From the top sticky.
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Warning! Snoozer follows!
Now can we calculate? Oh, goodie! On a short scope, the objective's focal length must be around 0.1 m considering that there is an erector lens in that tube also. The formula for the distances from a lens of the object and the image of that lens is:
O^-1 + I^-1 = F^-1
O = distance from object to lens
I = distance from image to lens
F = focal length of lens
What I'd like to know is how far we'd have to bring the objective lens in if we shift the parallax correction from 50 m to 100 m. Moving the objective lens relative to the scope body makes no essential change in the value of the variable, O. So how far is the image from the lens when the target is at
50 m? 100 m? 150 m?
I(50) = [(F^-1)- (O^-1)]^-1 = [(.1^-1)-(50^-1)]^-1 = .1002 m
I(100) = [(.1^-1)-(100^-1)]^-1 = .1001 m
I(150) = [(.1^-1)-(150^-1)]^-1 = .10007 m
We can now see that we're talking very small parallax correction movements here and that furthermore, the corrective movement required for an increment in target distance decreases rapidly as the distance to the target increases.
So the answer to my question is, if you move the target from a 50 m distance to a 100 m distance, the objective must be moved .1002-.1001= .0001 m to correct the parallax. In Marekin terms, this is .004". That sounds about right to me considering that the graduations on an objective bell are fairly close together and the objective bell's thread is very fine. This also explains
why it is difficult for the scope manufacturer to put the parallax marks on
the bell in exactly the right place. All eyes are closed? Have a nice sleep!
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I am not a machinist, but it appears that having the marks in calibration would be dependent upon getting the thread on the scope tube and on the AO ring to be started at exactly the same spot every time and for the thread cut to have exactly the same tolerances.