Re: Newbie in lith print
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- Subject: Re: Newbie in lith print
- From: Dean Kansky <firstname.lastname@example.org>
- Date: Sat, 09 Dec 2006 19:14:16 -0800 (PST)
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Sorry for the typo
Here is what someone wrote me in 1999 and it is in the
I have been student with the late Jean Pierre Sudre
and his wife Claudine (1of the best specialists in
Europe for 19th century processes) in the mid 80's.
I don't know if what you call crystalline images refer
to what Jean Pierre called his Cristallisations : very
large format colour-toned abstract photographs
obtained by enlarging 5x7 film on which JP had smashed
heated crystals of various photographic chemicals.
As far as I remember his first cristallisations had
been realized using Potassium Bromide; he also used
sodium sulfite and experimented a lot...
I have a text he wrote about this work. When I get
some time in the next days, I'll look at what he said
and I'll post you the details !
Hope it helps you
I used Lith chemicals and heated nothing. I think I
coated with fixer (by brush) and let it set (over
night). When it dried, I had crystals on lith.
Although I am not sure now, except it was some combo
of lith development chemicals. Or standard B&W
development chemicals or some combo of both lith and
B&W chemicals. I was playing around. If you go for
it, or know the proper way, let me know.
I am not a pro and was playing around while a Bridal
shower was being held at my apt for someone elses
bride. I wanted to stay away from that.
I found this e-mail on the web, but from 2002
However it is on the site of a photographer who took
his workshops, I think, and she may give the right
--- "Christina Z. Anderson" <email@example.com>
> Thank, Loris and Alberto, Ryuji,
> This makes more sense, because I got my decimal
> place wrong. It is 37.5g
> paraformaldehyde in a LITER, so that is only 3.75%
> solution, not 37.5% as I
> thought off the top of my head yesterday, so using
> close to 100ml is
> logical. DOH. I should NEVER post to the list in a
> Ryuji, I revisited that post of yours and it looks
> like a good formula in
> that the ingredients shouldn't be hard to find. I
> assume trisodium
> phosphate is the same as that at a hardware store or
> do you think it needs
> to be photo grade? The hot is problematic in our
> student darkroom, though,
> as is the short life.
> The Fotospeed, btw, was working in 5-15 minutes with
> my class so speed is
> not a problem. But I am intrigued to see the image
> tones you might get out
> of your lith developer as compared to Fotospeed. My
> guess is image tone is
> paper dependent, and would you agree?
> I did buy Tim Rudman's lith book and it is gorgeous
> but have only had time
> to look at pictures. But, last week of classes and
> only one more critique to
> Dean Kansky, what is the SUNDRE formula and what are
> ----- Original Message -----
> From: "Alberto Novo" <firstname.lastname@example.org>
> To: <alt-photo-process-L@usask.ca>
> Sent: Friday, December 08, 2006 10:32 AM
> Subject: Re: Newbie in lith print
> > Loris,
> > I don't know where you found that page...
> > Paraformaldehyde is a polymer of formaldehyde: the
> former is (CH2O)n (CAS
> > n.30525-89-4), the latter is CH2O. Hence, you can
> use the weight of
> > paraformaldehyde like it were pure (100%)
> formaldehyde, but you will need
> > to take into account the weight/weight percentage
> of the solution and its
> > density in order to calculete the corresponding
> liquid amount.
> > As an example, if 37.5 g of paraformaldehyde are
> required and you have a
> > 37% (w/w) formaldehyde solution @ 1.09 kg/dm3
> (this should be clearly
> > written in the label), you will need
> 37.5x100/37x1/1.09= 99 cc of
> > solution.
> > Alberto
> >> I found a page saying "...A 10% formalin solution
> should be equivalent
> >> to a 4% paraformaldehyde (by weight)
> solution...". Concentrated /
> >> saturated formalin is 37% AFAIK, not 40%. Anyway,
> I'll do the math: Since
> >> it says 100g 4% paraformaldehyde (4g
> paraformaldehyde) solution is
> >> equivalent to 100g 10% formalin (10g
> formaldehyde) then you have to use
> >> 2.5 units of formaldehyde per 1 unit of
> paraformaldehyde (10 / 4 = 2.5).
> >> Since Christina's formula calls for 37.5g
> paraformaldehyde, you have to
> >> use 37.5 x 2.5 = 93.75g formaldehyde. In 37%
> solution terms this should
> >> make 93.75 / 0.37 ~= 253ml 37% formalin, *if my
> logic is correct*... I
> >> don't know if the information I got from that
> page is correct and most
> >> importantly I'm not sure if my logic is correct.
> Therefore don't rely on
> >> this info ;) Regards,
> >> Loris.
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