Weigh your 1949 - 1954 Chevy (or any other vehicle).

This is an excercise I used to give my physical chemistry students when I was professor of Chemistry at South Dakota State University to demonstrate the gas laws.

Step 1. Drive your Chevy onto sheets of paper, one for each tire.

Step 2. Trace the outline of the contact between the tire and the sheet of paper. Be as accurate as you can be.

Step 3. Label the sheets of paper. For example, RF could mean "right front"


Step 4. Measure the air pressure in each tire and record it on the sheets of paper.

Step 5. Drive off the sheets of paper.

Step 6. Measure the area of the rectangles drawn on the papers. (area of a rectangle is the width times the length - use inches since your air pressure will probably be in lbs/sq. in.

Step 7. The weight of your vehicle is the sum of the weights that each tire supports. The weight that a tire is supporting is the pressure (lbs per square inch) times the area of the contact between the tire and the ground (paper). For example 16 square inches for the area of the rectange times 32 pounds per square inch is 512 pounds. If all 4 tires are the same, then the vehicle would weigh 2048 pounds.

Lee Prairie

I found this intriguing


I think tire size would weigh heavy on the figures

Yes, it does. For example, the larger the tire the heavier the vehicle is and that contributes to the weight of the vehicle. Very big tires are not elastic. By 'elastic' I mean deformable so that the air pressure is supporting the weight of the automobile and not the resistance to deformation as in a steel wheel.

There is a tacit assumption that the tire is made of an elastic material, which is a good approximation for materials that 'car' tires are made of. An example of the failure of this elasticity assumption is the case of completely steel wheels. Then it wouldn't work because a steel wheel is not elastic. Using the steel wheel as that extreme, one can see that the elasticity approximation neglects the energy or work ( = force (weight) x distance; force (weight) = pressure * area) required to deform the tire. But that introduces only a small error with rubber tires on a car or truck that weighs some hundreds (thousands) of pounds.

Also, a very coarse tread will interfere in the determination of the area of contact, and an extremely coarse tread, like tractor lugs, will make the tire non-elastic.

I first did this exercise on my ex-wife's Ford Windstar. The result was within 50 pounds of the spec sheet in the owner's manual. It will work well for cars and 1/2 ton pickups.

Lee Prairie

I've tried it on three of my vehicles. one figured 50% heavier than actual, one twice as heavy, and the other about 300 lbs heavier. each with factory specified tire size and pressure.

Ok Lee, now I'm not a scientist and I'm even a worse mathematician but this does sound interesting as Dad noted. Now the first thing that comes to mind is getting an accurate footprint. What if, you shot some blueing or paint on a section of the tire and let the car down on the sheet of paper. Wouldn’t this give you a much better picture of the contact area?? Like you could do the front first and then the back.
Denny Graham
Sandwich, IL

Would be easier to drive it on a scale and get the true weight.

Oregon leaves the scales on in the Truck Inspection Stations even if it is closed. This gives me an excellent place to check the three spot weight of the motor home and the car I am hauling at the time. If you want to spend the time on the scales you can play with the positioning spot for correct towing.

Thanks, guys. The largest source of error is measuring the footprint. One way to do it, similar to Denny's suggestion, but in a way that the marking would wash off the tire would be to use that powdered chalk for refilling a carpenters chalkline apparatus. It's a purple dust that comes in a squeeze bottle so it would be easy to 'dust' the contact between the tire and the paper. It would also wash off easily.

As as error analysis: consider 32 psi in all 4 tires. The difference between a 15 square inch and 16 square inch footprint (a 6% error in the footprint measurement) is 128 lbs in the vehicle weight. There's something wrong in the measurements if one gets a factor of 2 error.

Lee

my C5 has a footprint of 11"X5" (55 sq in) on each rear, 32psi
each front is 9.5X5 (47.5 sq in) at 32psi

that puts it at 6560 lbs. almost double it's advertised weight

super wide tires messes up the formula

hmmmmm..... let me think about this one. 55 sq. in. is a huge footprint. How thick is the tire rubber? Is it soft?

Please excuse that I don't know what is a "C5". Would you please describe it for me? I presume that it is a vehicle that weights about 3300 lbs. Is the 11 inch dimension perpendicular to the forward motion. I believe I may know how this could happen....

Lee

Lee,

"C5" is the designation for the 5th "generation" Corvette. The years of the C5 are 1997-2004.

Bill.

Very interesting excercise.

Some possible sources of the error can be the stiffness of the tire construction and possible cupping of the tread. Bias ply and radial tires will likely have much different calculation results if construction is an important factor. I have a 3/4 ton truck that has 80 psig in the rear tires and 60 psig in the front. I have not measured it nor my trailers that have from 50 psig to 85 psig rated tires. At that higher pressures and light loads I expect big errors measuring the ground contact of the tire.

Just thought that the "ground" surface needs to be perfectly flat as any imperfection will effect the calculation.

Regarding the ultra-wide tire, it is likely that the outer edges of the tire, even as they contact the ground, do not support much of the weight of the vehicle. So the assumption that the weight of the car is distributed approximately evenly over the tire footprint fails.

Lee

It appears the experiment itself failed ...Dads 31 did the test on 3 of his vehicles ....none came out correctly. I'll bet Don Herbert, known in the 50's as "Mr Wizard" never had this problem. LOL