Fwd: Iterative Solution

[Ian McCallion]

should have copied help-octave
---------- Forwarded message ---------
From: Ian McCallion <address@hidden>
Date: Tue, 28 Aug 2018 at 09:05
Subject: Re: Iterative Solution
To: <address@hidden>


This is simple geometry. Take the line between the centres at the
desired point of the tool. Starting with this line as the hypotenuse
complete a right-angled triangle using horizontal and vertical lines.
Apply Pythagoras theorem.

## 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 [y1] = dean()
   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

   hyplen=1.75+.125
   sidexlen=1.75-.74
   sideylen=sqrt(hyplen^2-sidexlen^2)
   y1=y0-sideylen
   ## maple answer y1 == 1.420276923
endfunction;

Cheers... Ian
On Tue, 28 Aug 2018 at 07:18, Thomas D. Dean <address@hidden> wrote:
>
> 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;
>
>