**Solutions:**

**1.** We can express the entire trajectory in terms of A and B. It will depend on the car speed b, the spinning speed s of the disks and the distances d, e and r.

See: http://cs.nyu.edu/cs/faculty/shasha/papers/eddyfig1.doc

Assuming angle A from the start point, we want to calculate a distance to the edge of the disk. Because the disk moves slowly compared to the car, we could tentatively approximate the distance from the start point to the edge of the disk to be d/cos(A). (This distance assumes that the triangle from the origin point to the top of the disk and then to the point where the trajectory from the origin hits the disk is a right triangle, whereas in fact it's a slightly oblique triangle.) Therefore, the time to travel from the start to the disk is approximately d/(b cos(A)).

We will correct this approximation later.

This result implies that C = arcsin ((d tan(A))/r).

Similarly, D = arcsin ((e tan(B))/r) and the distance to the destination from the disk is e/cos(B). The time to travel from the disk to the destination is e/(b cos(B)).

Now, using our chord approach, and with reference to the figure below, we will seek an angle H such that the time it takes to traverse the chord in the frame of the rotating disk equals the time it takes the disk to rotate appropriately.

http://cs.nyu.edu/cs/faculty/shasha/papers/eddyfig5car.doc

Thus, H is uniquely determined by the constraint that we want the car to displace itself in the absolute plane by (180 – (C + D)) degrees.

We'll call this the target angle displacement, or tad for short.

The tad will be equal to the angle subtended by the chord (achord = 180 – 2H) plus the amount of rotation the disk does as the chord is being traversed.

adisk = angle by which disk rotates while chord is traversed

= s * time to traverse chord

where time to traverse chord = length of chord/b

and where length of chord = 2r cos(H).

So, adisk = (2rs cos(H))/B and we have adisk + achord = tad.

This result constrains H.

So, the total time equals the time to go from the start to the disk plus the time to traverse the chord of the disk plus the time to go from the disk to the destination. All of these times can be found, given A and B.

Because the computer doesn't complain, I tried many values of A and B until I got the best value.

Now let's revisit that approximation.

The distance from the source to the disk is a bit more complicated, in fact. To figure it out, note that the distance from the center of the disk to Y is r.

**Let X be the point where the perpendicular from Y to the dashed line hits the dashed line.** *[see the figure eddyfig1.doc]* Therefore, r cos C is the distance from the center of the disk to point X. (We don't know C yet, but we will figure it out by iteration.)

Recall that the distance from the origin to the center of the disk is d + r, which is the distance from the origin to the topmost point of the disk plus the radius. Therefore, the distance from the origin to point X, denoted dsp for short, is dsp = d + (r – r cos C). This implies that the distance of the line segment from the source to the disk is [d + (r – r cos C)]/cos A.

To determine angle C by iteration, we start with our initial estimate of dsp (which was d/cos(A)). This estimate, together with A, gives an estimate of the distance between X and Y: dsp * sin(A). That in turn gives an initial estimate of C, which is arcsin(r/(dsp * sin(A))). This formula allows us to recalculate dsp as d + (r – r cos C).

The numerical conclusions are that the total time it takes, using the best route, is approximately 56.8 seconds. (It would be an even 60 seconds if the disk didn't move at all.) A and B are each 13 degrees. Coincidentally, C and D are each about 13.7 degrees. That means it is necessary to travel nearly 153 degrees around the disk, but because the disk is spinning quite fast, the angle subtended by the chord is only 114 degrees. The time on the disk is just under 25.2 seconds, in which time the disk spins a bit more than 37.7 degrees. So the numbers match up to within about 1 degree, which is good enough for a race.

2. The highest average value of A and B that I get is about 15 degrees. This occurs at spin rates of between 125 and 145 degrees per minute. Note that higher rates will shrink the A and B values. At 10,000 degrees per second, A and B decrease to about 1 degree.