OT? Weigh your car by checking tire pressure?

Even with normal weight loading, if one is driving around curves like speed racer, there is going to be excessive distortion introduced, reducing tire life.

This approach works well for automating fingerprint acquisition, I notice. But mapping the pressure would be interesting.

Brian Whatcott Altus OK

We did this experiment back in high school in Physics class. Everyone grouped up in teams of 3 I think and proceeded to drive our cars onto sheets of paper to get the imprint of the contact area and then took measurements of the tire pressure. Afterwards each vehicle was taken to the local grain elevator and scaled(results hidden from the teams). Considering I was a bit of a gearhead at the time it was silly because I knew exactly what my car weighed and the weight distribution, but in any case.....

I argued with our teacher that the experiment was flawed. Assuming perfectly elastic tires, that yes theoretically it should work, but that tires were not perfectly elastic. My prime counter example being an air tank. The pressure in a tank could be 20psia or 200psia and the contact area would not change. He mumbled something about me being a trouble maker and told us to finish our calculations.

FWIW: The results did work our "reasonably" close. I dont' remember exact, but I think most were within 10-15% of actual weight. Certainly proved there was a strong correlation. This did lead to a discussion of what errors were present and how we could have gotten closer to the right answer.

I was always just a little bit of a trouble maker in that class(well most of my science/math classes really). I challenged the assertions of the teachers, occasionally getting concessions that things were presented in greatly simplified means so "the rest" could understand them.

JW

What was the name of that one Donald Duck cartoon in which he destroys a car that was going to be given to him as a gift?

That's correct. But if you change the tire pressure, the footprint changes as well. The change in footprint isn't linear, being dependent on sidewall stiffness and other factors.

It depends on the change in volume of the tire as the tire's contact patch flattens out and takes up the vehicles load.

Think about this: I can over or under inflate my tires easily by a factor of two (or more) without the wheel's bead hitting bottom. The vehicle's weight didn't change by a factor of two.

Neither their theories nor their tires hold any air.

" snipped-for-privacy@hotmail.com" wrote in news: snipped-for-privacy@22g2000hsm.googlegroups.com:

You, sir, win a prize. That was rather funny.

Cheers

Greg Locock

ChairmanOfTheBored wrote in article ...

From the book, "THE RACING & HIGH PERFORMANCE TIRE - Using Tires to Tune for Grip and Balance" by Paul Haney (ISBN 0-9646414-2-9)

Haney is an engineer who works in various racing series including IndyCar.

-----------------------------------------------------

Page 70 -

"Load and Internal Pressure ........If you have a vertical force on a tire, the pressure inside the tire resists that force by developing a contact patch of a size so that the internal pressure acting on the ground through the area of the contact patch equals the load on the tire."

"The contact patch area (A) is equal to the vertical force (Fv) in pounds divided by the internal pressure (P) in pounds per square inch."

"I'll include these equations which all express this same relationship. Starting with the last sentence in the previous paragraph:

A (in²) = Fv (lb) / P (lb/in²)

Multiplying both sides of the equation by P and rearranging you get:

Fv (lb) = A (in²) X P (lb/in²)

Dividing each side by A and rearranging you get:

P (lb/in²) = Fv (lb) / A (in²)

------------------------------------------------------

He then goes on to give examples of a 1200 pound load being supported by 30 p.s.i. air pressure will result in a contact patch of 40 in². (1200/30=40).

And what would *he* know about it?

Ok, now you think about it some more. The pressure of the tire surfaces against the road isn't uniform as long as the tire remains effectively round. It's higher near the center of the footprint, and close to zero at the edge where the tire is barely touching the road. Deflate the tire, and as the tire squishes flatter, the pressure becomes more uniform over an area that hasn't increased a great deal. Deflate the tire enough, and the wheel has to bring the interior surfaces of the tire into contact, making the air pressure suddenly much less significant (the structure of the tire always supports some weight, just not much relatively speaking). With inflated tires, just about the only thing holding the car up is the air pressure (which of course has to measured while it's holding the car up, and which of course is strictly uniform inside the tire).

Overinflating by a factor of more than two eventually leads to the spectacular result I saw in an airline landing gear tire test. If you're not damaged that much by flying rubber, you'll now see the car sitting on its rims. Which will support all its weight. And may need replacing, along with other structures.

He has to troll, what with his big mouth. He scares off everyone who sees him. He sounds like he's even more desperate than usual.

You should watch them kick "Always Wrong" around on the electronics groups. They love him on alt.usenet.kooks. They have even given him several awards for trolling. He was so proud of them, he cross posted the threads to sci.electronics.design to show them off to everyone like some really homely girl who finally got an engagement ring.

Haney? Wasn't he the guy on Green Acres who invented the gasoline powered washing machine, built from a 55 gallon drum and an outboard motor.

Tune

Enough, apparently, for his book to be recognized and sold as an official publication of the Society of Automotive Engineers (SAE).

It depends on the tire.

It would not be the same on a ten speed bike tire as it would on a

There ARE other variables. The rubber compound is one major one.

A stiffer compound would NOT yield the same patch size as a less firm one would for the same tire size and inflation pressure. If you cannot see this is the case, you need to quit looking up relationships you do not fully understand, and touting the as golden rules.

Would you by any chance be from Germany?

On Sep 26, 1:47 am, ChairmanOfTheBored .

You are talking about second order effects which cause some inaccuracy. A stiffer compound would have almost no effect.

A ten speed bike tire and a 10.00 by 20 truck tire both loaded to say

Try measuring the pressure in a car tire, then jack the wheel off the ground and measure the pressure again. You will be surprised at how little difference there is in pressure. Post the numbers here so others can learn.

Dan

official

Now, THAT's funny!!!

I'm in North America - some say I'm in Canada, some say I'm in USA........

I think tread flexing may matter more than sidewall flexing, and taller wheels have less of it.

Suppose your tire has a 10" radius and, due to sidewall flexing, the distance from the center to the pavement is 9". The contact patch should be 8.7" long, and the tread must flex 26 degrees at the front and back of the patch.

If you get the same 1" of sidewall flexing on a tire with a 20" radius, the contact patch will be 13.5" long. A tread 2/3 as wide can carry the same weight at the same pressure, and it will have to flex only 18 degrees at each end of the contact patch.

My figures are simplified. The tread doesn't seem to grow shorter when pressed against pavement, so it's not quite the shape of a circle with a piece sliced off at the bottom.

That's not entirely true. In fact, the footprint would increase, and the pressure could never reach zero, unless you used a vacuum pump.

It certainly wouldn't be very accurate, however. ;-)

Cheers! Rich

Define "effectively round".

Not really. That will result in uneven tread wear. The pressure across the tire-road contact patch should be relatively uniform, dropping off at the edges, of course.

Take a look at the wear pattern of an under-inflated tire. Its greater near the edges where sidewall stiffness has more effect and less in the center. An over-inflated tire wears more in the middle due to non-uniform loading.