# Foot-pounds of torque?

I just looked through the list of unicycles at ebay.com for the first time and noticed that the DM has been “tested to 450 foot-pounds of torque”. What are foot-pounds of torque? Is there any way of estimating the amount of torque on my Profiles given a drop height and my mass (and the uni’s mass)?

Thanks,
Andrew

It’s been a while since I’ve had physics. I’ll do my best to explain off the top of my head.
Torque is a measure of force exerted on an object at some distance from its point of rotation. Think of a door. You push at the opposite end of the hinges and you can push the door open easily. If you push near the hinges its harder. How hard you have to push is the Force and how far (at a perpindicular) from the hinges you are is the distance. Torque is Force x distance. (x meaning cross product… simply the multiply if the distance and force are perpindicular)
In non-SI units, force is measured in pounds and distance in feet. I.E. torque is measured in foot*pounds. Force is mass * acceleration. Earth exerts a constant acceleration on things(approximately). So the force of your mass falling with acceleration 9.8m/s^2(apprx 30ft/s^2) is you’re weight in pounds.

Now the question is where is the rotation point. It could be considered the center of the axle at the point where the crank connects. In this case the distance would be the length of the crank. But the point could also be taken to be the exact center of the axle which would yield another distance.

There’s also the unicycle weight (which is mostly in the wheel) so there is an additional force at those points around the hub. But the force is directed straight into the hub(angled spokes horizontal forces basically cancel each other out) so there is no torque.

Another thing. When it was tested to 450 foot-pounds, does that mean it survived that amount of torque once? Or multiple times? Did it survive once and then immediately get tested at the next increment (455 foot-pounds ?) and break? Was it 450 on on side of the axle or 450 total on both simultaneously or asynchronously?
Lots of things to consider. Someone should give me a job! Anyway I think I analyzed this too much. I’ve got to get to class. Maybe someone else will be more coherent than I.

Thanks a lot Nick, that was really helpful.

Andrew

Been a while for me too, but I need to clarify a little.

The 450 ft-lbs is obtained by holding one crank stationary, then applying 450 lbs at one foot from the axle center on the other crank. If the weight is closer, then the actual poundage has to increase so that the product f*d is 450 ft-lbs. This puts a twisting force (torque) on the axle by the weight pulling down on one side, and the other side pressing up against something so that the axle doesn’t move. This is a static test; nothing moves.

When you do a drop, while it is true that acceleration is constant, it is not true that the resulting torque is a constant. It depends on how far you are dropping. When you drop, gravity accelerates you so that you are storing an increasing amount of kinetic energy in your velocity. This is m*v, mass times velocity, or momentum. When you hit the earth (assume that you land in balance with equal pedal pressure), you have to transfer all that kinetic energy to the earth. The earth stops the uni through the wheel, as Zim said. The earth stops you through the uni, through the pedals then cranks then hub then spokes and rim and tire. The forces the deceleration places on the hub is also a torque, but since the wheel is not rotating (by assumption), the torques from each side cancel themselves out. The hub flexes in a twisting way then springs back, if you have not passed the hub’s ability to twist without distortion.

So when you drop, the torque placed on the hub is dependent on how far you have dropped, because that determines how much kinetic energy you have built up and need to transfer, through the uni, to the earth.

If you are just standing on the uni, not moving or having dropped, the torque issues are the same but only depend on the force that gravity pulls on your mass with. This is a constant. ## There is another equation for the deceleration impulse; something to do with delta-p, but my books are buried somewhere in a box. That seems familiar. I said it had been a while. I did some calculations using a number of more or less reasonable assumptions.
The result was that even with a pretty bad landing this hub should be ok with an 8’ drop.

My assumptions where:

1. The rider weighs 200 pounds.
2. He uses 175mm cranks
3. The rider and the tire softens the landing in the same way an ideal spring would have done.
4. The “spring” is compressed 3" upon impact.

I think that these asumptions makes me underestimate the drop hight rather than overestimate it.

My guess is that the cranks or the rider will break before this hub.

Morten

Thanks Morten. Does anyone know what the Profile hub is rated to? Is it rated at all? I weigh abuot 65kg so I should be fine for anything up to about 10’ for now ;).

Andrew

Re: Foot-pounds of torque?

It doesn’t really matter what it’s rated to, not with a lifetime warranty! Dylan