fillet betwenn two lines

The perpendicular distance from a point(x, y) to a line y = a.x is (a.x-y)/sqrt(1+a**2).

If we have two lines, y = a.x and y = b.x, and distance r, we have two equations:

r = (a.x-y)/sqrt(1+a**2);

r = (b.x-y)/sqrt(1+b**2)

and solve for both x and y, giving:

x = r.(sqrt(1+a**2) - sqrt(1+b**2)) / (a - b);

y = a.x - r.sqrt(1+a**2);

this has four solutions, as can be seen from:

import java.awt.*;

import java.applet.*;

public class Fillets extends Applet {

double a = 0.5;

double b = 2.0;

double r = 40.0;

public void init () {

new Thread (new Runnable() {

public void run () {

try {

while(true) {

synchronized(this) {

Thread.sleep(250);

a+=0.2;

b-=0.1;

repaint();

}

}

} catch (InterruptedException ie) {

return;

}

}

}).start();

}

public void paint (Graphics g) {

int ox = getWidth()/2;

int oy = getHeight()/2;

g.setColor(Color.darkGray);

g.drawLine(0, oy, 2*ox, oy);

g.drawLine(ox, 0, ox, 2*oy);

g.setColor(Color.blue);

g.drawLine(0, oy - (int)(-ox*a), 2*ox, oy - (int)(ox*a));

g.setColor(Color.green);

g.drawLine(0, oy - (int)(-ox*b), 2*ox, oy - (int)(ox*b));

// maths bit here

double sqrt_1_plus_a2 = Math.sqrt(1.0+a*a);

double sqrt_1_plus_b2 = Math.sqrt(1.0+b*b);

// no solution if a==b

if (Math.abs(a-b)>1e-6) {

double x;

double y;

g.setColor(Color.cyan);

for (int sgna = 0; sgna<2; sgna++) {

for (int sgnb = 0; sgnb<2; sgnb++) {

// math bit here

x = r * (sqrt_1_plus_a2 - sqrt_1_plus_b2) / (a - b);

y = a * x - r * sqrt_1_plus_a2;

g.drawLine(ox + (int) x - 4, oy - (int) y,

ox + (int) x + 4, oy - (int) y);

g.drawLine(ox + (int) x, oy - (int) y - 4,

ox + (int) x, oy - (int) y + 4);

g.drawOval(ox + (int) (x - r), oy - (int) (y + r),

2*(int)r, 2*(int)r);

sqrt_1_plus_a2 = -sqrt_1_plus_a2;

}

sqrt_1_plus_b2 = -sqrt_1_plus_b2;

}

}

}

}

Pete

Pete_Kirkhama at 2007-7-15 0:26:21 >