Chamfer two lines by give distances using C# - c#

I am working on a simple tool in C#. I have three points which makes two lines meeting at point P. So that PP1 and PP2. I want to Chamfer the lines at meeting point such that distance d1 is trimmed from line PP1 and distance d2 is trimmed from line PP2 and then join the trimmed lines.
I have Problem as i can't get exact result. Any idea whats the problem in my code.
Thanks
Result1
Result2
private void Chamfer(Graphics g,PointF P,PointF P1,PointF P2,double d1,double d2)
{
//Vector 1 Length
double PP1 = Math.Sqrt((Math.Pow((P.X - P1.X), 2) + Math.Pow((P.Y - P1.Y), 2)));
//Vector 2 Length
double PP2 = Math.Sqrt((Math.Pow((P.X-P2.X), 2) + Math.Pow((P.Y - P2.Y), 2)));
//Slopes & Angles
double m1 = (P.Y-P1.Y) / (P.X - P1.X);
double angle1 = Math.Atan(m1) * (180 / Math.PI);
double m2 = (P.Y - P2.Y) / (P.X - P2.X);
double angle2 = Math.Atan(m2) * (180 / Math.PI);
//Coordinates of points of Chamfer
if(P1.X>P.X && P1.Y > P.Y)
{
int Pd1X = Convert.ToInt32(P.X + d1 * Math.Cos(angle1));
int Pd1Y = Convert.ToInt32(P.Y + d1 * Math.Sin(angle1));
Chp1 = new Point(Pd1X, Pd1Y);
}
else if (P1.X > P.X && P1.Y < P.Y)
{
int Pd1X = Convert.ToInt32(P.X + d1 * Math.Cos(angle1));
int Pd1Y = Convert.ToInt32(P.Y - d1 * Math.Sin(angle1));
Chp1 = new Point(Pd1X, Pd1Y);
}
else if (P1.X < P.X && P1.Y < P.Y)
{
int Pd1X = Convert.ToInt32(P.X - d1 * Math.Cos(angle1));
int Pd1Y = Convert.ToInt32(P.Y - d1 * Math.Sin(angle1));
Chp1 = new Point(Pd1X, Pd1Y);
}
else if (P1.X < P.X && P1.Y > P.Y)
{
int Pd1X = Convert.ToInt32(P.X - d1 * Math.Cos(angle1));
int Pd1Y = Convert.ToInt32(P.Y + d1 * Math.Sin(angle1));
Chp1 = new Point(Pd1X, Pd1Y);
}
if (P2.X > P.X && P2.Y > P.Y)
{
int Pd2X = Convert.ToInt32(P.X + d2 * Math.Cos(angle2));
int Pd2Y = Convert.ToInt32(P.Y + d2 * Math.Sin(angle2));
Chp2 = new Point(Pd2X, Pd2Y);
}
else if (P2.X > P.X && P2.Y < P.Y)
{
int Pd2X = Convert.ToInt32(P.X + d2 * Math.Cos(angle2));
int Pd2Y = Convert.ToInt32(P.Y - d2 * Math.Sin(angle2));
Chp2 = new Point(Pd2X, Pd2Y);
}
else if (P2.X < P.X && P2.Y < P.Y)
{
int Pd2X = Convert.ToInt32(P.X - d2 * Math.Cos(angle2));
int Pd2Y = Convert.ToInt32(P.Y - d2 * Math.Sin(angle2));
Chp2 = new Point(Pd2X, Pd2Y);
}
else if (P2.X < P.X && P2.Y > P.Y)
{
int Pd2X = Convert.ToInt32(P.X - d2 * Math.Cos(angle1));
int Pd2Y = Convert.ToInt32(P.Y + d2 * Math.Sin(angle1));
Chp2 = new Point(Pd2X, Pd2Y);
}
Pen penPre = new Pen(Color.Green);
penPre.DashStyle = System.Drawing.Drawing2D.DashStyle.Dot;
g.Clear(this.BackColor);
g.DrawLine(Pens.Black, P1, Chp1);
g.DrawLine(penPre, P1, P);
g.DrawLine(penPre, P2, P);
g.DrawString("P1", this.Font, Brushes.Red, new Point(Convert.ToInt32(P1.X + 2), Convert.ToInt32(P1.Y - 2)));
g.DrawLine(Pens.Black, P2, Chp2);
g.DrawString("P2", this.Font, Brushes.Red, new Point(Convert.ToInt32(P2.X + 2), Convert.ToInt32(P2.Y - 2)));
g.DrawString("P", this.Font, Brushes.Red, new Point(Convert.ToInt32( P.X + 3),Convert.ToInt32( P.Y - 2)));
g.DrawLine(Pens.Black, Chp1, Chp2);
}
what i got at a specific coordinates, my code is working as you can see from photo attached. I can't figure if i don't have these golden coordinate.
Result from golden coordinates

//Vector 1 Length
double PP1 = Math.Sqrt((Math.Pow((P.X - P1.X), 2) + Math.Pow((P.Y - P1.Y), 2)));
//unit direction vector
upx1 = (P1.X - P.X) / PP1
upy1 = (P1.Y - P.Y) / PP1
// chpoint coordinates
chx1 = P.X + upx1 * d;
chy1 = P.Y + upy1 * d;
// now the same for PP2

Related

How to find center point if startangle and sweep angle is changed

In android , I have drawn an arc based on startangle, sweepangle and radius. Let width be 400 and height be 500 as rectangle bounds in which radius is calculated as
var radius = Math.Min(Width,Height)/2;
Also if centre is calculated as
var x = (float)(Width * 0.5);
var y = (float)(Height * 0.5);
var centre = new PointF(x,y);
If above centre value is used, centre remains same for all start angle and sweepangle for rectangle. I need to change the centre if startangle and sweep angle changes
In the below image, rectangle bounds is 400,500 and the startangle is 0 and sweepangle is 360
If I change start angle to 180 and sweepangle to 180, centre remains same
I need the below image output,if I change startangle and sweepangle based on circle bounds, centre point should vary
I have done calculations for the above ,
private SystemPointF GetActualCenter(float x, float y, float radius)
{
SystemPointF actualCenter = new SystemPointF(x, y);
double startAngle1 = GetWrapAngle(StartAngle, -630, 630);
double endAngle1 = GetWrapAngle(EndAngle, -630, 630);
float[] regions = new float[] { -630, -540, -450, -360, -270, -180, -90, 0, 90, 180, 270, 360, 450, 540, 630 };
List<int> region = new List<int>();
if (startAngle1 < endAngle1)
{
for (int i = 0; i < regions.Length; i++)
{
if (regions[i] > startAngle1 && regions[i] < endAngle1)
region.Add((int)((regions[i] % 360) < 0 ? (regions[i] % 360) + 360 : (regions[i] % 360)));
}
}
else
{
for (int i = 0; i < regions.Length; i++)
{
if (regions[i] < startAngle1 && regions[i] > endAngle1)
region.Add((int)((regions[i] % 360) < 0 ? (regions[i] % 360) + 360 : (regions[i] % 360)));
}
}
double startRadian = 2 * Math.PI * (startAngle1) / 360;
double endRadian = 2 * Math.PI * (endAngle1) / 360;
SystemPointF startPoint = new SystemPointF((float)(x + radius * Math.Cos(startRadian)),
(float)(y + radius * Math.Sin(startRadian)));
SystemPointF endPoint = new SystemPointF((float)(x + radius * Math.Cos(endRadian)),
(float)(y + radius * Math.Sin(endRadian)));
switch (region.Count)
{
case 0:
float longX = Math.Abs(x - startPoint.X) > Math.Abs(x - endPoint.X) ? startPoint.X : endPoint.X;
float longY = Math.Abs(y - startPoint.Y) > Math.Abs(y - endPoint.Y) ? startPoint.Y : endPoint.Y;
SystemPointF midPoint = new SystemPointF(Math.Abs((x + longX)) / 2, Math.Abs((y + longY)) / 2);
actualCenter.X = x + (x - midPoint.X);
actualCenter.Y = y + (y - midPoint.Y);
break;
case 1:
SystemPointF point1 = new SystemPointF(), point2 = new SystemPointF();
float maxRadian = (float)(2 * Math.PI * region[0] / 360);
SystemPointF maxPoint = new SystemPointF((float)(x + radius * Math.Cos(maxRadian)),
(float)(y + radius * Math.Sin(maxRadian)));
switch (region[0])
{
case 270:
point1 = new SystemPointF(startPoint.X, maxPoint.Y);
point2 = new SystemPointF(endPoint.X, y);
break;
case 0:
case 360:
point1 = new SystemPointF(x, endPoint.Y);
point2 = new SystemPointF(maxPoint.X, startPoint.Y);
break;
case 90:
point1 = new SystemPointF(endPoint.X, y);
point2 = new SystemPointF(startPoint.X, maxPoint.Y);
break;
case 180:
point1 = new SystemPointF(maxPoint.X, startPoint.Y);
point2 = new SystemPointF(x, endPoint.Y);
break;
}
midPoint = new SystemPointF((point1.X + point2.X) / 2, (point1.Y + point2.Y) / 2);
actualCenter.X = x + ((x - midPoint.X) >= radius ? 0 : (x - midPoint.X));
actualCenter.Y = y + ((y - midPoint.Y) >= radius ? 0 : (y - midPoint.Y));
break;
case 2:
float minRadian = (float)(2 * Math.PI * region[0] / 360);
maxRadian = (float)(2 * Math.PI * (region[1]) / 360);
maxPoint = new SystemPointF((float)(x + radius * Math.Cos(maxRadian)),
(float)(y + radius * Math.Sin(maxRadian)));
SystemPointF minPoint = new SystemPointF((float)(x + radius * Math.Cos(minRadian)),
(float)(y + radius * Math.Sin(minRadian)));
if (region[0] == 0 && region[1] == 90 || region[0] == 180
&& region[1] == 270)
point1 = new SystemPointF(minPoint.X, maxPoint.Y);
else
point1 = new SystemPointF(maxPoint.X, minPoint.Y);
if (region[0] == 0 || region[0] == 180)
point2 = new SystemPointF(GetMinMaxValue(startPoint, endPoint, region[0]),
GetMinMaxValue(startPoint, endPoint, region[1]));
else
point2 = new SystemPointF(GetMinMaxValue(startPoint, endPoint, region[1]),
GetMinMaxValue(startPoint, endPoint, region[0]));
midPoint = new SystemPointF(Math.Abs(point1.X - point2.X) / 2 >= radius ? 0 : (point1.X + point2.X) / 2,
Math.Abs(point1.Y - point2.Y) / 2 >= radius ? 0 : (point1.Y + point2.Y) / 2);
actualCenter.X = x + (midPoint.X == 0 ? 0 : (x - midPoint.X) >= radius ? 0 : (x - midPoint.X));
actualCenter.Y = y + (midPoint.Y == 0 ? 0 : (y - midPoint.Y) >= radius ? 0 : (y - midPoint.Y));
break;
}
return actualCenter;
}
This works when startangle and sweep angle changed for all cases except the case startangle 179 and sweep angle changed to above 180. case 3 includes the region 180,270,0 . how to write calculations for regions 3.
Any help is really appreciated.
Thanks in advance
When you want to draw an object in the center of something you should do this :
Object.Point =
new Point((something.Width / 2) - (object.Widht /2) ,(something.Height / 2) - object.Height / 2));

Rasterizing curves algorithms

My current application needs an algorithm to drawing lines, circles and rotated ellipses of a given image. I have searched on the internet and found this article Algorithm for Drawing Curves explain how to achieve it. I'm trying to implement a method describe in the Page 40 in C#, but in the Line 40 has an operation between Int and Bool.
The operation in question is:
x1 = 2 * err > dy; // <- this
y1 = 2 * (err + yy) < -dy); // <- this
if (2 * err < dx || y1) // <- this
{
y0 += sy;
dy += xy;
err += dx += xx;
}
if (2 * err > dx || x1) // <- this
{
x0 += sx;
dx += xy;
err += dy += yy;
}
The variables x1, y1 are integers and err, yy and dy are double. I don't know which programming language the author used, but apparently seems to be in C.
How to translated this operation in C#?
The Complete method I tried to implemented in C#:
private static Point[] plotQuadRationalBezierSeg(int x0, int y0, int x1, int y1, int x2, int y2, double w)
{
// plot a limited rational Bezier segment, squared weight
var points = new List<Point>();
int sx = x2 - x1, sy = y2 - y1; // relative values for checks
double dx = x0 - x2, dy = y0 - y2, xx = x0 - x1, yy = y0 - y1;
double xy = xx * sy + yy * sx, cur = xx * sy - yy * sx, err; // curvature
//assert(xx * sx <= 0.0 && yy * sy <= 0.0); // sign of gradient must not change
if (cur != 0.0 && w > 0.0)
{ /* no straight line */
if (sx * (long)sx + sy * (long)sy > xx * xx + yy * yy)
{ // begin with longer part
x2 = x0;
x0 -= (int)dx;
y2 = y0;
y0 -= (int)dy;
cur = -cur; //swap P0 P2
}
xx = 2.0 * (4.0 * w * sx * xx + dx * dx); //differences 2nd degree
yy = 2.0 * (4.0 * w * sy * yy + dy * dy);
sx = x0 < x2 ? 1 : -1; // x step direction
sy = y0 < y2 ? 1 : -1; // y step direction
xy = -2.0 * sx * sy * (2.0 * w * xy + dx * dy);
if (cur * sx * sy < 0.0)
{
// negated curvature?
xx = -xx;
yy = -yy;
xy = -xy;
cur = -cur;
}
dx = 4.0 * w * (x1 - x0) * sy * cur + xx / 2.0 + xy; //differences 1st degree
dy = 4.0 * w * (y0 - y1) * sx * cur + yy / 2.0 + xy;
if (w < 0.5 && (dy > xy || dx < xy))
{
// flat ellipse, algorithm fails
cur = (w + 1.0) / 2.0;
w = Math.Sqrt(w);
xy = 1.0 / (w + 1.0);
sx = (int)Math.Floor((x0 + 2.0 * w * x1 + x2) * xy / 2.0 + 0.5); //subdivide curve in half
sy = (int)Math.Floor((y0 + 2.0 * w * y1 + y2) * xy / 2.0 + 0.5);
dx = (int)Math.Floor((w * x1 + x0) * xy + 0.5);
dy = (int)Math.Floor((y1 * w + y0) * xy + 0.5);
points.AddRange(plotQuadRationalBezierSeg(x0, y0, (int)dx, (int)dy, sx, sy, cur)); //plot separately
dx = Math.Floor((w * x1 + x2) * xy + 0.5); dy = Math.Floor((y1 * w + y2) * xy + 0.5);
points.AddRange(plotQuadRationalBezierSeg(sx, sy, (int)dx, (int)dy, x2, y2, cur));
return points.ToArray();
}
//error 1.step
err = dx + dy - xy;
do
{
points.Add(new Point(x0, y0)); // plot curve
if (x0 == x2 && y0 == y2)
return points.ToArray(); // last pixel -> curve finished
x1 = 2 * err > dy;
y1 = 2 * (err + yy) < -dy); // save value for test of x step
if (2 * err < dx || y1)
{
y0 += sy;
dy += xy;
err += dx += xx;
}/* y step */
if (2 * err > dx || x1)
{
x0 += sx;
dx += xy;
err += dy += yy;
}/* x step */
} while (dy <= xy && dx >= xy); // gradient negates -> algorithm fails
}
// plot remaining needle to end
points.AddRange(
new Point[] {
new Point(x0, y0),
new Point(x2, y2),
});
return points.ToArray();
}

vector point precision Error

Our application contains .Net FrameWork3.5 and DirectX For Making 2d drawing.
we are stuck with precision problem.Please give solution to solve the precision problem.
we are using vector3f(vector points) to store the values. while we store the result of point in vector3f only 2 precision value is matched.But we need 5 precision value.
Here i have mentioned my code below..
//Function To find ARC Parameters from Bulge value
polylineVertStart=6919.602,18951.51,0 polylineVertEnd=6916.602,18951.51,0
Arc StartPoint=6919.602,18951.5177,0 Endpoint=6916.602,18951.51,0
public static Arc BulgeToArc(PolylineVertex3d polylineVertStart, PolylineVertex3d polylineVertEnd)
{
PolylineVertex3d polylineVertex3d = polylineVertEnd;
PolylineVertex3d polylineVertex3d1 = polylineVertStart; //get previous point
double x1 = polylineVertex3d1.Position.X; //Assign start and end points
double y1 = polylineVertex3d1.Position.Y;
double x2 = polylineVertex3d.Position.X;
double y2 = polylineVertex3d.Position.Y;
if (y1 == y2)
{
y2 += (y1 * 0.0000001);
}
if (x1 == x2)
{
x2 += (x1 * 0.0000001);
}
double bulge = polylineVertex3d1.Bulge;
double incAngle = 4 * System.Math.Atan(System.Math.Abs(bulge)); //included Angle
double chord = System.Math.Sqrt((x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1)); //Chord length
double r = 0.5 * chord / System.Math.Cos(0.5 * incAngle - 0.5 * System.Math.PI); //Calculate radius
double Radius = r;
double dx = x2 - x1;
double dy = y2 - y1;
double slope = dy / dx; //slope of two points
double slopeAng = System.Math.Atan(slope);
if (System.Math.Sign(dy) == -1 && System.Math.Sign(dx) == -1)
{
slopeAng = System.Math.PI + slopeAng;
}
else if (System.Math.Sign(dy) == -1 && System.Math.Sign(dx) == 1)
{
slopeAng = 2 * System.Math.PI + slopeAng;
}
else if (System.Math.Sign(dy) == 1 && System.Math.Sign(dx) == -1)
{
slopeAng = System.Math.PI + slopeAng;
}
double d1 = System.Math.Sqrt((x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1));
double d2 = d1 / 2;
double startAng = System.Math.Acos(d2 / r); //calculate start angle
double sAngle;
if (System.Math.Abs(bulge) < 1) //get actual start angle based on bulge direction
{
if (System.Math.Sign(bulge) != -1)
{
sAngle = slopeAng + startAng;
}
else
{
sAngle = slopeAng - startAng;
}
}
else
{
if (System.Math.Sign(bulge) != -1)
{
sAngle = slopeAng - startAng;
}
else
{
sAngle = slopeAng + startAng;
}
}
double cx = x1 + r * System.Math.Cos(sAngle);
double cy = y1 + r * System.Math.Sin(sAngle);
Vector3F Center = new Vector3F((float)cx, (float)cy, 0f); //calculate center point
double dx1 = x1 - cx;
double dx2 = x2 - cx;
double dy1 = y1 - cy;
double dy2 = y2 - cy;
double sAng = System.Math.Atan(dy1 / dx1);
double eAng = System.Math.Atan(dy2 / dx2);
if (System.Math.Sign(dy1) == -1 && System.Math.Sign(dx1) == -1)
{
sAng = System.Math.PI + sAng;
}
else if (System.Math.Sign(dy1) == -1 && System.Math.Sign(dx1) != -1)
{
sAng = 2 * System.Math.PI + sAng;
}
else if (System.Math.Sign(dx1) == -1)
{
sAng = System.Math.PI + sAng;
}
if (System.Math.Sign(dy2) == -1 && System.Math.Sign(dx2) == -1)
{
eAng = System.Math.PI + eAng;
}
else if (System.Math.Sign(dy2) == -1 && System.Math.Sign(dx2) != -1)
{
eAng = 2 * System.Math.PI + eAng;
}
else if (System.Math.Sign(dx2) == -1)
{
eAng = System.Math.PI + eAng;
}
double StartAngle;
double EndAngle;
if (System.Math.Sign(bulge) != -1) //finalise start angle and end angle
{
StartAngle = sAng;
EndAngle = eAng;
}
else
{
StartAngle = eAng;
EndAngle = sAng;
}
Direction dir;
if (polylineVertStart.Bulge != 0)
dir = polylineVertStart.Bulge < 0 ? Direction.ClockWise : Direction.CounterClockWise;
else
dir = polylineVertEnd.Bulge < 0 ? Direction.ClockWise : Direction.CounterClockWise;
Arc arc = dir == Direction.ClockWise ? new Arc(Center, new Vector3F(0f, 0f, 1f), Radius, EndAngle, StartAngle, dir) : new Arc(Center, new Vector3F(0f, 0f, 1f), Radius, StartAngle, EndAngle, dir);
return arc;
}
//Function to find ARC from StartAngle,EndAngle,Radius
public RenderingEngine.Geometry GetGeomtry()
{
RenderingEngine.Geometry geometry = new RenderingEngine.Geometry
{
EntityPrimitiveType = EntityPrimitiveType.LineStrip
};
xPoints.Clear();
yPoints.Clear();
zPoints.Clear();
VertexList.Clear();
List<CustomVertex.PositionColored> vertices = new List<CustomVertex.PositionColored>();
if (direction == Direction.ClockWise)
{
#region clockwise
double WedgeAngle = 0.0; //clock wise
if (EndAngle != StartAngle)
{
if (EndAngle < StartAngle)
{
WedgeAngle = (StartAngle - EndAngle) / NUMPOINTS;
}
else
{
WedgeAngle = ((System.Math.PI * 2) - (EndAngle - StartAngle)) / NUMPOINTS;
}
}
double angle = StartAngle;
for (int i = 0; i <= NUMPOINTS; i++)
{
double theta = angle - (i * WedgeAngle);
double x = center.X + (radius * SystemMath.Cos(theta));
double y = center.Y + (radius * SystemMath.Sin(theta));
double z = center.Z;
CustomVertex.PositionColored positionColored = new CustomVertex.PositionColored
{
Position =
new Vector3((float) x, (float) y,
(float) z),
Color = Color.ToArgb()
};
vertices.Add(positionColored);
xPoints.Add((float)x);
yPoints.Add((float)y);
zPoints.Add((float)z);
VertexList.Add(new Vector3F((float)x, (float)y, (float)z));
}
double startx = Center.X + (Radius * System.Math.Cos(StartAngle));
double starty = Center.Y + (Radius * System.Math.Sin(StartAngle));
double startz = Center.Z;
double endx = Center.X + (Radius * System.Math.Cos(EndAngle));
double endy = Center.Y + (Radius * System.Math.Sin(EndAngle));
double endz = Center.Z;
startPoint = new Vector3F((float)startx, (float)starty, (float)startz);
endPoint = new Vector3F((float)endx, (float)endy, (float)endz);
#endregion clockwise
}
else
{
#region Counter ClockWise
double WedgeAngle = 0.0;
if (EndAngle != StartAngle)
{
if (EndAngle > StartAngle)
{
WedgeAngle = (EndAngle - StartAngle) / NUMPOINTS;
}
else
{
WedgeAngle = ((System.Math.PI * 2 - StartAngle) + EndAngle) / NUMPOINTS;
}
}
double angle = StartAngle;
for (int i = 0; i <= NUMPOINTS; i++)
{
double theta = angle + (i * WedgeAngle);
double x = center.X + (radius * SystemMath.Cos(theta));
double y = center.Y + (radius * SystemMath.Sin(theta));
double z = center.Z;
CustomVertex.PositionColored positionColored = new CustomVertex.PositionColored
{
Position =
new Vector3((float) x, (float) y,
(float) z),
Color = Color.ToArgb()
};
vertices.Add(positionColored);
xPoints.Add((float)x);
yPoints.Add((float)y);
zPoints.Add((float)z);
VertexList.Add(new Vector3F((float)x, (float)y, (float)z));
}
double startx = Center.X + (Radius * System.Math.Cos(StartAngle));
double starty = Center.Y + (Radius * System.Math.Sin(StartAngle));
double startz = Center.Z;
double endx = Center.X + (Radius * System.Math.Cos(EndAngle));
double endy = Center.Y + (Radius * System.Math.Sin(EndAngle));
double endz = Center.Z;
startPoint = new Vector3F((float)startx, (float)starty, (float)startz);
endPoint = new Vector3F((float)endx, (float)endy, (float)endz);
#endregion
}
geometry.Vertices.Add(vertices);
return geometry;
}
Here arc start point and endpoint x value always come correctly but y value is some precision problem.
Thanks in Advance..

Projection of a point on line defined by 2 points

I have been trying to figure this out for sometime now..
The problem to solve..
Say I have 3 Points..
P1 ---------- P2, and P3 can be anywhere around P1 and P2
What is the formula to calculate so that P3 is interpolated onto the line between P1 and P2?
I need a formula that calculates new X,Y coordinates for P3 that falls on the line between P1 and P2..
My code as of so far..
public Point lerp(Point P0, Point P1, Point P)
{
double y1 = P0.Y + (P1.Y - P0.Y) * ((P.X - P0.X) / (P1.X - P0.X));
double x1 = P.X;
double y2 = P.Y;
double x2 = P0.X + (P1.X - P0.X) * ((P.Y - P0.Y) / (P1.Y - P0.Y));
return new Point((x1 + x2) / 2, (y1 + y2) / 2);
}
And my reference..
http://en.wikipedia.org/wiki/Linear_interpolation
The above code gets it close, but its slightly off...
Here is the converted javascript code from Corey Ogburn
public Point _pointOnLine(Point pt1, Point pt2, Point pt)
{
bool isValid = false;
var r = new Point(0, 0);
if (pt1.Y == pt2.Y && pt1.X == pt2.X) { pt1.Y -= 0.00001; }
var U = ((pt.Y - pt1.Y) * (pt2.Y - pt1.Y)) + ((pt.X - pt1.X) * (pt2.X - pt1.X));
var Udenom = Math.Pow(pt2.Y - pt1.Y, 2) + Math.Pow(pt2.X - pt1.X, 2);
U /= Udenom;
r.Y = pt1.Y + (U * (pt2.Y - pt1.Y));
r.X = pt1.X + (U * (pt2.X - pt1.X));
double minx, maxx, miny, maxy;
minx = Math.Min(pt1.X, pt2.X);
maxx = Math.Max(pt1.X, pt2.X);
miny = Math.Min(pt1.Y, pt2.Y);
maxy = Math.Max(pt1.Y, pt2.Y);
isValid = (r.X >= minx && r.X <= maxx) && (r.Y >= miny && r.Y <= maxy);
return isValid ? r : new Point();
}
Here's some javascript code we've used here at work (a GIS company) to figure out the closest point on a line the mouse is next to in a situation where a user wants to split the line by adding a vertex to it. Should be easy to move over to C#:
function _pointOnLine(line1, line2, pt) {
var isValid = false;
var r = new Microsoft.Maps.Location(0, 0);
if (line1.latitude == line2.latitude && line1.longitude == line2.longitude) line1.latitude -= 0.00001;
var U = ((pt.latitude - line1.latitude) * (line2.latitude - line1.latitude)) + ((pt.longitude - line1.longitude) * (line2.longitude - line1.longitude));
var Udenom = Math.pow(line2.latitude - line1.latitude, 2) + Math.pow(line2.longitude - line1.longitude, 2);
U /= Udenom;
r.latitude = line1.latitude + (U * (line2.latitude - line1.latitude));
r.longitude = line1.longitude + (U * (line2.longitude - line1.longitude));
var minx, maxx, miny, maxy;
minx = Math.min(line1.latitude, line2.latitude);
maxx = Math.max(line1.latitude, line2.latitude);
miny = Math.min(line1.longitude, line2.longitude);
maxy = Math.max(line1.longitude, line2.longitude);
isValid = (r.latitude >= minx && r.latitude <= maxx) && (r.longitude >= miny && r.longitude <= maxy);
return isValid ? r : null;
}
line1 is a point with a latitude and longitude to represent one of the endpoints of the line, equivalent to your P1. line2 is the other endpoint: P2. pt is your P3. This will return the point on the line that P3 is perpendicular through. If P3 is past either end of the line, this will return null which means that one of the two end points is the closest point to P3.
For clarity:
The problem is that you Point has integer values for X and Y and therefore you are doing integer division. Try to cast your values into float or double, do the calculations and then return them back to the integers.
Note that when you are doing this:
(P1.Y - P0.Y) * ((P.X - P0.X) / (P1.X - P0.X))
you are actualy loosing the precision since the result of 5/2 is 2, not 2.5 but when your values are real numbers then 5.0/2.0 is indeed 2.5.
You should try this:
double y1 = P0.Y + (double)(P1.Y - P0.Y) * ((double)(P.X - P0.X) / (double)(P1.X - P0.X));
double x1 = P.X; //these two are implicit casts
double y2 = P.Y;
double x2 = P0.X + (double)(P1.X - P0.X) * ((double)(P.Y - P0.Y) / (double)(P1.Y - P0.Y));
return new Point((x1 + x2) / 2.0, (y1 + y2) / 2.0); //put 2.0 for any case even though x1+x2 is `double`
Also, then you are converting from double to int, decimal part of the number is automatically cut off so for instance 3.87 will become 3. Than your last line should be more precise if you could use this:
return new Point((x1 + x2) / 2.0 + 0.5, (y1 + y2) / 2.0 + 0.5);
which will effectively round double values to the closer integer value.
EDIT:
But if you just want to find the point p3 on the line between the two points, than it is easier to use this approach:
public Point lerp(Point P0, Point P1)
{
double x = ((double)P0.X + P1.X)/2.0;
double y = (double)P0.Y + (double)(P1.Y - P0.Y) * ((double)(x - P0.X) / (double)(P1.X - P0.X));
return new Point(x + 0.5, y + 0.5);
}
This is an old question and I found the Corey Ogburn solution quite useful. But I thought it might be helpful for others to post a less "map" version of the javascript code - which I used in canvas drawing.
export const pointOnLine = (p0, p1, q) => {
// p0 and p1 define the line segment
// q is the reference point (aka mouse)
// returns point on the line closest to px
if (p0.x == p1.x && p0.y == p1.y) p0.x -= 0.00001;
const Unumer = ((q.x - p0.x) * (p1.x - p0.x)) + ((q.y - p0.y) * (p1.y - p0.y));
const Udenom = Math.pow(p1.x - p0.x, 2) + Math.pow(p1.y - p0.y, 2);
const U = Unumer / Udenom;
const r = {
x: p0.x + (U * (p1.x - p0.x)),
y: p0.y + (U * (p1.y - p0.y))
}
const minx = Math.min(p0.x, p1.x);
const maxx = Math.max(p0.x, p1.x);
const miny = Math.min(p0.y, p1.y);
const maxy = Math.max(p0.y, p1.y);
const isValid = (r.x >= minx && r.x <= maxx) && (r.y >= miny && r.y <= maxy);
return isValid ? r : null;
}

How to find a point where a line intersects an ellipse in 2D (C#)

I need to find a point where a line (its origin is ellipse' center) intersects an ellipse in 2D... I can easily find a point on a circle, because I know an angle F and the circle' radius (R):
x = x0 + R * cosF
y = y0 + R * sinF
However I just can't figure how am I supposed to deal with an ellipse... I know it's dimensions (A & B), but what is the way of finding parameter T?!
x = x0 + A * cosT
y = y0 + B * sinT
From what I understand the parameter T (T angle) is not far from the F angle (approximately +-15 degrees in some cases), but I just can't figure how to calculate it!!!
If there is a kind hearted soul, please help me with this problem...
The standard equation of an ellipse, stationed at 0,0, is:
1 = (x)^2 / (a) + (y)^2 / (b)
Where a is 1/2 the diameter on the horizontal axis, and b is 1/2 the diameter on the vertical axis.
you have a line, assuming an equation:
y = (m)(x - x0) + y0
So, let us plug-and-play!
1 = (x)^2 / (a) + (m(x - x0) + y0)^2 / (b)
1 = x^2 / a + (mx + (y0 - mx0))^2 / b
1 = x^2 / a + (m^2 * x^2 + 2mx*(y0 - mx0) + (y0 - mx0)^2) / b
1 = x^2 / a + (m^2 x^2) / b + (2mx*(y0 - mx0) + (y0^2 - 2y0mx0 + m^2*x0^2)) / b
1 = ((x^2 * b) / (a * b)) + ((m^2 * x^2 * a) / (a * b)) + (2mxy0 - 2m^2xx0)/b + (y0^2 - 2y0mx0 + m^2*x0^2)/b
1 = ((bx^2 + am^2x^2)/(ab)) + (x*(2my0 - 2m^2x0))/b + (y0^2 - 2y0mx0 + m^2*x0^2)/b
0 = x^2*((b + a*m^2)/(ab)) + x*((2my0 - 2m^2x0)/b) + (((y0^2 - 2y0mx0 + m^2*x0^2)/b) - 1)
That last equation follows the form of a standard quadratic equation.
So just use the quadratic formula, with:
((b + a*m^2)/(ab))
((2my0 - 2m^2x0)/b)
and
(((y0^2 - 2y0mx0 + m^2*x0^2)/b) - 1)
to get the X values at the intersections; Then, plug in those values into your original line equation to get the Y values.
Good luck!
Don't do it this way. Instead check the equation that forms an ellipse and that forming a line and solve the set:
The ellipse: (x/a)^2 + (y/b)^2 = 1
Your line: y = cx
You know a, b and c, so finding a solution is going to be easy. You'll find two solutions, because the line crosses the ellipse twice.
EDIT: Note I moved your ellipse's center to (0,0). It makes everything easier. Just add (x0,y0) to the solution.
public Hits<float2> EllipseLineIntersection ( float rx , float ry , float2 p1 , float2 p2 )
{
Hits<float2> hits = default(Hits<float2>);
float2 p3, p4;
Rect rect = default(Rect);
{
rect.xMin = math.min(p1.x,p2.x);
rect.xMax = math.max(p1.x,p2.x);
rect.yMin = math.min(p1.y,p2.y);
rect.yMax = math.max(p1.y,p2.y);
}
float s = ( p2.y - p1.y )/( p2.x - p1.x );
float si = p2.y - ( s * p2.x );
float a = ( ry*ry )+( rx*rx * s*s );
float b = 2f * rx*rx * si * s;
float c = rx*rx * si*si - rx*rx * ry*ry;
float radicand_sqrt = math.sqrt( ( b*b )-( 4f * a * c) );
p3.x = ( -b - radicand_sqrt )/( 2f*a );
p4.x = ( -b + radicand_sqrt )/( 2f*a );
p3.y = s*p3.x + si;
p4.y = s*p4.x + si;
if( rect.Contains(p3) ) hits.Push( p3 );
if( rect.Contains(p4) ) hits.Push( p4 );
return hits;
}
public struct Hits<T>
{
public byte count;
public T point0, point1;
public void Push ( T val )
{
if( count==0 ) { point0 = val; count ++; }
else if( count==1 ) { point1 = val; count ++; }
else print("This structure can only fit 2 values");
}
}
I wrote a C# code for your problem and I hope you can find it helpful. the distance function inside this code calculates euclidean distance between two points in space.
wX denotes horizontal radios of ellipse and wY denotes vertical radios.
private PointF LineIntersectEllipse(PointF A, PointF B, float wX, float wY)
{
double dx = B.X - A.X;
double dy = B.Y - A.Y;
double theta = Math.Atan2(dy, dx);
double r = distance(A, B) - ((wX * wY) / Math.Sqrt(Math.Pow(wY * Math.Cos(theta), 2) + Math.Pow(wX * Math.Sin(theta), 2)));
return PointF((float)(A.X + r * Math.Cos(theta)), (float)(A.Y + r * Math.Sin(theta)));
}
Andrew Łukasik posted a good and useful answer, however it is not using regular C# types. As I wrote in the comments, I converted the code using System.Drawing objects PointF and RectangleF. I found out that if the points given as parameters are aligned as a vertical or horizontal line, then "rect" will have a width or a height equal to 0. Then, rect.Contains(point) will return false even if the point is on this line.
I also modified the "Hits" structure to check if the point pushed is not already existing, which is the case if the line is perfectly tangent, then p3 and p4 will have same coordinates, as the exact tangent point is the only crossing point.
Here is the new code taking care of all the cases :
public static Hits<PointF> EllipseLineIntersection0(float rx, float ry, PointF p1, PointF p2)
{
Hits<PointF> hits = default(Hits<PointF>);
PointF p3 = new PointF();
PointF p4 = new PointF();
var rect = default(RectangleF);
rect.X = Math.Min(p1.X, p2.X);
rect.Width = Math.Max(p1.X, p2.X) - rect.X;
rect.Y = Math.Min(p1.Y, p2.Y);
rect.Height = Math.Max(p1.Y, p2.Y) - rect.Y;
float s = (p2.Y - p1.Y) / (p2.X - p1.X);
float si = p2.Y - (s * p2.X);
float a = (ry * ry) + (rx * rx * s * s);
float b = 2f * rx * rx * si * s;
float c = rx * rx * si * si - rx * rx * ry * ry;
float radicand_sqrt = (float)Math.Sqrt((b * b) - (4f * a * c));
p3.X = (-b - radicand_sqrt) / (2f * a);
p4.X = (-b + radicand_sqrt) / (2f * a);
p3.Y = s * p3.X + si;
p4.Y = s * p4.X + si;
if (rect.Width == 0)
{
if (p3.Y >= rect.Y && p3.Y <= rect.Y + rect.Height) hits.Push(p3);
if (p4.Y >= rect.Y && p4.Y <= rect.Y + rect.Height) hits.Push(p4);
}
else if (rect.Height == 0)
{
if (p3.X >= rect.X && p3.X <= rect.X + rect.Width) hits.Push(p3);
if (p4.X >= rect.X && p4.X <= rect.X + rect.Width) hits.Push(p4);
}
else
{
if (rect.Contains(p3)) hits.Push(p3);
if (rect.Contains(p4)) hits.Push(p4);
}
return hits;
}
public struct Hits<T>
{
public byte Count;
public T P0, P1;
public void Push(T val)
{
if (Count == 0) { P0 = val; Count++; }
else if (Count == 1) { if (!P0.Equals(val)) { P1 = val; Count++; } }
else throw new OverflowException("Structure Hits can only fit 2 values.");
}
}

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