Unicycle Physics: Calculating Wheel Diameter & Maximum Hop Height

[smackay]

I have two unicycle-related questions that seem fairly straightforward.

1) Do smaller wheel diameters offer a mechanical advantage? For example, would you have more leverage on a unicycle with a 20 in wheel versus a 36 in if the crank length was the same? I know from experience that you do, but how could I show this with math?

Here's my stab at it.
Assuming a 20 in (0.508 m) wheel diameter, traveling at 8 mph (3.58 m/s), with 125 mm cranks, and 150 lb (68.04 kg) unicyclist:
A_c=v²/r
A_c = (3.58 m/s)²/0.254 m = 50.46 m/s²
F=ma
F=(68.04 kg)(50.46 m/s²)
F=3433 N
torque=rF
torque = (0.125 m)(3433 N)= 429.1 Joules

Assuming a 36 in (0.9144 m) wheel diameter, all other things constant.
A_c=v²/r
A_c= (3.58 m/s)²/0.457 m = 28.04 m/s²
F=ma
F=(68.04 kg)(28.04 m/s²)
F=1907 N
torque=rF
torque = (0.125 m)(1907 N) = 238.4 Joules

I am not even sure if this makes sense.

2) I didn't know how to approach this one. I want to find the maximum hop height for a decent unicyclist. People have been known to get over one meter using two techniques: http://www.youtube.com/watch?v=3TPT7vLl15s&feature=related". A quick calculation (using (1/2)mv²=mgh for the rolling hop) shows that less that a foot of the height could come from kinetic energy at typical unicycle speeds. Is there any way to calculate the height for either of the techniques?

 

[jack action]

First of all, your equations are misused and they don't make any physical sense.

Yes, smaller wheels offer a mechanical advantage.

The proper way to show it is by knowing that the torque at the crank (_c) is the same as the torque at the wheel (_w) (since they are part of the same unit):

T_c = T_w
F_c * R_c = F_w * R_w

or

F_w = F_c * R_c / R_w

Assuming the same force at the crank (F_c), the friction force acting between the ground and the tire (F_w) is inversely proportional to the wheel radius (R_w). The smaller the wheel, the larger the force.

Similarly, for speed (V), we know that the angular velocity (w) will be the same for the crank and the wheel:

w_c = w_w
V_c / R_c = V_w / R_w

or

V_w = V_c * R_w / R_c

Here, assuming the same speed at the crank (V_c), the velocity of the wheel at the ground (hence the unicycle velocity) is proportional to the wheel radius. The smaller the wheel, the lower the unicycle speed.

By changing the wheel or crank radius, you gain force and loose speed (or vice versa).

If you want to know how much torque and speed is possible with your unicycle, you are better using the power approach. The http://blog.mapawatt.com/2009/07/19/bicycle-power-watts/" is between 200 and 400 W. Power (P) is always constant throughout the mechanical system, so:

P = T * w --> torque and angular velocity is the same for the crank and the wheel
P = F_c * V_c
P = F_w * V_w

So assuming a power of 300 W, if you travel at 8 mph (3.58 m/s),the wheel friction force is 83.8 N. With a 20" wheel (0.508 m), the torque is 21.29 N.m (The proper unit for torque is not «joule», which is an energy unit). The angular velocity is 14.1 rad/s (135 rpm). You can find the foot speed and force, knowing the crank radius. These equations are true for instantaneous values or average values.

As for the wheel hop question, it is more complicated as I think there is some energy storage in the tire involved (through multiple jumps). Although my best bet to evaluate how high you can go would be:

(energy stored in tire) = (potential energy)

½kx² = mgh

Where k is the tire vertical spring stiffness (N/m) and x is the maximum tire deflection achievable.

But I'm really not sure about how to evaluate this one.

 

[tvavanasd]

2) What is being raised over one meter during a hop? Is the rider actually raising their center of gravity by a meter or just their wheel (or lowest point)?

 

[smackay]

2) What is being raised over one meter during a hop? Is the rider actually raising their center of gravity by a meter or just their wheel (or lowest point)?

The bottom of the tire is raised over one meter; the center of gravity is not raised as much because (I think) the tuck during the jump accounts for much of the height.

With some additional thought, this description seems plausible:
Upward forces = (spring force of tire)+(force applied to pedals from feet)
After using mechanics to figure out the height from these forces, I could add additional height due to the tuck and kinetic energy->potential energy transfer(the latter mainly in the "rolling hop" scenario).

If someone would check my reasoning, I would be very grateful.

First of all, your equations are misused and they don't make any physical sense.

I figured as much. I used the same method as a sketchy site I found and didn't think it through. Thanks for the incredibly cogent response.

 

[jack action]

Energy (E) is the force (F_w) times the distance traveled (d). Since we already determined that P = F_w * V_w, if we assume constant power input and constant speed, we get:

E = P_w / V_w * d

To which you might add some losses due to the friction in the mechanical components like bearings (although, very small).

 

[tvavanasd]

The majority of the energy demand (in many cases) comes from the effort required to maintain stability. I think that the problem is similar to that of an inverted pendulum.