Allen's alignement points are 57.8mm and 129mm

Posted by Mark Kelly (M) on September 30, 2007 at 15:55:16

In Reply to: Pinging Allen Wright - your "Cartridge Alignment Guide"? ... posted by andyr on September 30, 2007 at 05:23:50:

If you recalculate the alignment points from Allen's data they come out to 57.8mm and 129mm.

I checked this with Brian Kearns (who knows more about alignment than anyone else I know). Brian corrected my calculation and sent me a spreadsheet that generates the new alignment points for any set back. His method involves calculating a null point difference which is subtracted from the inner null point and added to the outer. The difference length is given by

D = - (n2-n1)/2 + SQRT (((n2-n1)/2)^2 - s^2 + 2s.SQRT(Le^2 - ((n1+n2)/2)^2)) Where n1 and n2 are the original null radii, Le is the original effective length and s is the setback length. For an arm of 225mm length the delta is 8.2mm when setback is 1.27mm so the null points are 57.8mm and 129.1mm.

Mark Kelly

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Follow Ups Full Thread

Topic - Pinging Allen Wright - your "Cartridge Alignment Guide"? ... - andyr 05:23:50 09/30/07 (7)

Allen's alignement points are 57.8mm and 129mm - Mark Kelly 15:55:16 09/30/07 (2)

RE: Allen's alignement points are 57.8mm and 129mm - John Elison 21:15:24 09/30/07 (0)

Thanks, Mark. :-)) nt - andyr 16:16:51 09/30/07 (0)

Very Interesting Post! - John Elison 11:51:45 09/30/07 (1)

Thanks, John ... - andyr 14:30:55 09/30/07 (0)

Baerwald Nulls - jsusky 11:36:06 09/30/07 (1)

RE: Baerwald Nulls - John Elison 12:13:33 09/30/07 (0)

Follow Ups

Allen's alignement points are 57.8mm and 129mm - Mark Kelly 15:55:16 09/30/07 (2)

RE: Allen's alignement points are 57.8mm and 129mm - John Elison 21:15:24 09/30/07 (0)

Thanks, Mark. :-)) nt - andyr 16:16:51 09/30/07 (0)

Allen's alignement points are 57.8mm and 129mm

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