# Archive for category Math

### Equation for the Line of Night on a Map of Earth

Recently a friend of mine expressed an interest in knowing what the equation was for the night/day line on a rectangular map of the earth. For example:

Line of Night

The line of night is generated by a projection of the Sun’s light onto the Earth, which we will consider to be a perfect spheroid with semi-major axis of $R_1$ and semi-minor axis of $R_2$. If viewed from the “side” the line of night will appear to be a line with slope equal to that of the axial tilt (a.k.a obliquity) of earth. This is the inclination angle of Earth’s planetary rotation axis in relation to its orbital plane, usually denoted by $\epsilon$

Earth / Sun System

A typical map of the earth is a projection from spherical coordinates onto a cylindrical plane. This is done by drawing a rectangular image where the latitude ($\lambda$) is mapped to the x-axis and the longitude ($\phi$) is mapped to the y-axis (contrary to the coordinate system used by the digital imaging industry, here we describe the origin as being in the center of the image, with the positive x-axis being to the right, and positive y-axis being up).

Consider then a plane that is perpendicular to the earth’s plane of orbit, and perpendicular to the vector pointing from the center of the earth toward the sun, and that intersects the earth at it’s center. The line of night is the intersection of this plane with the Earth. From this observation we can develop an equation for the line of night in terms of $\phi$ and $\lambda$.

Lattitude and Longitude

Line of Night

If we place a coordinate system on the surface of the earth, centered at the intersection of the line of night with the equator with the positive x-axis corresponding to positive $\lambda$ and positive y-axis corresponding to positive $\phi$ we can describe the line-of-night plane as

$y = s x$

where the slope $s$ is related to the obliquity as

$s = \frac{1}{\tan(\epsilon)}$

where $\epsilon$ is the inclination of the axial tilt of the Earth’s axis at solstice ($\epsilon_0$) scaled according to the position of the Earth in it’s orbit

$\epsilon = \epsilon_0 \cos( \frac{day}{360} )$

The inclination of the earth’s axis at solstice is $\epsilon_0 = 23.5^\circ$

From the parameterization of a spheroid, we know that

$x = R_1 \sin( \lambda ) \cos( \phi )$
$y = R_2 \sin( \phi )$

Combining these with the equation above for the line-of-night plane we see that

$y = \frac{1}{\tan( \epsilon )} x$
$R_2 \sin( \phi ) = \frac{1}{\tan( \epsilon )} R_1 \sin( \lambda ) \cos( \phi )$
$\tan( \phi ) = \frac{1}{\tan( \epsilon )} \frac{R_1}{R_2} \sin( \lambda )$
$\phi = \text{tan}^-1 \left( \frac{1}{\tan( \epsilon )} \frac{R_1}{R_2} \sin( \lambda ) \right)$

Plotting the value of the longitude against latitude over a range of axial inclinations from $\epsilon = 0$ to $\epsilon = 23.5^\circ$, and assuming $R_1 = R_2$ results in the following:

Plot of Resulting Line-of-Night Equation

Which has exactly the shape we were looking for.

### Point on an Ellipsoid of Minimum Distance to Another Point in Space

This is a response to a question posted on the irrlicht forums at here.

PDF file:
Closest point on an ellipsoid to a point in space (to come)

If one wants to know the closest point on an ellipsoid to another point in space, it can be found using constrained nonlinear optimization. It was stated in the forum thread that there is no closed-form solution to this problem, though I find that rather surprising. I will update this post if I can confirm (or confirm an opposition) to that fact.

In any case, given a point $\vec{p}$ in space, where

$\vec{p} = \begin{bmatrix} p_x && p_y && p_z \end{bmatrix}$

we want to find a point $\vec{q}$ that minimizes the euclidean distance to
$\vec{p}$ subject to the contraint that it is on an ellipsoid with the
following functional description

$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$

This corresponds to the minimization of the following cost function (the “cost” is the distance to the target point)

$J(\vec{q}) = \sqrt{ (p_x-q_x)^2 + (p_y-q_y)^2 + (p_z-q_z)^2 }$

subject to the following constraint (which just says that it must be on the ellipsoid)

$f(\vec{q}) = \frac{q_x^2}{a} + \frac{q_y^2}{b} + \frac{q_z^2}{c} - 1 = 0$

Now there are many ways to solve this problem. I will outline two of them here known as steepest descent (or gradient search) and the leGrange method.

# Steepest Descent

First comes steepest decent, because it is the easiest to understand graphically. The goal here is that, starting at some point on the ellipse, we will look at how the value of the cost function changes depending on the direction we move, and choose to move in the direction that most improves the cost. To begin with, let us redefine the problem in terms of two independent variables, and let the third be dependent. In the following formulation, I will choose $q_z$ to be the dependent variable, but you should note that this is not always the best choice. A good choice of dependent variable is to look at the coordinates of the target point and pick the one that is furthest from the origin (to avoid numerically ill-conditioned situations).

Using $q_z$ as the dependent variable we can reformulate the constraint as the following

$q_z = \pm c \sqrt{ 1 - \frac{q_x^2}{a^2} - \frac{q_y^2}{b^2} }$

Given this, we exam the effect that changing each of our independent variables has on our cost function, by calculating the partial derivatives. First though, we note that the minimum of a square root expression will occur at the same point as the minimum of the argument, so we will throw away the square root part and use

$J = (p_x-q_x)^2 + (p_y-q_y)^2 + (p_z-q_z)^2$

for the derivatives instead.

$\frac{ \partial J }{ \partial q_x } = -2(p_x - q_x) + -2(p_z - q_z)\frac{ \partial q_z }{ \partial q_x }$
$\frac{ \partial J }{ \partial q_y } = -2(p_y - q_y) + -2(p_z - q_z)\frac{ \partial q_z }{ \partial q_y }$

And given the equation above for $q_z$ we can determine the following

$\frac{ \partial q_z }{ \partial q_x } = \pm c \frac{1}{2} \sqrt{ 1 - \frac{q_x^2}{a^2} - \frac{q_y^2}{b^2} }^{-\frac{1}{2}} \left( -2 \frac{q_x}{a^2}\right)$

which simplifies to

$\frac{ \partial q_z }{ \partial q_x } = \mp \frac{c}{a^2} \frac{q_x}{q_z}$

and similarly

$\frac{ \partial q_z }{ \partial q_y } = \mp \frac{c}{b^2} \frac{q_y}{q_z}$

Note that the sign of these derivatives is opposite the sign of $q_z$. Therefore we see the following

$\frac{ \partial J }{ \partial q_x } = -2(p_x - q_x) + 2 sign(q_z) (p_z - q_z) \left( \frac{c}{a^2} \frac{q_x}{q_z} \right)$
$\frac{ \partial J }{ \partial q_y } = -2(p_y - q_y) + 2 sign(q_z) (p_z - q_z) \left( \frac{c}{b^2} \frac{q_y}{q_z} \right)$

where $sign(x)$ is the signum function and returns $-1$ if the argument is less than zero ($x < 0[/latex]) or $1$ if the argument is greater than zero ([latex]x > 0$).

Now we solve for the minimum point using the following algorithm

1. start at some $q_x$ and $q_y$
2. calculate $\pm q_z$
3. pick whichever of the two is closer by picking the one with the same sign as $p_z$
4. calculate the cost $J$
5. if $J$ of the current iteration is within some tolerance of $J$ at the previous iteration, stop here
6. calcluate $\frac{\partial J}{\partial q_x}$ and $\frac{\partial J}{\partial q_y}$
7. calculate $\Delta x$ and $\Delta y$ from $J$, $\frac{\partial J}{\partial q_x}$, and $\frac{\partial J}{\partial q_y}$
8. update $q_x$ and $q_y$ using $\Delta x$ and $\Delta y$

Here is a matlab code that demonstrates this algorithm. Note that in this code, for the initial guess I use the intersection of the line segment between the origin of the ellipse and $\vec{p}$ with the surface of the ellipse.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 clc; clear;   % define the ellipsoid parameters a = 3; b = 7; c = 9;   % createa a 100 x 100 point mesh for displaying the ellipsoid x = linspace( -a, a, 50 ); y = linspace( -b, b, 50 );   % calculate the z coordinates [X,Y] = meshgrid(x,y); Zp = c*sqrt(1 - X.*X/(a*a) - Y.*Y/(b*b) ); Zm = -Zp;   % create the point we're going to calculate the closest too px = 3; py = 4; pz = 8;     % create an initial guess by finding the intersection of the ray from the % centroid to point with the ellipsoid surface k = sqrt( 1 / ( (px^2)/(a^2) + (py^2)/(b^2) + (pz^2)/(c^2) ) );   qx = k*px; qy = k*py; qz = k*pz;   % this is the scale factor for our movement along the ellipse, this can % start off quite large because it will be reduced as needed s = 1;     % this is the main loop, you'll want to put some kind of tolerance based % terminal condition, i.e. when the cost function (distance) is only % bouncing back and forth between a tolerance or something more tailored to % your application, I will stop when the distance is +/- 1% i = 1; while 1 % calculate the z for our given x and y qz_plus = c * sqrt( 1 - qx*qx/(a*a) - qy*qy/(b*b) ); qz_minus = - qz_plus;   % we want the one that get's us closest to the target so if( abs(pz - qz_plus) < abs(pz - qz_minus) ) qz = qz_plus; else qz = qz_minus; end       % calculate the current value of the cost function J = sqrt( (px - qx)^2 + (py - qy)^2 + (pz - qz)^2 );   % store the current values for the plots qxplot(i) = qx; qyplot(i) = qy; qzplot(i) = qz; Jplot(i) = J;   % check to see if we overshot the goal or jumped off the ellipsoid if( i > 1 ) % if we jumped off the ellipsoid or overshot the minimal cost if( imag(qz) ~= 0 || J > Jplot(i-1) ); % then go back to the previous position and use a finer % step size qx = qxplot(i-1); qy = qyplot(i-1); qz = qzplot(i-1); J = Jplot(i-1); s = s/10; i = i-1;   % if we did just jump over the actual minimum, let's check to % see how confident we are at this point, if we're with in % 1% there's no need to continue if( i > 3 ) if( abs( (Jplot(i-1) - Jplot(i-3)) / Jplot(i-3) ) < 0.001 ) break; end end end end   % calculate the gradient of the cost with respect to our control % variables; since we divide by qz in the second term we first need % to check whether or not qz is too close to zero; if it is, we know % that the second term should be zero if( qz < 1e-10 ) dJdx = -2*(px - qx); dJdy = -2*(py - qy); else dJdx = -2*(px - qx) + 2*sign(qz)*(pz - qz)*( c/(a*a) * qx/qz ); dJdy = -2*(py - qy) + 2*sign(qz)*(pz - qz)*( c/(b*b) * qy/qz ); end   % calculate the update \vector that we will move along dx = -J/dJdx; dy = -J/dJdy;   % calculate the magnitude of that update \vector magnitude = sqrt( dx*dx + dy*dy );   % normalize our update \vector so that we don't shoot off at an % uncontrolable rate dx = s * (1/magnitude) *dx; dy = s * (1/magnitude) *dy;   % update the current position qx = qx + dx; qy = qy + dy;   % increment the index i = i+1; end;   index = 1:i+1;   % generate a little report message = sprintf( ... ['The closest point was found at [%f, %f, %f]', ... 'with a distance of %f and a confidence of +/- %f %%'], ... qx, qy, qz, J, 100*abs((Jplot(i+1)-Jplot(i-1)) / Jplot(i-1)) ); disp(message);   % now we'll verify that the one we found is the closest Jmap_plus = sqrt( (px - X).*(px - X) + (py - Y).*(py - Y) + (pz - Zp).*(pz - Zp ) ); Jmap_minus = sqrt( (px - X).*(px - X) + (py - Y).*(py - Y) + (pz - Zp).*(pz - Zm ) );   [Jmin_plus_row, imin_plus_row] = min( Jmap_plus ); [Jmin_plus_col, imin_plus_col] = min( Jmin_plus_row );   imin = imin_plus_row(imin_plus_col); jmin = imin_plus_col;   xmin = X(imin, jmin); ymin = Y(imin, jmin); zmin = Zp(imin, jmin); Jmin = Jmin_plus_col; Jmin2 = Jmap_plus(imin,jmin);   message = sprintf( ... ['The closest point by searching the mesh was at [%f, %f, %f]' ... 'and has a distance of %f (%f)'], xmin, ymin, zmin, Jmin, Jmin2 ); disp(message);     % and draw some pretty pictures   close(figure(1)); figure(1); grid on; hold on; mesh(X,Y,real(Zp), 'FaceAlpha', 0.5); mesh(X,Y,real(Zm), 'FaceAlpha', 0.5); plot3(px,py,pz,'ko'); plot3(qxplot,qyplot,qzplot,'ko'); plot3(qxplot,qyplot,qzplot,'k-'); plot3([px qx], [py qy], [pz qz], 'b-'); xlabel('X'); ylabel('Y'); zlabel('Z'); title('steepest decent method'); hold off;   close(figure(2)); figure(2); plot(index, Jplot, 'g-'); xlabel( 'iteration' ); ylabel( 'distance (cost)' );   close(figure(3)); figure(3); mesh(X,Y,real(Jmap_plus), 'FaceAlpha', 0.5); xlabel('X'); ylabel('Y'); zlabel('Z'); title('distance to target');

The output of this script is

 The closest point was found at [1.321810, 3.187960, 6.962411]with a distance of 2.133617 and a confidence of +/- 0.000000 % The closest point by searching the mesh was at [1.285714, 3.285714, 6.948103]and has a distance of 2.134354 (2.134354) 

You can see the work done by the gradient search in the first figure, which is shown below:

Result of Steepest Descent Algorithm

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Detail of Steepest Descent

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I know this doesn’tlook like the closest point, but, as you can see, the script actually searches the mesh for the lowest value point and comes up with essentially the same point. To demonstrate, here is a 3d plot of the distance of each point on the upper half of the ellipse to the target point, with the minimum point labeled. This is the output of the script to figure 3. The reason it doesn’t look right is because of how matlab squashes everything to fit the window.

Distance of Point on Ellipse to Target Point

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And finally here is a plot of the distance calculated at each iteration of the loop. As you can see, it doesn’t take many to find a value that is pretty damn good.

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Steepest decent works pretty well right? Well, not always. The initial guess used in the code above will save your ass, but steepest descent can suffer from problems of local minima. Look what happens when I try an initial guess on the other side of the ellipsoid.

Failure of Steepest Descent Algorithm

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The algorithm get’s caught in a local minimum, and converges to a very precise, very incorrect solution.

# LeGrange Method

This method is a little more elegant, but still ends up with a search algorithm. However, in this case we’re only doing a one dimensional search for root-finding of a polynomial, which is a quite well documented problem.

In this method, we note that, since the constraint must be satisfied everywhere, if we know the minimal cost $J^*$ is achieved at a point $\vec{q}^*$, then we know that

$J^* = \sqrt{ (p_x-q^*_x)^2 + (p_y-q^*_y)^2 + (p_z-q^*_z)^2 }$

and we also know that if the augmented cost function is defined as

$L = J(\vec{q}) + \lambda f(\vec{q})$

which expands to

$L = \sqrt{ (p_x-q_x)^2 + (p_y-q_y)^2 + (p_z-q_z)^2 } + \lambda \left[ \frac{q_x^2}{a^2} + \frac{q_y^2}{b^2} + \frac{q_z^2}{c^2} - 1 \right]$

then $L^* = J^*$ and is also at $\vec{q}^*$ since the latter term must be zero if the constraint is to be satisfied. We can then minimize the cost function as we usually do, by taking the derivative with respect to the independent variables, setting them to zero, and solving the system for the unknowns.

First though, we note that the minimum of a square root expression will occur at the same point as the minimum of the argument, so we will throw away the square root part and solve instead for the minimum of

$L = (p_x-q_x)^2 + (p_y-q_y)^2 + (p_z-q_z)^2 + \lambda \left[ \frac{q_x^2}{a} + \frac{q_y^2}{b} + \frac{q_z^2}{c} - 1 \right]$

We take the partial derivative of the augmented cost with respect to
the four unknowns and see that

$\frac{\partial{L}}{\partial{q_x}} = -2(p_x - q_x) + \frac{2 \lambda q_x}{a^2}$
.
$\frac{\partial{L}}{\partial{q_y}} = -2(p_y - q_y) + \frac{2 \lambda q_y}{b^2}$
.
$\frac{\partial{L}}{\partial{q_z}} = -2(p_z - q_z) + \frac{2 \lambda q_z}{c^2}$
.
$\frac{\partial{L}}{\partial{\lambda}} = \frac{q_x^2}{a^2} + \frac{q_y^2}{b^2} + \frac{q_z^2}{c^2} - 1$

Setting these all to zero we can solve for the following

$q_x = \frac{p_x}{\lambda + \frac{1}{a^2}} = \frac{a^2 p_x}{\lambda + 1} = 0$
.
$q_y = \frac{p_y}{\lambda + \frac{1}{b^2}} = \frac{b^2 p_y}{\lambda + 1} = 0$
.
$q_z = \frac{p_z}{\lambda + \frac{1}{c^2}} = \frac{c^2 p_z}{\lambda + 1} = 0$

Substituting back into the fourth of the differentials we get that

$\frac{\partial{L}}{\partial{\lambda}} = \frac{a^2 p_x^2}{ \left(\lambda + a^2 \right)^2} + \frac{b^2 p_y^2}{ \left(\lambda + b^2 \right)^2} + \frac{c^2 p_z^2}{ \left(\lambda + c^2 \right)^2} - 1 = 0$

which simplifies to

$\frac{\partial{L}}{\partial{\lambda}} = \begin{matrix} a^2 p_x^2 \left( \lambda + b^2 \right)^2 \left( \lambda + c^2 \right)^2 + \ b^2 p_y^2 \left( \lambda + a^2 \right)^2 \left( \lambda + c^2 \right)^2 + \ c^2 p_z^2 \left( \lambda + a^2 \right)^2 \left( \lambda + b^2 \right)^2 - \left( \lambda + a^2 \right)^2 \left( \lambda + b^2 \right)^2 \left( \lambda + c^2 \right)^2 = 0 \end{matrix}$

Note that this is a sixth order polynomial in $\lambda$. Use your favorite zero-finding algorithm (same as a root finding algorithm) to solve for the zeros of this polynomial, (of which there should only be two as there can only be one closest point, and one furthest point on an ellipsoid unless the target point is along one of the axes), and you will find the minimum and the maximum of your cost function. Once the solutions of $\lambda$ are known, plug them into the equations above to solve for $q_x$, $q_y$, and $q_z$. Then plug those into the cost function, and pick the smaller of the two.

For the example problem used in the steepest descent code above, the polynomial looks like this:

Polynomial For Legrange Multiplier

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Here is a detail view:

Polynomial for LeGrange Multiplier Detail

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As you can see, there are two solutions. Below is a matlab code that generates the above graphs, and solves for the minimum point using a binary search. Note that the search assumes that one of the values of \lambda that makes the polynomial zero is to the left of the origin, and the other is to the right. That is true for this example buy may not be in general (I didn’t take the time to work out why that may be the case ) so you may have to modify the search to look for both roots on both sides of the origin.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 clc; clear;   % define the ellipsoid parameters a = 3; b = 7; c = 9;   % createa a 100 x 100 point mesh for displaying the ellipsoid x = linspace( -a, a, 50 ); y = linspace( -b, b, 50 );   % calculate the z coordinates [X,Y] = meshgrid(x,y); Zp = c*sqrt(1 - X.*X/(a*a) - Y.*Y/(b*b) ); Zm = -Zp;   % create the point we're going to calculate the closest too px = 3; py = 4; pz = 8;   \lambda = -160:0.01:20; a2 = a*a; b2 = b*b; c2 = c*c;   px2 = px*px; py2 = py*py; pz2 = pz*pz;   lpa = \lambda + a*a; lpb = \lambda + b*b; lpc = \lambda + c*c;   P = a2*px2*lpb.*lpb.*lpc.*lpc + ... b2*py2*lpa.*lpa.*lpc.*lpc + ... c2*pz2*lpa.*lpa.*lpb.*lpb - ... lpa.*lpa.*lpb.*lpb.*lpc.*lpc;   x = a2*px./lpa; y = b2*py./lpb; z = c2*pz./lpc;   J = sqrt( (px-x).*(px-x) + (py-y).*(py-y) + (pz-z).*(pz-z) );   close(figure(1)); figure(1); plot( \lambda, P, 'b-', 'linewidth', 2 ); xlabel('$\lambda$', 'Interpreter', 'LaTex'); ylabel('$\frac{\partial L}{ \partial \lambda }$', 'Interpreter', 'LaTex'); title('value of the polynomial');   close(figure(2)); figure(2); plot( \lambda, P, 'b-', 'linewidth', 2 ); axis([-160,20,0, 1e10]); xlabel('$\lambda$', 'Interpreter', 'LaTex'); ylabel('$\frac{\partial L}{ \partial \lambda }$', 'Interpreter', 'LaTex'); title('detail of value of the polynomial');     % start search by looking at the value of the polynomial at zero \lambda = 0;   lpa = \lambda + a*a; lpb = \lambda + b*b; lpc = \lambda + c*c;   P = a2*px2*lpb.*lpb.*lpc.*lpc + ... b2*py2*lpa.*lpa.*lpc.*lpc + ... c2*pz2*lpa.*lpa.*lpb.*lpb - ... lpa.*lpa.*lpb.*lpb.*lpc.*lpc;   % we need to store the start sign so that we know what sign we're looking % for in order to determine a zero crossing startSign = sign(P);   % start our search in one direction i = 1;   d\lambda = 1; lhs\lambda = 0; % start the forward search, we're looking for the first value we can % find that has a sign opposite that at the zero point while d\lambda < realmax/10 rhs\lambda = d\lambda; lpa = rhs\lambda + a*a; lpb = rhs\lambda + b*b; lpc = rhs\lambda + c*c;   P = a2*px2*lpb.*lpb.*lpc.*lpc + ... b2*py2*lpa.*lpa.*lpc.*lpc + ... c2*pz2*lpa.*lpa.*lpb.*lpb - ... lpa.*lpa.*lpb.*lpb.*lpc.*lpc;   % if we've discovered a sign change we can stop searching if( sign(P) ~= startSign ) lhs\lambda = d\lambda/10; break;   % if not, we need to grow the increment else d\lambda = d\lambda * 10; end end     % now we can start a bisection search using the lhs and rhs that we've % just determined, which are exactly one order of magnitude apart error = (rhs\lambda - lhs\lambda)/lhs\lambda; while( abs(error) > 1e-2) \lambda = (lhs\lambda + rhs\lambda)/2; lpa = \lambda + a*a; lpb = \lambda + b*b; lpc = \lambda + c*c;   P = a2*px2*lpb.*lpb.*lpc.*lpc + ... b2*py2*lpa.*lpa.*lpc.*lpc + ... c2*pz2*lpa.*lpa.*lpb.*lpb - ... lpa.*lpa.*lpb.*lpb.*lpc.*lpc;   if( sign(P) ~= startSign ) rhs\lambda = \lambda; else lhs\lambda = \lambda; end   error = (lhs\lambda - rhs\lambda)/lhs\lambda; end   % store the found value \lambda1 = (lhs\lambda + rhs\lambda)/2;   % now we search in the other direction; d\lambda = 1; rhs\lambda = 0; % start the forward search, we're looking for the first value we can % find that has a sign opposite that at the zero point while d\lambda < realmax/10 lhs\lambda = -d\lambda; lpa = lhs\lambda + a*a; lpb = lhs\lambda + b*b; lpc = lhs\lambda + c*c;   P = a2*px2*lpb.*lpb.*lpc.*lpc + ... b2*py2*lpa.*lpa.*lpc.*lpc + ... c2*pz2*lpa.*lpa.*lpb.*lpb - ... lpa.*lpa.*lpb.*lpb.*lpc.*lpc;   % if we've discovered a sign change we can stop searching if( sign(P) ~= startSign ) rhs\lambda = -d\lambda/10; break;   % if not, we need to grow the increment else d\lambda = d\lambda * 10; end end   % now we can start a bisection search using the lhs and rhs that we've % just determined, which are exactly one order of magnitude apart error = (rhs\lambda - lhs\lambda)/lhs\lambda; while( abs(error) > 1e-2) \lambda = (lhs\lambda + rhs\lambda)/2; lpa = \lambda + a*a; lpb = \lambda + b*b; lpc = \lambda + c*c;   P = a2*px2*lpb.*lpb.*lpc.*lpc + ... b2*py2*lpa.*lpa.*lpc.*lpc + ... c2*pz2*lpa.*lpa.*lpb.*lpb - ... lpa.*lpa.*lpb.*lpb.*lpc.*lpc;   if( sign(P) ~= startSign ) lhs\lambda = \lambda; else rhs\lambda = \lambda; end   error = (lhs\lambda - rhs\lambda)/lhs\lambda; end   % store the found value \lambda2 = (lhs\lambda + rhs\lambda)/2;   x1 = a2*px/(\lambda1+a2); y1 = b2*py/(\lambda1+b2); z1 = c2*pz/(\lambda1+c2);   x2 = a2*px/(\lambda2+a2); y2 = b2*py/(\lambda2+b2); z2 = c2*pz/(\lambda2+c2);   J1 = sqrt( (px-x1)*(px-x1) + (py-y1)*(py-y1) + (py-y1)*(py-y1) ); J2 = sqrt( (px-x2)*(px-x2) + (py-y2)*(py-y2) + (py-y2)*(py-y2) );   % print a little report message = sprintf( ['found zero crossings at \lambda = %f and %f', ... 'which corresponds to the points [%f, %f, %f]', ... 'and [%f, %f, %f] with costs of %f and %f'], ... \lambda1, \lambda2, x1, y1, z1, x2, y2, z2, J1, J2 ); disp(message);   if( J1 < J2 ) message = sprintf( 'the point of minimum distance is [%f, %f, %f]', ... x1, y1, z1 ); else message = sprintf( 'the point of minimum distance is [%f, %f, %f]', ... x2, y2, z2 ); end   disp(message);

The output of this script is

 found zero crossings at \lambda = 11.801758 and -155.810547which corresponds to the points [1.297967, 3.223591, 6.982626]and [-0.183910, -1.835025, -8.661880] with costs of 2.025472 and 8.844903 the point of minimum distance is [1.297967, 3.223591, 6.982626] 

Which matches with the previous method. Sweet.

As a brief note, if the target point is inside the ellipse and along one of it’s axes, there may be an infinite number of solutions (all points on a ring around the ellipse located at the coordinate of the target point will be the same distance away). If this is a possibility in your application, make sure you watch out for it and decide what the “right” solution is.

I hope this helps. Sorry I didn’t write the code in c++ but it took long enough to generate the results without doing that. Hopefully you can figure out from the matlab code exactly what you need to. Matlab script isn’t too different from C++.

-Cheshirekow