Here is one way to calculate the intersection points of two circles: Given circle of radius r1 at origin: x^2 + y^2 = r1^2
Circle at (x0,y0) with radius r2: (x-x0)^2 + (y-y0)^2 = r2^2 Intersection points between the circles lie on a line: y = mx + b Where: m = -x0/y0 b = - 1/2y0 (r2^2 - r1^2 - y0^2 - x0^2) x = (-mb +/- SQR(m^2 b^2 - (m^2+1)(b^2-r1^2)) / (m^2 + 1) If the roots of x are imaginary (i.e. the radical is negative) then there is no intersection. Adjust coordinates of intersecting circle relative to base circle. I have this function coded in C, QBASIC and assembler if anyone wants it. Best regards, Jim James E. Morrison Astrolabe web pages at: http://myhouse.com/mc/planet/astrodir/astrolab.htm ----- Original Message ----- From: John Carmichael <[EMAIL PROTECTED]> To: Ron Anthony <[EMAIL PROTECTED]> Cc: <sundial@rrz.uni-koeln.de> Sent: Wednesday, May 26, 1999 10:43 AM Subject: Re: frame & grid method > Ron: > > Very cool! Point C is the intersection of two offset circles with centers > at A and B, whose radii is known. I wonder what the exact mathematical > formula for this is? Wouldn't it come out as an (x,y) coordinate? > > John Carmichael > http://www.azstarnet.com/~pappas > > >All, > > > >I'm sorry I was only half awake when this thread started so forgive me if > >I'm off course. If I had to lay out a large dial (say 100 ft) to a high > >degree of accuracy (say .1 of an inch) I would plot all the points not as > >x,y co-ordinates. I would plot them all out as the intersection of two > >lines from two fixed points. > > > >To see what I mean pick 2 points that are well established, e.g., point A > >where the gnomom meets the dial face, and point B some number of feet due > >north (in line with the gnomon base) of point A. Every point on the dial > >face is now at the intersection of two tape measures that start at points A > >and B. Assuming that the dial face is flat the accuracy would be good as > >the tape measures used. For the points that are almost inline with the AB > >line, a third point C could be used as one of the points. Point C could be > >calculated from points A and B. Of course the computer would have to > >calculate all of the points for you. > > > >As a crude ASCII art: Point X is 30" 1 1/4" from point A, and 22" 3 7/8" > >from point B. (A metric tape measure would be a lot handier) > > > > > > B > > \ > > \ > > \ > > \ > > \ > > / X > > / C > > / > > / > > / > > / > > / > > / > >A > > > >