- #1

- 3

- 1

This is to help give an idea of stress on a transmission.

I have searched an hour hear as well as Google looking for a formula or calculator, but have been unsuccessful.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- #1

- 3

- 1

This is to help give an idea of stress on a transmission.

I have searched an hour hear as well as Google looking for a formula or calculator, but have been unsuccessful.

- #2

Mentor

- 22,323

- 9,545

Your question is not at all clear. Force from what? Why would this stress the transmission? Get the tire spinning under what conditions? Why would tire brand or size matter?

This is to help give an idea of stress on a transmission.

I have searched an hour hear as well as Google looking for a formula or calculator, but have been unsuccessful.

Please describe exactly the scenario you are trying to analyze.

- #3

- 3

- 1

Using the weight of the tire and the diameter of the tire and maybe the rolling resistance, if published.

I want to compare how weight and size of tire affect the force needed to start a tire rolling. This will allow me to compare say, BF Goodrich/Tusk/Tensor/Sedona tires of 28" and 30" diameters in a standard way to help aid my decision in which tire to purchase.

- #4

- 228

- 103

The key factors in making this calculation are coefficient of friction, weight on the wheel and wheel diameter.

Here are some hints:

1. If you assume a coefficient of friction between the tire and the surface of the road to be 1 then it will take as much force to slid (spin) the wheel as is pushing down on that wheel.

If you assume a coefficient of 0.50 then it would take only 1/2 the weight on that wheel to slid the tire. BTW-This force is perpendicular to the applied weight and right at the road surface.

2. How do you get this force to be applied right at the interface of the wheel and the road surface?

3. How does the diameter of the wheel fit into the equation?

4. How does the tires’ rubber compound effect the equation?

5. How does the footprint made by the tire on the road effect the equation?

What do you think?

BTW - Do you mean rolling or sliding (spinning)?

Here are some hints:

1. If you assume a coefficient of friction between the tire and the surface of the road to be 1 then it will take as much force to slid (spin) the wheel as is pushing down on that wheel.

If you assume a coefficient of 0.50 then it would take only 1/2 the weight on that wheel to slid the tire. BTW-This force is perpendicular to the applied weight and right at the road surface.

2. How do you get this force to be applied right at the interface of the wheel and the road surface?

3. How does the diameter of the wheel fit into the equation?

4. How does the tires’ rubber compound effect the equation?

5. How does the footprint made by the tire on the road effect the equation?

What do you think?

BTW - Do you mean rolling or sliding (spinning)?

Last edited:

- #5

- 3

- 1

The key factors in making this calculation are coefficient of friction, weight on the wheel and wheel diameter.

Here are some hints:

1. If you assume a coefficient of friction between the tire and the surface of the road to be 1 then it will take as much force to slid (spin) the wheel as is pushing down on that wheel.

If you assume a coefficient of 0.50 then it would take only 1/2 the weight on that wheel to slid the tire. BTW-This force is perpendicular to the applied weight and right at the road surface.

2. How do you get this force to be applied right at the interface of the wheel and the road surface?

3. How does the diameter of the wheel fit into the equation?

4. How does the tires’ rubber compound effect the equation?

5. How does the footprint made by the tire on the road effect the equation?

What do you think?

I think I’m not wanting that much specificity. Which was a concern when I posted. It might not be possible to get the information I’m wanting, and that’s ok. I’m just looking for some how to add a little science into my decision process. Everyone has an opinion on which tire is best, but objective facts would be nice.

My physics classes were A LONG time ago. So, I came here for a little help. If that help ends up being that I can’t get the info I’m looking for, then that’s fine. Reading through the posts during my search, you are some smart folks here, so anything helps.

- #6

- 228

- 103

We expect you to invest a little into the process of finding the answer. Just like in school!

- #7

Mentor

- 22,323

- 9,545

Is this tire on the ground or spinning in air? You said "spinning" in your first post. Is the vehicle in gear? What force is making it "spin" - the engine torque? Are you just trying to get the tire to turn or are you really trying to find out what it takes to get the vehicle to move?I am interested in the force required to start a tire moving from a dead stop on my side by side/UTV.

Why do you think any of that matters?Using the weight of the tire and the diameter of the tire and maybe the rolling resistance, if published.

I want to compare how weight and size of tire affect the force needed to start a tire rolling. This will allow me to compare say, BF Goodrich/Tusk/Tensor/Sedona tires of 28" and 30" diameters in a standard way to help aid my decision in which tire to purchase.

Again, you need to put a lot more effort into this, not just so you can learn like in school, but also because with the information you've given us, we literally cannot help you. We have no idea what you are trying to do if you won't tell us. But if making a purchasing decision on the right tire for your vehicle is what you are ultimately after, then the short answer is that none of what you are saying matters.

- #8

- 228

- 103

- #9

- 12

- 0

Here are some hints:

1. If you assume a coefficient of friction between the tire and the surface of the road to be 1 then it will take as much force to slid (spin) the wheel as is pushing down on that wheel.

If you assume a coefficient of 0.50 then it would take only 1/2 the weight on that wheel to slid the tire. BTW-This force is perpendicular to the applied weight and right at the road surface.

2. How do you get this force to be applied right at the interface of the wheel and the road surface?

3. How does the diameter of the wheel fit into the equation?

4. How does the tires’ rubber compound effect the equation?

5. How does the footprint made by the tire on the road effect the equation?

What do you think?

BTW - Do you mean rolling or sliding (spinning)?

hey azfireball,

how to proceed if i want to calculate the starting torque required for a bike weighing 120 kg and a load of 150kg with tyre specification of 120/70 r12.

- #10

- 228

- 103

- #11

Mentor

- 22,323

- 9,545

Share:

- Replies
- 20

- Views
- 1K

- Replies
- 20

- Views
- 690

- Replies
- 5

- Views
- 817

- Replies
- 16

- Views
- 861

- Replies
- 29

- Views
- 2K

- Replies
- 3

- Views
- 403

- Replies
- 6

- Views
- 333

- Replies
- 1

- Views
- 613

- Replies
- 3

- Views
- 414

- Replies
- 10

- Views
- 824