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| Trig. is your friend... http://www.1728.com/circpart.htm ![]() whenever 2 chords of a circle intersect at a point E, then AE EB = CE ED if we use the diameter of the circle through the center as one chord then let H be the hatch distance, and D the diameter, while AE=EB AE EB = CE ED AE AE = [D+H] [D-H] AE = SQAREROOT( [D+H] [D-H] ) this will get you started Code: (This code was tested in Mach3) #1=10 (diameter) #2=.5 (hatch) #3=FIX[#1/#2] #4=0 m98 p1 l#3 m30 o1 #5=[sqrt[#4*[#1-#4]]/2] g0 x[#4*#2] y#5 g1 y[-1*#5] #4=[#4+#2] m99 |
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