Projection Lens

[seawolf66]

Is There a simple way to determine the F-stop of a projection lens
its 6-1/2 EF label on the lens CINDO that's all , but it came with its own lens shade
Thanks for any help on this:

[Mark Sawyer]

Measure the diameter of the aperture in inches through the front of the lens. Divide 6.5 by that number, and that's your f/stop.

Example: If the aperture is 2 inches in diameter, then 6.5/2 = 3.25. You have an f/3.25 lens when focused at infinity.

[seawolf66]

Measure the diameter of the aperture in inches through the front of the lens. Divide 6.5 by that number, and that's your f/stop.

Example: If the aperture is 2 inches in diameter, then 6.5/2 = 3.25. You have an f/3.25 lens when focused at infinity.

Thank you very much just my life simpler : Lauren

[seawolf66]

Mark I tried that formula and 6.5 x 2 = equals 13. on my calculator ??
and since my has a opening of 1,5 inch's x 6.5 = 9.75

I am now lost :

[imagedowser]

I think Mark's word was divide not multiply... divide 2 into 6.5= 3.25 Be Well, Bill

[seawolf66]

I think Mark's word was divide not multiply... divide 2 into 6.5= 3.25 Be Well, Bill

Thank for your input here;

[Richard Rankin]

A lot of the cine and projection lenses have an internal ring that restricts the inside diameter of the barrel, and also makes them slower, to sharpen them up for their original purposes. So, if you have an internal ring smaller than the front or rear glass, you should use that diameter to calculate the f-stop.

Richard

[seawolf66]

A lot of the cine and projection lenses have an internal ring that restricts the inside diameter of the barrel, and also makes them slower, to sharpen them up for their original purposes. So, if you have an internal ring smaller than the front or rear glass, you should use that diameter to calculate the f-stop.

Richard

Richard I Thank you for your input on this matter it turns out there are two rings inside as you said but in this case the lenses are the same in this case
I took one lens out to check it out and it was great news for me: Thanks again: