Bellows factor conversion method????

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RalphLambrecht

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Oh, let's see. 13.5 is about 1/3 stop more than 11, and 18.0 is about 1/3 stop more than 16. I'll bet one stop would work just fine. Spoiling the fun you all want to have with your gizmos. I know.

I'm not clever enough to follow your math but my gizmo tells me about 0.8 stops.
 

tomalophicon

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Hi, Mike.

That's right. The smaller number is the FL of the lens at infinity. The large number will be the extended distance measured from the film plane to the same point on the lens that corresponded to the FL when the lens was focused at infinity. This will usually be at the plane of the aperture.

(450/210)² = easy

With telephoto lenses, you are better off using magnification as a guide by using an object of a known size (such as a ruler) in the shot and measuring it on the ground glass, then using those two numbers to do the math.


2F/2F,

When you multiply the shutter speed, are you multiplying it as a fraction?

I tried it as eg. 4.6 x 1/125 and got .0368. Does that mean 1/30th would be the most logical shutter speed? I do the same for 1/500th and get 0.93 or something. Would that mean the closest speed would be 1/125th?
Please correct me if I'm doing the conversion incorrectly...
Tom.
 

RalphLambrecht

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Tom

You are better off calculating (or measuring) the exposure correction in f/stops first. Then you have a choice to correct with aperture or shutter speed, or a combination of both. With most analog cameras, you are limited to set the shutter speed in full stops, however. So, correcting with shutter speed alone rarely works.
 

tomalophicon

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Tom, another method to try is to figure the factor and then divide your film speed by that factor. For example, if you use a 150mm lens drawn out to 300mm then (300/150)²=4, then (ISO/4)=new ISO. If your film is TMY-2 and you rate it at 400, divide that by your factor: 400/4=100. Measure your scene at 100 to get the proper exposure.

Cool, that makes sense.
 
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