Dave Finkelnburg on tue 20 jan 04
Brogniart and the long way to understanding...long, technical...
Dear Phil,
You are thinking about the right stuff, but you have overlooked that in
the glaze the dry ingredients have volume. In fact, the glaze dry
ingredients have a density, in grams per cubic centimeter (g/cc) of about
2.65. At least, this is the average I calculated for mid-range and
high-fire stoneware and porcelain glaze ingredients using density
information from my edition of the Chemical Rubber Company Handbook. For
more, see Harry Fraser, "Glazes for the Craft Potter," page 130.
By assuming this density we can say the Volume (V) in cc's of dry solids
in a glaze sample is equal to the Weight (W) in grams of dry solids in the
glaze sample divided by 2.65 g/cc, or V = W/2.65.
Also, as you state, one cubic centimeter (cc) of water weights one gram
(g). Thus, the volume of the water, Vw, in cc's, equals the weight of the
water Ww, in grams.
To solve the problem you pose one can work many ways. I chose to write
two equations because there are two unknowns, the weight of the solids and
the volume of the solids.
Equation 1) Weight of solids (W) in sample equals Weight of slurry (Wsl)
minus weight of water, Ww, all in grams, or
W = Wsl - Ww
Equation 2) Volume of solids (V) in sample equals Volume of slurry (Vsl)
minus volume of water, Vw, all in cc's. However, as we said above, Vw = Ww,
and V = W/2.65, so lets write equation 2) this way,
W/2.65 = Vsl - Ww
Now, let's subtract equation 2) from equation 1), which gives us
Equation. 3) W - W/2.65 = Wsl - Vsl
You told us you had 50 cc's of slurry that weighed 65 grams, so we can
rewrite equation 3) this way:
W(1-1/2.65) = 65-50 Then, if we do the arithmetic,
W = 1.6 x (65-50) which is in the form of this equation I posted
yesterday. But as a reminder, that simplified equation is, "The weight of
dry solids in any sample of glaze is equal to the (Weight of the sample in
grams minus the volume of the sample in cc's) times 1.6."
So you have about 15 x 1.6 or 24 grams of dry glaze ingredients in your
particular 65 gram sample of glaze. Your approach in the rest of your
example below appears correct, so once you use this new information you'll
get the right number for the grams of cobalt.
Good glazing!
Dave
Note: I sent a brief e-mail about this to Phil personally late last night.
I am very sensitive about pointing out an error to someone in front of more
than 3,000 list members and potentially the rest of the interested Internet!
Phil, to his credit, replied on list. He is clearly concerned, not about
his ego, but about his understanding. In the room when I type this I am
blessed to have one of Phil's gorgeous copper red vases gracing a shelf that
holds some of my small pottery collection. He does very, very nice work.
That's especially creditable considering he hasn't been potting all that
long. But it's understandable, knowing how hard Phil works, how disciplined
he is in that work, and how open his mind is. These are some good
behaviors. I am trying to learn from Phil.
----- Original Message -----
From: "Phil Smith"
To:
> If I weigh 50 cc's of glaze and it weighs 65 grams I should have 15
> grams of glaze materials in there. 1.3.
> Water divided by total weight = .76923 percent water.
>
> If I were to weigh out 1814.369(4 lbs) wet glaze.
>
> 1814.369 x .7692 = 1395.613 water
>
> 418.7 should be glaze materials.
>
> Recipe calls for 2 percent cobalt.
>
> .02 x 418.7 = 8.374 grams.
>
> I would add 8.37 grams cobalt to my 4lbs wet glaze.
>
>
> Phil...
| |
|
