rider's weight and hence a
bigger footprint where the tire rubber meets the road?

Quoted message said:

You miss two of Carl's points. The first has to do with
the nature of the materials. In the railway wheel analogy
the steel is behaving elastically at the contact patch? it
is being loaded and unloaded along the same (linear) path
in the load versus elongation plot. As work is force times
the distance the force moves, no net work is being done.

Steel isn't entirely without losses and that is why, for
instance, a loaded rail car cannot be pushed by hand on a
level track with trivial. Part of that is cause by sag in
the rail, placing the car's wheels in depressions and the
rest is hysteretic losses in steel. There are losses in
steel just as there are looses in tires.

Quoted message said:

In the case of the bicycle tyre, you are dealing with an
elastic material. Because its force vs elongation
characteristic exhibits hysteresis, work must be done on
it during loading and unloading. The rate of doing work is
power and this power comes from the rider's legs.

Let's not get the rider involved in RR of bicycle tires.
That only clouds the matter. It might be worth mentioning
that hysteresis is the characteristic by which some of the
force that deformed the material is not returned as force on
rebound but generates heat. High temper steels have little
hysteresis compared to mild steel while rubber has far
greater losses. Typically a toy monkey hanging on a steel
spring and a rubber band, the one on the rubber band will
stop bouncing long before the one on steel comes to rest.

Quoted message said:

The area of the contact patch of a tyre is almost solely a
function of force on the tyre (the weight) and its
pressure. This just comes from the definition of pressure
(force/area). The force supported by the sidewalls is
negligible for a feasible bicycle tyre. Try pressing a
fully deflated tyre with your thumb. So the thick walled
tyre will have essentially the same contact patch area as
its thin walled counterpart and will absorb more of the
rider's power in hysteresis loss as its walls deform as it
rolls along. Just like Carl said.

Well, "AS Carl said" but the reasons for low TPI tires
having greater losses than high TPI tires has been discussed
here at great length, recently. Maybe a GOOGLE search would
help. In any case, there is an FAQ abut this at:

draco.acs.uci.edu8b.14.htmlOpen ↗

Jobst Brandt [email hidden]

John Retchford said:

Quoted message said:

I like this line of reasoning. However, if you are saying
that stiffer tires = less rolling resistance due to less
footprint where the tire rubber meets the road, then why
would lighter thinner casing tires such as those with
higher TPI count = less rolling resistance? Wouldn't you
get more rolling resistance due to the thinner tire
deforming more under the rider's weight and hence a
bigger footprint where the tire rubber meets the road?

Quoted message said:

You miss two of Carl's points. The first has to do with
the nature of the materials. In the railway wheel analogy
the steel is behaving elastically at the contact patch? it
is being loaded and unloaded along the same (linear) path
in the load versus elongation plot. As work is force times
the distance the force moves, no net work is being done.

Steel isn't entirely without losses and that is why, for
instance, a loaded rail car cannot be pushed by hand on a
level track. Part of that is cause by sag in the rail,
placing the car's wheels in depressions while the rest is
hysteretic losses in steel. There are losses in steel just
as there are looses in tires.

Quoted message said:

In the case of the bicycle tyre, you are dealing with an
elastic material. Because its force vs elongation
characteristic exhibits hysteresis, work must be done on
it during loading and unloading. The rate of doing work is
power and this power comes from the rider's legs.

Let's not get the rider involved in RR of bicycle tires.
That only clouds the matter. It might be worth mentioning
that hysteresis is the characteristic by which some of the
force that deformed the material is not returned as force on
rebound but generates heat. High temper steels have little
hysteresis compared to mild steel while rubber has far
greater losses. Typically a toy monkey hanging on a steel
spring and a rubber band, the one on the rubber band will
stop bouncing long before the one on steel comes to rest.

Quoted message said:

The area of the contact patch of a tyre is almost solely a
function of force on the tyre (the weight) and its
pressure. This just comes from the definition of pressure
(force/area). The force supported by the sidewalls is
negligible for a feasible bicycle tyre. Try pressing a
fully deflated tyre with your thumb. So the thick walled
tyre will have essentially the same contact patch area as
its thin walled counterpart and will absorb more of the
rider's power in hysteresis loss as its walls deform as it
rolls along. Just like Carl said.

Well, "AS Carl said" but the reasons for low TPI tires
having greater losses than high TPI tires has been discussed
here at great length, recently. Maybe a GOOGLE search would
help. In any case, there is an FAQ abut this at:

draco.acs.uci.edu8b.14.htmlOpen ↗

Jobst Brandt [email hidden]

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.

--
Ted Bennett Portland OR

Quoted post said:

Originally posted by Ted Bennett

BBaCardi 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.

--
Ted Bennett Portland OR [/B]

OK, and your premise is based on the area of contact patch IS determined by the pressure in the tire and weight. Right? So, so basically what you and Doug are saying is not the same as what Jobst patiently explained. I think you ought to go back and read what Jobst is saying.

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 +-----------------------------------------------
+
| I=92m appearing as Preposteros in | Gilbert &
| Sullivan's Thespis at M.I.T. | April 9-10,15-17
| web.mit.eduwwwOpen ↗ |
| sheldonbrown.commusic.htmlOpen ↗ | Photos: sheldonbrown.orggOpen ↗
| s/thespis |
+-----------------------------------------------+ Harris
Cyclery, West Newton, Massachusetts Phone 617-244-9772 FAX
617-244-1041 harriscyclery.comharriscyclery.comOpen ↗ Hard-to-find parts
shipped Worldwide captainbike.comcaptainbike.comOpen ↗
sheldonbrown.comsheldonbrown.comOpen ↗

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 bike.terrymorse.combike.terrymorse.comOpen ↗

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