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In Reply to: Allen's alignement points are 57.8mm and 129mm posted by Mark Kelly on September 30, 2007 at 15:55:16:
Very nice equation. Brian is a true genius. He taught me much of what I know about tonearm alignment geometry.I attacked the problem a bit differently, and when I finally figured out how to do it properly I came up with similar numbers to Brian's equation. I used the law of cosines to analyze the pivot-stylus-spindle triangle for the tracking error at the new alignment points based on the 1.27-mm set-back from Baerwald's geometry. I also used a different effective length based on measurements from Allen's protractor.
I'm rather curious as to why you settled on 225-mm for your effective length. I measured the print-out of Allen's protractor and came up with the following parameters.
Then I calculated the tonearm effective length based on the angle between the grids for Baerwald's null-points.
Based on these measurements and calculatons, I decided to take an approximate average effective length of 232-mm for Baerwald's alignment, which yielded a pivot-to-spindle distance of 214.1154-mm.
Since Brian's equation is a function of effective length, I'm wondering why you decided to choose 225-mm for your effective length when the angle between the grids suggest a 232-mm effective length. Your calculations should be based on a mounting distance of 214.1154-mm for Baerwald's alignment, which translates into an effective length for the set-back alignment of 230.8-mm.
Anyway, here is a graph of Allen's Guru's Set-back alignment.
As a result, I think if you correct your effective length based on the angle of the grids for Allen's protractor, we will have very similar alignment null-points.
Best regards,
John Elison
Edits: 09/30/07
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Follow Ups
- RE: Allen's alignement points are 57.8mm and 129mm - John Elison 21:15:24 09/30/07 (0)