Thumb test

Does the "thumb test" (squeezing a tyre to see if it's hard enough)
measure tyre pressure or casing tension?

In article ,
Ben C wrote:

Does the "thumb test" (squeezing a tyre to see if it's hard enough)
measure tyre pressure or casing tension?

Directly the latter and indirectly the former, I think.

On 2006-10-02, Tim McNamara wrote:
In article ,
Ben C wrote:

Does the "thumb test" (squeezing a tyre to see if it's hard enough)
measure tyre pressure or casing tension?

Directly the latter and indirectly the former, I think.

This is what I was starting to think too. What this means of course is
that if you have a fat tyre and a thin tyre that feel the same, the fat
tyre will actually be at a lower pressure (as it typically should be),
and have higher rolling resistance.

Ben C wrote:
On 2006-10-02, Tim McNamara wrote:
In article ,
Ben C wrote:

Does the "thumb test" (squeezing a tyre to see if it's hard enough)
measure tyre pressure or casing tension?

Directly the latter and indirectly the former, I think.

This is what I was starting to think too. What this means of course is
that if you have a fat tyre and a thin tyre that feel the same, the fat
tyre will actually be at a lower pressure (as it typically should be),
and have higher rolling resistance.

Hm. I'm not sure, and I'm just a fly on the wall in this forum, but
this makes little sense to me. Doesn't the larger tire/tyre require
more pressure to get to the same hardness as a small tire? The large
tire has more rubber, so it's got more "stretchiness" or elasticity
across it's cross section. Further, wouldn't a large tire with the same
"thumb feel" as a thin tire (ie, really hard), have less rolling
resistance than a large tire with less pressure? I must be
misunderstanding your statement, sorry....

On 2006-10-02, wrote:
Ben C wrote:
On 2006-10-02, Tim McNamara wrote:
In article ,
Ben C wrote:

Does the "thumb test" (squeezing a tyre to see if it's hard enough)
measure tyre pressure or casing tension?

Directly the latter and indirectly the former, I think.

This is what I was starting to think too. What this means of course is
that if you have a fat tyre and a thin tyre that feel the same, the fat
tyre will actually be at a lower pressure (as it typically should be),
and have higher rolling resistance.

Hm. I'm not sure, and I'm just a fly on the wall in this forum, but
this makes little sense to me. Doesn't the larger tire/tyre require
more pressure to get to the same hardness as a small tire?

Well, the larger tyre requires less pressure for a given casing tension.
I read that here recently and am still getting my head around it. It's
also mentioned here by Jobst Brandt:

http://www.sheldonbrown.com/brandt/rim-support.html

"[...] unit casing tension is equivalent to inflation pressure times the
radius of curvature divided by pi [...]".

I was a bit surprised by this at first, but then if you think pressure
is force per unit area, if you increase the area of the inside of the
casing, you need more force for a given pressure. Not sure if this
reasoning is bogus or not though.

So if the thumb test measures casing tension, then the larger tyre does
require less pressure to get to the same thumb-hardness.

The large tire has more rubber, so it's got more "stretchiness" or
elasticity across it's cross section. Further, wouldn't a large tire
with the same "thumb feel" as a thin tire (ie, really hard), have less
rolling resistance than a large tire with less pressure?

A hard tyre basically has less rolling resistance than if it's soft. But
rolling resistance depends on pressure rather than on casing tension, or
at least, most of the graphs you see are of pressure against RR.

I think this also starts to explain the counter-intuitive result that
fatter tyres have lower RR than thinner tyres at the same pressure.
There are some explanations given in terms of fatter tyres needing to
deform less which I don't fully understand. But if at equal pressures
the fat tyre actually feels harder, it's easier to see intuitively that
it would roll better.

I might make a graph of RR against casing tension (as opposed to
pressure) for different tyre diameters...

E C McDougall writes:

Does the "thumb test" (squeezing a tyre to see if it's hard
enough) measure tyre pressure or casing tension?

Directly the latter and indirectly the former, I think.

This is what I was starting to think too. What this means of
course is that if you have a fat tyre and a thin tyre that feel the
same, the fat tyre will actually be at a lower pressure (as it
typically should be), and have higher rolling resistance.

Hm. I'm not sure, and I'm just a fly on the wall in this forum, but
this makes little sense to me. Doesn't the larger tire/tyre require
more pressure to get to the same hardness as a small tire? The large
tire has more rubber, so it's got more "stretchiness" or elasticity
across it's cross section. Further, wouldn't a large tire with the
same "thumb feel" as a thin tire (ie, really hard), have less
rolling resistance than a large tire with less pressure? I must be
misunderstanding your statement, sorry...

The thumb test accurately exposes the thumb to tire pressure. How
that gets to the rim is another problem, but to flatten a portion of
the tire by pressing on it is directly related to pressure (assuming
the tire is a 23mm cross section). I'm assuming the thumb is not
that of a giant and has a width of 10-15mm contact.

The size of the tire does not affect that sensation, just as it
doesn't affect the contact patch area. What is different is the
compliance of the tire when loaded. A fat tire can absorb more
deflection than a narrow one.

Jobst Brandt

On Mon, 02 Oct 2006 03:38:16 -0500, Ben C wrote:

Does the "thumb test" (squeezing a tyre to see if it's hard enough)
measure tyre pressure or casing tension?

Dear Ben,

Mostly, it measures the operator's belief in his sensitivity to
pressing one squashy pad against a less squashy pad as the contact
surface broadens.

(It's good enough for riding around.)

But you were asking about casing tension versus air pressure.

Thumb pressure measures casing tension.

Yes, there's 100 psi pushing the casing outward against your thumb.

But when you push the tire a quarter of an inch inward locally,
there's still 100 psi pushing against your thumb.

The 100 psi didn't go away. It didn't increase significantly due to
the tiny change in tire volume. But something stopped you from pushing
any further.

Think of a trampoline.

Air pressure does not hold the trampoline up, any more than it holds
the trampoline down. What holds the tramoline taut is the springs
pulling it tight at the edges. Push down, and the trampoline dents
easily. Push a little further, and more force is needed. The
resistance is nicely progressive . . .

Just like a tire.

A tire is a doughnut-shaped trampoline. What stretches the tire tight
in all directions is the expanding spring of the air pressure.

To dent a tire, you must overcome tension in every direction around
the dent. The deeper you try to push the casing in, the more tension
you must overcome.

_________________________ _ = tire casing
||||||||||||||||||||||||| | = 100 psi upward force


- ______ ______ - exaggerated shortening for dent
\/ pulling casing in against tension

Since 100 psi of air pressure is trying to keep the rest of the tire
casing tight everywhere else, you won't get very far.

Here's a simple demonstration that everyone knows--push on a tire
valve. There's no side tension holding the metal valve stem up against
your finger because the metal valve is a special case in the otherwise
uniform doughnut.

Notice that the valve doesn't move (dent) as you slowly apply more and
more pressure. It just sits there until your thumb pressure matches
the air pressure resisting it.

At that point, the valve moves dramatically because almost all the
resisting pressure is no longer being applied--the end of the valve is
just waving around inside the 100 psi air chamber.

But if you were simply pushing a tiny metal piston down into a deep
metal tube set into the tire, you'd push harder and harder without any
movement--and then the piston would begin to move steadily inward and
keep moving steadily because the air pressure resisting your thumb had
been overcome.

Here's another simple demonstration with a bicycle.

Toss a metal tool with a rounded handle less than an inch thick on the
garage floor--a slim socket wrench handle will work well.

Now roll your roughly 1-inch wide 700c front tire at 100 psi onto the
thin round handle.

Lean on the handlebars.

Presumably you weigh well over 100 pounds, but it's unlikely that you
can make the tire touch the garage floor on both sides of the round
metal handle.

A local pressure of 100 psi is not enough to overcome the casing
tension of a tire inflated all the way around to 100 psi.

Cheers,

Carl Fogel

Carl Fogel writes:

Here's another simple demonstration with a bicycle.

Toss a metal tool with a rounded handle less than an inch thick on
the garage floor--a slim socket wrench handle will work well.

Now roll your roughly 1-inch wide 700c front tire at 100 psi onto
the thin round handle.

Lean on the handlebars.

Presumably you weigh well over 100 pounds, but it's unlikely that
you can make the tire touch the garage floor on both sides of the
round metal handle.

A local pressure of 100 psi is not enough to overcome the casing
tension of a tire inflated all the way around to 100 psi.

Invalid experiment!

When a tire is pressed against a flat surface, the tire flattens until
the flat contact area times inflation pressure equal the load.

Remember, how tight must a wire be pulled so that it doesn't sag at
midspan? Casing tension does not come into play at tire-to-ground
contact where it has no curvature, only inflation pressure. In
contrast your complex experiment, involves pressure, casing curvature
in two directions with cord angle and more.

Jobst Brandt

On 02 Oct 2006 20:33:52 GMT, wrote:

Carl Fogel writes:

Here's another simple demonstration with a bicycle.

Toss a metal tool with a rounded handle less than an inch thick on
the garage floor--a slim socket wrench handle will work well.

Now roll your roughly 1-inch wide 700c front tire at 100 psi onto
the thin round handle.

Lean on the handlebars.

Presumably you weigh well over 100 pounds, but it's unlikely that
you can make the tire touch the garage floor on both sides of the
round metal handle.

A local pressure of 100 psi is not enough to overcome the casing
tension of a tire inflated all the way around to 100 psi.

Invalid experiment!

When a tire is pressed against a flat surface, the tire flattens until
the flat contact area times inflation pressure equal the load.

Remember, how tight must a wire be pulled so that it doesn't sag at
midspan? Casing tension does not come into play at tire-to-ground
contact where it has no curvature, only inflation pressure. In
contrast your complex experiment, involves pressure, casing curvature
in two directions with cord angle and more.

Jobst Brandt

Dear Jobst,

The point is that the casing tension must be overcome in all
directions to make a dent in the doughnut-like surface.

The dent does not raise the air pressure significantly.

But the dent does require dragging material inward in all directions,
material that's being held in tension over the rest of the doughnut by
100 psi.

Consider the reverse case, an imaginary thumb pushing outward from
inside the tire. What would stop it from pulling the tire wildly out
of shape?

Stick a nail inside a tubeless tire, a nail large enough to make a
bulge in the uninfalted tire.

When you inflate the tire, the air pressure will pushe the tire
outward in all directions.

The increasing tension, however, will pull the tire onto the nail and
puncture the tire.

Cheers,

Carl Fogel

Carl Fogel writes:

Here's another simple demonstration with a bicycle.

Toss a metal tool with a rounded handle less than an inch thick on
the garage floor--a slim socket wrench handle will work well.

Now roll your roughly 1-inch wide 700c front tire at 100 psi onto
the thin round handle.

Lean on the handlebars.

Presumably you weigh well over 100 pounds, but it's unlikely that
you can make the tire touch the garage floor on both sides of the
round metal handle.

A local pressure of 100 psi is not enough to overcome the casing
tension of a tire inflated all the way around to 100 psi.

Invalid experiment!

When a tire is pressed against a flat surface, the tire flattens
until the flat contact area times inflation pressure equal the
load.

Remember, how tight must a wire be pulled so that it doesn't sag at
midspan? Casing tension does not come into play at tire-to-ground
contact where it has no curvature, only inflation pressure. In
contrast your complex experiment, involves pressure, casing
curvature in two directions with cord angle and more.

The point is that the casing tension must be overcome in all
directions to make a dent in the donut-like surface.

It may do that but for bicycle tires that have little curvature with
respect to the minor diameter even these distortions are small. This
has no effect on pressing against a flat surface as I pointed out.

The dent does not raise the air pressure significantly.

....and who said it did?

But the dent does require dragging material inward in all
directions, material that's being held in tension over the rest of
the donut by 100 psi.

That may be a dynamic effect, the one that causes rolling resistance,
but it has no effect on contact pressure between tire and road.

Consider the reverse case, an imaginary thumb pushing outward from
inside the tire. What would stop it from pulling the tire wildly
out of shape?

The reverse case does not apply. There is no curvature at the ground
contact and thumb pushing is not a dynamic effect.

Stick a nail inside a tubeless tire, a nail large enough to make a
bulge in the uninfalted tire.

You're grasping at straws. All this does not apply to the contact
pressure with the road or a thumb flattening the tire locally.

When you inflate the tire, the air pressure will push the tire
outward in all directions.

The increasing tension, however, will pull the tire onto the nail
and puncture the tire.

And how does this apply to contact pressure with an essentially flat
surface?

Jobst Brandt