Thread: Equilibrium of a dice View Single Post
 P: 613 I did a bunch of triangulation and found two things you need: 1.) If the center of mass of the square is traced down to where it crosses the line between the two pegs, the distance from the left peg to that crossing is (L + (L - L COS a) COS a) - (2 ^ .5 * L / 2 * SIN (a + PI/4)) And the distance from the right peg to that crossing is (L - distance to left peg). I don't know how much that expression can be simplified--I didn't even try. You need those distances to find how much of the weight rests on each peg--it's not an equal split except when "a" is 45 degrees. I'm going to call the distance from the left peg to the crossing X and the distance from the right peg to the crossing L - X so that I don't have to say that whole thing out again. 2.) First, the distance from the center of the square to the crossing is (2 ^ .5 * L / 2 * SIN (a + PI/4)) - L * COS a SIN a I will call that distance Y. The distance from the left peg to the center of mass is therefore (X ^ 2 + Y ^ 2) ^ .5 and the distance from the right peg to the center of mass is ((L - X) ^ 2 + Y ^ 2) ^ .5 Those two distances are your lever arms.