| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
This problem came up recently when my friend Geoff Schultz was attempting to design and implement the board game 'Speed Circuit'. His design specified that the race course should be derived from a set of turn centers and corresponding radii. The course would then be displayed graphically. To derive the race course, each pair of adjacent turns need to be connected by straight-aways. Each straight-away corresponds to a tangent common to the two circles given by the turn centers and radii. The problem is to calculate the coordinates of the tangent points.
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 65.1 | TURTLE::GILBERT | Thu May 10 1984 18:21 | 17 | ||
Let r0 and r1 be the radii, and let d be the distance between the centers of the circles. Note that there may be 0, 1, 2, 4, or an infinite number of tangent lines between the two circles. If d=0, there are either an infinite number of solutions, or no solutions, depending on whether r0=r1. In the remaining solutions, the angle (theta) between the tangent line and the line connecting the centers of the circles must satisfy either: sin(theta) = (r1-r0)/d, or (1) sin(theta) = (r1+r0)/d. (2) In equation (1), the race course would go around the two circles in the same direction (both clockwise, or both counter-clockwise), and in equation (2), the course would go around the circles in different directions (one clockwise, and one counter-clockwise). Given a value of theta, the points of tangency are easily computed. | |||||
| 65.2 | Left out one possible case | DSSDEV::CROCKER | Tue Jul 13 1993 16:58 | 5 | |
There can also be exactly three common tangent lines
between the two circles.
- Ben
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| 65.3 | re .2: Better late than never! | AUSSIE::GARSON | nouveau pauvre | Wed Jul 14 1993 23:47 | 0 |