distance from given point to given ellipse

Question

I have an ellipse, defined by Center Point, radiusX and radiusY, and I have a Point. I want to find the point on the ellipse that is closest to the given point. In the illustration below, that would be S1.

Now I already have code, but there is a logical error somewhere in it, and I seem to be unable to find it. I broke the problem down to the following code example:

#include <vector>
#include <opencv2/core/core.hpp>
#include <opencv2/highgui/highgui.hpp>
#include <math.h>

using namespace std;

void dostuff();

int main()
{
    dostuff();
    return 0;
}

typedef std::vector<cv::Point> vectorOfCvPoints;

void dostuff()
{

    const double ellipseCenterX = 250;
    const double ellipseCenterY = 250;
    const double ellipseRadiusX = 150;
    const double ellipseRadiusY = 100;

    vectorOfCvPoints datapoints;

    for (int i = 0; i < 360; i+=5)
    {
        double angle = i / 180.0 * CV_PI;
        double x = ellipseRadiusX * cos(angle);
        double y = ellipseRadiusY * sin(angle);
        x *= 1.4;
        y *= 1.4;
        x += ellipseCenterX;
        y += ellipseCenterY;
        datapoints.push_back(cv::Point(x,y));
    }

    cv::Mat drawing = cv::Mat::zeros( 500, 500, CV_8UC1 );

    for (int i = 0; i < datapoints.size(); i++)
    {
        const cv::Point & curPoint = datapoints[i];
        const double curPointX = curPoint.x;
        const double curPointY = curPoint.y * -1; //transform from image coordinates to geometric coordinates

        double angleToEllipseCenter = atan2(curPointY - ellipseCenterY * -1, curPointX - ellipseCenterX); //ellipseCenterY * -1 for transformation to geometric coords (from image coords)

        double nearestEllipseX = ellipseCenterX + ellipseRadiusX * cos(angleToEllipseCenter);
        double nearestEllipseY = ellipseCenterY * -1 + ellipseRadiusY * sin(angleToEllipseCenter); //ellipseCenterY * -1 for transformation to geometric coords (from image coords)


        cv::Point center(ellipseCenterX, ellipseCenterY);
        cv::Size axes(ellipseRadiusX, ellipseRadiusY);
        cv::ellipse(drawing, center, axes, 0, 0, 360, cv::Scalar(255));
        cv::line(drawing, curPoint, cv::Point(nearestEllipseX,nearestEllipseY*-1), cv::Scalar(180));

    }
    cv::namedWindow( "ellipse", CV_WINDOW_AUTOSIZE );
    cv::imshow( "ellipse", drawing );
    cv::waitKey(0);
}

It produces the following image:

You can see that it actually finds "near" points on the ellipse, but it are not the "nearest" points. What I intentionally want is this: (excuse my poor drawing)

would you extent the lines in the last image, they would cross the center of the ellipse, but this is not the case for the lines in the previous image.
I hope you get the picture. Can anyone tell me what I am doing wrong?

Answer 1

Here's a more concise, numpy-vectorized version of user364952's answer, in Python:

def find_closest_points_on_ellipse(
    a: float, 
    b: float, 
    p: tuple[np.ndarray, np.ndarray],
    t=np.sqrt(2)/2, unit_vector=lambda v: v / np.hypot(*v)
) -> np.ndarray:
    p_abs = np.abs(p)
    axes = np.reshape((a, b), (2,)+(1,)*(len(p_abs.shape) - 1))
    factor = np.reshape((a - b*b/a, b - a*a/b), axes.shape)
    
    for _ in range(0, 3):
        e = t*t*t*factor
        t = unit_vector(((np.hypot(*(t*axes - e)) * unit_vector(p_abs - e) + e) / axes).clip(0, 1))
        
    return np.copysign(t * axes, p)

Inputs:

a: the radius at the x-axis (a.k.a. semi-major axis)

b: the radius at the y-axis (a.k.a. semi-minor axis)

p: a tuple of x-coordinates, y-coordinates, both of which must have the same, but arbitrary, shape S. (Or any array-like object with shape (2,) + S.)

Returns:

A numpy array of shape (2,) + S containing the closest point on the ellipse to each of the input points p.

Answer 2

Consider a bounding circle around the given point (c, d), which passes through the nearest point on the ellipse. From the diagram it is clear that the closest point is such that a line drawn from it to the given point must be perpendicular to the shared tangent of the ellipse and circle. Any other points would be outside the circle and so must be further away from the given point.

So the point you are looking for is not the intersection between the line and the ellipse, but the point (x, y) in the diagram.

Gradient of tangent:

Gradient of line:

Condition for perpedicular lines - product of gradients = -1:

When rearranged and substituted into the equation of your ellipse...

...this will give two nasty quartic (4th-degree polynomial) equations in terms of either x or y. AFAIK there are no general analytical (exact algebraic) methods to solve them. You could try an iterative method - look up the Newton-Raphson iterative root-finding algorithm.

Take a look at this very good paper on the subject: https://web.archive.org/web/20170829172516/https://www.spaceroots.org/documents/distance/distance-to-ellipse.pdf

Sorry for the incomplete answer - I totally blame the laws of mathematics and nature...

EDIT: oops, i seem to have a and b the wrong way round in the diagram xD

Answer 3

Here is the code translated to C# implemented from this paper to solve for the ellipse: http://www.geometrictools.com/Documentation/DistancePointEllipseEllipsoid.pdf

Note that this code is untested - if you find any errors let me know.

//Pseudocode for robustly computing the closest ellipse point and distance to a query point. It
//is required that e0 >= e1 > 0, y0 >= 0, and y1 >= 0.
//e0,e1 = ellipse dimension 0 and 1, where 0 is greater and both are positive.
//y0,y1 = initial point on ellipse axis (center of ellipse is 0,0)
//x0,x1 = intersection point

double GetRoot(double r0, double z0, double z1, double g) {
    double n0 = r0 * z0;
    double s0 = z1 - 1;
    double s1 = (g < 0 ? 0 : Math.Sqrt(n0 * n0 + z1 * z1) - 1);
    double s = 0;
    for (int i = 0; i < maxIter; ++i) {
        s = (s0 + s1) / 2;
        if (s == s0 || s == s1) {
            break;
        }
        double ratio0 = n0 / (s + r0);
        double ratio1 = z1 / (s + 1);
        g = ratio0 * ratio0 + ratio1 * ratio1 - 1;
        if (g > 0) {
            s0 = s;
        } else if (g < 0) {
            s1 = s;
        } else {
            break;
        }
    }
    return s;
}

double DistancePointEllipse(double e0, double e1, double y0, double y1, out double x0, out double x1) {
    double distance;
    if (y1 > 0) {
        if (y0 > 0) {
            double z0 = y0 / e0;
            double z1 = y1 / e1;
            double g = z0 * z0 + z1 * z1 - 1;
            if (g != 0) {
                double r0 = (e0 / e1) * (e0 / e1);
                double sbar = GetRoot(r0, z0, z1, g);
                x0 = r0 * y0 / (sbar + r0);
                x1 = y1 / (sbar + 1);
                distance = Math.Sqrt((x0 - y0) * (x0 - y0) + (x1 - y1) * (x1 - y1));
            } else {
                x0 = y0;
                x1 = y1;
                distance = 0;
            }
        } else // y0 == 0
            x0 = 0;
        x1 = e1;
        distance = Math.Abs(y1 - e1);
    } else { // y1 == 0
        double numer0 = e0 * y0, denom0 = e0 * e0 - e1 * e1;
        if (numer0 < denom0) {
            double xde0 = numer0 / denom0;
            x0 = e0 * xde0;
            x1 = e1 * Math.Sqrt(1 - xde0 * xde0);
            distance = Math.Sqrt((x0 - y0) * (x0 - y0) + x1 * x1);
        } else {
            x0 = e0;
            x1 = 0;
            distance = Math.Abs(y0 - e0);
        }
    }
    return distance;
}

Answer 4

The fastest method I guess is http://wwwf.imperial.ac.uk/~rn/distance2ellipse.pdf Which has been mentioned also by Matt but as he found out the method doesn't work very well inside of ellipse.

The problem is the theta initialization. I proposed an stable initialization:

1. Find the intersection of ellipse and horizontal line passing the point.
2. Find the other intersection using vertical line.
3. Choose the one that is closer the point.
4. Calculate the initial angle based on that point.

I got good results with no issue inside and outside:

As you can see in the following image it just iterated about 3 times to reach 1e-8. Close to axis it is 1 iteration.

The C++ code is here:

double initialAngle(double a, double b, double x, double y) {
    auto abs_x = fabs(x);
    auto abs_y = fabs(y);
    bool isOutside = false;
    if (abs_x > a || abs_y > b) isOutside = true;

    double xd, yd;
    if (!isOutside) {
        xd = sqrt((1.0 - y * y / (b * b)) * (a * a));
        if (abs_x > xd)
            isOutside = true;
        else {
            yd = sqrt((1.0 - x * x / (a * a)) * (b * b));
            if (abs_y > yd)
                isOutside = true;
        }
    }
    double t;
    if (isOutside)
        t = atan2(a * y, b * x); //The point is outside of ellipse
    else {
        //The point is inside
        if (xd < yd) {
            if (x < 0) xd = -xd;
            t = atan2(y, xd);
        }
        else {
            if (y < 0) yd = -yd;
            t = atan2(yd, x);
        }
    }
    return t;
}


double distanceToElipse(double a, double b, double x, double y, int maxIter = 10, double maxError = 1e-5) {
    //std::cout <<"p="<< x << "," << y << std::endl;

    
    auto a2mb2 = a * a - b * b;
    double t = initialAngle(a, b, x, y);
    auto ct = cos(t);
    auto st = sin(t);

    int i;
    double err;
    for (i = 0; i < maxIter; i++) {
        auto f = a2mb2 * ct * st - x * a * st + y * b * ct;
        auto fp = a2mb2 * (ct * ct - st * st) - x * a * ct - y * b * st;
        auto t2 = t - f / fp;
        err = fabs(t2 - t);
        //std::cout << i + 1 << " " << err << std::endl;
        t = t2;
        ct = cos(t);
        st = sin(t);
        if (err < maxError) break;
    }
    auto dx = a * ct - x;
    auto dy = b * st - y;
    //std::cout << a * ct << "," << b * st << std::endl;
    return   sqrt(dx * dx + dy * dy);
}

Answer 5

As propposed by user3235832 you shall solve quartic equation to find the normal to the ellipse (https://www.mathpages.com/home/kmath505/kmath505.htm). With good initial value only few iterations are needed (I use it myself). As an initial value I use S1 from your picture.

Answer 6

There is a relatively simple numerical method with better convergence than Newtons Method. I have a blog post about why it works http://wet-robots.ghost.io/simple-method-for-distance-to-ellipse/

This implementation works without any trig functions:

def solve(semi_major, semi_minor, p):  
    px = abs(p[0])
    py = abs(p[1])

    tx = 0.707
    ty = 0.707

    a = semi_major
    b = semi_minor

    for x in range(0, 3):
        x = a * tx
        y = b * ty

        ex = (a*a - b*b) * tx**3 / a
        ey = (b*b - a*a) * ty**3 / b

        rx = x - ex
        ry = y - ey

        qx = px - ex
        qy = py - ey

        r = math.hypot(ry, rx)
        q = math.hypot(qy, qx)

        tx = min(1, max(0, (qx * r / q + ex) / a))
        ty = min(1, max(0, (qy * r / q + ey) / b))
        t = math.hypot(ty, tx)
        tx /= t 
        ty /= t 

    return (math.copysign(a * tx, p[0]), math.copysign(b * ty, p[1]))

Credit to Adrian Stephens for the Trig-Free Optimization.

Answer 7

Given an ellipse E in parametric form and a point P

the square of the distance between P and E(t) is

The minimum must satisfy

Using the trigonometric identities

and substituting

yields the following quartic equation:

Here's an example C function that solves the quartic directly and computes sin(t) and cos(t) for the nearest point on the ellipse:

void nearest(double a, double b, double x, double y, double *ecos_ret, double *esin_ret) {
    double ax = fabs(a*x);
    double by = fabs(b*y);
    double r  = b*b - a*a;
    double c, d;
    int switched = 0;

    if (ax <= by) {
        if (by == 0) {
            if (r >= 0) { *ecos_ret = 1; *esin_ret = 0; }
            else        { *ecos_ret = 0; *esin_ret = 1; }
            return;
        }
        c = (ax - r) / by;
        d = (ax + r) / by;
    } else {
        c = (by + r) / ax;
        d = (by - r) / ax;
        switched = 1;
    }

    double cc = c*c;
    double D0 = 12*(c*d + 1);      // *-4
    double D1 = 54*(d*d - cc);     // *4
    double D  = D1*D1 + D0*D0*D0;  // *16

    double St;
    if (D < 0) {
        double t = sqrt(-D0);             // *2
        double phi = acos(D1 / (t*t*t));
        St = 2*t*cos((1.0/3)*phi);        // *2
    } else {
        double Q = cbrt(D1 + sqrt(D));    // *2
        St = Q - D0 / Q;                  // *2
    }

    double p    = 3*cc;                          // *-2
    double SS   = (1.0/3)*(p + St);              // *4
    double S    = sqrt(SS);                      // *2
    double q    = 2*cc*c + 4*d;                  // *2
    double l    = sqrt(p - SS + q / S) - S - c;  // *2
    double ll   = l*l;                           // *4
    double ll4  = ll + 4;                        // *4
    double esin = (4*l)    / ll4;
    double ecos = (4 - ll) / ll4;

    if (switched) {
        double t = esin;
        esin = ecos;
        ecos = t;
    }

    *ecos_ret = copysign(ecos, a*x);
    *esin_ret = copysign(esin, b*y);
}

Try it online!

Answer 8

The following python code implements the equations described at "Distance from a Point to an Ellipse" and uses newton's method to find the roots and from that the closest point on the ellipse to the point.

Unfortunately, as can be seen from the example, it seems to only be accurate outside the ellipse. Within the ellipse weird things happen.

from math import sin, cos, atan2, pi, fabs


def ellipe_tan_dot(rx, ry, px, py, theta):
    '''Dot product of the equation of the line formed by the point
    with another point on the ellipse's boundary and the tangent of the ellipse
    at that point on the boundary.
    '''
    return ((rx ** 2 - ry ** 2) * cos(theta) * sin(theta) -
            px * rx * sin(theta) + py * ry * cos(theta))


def ellipe_tan_dot_derivative(rx, ry, px, py, theta):
    '''The derivative of ellipe_tan_dot.
    '''
    return ((rx ** 2 - ry ** 2) * (cos(theta) ** 2 - sin(theta) ** 2) -
            px * rx * cos(theta) - py * ry * sin(theta))


def estimate_distance(x, y, rx, ry, x0=0, y0=0, angle=0, error=1e-5):
    '''Given a point (x, y), and an ellipse with major - minor axis (rx, ry),
    its center at (x0, y0), and with a counter clockwise rotation of
    `angle` degrees, will return the distance between the ellipse and the
    closest point on the ellipses boundary.
    '''
    x -= x0
    y -= y0
    if angle:
        # rotate the points onto an ellipse whose rx, and ry lay on the x, y
        # axis
        angle = -pi / 180. * angle
        x, y = x * cos(angle) - y * sin(angle), x * sin(angle) + y * cos(angle)

    theta = atan2(rx * y, ry * x)
    while fabs(ellipe_tan_dot(rx, ry, x, y, theta)) > error:
        theta -= ellipe_tan_dot(
            rx, ry, x, y, theta) / \
            ellipe_tan_dot_derivative(rx, ry, x, y, theta)

    px, py = rx * cos(theta), ry * sin(theta)
    return ((x - px) ** 2 + (y - py) ** 2) ** .5

Here's an example:

rx, ry = 12, 35  # major, minor ellipse axis
x0 = y0 = 50  # center point of the ellipse
angle = 45  # ellipse's rotation counter clockwise
sx, sy = s = 100, 100  # size of the canvas background

dist = np.zeros(s)
for x in range(sx):
    for y in range(sy):
        dist[x, y] = estimate_distance(x, y, rx, ry, x0, y0, angle)

plt.imshow(dist.T, extent=(0, sx, 0, sy), origin="lower")
plt.colorbar()
ax = plt.gca()
ellipse = Ellipse(xy=(x0, y0), width=2 * rx, height=2 * ry, angle=angle,
                  edgecolor='r', fc='None', linestyle='dashed')
ax.add_patch(ellipse)
plt.show()

Which generates an ellipse and the distance from the boundary of the ellipse as a heat map. As can be seen, at the boundary the distance is zero (deep blue).

Answer 9

You just need to calculate the intersection of the line [P1,P0] to your elipse which is S1.

If the line equeation is:

and the elipse equesion is:

than the values of S1 will be:

Now you just need to calculate the distance between S1 to P1 , the formula (for A,B points) is: