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khdani

Help Drawing Lines

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Second hit on Google.

Also, you tend not to specify lines as position and direction, more start and end points. You can calculate the end point from an angle with some basic trigonometry though.

EDIT: With GDI in C++, the function is LineTo. I imagine the C# / GDI+ equivalent will be similar and easy to find on MSDN / Google.

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you mean dy=dx*tan(alpha) ?
anyway i don't have the x coordinate.
I've a rectangle with its (left,top) and (right, bottom), i've the center of the rectangle. I need to draw 24 lines, one every 15 degrees, from the center to the edge of the rectangle.

i can do:
line(cx,cy,cx+cos(alpha),cy+sin(alpha))
it will make the lines which i need but they won't reach the edges of the rectangle but the perimeter of the surrounded circle. i need them to reach the edges of the rectangle.

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Quote:
Original post by khdani
you mean dy=dx*tan(alpha) ?
anyway i don't have the x coordinate.
I've a rectangle with its (left,top) and (right, bottom), i've the center of the rectangle. I need to draw 24 lines, one every 15 degrees, from the center to the edge of the rectangle.

i can do:
line(cx,cy,cx+cos(alpha),cy+sin(alpha))
it will make the lines which i need but they won't reach the edges of the rectangle but the perimeter of the surrounded circle. i need them to reach the edges of the rectangle.


C++ code, but should convert to C# easily enough:

POINT ptTopLeft{10, 20}; // Top left corner of rect
POINT ptBottomRight{50, 200}; // Bottom right corner of rect


SIZE sizeRect;
sizeRect.cx = ptBottomRight.x - ptTopLeft.x;
sizeRect.cy = ptBottomRight.y - ptTopLeft.y;
POINT ptCenter;
ptCenter.x = sizeRect.cx/2 + ptTopLeft.x;
ptCenter.y = sizeRect.cy/2 + ptTopLeft.y;

for(int nAngle=0; nAngle<=360; nAngle += 15)
{
POINT ptTarg;

// Special case for vertical lines
if(nAngle == 0)
{
ptTarg.x = ptCenter.x;
ptTarg.y = ptTopLeft.y;
}
else if(nAngle == 180)
{
ptTarg.x = ptCenter.x;
ptTarg.y = ptBottomRight.y;
}
else
{
// Find Y-intersection point with edge
int y = tan(nAngle) * sizeRect.cx/2;

// Out of bounds?
if(y > sizeRect.cy)
{
// Intersection happens on top or bottom edge
ptTarg.x = (sizeRect.cy/2) / tan(nAngle);
ptTarg.y = sizeRect.cy/2;
}
else
{
// Intersection happens on left or right edge
ptTarg.x = sizeRect.cx/2;
ptTarg.y = y;
}
}
}

That's just a very rough stab at it, I don't have time to properly test it just now. You'll also want to convert all angles to radians.

Essentially you know the X or Y value and the angle, so you can work out the missing value.

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That's C++ code...try porting it in C#

void Graphic::FillRect(RECT Rect, float radinc)
{
POINT Pt;
POINT Center;
float alfa = 0.0f;
float archeight = atan(( (Rect.bottom-Rect.top)/2 ) / ( (Rect.right-Rect.left)/2));

//point at the center of the rectangle
Center.x = LONG(Rect.left + ((Rect.right-Rect.left)/2));
Center.y = LONG(Rect.top + ((Rect.bottom-Rect.top)/2));

Pt.x = Rect.right;
for (; alfa < archeight; alfa += radinc)
{
Pt.y = LONG(floor(tan(alfa) * Rect.right));
Graphic.DrawLine(Center, Pt);
}
Pt.y = Rect.top;
for (; alfa < (Math.Pi-archeight); alfa += radinc)
{
Pt.x = LONG(floor(Rect.top / tan(alfa)));
Graphic.DrawLine(Center, Pt);
}
Pt.x = Rect.left;
for (; alfa < (Math.Pi+archeight); alfa += radinc)
{
Pt.x = LONG(floor(Rect.top / tan(alfa)));
Graphic.DrawLine(Center, Pt);
}
Pt.y = Rect.bottom;
for (; alfa < ((2*Math.Pi)-archeight); alfa += radinc)
{
Pt.x = LONG(floor(tan(alfa) * Rect.right));
Graphic.DrawLine(Center, Pt);
}
Pt.x = Rect.right;
for (; alfa < (2*Math.Pi); alfa += radinc)
{
Pt.y = LONG(floor(tan(alfa) * Rect.right));
Graphic.DrawLine(Center, Pt);
}
}

radinc is the angular increment (in radians!)

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