I’m doing physics in high school and I have devised a problem for myself I can’t find the answer to:
What is the torque on a unicycle hub with 7 inch cranks and a 160 pound rider who endures a 4 foot free fall landing in rotational equilibrium?
Rider weight: 712 Newtons
Each Lever arm: .1778 Meters
Fall distance: 1.23 Meters
Acceleration: 9.8 M/sec2
Max velocity: 4.91 M/Sec?
Any help would be welcome,
Have fun!
Cameron
swings arms violently at the cobwebs 
So torque is the crank length, assuming your cranks are at 3 and 9, times the orthogonal force, that is, the force pushing straight down into the ground. But that force depends on the impulse time, that is, how long your tire compresses to absorb the force.
How do you calculate the force, that isn’t just the riders weight is it?
The impulse time would be approx. 15 milliseconds. (Three frames at aprox 20 fps. video)
~Cameron
The true answer will vary depending on a number of variables. The problem as you pose it is: how much torque would be applied to the hub if a rider of a given mass fell a given distance and landed suddenly. That isn’t what happens in real life.
You express the weight of the rider in Newtons. Newtons are a unit of force. The rider has a mass in Kilogrammes (why say Pounds when Newtons and Millimetres etc. are metric units?) The force (Newtons) that the rider exerts will vary throughout the manoeuvre. You could only legitimately say he has a weight of X Newtons when he is at rest or in another absolutely stable state.
So, the rider drops and lands. During the fall, he is exerting minimal torque on the cranks.
At the moment of landing, the tyre starts to compress, meaning that the energy of the fall is dissipated slowly. At the same time, the rider’s ankles and knees flex in synchrony to soften the impact, and the rider may well bend at the waist too. The maximum torque in a real life situation will therefore be very different indeed from this scenario:
A 160 dummy with rigid legs is placed on a unicycle with a solid wheel and no tyre, and dropped 4 feet. And this is the simplified version of the question that you have asked.
This might be an interesting calculation in its own right, but the only way to answer the real question (how much torque does the unicyclist really apply) would be better answered with measuring equipment. I believe that DM Unicycles did som experiments with forces on cranks during normal pedaling.
I have seen a street performer jump as high as he could manage, and land barefoot on a bed of nails. If he had fallen like a dead weight with his legs straight, his feet would have been pierced. This illustrates that the maximum force is hugely dependant on technique. The amount of energy in the system is a constant. The trick is to dissipate that energy over as long a time as possible, and at as constant a rate as possible.
How do you know what you are saying is true? 
Actually, this I meant this to be a kind of serous thread. 
Mikefule: “A 160 dummy with rigid legs is placed on a unicycle with a solid wheel and no tyre, and dropped 4 feet. And this is the simplified version of the question that you have asked.”
I’ve seen videos of 160lb dummies landing four foot drops on a unicycle with rigid legs" I was hoping to calculate how much force the hub experiences with a rider without very much technique. A “160lb dummy” However, calculating in the tire as a cushion.
~Cameron
I think the difficultly is that there are two forces acting at roughly the same time. There is the downward force of the dummy on the pedals, and there is the upward force of the ground pushing up through the unicycle, ultimately concentrated on the hub. Both of these forces are felt throughout the cranks and hub assembly. There is oscillation in the tire and the cranks, and in reality, much dampening does happen within the rider. Also, depending on how the tire oscillation and the crank oscillation line up, the actual torque could vary quite a bit.
I fear that even our best attempts to apply physics forumulas will still only provide rough estimates of the actual force. I am reminded of the “water balloon slingshot” lab project I did in my high school physics class. I derived a formula from basic kinematics formulas to predict how far and at what angle I needed to stretch the slingshot to shoot the water balloon of a certain mass at a target at a given distance. I also experimentally determined a spring constant that accounted for the elasticity of the particular slingshot. However, in the end, we found that I could hit the target more often using my gut feeling and months of experience than by using the physics formula.
Here’s one approach with some handy simplifications:
Calculate the kinetic energy of the rider + uni on impact.
Assume the dummy is really rigid and the tire is the only thing that absorbs the shock.
Assume the tire acts like an ideal spring and that the tirepressure is such that the tire compresses e.g. 2 inches upon impact.
The tire/spring gets compressed until it has absorbed the kinetic energy the rider + uni had at impact, turning it into potential energy in the tire/spring.
Using a formula for the potential energy of a spring, and the fact that it is compressed 2" you should be able to calculate the force the tire/spring excerts on the uni. This will be the same force the dummy excerts on the pedals (assuming the dummy is at least smart enough not to sit on the saddle).
Knowing the force on the pedals and the length of the crank arms you can calculate the torque.
Hmm, but don’t you need a spring coefficient to find the potential energy of a spring? And isn’t the most reliable way to derive a spring coefficient by experiment? And if you’re going to experiment anyway, why not just measure the torque? 
The trick is in the assumption that the tire will compress at least 2 inches.
Epot=0.5kX^2. Epot is know and we assume a lower limit for X. Then we can calculate an upper limit for the spring coefficient.
But you’re right, you still have to experiment make sure the tire doesn’t compress less than 2 inches.
I see, I was under the assumption that we could treat the unicyclist and unicycle together as one big spring. Then everything that flexes and dampens would be accounted for, at least on average. If only I were a mere point in space and time when I ride my unicycle, then I could give the answer.
You could say that the riders legs where one spring and the tire another and do calculations on the two masses, rider and uni seperately. I remember calculating stuff like that at university, but I don’t remember how anymore.
Anyway it would still be a simplification. If you want exact results you’ll just have to turn into a point in vacuum and land exactly straight and rigid and invent a tire that acts like an ideal spring etc.
Btw. Scratch the upper limit stuff I said earlier. It needs to be exact. There may be even more errors. I think I’ll just stop pretending I can still think. It’s almost midnight here. Goodnight.