Mechanical engineering question.

[JohnnieB]

Out of curiosity, I was trying to figure out how much torque is applied where rubber meets road and ran into a stumbling block.

I know you multiply the torque generated by the motor by the gear ratio of the gears in the differential to get the torque at the axle, but not quite sure of how to calculate what portion of that is transmitted to the tire's footprint.

For example: My motor develops about 10.3HP at 1450RPM (@42V), so it produces about 37.4 ft/lb of torque. The gear ratio is 12.44:1, so there is about 465.2 ft/lb of torque present at the axles, disregarding mechanical losses.

My question is: How much of that torque is applied to the road through tires that are 17" in diameter?

[rlw]

JohnnieB,

I'm not an engineer, nor do I play one on TV.

I'm guessing you're looking for the force that the tire applies to the (for illustration purposes) road.

Is torque the applicable force? I thought torque was the measure of rotational force - a wheel around an axle, a lever at the point of its fulcrum, etc.

If it does apply, wouldn't it be a function of the surface area of the tire's contact with the road and the elasticity of the tire itself (although intuitively, I think you'd eventually get that back as the tire's rubber rebounds from the initial stretch)?

There's a pretty dense discussion of torque on Wikipedia that might shed some light: http://en.wikipedia.org/wiki/Torque

Just wondering...

RLW

[sportcoupe]

I also am not a mech eng but I do have The Google. I searched for torque vs tire size and found this statement.

"As a rule of thumb you can expect to lose about 3.5% of torque/gear ratio for every inch of tire size increase."

Source: http://www.4x4abc.com/4WD101/math_wheels.html

[JohnnieB]

Originally Posted by rlw

JohnnieB,

I'm not an engineer, nor do I play one on TV.

I'm guessing you're looking for the force that the tire applies to the (for illustration purposes) road.

Is torque the applicable force? I thought torque was the measure of rotational force - a wheel around an axle, a lever at the point of its fulcrum, etc.

If it does apply, wouldn't it be a function of the surface area of the tire's contact with the road and the elasticity of the tire itself (although intuitively, I think you'd eventually get that back as the tire's rubber rebounds from the initial stretch)?

There's a pretty dense discussion of torque on Wikipedia that might shed some light: http://en.wikipedia.org/wiki/Torque

Just wondering...

RLW

What the force is called and how it is measured, does gets a little gray at the point were the tire touches the road.

A tire is effectively a lever rotating around a fulcrum, so a 17" tire applies twice as much force at its point of contact with the road than a 34" tire does.

What I'm looking for is a formula (simplified version if possible) that tells me how many pounds of force is applied at the end of the lever (tire tread) when the ft/lb of torque at the fulcrum (axle) and length of the lever arm (Tire radius) are known.

It looks like I'll have to break out the trig tables and scientific calculator.

However, I think I'll watch NASCAR at Martinsville instead.

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For my purposes, the rolling loses of the tires are not part of it.

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Full disclosure: I have a spreadsheet I use to calculate things like relative gains and losses in speed, motor rpm, and torque with different sized tires and gear ratios and was wondering if the suppositions I've been making are valid.

[JohnnieB]

Originally Posted by sportcoupe

I also am not a mech eng but I do have The Google. I searched for torque vs tire size and found this statement.

"As a rule of thumb you can expect to lose about 3.5% of torque/gear ratio for every inch of tire size increase."

Source: http://www.4x4abc.com/4WD101/math_wheels.html

Thanks, a quick peek tells me my answer might be buried in there someplace.

I'll check that out after the race.

[mcdave71]

Don't you also have to factor in for the ratio from the clutches? It's not straight one to one from motor to rear end gearing.

[LVCJ]

Originally Posted by mcdave71

Don't you also have to factor in for the ratio from the clutches? It's not straight one to one from motor to rear end gearing.

I don't think he has "clutches". He's dealing with an electric motor.

[JohnnieB]

Originally Posted by LVCJ

I don't think he has "clutches". He's dealing with an electric motor.

Yes, my electric motor is directly coupled to the input shaft of the differential, so it is a 1:1 ratio.

If the Motor/Engine is indirectly coupled to the differential, that ratio must be included also.

[mcdave71]

Missed that..

Originally Posted by LVCJ

I don't think he has "clutches". He's dealing with an electric motor.

[David Maner]

Originally Posted by JohnnieB

Out of curiosity, I was trying to figure out how much torque is applied where rubber meets road and ran into a stumbling block.

I know you multiply the torque generated by the motor by the gear ratio of the gears in the differential to get the torque at the axle, but not quite sure of how to calculate what portion of that is transmitted to the tire's footprint.

For example: My motor develops about 10.3HP at 1450RPM (@42V), so it produces about 37.4 ft/lb of torque. The gear ratio is 12.44:1, so there is about 465.2 ft/lb of torque present at the axles, disregarding mechanical losses.

My question is: How much of that torque is applied to the road through tires that are 17" in diameter?

Plug the radius of the tire in your equation. What you're working with is foot pounds so you know that you start with 12 inches. if your tires are smaller than 24 OD then add to the figure you have, if larger then subtract whatever percentage of the radius to 12 inches is.

Heck now I'm confused too LOL