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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
- 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)