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# Thread: Bellows factor conversion method????

1. Originally Posted by ralnphot
Ralph, I made two of your calculators and laminated them. One lives in my LF gear bag, my daughter grabbed the other. This is the handiest no brainer accessories I own. Thank you.
Thanks.

The ruler works best with large-format cameras but I have used it successfully with my Hasselblad. I need to take the hood off, but then, I can easily use the ruler to measure the target.

2. Mr. Adams explains Lens Extension Factor (bellows factor) on pages 68 & 69 in his book "The Camera". If you decide to add this book to your library, then you may want to pick-up "The Negative" and "The Print" while you're at it, to make a set. These books can be purchased used and fairly inexpensively through sites like Barnes & Noble and/or eBay. IMO all three are must-haves for a photographer's bookshelf.

3. Ditto for the ruler. I sat down with a calculator one day and figured the factors backwards by 1/2 stops so I would know the amount of extension. I marked the factors for each of my lenses on a small retractable metal tape measure that I attached to the strap of my light meter. It's easy to measure from the lens to the film plane and immediately have the compensation in f/stops.

http://www.jeffreyglasser.com/

4. For those who don't want to carry a calculator or a slide rule into the field with them, the original poster's method is spot on. 2 stops.

5. How about a 135 mm lens and a 180 mm draw? I think the ruler weighs 10 g, less than any calculator, but about as much as the piece of paper to remember the equation or any mysterious shortcut.

6. Originally Posted by RalphLambrecht
How about a 135 mm lens and a 180 mm draw? I think the ruler weighs 10 g, less than any calculator, but about as much as the piece of paper to remember the equation or any mysterious shortcut.
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.

7. Originally Posted by jimgalli
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.

8. I have made a copy of Ralph's no-brainer calculator and the is no math involved. I must say that the f/stop concept is the easiest of the math methods.

9. Originally Posted by 2F/2F
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.

10. 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.

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