On what kind of surface would a square wheel function the same as a round wheel?

Imagine a circle with an inscribed square in it, when the circle rotates on a planar surface, the center never moves, but the vertex of the square are rotating and at the same time moving in an x-axis direction. The sine function relates one point in the circunference rotating and moving in X direction. In this case, we have to use all angle measurement in radians in order to relate rotation and distance. As the square rotates and moves, one vertex of the equare respect to the floor, moves in a way like a curve, in order to obtain the behavior of this curve, we first relate the distance of the center of the square to one of it's sides L. We make an isoceles triangle of two sides √2L/2 and the other would be L, in this triangle we make a random line that touches the center of the square and one random point in it's opposite side L (this new line would be M). This relation is proportional with the rotation of the square, so the angle of aperture between side √2L/2 and the line randomly drawed (M) is X radians. With this relationship we use the law of sines and the angles inside the new triangle in order to leave M in therms of X radians. This relation would give M=√2*L*sin(Pi/4)/(2*sin((3*Pi/4)-X)). M represents the distance that the mentioned curve would have from the center of the square to the curve, so to get the height of such curve from the x-axis to any point in the curve we subtract M from √2*L/2. This subtraction would be the amplitude A of a new sine equation that we must find in order to know the behavior of the curve. Simplifying the amplitude A gives A=√2*L*(2*sin((3*Pi-4*X)/4)-√2)/(4*sin((3*Pi-4*X)/4)-√2)). With the amplitude A we finally observe that the curve that describes the movement of the side L rotating like a wheel is a sine ecuation of amplitude A, but with the diference that in every cycle of the original square, the curve formed in the floor makes two cycles (that is 2*X) only in positive values (the curve is never negative in the y-axis). With this observation, the new equation must be evaluated in absolute values (always positive), so the solution to the problem would be Surface=Y=ABSOLUTE[A*sin(2*X)]=ABSOLUTE[(√2*L*(2*sin((3*Pi-4*X)/4)-√2)/(4*sin((3*Pi-4*X)/4)-√2)))*sin(2*X)]. It would plot a series of bump-like curves that is only a sine function showing always absolute or positive values.

Posted by Antonio on 2003-09-04 04:25:55