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SKATEway, week 3

Hi,

This week was a rather strenuous one. Indeed, what we thought to be a reasonably difficult problem has turned out to be a bit too much for us. Indeed, calculating the necessary torque for our skateway should be possible, especially since it is quite like the classical exercise of the dynamical study of an inverted pendulum. And yet, the equations seem to always elude our grasp.

Newton’s 2nd Law yields this system of equations:

\left\{\begin{matrix}  (m + M)\ddot{x}-ml\ddot{\alpha }cos(\alpha )+ml\dot{\alpha }^{2}sin(\alpha) = \frac{C}{r} \\  l\ddot{\alpha }-gsin(\alpha ) = \ddot{x}cos(\alpha )  \end{matrix}\right.

In which:

  • m : mass of the user
  • M : mass of the SKATEway
  • l : height of the center of gravity of the user
  • alpha : angle at which the user is tilted (0 if standing straight)
  • g : gravitational acceleration
  • r : radius of the wheels
  • C : torque of the engine

However, how do we extract the necessary torque from this? What exactly is our scenario? Is it: going from the maximum value of alpha to standing straight? Is the skateway moving or not? Are we simply trying to keep the user in the unstable equilibrium in which the acceleration of the skateway is compensating gravity?

Many questions with few answers.

A very crude approximation is to consider the device as a lever instead of an inverted pendulum: it is equivalent to blocking the wheels. In this scenario, the torque is independent from the radius of the wheels, and is given by the formula: C = m(l+r)gsin(\alpha). However crude this may look, it is the best we could do.

I feel like I am missing something obvious. It should not be that hard, it is a very common problem, the tools to use are well-known, and yet, each and every time, the answer slips away from my mind. Any help on this would be immensely appreciated.

Where does that leave us? One week behind schedule. We did find DIY LiFePo batteries as well as what appears to be a satisfying hub engine, although they were not vetted by our teachers yet. Because we do not have a torque value precisely calculated, we are not too sure of our estimation. Since we need to move on, we will probably use this crude approximation; we did compare it to existing projects, such as gyropodes, and it is of the same order of magnitude. And now, onward with the vetting and then, the liaison.

Edit: with the crude approximation of C = m(l+r)gsin(\alpha), we get: C ~= 311 N.m.

If we assume a maximum speed of 25 km/h (the same as the oneWheel), we need 600 RMP.

Such an engine would require about 20kW. This is a riduculously high value which we cannot use.

Alban

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