Drive Wheel Force...

[trueforensics]

I'm trying to calculate the drive wheel tangential output force of a car and ran into some trouble.

Here is the situation. The car is basically locked into place (ie: tires strapped down). I know the horsepower, the drive wheel torque of the car, and the weight on each wheel. I'm trying to figure out how large of a tengential force the tire could possible put out(to the ground). My first instinct was to simply multiply the maximum drive wheel torque by the radius of the tire to get the output force. But, that turned out to be:

185ft-lbs / 1 ft = 185 lbs. Which seems much too low.

Any advice is greatly appreciated. Thanks.

 

[trueforensics]

Oh. I think I just reaized part of my problem. 185 ft-lbs must be the engine output torque not the drive wheel torque.

 

[NickE]

Yep now youve started to figure it out. And of course the torque multiplication in 1st gear is going to be around 3:1 for 185x3 (555ft-lbs).. Then there's the final drive... another 3:1 or so... (1665ft-lbs) then divide by ft from teh centerline of the axel to the road: (225/50x16 is 24.8dia, and r= 12.4, or 1.03ft) 1665/1.03 = 1616.5 lbs of force (with various assumptions)

 

[trueforensics]

Thanks a lot NickE I appreciate your help.

So the final drive reduction IS multiplied by the 1st gear ratio.

I was unsure if the final gear ratio was the 1st gear reduction (already) multiplied by the axle ratio or if it was a completely seperate reduction.

You were right on. I got almost the exact same result as you (using the actual numbers), except it took me half a day!

1st gear reduction = 2.84
Final Drive Ratio = 3.18
Max Torque = 185 ft-bs
Tire Radius = 24.925/2=12.463"

So:

Force = (2.84*3.18*185)/(12.463/12)=1609 lbs.

Thanks again for your help.

 

[EdDanzer]

Be aware that the torque value for the engine you are using for your calculation may not be available during all portions of a drive cycle.

 

[NormPeterson]

That might depend on whether the end product of this is to simply determine enveloping values for traction requirements and/or loads applied to driveline and suspension components - or if it represents the beginnings of an acceleration or top speed simulation of some sort.

tf?


Norm

 

[trueforensics]

Good point Ed. That max torque is available at 4000rpm, which, I'm assuming, is attainable in this specific situation.

Norm, I am indeed determining the envelope. In short: I'm trying to figure out if a car strapped to a tow truck can produce enough force to break the tow straps (which go around the tire). Obviously the next step is determining the tensile strength of the straps, which I hope to do tomorrow.

Thanks for all of your input. It helps a lot to hear different opinions and cautions.

 

[EdDanzer]

This is a simple problem to solve. Use the maximum weight on the drive axle times the coefficient of friction between the tires and the deck material to obtain the maximum drawbar pull possible.

Or use 80% of the drive axle weight as a maximum approximation. Typically, less than 30% of the axle weight can be converted to drawbar pull.

 

[trueforensics]

Ed,

The method you described was my first reaction too. But that would only accurately model the situation if the tire was not strapped to the tow truck. In this situation the tension in the strap helps the car gain traction by virtually adding weight. The normal force the tire applies is the weight on the axle plus the normal force contribution from the strap tension.

Thanks a lot for your post Ed.

 

[EdDanzer]

The strap tension will be the same as added weight.

 

[trueforensics]

ohhhhhhhhhhhhhh. Now I see. You're right. You just saved me a few hours of figuring. Thanks!!

 

[JayMaechtlen]

Stick or automatic trans?
Is someone revving the engine and engaging the clutch quite suddenly? If so, there is stored momentum in the flywheel assembly (and engine) which will appear as torque/force at the rear wheels.
If automatic, it probably has a torque converter.
What is multiplication in the converter? What is stall speed, which will tell you engine speed which tells you max steady-state torque that the engine will produce.
Oh- you won't have anyone revving the engine and dropping it in gear, will you?

What's trans ratio in reverse?
Is the strap deployment equally strong/effictive in the reverse direction?

I ass/u/me that there's a differential on the drive axle.
Should you ass/u/me that only one half the torque will ever go to either axle?

If nothing else, a nice thought exercise...

Regards
Jay



Jay Maechtlen

http://home.covad.net/~jmaechtlen/