Thoughts on SKINNY TIRES VS THICK TIRES

https://www.youtube.com/watch?v=PfJP-3CVUSM

 

Oh my, I'm so sorry. I neglected to see everyone was just arguing with each other in here so I tried to help the OP. Again. My sincerest apologies. I hope I am getting back on track here.

 

So today 10/24 Saturday decided to take the taco with the stockie tires up to Pismo and check out the dunes. Also grub on the $40 for 40 Oysters at the Jolly oyster. Made it okay for the most part then found myself in soft sand, little acceleration and "Fack" .... Thanks helpful lady in the Rubicon

 

4x4?

 

Given a choice I would run a tall skinny tire. In my off roading experience I have needed ground clearance most of all. I have run 285/75/16's and 265/75/16. Currently on 285's but only because I got a smoking deal on them. My next tires will be 255/85/16's.

 

RWD

 

Ah

 

Did you try crawl control?

 

2wd...

 

What air pressure in the tires? Doesn't look like you aired down.

 

Yeah no air down I tried to live on the edge

 

Coming from a science perspective. Traction is proportional to the amount of friction between the tire and the ground. Given the Frictional Force formula --> Friction= (Coefficient of Friction) x (Normal force). Coefficient of friction is a constant depending on the two surfaces. Rubber and asphalt is .71. Coefficient of friction for wet snow is .1. The normal force is the force perpendicular to the ground. According to newtons 3rd law (Action Reaction) the force exerted toward the ground will receive an equal amount of force in the opposite direction. If you multiply the gradational force times the mass of the tire this will give you the normal force (on level ground).

50lb tire weighs about 22kilograms
gradational force=9.81meters per second
Coefficient of friction for Rubber and wet Snow is .1----> Rubber and asphalt .71

For example F=(.71)x(9.81m/s x 22kg) = 153 Newtons
For example F=(.1)x(9.81m/s x 22kg) = 21.58 Newtons

If you add a heavier tire in the same scenarios then it changes.

60lb tire is about 27 kilograms
For example F=(.71)x(9.81m/s x 27kg) = 188 Newtons
For example F=(.1)x(9.81m/s x 27kg) = 26 Newtons

The heavier tire has more traction than the lighter tire in both scenarios. The thinner tires are better in snow because they are able to slice down and make contact with a surface with a higher coefficient of friction. The wider tires are better for sand because they are able to stay on top of the sand instead of digging in.

So weight of the tire might be something to consider as well.

 

It didn't load! maybe because the website no longer has it published. I was very interested though.

 

Super old thread... But decent topic so I'll re-engage .

You're really close. The problem with your analysis is that the force between the tire and ground is not only from the weight of the tire, but the whole corner of the truck. There's roughly 1500lbs on each front corner of my Taco. You are technically correct that a heavier tire will increase the normal force, but not as much as you think. Going from a 50 to 60lb tire is going to increase your normal force from 1500 to 1510lbs... Which is pretty insignificant. It will increase the rotational mass and moment of inertia waaaaay more significantly, making your engine work a lot harder to get things moving. If I could get a 20lb 35" mud tire with similar performance (loosely interpreted as coefficient of friction) and reliability as their 70lb counterparts, I'd be all over it!

A second point, you can increase your total friction force by adding weight but that extra weight will require more force to move. If you do the math they exactly cancel out and the system is mass independent. You end up with force = (mass * acceleration), but your force is limited by friction so it's (mass * coefficient of friction) = (mass * acceleration). Mass on both sides, cancels out.