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Bujor - 22 Aug 2003 18:50 GMT

As far as I can observe, all consumer enlargers work in CYM. I built one which
works with three sources of light in RGB. My analyzer reads CYM. DOes anybody
have any idea on a simple way of equivalency?

Thank you,
the rookie

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Dan Dunphy - 28 Aug 2003 01:15 GMT

Using primary (RGB) filters, you must make 3 exposures, one of each
color. If you put all three filters in the light source, you turn
white into gray. I believe this is the way the Minolta works.
Dan

>As far as I can observe, all consumer enlargers work in CYM. I built one which
>works with three sources of light in RGB. My analyzer reads CYM. DOes anybody
>have any idea on a simple way of equivalency?
>
>Thank you,
>the rookie

Colorado Springs, CO
My advice may be worth what you paid for it.

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Bujor - 28 Aug 2003 20:46 GMT

Thanks for your input.

I know what happens when you mix filters on the same light source. What I
didn't tell you is that I have three different sources, each having it's own
filter (25, 58 and 47B. If you are interested, I have the exact frequencies and
%Ts, as I scanned the filters on my UV-VIS spectrometer). I think this detail
changes basis for discussion a bit... However, my mean color analyzer provides
readings only in CYM.

If you have color related experience (as I don't have ANY, and obviously I
didn't read the right articles yet), help me interpret the readings in terms of
time of exposure given to the three sources.

Simplier put: could you please explain what happens in two different instances,
when you have two readings, say, 90Y and 10Y on your Y pontentiometer after you
zero the analyzer. You dial those values on your enlarger head. Which setting
appears more yellow when reflected on your white enlarger table: the one for
dialing 90Y or the one for 10Y? Maybe I can then figure out how long must I
expose (or correctly cut exposure, because Yellow is negative Blue) with the
blue source. And so on...

I can't think what the final outcome would look like. Maybe I'll get a negative
from a negative... I don't see it now... In film you have three layers sensible
to CYM. Nevertheless, your camera exposes it to RGB light, and you get a
negative... I have to read more on this.

I know, it seems complicated, but I'm sure I can get it in the end. Afterall,
this is the way the process should have been physically done, but people
engineered it more economically.

the rookie

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fnovau - 29 Aug 2003 12:27 GMT

I understand you are mixing RGB additive and CMY subtractive printing
systems. If you have RGB lights sources forget about measuring CMY
with your color analyser. I believe you could try to measure RGB on
the easel and to compute directly your RGB exposure times, the way
modern automatic color printing systems do. For this you must measure
LATD (large area transmission density) of your negative and apply a
simple equation to compute your RGB exposure times. Equation is:
log(t/tref)=S*(LATD-LATDref) where t is exposure time, tref is
exposure time of reference negative, S is slope constant, LATD is
density of negative to be printed and LATDref is density of reference
negative. If interested in this I can describe process.
I got your e-mail with some details of your system. You control power
trough triac but it's not clear for me if you pretend to control power
applied to light source and exposure time or both. Take care because
if you apply a variable power to your light source you will get
different color temperatures and so different ratios of blue to red
light
Regards
-Francesc Novau

> Thanks for your input.
>
[quoted text clipped - 27 lines]
>
> the rookie

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Bujor - 29 Aug 2003 14:32 GMT

First to answer your last question: I can control the voltage of each source
separately (as you said, to a certain extend, because of bulbs color shift),
and I may control the time of exposure for each source too, through simple 555
circuitry timers (I don't want to introduce this though, unless I have to). The
filters that I use now have transmission patterns pretty much juxtaposed on
film sensitivities ( if Blue for film is 80%, then filter will give a 20%T).

Sure I would like to knor the process with LATD, please send it if you can

the rookie

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