Forums
New posts
Search forums
What's new
Articles
Latest reviews
Author list
Classifieds
Log in
Register
What's new
Search
Search
Search titles and first posts only
Search titles only
By:
New posts
Search forums
Menu
Log in
Register
Install the app
Install
Forums
Rifles, Reloading, Optics, Equipment
Long Range Scopes and Other Optics
Accuracy of ParallexMarks on SF or AO
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Reply to thread
Message
<blockquote data-quote="Buffalobob" data-source="post: 99067" data-attributes="member: 8"><p>From the top sticky. </p><p></p><p>[ QUOTE ]</p><p> Warning! Snoozer follows!</p><p></p><p>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:</p><p>O^-1 + I^-1 = F^-1</p><p>where:</p><p>O = distance from object to lens</p><p>I = distance from image to lens</p><p>F = focal length of lens</p><p></p><p>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</p><p>50 m? 100 m? 150 m?</p><p></p><p>I(50) = [(F^-1)- (O^-1)]^-1 = [(.1^-1)-(50^-1)]^-1 = .1002 m</p><p>I(100) = [(.1^-1)-(100^-1)]^-1 = .1001 m</p><p>I(150) = [(.1^-1)-(150^-1)]^-1 = .10007 m</p><p></p><p>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.</p><p>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</p><p>why it is difficult for the scope manufacturer to put the parallax marks on</p><p>the bell in exactly the right place. All eyes are closed? Have a nice sleep!</p><p></p><p><a href="mailto:JHBercovitz@lbl.gov">JHBercovitz@lbl.gov</a> (John Bercovitz)</p><p></p><p> </p><p></p><p>[/ QUOTE ] </p><p></p><p></p><p></p><p>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.</p><p></p><p>I</p></blockquote><p></p>
[QUOTE="Buffalobob, post: 99067, member: 8"] From the top sticky. [ QUOTE ] 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 where: 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! [email="JHBercovitz@lbl.gov"]JHBercovitz@lbl.gov[/email] (John Bercovitz) [/ QUOTE ] 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. I [/QUOTE]
Insert quotes…
Verification
Post reply
Forums
Rifles, Reloading, Optics, Equipment
Long Range Scopes and Other Optics
Accuracy of ParallexMarks on SF or AO
Top