- #51
jbriggs444
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Let me try to flesh out some equations for this.
Position one wheel of radius R with its center at the origin, (0,0). Position a second wheel of radius R at a distance d to the right so that its center is at (d,0). Paint a dot on the inner surface of the track. At time t=0 this dot will be on the top of the first wheel at coordinate (0,R). Refer to the position of the inner dot as ##(x_i(t), y_i(t))##. Paint a dot on the outer surface. At time t=0 this dot will be on a distance R above the top of the first wheel at coordinate (0,R+r). Refer to the position of the outer dot as ##(x_o(t), y_o(t))##
As per the simple model, the dots move in lock step so that a line perpendicular to the track through the inner dot runs through the outer dot as well. As per the simple model, the distance between the dots is always equal to R.
Assume that the track has a velocity v in the clockwise direction. The dots begin by moving to the right on a stright line path from the top of the first wheel to the top of the second wheel. The position functions for the two dots are given by:
##(x_i(t), y_i(t)) = (vt, r)## for 0 ≤ t ≤ d/v
##(x_o(t), y_o(t)) = (vt, r+R)## for 0 ≤ t ≤ d/v
It is clear that differentiating either position function with respect to time will yield a constant speed equal to v for 0 ≤ t ≤ d/v.
At time t=d/v, the dots will have reached a point directly above the axle for the second wheel and will transition from the flat portion of their trajectory to a curved portion. Match the angular rotation rate to preserve the speed of the outer dot by setting ##\omega = \frac{v}{R+r}##
NOTE: This is an arbitrary choice. Nothing in the simple model requires this choice. One could equally well choose to set ##\omega = \frac{v}{R}## or use any other arbitrary depth within the track by setting the divisor equal to something in the range [R, R+r]. One could even go a little crazy and imagine scenarios where depths outside that range would be appropriate, but let's keep things simple.
Substituting for omega and putting in an appropriate shift to match positions at t=d/v, the position functions for the two dots are then given by:
##(x_i(t), y_i(t)) = (d+R\sin(\frac{v(t-\frac{d}{v})}{R+r}), R\cos(\frac{v(t-\frac{d}{v})}{R+r}))##
##(x_o(t), y_o(t)) = (d+(R+r)\sin(\frac{v(t-\frac{d}{v})}{R+r}), (R+r)\cos(\frac{v(t-\frac{d}{v})}{R+r}))##
Both valid for ##\frac{d}{v} \leqslant t \leqslant \frac{d}{v} + \pi\frac{R+r}{v}##
[I think I have those equations right]
Feel free to differentiate the above with respect to time and take the magnitude of the result to obtain speeds ##v_i(t)## and ##v_o(t)##. Everything simplifies dramatically. As expected from the physical situation, the result should be a pair of constant functions:
##v_i(t)=v\frac{R}{R+r}##
##v_o(t)=v##
NOTE: the speed of the inner dot ##v_i(t)## has a jump discontinuity at the time of the transition from flat to curved from a speed equal to ##v## before the transition to a speed of ##v\frac{R}{R+r}## afterward.
Edit: Note that the relevant observation has been made previously in this thread, notably at post #32 by A.T.
Position one wheel of radius R with its center at the origin, (0,0). Position a second wheel of radius R at a distance d to the right so that its center is at (d,0). Paint a dot on the inner surface of the track. At time t=0 this dot will be on the top of the first wheel at coordinate (0,R). Refer to the position of the inner dot as ##(x_i(t), y_i(t))##. Paint a dot on the outer surface. At time t=0 this dot will be on a distance R above the top of the first wheel at coordinate (0,R+r). Refer to the position of the outer dot as ##(x_o(t), y_o(t))##
As per the simple model, the dots move in lock step so that a line perpendicular to the track through the inner dot runs through the outer dot as well. As per the simple model, the distance between the dots is always equal to R.
Assume that the track has a velocity v in the clockwise direction. The dots begin by moving to the right on a stright line path from the top of the first wheel to the top of the second wheel. The position functions for the two dots are given by:
##(x_i(t), y_i(t)) = (vt, r)## for 0 ≤ t ≤ d/v
##(x_o(t), y_o(t)) = (vt, r+R)## for 0 ≤ t ≤ d/v
It is clear that differentiating either position function with respect to time will yield a constant speed equal to v for 0 ≤ t ≤ d/v.
At time t=d/v, the dots will have reached a point directly above the axle for the second wheel and will transition from the flat portion of their trajectory to a curved portion. Match the angular rotation rate to preserve the speed of the outer dot by setting ##\omega = \frac{v}{R+r}##
NOTE: This is an arbitrary choice. Nothing in the simple model requires this choice. One could equally well choose to set ##\omega = \frac{v}{R}## or use any other arbitrary depth within the track by setting the divisor equal to something in the range [R, R+r]. One could even go a little crazy and imagine scenarios where depths outside that range would be appropriate, but let's keep things simple.
Substituting for omega and putting in an appropriate shift to match positions at t=d/v, the position functions for the two dots are then given by:
##(x_i(t), y_i(t)) = (d+R\sin(\frac{v(t-\frac{d}{v})}{R+r}), R\cos(\frac{v(t-\frac{d}{v})}{R+r}))##
##(x_o(t), y_o(t)) = (d+(R+r)\sin(\frac{v(t-\frac{d}{v})}{R+r}), (R+r)\cos(\frac{v(t-\frac{d}{v})}{R+r}))##
Both valid for ##\frac{d}{v} \leqslant t \leqslant \frac{d}{v} + \pi\frac{R+r}{v}##
[I think I have those equations right]
Feel free to differentiate the above with respect to time and take the magnitude of the result to obtain speeds ##v_i(t)## and ##v_o(t)##. Everything simplifies dramatically. As expected from the physical situation, the result should be a pair of constant functions:
##v_i(t)=v\frac{R}{R+r}##
##v_o(t)=v##
NOTE: the speed of the inner dot ##v_i(t)## has a jump discontinuity at the time of the transition from flat to curved from a speed equal to ##v## before the transition to a speed of ##v\frac{R}{R+r}## afterward.
Edit: Note that the relevant observation has been made previously in this thread, notably at post #32 by A.T.
Again, OldYat is still correct.....
