I'm not sure exactly what "ground pressure" means. Not because I don't understand the forces involved, but because it's not a real term. My interpretation of Skid Mark's original post seems to indicate that it's the weight of the vehicle (at each tire) divided by the contact patch area. My assertion is that you could save yourself a lot of drawing, measuring, weighing, and calculating by just using a tire pressure gauge.
Dang ChopShop, I tried, but I couldn't beat that one. That was funny:haha:
Whats the footprint on a tire like this make for ground pressure?
Ground Pressure is not a real term? Someone better tell CAT, they list ground pressure specs on lots of their equipment, I think they'd want to know its not a real term.
http://www.cat.com/cda/components/fullArticle?m=38580&x=7&id=1012259
http://en.wikipedia.org/wiki/Ground_pressure
:haha:
Its gets tiresome when the Pressure is on!!:haha::haha:
This is a legitimate question, but there are so many variables, that it is just way to hard to collect the data. With machinary, you generally have specific sized tracks or tires designed for a machine, so it is easy to measure. We are going against what our machines are designed for:;
No variables at all, in soft ground like we have in the PNW the WHOLE tire foot print is carrying weight, everyone is over thinking this question, just figure out what your footprint is on each tire, then divide it by the weight, simple math guys.
And if ground pressure didn't matter nobody would air down in the snow :;
Ground Pressure is not a real term? Someone better tell CAT, they list ground pressure specs on lots of their equipment, I think they'd want to know its not a real term.
And to reference the tire psi issue 64fj40 was having A tire is round with equal sides on in it doesn't matter what air pressure your at the pressure is pushing on ALL sides evenly thus has NO direct effect on ground pressure other then changing the size of the foot print of the said tire.
Holy cow. Do you really believe this? Why do you think the footprint changes size with different internal air pressures?
816 inches of contact with my 33/12.5 @12 psi. My jeep prepared to go is about 5500lbs so ON AVERAGE I have a 6.74 gppsi
Why do you think the footprint changes size with different internal air pressures?
I'm going to make this as simple as possible without breaking out my crayons and drawing you a picture. If anybody else following along disagrees with me or doesn't understand, feel free to explain why and I'll try to address those with more specific examples.
It is not possible to have a 6.74psi contact pressure when your tires are inflated to 12psi unless you're sitting on a surface that can only support 6.74psi. With 1375 pounds on a tire at 12psi, the contact patch area is 114.6 square inches when on surface capable of supporting weight at 12psi or greater. The air pressure inside the tire is what supports that weight of 1375 pounds, and it does that by exerting a force against the ground. If the ground is unable to support weight at 12psi, you sink until you have enough contact patch to support the weight being placed on that tire. Just for illustration, pretend some sand area can support 8psi. Your tires will sink in a little bit. This increases your contact patch area until the area reaches 172 square inches.
Think of it like a latex balloon. Pretend that inflated balloon has 1psi of internal air pressure. If you push the balloon against the wall with 5 pounds of force, it the contact area will be 5 square inches (that's why it's called pounds per square inch). Now push with 10 pounds of force. The contact pressure is still only 1 psi, but that pressure is spread out over 10 square inches. The contact patch just keeps getting bigger (until you reach the elastic capacity of the balloon) in order to support more weight.
EPIC FAIL
:haha::haha:
Exactly, as the footprint gets bigger your ground pressure is decreased
BTW..You are being way too technical for this crowd, this is a 4x4 forum, not an MIT discussion board, opposite ends of the spectrum :redneck:
Your statement is incorrect. The tire pressure (air pressure inside the tire) may not be 40 psi. It could even be ZERO psi for that matter. Your statement does not account for the tire construction. For example, take the same 5,000 lb. truck and fit semi tractor tires to it, with an 8,000lb load rating per tire. You could run zero psi in those tires and support the weight of the truck with almost no deformation of the carcass.
Consider Newtons third law. In the case of the tire, the opposite reaction is a combination of tire pressure AND carcass construction.