I used my jig and projected a test slide on the wall from
a fixed distance with each of my five lenses. I marked (and
then later measured) the the four corner points of a 50x50mm
square and the vertical and horizontal distances across the edges
of the square.
| Lens No. | Serial No. | Condition | Diag.1/Diag.2 | Sum Diags | Horiz./Vert. | Sum H+V | Relative FLs (diag) | Relative FLs (h+v) | Distortion* |
| 1 | 35915 | EX, f3.2 viewer | 619.0/618.0 mm | 1237.0 | 436.5/433.0 | 869.5 | Reference lens (0.0%) | Reference lens (0.0%) | 0.60% |
| 2 | 29432 | EX, f3.2 viewer | 622.0/621.5 mm | 1243.5 | 438.0/434.5 | 872.5 | 0.53% | 0.34% | 0.78% |
| 3 | 21316 | EX, f3.5 viewer | 621.5/620.0 mm | 1241.5 | 437.5/435.5 | 873.0 | 0.36% | 0.23% | 0.56% |
| 4 | 14469 | rear element separation | 626.0/625.0 mm | 1251.0 | 440.5/438.0 | 878.5 | 1.13% | 0.92% | 0.69% |
| 5 | 26372 | front coating scratched | 620.5/619.5 mm | 1240.0 | not avail./435.5 | 435.5 (V only) | 0.24% | 0.58% (based on Vert. only) | 0.67% (based on Vert. only) |
* Distortion is the amount the diagonal is longer than you would expect from Diag^2 = H^2 + V^2.Marking and measuring the distances is probably accurate to with 1mm (+/- 0.5 mm). Thus diagonal measurements are accurate to 0.08% and comparisons between diagonals are accurate to 0.2% or so. Thus I am fairly confident that the top 3 lenses are close matches. The measurements of both diagonals and the horizontal and verticals increases the confidence since they agree to with the 0.2% error.
The distortion measurments are most error-prone since you are basically subtracting two measurements. Thus their error rate is likely higher.
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Email: greg.erker@home.com
