Time calculation when switching print size once again

Chan Tran · Jul 18, 2006

After going thru the other thread it made me thinking once again. What I often do is to make 4x5 prints from 35mm using the 80mm lens and when I switch to 8x10 I use the 50mm lens. Assuming that the f/stop marking on the lens is accurate and the magnification for the 4x5 prints is 4x and 8x10 prints is 8x. Also assuming that the same aperture is to be used (I actually use 1 stop larger when go to 8x10). I always use a color analyzer to determine printing time which is simple. However, I wonder how do one calculate? I think I would find the answer in the next few days but I wonder what you're all thinking?

Chan Tran · Jul 18, 2006

Well, I think the exposure time is proportional to the squared of the magnification. In this case it's simply 4x the original printing time. So I guess the neg to paper distance won't work if you change lens as in this case they are 400mm with the 80mm lens at 4x and 450mm with the 50mm lens at 8x.

Loose Gravel · Jul 18, 2006

Chan's got it exactly right. Exposure is proportional to the square of magnification and directly proportional to the area of the print. When you did the lens change, you screwed up the negative to paper distance calculations unless you want to get complicated and there's no reason for that.

Max Power · Jul 18, 2006

There's actually a PDF floating around in cyberspace which is dead simple to use. I will try to find it and post the link back here.

MTF

Kent

hortense · Jul 18, 2006

Time Calculation When Changing Column Height
Where Enlarger Column Height was = 16 in.; then moved up to = 20 in., the following formula would be used:
Excel: = ((10^2)/(8^2))*20 = 31.25 seconds
Cursive: = (102/82) x 20 = 31.25 seconds

hortense · Jul 18, 2006

Time Calculation When Changing Column Height
Where Enlarger Column Height was = 16 in.; then moved up to = 20 in., the following formula would be use:
Excel: =((10^2)/(8^2))*20 = 31.25 seconds
Cursive: (102/82) x 20 = 31.25 seconds

Jim Jones · Jul 18, 2006

Chan Tran said:
Well, I think the exposure time is proportional to the squared of the magnification. In this case it's simply 4x the original printing time. So I guess the neg to paper distance won't work if you change lens as in this case they are 400mm with the 80mm lens at 4x and 450mm with the 50mm lens at 8x.

Exposure is proportional to the square of the magnification if the lens isn't refocused. I do use the squared magnification rule for moderate changes in big enlargements. It doesn't work well for smaller magnifications.

Clueless · Jul 19, 2006

Or, just take a light level reading when through printing the small print. Duplicate "that" light-level when printing the larger print and use the same time; bypassing the calculations. Granted there are some contingencies but they can be overcome with the common solution of "Double and half".

MichaelBriggs · Jul 19, 2006

As pointed out by Jim above, and by others on previous threads (e.g., (there was a url link here which no longer exists) and (there was a url link here which no longer exists)), exposure time does NOT go as the print magnification squared. The correct rule for changing the exposure time with the same lens is time goes as (M2 + 1)^ 2 / (M1 + 1)^2. This rule takes care of refocusing the lens. (I'm not sure if it is applicable to Chan's question about switching focal lengths -- I'll have to think some.) If both magnifcations are large, the "+1" terms become unimportant, and the rule reduces the to simple rule that many people cite. The simple rule will break down if one of the prints is small compared to the negative.

As the previous threads discussed, another accurate rule is to use the image distance (lens to paper) squared. Using the column height will be approximate. Again, this is for refocusing with the same lens. This rule will not work if you switch focal lengths.

Chan Tran · Jul 19, 2006

In the previous thread I have come up with a way to figure out the lens to paper distance from the column height. Also we found way to figure out the magnification from the lens to paper column. So if your formula is correct (which I think so) we can calculate the new printing time if we know the following:
1. the original printing time
2. the column height of the original print
3. the column height of the new print
4. the focal length of the lens used to make the original print
5. the focal length of the lens used to make the new print

I want to use the column height because you can read it off the marking on the enlarger column. Measuring the lens to paper distance is a more difficult task if one wants to do it accurately.

pnance · Jul 19, 2006

Measuring the lens to paper distance is a trick, if you use a regular tape measure you have to convert fractions to decimal. I use a metric tape measure, no conversions. Works like a charm. When using 35mm, I have never wanted to make a print smaller than the negative, not even with 2-1/4 either. So if you into roll film, the simpler lens to paper distance is easier and faster, just use a metric measure.

Chan Tran · Jul 19, 2006

The problem is not metric or fractional etc... To accurately measure the distance you must have a squrare or something on the baseboard to make sure you're not measuring at an angle. You also need to know the nodal point of the lens.

MichaelBriggs · Jul 20, 2006

If the column distance gives you the distance between the negative and the paper, then you have the have the quantity d_o + d_i, when d_o is the object distance (negative to lens) and d_i is the image distance (lens to paper). But perhaps the column distance measures some arbitary point on the head. (This also neglects the separation of the principal planes of the lens, but that shoud only be a couple of mm.)

The focusing equation is 1/d_o + 1/d_i = 1/f, where f is the focal length. Using the equations for magnifcation (e.g., d_i = f(m+1)), and some algebra, one can obtain the equation:

d_o + d_i =f/m (m+1)^2 = f (m + 2 + 1/m)

So from the column height, hopefully a measurement of d_o + d_i, you could solve the above equation for m for both print sizes, then use the (m+1)^2 equation to calculate a new exposure time. Obviously the above equation is not so easy to solve.

Possible approaches: 1) measure the lens position and not bother with the above equation, 2) program a scientific calculator or computer to solve the equation, 3) measure the film and prints to obtain m, 4) use a light meter, or 5) test strips.

To measure the lens distance accurately, you could just mark the optical center line on the baseboard and extend a tape measure from the lens to that point. I would't worry about the nodal points of the lens -- measuring to the aperture or middle of the lens should be sufficiently accurate.

pnance · Jul 20, 2006

My testing has shown if you measure from the same reference points, it doesn't make that much of a difference. If you use the markings on the column, and you change the easel, and its thicker, or thiner, you have a problem. If you just consistently measure from the paper position to the same spot on the lens I find the exposure is close enough for me.

Jerevan · Jul 20, 2006

Max Power said:
There's actually a PDF floating around in cyberspace which is dead simple to use. I will try to find it and post the link back here.

MTF

Kent

Is it this one? -> http://www.bonavolta.ch/hobby/files/ratiomod.pdf

RalphLambrecht · Jul 20, 2006

Chan Tran

I thought we went through this already. The inverse-square law is based on the lens-to-paper distance, not the magification.

RalphLambrecht · Jul 20, 2006

MichaelBriggs said:
As pointed out by Jim above, and by others on previous threads (e.g., (there was a url link here which no longer exists) and (there was a url link here which no longer exists)), exposure time does NOT go as the print magnification squared. The correct rule for changing the exposure time with the same lens is time goes as (M2 + 1)^ 2 / (M1 + 1)^2. This rule takes care of refocusing the lens. (I'm not sure if it is applicable to Chan's question about switching focal lengths -- I'll have to think some.) ...snip

It does. Your equation works with the magnification factor, consequently, you can switch the focal length. It won't matter.

dancqu · Jul 20, 2006

RalphLambrecht said:
The inverse-square law is based on the
lens-to-paper distance...

Did you forget the lens to negative distance? Any
computation which does not factor in the change in
lens to negative distance will not be correct. Have
any of those interested in this subject found or
derived A formula integrating the two? Dan

RalphLambrecht · Jul 20, 2006

You don't need the lens-to-negative distance to calculate the change in exposure. Using the lens-to-paper distance alone will yield accurate results.

dancqu · Jul 20, 2006

RalphLambrecht said:
You don't need the lens-to-negative distance to
calculate the change in exposure.

You can not ignore the change in lens to negative
distance when it is that distance which determines
the speed of the lens. You do recall the formula by
which the speed of a lens is determined? Dan

Chan Tran · Jul 20, 2006

Thanks Ralph and MichealBriggs! I think I got it now. The formula by Micheal would work for all focal lengths. Now I am going to work on my enlarger on the procedure to calculate exposure time based on column height and focal length change. I think one more thing I have to address is the neg to paper distance is the sum of neg to rear nodal point, front nodal point to paper and the distance between the two nodal points as I think most lenses do not have these two points the same.

MichaelBriggs · Jul 20, 2006

dancqu said:
Did you forget the lens to negative distance? Any
computation which does not factor in the change in
lens to negative distance will not be correct. Have
any of those interested in this subject found or
derived A formula integrating the two? Dan

As mentioned previously ((there was a url link here which no longer exists)), d_i = focal length * (m + 1), so the (m+1) squared equation and the image distance squared equation are equivalent. Both are correct for changing print size with the same lens.

Helen B · Jul 20, 2006

Chan Tran said:
... I think one more thing I have to address is the neg to paper distance is the sum of neg to rear nodal point, front nodal point to paper and the distance between the two nodal points as I think most lenses do not have these two points the same.

As you have the lenses you can satisfy yourself of the importance or unimportance of allowing for the location of the nodal points: just use the lenses both ways round to form an image of a distant object and measure the nominal focal length from the image plane in each case.

Best,
Helen

dancqu · Jul 21, 2006

MichaelBriggs said:
As mentioned previously...

As I've mentioned two or three times previously
and your self as well as many times, the change in
the effective speed of the lens is NOT to be ignored.
By your example I see it can make a BIG difference.

For myself the matter is academic. I use an
EM-10 which, BTW, I've calibrated for use as
a densitometer. Dan

RalphLambrecht · Jul 21, 2006

I agree, but I'm working on the assumption that this is taken care of by the lens manufacturer through the aperture markings. You got to rely on something.

Time calculation when switching print size once again

Privacy & Transparency

Privacy & Transparency