Find angle of point on circle

问题

Imagine I have drawn a circle with center coordinates (cx,cy) on the screen and a random point (A) is selected on the circle.

By having the the coordinates of point A, I need to find the angle of (a).

Update:

I have tried using the following formula:

Math.toDegrees(Math.asin(((x - cx) / radius).toDouble()))

which is actually the reverse of (the circle is created by feeding angles to this one):

x = radius * Math.sin(Math.toRadians(angle.toDouble())) + cx
y = radius * Math.cos(Math.toRadians(angle.toDouble())) + cy

But since the y coordinate is not present in the formula the answer could be wrong.

回答1:

If you know the cartesian coordinates of the point A(x,y), then you can find the angle theta by converting it into polar coordinates as below:

double theta = Math.toDegrees(Math.atan2(y - cy, x - cx));

This formula works if your X axis is in 0 degrees, otherwise you need to consider the offset.

回答2:

I think the method you are looking for i Math.atan2 which computes the angle to an x and y cordinate. I have now modified the code to adjust for putting 0 degrees downwards. I have also flipped the y-axis to put 0, 0 cordinate in the upper left corner (screen coordinates), and adjusted degrees above 180 to be reported as negative degrees:

public double theta(int cx, int cy, int x, int y)
{
    double angle = Math.toDegrees(Math.atan2(cy - y, x - cx)) + 90;
    return angle <= 180? angle: angle - 360;
}

A small test to verify some angles...

@Test
public void test()
{
    assertThat(theta(50, 50, 60, 50), is(90.0));
    assertThat(theta(50, 50, 50, 60), is(0.0));
    assertThat(theta(50, 50, 40, 50), is(-90.0));
    assertThat(theta(50, 50, 50, 40), is(180.0));
}

回答3:

You can find the tangent angles and add to 90 or substruct from 270 that angle and find the result, I believe. I design the code like your drawing. You can make it more generic, I guess. You have 4 area:

0 degree to 90 degree
90 degree to 180 degree
90 degree to -90(270) degree
-90(270) degree to 0(360) degree

Code:

public static double findAngle(double x,  double y,
                               double cx, double cy, double radius){
    double beta, alfa;

    double distanceX = Math.abs(Math.abs(x) - Math.abs(cx));
    double distanceY = Math.abs(Math.abs(y) - Math.abs(cy));

    // check the point is on the circle or not
    // with euchlid
    if (radius != Math.sqrt(x * x + y * y)) {
        return -1;
    }

    if (x >= cx && y <= cy) {
        // find tangent
        beta = Math.atan(distanceY / distanceX);
        alfa = 90 - beta;
        return alfa;
    }
    // 90-180 -> second area
    else if (x >= cx && y >= cy) {
        beta = Math.atan(distanceY / distanceX);
        alfa = 90 + beta;
        return alfa;
    }
    // 180 - -90 -> third area
    else if (x <= cx && y >= cy) {
        beta = Math.atan(distanceY / distanceX);
        alfa = 270 - beta;
        return alfa;
    }
    // -90 - 0 -> forth area
    else if (x <= cx && y <= cy) {
        beta = Math.atan(distanceY / distanceX);
        alfa = 270 + beta;
        if (alfa == 360) {
            alfa = 0;
        }
        return alfa;    
    } 
    else {
        return -1;
    }
}

回答4:

The main catch with Atan2 is the center point. If your center is offset (Cx, Cy) and not at the origin and you want to find the arc angle between (X1, Y1) and (X2, Y2) then I would suggest this approach:

double startAngle = Math.Atan2(Y1-Cy, X1-Cx);
double endAngle = Math.Atan2(Y2-Cy, X2-Cx);
double ArcAngle = endangle - startAngle;

来源：https://stackoverflow.com/questions/46387747/find-angle-of-point-on-circle

标签

java

android

math

trigonometry