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: <[email protected]>
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
> >
>
>
