RE: [CAD_CAM_EDM_DRO] Re: 7 square holes --was,lost post
Posted by
Kevin P. Martin
on 2000-12-14 08:15:34 UTC
The formula is a simple problem in spherical trigonometry (simple, that is, once
you can visualize all the angles).
Imagine you've already cut part of the inside corner. The tip of the cutter is
at O, the origin for all the spherical angles.
Now imagine the spherical triangle ABC defined by the directions:
One corner of the spherical triangle A is back along the feed direction, along
the cut inside corner.
One corner B is along the axis of the cutter spindle.
The third corner C is in the direction of the point where the cutter is
currently cutting the final surface. Because of the geometry of milling
processes, the corner angle at C is 90 degrees so you have a right spherical
triangle.
The angle at A is half of the inside corner angle you want to produce.
The *sides* of the spherical triangle are:
a (opposite corner A) is half the cutter's tip angle,
c (opposite corner C) is the tilt angle of your milling spindle
b doesn't mean much useful
The solution for such a triangle is: sin a = sin A sin c.
In your example, you want a 90 degree inside corner, so A is 45 degrees and sin
A is sqrt(1/2).
If, for instance, you tilt your spindle 30 degrees, A is 30 degrees, sin A is
1/2, so sin a = sin A sin c = sqrt(1/8), and c = 20.7 degrees, and your cutter
must have a total tip angle twice this, or 41.4 degrees.
If you tilt your spindle 45 degrees, sin a = 1/2 and a = 30 degrees, so you need
a 60 degree cutter, which you might actually be able to buy.
As an extreme case, if you tilt your spindle 90 degrees, a = 45 degrees, and you
need a 90 degree cutter. This is just the usual geometry for using a 90 deg.
cutter to cut a 90 degree vee groove.
-Kevin Martin
you can visualize all the angles).
Imagine you've already cut part of the inside corner. The tip of the cutter is
at O, the origin for all the spherical angles.
Now imagine the spherical triangle ABC defined by the directions:
One corner of the spherical triangle A is back along the feed direction, along
the cut inside corner.
One corner B is along the axis of the cutter spindle.
The third corner C is in the direction of the point where the cutter is
currently cutting the final surface. Because of the geometry of milling
processes, the corner angle at C is 90 degrees so you have a right spherical
triangle.
The angle at A is half of the inside corner angle you want to produce.
The *sides* of the spherical triangle are:
a (opposite corner A) is half the cutter's tip angle,
c (opposite corner C) is the tilt angle of your milling spindle
b doesn't mean much useful
The solution for such a triangle is: sin a = sin A sin c.
In your example, you want a 90 degree inside corner, so A is 45 degrees and sin
A is sqrt(1/2).
If, for instance, you tilt your spindle 30 degrees, A is 30 degrees, sin A is
1/2, so sin a = sin A sin c = sqrt(1/8), and c = 20.7 degrees, and your cutter
must have a total tip angle twice this, or 41.4 degrees.
If you tilt your spindle 45 degrees, sin a = 1/2 and a = 30 degrees, so you need
a 60 degree cutter, which you might actually be able to buy.
As an extreme case, if you tilt your spindle 90 degrees, a = 45 degrees, and you
need a 90 degree cutter. This is just the usual geometry for using a 90 deg.
cutter to cut a 90 degree vee groove.
-Kevin Martin
Discussion Thread
ballendo@y...
2000-12-12 05:59:49 UTC
Help! I just lost another long, detailed post
Smoke
2000-12-12 07:11:30 UTC
Re: [CAD_CAM_EDM_DRO] Help! I just lost another long, detailed post
Jeff Barlow
2000-12-12 11:02:56 UTC
Re: [CAD_CAM_EDM_DRO] Help! I just lost another long, detailed post
Brian Pitt
2000-12-12 11:37:03 UTC
Re: [CAD_CAM_EDM_DRO] Help! I just lost another long, detailed post
Fred Smith
2000-12-12 12:04:58 UTC
Re: Help! I just lost another long, detailed post
ballendo@y...
2000-12-12 13:47:31 UTC
re:Help! I just lost another long, detailed post
Tim Goldstein
2000-12-12 13:54:19 UTC
RE: [CAD_CAM_EDM_DRO] re:Help! I just lost another long, detailed post
Smoke
2000-12-12 14:38:53 UTC
Re: [CAD_CAM_EDM_DRO] re:Help! I just lost another long, detailed post
Jon Elson
2000-12-12 16:10:57 UTC
Re: [CAD_CAM_EDM_DRO] Help! I just lost another long, detailed post
Brian Pitt
2000-12-12 22:22:06 UTC
Re: [CAD_CAM_EDM_DRO] re:Help! I just lost another long, detailed post
Brian Pitt
2000-12-12 23:33:39 UTC
7 square holes --was,lost post
Smoke
2000-12-13 00:08:55 UTC
Re: [CAD_CAM_EDM_DRO] 7 square holes --was,lost post
Brian Pitt
2000-12-13 00:41:19 UTC
Re: [CAD_CAM_EDM_DRO] 7 square holes --was,lost post
Matt Shaver
2000-12-13 05:37:29 UTC
Re: [CAD_CAM_EDM_DRO] 7 square holes --was,lost post
Jon Anderson
2000-12-13 09:23:32 UTC
Re: [CAD_CAM_EDM_DRO] 7 square holes --was,lost post
Smoke
2000-12-13 09:58:01 UTC
Re: [CAD_CAM_EDM_DRO] 7 square holes --was,lost post
Greg Nuspel
2000-12-13 10:04:37 UTC
Re: [CAD_CAM_EDM_DRO] 7 square holes --was,lost post
Smoke
2000-12-13 10:26:24 UTC
Re: [CAD_CAM_EDM_DRO] 7 square holes --was,lost post
Brian Pitt
2000-12-13 11:24:21 UTC
Re: [CAD_CAM_EDM_DRO] 7 square holes -square drills
Fred Smith
2000-12-13 12:13:01 UTC
Re: 7 square holes --was,lost post
Smoke
2000-12-13 14:12:42 UTC
Re: [CAD_CAM_EDM_DRO] Re: 7 square holes --was,lost post
Fred Smith
2000-12-13 16:20:08 UTC
Re: 7 square holes --was,lost post
Smoke
2000-12-13 20:36:48 UTC
Re: [CAD_CAM_EDM_DRO] Re: 7 square holes --was,lost post
Fred Smith
2000-12-14 06:25:08 UTC
Re: 7 square holes --was,lost post
Smoke
2000-12-14 07:41:50 UTC
Re: [CAD_CAM_EDM_DRO] Re: 7 square holes --was,lost post
Kevin P. Martin
2000-12-14 08:15:34 UTC
RE: [CAD_CAM_EDM_DRO] Re: 7 square holes --was,lost post
Smoke
2000-12-14 10:55:32 UTC
Re: [CAD_CAM_EDM_DRO] Re: 7 square holes --was,lost post