# Change in Enlarger Head Height corresponds to Change in Exposure Time, but by how much?

Discussion in 'Darkroom Developing and Printing' started by Gregory W Blank, Jul 20, 2004.

1. ### Gregory W BlankGuest

My experience tells me it doubles as you double the size, yet it is approximate.
In other words.....going from an 8x10 enlargement exposed at 10 seconds
to a 16x20 would give you an exposure time around 20-22 seconds at the same aperture.
One factor you may encounter is that a larger print seems to require a little more contrast
filteration than a smaller print to look the same......never the less that could be deemed
subjective. If that is the case it may make the exposure a little different >>>>towards the plus side.

Gregory W Blank, Jul 20, 2004

2. ### Mike KingGuest

I don't have the formula handy right now but it's not quite double the size
double the time. BTW it's much more direct to measure image size on the
baseboard or easel than to measure the height of the enlarger head for such
changes. I have an old slide rule that GraLab used to sell that calculates
such things easily but it's probably easier still to get an Ilford EM-10
(\$20 US) and calculate exposures. Since what you meter affects your results
I always pull my carrier and measure the intensity of the raw light on the
baseboard and "null" the meter using the aperture ring exposure time is then
constant.

Mike King, Jul 20, 2004

3. ### TomGuest

Inverse square law applies here. 2x bigger requires 4x exposure. 1.4x bigger
requires 2x exposure. Etc. Ge ahold of a Kodak Darkroom Dataguide, it has a
computer wheel in it that calculates this for any change of magnification.

Tom, Jul 20, 2004
4. ### Francis A. MiniterGuest

Roughly, light intensity decreases in proportion to the square of the
distance. So, doubling the height quadruples the exposure. There are
(1) formulas which you will find in the archives of this newsgroup (see
Google groups). Also, Kodak publishes a Darkroom Guide that has an
exposure wheel that enables you to quickly ascertain the new time
without computation.

Francis A. Miniter

Francis A. Miniter, Jul 20, 2004
5. ### AlparslanGuest

Hi,
I am planning to make small B&W prints until I find my desired print and
than print it on a bigger scale. Is there a formula I can convert the
printing time for the new scale? It sounds logical to me that if I double
the height of the enlarger head I should double the exposure time. Is it
this simple? What if I additionally raise the enlarger head less (say 2/3)
than the previous height of the head? Regards

Alparslan, Jul 20, 2004
6. ### Craig SchroederGuest

Try this....

http://www.chibardun.net/~craigclu/enlargecalc.xls

I did this some time back and believe it is accurate.

Craig Schroeder
craig nospam craigschroeder com

-Eschew Obfuscation-

Craig Schroeder, Jul 20, 2004
7. ### Claudio BonavoltaGuest

If you've a computer handy, you can check my software, version 3 is pretty
stable now:
http://www.bonavolta.ch/hobby/en/photo/labsoftV3.htm
Although, it drives directly the enlarger, you don't need absolutely the
hardware and can use it just for the calculations.
So, if electronics makes you sick, just go down to software and download the
english version.
Documentation is still poor and will follow later.

Check also the wet side that manages the processes:
http://www.bonavolta.ch/hobby/en/photo/labsoft.htm

Regards,

Claudio Bonavolta, Jul 20, 2004
8. ### f/256Guest

Assuming that what you call "original and new positions" are the distances
from lens stage to baseboard, I believe the formula in cell "B7" is not
quite correct, I think it should be just B6*((B4)/(B3))^2 But then
again, I've been found confused and incorrect a number of times!!. Care to
explain how formula in cell "B7" came about?

If you were dealing with magnifications instead of distances to baseboard,
then, to find the exposure time factor, you would add 1 to each
magnification before dividing them and squaring them: exposure factor =
[(M+1)/(m+1)]^2

Guillermo

f/256, Jul 21, 2004
9. ### Michael ScarpittiGuest

Divide the area of the larger print by the smaller, then use that factor.

8x10 = 80 sq in.

11x14 = 154 sq in.

Going from 8x10 to 11x14 would require roughly double the exposure.

Michael Scarpitti, Jul 21, 2004
10. ### Donald QuallsGuest

Doubling the enlarger height will double the print dimensions, spreading
your light over four times the area -- so you'll need to increase
exposure by two stops. You can do that by opening the lens two stops,
if you were already stopped down beyond optimal to get a longer
exposure, but more likely you'll have to add time -- which means you'll
also have to account for reciprocity failure, and will need somewhat
more than four times the exposure time at the same aperture. How much
more than 4x? Depends on the paper; it's been twenty-five years since
I've done much printing, and in those days it would have been about
double the 4x figure. Modern papers may have less effect -- but bottom
line is you'll have to test the paper you're using, but once you have a
ratio between (say) a 4x6 print and an 8x12, it should remain pretty
constant (though a denser negative may require still more additional
time on the larger print, again because of reciprocity failure).

--
I may be a scwewy wabbit, but I'm not going to Alcatwaz!
-- E. J. Fudd, 1954

Donald Qualls, aka The Silent Observer
Lathe Building Pages http://silent1.home.netcom.com/HomebuiltLathe.htm
Speedway 7x12 Lathe Pages http://silent1.home.netcom.com/my7x12.htm

Opinions expressed are my own -- take them for what they're worth
and don't expect them to be perfect.

Donald Qualls, Jul 21, 2004
11. ### Dan QuinnGuest

"...not quite double..." The lens becomes faster as enlargement size
increases. I think the area method one I'd go with. Dan

Dan Quinn, Jul 21, 2004
12. ### f/256Guest

I think the opposite is true, the lens becomes slower as enlargement size
increases.

Guillermo

f/256, Jul 21, 2004
13. ### Gregory W BlankGuest

I think your both wrong,....b ut - Tu' es mas correctamundo ;-). The slowing is more
likely reciprocity,......or the need for a greater degree of filteration.

Gregory W Blank, Jul 21, 2004
14. ### f/256Guest

Gregorio, me think tu no es correctamundo

Reciprocity does not affect the speed of a lens (where speed is expressed in
f/stops), reciprocity does affect how long you have to expose the emulsion,
though.

The lens "slows" down as the enlargement increases because the effective
focal length of the lens increases directly proportional with the
enlargement/magnification (actually proportional to the infinity focal
length of the lens multiplied by the sum of the magnification plus 1), so if
you increase the enlargement/magnification you then have a larger distance
lens to image (baseboard) but you still have the same lens diaphragm
aperture diameter, consequently, the numeric value of the lens' effective
f/stop would be greater (f/stop = distance lens to baseboard divided by the
diameter of the aperture, assuming a thin lens, for the sake of simplicity),
the greater the numeric value of the f/stop the slower a lens is. The
lens "slows" down as the enlargement increases and that may or may not cause
the exposure time fall into the realm of reciprocity. The above is not
different than what happens when you use bellows extensions greater than the
focal length of the lens. Comprende compadre Gregorio?

Guillermo

f/256, Jul 22, 2004
15. ### Gregory W BlankGuest

Guillermo/William

Posible' mi amigo.
My point,.........which is why I stated it as such. (Maybe not too clearly however).
Si Yo comprendo,.....lo es un funcion del fuelle elemento. As well as perhaps
running into possible paper reciprocity (to a small degree).

Syntax Ok?

Gregory W Blank, Jul 22, 2004
16. ### John StockdaleGuest

Even if there is no reciprocity effect, the inverse square law is not
correct, although it works tolerably well for small adjustments. For
major changes in enlargement size, do a search here for:

gudzinowicz enlarger height factor

and you will find the correct equation.

Reciprocity is probably ignorable if the specs of the paper
manufacturers are to be belived.

John Stockdale, Jul 22, 2004
17. ### Craig SchroederGuest

I don't recall.... It was awhile back and about that time I began
using a whole different meter/timer combo that worked off of maximum
black and never got back to it. Sorry if I sent anyone down the wrong
trail! It may have been a foggy, late night project that adapted your
referenced formula below into the wrong interpretation...?

Craig Schroeder
craig nospam craigschroeder com

-Eschew Obfuscation-

Craig Schroeder, Jul 22, 2004
18. ### f/256Guest

Si !!

f/256, Jul 22, 2004
19. ### f/256Guest

I don't see why inverse square law is not correct.

M.Gudzinowicz writes:

new_time = old_time x (new_M +1)^2 / (old_M+1)^2

So if M = new magnification and m = old magnification, the exposure time
factor would be:

Factor = (M + 1)^2 / (m + 1)^2

Let's assume M = 4 and m = 2

Factor = 25 / 9 => Factor = 2.77

Now lets see what inverse square law says:

if N = distance negative to lens, B = distance lens to baseboard, F =
"thin" lens' focal length and E = magnification or enlargement

E = B / N
N = B / E
1 / N + 1 / B = 1 / F
1 / (B / E) + 1 / B = 1 / F
(E + 1) / B = 1 / F

Distance lens to baseboard = B = F ( E + 1)

When Enlargement = m = 2
Distance lens to baseboard = B = F ( 2 + 1) = 3F

When Enlargement = M = 4
Distance lens to baseboard = B = F ( 4 + 1) = 5F

Inverse square law tell us that the exposure time factor would be:

Factor = new_distance_lens_to_baseboard^2 /
old_distance_lens_to_baseboard^2

Factor = (5F)^2 / (3F)^2

Factor = 25 / 9

Factor = 2.77 Which is same as calculated with M.Gudzinowicz formula
above.

Guillermo

f/256, Jul 22, 2004
20. ### John StockdaleGuest

Guillermo wrote (edited):
You are quite right. I should have been more careful with my
description. I was referring to the common incorrect expression of
the inverse square law that goes something like:

"new size is 2 x old size, so light is reduced by a factor of 2^2 = 4,
so new exposure is 4 x old exposure."

referred to the use of (M+1)^2 / (m+1)^2

Interestingly, and unfortunately, the use of this correct formula
makes it not possible to prepare a single table with exposure factors
for old size and new size unless a fixed negative size is specified. I
mean, going from 4"x5" print size to 8"x10" print size will require a
different factor for 35mm and 120 size film because the magnification
is different.

A table would work if the parameters were old_magnification and
new_magnification, and Kodak used to publish one.

John Stockdale, Jul 22, 2004