1 stop would be at 212 (and a bit) mm.
perhaps he meant 225mm
As the lens gets further from the film the light passing though the lens is spread over a wider area and proportionately less falls on any given area. The proportion is based on the square of the change in distance. Thus if the lens were twice as far away the light would be diffused over an area twice as high and twice as wide so the adjustment would be based on one quarter of the light hitting any area of film and the adjustment would be 2 f stops.
I .tried measuring my bellows extension for this shot the bellows from the film plane to the lensboard is 16". My lens is 180mm. If I square the extension and divid by the square of my lens length I come up with 5 stop bellows factor??? Is this right?
Obviously this is a very close up shot. This is the total extension of the bellows.
that's the bellows FACTOR...not the stops--
FACTOR is used to multiply exposure TIME...
you want STOPS, what you really want is to know how many doublings it takes to get to that number
so if you get 5 for the FACTOR, you want to solve
2^n = 5 (where n=the Number of stops)
take logarithm of both sides: n*log2 = log5--this gives n = log5/log2 = 2.3
so you need 2 and one third stops compensation for this setup
The Kodak Master Photoguide has a very comprehensive f number computer and simple way of determining exposure correction I have one taped to the back of my view camera glass cover.That and a pocket tape measure and you can figure the correction in seconds with out all the math. If I have a 6 inch lens 8 inches from the film plane my exposure should be increased by one stop.
I did the math and keep a copy in my bag
FILTER FACTOR & STOPS
the format did not hold up.
The FACTOR should read as a number and decimal point, IE: factor of 1 is 1.0, 1-1/2 is 1.5.
The STOP numbers are one decimal point also, IE: STOP of 1 is 0, STOP of 1.5 is 0.6
Sorry the column spacing between the numbers didn't work.
well....that's basically a table of logaritms base 2....I prefer to just do something I call the "effective f stop"
you measure the lens extension then you divide by the lens focal length (the infinity "marked" focal length)
this give you a multiplier for the fstops on the lens...for instance if you have an extension of 140mm and the lens is 100mm then your muliplier is 1.4...so f5.6 as marked on the lens is now an "effective fstop" of 5.6*1.4 = 8
so I call it "f8 effective" when the lens is set at f5.6 at that bellows extension....