Stuart Altmann on fri 30 jan 98
Brongniart's formula enables one to estimate the amount of dry material in a
given volume of "slop," that is, of a glaze or slip already mixed with
water. This formula has a number of uses, as will be described below. But
first, let me present the formula, here expressed in convenient metric
units. The only measurement that you need to make is to determine the
density of your slop, by weighing one liter of it, in grams. Then plug that
density value into Brongniart's formula:
grams of dry matter per liter of slop
(density of slop - density of water) (density of dry matter)
= ____________________________________________________________
(density of dry matter - density of water)
Because the density of dry matter in most non-lead glazes is about 2500 g/l
and the density of water is about 1000 g/l, the formula simplifies to:
g dry matter per liter of slop = (density of slop - 1000) 5/3
So, for example, suppose that a liter of your glaze slop weighs 1500 g.
Then the amount of dry material in it is (1500-1000)5/3 = 833.3 g dry matter
per liter. And since the total slop mass is 1500 g, we also know that the
glaze slop contains 1500-833.3 = 666.7 g water per liter of slop.
For most purposes, this formula is accurate enough. For greater accuracy,
you (or your glaze program) can calculate the actual density of your dry
glaze and you can determine the actual density of your local water at your
room temperature.
(In various common references, including the standard dictionaries and
encyclopedias of ceramics, you will find more obscure versions of
Brongniart's formula, with terms like "apparent dry weight" that need to be
explained. By sticking with standard metric measures, one obtains the above
simple version.)
Here are three uses I have found for Brongniart's formula. I would be
interested to hear of others.
1. Suppose you have a bucket of glaze slop--maybe not even a full recipe's
worth and perhaps a bit thicker now as a result of evaporation. You would
like to try adding, say, 1% cobalt carbonate to a sample of it, to tint the
glaze blue. But that's 1% of the dry glaze, and you don't know how much dry
matter there is in a given volume of slop. Mix your glaze very well,
including scrapings from the sides of the bucket, weigh a liter of it, apply
Brongniart's formula, then add cobalt carb at 1% of the dry matter. Simple.
2. Most stoneware glazes that are applied by dipping work best at a density
of about 1500 g/l (i.e. a specific gravity of 1.5) Suppose you are making
up 14 kg of a glaze and want to know how much water would bring the slop to
that density. The amount of water that would make a liter of slop with a
density of 1500 g/l can be found by using Brongniart's formula, and in fact
we have also done so, in the illustration above, because we assumed for
purposes of illustration that slop density was 1500. The result was a
mixture that for every 833.3 g dry matter contains 666.7 g (666.7 ml) water.
That is, the water added per 833.3 g dry glaze material is 666.7/833.3 =
80%. So, our hypothetical batch of 14 kg dry matter would require an
addition of 80% of 14 kg water, i.e. 11.2 kg ( 11.2 liters) of water.
3. I recently made up a new glaze with the amount of water calculated to
bring the density to 1500 g/l, but what I got was something too thick to be
a normal glaze, and sure enough its density was too high, well above 1500.
Assuming that I had weighed out one ingredient twice (the ingredients were
weighed out over a two-day period), I calculated the total amount of dry
material in my bucket. It was very close to 600 g above what it should have
been; only one of the ingredients in my glaze was called for in that
amount.
Stuart Altmann
tgschs10 on sat 27 nov 99
In Ceramic Review Nov/Dec 1999 [Technical Section], there was an article
about Alexandre Brongniart formula to calculate the about of dry ingredients
in a slop/slip. I thought HORRAY I found the answer to a problem that has
been perplexing me for quite a while. However, I find the formula confusing;
it is given below:
Brongniart's formula is as follows:
Dry weight of the ingredients in a glaze
= (weight of slop ?100) x RD = (RD ?1)
where RD = Relative Density.
= (weight of slop ?100) x 1.667 for
stoneware glazes (average RD = 2.5)
= (weight of slop ?100) x 1.5 for low
solubility glazes (average RD = 3.0)
Also, in the article there is a statement "Where oz per pint are quoted they
can be converted into gm per 100ml by dividing by 0.2. To convert gm per
100mL into ounces per pint multiply by 0.2." It is my understanding that
there are 30 gms to an ounce and 480 gms to a pint so this doesn't make
sense. This makes me wonder about the whole concept. To make matters worse
it seems that the author uses an approximation of grams per pint for various
clays; obviously, I have different clays at hand. I had a question how to
calculate the dry weight in a slop/slip some months ago and asked about
using specific gravity; however, I got a message saying this was incorrect;
then the other day I saw a message from a clayarter saying its easy just
weight out equal volumes of water and slip and the difference is the dry
weight. I'm certain this is wrong because the particulate material uses
volume displacing some water. I have mixed up large amounts of Goldart terra
sigillata, stoneware and porcelain white slips; I would like to take a small
amount of these and mix in different stains and oxides. My question is if I
wish to add 2% cobalt, for example, how do I calculate the amount needed? Is
the best answer just to approximate the particulate weight, I did let some
slip dry out after pre-weighting and found that the dry weight was
approximately 40%; however, it is obvious as the slip drys out or new slip
is mixed different amounts of water and particulate matter compose the
solution. HELP.
Tom Sawyer
Orlando, Florida
tgschs10@classic.msn.com
Carenza Hayhoe on sun 28 nov 99
Charles Stileman the author of the article about Brongniart's Formula is a
good friend of mine - his ability to juggle figures leaves me breathless. I
will send Tom Sawyer's message to him by snail mail and as soon as I get an
answer I will post it to the list. Sorry Tom - I wish I could supply an
answer - but at least I know a man who can!
Carenza Hayhoe
www.mochaware.com
----- Original Message -----
From: tgschs10
To:
Sent: Saturday, November 27, 1999 3:38 PM
Subject: Brongniart's Formula
> ----------------------------Original message----------------------------
> In Ceramic Review Nov/Dec 1999 [Technical Section], there was an article
> about Alexandre Brongniart formula to calculate the about of dry
ingredients
> in a slop/slip. I thought HORRAY I found the answer to a problem that has
> been perplexing me for quite a while. However, I find the formula
confusing;
> it is given below:
>
> Brongniart's formula is as follows:
> Dry weight of the ingredients in a glaze
> = (weight of slop ?100) x RD = (RD ?1)
> where RD = Relative Density.
> = (weight of slop ?100) x 1.667 for
> stoneware glazes (average RD = 2.5)
> = (weight of slop ?100) x 1.5 for low
> solubility glazes (average RD = 3.0)
>
> Also, in the article there is a statement "Where oz per pint are quoted
they
> can be converted into gm per 100ml by dividing by 0.2. To convert gm per
> 100mL into ounces per pint multiply by 0.2." It is my understanding that
> there are 30 gms to an ounce and 480 gms to a pint so this doesn't make
> sense. This makes me wonder about the whole concept. To make matters worse
> it seems that the author uses an approximation of grams per pint for
various
> clays; obviously, I have different clays at hand. I had a question how to
> calculate the dry weight in a slop/slip some months ago and asked about
> using specific gravity; however, I got a message saying this was
incorrect;
> then the other day I saw a message from a clayarter saying its easy just
> weight out equal volumes of water and slip and the difference is the dry
> weight. I'm certain this is wrong because the particulate material uses
> volume displacing some water. I have mixed up large amounts of Goldart
terra
> sigillata, stoneware and porcelain white slips; I would like to take a
small
> amount of these and mix in different stains and oxides. My question is if
I
> wish to add 2% cobalt, for example, how do I calculate the amount needed?
Is
> the best answer just to approximate the particulate weight, I did let some
> slip dry out after pre-weighting and found that the dry weight was
> approximately 40%; however, it is obvious as the slip drys out or new slip
> is mixed different amounts of water and particulate matter compose the
> solution. HELP.
>
> Tom Sawyer
> Orlando, Florida
> tgschs10@classic.msn.com
>
>
Tom Buck on sun 28 nov 99
Mr. Sawyer:
I did the following for the local Guild, so I guess I should
post it to the list:
------------------------
Ask Tom
a column on technical matters by Tom Buck (550 words)
Adjusting slips/glazes
Years ago, a technical specialist, M. Brongniart, was hired by the
French pottery that made (still does) the famous Limoges porcelain
foodware. Since slip-casting was the dominant method of forming pots, he
came up with a formula to determine/measure indirectly the amount of
solids in a clay slip so he could know how much water or clay to add. It
also works for glaze slurries. Cardew cites it as follows:
W = P-20 (g/g-1)
where W = dry weight of clay/glaze solids (raw materials);
P = the "pint" weight of the clay slip or glaze slurry in ounces;
20 is weight in ounces of a pint of water;
and g is the density of the raw materials (the old term was specific
gravity now declared obsolete by the Chemists' congress).
Cardew goes onto explain (see Pg 269, Pioneer Pottery) that most
clay and glaze materials have a density of 2.5-2.6 grams per millilitre
(g/mL) and this assumption allows use of the formula with enough
accuracy for most purposes.
One can convert the formula to SI (metric) thusly:
W = [Wt - Ww] x (d/d-1)
where W is the weight of suspended solids (ie, raw materials) in the
weighed sample;
Wt is the total weight of a sample of the clay slip or glaze slurry;
Ww is the weight of water equivalent to the volume of the sample;
and d is the density of the raw materials.
The units of weight would be grams or kilograms; and the density units
would be any as long as one uses the correct/appropriate value for the
density of water ...the usual density unit is grams/millilitre since
this leads to a density of water at 1 g/mL. (One could also use
kilograms/Litre and the value for water is still 1).
So one weighs say 200 mL of slip or slurry, determining the
weight to be 310 grams (+ or - 10).
(As an aside, this measurement yields a glaze slurry a density of
310/200 or 1.55 g/mL, the typical value for dipping; for spraying, a
density is 1.4-1.45 g/mL may be preferred).
Hence, the suspended solids content of the 200 mL sample is
W (in grams) = [310-200] x (2.5/2.5-1) = 110x1.7 = 187 rounded to 190
grams. By difference the weight of water is 120 grams.
Proving it to be ballpark:
190 g divided by 2.5 g/mL density = 76 mL (volume) of solids;
120 g divided by 1.0 g/mL density = 120mL (volume) of water;
and their sum is 196, rounding, 200 mL. This is the level of accuracy one
can get by assuming the powdered raw materials all have a density of 2.5
g/mL (particle density not "bulk" density where air is included).
In a factory situation where one works with tons of materials, one
would determine each individual solid's density value, and then using the
pecentages of the recipe one would compute the "real" density of the
combined solids (slip or glaze). And use this value in the Brongniart
Formula. The B Formula really does help when a studio potter is recycling
some clay slip gone too thick or a glaze slurry needing additions or
adjustments after it has been used/tested for awhile.
------------end------------
Does this make it clearer Mr Sawyer? Til later. Peace. Tom B.
Tom Buck ) tel: 905-389-2339
(westend Lake Ontario, province of Ontario, Canada).
mailing address: 373 East 43rd Street,
Hamilton ON L8T 3E1 Canada
Nikom Chimnok on sun 28 nov 99
Hello Tom,
From an Industrial Ceramics seminar I attended not so long ago, I
have Brongniart's Formula in a slightly different form. It goes as follows:
W=(P-A)*S/(S-1).
To say that again, since it's hard to write formulas in this format: W
equals (P minus A) times S over (S minus 1).
Where: P is the weight of ANY volume of slip in ANY units.
W is the weight of the dry material in THAT volume of slip in the
SAME weight units.
A is the weight of the SAME volume of water in the SAME weight units.
S is the specific gravity of the dry material.
So, assume we have a triple beam balance which weighs in grams. Suppose you
find a large jar and fill it with water till it weighs a thousand grams.
That's a liter. Use a broken hacksaw blade to make a fine mark on the jar.
Write over it with Magic Marker, and you'll be able to see the mark, not
only feel it. All this part could be avoided if you had a graduated
cylinder, but I never have had. My standard is actually a water glass marked
to 200 grams, but for heuristic purposes I'll use 1000 grams.
Now fill the jar to the mark with the slip you want to measure. If it's
casting slip it should weigh about 1800 grams. That's "P" in the formula.
"A" of course is 1000 grams-the original liter of water.
So, (P - A) = 800 grams.
Now you've got the know the specific gravity of the dry material. For most
clays it is about 2.6. The clay I use is 2.7.
Thus, the 2nd part of the equation (S over (S-1)), providing the specific
gravity is 2.6, will equal 2.6/(2.6-1), or 2.6/1.6, which equals 1.625.
So the answer we are after, 'W", equals 800 times 1.625, or 1300 grams. Out
of the 1800 grams of slip, 1300 are clay, and 500 are water. This says that
the the slip is 72.2% clay and 27.8% water, which is about right.
The hard part of all this is determining the Specific Gravity of the dry
material in question. The professor who gave the seminar explained that you
need special apparatus--a bottle with a stopper with a hole in it. You put
any amount of the powder of the dry material into the bottle, then fill the
remainder with parrafin, weighing all the while. The hole in the stopper
insures the parrafin is filled to the right level. I don't really understand
this part of it, never having seen it done. I got the number for the SG of
my clay from a lab, where they used a mass spectrometer or some such, and so
have never needed to do the test, which sounds quite difficult--unless you
have the appartatus and have done it once. However, the method outlined
above I have used for making slip and terra sig, and it works. As to
measuring SG, I'm sure someone like Michael Banks or Gavin Stairs or
numerous others with more science education than I can explain it quite easily.
As far as I can go,
Nikom
And yeah, tho I don't follow the formula you mention below, it's obviously
sloppy, for as we all know, a pint (of water)'s a pound the year around, and
a pound weighs 454 grams. And so far as the weight of an ounce, there ain't
a drug dealer anywhere will give you 30 grams when you buy one. They're more
likely to give you 28 and smoke the decimals themselves. Odd, isn't it, that
that segment of the population should know more about metric conversions
than anyone but scientists?
********************************************************
At 10:38 27/11/99 EST, you wrote:
>----------------------------Original message----------------------------
>In Ceramic Review Nov/Dec 1999 [Technical Section], there was an article
>about Alexandre Brongniart formula to calculate the about of dry ingredients
>in a slop/slip. I thought HORRAY I found the answer to a problem that has
>been perplexing me for quite a while. However, I find the formula confusing;
>it is given below:
>
>Brongniart's formula is as follows:
>Dry weight of the ingredients in a glaze
>= (weight of slop ?100) x RD = (RD ?1)
>where RD = Relative Density.
>= (weight of slop ?100) x 1.667 for
>stoneware glazes (average RD = 2.5)
>= (weight of slop ?100) x 1.5 for low
>solubility glazes (average RD = 3.0)
>
>Also, in the article there is a statement "Where oz per pint are quoted they
>can be converted into gm per 100ml by dividing by 0.2. To convert gm per
>100mL into ounces per pint multiply by 0.2." It is my understanding that
>there are 30 gms to an ounce and 480 gms to a pint so this doesn't make
>sense. This makes me wonder about the whole concept. To make matters worse
>it seems that the author uses an approximation of grams per pint for various
>clays; obviously, I have different clays at hand. I had a question how to
>calculate the dry weight in a slop/slip some months ago and asked about
>using specific gravity; however, I got a message saying this was incorrect;
>then the other day I saw a message from a clayarter saying its easy just
>weight out equal volumes of water and slip and the difference is the dry
>weight. I'm certain this is wrong because the particulate material uses
>volume displacing some water. I have mixed up large amounts of Goldart terra
>sigillata, stoneware and porcelain white slips; I would like to take a small
>amount of these and mix in different stains and oxides. My question is if I
>wish to add 2% cobalt, for example, how do I calculate the amount needed? Is
>the best answer just to approximate the particulate weight, I did let some
>slip dry out after pre-weighting and found that the dry weight was
>approximately 40%; however, it is obvious as the slip drys out or new slip
>is mixed different amounts of water and particulate matter compose the
>solution. HELP.
>
>Tom Sawyer
>Orlando, Florida
>tgschs10@classic.msn.com
>
>
Chris Schafale on sun 28 nov 99
Below is a copy of a post by Stuart Altmann on this subject. I have
found it very helpful when experimenting with adding colorants to
already mixed glazes. It makes some assumptions, but my experience
is, it works. Thanks Stuart!
Chris
Stuart wrote:
Brongniart's formula enables one to estimate the amount of dry
material in a given volume of "slop," that is, of a glaze or slip
already mixed with water. This formula has a number of uses, as will
be described below. But first, let me present the formula, here
expressed in convenient metric units. The only measurement that you
need to make is to determine the density of your slop, by weighing one
liter of it, in grams. Then plug that density value into Brongniart's
formula:
grams of dry matter per liter of slop
(density of slop - density of water) (density of dry
matter)
=
____________________________________________________________
(density of dry matter - density of
water)
Because the density of dry matter in most non-lead glazes is about
2500 g/l and the density of water is about 1000 g/l, the formula
simplifies to:
g dry matter per liter of slop = (density of slop - 1000) 5/3
So, for example, suppose that a liter of your glaze slop weighs 1500
g. Then the amount of dry material in it is (1500-1000)5/3 = 833.3 g
dry matter per liter. And since the total slop mass is 1500 g, we
also know that the glaze slop contains 1500-833.3 = 666.7 g water per
liter of slop.
For most purposes, this formula is accurate enough. For greater
accuracy, you (or your glaze program) can calculate the actual density
of your dry glaze and you can determine the actual density of your
local water at your room temperature.
(In various common references, including the standard dictionaries and
encyclopedias of ceramics, you will find more obscure versions of
Brongniart's formula, with terms like "apparent dry weight" that need
to be explained. By sticking with standard metric measures, one
obtains the above simple version.)
Here are three uses I have found for Brongniart's formula. I would be
interested to hear of others.
1. Suppose you have a bucket of glaze slop--maybe not even a full
recipe's worth and perhaps a bit thicker now as a result of
evaporation. You would like to try adding, say, 1% cobalt carbonate
to a sample of it, to tint the glaze blue. But that's 1% of the dry
glaze, and you don't know how much dry matter there is in a given
volume of slop. Mix your glaze very well, including scrapings from
the sides of the bucket, weigh a liter of it, apply Brongniart's
formula, then add cobalt carb at 1% of the dry matter. Simple.
2. Most stoneware glazes that are applied by dipping work best at a
density of about 1500 g/l (i.e. a specific gravity of 1.5) Suppose
you are making up 14 kg of a glaze and want to know how much water
would bring the slop to that density. The amount of water that would
make a liter of slop with a density of 1500 g/l can be found by using
Brongniart's formula, and in fact we have also done so, in the
illustration above, because we assumed for purposes of illustration
that slop density was 1500. The result was a mixture that for every
833.3 g dry matter contains 666.7 g (666.7 ml) water. That is, the
water added per 833.3 g dry glaze material is 666.7/833.3 = 80%. So,
our hypothetical batch of 14 kg dry matter would require an addition
of 80% of 14 kg water, i.e. 11.2 kg ( 11.2 liters) of water.
3. I recently made up a new glaze with the amount of water calculated
to bring the density to 1500 g/l, but what I got was something too
thick to be a normal glaze, and sure enough its density was too high,
well above 1500. Assuming that I had weighed out one ingredient twice
(the ingredients were weighed out over a two-day period), I calculated
the total amount of dry material in my bucket. It was very close to
600 g above what it should have been; only one of the ingredients in
my glaze was called for in that amount.
Stuart Altmann
> ----------------------------Original message----------------------------
> In Ceramic Review Nov/Dec 1999 [Technical Section], there was an article
> about Alexandre Brongniart formula to calculate the about of dry ingredients
> in a slop/slip. I thought HORRAY I found the answer to a problem that has
> been perplexing me for quite a while. However, I find the formula confusing;
> it is given below:
>
> Brongniart's formula is as follows:
> Dry weight of the ingredients in a glaze
> = (weight of slop ?100) x RD = (RD ?1)
> where RD = Relative Density.
> = (weight of slop ?100) x 1.667 for
> stoneware glazes (average RD = 2.5)
> = (weight of slop ?100) x 1.5 for low
> solubility glazes (average RD = 3.0)
>
> Also, in the article there is a statement "Where oz per pint are quoted they
> can be converted into gm per 100ml by dividing by 0.2. To convert gm per
> 100mL into ounces per pint multiply by 0.2." It is my understanding that
> there are 30 gms to an ounce and 480 gms to a pint so this doesn't make
> sense. This makes me wonder about the whole concept. To make matters worse
> it seems that the author uses an approximation of grams per pint for various
> clays; obviously, I have different clays at hand. I had a question how to
> calculate the dry weight in a slop/slip some months ago and asked about
> using specific gravity; however, I got a message saying this was incorrect;
> then the other day I saw a message from a clayarter saying its easy just
> weight out equal volumes of water and slip and the difference is the dry
> weight. I'm certain this is wrong because the particulate material uses
> volume displacing some water. I have mixed up large amounts of Goldart terra
> sigillata, stoneware and porcelain white slips; I would like to take a small
> amount of these and mix in different stains and oxides. My question is if I
> wish to add 2% cobalt, for example, how do I calculate the amount needed? Is
> the best answer just to approximate the particulate weight, I did let some
> slip dry out after pre-weighting and found that the dry weight was
> approximately 40%; however, it is obvious as the slip drys out or new slip
> is mixed different amounts of water and particulate matter compose the
> solution. HELP.
>
> Tom Sawyer
> Orlando, Florida
> tgschs10@classic.msn.com
>
>
Light One Candle Pottery
Fuquay-Varina, NC
candle@intrex.net
David Hewitt on sun 28 nov 99
Tom,
I don't know if it is some computer quirk, but all '?' that you give
below should be minus signs. Also you have to take into account that the
author is in the UK and the conversion units are for the UK. 1oz is
28.3grms and 1 UK pint of water is 568grms. I don't use Brongniart's
formula myself, but I think that will make it clearer to you.
David
In message , tgschs10 writes
>----------------------------Original message----------------------------
>In Ceramic Review Nov/Dec 1999 [Technical Section], there was an article
>about Alexandre Brongniart formula to calculate the about of dry ingredients
>in a slop/slip. I thought HORRAY I found the answer to a problem that has
>been perplexing me for quite a while. However, I find the formula confusing;
>it is given below:
>
>Brongniart's formula is as follows:
>Dry weight of the ingredients in a glaze
>= (weight of slop ?100) x RD = (RD ?1)
>where RD = Relative Density.
>= (weight of slop ?100) x 1.667 for
>stoneware glazes (average RD = 2.5)
>= (weight of slop ?100) x 1.5 for low
>solubility glazes (average RD = 3.0)
>
>Also, in the article there is a statement "Where oz per pint are quoted they
>can be converted into gm per 100ml by dividing by 0.2. To convert gm per
>100mL into ounces per pint multiply by 0.2." It is my understanding that
>there are 30 gms to an ounce and 480 gms to a pint so this doesn't make
>sense. This makes me wonder about the whole concept. To make matters worse
>it seems that the author uses an approximation of grams per pint for various
>clays; obviously, I have different clays at hand. I had a question how to
>calculate the dry weight in a slop/slip some months ago and asked about
>using specific gravity; however, I got a message saying this was incorrect;
>then the other day I saw a message from a clayarter saying its easy just
>weight out equal volumes of water and slip and the difference is the dry
>weight. I'm certain this is wrong because the particulate material uses
>volume displacing some water. I have mixed up large amounts of Goldart terra
>sigillata, stoneware and porcelain white slips; I would like to take a small
>amount of these and mix in different stains and oxides. My question is if I
>wish to add 2% cobalt, for example, how do I calculate the amount needed? Is
>the best answer just to approximate the particulate weight, I did let some
>slip dry out after pre-weighting and found that the dry weight was
>approximately 40%; however, it is obvious as the slip drys out or new slip
>is mixed different amounts of water and particulate matter compose the
>solution. HELP.
>
>Tom Sawyer
>Orlando, Florida
>tgschs10@classic.msn.com
>
--
David Hewitt
David Hewitt Pottery ,
7 Fairfield Road, Caerleon, Newport,
South Wales, NP18 3DQ, UK. Tel:- +44 (0) 1633 420647
FAX:- +44 (0) 870 1617274
Own Web site http://www.dhpot.demon.co.uk
IMC Web site http://digitalfire.com/education/people/hewitt.htm
Dave Finkelnburg on mon 29 nov 99
Technically inclined:
Tom and I have corresponded about this. When calculating the % by
weight dry solids in a glaze, I have been assuming the average density of
the dry glaze is about 2.65. Is that a reasonable approximation?
From his work, it's obvious Brongniart assumed 2.5 for the slip he was
working with.
Thanks for any input.
Dave Finkelnburg
dfinkeln@cyberhighway.net
-----Original Message-----
From: tgschs10
To: CLAYART@LSV.UKY.EDU
Date: Saturday, November 27, 1999 8:39 AM
Subject: Brongniart's Formula
----------------------------Original message----------------------------
In Ceramic Review Nov/Dec 1999 [Technical Section], there was an article
about Alexandre Brongniart formula to calculate the about of dry ingredients
in a slop/slip. I thought HORRAY I found the answer to a problem that has
been perplexing me for quite a while. However, I find the formula confusing;
it is given below:
Brongniart's formula is as follows:
Dry weight of the ingredients in a glaze
= (weight of slop ?100) x RD = (RD ?1)
where RD = Relative Density.
= (weight of slop ?100) x 1.667 for
stoneware glazes (average RD = 2.5)
= (weight of slop ?100) x 1.5 for low
solubility glazes (average RD = 3.0)
Also, in the article there is a statement "Where oz per pint are quoted they
can be converted into gm per 100ml by dividing by 0.2. To convert gm per
100mL into ounces per pint multiply by 0.2." It is my understanding that
there are 30 gms to an ounce and 480 gms to a pint so this doesn't make
sense. This makes me wonder about the whole concept. To make matters worse
it seems that the author uses an approximation of grams per pint for various
clays; obviously, I have different clays at hand. I had a question how to
calculate the dry weight in a slop/slip some months ago and asked about
using specific gravity; however, I got a message saying this was incorrect;
then the other day I saw a message from a clayarter saying its easy just
weight out equal volumes of water and slip and the difference is the dry
weight. I'm certain this is wrong because the particulate material uses
volume displacing some water. I have mixed up large amounts of Goldart terra
sigillata, stoneware and porcelain white slips; I would like to take a small
amount of these and mix in different stains and oxides. My question is if I
wish to add 2% cobalt, for example, how do I calculate the amount needed? Is
the best answer just to approximate the particulate weight, I did let some
slip dry out after pre-weighting and found that the dry weight was
approximately 40%; however, it is obvious as the slip drys out or new slip
is mixed different amounts of water and particulate matter compose the
solution. HELP.
Tom Sawyer
Orlando, Florida
tgschs10@classic.msn.com
tgschs10 on fri 3 dec 99
Thanks to all who replied to my earlier message regarding Brongniart's
Formula. I waded through all of the calculations and believe that I have
come to some limited understanding although I'm not completely comfortable.
One of the respondents suggested measuring out 10 ml of slip/slop, place it
in the oven until dry, and weight out the results. This I am comfortable
with. Indeed I measured out 50 ml and microwaved it over about 5 minutes.
This is simple merely multiply by 2 to get gms/100 ml. Its really odd that
simple solutions to practical problems are so damn obvious when someone
points the way.
Thanks to everyone
Tom Sawyer
Orlando, Florida
tgschs10@classic.com
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