OT? Weigh your car by checking tire pressure?

[Rich Grise]

Saturday morning TV is mostly cartoons, yes, but there's one show that's
not a cartoon, but it's still fun:
http://www.beakmansworldtv.com/

It's like Stealth Educational TV - like Mr. Wizard for the 21st century
kid. And they do Real Science.

But today, they said that you can weigh your car by using your tire
pressure. What you do is measure the footprint of each tire, take its
pressure, and then the footprint of the tire times the PSI equals
the number of pounds that the tire is supporting, and their sum is the
weight of the car. (except for the tires themselves, I presume).

But they did the experiment - they measured the footprints of the four
tires on Some very well-preserved Nash Rambler, and their PSI, and did
the arithmetic, and came up with a number that was within 10% of the car's
"official" weight.

Well, I'm a little uncomfortable with that. What about the pressure that's
in a tire when it's not on the car? Where does that pressure go? Is that
that 10% fudge factor that they admitted to in the show?

How much does the pressure change when you take a standalone tire, mount
it on the car, and let the car down on it?

Or do they get around that by saying, "Well, you can ignore that, because
the tire's not supporting any weight". Or is it entirely ( or mostly) due
to the flattening of the bottom of the tire? Does the "bias pressure" (a
term I just made up now, for the pressure that's there when it's not on
the car) get lost below the noise floor?

Thanks,
Rich

 

[[email protected]]

But today, they said that you can weigh your car by using your tire
pressure. What you do is measure the footprint of each tire, take its
pressure, and then the footprint of the tire times the PSI equals
the number of pounds that the tire is supporting, and their sum is the
weight of the car. (except for the tires themselves, I presume).

Well, I'm a little uncomfortable with that. What about the pressure that's
in a tire when it's not on the car? Where does that pressure go? Is that
that 10% fudge factor that they admitted to in the show?

It took me a while before I decided that they are right. When the
tire is not on the ground you multiply the pressure ( Say 30 psi ) by
the footprint ( 0 ) and get zero. The big problem is measuring the
footprint and the stiffness of the tire sidewalls and tread.

Dan

 

[Dave Baker]

Saturday morning TV is mostly cartoons, yes, but there's one show that's
not a cartoon, but it's still fun:
http://www.beakmansworldtv.com/

It's like Stealth Educational TV - like Mr. Wizard for the 21st century
kid. And they do Real Science.

But today, they said that you can weigh your car by using your tire
pressure. What you do is measure the footprint of each tire, take its
pressure, and then the footprint of the tire times the PSI equals
the number of pounds that the tire is supporting, and their sum is the
weight of the car. (except for the tires themselves, I presume).

But they did the experiment - they measured the footprints of the four
tires on Some very well-preserved Nash Rambler, and their PSI, and did
the arithmetic, and came up with a number that was within 10% of the car's
"official" weight.

Well, I'm a little uncomfortable with that. What about the pressure that's
in a tire when it's not on the car? Where does that pressure go? Is that
that 10% fudge factor that they admitted to in the show?

How much does the pressure change when you take a standalone tire, mount
it on the car, and let the car down on it?

Or do they get around that by saying, "Well, you can ignore that, because
the tire's not supporting any weight". Or is it entirely ( or mostly) due
to the flattening of the bottom of the tire? Does the "bias pressure" (a
term I just made up now, for the pressure that's there when it's not on
the car) get lost below the noise floor?

There are two sources of error in that method. The first, which is the one I
think you are referring to is the increase in pressure as you put weight on
the tyre. This will be the inverse of the reduction in volume of the tyre as
the bottom of the tread flattens and will be very small.

The second, which is the significant one, is the amount of the load which is
being supported by the stiffness of the tyre tread and sidewalls. If the
tyre were a balloon, i.e. perfectly elastic, the measurement as above would
be exactly correct.

So the result they should have obtained is a calculated weight somewhat less
than the actual weight of the car.

 

[Brian Whatcott]

Well, I'm a little uncomfortable with that. What about the pressure that's
in a tire when it's not on the car? Where does that pressure go? Is that
that 10% fudge factor that they admitted to in the show?

How much does the pressure change when you take a standalone tire, mount
it on the car, and let the car down on it?

Or do they get around that by saying, "Well, you can ignore that, because
the tire's not supporting any weight". Or is it entirely ( or mostly) due
to the flattening of the bottom of the tire? Does the "bias pressure" (a
term I just made up now, for the pressure that's there when it's not on
the car) get lost below the noise floor?

Thanks,
Rich

In the days of cross-ply tires, it took energy to flex the sidewall as
the tire rotated and deflected to bear the vehicle weight.

With regular radial tires, this energy loss is minimized both when
rotating, and as weight is applied to the tire - hence the cautions
not to over inflate them.

You can verify this for yourself by taking tire pressures at one end
of your car, then jacking that end up.
You should not see much change at all.

Brian W

 

[Phil Allison]

"Brian Whatcott"

In the days of cross-ply tires, it took energy to flex the sidewall as
the tire rotated and deflected to bear the vehicle weight.

With regular radial tires, this energy loss is minimized both when
rotating, and as weight is applied to the tire - hence the cautions
not to over inflate them.

** Classic example of a non-sequitur.

You can verify this for yourself by taking tire pressures at one end
of your car, then jacking that end up.
You should not see much change at all.

** Perfect example of a illogical conclusion drawn from a non- sequitur.

Brian Wanker...

....... Phil

 

[Brian Whatcott]

** Classic example of a non-sequitur.
** Perfect example of a illogical conclusion drawn from a non- sequitur.

...... Phil

There you go folks, that's an Australian contribution for you:
pithy, to the point, incisive, witty, analytical, carefully reasoned.

And they still cling to men-only bars there, isn't it Dai?



Brian W

 

[Phil Allison]

"Brian Whatcott"

( snip this utter moron's' ASD fucked drivel )


** Hey - Mr Autistic Shit for Brains.


Go **** your mother !!.




...... luv, Phil

 

[E Z Peaces]

Saturday morning TV is mostly cartoons, yes, but there's one show that's
not a cartoon, but it's still fun:
http://www.beakmansworldtv.com/

It's like Stealth Educational TV - like Mr. Wizard for the 21st century
kid. And they do Real Science.

But today, they said that you can weigh your car by using your tire
pressure. What you do is measure the footprint of each tire, take its
pressure, and then the footprint of the tire times the PSI equals
the number of pounds that the tire is supporting, and their sum is the
weight of the car. (except for the tires themselves, I presume).

But they did the experiment - they measured the footprints of the four
tires on Some very well-preserved Nash Rambler, and their PSI, and did
the arithmetic, and came up with a number that was within 10% of the car's
"official" weight.

Well, I'm a little uncomfortable with that. What about the pressure that's
in a tire when it's not on the car? Where does that pressure go? Is that
that 10% fudge factor that they admitted to in the show?

How much does the pressure change when you take a standalone tire, mount
it on the car, and let the car down on it?

Or do they get around that by saying, "Well, you can ignore that, because
the tire's not supporting any weight". Or is it entirely ( or mostly) due
to the flattening of the bottom of the tire? Does the "bias pressure" (a
term I just made up now, for the pressure that's there when it's not on
the car) get lost below the noise floor?

Thanks,
Rich

Well, if you measure the tire radius in a horizontal plane, then measure
the distance from the center of the wheel to the ground, you will know
how much the tire is flexing in mm. Then if you unscrew the valve core
from your spare, you can stand the wheel up and see how much weight it
takes to push the center of the wheel down to that height. With a
radial, I wonder if the weight of the rim would flex the tire that much.

It seems to me that except those few pounds, it's air pressure that
holds up an inflated tire. In ordinary circumstances I wouldn't know
how to measure the footprint accurately. The car would have to be on a
very smooth, flat surface and I would need a thin "feeler gage."

I've read that tires of larger diameter have less rolling resistance,
but for some reason tires of smaller diameter hold better on snow. The
contact patches are the same area.

 

[jk]

It seems to me that except those few pounds, it's air pressure that
holds up an inflated tire. In ordinary circumstances I wouldn't know
how to measure the footprint accurately. The car would have to be on a
very smooth, flat surface and I would need a thin "feeler gage."

The method also assumes (I think) that the pressure distribution
across the footprint is uniform.
jk

 

[Nick Mueller]

Well, I'm a little uncomfortable with that. What about the pressure that's
in a tire when it's not on the car? Where does that pressure go? Is that
that 10% fudge factor that they admitted to in the show?

Hahaha! They pulled your leg!
The reason why this works is, that there is an efficient tire pressure for
each load and tire. The tire manufacturers tell what pressure for what
load. And that number (psi per lb) is in the same ballpark. It depends on
the construction of the tire (stiff/soft side walls, width, diameter, etc).


Nick

 

[[email protected]]

It took me a while before I decided that they are right. When the
tire is not on the ground you multiply the pressure ( Say 30 psi ) by
the footprint ( 0 ) and get zero. The big problem is measuring the
footprint and the stiffness of the tire sidewalls and tread.

I'm tempted to come round and let your tyres down, and watch you try
to catch your now weightless car as it drifts away across the sky...

Roy

 

[Brian Whatcott]

... if you measure the tire radius in a horizontal plane, then measure
the distance from the center of the wheel to the ground, you will know
how much the tire is flexing in mm. Then if you unscrew the valve core
from your spare, you can stand the wheel up and see how much weight it
takes to push the center of the wheel down to that height. With a
radial, I wonder if the weight of the rim would flex the tire that much.

It seems to me that except those few pounds, it's air pressure that
holds up an inflated tire. In ordinary circumstances I wouldn't know
how to measure the footprint accurately. The car would have to be on a
very smooth, flat surface and I would need a thin "feeler gage."

Here's a question on an engineering group that has an
engineering answer!

There is a pressure sensitive material available (for gaskets etc.)
that color codes the pressure that it experiences.
That would be one analytical approach.

Brian W

 

[Joseph Gwinn]

Saturday morning TV is mostly cartoons, yes, but there's one show that's
not a cartoon, but it's still fun:
http://www.beakmansworldtv.com/

It's like Stealth Educational TV - like Mr. Wizard for the 21st century
kid. And they do Real Science.

But today, they said that you can weigh your car by using your tire
pressure. What you do is measure the footprint of each tire, take its
pressure, and then the footprint of the tire times the PSI equals
the number of pounds that the tire is supporting, and their sum is the
weight of the car. (except for the tires themselves, I presume).

But they did the experiment - they measured the footprints of the four
tires on Some very well-preserved Nash Rambler, and their PSI, and did
the arithmetic, and came up with a number that was within 10% of the car's
"official" weight.

Well, I'm a little uncomfortable with that. What about the pressure that's
in a tire when it's not on the car? Where does that pressure go? Is that
that 10% fudge factor that they admitted to in the show?

How much does the pressure change when you take a standalone tire, mount
it on the car, and let the car down on it?

Or do they get around that by saying, "Well, you can ignore that, because
the tire's not supporting any weight". Or is it entirely ( or mostly) due
to the flattening of the bottom of the tire? Does the "bias pressure" (a
term I just made up now, for the pressure that's there when it's not on
the car) get lost below the noise floor?

It is true that one can weigh a car this way, using the principle of the
tonometer, which is most often used in medicine to measure the pressure
within the eyeball.

The basic principle is that in the flat section (the footprint), tension
in the membrane (the tire) is perpendicular to the surface, and thus
cancels out, so the net force generated is the internal pressure times
the area of the flat section.

The tonometer was invented a century ago, so most web references assume
that one already knows how it works, and tries to sell you one. But
here is one article with a good explanation:

<http://brain.berkeley.edu/~minerva1/publications/bending/>

(Reprinted from the Archives of Ophtalmology,
January 1961, Vol. 65, pp. 67-74, Copyright 1961, by American Medical
Association, "Corneal Bending and Buckling in Tonometry",
ELWIN MARG, Ph.D.; R. STUART MACKAY, Ph.D., and RAYMOND OECHSLI, A.B.,
Berkeley, Calif.)

Joe Gwinn

 

[ChairmanOfTheBored]

It is true that one can weigh a car this way, using the principle of the
tonometer, which is most often used in medicine to measure the pressure
within the eyeball.

The basic principle is that in the flat section (the footprint), tension
in the membrane (the tire) is perpendicular to the surface, and thus
cancels out, so the net force generated is the internal pressure times
the area of the flat section.

The tonometer was invented a century ago, so most web references assume
that one already knows how it works, and tries to sell you one. But
here is one article with a good explanation:

<http://brain.berkeley.edu/~minerva1/publications/bending/>

(Reprinted from the Archives of Ophtalmology,
January 1961, Vol. 65, pp. 67-74, Copyright 1961, by American Medical
Association, "Corneal Bending and Buckling in Tonometry",
ELWIN MARG, Ph.D.; R. STUART MACKAY, Ph.D., and RAYMOND OECHSLI, A.B.,
Berkeley, Calif.)

Joe Gwinn

There are far too many variables involved in getting an accurate
reading in this manner for it to EVER be feasible.

What DUMB BASTARD that doesn't know a fucking thing about Usenet made
this stupid, cross-posted SHIT?

 

[E Z Peaces]

The method also assumes (I think) that the pressure distribution
across the footprint is uniform.
jk

When I posted, it occurred to me that the contact patch was probably
larger than weight/pressure because at least at the front and back of
the patch, the tread would press down with less pressure. Now I've
found this article by a BMW motorcycle rider explaining that the subject
is very complicated:
http://www.ibmwr.org/otech/tirestuff.html

Because the pressure of the tread approaches zero at some points, it
seems a water film will carry some of the weight any time you roll on a
wet road. Traction is diminished, and you hydroplane when you go fast
enough and the fluid is deep and viscous enough that the film supports
even the parts of the tread with the most pressure.

 

[Gunner Asch]

What DUMB BASTARD that doesn't know a fucking thing about Usenet made
this stupid, cross-posted SHIT?

Evidently it was you.

Gunner

 

[CBFalconer]

.... snip ...

What DUMB BASTARD that doesn't know a fucking thing about Usenet
made this stupid, cross-posted SHIT?

PLONK

 

[mpm]

Saturday morning TV is mostly cartoons, yes, but there's one show that's
not a cartoon, but it's still fun:http://www.beakmansworldtv.com/

It's like Stealth Educational TV - like Mr. Wizard for the 21st century
kid. And they do Real Science.

But today, they said that you can weigh your car by using your tire
pressure. What you do is measure the footprint of each tire, take its
pressure, and then the footprint of the tire times the PSI equals
the number of pounds that the tire is supporting, and their sum is the
weight of the car. (except for the tires themselves, I presume).

But they did the experiment - they measured the footprints of the four
tires on Some very well-preserved Nash Rambler, and their PSI, and did
the arithmetic, and came up with a number that was within 10% of the car's
"official" weight.

Well, I'm a little uncomfortable with that. What about the pressure that's
in a tire when it's not on the car? Where does that pressure go? Is that
that 10% fudge factor that they admitted to in the show?

How much does the pressure change when you take a standalone tire, mount
it on the car, and let the car down on it?

Or do they get around that by saying, "Well, you can ignore that, because
the tire's not supporting any weight". Or is it entirely ( or mostly) due
to the flattening of the bottom of the tire? Does the "bias pressure" (a
term I just made up now, for the pressure that's there when it's not on
the car) get lost below the noise floor?

Thanks,
Rich

Take the car and launch it into orbit.
Then, measure the tire pressure and determine the weight.

 

[Matthew T. Russotto]

I've read that tires of larger diameter have less rolling resistance,
but for some reason tires of smaller diameter hold better on snow. The
contact patches are the same area.

I think the important factor is width, not diameter.

The contact patches are the same size, but a different shape. After
the first part of the contact patch has displaced the snow, there's
more contact patch left in a narrow tire to contact the pavement. The
amount of snow which needs to be displaced is proportional to the
width of the contact patch, not the area.

I don't know why a tire of larger width would have less rolling
resistance, unless it's just that tires of larger width usually have a
smaller sidewall (proportionally) when used on the same car.

 

[ChairmanOfTheBored]

Evidently it was you.

Gunner

Responses don't count, dipshit. I have to ensure that my post gets read
by the twit that posted across several groups, many of which he may not
even visit or pull headers from. Get a clue.