Iterative Solution

[Thomas D. Dean]

I was watching a machinist make a part.  The portion of the part I was 
interested in consisted of a circle and a line.  The machinist wanted to 
locate a tool such that it was tangent to the line and the circle.

I used an iterative method to calculate the location of the center of 
the tool circle.  This took 71 iterations.  Must be a better method.

Tom Dean

## given a circle, C1,
##     center (3+1/2)/2, (3+1/2)/2+1+1/4, radius (3+1/2)/2
## and a line, L1,
##     x=(3+1/2)/2 - (3/4+20/1000)/2 - 1/2,
## find a circle, CT1, of radius 1/8 such that it is tangent to both
## L1 and C1.  CT1 is outside C1 and between L1 and the origin.

## The x value of center(CT1) is
##      (3+1/2)/2 - (3/4+20/1000)/2 - 1/2 - 1/8.

## The y value of center(CT1) is that value that results in the
## distance from center(C1) to center(CT1) is equal to
##      radius(C1) + radius(CT1)
## Start with y = 0 and increase y until
##    d = distance( center(C1) - center(CT1) )
##      = sqrt( ((3+1/2)/2 - 1/8)^2 + ((3+1/2)/2+1+1/4 - x_value)^2)
##    d - ( (3+1/2)/2 + 1/8 ) = 0

function [x1, y1] = find_loc()
  x0 = (3+1/2)/2; y0 = (3+1/2)/2+1+1/4; r0 = (3+1/2)/2;
  x1 = (3+1/2)/2 - (3/4+20/1000)/2 - 1/2 - 1/8; y1 = 0; r1 = 1/8;
  ## maple answer y1 == 1.420276923
  dy = 1;
  d = sqrt( (x0 - x1)^2 + (y0 - y1)^2);
  err = d - (r0 + r1);
  s = sign(err);

  iters = 1;

  while (abs(err) > 1e-15)
    if (sign(err) == s);
      y1 += s*dy;
    else
      s = sign(err);
      dy /= 10;
      y1 += s*dy;
    endif
    d = sqrt( (x0 - x1)^2 + (y0 - y1)^2);
    err = d - (r0 + r1);
    iters++;
  endwhile

  printf("iterations %d\n", iters);

endfunction;