Subject:
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Re: Part: Plate 2 x 2 with hand rail
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Newsgroups:
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lugnet.cad
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Date:
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Thu, 1 Apr 1999 13:24:53 GMT
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Viewed:
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904 times
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On Thu, 1 Apr 1999 03:37:22 GMT, "Tim McSweeney" <tim##NO_SPAM##@ams.co.nz>
wrote:
> Ok I think I'm with you, (unfortunately I'm at work so I have to do all
> this in my head :) The result of the above transformation
> should be a circle skewed upwards into an ellipse, Seen from above it
> looks like a circle and seen form along the Z axis it looks
> like a line at 45 deg to the horizontal plane?
>
> Am I correct so far?
Right. Well, I'm not sure it is *technically* an ellipse, but otherwise,
your exactly right.
> This would be perfect if the diagonal corner arms intersected with a horizontal tube but what actually happens is that at the
> corners of the rail there is a 1/4 Torus and the Diagonal Arms intersect at the midpoint of the Torus, ie there is about 1/8th of a
> torus on either side of the arm.
Oops. Sorry.
> Damn this is hard to explain.
> Have you seen this part?
Yes.
You could do this by brute force -- generate an octer[1]-torus, then find
the segment that intersects the side-arm. Interpolate X and Z to get the
points along the actual edge of intersection, and you've got it.
Assuming your side-arms are 6LDU wide, centered, and at a 45-degree angle
(all of which looks to be right), you'd want to delete the bits where Z > X
- 3*sqrt(2).
Then reflect and rotate to use that solution 8 times in your part.
That's how I would do it.
Or get some 3D modeling software that will figure the difference for you,
and you can convert the result to DAT code.
Steve
--
1- 'Octer' is the word for 1/8, right?
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Message is in Reply To:
 | | Re: Part: Plate 2 x 2 with hand rail
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| (...) Ok I think I'm with you, (unfortunately I'm at work so I have to do all this in my head :) The result of the above transformation should be a circle skewed upwards into an ellipse, Seen from above it looks like a circle and seen form along (...) (26 years ago, 1-Apr-99, to lugnet.cad)
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