intersection of an arc and a line?

Question

An arc with center of Xc, Yc and Radius of R, it begins at angle A1 and ends at angle A2, where angel 0 is the horizon toward the right.

A line passes at point Xa, Ya with angle Aa.

what is the coordinates of the intersection (if it exisit) and what is the angle of the tangent to arc at that intersection. thanks

Answer 1

To find intersection, let say intersection point is (x,y)

1) The intersection line passes at point Xa, Ya with angle Aa.
So, (Ya-y)/(Xa-x) = tan Aa.

2) The intersection arc with center of Xc, Yc and radius of R.
So, (y-Yc)^2 + (x-Xc)^2 = R^2

3) The intersection arc begins at angle A1 and ends at angle A2 where angle 0 is horizon towards the right.
So, A1< ArcSin (y-Yc)/R < A2

From (1),
(Ya-y)/(Xa-x) = tan Aa
Ya-y = (Xa-x)tanAa
y= Ya-(Xa-x)tanAa

Insert y into (2)
(y-Yc)^2 + (x-Xc)^2 = R^2
(Ya-(Xa-x)tanAa-Yc)^2 + (x-Xc)^2 = R^2
((Ya-Yc)-(Xa-x)tanAa)^2 + (x-Xc)^2 = R^2
If you know actual value for Ya, Xa, Aa, Xc, Yc, R,
then you can get intersection x
By insert value of x, you can get y from (1)

Finally, check A1< ArcSin (y-Yc)/R < A2 is valid, otherwise, there is no intersection!

To find angle of tangent,
let say angle C is angle of intersection point from horizon.
C = ArcSin (y-Yc)/R
The angle of the tangent to arc at intersection point should be
C + 90 degree!!!

Answer 2

I can set up the equations but don't yet know whether I can solve them.

The line is Y - Ya = (tan Aa)*(X - Xa)

The arc is part of the circle (X - Xc)^2 + (Y -Yc)^2 = R^2, however I'm thinking it might be better to use the parametric form
X = Xc + R cos A
Y = Yc + R sin A where A1 < A < A2

Sub this in the equation of the line, giving
Yc + R sin A - Ya = (Xc + R cos A - Xa)*tan Aa

The unknown to be solved here is A. We can rearrange the equation into the form
P cos A + Q sin A = K, where P, Q and K are constants. It has solutions for A if P^2 + Q^2 < K^2, but then A has to be between A1 and A2. I think the solution is

A = arcos(P/sqrt(P^2 + Q^2)) + arcos(K/sqrt(P^2 + Q^2)), obtained using the auxiliary angle method.

If the intersection exists, then the angle you want is 90 deg + A. It's a bit sloppy to say that, because the expression above that I've written for A is, strictly speaking, in radians.

email me [e-mail] if you want to discuss it further. I'll keep trying it but am leaving Yahoo answers now.

Answer 3

The equation of the line is,
Y = tan Aa (X - Xa) + Ya

The equation of the arc is,
(X - Xc)^2 + (Y - Yc)^2 = R^2

Assume the line does intersect with the arc. Solving the above two equations for (X, Y) gives the coordinates of the intersection. Can you finish this part?

Let At be the angle of the tangent to the arc at the intersection.

At = - arctan [(Xc - X)/(Yc - Y)]
(Why? I got it directly based on the slope of its perpendicular line.)

The angle between the line and the tangent to arc at that intersection is |Aa-At|.

Answer 4

sounds like a perfect strait line to me. then again i dont have a clue. good luck. you may want to email norte dame about this one.