OT? Weigh your car by checking tire pressure?

Saturday morning TV is mostly cartoons, yes, but there's one show that's not a cartoon, but it's still fun:

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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

Reply to
Rich Grise
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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

Reply to
dcaster

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.

--
Dave Baker - Puma Race Engines
Reply to
Dave Baker

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

Reply to
Brian Whatcott

"Brian Whatcott"

** Classic example of a non-sequitur.

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

...... Phil

Reply to
Phil Allison

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

Reply to
Brian Whatcott

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

Reply to
jk

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

--
The lowcost-DRO:
Reply to
Nick Mueller

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

Reply to
rthearle

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

Reply to
Brian Whatcott

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:

(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

Reply to
Joseph Gwinn

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

Reply to
mpm

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.

--
  There\'s no such thing as a free lunch, but certain accounting practices can
  result in a fully-depreciated one.
Reply to
Matthew T. Russotto

As rolling resistance would be a function of the weight they bear, one would think that the larger distributed weight area (contact patch) would result in a lower rolling resistance, as there would be less overall rubber sidewall distortion.

However, they may have a lower fill pressure as well, which might be a reason for a claim of higher rolling resistance.

Reply to
ChairmanOfTheBored

They tried to do that on Top Gear:

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Nice try!

Dave.

Reply to
David L. Jones

If the tire pressures are the same, the contact patch is the same (approximately), regardless of the tire width.

Typically lower pressures result in higher rolling resistance, due to increased sidewall flex.

--
  There\'s no such thing as a free lunch, but certain accounting practices can
  result in a fully-depreciated one.
Reply to
Matthew T. Russotto

Increasing contact patch length, whether by increasing weight or decreasing pressure or decreasing tire width, increases sidewall distortion.

If tire width is increased but pressure and weight are unchanged, the contact patch length will decrease. I think that would decrease rolling resistance.

One factor that appears significant is ratio of radii from axle center to tire surface, at center of contact patch and at forward/rearward extremes of the contact patch. The closer this ratio is to 1, the less rolling resistance will be. One reason I see: Less rubbing of tread against the road surface upon meeting and leaving the road surface. Also less deformation counts at least a little, even other than by actual scraping/rubbing of tread on road surface. It appears to me that 1 minus this radius ratio strongly influences the rolling resistance coefficent.

And tread duration goes largely as:

  • Directly proportionately with contact patch width
  • Directly proportionately with depth of rubber to wear away
  • Inversely proportionately with pressure
  • Inversely proportional to either the above {1 minus radius ratio}, or to actual rolling resistance coefficient (ratio of rolling resistance to weight supported by the tire including its own)

As pressure increases, tire life tends to increase as long as product of rolling resistance coefficient (or the above radius ratio) and pressure decreases more than contact patch width decreases.

In general, tires for vehicles having 4 or more wheels have contact patch width varying only slightly inversely with pressure, unless the contact patch width is narrowed by a large pressure uptick, and then maybe only after the tire has been "broken in" at the lower pressure. That problem tends to require either overinflation or proper pressure after experiencing significant wear while underinflated.

One more thing: Assuming tire design that has contact patch width not varying much with pressure (or ratio of pressure to weight loading), the above radius ratio is approximating the cosine of the angle between straight down and from axle center to either forward or rearward end of the contact patch. This (1-minus-radius_ratio) in such case tends to be close to inversely proportional to square of tire pressure (above atmospheric pressure). That appears to me to indicate that things tend to get better with such tires as pressure gets higher, as long as the pressure is not exsceeding the pressure capability of the tire.

One more thing: Tire pressure is supposed to be measured "cold" - when the tires are not heated up by using them. If the "cold" pressure does not exceed the maximum rating for the tire, then the "warmed up pressure" is supposed to be unable to be "excessive" unless the weigt loading exceeds the tire rating for that, vehicle speed is excessive, or the ambient temperature takes a jump big enough to be a considerable factor.

If you have excessive weight loading on your tires, then they are unsafe at any pressure - their deformation varies directly with ratio of weight loading to pressure, and any fatigue effects of that get worse when the deformation is greater or the when "same deformation" occurs at higher pressure - excessive weight loading is bad no matter what you do. However, in my experience of overloading bicycle tires (by being adventurous as a bicycle messenger by carrying heavier packages or clusters thereof), I have found "least-worst" (still "living dangerously") results from having pressure measured-cold at or a little above the pressure marking on the sidewall of the tire - I don't see tires having tolerance of excessive weight load improved by lower pressure; lower pressure with excessive weight increases bigtime flexing of the sidewall, and fatigue in the sidewall towards the bead.

- Don Klipstein ( snipped-for-privacy@misty.com)

Reply to
Don Klipstein

Like this?

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--
Mike Williams
Gentleman of Leisure
Reply to
Mike Williams

Brian Whatcott snipped-for-privacy@sbcglobal.net posted to sci.electronics.design:

If you use structural strength glass (or other structural transparent material) with side (edge) lighting the contact area (non-moving) could be directly observed. Perhaps with a "window" smoothly embedded in pavement and careful photograph timing the contact patch could be measured in a variety of circumstances.

Reply to
JosephKK

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

Easy:

The contact patch with the ground is zero. Multiply this by the tire pressure and you get a weight of, you guessed it, zero.

(Determining the mass, however, is an entirely different matter.) :-)

SaSW, Willem

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Reply to
Willem

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