Develop Reference

Angle offsets and positive/negative angles

I've been stuck with rotating an asset with an offset.
Sstart = e.GetPosition(dial);
if (dial.IsStylusCaptured)
{
AngleRot = Math.Atan2((Y - Sstart.Y) , (X - Sstart.X));
radAngle = AngleRot / Math.PI * 180 + 180;
radAngle = radAngle - AthetaD;
di.RenderTransform = new RotateTransform(radAngle + 90);
}
Using this I was able to rotate my object from 0° with an offset of x to angle theta. But when I make a second rotation, instead of it rotating from angle theta with an offset x it resets the object back to 0°. How can I make it so the offset is always from theta and not 0°?
Here I Rotate with an offset angle
Here my angle resets back to 0° instead of moving from -56°

Let current direction vector is (cx, cy).
In the beginning set it into (1,0) or another needed value.
When you rotated dial, you have new direction
nx = X - Sstart.X
ny = Y - Sstart.Y
To rotate current direction vector to new one, you need to calculate relative angle. This approach uses cross product and dot product of vectors. Perhaps you'll need to change sign of the first expression
Angle = Math.Atan2(nx * cy - ny * cx, cx * nx + cy * ny)
..apply rotation by Angle
After rotation remember current direction to use it later
cx = nx
cy = ny

Finding the true anomaly from state vectors

I'm attempting to convert from state vectors (position and speed) into Kepler elements, however I'm running into problems where a negative velocity or position will give me wrong results when trying to calculate true anomaly.
Here are the different ways I'm trying to calculate the True Anomaly:
/// <summary>
/// https://en.wikipedia.org/wiki/True_anomaly#From_state_vectors
/// </summary>
public static double TrueAnomaly(Vector4 eccentVector, Vector4 position, Vector4 velocity)
{
var dotEccPos = Vector4.Dot(eccentVector, position);
var talen = eccentVector.Length() * position.Length();
talen = dotEccPos / talen;
talen = GMath.Clamp(talen, -1, 1);
var trueAnomoly = Math.Acos(talen);
if (Vector4.Dot(position, velocity) < 0)
trueAnomoly = Math.PI * 2 - trueAnomoly;
return trueAnomoly;
}
//sgp = standard gravitational parameter
public static double TrueAnomaly(double sgp, Vector4 position, Vector4 velocity)
{
var H = Vector4.Cross(position, velocity).Length();
var R = position.Length();
var q = Vector4.Dot(position, velocity); // dot product of r*v
var TAx = H * H / (R * sgp) - 1;
var TAy = H * q / (R * sgp);
var TA = Math.Atan2(TAy, TAx);
return TA;
}
public static double TrueAnomalyFromEccentricAnomaly(double eccentricity, double eccentricAnomaly)
{
var x = Math.Sqrt(1 - Math.Pow(eccentricity, 2)) * Math.Sin(eccentricAnomaly);
var y = Math.Cos(eccentricAnomaly) - eccentricity;
return Math.Atan2(x, y);
}
public static double TrueAnomalyFromEccentricAnomaly2(double eccentricity, double eccentricAnomaly)
{
var x = Math.Cos(eccentricAnomaly) - eccentricity;
var y = 1 - eccentricity * Math.Cos(eccentricAnomaly);
return Math.Acos(x / y);
}
Edit: another way of doing it which Spectre pointed out:
public static double TrueAnomaly(Vector4 position, double loP)
{
return Math.Atan2(position.Y, position.X) - loP;
}
Positions are all relative to the parent body.
These functions all agree if position.x, position.y and velocity.y are all positive.
How do I fix these so that I get a consistent results when position and velocity are negitive?
Just to clarify: My angles appear to be sort of correct, just pointing in the wrong quadrant depending on the position and or velocity vectors.
Yeah so I was wrong, the above all do return the correct values after all.
So I found an edge case where most of the above calculations fail.
Given position and velocity:
pos = new Vector4() { X = -0.208994076275941, Y = 0.955838328099748 };
vel = new Vector4() { X = -2.1678187689294E-07, Y = -7.93096769486992E-08 };
I get some odd results, ie ~ -31.1 degrees, when I think it should return ` 31.1 (non negative). one of them returns ~ 328.8.
However testing with this position and velocity the results apear to be ok:
pos = new Vector4() { X = -0.25, Y = 0.25 };
vel = new Vector4() { X = Distance.KmToAU(-25), Y = Distance.KmToAU(-25) };
See my answer for extra code on how I'm testing and the math I'm using for some of the other variables.
I'm going around in circles on this one. this is a result of a bug in my existing code that shows up under some conditions but not others.
I guess the real question now is WHY am I getting different results with position/velocity above that don't match to my expectations or each other?

Assuming 2D case... I am doing this differently:
compute radius of semi axises and rotation
so you need to remember whole orbit and find 2 most distant points on it that is major axis a. The minor axis b usually is 90 deg from major axis but to be sure just fins 2 perpendicularly most distant points on your orbit to major axis. So now you got both semi axises. The initial rotation is computed from the major axis by atan2.
compute true anomaly E
so if center is x0,y0 (intersection of a,b or center point of both) initial rotation is ang0 (angle of a) and your point on orbit is x,y then:
E = atan2(y-y0,x-x0) - ang0
However in order to match Newton/D'Alembert physics to Kepler orbital parameters you need to boost the integration precision like I did here:
Is it possible to make realistic n-body solar system simulation in matter of size and mass?
see the [Edit3] Improving Newton D'ALembert integration precision even more in there.
For more info and equations see:
Solving Kepler's equation
[Edit1] so you want to compute V I see it like this:
As you got your coordinates relative to parent you can assume they are already in focal point centered so no need for x0,y0 anymore. Of coarse if you want high precision and have more than 2 bodies (focal mass + object + proximity object(s) like moons) then the parent mass will no longer be in focal point of orbit but close to it ... and to remedy you need to use real focal point position so x0,y0 again... So how to do it:
compute center point (cx,cy) and a,b semi axises
so its the same as in previous text.
compute focal point (x0,y0) in orbit axis aligned coordinates
simple:
x0 = cx + sqrt( a^2 + b^2 );
y0 = cy;
initial angle ang0 of a
let xa,ya be the intersection of orbit and major axis a on the side with bigger speeds (near parent object focus). Then:
ang0 = atan2( ya-cy , xa-cx );
and finally the V fore any of yours x,y
V = atan2( y-y0 , x-x0 ) - ang0;

Ok so on further testing it appears my original calcs do all return the correct values, however when I was looking at the outputs I was not taking the LoP into account and basically not recognizing that 180 is essentially the same angle as -180.
(I was also looking at the output in radians and just didn't see what should have been obvious)
Long story short, I have a bug I thought was in this area of the code and got lost in the weeds.
Seems I was wrong above. see OP for edge case.
Here's some code I used to test these,
I used variations of the following inputs:
pos = new Vector4() { X = 0.25, Y = 0.25 };
vel = new Vector4() { X = Distance.KmToAU(-25), Y = Distance.KmToAU(25) };
And tested them with the following
double parentMass = 1.989e30;
double objMass = 2.2e+15;
double sgp = GameConstants.Science.GravitationalConstant * (parentMass + objMass) / 3.347928976e33;
Vector4 ev = OrbitMath.EccentricityVector(sgp, pos, vel);
double e = ev.Length();
double specificOrbitalEnergy = Math.Pow(vel.Length(), 2) * 0.5 - sgp / pos.Length();
double a = -sgp / (2 * specificOrbitalEnergy);
double ae = e * a;
double aop = Math.Atan2(ev.Y, ev.X);
double eccentricAnomaly = OrbitMath.GetEccentricAnomalyFromStateVectors(pos, a, ae, aop);
double aopD = Angle.ToDegrees(aop);
double directAngle = Math.Atan2(pos.Y, pos.X);
var θ1 = OrbitMath.TrueAnomaly(sgp, pos, vel);
var θ2 = OrbitMath.TrueAnomaly(ev, pos, vel);
var θ3 = OrbitMath.TrueAnomalyFromEccentricAnomaly(e, eccentricAnomaly);
var θ4 = OrbitMath.TrueAnomalyFromEccentricAnomaly2(e, eccentricAnomaly);
var θ5 = OrbitMath.TrueAnomaly(pos, aop);
double angleΔ = 0.0000001; //this is the "acceptable" amount of error, really only the TrueAnomalyFromEccentricAnomaly() calcs needed this.
Assert.AreEqual(0, Angle.DifferenceBetweenRadians(directAngle, aop - θ1), angleΔ);
Assert.AreEqual(0, Angle.DifferenceBetweenRadians(directAngle, aop - θ2), angleΔ);
Assert.AreEqual(0, Angle.DifferenceBetweenRadians(directAngle, aop - θ3), angleΔ);
Assert.AreEqual(0, Angle.DifferenceBetweenRadians(directAngle, aop - θ4), angleΔ);
Assert.AreEqual(0, Angle.DifferenceBetweenRadians(directAngle, aop - θ5), angleΔ);
and the following to compare the angles:
public static double DifferenceBetweenRadians(double a1, double a2)
{
return Math.PI - Math.Abs(Math.Abs(a1 - a2) - Math.PI);
}
And eccentricity Vector found thus:
public static Vector4 EccentricityVector(double sgp, Vector4 position, Vector4 velocity)
{
Vector4 angularMomentum = Vector4.Cross(position, velocity);
Vector4 foo1 = Vector4.Cross(velocity, angularMomentum) / sgp;
var foo2 = position / position.Length();
return foo1 - foo2;
}
And EccentricAnomaly:
public static double GetEccentricAnomalyFromStateVectors(Vector4 position, double a, double linierEccentricity, double aop)
{
var x = (position.X * Math.Cos(-aop)) - (position.Y * Math.Sin(-aop));
x = linierEccentricity + x;
double foo = GMath.Clamp(x / a, -1, 1); //because sometimes we were getting a floating point error that resulted in numbers infinatly smaller than -1
return Math.Acos(foo);
}
Thanks to Futurogogist and Spektre for their help.

I am assuming you are working in two dimensions?
Two dimensional vectors of position p and velocity v. The constant K is the the product of the gravitational constant and the mass of the gravity generating body. Calculate the eccentricity vector
eccVector = (dot(v, v)*p - dot(v, p)*v) / K - p / sqrt(dot(p, p));
eccentricity = sqrt(dot(eccVector, eccVector));
eccVector = eccVector / eccentricity;
b = { - eccVector.y, eccVector.x}; //unit vector perpendicular to eccVector
r = sqrt(dot(p, p));
cos_TA = dot(p, eccVector) / r; \\ cosine of true anomaly
sin_TA = dot(p, b) / r; \\ sine of true anomaly
if (sin_TA >= 0) {
trueAnomaly = arccos(cos_TA);
}
else if (sin_TA < 0){
trueAnomaly = 2*pi - arccos(cos_TA);
}

EmguCV - Motion detection not returning angles

I am runnng the motion detection algorithm against a video (file) and following the code sample motion detection, and trying to find the angle of each component and the overall motion. I do get a motion value back, with blobs etc., but the motion direction of each component always is always 0 degrees or 360 degrees and make no sense. What could I be doing wrong? Please help, thanks.
This is the constructor
_motionHistory = new MotionHistory(
10.0, //in second, the duration of motion history you wants to keep
0.05, //in second, parameter for cvCalcMotionGradient
0.5); //in second, parameter for cvCalcMotionGradient
The following is the code for looping through the motion components:
foreach (MCvConnectedComp comp in motionComponents)
{
//reject the components that have small area;
if (comp.area < 1) continue;
// find the angle and motion pixel count of the specific area
double angle, motionPixelCount;
_motionHistory.MotionInfo(comp.rect, out angle, out motionPixelCount);
string motion_direction = GetMotionDescriptor(comp.rect);
Console.writeline (motion_direction);
}
// find and draw the overall motion angle
double overallAngle, overallMotionPixelCount;
_motionHistory.MotionInfo(motionMask.ROI, out overallAngle, out overallMotionPixelCount);
And this where I get my motion descriptor angle
private string GetMotionDescriptor(Rectangle motionRegion)
{
float circleRadius = (motionRegion.Width + motionRegion.Height) >> 2;
Point center = new Point(motionRegion.X + motionRegion.Width >> 1, motionRegion.Y + motionRegion.Height >> 1);
int xDirection = (int)(Math.Cos(angle * (Math.PI / 180.0)) * circleRadius);
int yDirection = (int)(Math.Sin(angle * (Math.PI / 180.0)) * circleRadius);
//double movementAngle = Math.Atan(xDirection / yDirection) * 180 / Math.PI;
Point pointOnCircle = new Point(center.X + xDirection, center.Y - yDirection);
double slope = (double)(pointOnCircle.Y - center.Y)/(double)(pointOnCircle.X - center.X);
double ang = Math.Atan(slope) * 180/Math.PI;
return (ang).ToString() + " degrees";
}

aha! I figured out the reason and posting here if anyone is running into the same problem. The
_motionHistory = new MotionHistory(mhi, maxDelta, minDelta);
Should be adjusted to the frame rate and the motion. The trick lies in the 3 params
(1) motion history to keep, (2) max time delta, (3) min time delta.
They need to be adjusted in some way to reflect the motion you wish to capture.
Hope that helps.

Calculate coordinates of a regular polygon's vertices

I am writing a program in which I need to draw polygons of an arbitrary number of sides, each one being translated by a given formula which changes dynamically. There is some rather interesting mathematics involved but I am stuck on this probelm.
How can I calculate the coordinates of the vertices of a regular polygon (one in which all angles are equal), given only the number of sides, and ideally (but not neccessarily) having the origin at the centre?
For example: a hexagon might have the following points (all are floats):
( 1.5 , 0.5 *Math.Sqrt(3) )
( 0 , 1 *Math.Sqrt(3) )
(-1.5 , 0.5 *Math.Sqrt(3) )
(-1.5 , -0.5 *Math.Sqrt(3) )
( 0 , -1 *Math.Sqrt(3) )
( 1.5 , -0.5 *Math.Sqrt(3) )
My method looks like this:
void InitPolygonVertexCoords(RegularPolygon poly)
and the coordinates need to be added to this (or something similar, like a list):
Point[] _polygonVertexPoints;
I'm interested mainly in the algorithm here but examples in C# would be useful. I don't even know where to start. How should I implement it? Is it even possible?!
Thank you.

for (i = 0; i < n; i++) {
printf("%f %f\n",r * Math.cos(2 * Math.PI * i / n), r * Math.sin(2 * Math.PI * i / n));
}
where r is the radius of the circumsribing circle. Sorry for the wrong language No Habla C#.
Basically the angle between any two vertices is 2 pi / n and all the vertices are at distance r from the origin.
EDIT:
If you want to have the center somewher other than the origin, say at (x,y)
for (i = 0; i < n; i++) {
printf("%f %f\n",x + r * Math.cos(2 * Math.PI * i / n), y + r * Math.sin(2 * Math.PI * i / n));
}

The number of points equals the number of sides.
The angle you need is angle = 2 * pi / numPoints.
Then starting vertically above the origin with the size of the polygon being given by radius:
for (int i = 0; i < numPoints; i++)
{
x = centreX + radius * sin(i * angle);
y = centreY + radius * cos(i * angle);
}
If your centre is the origin then simply ignore the centreX and centreY terms as they'll be 0,0.
Swapping the cos and sin over will point the first point horizontally to the right of the origin.

Sorry, I dont have a full solution at hand right now, but you should try looking for 2D-Rendering of Circles. All classic implementations of circle(x,y,r) use a polygon like you described for drawing (but with 50+ sides).

Say the distance of the vertices to the origin is 1. And say (1, 0) is always a coordinate of the polygon.
Given the number of vertices (say n), the rotation angle required to position the (1, 0) to the next coordinate would be (360/n).
The computation required here is to rotate the coordinates. Here is what it is; Rotation Matrix.
Say theta = 360/n;
[cos(theta) -sin(theta)]
[sin(theta) cos(theta)]
would be your rotation matrix.
If you know linear algebra you already know what i mean. If dont just have a look at Matrix Multiplication

One possible implementation to generate a set of coordinates for regular polygon is to:
Define polygon center, radius and first vertex1. Rotate the vertex n-times2 at an angle of: 360/n.
In this implementation I use a vector to store the generated coordinates and a recursive function to generate them:
void generateRegularPolygon(vector<Point>& v, Point& center, int sidesNumber, int radius){
// converted to radians
double angRads = 2 * PI / double(sidesNumber);
// first vertex
Point initial(center.x, center.y - radius);
rotateCoordinate(v, center, initial, angRads, sidesNumber);
}
where:
void rotateCoordinate(vector<Point>& v, Point& axisOfRotation, Point& initial, double angRads, int numberOfRotations){
// base case: number of transformations < 0
if(numberOfRotations <= 0) return;
else{
// apply rotation to: initial, around pivot point: axisOfRotation
double x = cos(angRads) * (initial.x - axisOfRotation.x) - sin(angRads) * (initial.y - axisOfRotation.y) + axisOfRotation.x;
double y = sin(angRads) * (initial.x - axisOfRotation.x) + cos(angRads) * (initial.y - axisOfRotation.y) + axisOfRotation.y;
// store the result
v.push_back(Point(x, y));
rotateCoordinate(v, axisOfRotation, Point(x,y), angRads, --numberOfRotations);
}
}
Note:
Point is a simple class to wrap the coordinate into single data structure:
class Point{
public:
Point(): x(0), y(0){ }
Point(int xx, int yy): x(xx), y(yy) { }
private:
int x;
int y;
};
1 in terms of (relative to) the center, radius. In my case the first vertex is translated from the centre up horizontally by the radius lenght.
2 n-regular polygon has n vertices.

The simple method is:
Let's take N-gone(number of sides) and length of side L. The angle will be T = 360/N.
Let's say one vertices is located on origin.
* First vertex = (0,0)
* Second vertex = (LcosT,LsinT)
* Third vertex = (LcosT+Lcos2T, LsinT+Lsin2T)
* Fourth vertex = (LcosT+Lcos2T+Lcos3T, LsinT+Lsin2T+Lsin3T)
You can do in for loop

hmm if you test all the versions that are listed here you'll see that the implementation is not good. you can check the distance from the center to each generated point of the polygon with : http://www.movable-type.co.uk/scripts/latlong.html
Now i have searched a lot and i could not find any good implementation for calculating a polyogon using the center and the radius...so i went back to the math book and tried to implement it myself. In the end i came up with this...wich is 100% good:
List<double[]> coordinates = new List<double[]>();
#region create Polygon Coordinates
if (!string.IsNullOrWhiteSpace(bus.Latitude) && !string.IsNullOrWhiteSpace(bus.Longitude) && !string.IsNullOrWhiteSpace(bus.ListingRadius))
{
double lat = DegreeToRadian(Double.Parse(bus.Latitude));
double lon = DegreeToRadian(Double.Parse(bus.Longitude));
double dist = Double.Parse(bus.ListingRadius);
double angle = 36;
for (double i = 0; i <= 360; i += angle)
{
var bearing = DegreeToRadian(i);
var lat2 = Math.Asin(Math.Sin(lat) * Math.Cos(dist / earthRadius) + Math.Cos(lat) * Math.Sin(dist / earthRadius) * Math.Cos(bearing));
var lon2 = lon + Math.Atan2(Math.Sin(bearing) * Math.Sin(dist / earthRadius) * Math.Cos(lat),Math.Cos(dist / earthRadius) - Math.Sin(lat) * Math.Sin(lat2));
coordinates.Add(new double[] { RadianToDegree(lat2), RadianToDegree(lon2) });
}
poly.Coordinates = new[] { coordinates.ToArray() };
}
#endregion
If you test this you'll see that all the points are at the exact distance that you give ( radius ). Also don't forget to declare the earthRadius.
private const double earthRadius = 6371.01;
This calculates the coordinates of a decagon. You see the angle used is 36 degrees. You can split 360 degrees to any number of sides that you want and put the result in the angle variable.
Anyway .. i hope this helps you #rmx!

How do I find the lat/long that is x km north of a given lat/long?

I have some C# code that generates google maps. This codes looks at all the Points I need to plot on the map and then works out the Bounds of a rectangle to include those points. It then passes this bounds to the Google Maps API to set the zoom level appropriately to show all of the points on the map.
This code is working fine however I have a new requirement.
One of the points may have a precision associated with it. If this is the case then I draw a circle around the point with the radius set to the precision value. Again this works fine however my bounds checking is now not doing what I want it to do. I want to have the bounding box include the complete circle.
This requires an algorithm to take a point x and calculate the point y that would be z metres north of x and also z metres south of x.
Does anyone have this algorithm, preferably in C#. I did find a generic algorithm here but I appear to have not implemented this correctly as the answers I am getting are 1000s of km adrift.
This is the Generic example
Lat/lon given radial and distance
A point {lat,lon} is a distance d out on the tc radial from point 1 if:
lat=asin(sin(lat1)*cos(d)+cos(lat1)*sin(d)*cos(tc))
IF (cos(lat)=0)
lon=lon1 // endpoint a pole
ELSE
lon=mod(lon1-asin(sin(tc)*sin(d)/cos(lat))+pi,2*pi)-pi
ENDIF
And this is my C# translation.
// Extend a Point North/South by the specified distance
public static Point ExtendPoint(Point _pt, int _distance, int _bearing )
{
Decimal lat = 0.0;
Decimal lng = 0.0;
lat = Math.Asin(Math.Sin(_pt.Lat) * Math.Cos(_distance) + Math.Cos(_pt.Lat) *
Math.Sin(_distance) * Math.Cos(_bearing));
if (Math.Cos(lat) == 0)
{
lng = _pt.Lng; // endpoint a pole
}
else
{
lng = (
(_pt.Lng - Math.Asin(Math.Sin(_bearing) * Math.Sin(_distance) / Math.Cos(lat))
+ Math.PI) % (2 * Math.PI)) - Math.PI;
}
ret = new Point(lat,lng);
return ret;
}
I am calling this function with a bearing of 0 to calculate the new northerly position and a value of 180 to calculate the new southerly position.
Can anyone either see what I have done wrong or perhaps provide a known working algorithm?

I have a very similar piece of code. It got me very close results when compared to another implementation.
I think the problem with yours is that you are using "distance" as linear distance in meters instead of angular distance in radians.
/// <summary>
/// Calculates the end-point from a given source at a given range (meters) and bearing (degrees).
/// This methods uses simple geometry equations to calculate the end-point.
/// </summary>
/// <param name="source">Point of origin</param>
/// <param name="range">Range in meters</param>
/// <param name="bearing">Bearing in degrees</param>
/// <returns>End-point from the source given the desired range and bearing.</returns>
public static LatLonAlt CalculateDerivedPosition(LatLonAlt source, double range, double bearing)
{
double latA = source.Latitude * UnitConstants.DegreesToRadians;
double lonA = source.Longitude * UnitConstants.DegreesToRadians;
double angularDistance = range / GeospatialConstants.EarthRadius;
double trueCourse = bearing * UnitConstants.DegreesToRadians;
double lat = Math.Asin(
Math.Sin(latA) * Math.Cos(angularDistance) +
Math.Cos(latA) * Math.Sin(angularDistance) * Math.Cos(trueCourse));
double dlon = Math.Atan2(
Math.Sin(trueCourse) * Math.Sin(angularDistance) * Math.Cos(latA),
Math.Cos(angularDistance) - Math.Sin(latA) * Math.Sin(lat));
double lon = ((lonA + dlon + Math.PI) % UnitConstants.TwoPi) - Math.PI;
return new LatLonAlt(
lat * UnitConstants.RadiansToDegrees,
lon * UnitConstants.RadiansToDegrees,
source.Altitude);
}
Where
public const double EarthRadius = 6378137.0; // WGS-84 ellipsoid parameters
and LatLonAlt is in degrees/meters (conversion takes place internally).
Adjust as needed.
I assume you can figure out what the value for UnitConstants.DegreesToRadians is :)

For lazy people, (like me ;) ) a copy-paste solution, Erich Mirabal's version with very minor changes:
using System.Device.Location; // add reference to System.Device.dll
public static class GeoUtils
{
/// <summary>
/// Calculates the end-point from a given source at a given range (meters) and bearing (degrees).
/// This methods uses simple geometry equations to calculate the end-point.
/// </summary>
/// <param name="source">Point of origin</param>
/// <param name="range">Range in meters</param>
/// <param name="bearing">Bearing in degrees</param>
/// <returns>End-point from the source given the desired range and bearing.</returns>
public static GeoCoordinate CalculateDerivedPosition(this GeoCoordinate source, double range, double bearing)
{
var latA = source.Latitude * DegreesToRadians;
var lonA = source.Longitude * DegreesToRadians;
var angularDistance = range / EarthRadius;
var trueCourse = bearing * DegreesToRadians;
var lat = Math.Asin(
Math.Sin(latA) * Math.Cos(angularDistance) +
Math.Cos(latA) * Math.Sin(angularDistance) * Math.Cos(trueCourse));
var dlon = Math.Atan2(
Math.Sin(trueCourse) * Math.Sin(angularDistance) * Math.Cos(latA),
Math.Cos(angularDistance) - Math.Sin(latA) * Math.Sin(lat));
var lon = ((lonA + dlon + Math.PI) % (Math.PI*2)) - Math.PI;
return new GeoCoordinate(
lat * RadiansToDegrees,
lon * RadiansToDegrees,
source.Altitude);
}
private const double DegreesToRadians = Math.PI/180.0;
private const double RadiansToDegrees = 180.0/ Math.PI;
private const double EarthRadius = 6378137.0;
}
Usage:
[TestClass]
public class CalculateDerivedPositionUnitTest
{
[TestMethod]
public void OneDegreeSquareAtEquator()
{
var center = new GeoCoordinate(0, 0);
var radius = 111320;
var southBound = center.CalculateDerivedPosition(radius, -180);
var westBound = center.CalculateDerivedPosition(radius, -90);
var eastBound = center.CalculateDerivedPosition(radius, 90);
var northBound = center.CalculateDerivedPosition(radius, 0);
Console.Write($"leftBottom: {southBound.Latitude} , {westBound.Longitude} rightTop: {northBound.Latitude} , {eastBound.Longitude}");
}
}

I'm not sure if I'm missing something here, but I think the question could be rephrased as, "I have a lat/lon point, and I want to find the point x meters north and x meters south of that point."
If that's the question then you don't need to find a new longitude (which makes things simpler), you just need a new latitude. A degree of latitude is roughly 60 nautical miles long anywhere on Earth, and a nautical mile is 1,852 meters. So, for new latitudes x meters north and south:
north_lat = lat + x / (1852 * 60)
north_lat = min(north_lat, 90)
south_lat = lat - x / (1852 * 60)
south_lat = max(south_lat, -90)
This is not completely accurate because the Earth is not a perfect sphere with exactly 60 nautical miles between each degree of latitude. However, the other answers assume that lines of latitude are equidistant, so I'm assuming you don't care about that. If you're interested in how much error that might introduce, there is a nice table on Wikipedia that shows "Surface distance per 1° change in latitude" for different latitudes at this link:
http://en.wikipedia.org/wiki/Latitude#Degree_length

If you have a given latitude and longitude you can calculate the correct latitude and longitude of an x-km change in latitude like so:
new-lat = ((old-km-north + x-km-change)/40,075) * 360)
^ is the ratio of the ^ times the ratio of the circle
earth the change by 360 to get the total ratio
covers. covered in degrees.
The same can apply to longitude. If you have the total distance plus the change you can calculate the total degrees in a similar fashion.
new-long = ((old-km-east + x-km-change)/40,075) * 360)
^ is the ratio of the ^ times the ratio of the circle
earth the change by 360 to get the total ratio
covers. covered in degrees.
Again, these calculations should work, but I'm running off pure intuition here, but the logic does seem to hold true.
Edit: As pointed out by Skizz 40,075 needs to be adjusted to the circumference of the earth at any given latitude using 2.pi.r.cos(lat) or 40074.cos(lat)

There are problems with the two equations on Ed William's rather awesome site... but I didn't analyze them to see why.
A third equation that I found here seems to give proper results.
Here is the test case in php... the third equation is correct, the first two give wildly incorrect values for longitude.
<?php
$lon1 = -108.553412; $lat1 = 35.467155; $linDistance = .5; $bearing = 170;
$lon1 = deg2rad($lon1); $lat1 = deg2rad($lat1);
$distance = $linDistance/6371; // convert dist to angular distance in radians
$bearing = deg2rad($bearing);
echo "lon1: " . rad2deg($lon1) . " lat1: " . rad2deg($lat1) . "<br>\n";
// doesn't work
$lat2 = asin(sin($lat1) * cos($distance) + cos($lat1) * sin($distance) * cos($bearing) );
$dlon = atan2(sin($bearing) * sin($distance) * cos($lat1), cos($distance) - sin($lat1) * sin($lat2));
$lon2 = (($lon1 - $dlon + M_PI) % (2 * M_PI)) - M_PI; // normalise to -180...+180
echo "lon2: " . rad2deg($lon2) . " lat2: " . rad2deg($lat2) . "<br>\n";
// same results as above
$lat3 = asin( (sin($lat1) * cos($distance)) + (cos($lat1) * sin($distance) * cos($bearing)));
$lon3 = (($lon1 - (asin(sin($bearing) * sin($distance) / cos($lat3))) + M_PI) % (2 * M_PI)) - M_PI;
echo "lon3: " . rad2deg($lon3) . " lat3: " . rad2deg($lat3) . "<br>\n";
// gives correct answer... go figure
$lat4 = asin(sin($lat1) * cos($linDistance/6371) + cos($lat1) * sin($linDistance/6371) * cos($bearing) );
$lon4 = $lon1 + atan2( (sin($bearing) * sin($linDistance/6371) * cos($lat1) ), (cos($linDistance/6371) - sin($lat1) * sin($lat2)));
echo "lon4: " . rad2deg($lon4) . " lat4: " . rad2deg($lat4) . "<br>\n";
?>
Note I recieved by email from the author (Ed Williams) of the first two equations:
From my "implementation notes":
Note on the mod function. This appears to be implemented differently in
different languages, with differing conventions on whether the sign of the
result follows the sign of the divisor or the dividend. (We want the sign
to follow the divisor or be Euclidean. C's fmod and Java's % do not work.)
In this document, Mod(y,x) is the remainder on dividing y by x and always
lies in the range 0 <= mod < x. For instance: mod(2.3,2.)=0.3 and
mod(-2.3,2.)=1.7
If you have a floor function (int in Excel), that returns floor(x)=
"largest integer less than or equal to x" e.g. floor(-2.3)=-3 and
floor(2.3) =2
mod(y,x) = y - x*floor(y/x)
The following should work in the absence of a floor function- regardless of
whether "int" truncates or rounds downward:
mod=y - x * int(y/x)
if ( mod < 0) mod = mod + x
php is like fmod in C and does it "wrong" for my purposes.

It is more accurate if you first reproject it to UTM and then check the distance.
Hope this helps

For people who want a java version Eirch's code
/**
* move latlng point by rang and bearing
*
* #param latLng point
* #param range range in meters
* #param bearing bearing in degrees
* #return new LatLng
*/
public static LatLng moveLatLng(LatLng latLng, double range, double bearing) {
double EarthRadius = 6378137.0;
double DegreesToRadians = Math.PI / 180.0;
double RadiansToDegrees = 180.0 / Math.PI;
final double latA = latLng.latitude * DegreesToRadians;
final double lonA = latLng.longitude * DegreesToRadians;
final double angularDistance = range / EarthRadius;
final double trueCourse = bearing * DegreesToRadians;
final double lat = Math.asin(
Math.sin(latA) * Math.cos(angularDistance) +
Math.cos(latA) * Math.sin(angularDistance) * Math.cos(trueCourse));
final double dlon = Math.atan2(
Math.sin(trueCourse) * Math.sin(angularDistance) * Math.cos(latA),
Math.cos(angularDistance) - Math.sin(latA) * Math.sin(lat));
final double lon = ((lonA + dlon + Math.PI) % (Math.PI * 2)) - Math.PI;
return new LatLng(lat * RadiansToDegrees, lon * RadiansToDegrees);
}