Scope formula???


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Rpriestlyjr
February 4, 2006, 07:21 PM
I'm not too sure on this, but I was at a local gunshop BS'ing and some guy tells me and the owner of a simple mathematical formula for scopes.
Apparently, the size of the objective and the power of the scope, somehow multiplied by 5 (or something like that) will let you know if the scope is what is needed.
This was new to everyone in the shop.
Is there some truth to this?
If so, how does this work?

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rbernie
February 4, 2006, 07:53 PM
Sounds like he's trying to calculate the exit pupil (http://www.celestron.com/education/binbasic.htm), to determine objective lens size.... It's commonly accepted (translated to mean that it's what I've been told) that the human eye needs an exit pupil of at least 5mm for low-light visibility; you calculate this by dividing the objective lense size by the highest magnification.

For example, a 3x9/40 would have an exit pupil of 13.3mm at 3x and an exit pupil of 4.4mm at 9x. This menas that this scope would be fine in daylight or on lower magnifications, but would not potentially be as good a light-catcher in twilight on high magification as would a scope with a larger objective lens, like say a 3x9/50.

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