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Nightforce mirage insert

Ajwilly96

Gold $$ Contributor
I’m breaking out the old 308 that I haven’t shot in several years. I shot with a vortex viper and decided to swap it out for a nightforce comp I had laying around. At 100yds I was able to see a ton of mirage

I am unsure of the Mirage cap purpose, but if I were to use one would it cut the mirage at shorter distances?
 
Did you try to turn the NF down until the mirage went away?
The Vipers max power is 25 and the NF is 55, you have a lot of room
to play with.
 
Did you try to turn the NF down until the mirage went away?
The Vipers max power is 25 and the NF is 55, you have a lot of room
to play with.
I did not, I had it set at 40X. I was more curious if I could have the power turned up and cut the mirage at short range
 
You can use painters tape and make a temp one, to see if it works for you. just leave a hole about an inch or so square in the center.( of course in the center)
 
I used mine for the 1st time last weekend and it really helped. It cut the mirage down just by installing it, then even more when refocused. I'm just going to leave it in there all the time now.
 
For example, what does the center hole diameter measure on these aperture reducers?
In photography, the concept of increasing or decreasing relies on specific values. The unit is called a f-number and it's a ratio, not a unit of measure. The maximum f-number (f for focal) of a lens is its focal length divided by the diameter of the objective lens. When we want to control the amount of light, the equivalent of 1 f-number, what we call an f-stop is equivalent to doubling the amount of light or cutting it in half. For exposure purposes, 1 f-stop is also the equivalent of doubling or halving the exposure time (shutter speed). Of course, there are intermediary steps to both. So you have half f-stops or even third f-stop.

But what about riflescopes? Well, it's the same principle but we can be a little sloppy. The equivalent of 1 f-stop is about the size of a change that can be reliably discerned by the human eye, so we want to be able to reduce the light by 50%.

The size of the reduced aperture has to represent 50% of the size of the objective lens.
If the objective lens if 56mm in diameter, that means its area is (56/2)^2 * pi() => 28*28*3.14159 => 2463 mm2. Half of that is 1231 mm2. 1231/pi() = 392 mm2. Take the square root of 392 => 19.8 mm. That is the radius of the reduced hole, call it 20mm, or 40mm diameter.
 
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