I am working on an application for the past few weeks which involves some trigonometry and am currently stuck. As shown in the diagram below, I have a circular item (green circle at position #1) which I know the center point (let's call that X1,Y1). The circle has another point (orange circle) that is off-centered a bit - midway between two other marks (blue circles). These marks can move around. The coordinates of the orange point are calculated (let's call it X2, Y2) and the angle of the blue line is calculated (call it Angle) in relation to the horizontal of the circle.
I can calculate the difference between the center of the circle and the point by:
deltaX = X2-X1
deltaY = Y2-Y1
I need to move and rotate the green circle (either CW or CCW - whichever is shorter) from it's start location (position 1) over to position 2. This means the angle could be negative or positive. The blue line must end up vertical and the orange dot at the center of position 2 (red square). I know the coordinates for the center of position 2 (let's call this point X3,Y3). Position #1 and position #2 are exactly 90 degrees from each other.
I thought I could use some trig identity formulas that calculate the rotation of a point, as such:
offsetX = deltaX * cos(90-Angle) - deltaY * sin(90-Angle)
offsetY = deltaX * sin(90-Angle) + deltaY * cos(90-Angle)
I was hoping these offsets would be what I need to adjust the circle to it's new center when it moves/rotates over to position 2.
X3 = X3 + offsetX
Y3 = Y3 + offsetY
However, when I try use this math, it's not placing the orange mark of the circle in the center of the square. Not sure if my equations and calculations are correct based on the angle of rotation (positive or negative, CW or CCW) or if I'm using the angle correctly (where I subtract the known angle from 90 degrees). How do I correctly calculate the final point/position? Any help and examples would be greatly appreciated!
Thank you very much for your time!
So you need to rotate your circle by 90 - Angle and then move orange point to (X3, Y3)?
First you need to find orange point coordinate after rotation:
newX = X2 * cos(90 - Angle) - Y2 * sin(90 - Angle);
newY = X2 * sin(90 - Angle) + Y2 * cos(90 - Angle);
newX and newY are orange point coordinates after rotation. To find move transformation simply substract:
moveX = X3 - newX;
moveY = Y3 - newY;
Now if you rotate circle by 90 - Angle and move it by (moveX, moveY) orange point will move to (X3, Y3). That is if you rotate circle around (0, 0) point. If you rotating around some (X, Y) point, you first need to substract X from X2, Y from Y2 and then add X to newX, Y to newY. That substraction 'moves' your rotation base point to (0, 0), so after rotation you need to move it back:
newX = (X2 - X) * cos(90 - Angle) - (Y2 - Y) * sin(90 - Angle) + X;
newY = (X2 - X) * sin(90 - Angle) + (Y2 - Y) * cos(90 - Angle) + Y;
Be aware that your code is using a counter clockwise rotation, (conventionally angles are measured counter clockwise) which may be why you're not getting the results you expect. if you want a clockwise rotation try:
offsetX = deltaX * cos(angle) + deltaY * sin(angle)
offsetY = -deltaX * sin(angle) + deltaY * cos(angle)
Ensure your angles are in radians not degrees.
Drawing some lines may help you debug things too.
Related
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
Given an Point array and an arbitrary x,y coordinate, find the index for _points that is closest to the given coordinate.
PointD[] _points
//create a list of x,y coordinates:
for (int i = 0; i < _numberOfArcSegments + 1; i++)
{
double x1 = _orbitEllipseSemiMaj * Math.Sin(angle) - _focalDistance; //we add the focal distance so the focal point is "center"
double y1 = _orbitEllipseSemiMinor * Math.Cos(angle);
//rotates the points to allow for the LongditudeOfPeriapsis.
double x2 = (x1 * Math.Cos(_orbitAngleRadians)) - (y1 * Math.Sin(_orbitAngleRadians));
double y2 = (x1 * Math.Sin(_orbitAngleRadians)) + (y1 * Math.Cos(_orbitAngleRadians));
angle += _segmentArcSweepRadians;
_points[i] = new PointD() { x = x2, y = y2 };
}
I'm drawing an ellipse which represents an orbit. I'm first creating the point array above, then when I draw it, I (attempt) to find the point closest to where the orbiting body is.
To do this I've been attempting to calculate the angle from the center of the ellipse to the body:
public void Update()
{
//adjust so moons get the right positions (body position - focal point position)
Vector4 pos = _bodyPositionDB.AbsolutePosition - _positionDB.AbsolutePosition;
//adjust for focal point
pos.X += _focalDistance;
//rotate to the LonditudeOfPeriapsis.
double x2 = (pos.X * Math.Cos(-_orbitAngleRadians)) - (pos.Y * Math.Sin(-_orbitAngleRadians));
double y2 = (pos.X * Math.Sin(-_orbitAngleRadians)) + (pos.Y * Math.Cos(-_orbitAngleRadians));
_ellipseStartArcAngleRadians = (float)(Math.Atan2(y2, x2)); //Atan2 returns a value between -180 and 180;
}
then:
double unAdjustedIndex = (_ellipseStartArcAngleRadians / _segmentArcSweepRadians);
while (unAdjustedIndex < 0)
{
unAdjustedIndex += (2 * Math.PI);
}
int index = (int)unAdjustedIndex;
The ellipse draws fine, (the point array is correct and all is good once adjusted for viewscreen and camera offsets and zoom)
But does not start at the correct point (I'm decreasing the alpha in the color so the resulting ellipse fades away the further it gets from the body)
I've spend days trying to figure out what I'm doing wrong here and tried a dozen different things trying to figure out where my math is wrong, but I'm not seeing it.
I assume that _points should be an array of PointD;
This is the shortest way to get the closest point to your array (calcdistance should be a simple function that calculate the euclidean distance):
PointD p = _points.OrderBy(p => CalcDistance(p, gievnPoint)).First();
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.
I have locations of three points along the circle. pt1(x1, y1,z1), pt2(x2, y2, z2), pt3(x3, y3,z3). and want to find the radius of the circle. Already I have a function to compute radius in 2d space, which I am copying it here
public static double ComputeRadius(Location a, Location b, Location c)
{
double x1 = a.x;
double y1 = a.y;
double x2 = b.x;
double y2 = b.y;
double x3 = c.x;
double y3 = c.y;
double mr = (double)((y2 - y1) / (x2 - x1));
double mt = (double)((y3 - y2) / (x3 - x2));
double xc = (double)((mr * mt * (y3 - y1) + mr * (x2 + x3) - mt * (x1 + x2)) / (2 * (mr - mt)));
double yc = (double)((-1 / mr) * (xc - (x1 + x2) / 2) + (y1 + y2) / 2);
double d = (xc - x1) * (xc - x1) + (yc - y1) * (yc - y1);
return Math.Sqrt(d);
}
If you know the order of points pt1,pt2,pt3 along the circle then you can use graphical method:
cast normal axises from middle of each line segment in the plane of circle
your circle plane is defined by your 3 points. so the normal vector is
n = (pt2-pt1) x (pt3-pt2)
where the x is cross product so you have 2 lines (pt1,pt2) and (pt2,pt3) in black. The mid points are easy
p0=0.5*(pt1+pt2)
p1=0.5*(pt2+pt3)
the axis directions can be obtained also by cross product
dp0=(pt2-pt1) x n
dp1=(pt3-pt2) x n
so you got 2 axises:
pnt0(t)=p0+dp0*t
pnt1(u)=p1+dp1*u
Where t,u are scalar parameters t,u=(-inf,+inf) it is just position in axis from the starting mid point ...
the intersection is center of circle
So find the intersection of 2 axises and call it pt0
compute distance between center and any of your points
r=|pt1-pt0|
Sorry the image is for any curve (too lazy to repaint for circle as it is almost the same). If you do not know the order of points then the 2 points that are most distant to each other are the outer points ... In case they are equidistant the order does not matter any is OK
I need to draw an arc using GraphicsPath and having initial, median and final points. The arc has to pass on them.
I tried .DrawCurve and .DrawBezier but the result isn't exactly an arc.
What can I do?
SOLUTION:
After a couple of hours of code writing I managed to draw what I wanted with this algorithm (give 3 Point a,b,c and a GraphicsPath path):
double d = 2 * (a.X - c.X) * (c.Y - b.Y) + 2 * (b.X - c.X) * (a.Y - c.Y);
double m1 = (Math.Pow(a.X, 2) - Math.Pow(c.X, 2) + Math.Pow(a.Y, 2) - Math.Pow(c.Y, 2));
double m2 = (Math.Pow(c.X, 2) - Math.Pow(b.X, 2) + Math.Pow(c.Y, 2) - Math.Pow(b.Y, 2));
double nx = m1 * (c.Y - b.Y) + m2 * (c.Y - a.Y);
double ny = m1 * (b.X - c.X) + m2 * (a.X - c.X);
double cx = nx / d;
double cy = ny / d;
double dx = cx - a.X;
double dy = cy - a.Y;
double distance = Math.Sqrt(dx * dx + dy * dy);
Vector va = new Vector(a.X - cx, a.Y - cy);
Vector vb = new Vector(b.X - cx, b.Y - cy);
Vector vc = new Vector(c.X - cx, c.Y - cy);
Vector xaxis = new Vector(1, 0);
float startAngle = (float)Vector.AngleBetween(xaxis, va);
float sweepAngle = (float)(Vector.AngleBetween(va, vb) + Vector.AngleBetween(vb, vc));
path.AddArc(
(float)(cx - distance), (float)(cy - distance),
(float)(distance * 2), (float)(distance * 2),
startAngle, sweepAngle);
I would use DrawArc() as suggested by ANC_Michael. To find an arc that passes through 3 points you want to calculate the circumcircle of the triangle formed by the points.
Once you have the circumcircle calculate a bounding box for the circle to use with DrawArc using the min/max dimensions (center +/- radius). Now calculate your start and stop angles by translating the points so that the circumcircle is centered on the origin (translate by -circumcenter) and take the dot-product of the normalized start and end vectors with the X-axis:
double startAngle = Math.Acos(VectorToLeftPoint.Dot(XAxis));
double stopAngle = Math.Acos(VectorToRightPoint.Dot(XAxis));
Note that DrawArc expects angles clockwise from the X-axis so you should add Math.PI if the calculated vector is above the x-axis. That should be enough information to call DrawArc().
Edit: This method will find a circular arc and not necessarily the 'best fit' arc depending on your expected endpoint behavior.
Have you tried the DrawArc method and seeing if u can manipulate your 3 points somehow?
maybe
Pen blackPen= new Pen(Color.Black, 3);
// Create rectangle to bound ellipse.
Rectangle rect = new Rectangle(initial x, initial y, final x, median y);
// Create start and sweep angles on ellipse.
float startAngle = 0F;
float sweepAngle = 270.0F;
// Draw arc to screen.
e.Graphics.DrawArc(blackPen, rect, startAngle, sweepAngle);
http://msdn.microsoft.com/en-us/library/system.drawing.graphics.drawarc%28VS.71%29.aspx