carsarecool

05-12-2004, 10:29 AM

this is for school, does anybody know how to calculate the weight of a car using just the contact area of the tires and the PSI its at? its a math thing

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carsarecool

05-12-2004, 10:29 AM

this is for school, does anybody know how to calculate the weight of a car using just the contact area of the tires and the PSI its at? its a math thing

CoopGT

05-12-2004, 04:40 PM

I don't think that can be done , I could be wrong though. There are way to many variables.

Mike_99GAGT

05-12-2004, 07:38 PM

PSI of what???

AaronGAGT

05-12-2004, 07:51 PM

PSI of what???

tires... what else????

tires... what else????

cavingman

05-12-2004, 07:54 PM

PSI of what???

headlight fluid

headlight fluid

2002AmSE

05-12-2004, 08:04 PM

No not the headlight fluid, it can be done with our blinker fluid though, our headlight fluids vary too much in viscosity with age, but our blinker fluid would be the best bet, the amount of CC's of blinker fluid you have in the reservoir is equivelent to the amount of weight the car is when you take that amount multiplied by 0.01. So basically, 30cc's of fluid is a 3000 pound car.

The tires won't work because the weight distribution is towards the front. Me and a friend of mine picked the back end of my car up without a whole pile of effort, but that sure is never going to happen with the front. So basically its not possible because the sidewall stregth and age of the tire is something that you can't just take a footprint and put it together with PSI to get a weight.

The tires won't work because the weight distribution is towards the front. Me and a friend of mine picked the back end of my car up without a whole pile of effort, but that sure is never going to happen with the front. So basically its not possible because the sidewall stregth and age of the tire is something that you can't just take a footprint and put it together with PSI to get a weight.

AndyVTek

05-12-2004, 08:04 PM

this is for school, does anybody know how to calculate the weight of a car using just the contact area of the tires and the PSI its at? its a math thing

It shouldn't be too hard... You have four tires, each of which exert a pressure upwards on the car, downwards on the ground, and vice versa (Newton's 3rd Law)...

Pressure ----> Force P= F / A

In this problem, you're probably asked (or atleast, you need to) guesstimate the area of the tire which is in contact with the ground. Then use the above equation to determine the +y force on the car.

Multiply by 4 (4 tires).... Set that number equal to the weight of the car (net force equation).

Final... Weight = 4 *( P * A )

'Er somethin' like that. lol

EDIT: 2002AmSe... It's a simple math problem. Yes, it's possible... :boogie:

It shouldn't be too hard... You have four tires, each of which exert a pressure upwards on the car, downwards on the ground, and vice versa (Newton's 3rd Law)...

Pressure ----> Force P= F / A

In this problem, you're probably asked (or atleast, you need to) guesstimate the area of the tire which is in contact with the ground. Then use the above equation to determine the +y force on the car.

Multiply by 4 (4 tires).... Set that number equal to the weight of the car (net force equation).

Final... Weight = 4 *( P * A )

'Er somethin' like that. lol

EDIT: 2002AmSe... It's a simple math problem. Yes, it's possible... :boogie:

Mike_99GAGT

05-12-2004, 09:02 PM

tires... what else????

The pressure of the tires on the ground........its measured in PSI also

The pressure of the tires on the ground........its measured in PSI also

carsarecool

05-14-2004, 10:30 AM

yeah that was right Andy, he gave me 2.6 x 6 inches for the contact patch so the area is 15.6 * 4 tires is 62.4in^2 *42 PSI is 2,620.8 lbs :) thanks

Hexx

05-14-2004, 11:25 AM

Interesting stuff!

AndyVTek

05-14-2004, 11:36 AM

yeah that was right Andy, he gave me 2.6 x 6 inches for the contact patch so the area is 15.6 * 4 tires is 62.4in^2 *42 PSI is 2,620.8 lbs :) thanks

Anytime... :boogie:

Anytime... :boogie:

Slim

05-17-2004, 10:04 AM

Ummm... I might be missing something here, but I think you would need to know the unladed pressure in the tires. I'm sure the equation needs some sort of difference in pressure. I could be wrong though.

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