Co-Ord Arc
Richard
Would anybody help me with this one, I'm stuck here!
I have two co-ord positions for an ARC, the start the end and
I also have the radius of the ARC.
My question is - Is there a formula to work out the position of the
centre point.
example :
ARC START
X = 86.27mm Y = 73.94mm
X = 92.84mm Y = 60.14mm
With a radius of 10mm
What is the centre?
TIA to anyone who can help.
Best Regards
Richard
Virgil
<ric...@softsite.co.uk> wrote:
Given an arc with ends at P and Q and radius r, the center, C, will lie
on the perpendicular bisector of segment PQ and the distances |PC| and
|QC| will equal r. This will give 2 possible positions for C, one on
each side of the line PQ.
--
Virgil
vm...@frii.com
David Fleming
As Virgil noted, two possible centers exist. The correct one will be
obviously the one on the concave side of the arc.
co-ord (x,y) = (a,b) and (c,d) = (86.27,73.94) and (92.84,60.14)
radius = 10
MidpointOfChord = ((a+c)/2,(b+d)/2) = (89.555,67.04) = (e,f)
ChordSlope = (b-d)/(a-c)
BisectorSlope = -1/ChordSlope = (c-a)/(b-d) = .4761
BisectorYIntercept = f-e(BisectorSlope) = 24.40
Equation of Bisector: y = .4761x + 24.40
Centers must lie on Bisector and circles of radius 10 centered at your
given points. Pick a circle and solve. two equations in two unknowns.
Circle: (x-a)^2 + (y-b)^2 = radius^2
(x-86.27)^2 + (y-73.94)^2 = 10^2 = 100
substitute for y: (x-86.27)^2 + (.4761x + 24.40 - 73.94)^2 = 100
expand & simplify: 1.22667x^2 - 219.712x + 9796.72 = 0
simple quadratic. x = 95.38 and 83.73
solve in Bisector. y = 69.81 and 64.26
Two possible centers: (x,y) = (95.38,69.81) and (83.73,64.26)
Q.E.D.
Regards - Dave Fleming