找到一个圆的中心（x和y位置），只有2个随机点和凸起

Question

im trying to find the center of a circle. the only information I have is: Two random points in the circle and the circle bulge. So far i've manage to calculate the radius of the circle (at least i think i did). Ill post bellow the equasions ive used so far.

these are just random values and will change on user input)

PointA(x = 10, y = 15) PointB(x = 6, y = 12)

circle_bulge = 0.41

distance = PointB - PointA

radius = (distance / 4) * (circle_bulge + (1 / circle_bulge ))

if this math is incorrect, please let me know, but keep in mind that i need to find the X and Y coordinates of the center of the circle

Answer 1

Here is a picture of the problem:

By definition the bulge is b = tg( Alpha /4)

From the trigonometric formula: tg(2 angle ) = 2tg( angle )/(1-tg ² ( angle ))

applied to angle = Alpha /4 and using the definition of bulge:

tg( Alpha /2) = 2 b /(1- b ² )

On the other hand

tg( Alpha /2) = s / d

Then

s / d = 2 b /(1- b ² ) and

d = s (1- b ² )/(2 b )

which allows us to calculate d because b is known and s = || B - A ||/2, where || B - A || denotes the norm of the vector B - A .

Now, let's calculate

( u , v ) = ( B - A )/|| B - A ||

Then ||( u , v )|| = 1, ( v ,- u ) is orthogonal to B - A , and we have

C = ( v ,- u ) d + ( A + B )/2

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UPDATE

Pseudo code to compute the center

Inputs:

A = (a1, a2), B = (b1, b2) "two points"; b "bulge"

Calculations:

"lengths"
norm := sqrt(square(b1-a1) + square(b2-a2)).
s := norm/2.
d := s * (1-square(b))/(2*b)

"direction"
u := (b1-a1)/ norm.
v := (b2-a2)/ norm.

"center"
c1 := -v*d + (a1+b1)/2.
c2 := u*d + (a2+b2)/2.
Return C := (c1, c2)

Note: There are two solutions for the Center, the other one being

c1 := v*d + (a1+b1)/2.
c2 := -u*d + (a2+b2)/2.
Return C := (c1, c2)

Answer 2

I believe that you have all the required math in this reference: http://autocad.wikia.com/wiki/Arc_relationships

The center of the circunference has to be in the straight line that passes trough the middle point of the segment with perpendicular vector AB. You know the angle of the triangle formed by the three points because you know the bulge so you have to solve a simple equation. Try it out, if you can not I will try to help you.

Ignacio is right pointing out that there will be two solutions.

EDIT

The middle point is given by:

M = (A + B) / 2

The AB vector is gien by:

AB = A-B

The center of the circle C with coordinates (X, Y) has to be in the line given by:

((X, Y) - M) * AB = 0 //where * is the scalar vector product

The isosceles triangle generated by the points A, B and C has an angle opposite of the AB segment:

Angle = 4 arctan(bulge)

now we can compute the distance from the center C to A that we call d1 and half the distance between A and B that we call d2 we know then that

sin (Angle/2) = d2/d1

This will give you the second equation for X and Y that is quadratic (it has two solutions).

Excuse the notation but I do not know how to insert math here :-)