I have made this case in multiple other threads (mostly in regard to hops), however I’ll make it here as well. Physics and unicycling should not be mixed. They do not work well together. A unicycle and a rider are too complicated of a system for basic Newtonian mechanics as applied by joe schmoe to solve. Read the second most recent “Physics Today” magazine, and you will see a research paper on the physics of walking. These re post-docs doing that work, to boot. I’m an undergrad physics major and have a mechanics class under my belt. I still don’t tell myself I can solve problems this hard.
AscenXion, what you are referring to is a larger TORQUE required to accelerate a more massive wheel (remember, accelerate means both in speed and direction). We all remember our basic newtonian kinematics equations, right? Force=MassAcceleration. Well, the same goes for gyroscopes, except replace force with torque, mass with “moment of inertia” and acceleration “angular acceleration” (units: radians per second per second, aka inverse seconds squared). Moment of inertia has units of mass(distance squared).
Well, I said physics and unicycles shouldn’t be mixed, but here I go anyways. Bear with me, this does lead somewhere.
So, we can model a unicycle wheel as (approximately) an ideal ring with all of the mass concentrated at the rim (if it were a uniform disc it would be 1/2 of what it is. Really it’s somewhere in between). Thus, a wheel has a moment of inertia of M(R^2) where m=mass and r=radius.
Now, what is important is angular momentum (we will call it “L”), which is moment of inertia (commonly called “I”) times angular velocity (radians per second, which we will call “O”). Thus, L=I*O. Now, Torque (“T”) can also be described as the change in angular momentum, L. Thus, T=dL/Dt (t=time). This means that the instantaneous torque on a system=the change in angular momentum with respect to time (rather than position, for example). Thus, a greater L means a greater T must be applied in order to get the same change in angular momentum (dL/dt). Follow? Note: Angular momentum, unlike regular momentum, is a VECTOR which means it has a direction and a value associated with it. Thus, a gyroscope can spin at the same rate but have its axle rotated perpendicular to the direction of its spin, and the angular momentum will have changed.
So, ultimately, it takes more energy for a rider to change the direction of a heavier wheel (where the mass is at the rim) than a lighter one, as well as speed it up or slow it down. however, if we model a unicycle rider as riding in a straight line at a constant speed, all of this falls out, since the only issue is friction in the system, which remains constant. The problem is that we don’t ride in straight lines at constant speeds.
All else being equal, the rider with the heavier wheel should ride a straighter line, since the heavier wheel will be more resistant to the rider’s slight wobble. This indicates that the heavier wheel will take LESS energy to ride, since because it wobbles less, the rider will ultimately cover less distance and thus do less work (int he technical sense) to fight friction. Furthermore, the heavier wheel will be more stabler (the heavier the gyro, the more stable it is, and a spinning wheel is a gyro). At the same time, the rider with the lighter wheel will have to exert less force to make a turn. This will use less energy. He or she will also have to use less energy to get up to speed and slow down. Ultimately, these four effects balance out to some degree, and you should ride whatever you prefer. And once again, this is a VERY simplified model of a unicyclist, and has little if any bearing on reality.
AscenXion, please, lets not criticize people’s grammar. That would be like me, instead of taking the time to explain the nuances of why you’re wrong, just calling you a dumbass for not being able to infer what to me is intuitive basic physics (which it is).
Edit: Centripetal force has absolutely nothing to do with this problem. Centripetal force would be a problem if we wanted to figure out (to an ungodly precision) how much tension is in the spokes of the wheel at various speeds. Even then, my physics professor would slap me for using it instead of “centripetal acceleration”. Once again, if we assume the rider is going in a straight line at a constant speed, the ground will exert the same friction no matter what the mass of the wheel (don’t tell me you can tell the difference of the friction due to a few grams of extra mass) and thus the rider will have to exert the same force to keep the wheel spinning.
Honestly, I think AscenXion spends too much time posting about equipment choices based on minor details rather than just going out and riding to see what he prefers. I’ve ridden the gazz and duro, and I really see little, if any difference between the two. Neither takes more energy to ride (just from experience I say that), and whomever thinks that is projecting based on pre-determined conclusions.