Path: utzoo!utgpu!news-server.csri.toronto.edu!cs.utexas.edu!wuarchive!uunet!timbuk!cs.umn.edu!chan From: chan@umn-cs.cs.umn.edu (Tak-Yuen Melinda Chan) Newsgroups: comp.graphics Subject: Question on geometry problem (line and curve) Message-ID: <1990Dec4.210527.967@cs.umn.edu> Date: 4 Dec 90 21:05:27 GMT Sender: chan@cs.umn.edu (Tak-Yuen Melinda Chan) Organization: University of Minnesota, Minneapolis - CSCI Dept. Lines: 63 1. I am tasked with writing a computer routine that renders a series of vertices into a series of straight lines and circular arcs. The resulting lines will reduce the amount of computer storage space if a number of vertices can be represented by each line or arc. 2. To date I've succeeded in writing the routines that combine these vertices into lines and arcs. I then use another routine which links these lines and arcs into a continuous stream while attempting to maintain tangential continuity. I link two adjacent straight lines with short arcs. I link lines and arcs by adjusting their equations (staying within a given tolerance of the vertices). One problem I fail to solve involves linking three adjacent arcs together while maintaining tangential continuity. 3. The Problem: Given: x1, {the x coordinate of the first arc radi y1, {the y coordinate of the first arc radi r1, {the radius length of the first arc x2, {the x coordinate of the second arc radi y2, {the y coordinate of the second arc radi r2, {the radius length of the second arc x3, {the x coordinate of a point on the mid arc y3 {the y coordinate of a point on the mid arc Find the points (there is at least one point and sometimes two) equal distant from the point (x3,y3) and the two arcs. 4. Possible solution techniques include intersecting the arcs by adding an offset distance to each radius length, thereby finding the radical axis. the first arc equation: (x1)**2 + (y1)**2 = (r1)**2 the equal distance is referred to as the offset distance (OFF). adding OFF to the first arc equation: (x1)**2 + (y1)**2 = (r1+ OFF)**2 {EQ 1 the second arc equation (x2)**2 + (y2)**2 = (r2)**2 with the same offset this equation becomes (x2)**2 + (y2)**2 = (r2+OFF)**2 {EQ 2 subtracting eq 2 from eq 1 produces the radical axis equation (x1)**2 + (y1)**2 - (r1+ OFF)**2 - (x2)**2 - (y2)**2 + (r2+OFF)**2 = 0 this will produce a series of points equal distance from the two arcs. The additional requirement of being the same distance away from x3, y3 as arc 1 and arc 2 can be represented by using the (x3, y3) point as the radi of another arc with a radius length of OFF. It's equation (x3)**2 + (y3)**2 = (OFF)**2 {EQ 3 Intersecting this equation with the (eq 1 - eq 2 = 0) radical axis produces (eq 1 - eq 2) - eq 3 = 0 which I believe would produce the answer I seek, however, in the several attempts made all the substitutions used cancelled all the variables out leaving me with 0 = 0 Thank you for any assistance provided Brought to you by Super Global Mega Corp .com