Re: G02 - G03
Posted by
ballendo@y...
on 2000-08-21 17:59:04 UTC
Art,
If you've done a Win9X VXD, IJK will be childs play!
BTW, would you be willing to send source of your s/w?
Reply off list to Ballendo@...
Now to IJK, etc.
The IJK are simply the distances from where you are(currently) to the
arc center. They are usually signed (this is where most of the newbie
problems are) and can be either absolute or incremental (from zero or
from current position, respectively)
Use pythagorus to get the radius of the arc. You have the legs of the
triangle as IJ,IK,JK,etc. Radius will be the hypotenuse. Using SIN,
COS with IJK you get the center point of the arc.
Now you find the starting and ending angle of the arc. Atan makes
this easy,remember that by convention, X+ (to the right) is 0
degrees,and ccw is positive, cw negative.
From here there are two paths.
Use your linear routine to incrementally step around the path of the
arc, finding XY of the next point with trig,or:
Calc the "true" points on the circle using x*x + y*y = 0.
Both ways work, most guys start with the "true" circle, then find out
using lots of little line segments is ok, IF the segments are short
enough for your users needs, and IF you can process those short
segments quickly and smoothly enough for a good finished workpiece.
Hope this helps
Ballendo
If you've done a Win9X VXD, IJK will be childs play!
BTW, would you be willing to send source of your s/w?
Reply off list to Ballendo@...
Now to IJK, etc.
The IJK are simply the distances from where you are(currently) to the
arc center. They are usually signed (this is where most of the newbie
problems are) and can be either absolute or incremental (from zero or
from current position, respectively)
Use pythagorus to get the radius of the arc. You have the legs of the
triangle as IJ,IK,JK,etc. Radius will be the hypotenuse. Using SIN,
COS with IJK you get the center point of the arc.
Now you find the starting and ending angle of the arc. Atan makes
this easy,remember that by convention, X+ (to the right) is 0
degrees,and ccw is positive, cw negative.
From here there are two paths.
Use your linear routine to incrementally step around the path of the
arc, finding XY of the next point with trig,or:
Calc the "true" points on the circle using x*x + y*y = 0.
Both ways work, most guys start with the "true" circle, then find out
using lots of little line segments is ok, IF the segments are short
enough for your users needs, and IF you can process those short
segments quickly and smoothly enough for a good finished workpiece.
Hope this helps
Ballendo
Discussion Thread
Art Fenerty
2000-08-21 10:23:46 UTC
G02 - G03
Kevin P. Martin
2000-08-21 10:51:01 UTC
G02 - G03 - an aside
Fred Smith
2000-08-21 12:18:57 UTC
Re: G02 - G03
ballendo@y...
2000-08-21 15:58:03 UTC
re:G02 - G03 - an aside
Ray
2000-08-21 17:32:50 UTC
Re: G02 - G03
ballendo@y...
2000-08-21 17:59:04 UTC
Re: G02 - G03
Alan Marconett KM6VV
2000-08-21 22:43:27 UTC
Re: G02 - G03