Develop Reference

Lines and shapes drawn at incorrect angles

I have a c# program where I need to draw some simple 2D objects on the canvas.
One of these involves drawing a rectangle and lines where I know the start point, the length and I have to calculate the end position. So I have the following code;
private void CalculateEndPoint()
{
double angle = Helper.deg2rad((double)this.StartAngle);
int x = this.StartPoint.X + (int)(Math.Cos(angle) * this.Length * -1);
int y = this.StartPoint.Y + (int)(Math.Sin(angle) * this.Length);
this.EndPoint = new Point(x, y);
}
Now this seems to work OK to calculate the end points. The issue I have is with the angle (this.StartAngle), the value I specify seems not to be how it is drawn and I seem to have the following;
Where as I'm expecting 0 at the top, 90 on the right, 180 at the bottom etc.
So to get a shape to draw straight down the canvas I have to specify 90 degrees, where as I would expect to specify 180.
Have I done something wrong? Or is it just a lack of understanding?

You should change your CalculateEndPoint function to have that:
private static void CalculateEndPoint(double dec)
{
double angle = (Math.PI / 180) * (this.StartAngle + 90); // add PI / 2
int x = StartPoint.X + (int)(Math.Cos(angle) * Length * -1);
double angle2 = (Math.PI / 180) * (this.StartAngle - 90); // minus PI / 2
int y = StartPoint.Y + (int)(Math.Sin(angle2) * Length);
EndPoint = new Point(x, y);
}

Actually, 0 should be on the right. You are multiplying the x-coordinate by -1, so you're moving it to the left.
Just remember these 2 rules:
- The cosine of the angle is the x-coordinate of the unit circle.
- The sine of the angle is the y-coordinate of the unit circle.
Since cos(0) = 1 and sin(0) = 0, the coordinate corresponding to angle 0 is (1, 0).
Whether 90 is on top or on the bottom depends on the canvas.
Some applications/frameworks consider y-coordinate 0 to be at the top of the canvas. That means you go clockwise around the circle and 90 will be at the bottom.
If y-coordinate 0 is at the bottom of the canvas, you go counter-clockwise and 90 will be at the top.

how to find edge of an image with most distance from center?

i writing a sample program in c# who drawing some point in page
i set center point with calculate distance of point
but how can found most distance point from center point ?
sample code :
void draw(string label,float x,float y)
{
Graphics g = panel1.CreateGraphics();
Pen p = new Pen(Color.YellowGreen, 5);
Random randomGen = new Random(Convert.ToInt32(label));
KnownColor[] names = (KnownColor[])Enum.GetValues(typeof(KnownColor));
KnownColor randomColorName = names[randomGen.Next(names.Length)];
Color randomColor = Color.FromKnownColor(randomColorName);
SolidBrush s = new SolidBrush(randomColor);
g.FillEllipse(s, x * 1, y * 1, 10, 10);
}

The best and simplest approach for this problem what I can think of is:
1) scan the image/coordinate system horizontally and vertically
2) For each row/column store the lowest and the highest coordinates with non-zero intensity
that will be your boundary points

You can define the furthest point of an angle. This can solve the problem of convex hull, but this technique fits well only in a circle space, very like yours.
for each angle you have to find the furthest point and then assign it the red color.
You can Use as much angle as you want.
PSEUDO...
npoints = 10;
furthestPts = zeros(npoints );//Initialize vectors with 0
distances = zeros(npoints );
for each pt in points
angle = atan((pt.y - c.y)/pt.x - c.x) * 360 / ( 2 * pi); //degres
angle = (int) (angle/npoints); //we will have only 10 points separated by 36 degrees
d = distance(pt,center);
if(distances[angle] < d){
distances[angle] = d; //Updating furthest point
furthestPts.[angle] = (pt); //Updating furthest point
}
You will see that this algoritm has a few problems if the point are to far from center or if the points are to far from each other.

Problems with centering scene

I have to display stl models with openGL. (SharpGL.) I'd like to set the initial view, so that the model is at the center of the screen and approximately fills it. I've calculated the bounding cube of the models and set the view like this: (sceneBox is a Rect3D - it stores the location of the left-back-bottom corner and the sizes)
// Calculate viewport properties
double left = sceneBox.X;
double right = sceneBox.X + sceneBox.SizeX;
double bottom = sceneBox.Y;
double top = sceneBox.Y + sceneBox.SizeY;
double zNear = 1.0;
double zFar = zNear + 3 * sceneBox.SizeZ;
double aspect = (double)this.ViewportSize.Width / (double)this.ViewportSize.Height;
if ( aspect < 1.0 ) {
bottom /= aspect;
top /= aspect;
} else {
left *= aspect;
right *= aspect;
}
// Create a perspective transformation.
gl.Frustum(
left / ZoomFactor,
right / ZoomFactor,
bottom / ZoomFactor,
top / ZoomFactor,
zNear,
zFar);
// Use the 'look at' helper function to position and aim the camera.
gl.LookAt(
0, 0, 2 * sceneBox.SizeZ,
sceneBox.X + 0.5 * sceneBox.SizeX, sceneBox.Y + 0.5 * sceneBox.SizeY, sceneBox.Z - 0.5 * sceneBox.SizeZ,
0, 1, 0);
This works nice with my small, hand-made test model: (it has a box size of 2*2*2 units)
This is exactly what I want. (The yellow lines show the bounding box)
But, when I load an stl model, which is about 60*60*60 units big, I get this:
It's very small and too far up.
What should I change to make it work?
Here's the full thing: https://dl.dropbox.com/u/17798054/program.zip
You can find this model in the zip as well. The quoted code is in KRGRAAT.SZE.Control.Engine.GLEngine.UpdateView()

Apparently the problem are the arguments you are using in lookAt function. If you have calculated bounding cube all you need to do is to place it in the distance (eyeZ) from the camera of
sizeX/tan(angleOfPerspective)
where sizeX is width of Quad of which cube is built, angleOfPerspective is first parameter of GlPerspective of course having centerX == posX == centreX of the front quad and centerY == posY == centreY of the front quad and frustum is not necessary
lookAt reference http://www.opengl.org/sdk/docs/man2/xhtml/gluLookAt.xml

So, to clarify Arek's answer, this is how I fixed it:
// Calculate viewport properties
double zNear = 1.0;
double zFar = zNear + 10 * sceneBox.SizeZ; // had to increase zFar
double aspect = (double)this.ViewportSize.Width / (double)this.ViewportSize.Height;
double angleOfPerspective = 60.0;
double centerX = sceneBox.X + 0.5 * sceneBox.SizeX;
double centerY = sceneBox.Y + 0.5 * sceneBox.SizeY;
double centerZ = sceneBox.Z + 0.5 * sceneBox.SizeZ;
// Create a perspective transformation.
gl.Perspective( // swapped frustum for perspective
angleOfPerspective / ZoomFactor, // moved zooming here
aspect,
zNear,
zFar);
// Use the 'look at' helper function to position and aim the camera.
gl.LookAt(
centerX, centerY, sceneBox.SizeX / Math.Tan(angleOfPerspective), // changed eye position
centerX, centerY, -centerZ,
0, 1, 0);

How to Draw an Arc using just 2 known points without DrawPie or DrawArc

I've question to ask you guys...
Let's say I have 2 lines as shown below:
What I want is to draw an arc that complete the shape as shown below:
Notes:
I know the start and end points of the lines and the angle between them.
The length of the lines are equall.
I don't use DrawPie method. Please don't suggest. I must use the end points of the line.
How can I achieve this?
My best regards...

//I use here an arbitrarily angle of 60 degrees, 30 above x-axis and 30 under it.
//If you have the end-points of each line, all you have to do is to implement it
//instead and do some adaptation to extend the arc properly:
private void btnArc_Click(object sender, EventArgs e)
{
int startX1, startX2, endX1, endX2, startY1, startY2, endY1, endY2, LineLength;
double angle, startAngle, sweepAngle;
const double RADIANS_TO_DEGREES = 180.0 / Math.PI;
const double DEGREES_TO_RADIANS = Math.PI / 180.0;
startX1 = startX2 = 350; //Set arbitrarily as the middle of a pictureBox, can be changed to pictureBox1.Width/2
startY1 = startY2 = 350; //Set arbitrarily as the middle of a pictureBox, can be changed to pictureBox1.Height/2
LineLength = 100; // length of 100 pixels set arbitrarily
angle = 30; // measured in degrees, starts counting from x-axis, hour 3 position clockwise
endX1 = startX1 + (int)(LineLength * Math.Cos(angle * DEGREES_TO_RADIANS)); //60 is an arbitary angle of mine
endY1 = startY1 + (int)(LineLength * Math.Sin(angle * DEGREES_TO_RADIANS));
endX2 = startX2 + 87; // Point equal to 30 degrees from x-axis set arbitrarily for demo purposes
endY2 = startY2 - 50;
//following multiplied by 2 since arbitrarily angle of 30 degrees from above axis to 30 deg. under axis equals 60 degrees
sweepAngle = 2 * RADIANS_TO_DEGREES*(Math.Asin(((double)endY1 - (double)startY1) / (double)LineLength));
startAngle = RADIANS_TO_DEGREES*(Math.Asin(((double)endY2 - (double)startY2) / (double)LineLength));
System.Drawing.Graphics surface3 = pictureBox1.CreateGraphics();
Pen pen1 = new Pen(Color.Blue, 2.0f);
Pen pen2 = new Pen(Color.Red, 1.0f);
//Assuming the arc should be symmetrical, as you didn't mention otherwise
//A surrounding rectangle would have length and height equal to twice the line-length
//The surrounding rectabgle middle point would be the common starting point of your lines
int RecX = startX1 - LineLength;
int RecY = startY1 - LineLength;
int RecWidth = LineLength * 2;
int RecHeight = LineLength * 2;
Rectangle rect3 = new Rectangle(RecX, RecY, RecWidth, RecHeight);
surface3.SmoothingMode = SmoothingMode.AntiAlias;
surface3.DrawLine(pen1, startX1, startY1, endX1, endY1);
surface3.DrawLine(pen1, startX2, startY2, endX2, endY2);
surface3.DrawRectangle(pen1, rect3);
surface3.DrawArc(pen2, rect3, (Single)startAngle, (Single)sweepAngle);
}

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!