determining weight of dry ingredients v. water in a glaze

[Pir]

Is this the correct mathematical procedure? I got it from this (rather confusing) thread: http://www.potters.org/subject72185.htm and it seems to work.

I want to add, say, 3% of an oxide to a glaze that's already made. In order to know how many grams to add, I need to do this:

1.     100/Specific Gravity       100/145 = 0.6896 (or 69% water)

2.     Weight of glaze batch, say, 350 grams  X  % water    350 X 0.6896 = 241 grams of water

3.     Weight of glaze batch - weight of water     350-241 = 109 grams dry materials

4.     Dry weight  X  % oxide to be supplied     109 grams  X  0.03 = 3.27 grams of oxide.

I was confused about step 1 when I saw this on Digitalfire: "This deflocculated slurry of 1.79 specific gravity (only 28% water)." By my calculations, 100/179 =  0.5586, or 56% water, not 28%. Am I getting something wrong here? I am not mathematically inclined...

Thanks,

Pir

 

[PeterH]

You want to use  Brongniart's formula, which allows for the volume taken up by the  (insoluble) glaze ingredients.

You might start with http://www.potteryatoldtoolijooaschool.com/brongniarts_formula_made_easy.pdf
... a syringe is an easy way to measure specific gravity https://ratcitystudios.com/blog/2017/11/14/specific-gravity

There is a calculate about https://pietermostert.github.io/SG_calc/html/brongniart.html 

[Pir]

Thanks, Peter... Wow, that's even more confusing (for me). I didn't expect that the mass or the specific gravity of dry materials came into play.

The other day I wanted to add 5% black stain to a high-calcium semi-mat clear--it makes a deep, glossy black (somehow). Here's a comparison.

The method I used was:

SG of glaze 142... 100/142 = .70

2640 grams in the bucket x .70 = 1859 grams wet 

subtracted from total = 780 grams dry x .05 grams of stain = 39 grams of black stain to be added.

But when I plugged those #s into the Brongniart calculator I get 1268.9 grams of dry, which multiplied by .05 = 63.4 grams of black stain to be added.

I guess I'll see how the test tiles turn out!

 

 

[PeterH]

780

6 hours ago, Pir said:

I guess I'll see how the test tiles turn out!

That's what the calculator gives me as well.

SG of glaze 1.42, Weight of glaze in the bucket.

Calculator gives weight of dry ingredients 1268.9g

5% of 1268.9 = 1268.9*5/100 = 63.444g stain

-----------
So what's going on?

First Brongniart's calculator gives results consistent with its model (glaze-powder is insoluble and has a typical SG of 2.6).
Your figures are weight-of-glaze-powder= 1268.9, so weight of water =  2640 - 1268.9 = 1371.1 and SG = 1.42

Now volume-of-glaze = weight-of-water/SG-of water +  weight-of-glaze-powder/SG-of-glaze-powder 
                                                = 1371.1/1.0 + 1268.9/2.6 = 1859.1
So SG-of-glaze = weight/volume = 2640/1859.1 =  1.42

-----

Now you calculations gave weight-of-water = 1859, and weight-of-glaze-powder = 780
If this were true then ...
volume-of-glaze = weight-of-water/SG-of water +  weight-of-glaze-powder/SG-of-glaze-powder 
                                      = 1859/1.0 + 780/SG-of-glaze-powder
                                      = 1859 + 780/SG-of-glaze-powder
and SG-of-glaze = weight/volume = 2640/(1859 + 780/SG-of-glaze-powder)
substituting  SG-of-glaze-powder = infinity gives SG-of-glaze = 1.42 (the measured value)

So the calculations you are performing assume that the insoluble glaze powder has an  infinite SG (ie  zero volume).  Which obviously isn't true.

--------------------

Double checking, if you apply the calculator with a stupidly high SG for for the dry ingredients it approaches you estimate for the dry weight.

Edited November 15, 2022 by PeterH

[Pir]

Wow, huh, hmm, man, that was complex! Thanks for showing me that.

Have you used this calculator and gotten correct results, I mean, you actually applied it to an amount of some addition you wanted in a glaze and it came out properly? 

I suppose there must also be a way to test it with a simple 100 gram dry recipe + a known weight of water (before adding). As I said, I'm not mathematically inclined, so this reverse engineering might be terribly obvious, or maybe you've shown that already... I'll have to actually go thru the steps, though!

[PeterH]

No I haven't used the calculator "in anger". 

11 hours ago, Pir said:

As I said, I'm not mathematically inclined, so this reverse engineering might be terribly obvious,

It's not you maths, it's your modelling/physics that's leading you astray.

IMHO it's a tricky and superficially misleading problem. After all a glaze is a "liquid" containing heavier-than water particles that aren't setting out (or at least only doing so slowly) and you don't meet too many of them.

I've got a (very old) degree in maths & physics and still cannot just "see" what the formula should be, I have to go back to basics and start with:

SG-of-glaze = weight-of-glaze/volume-of-glaze

weight-of-glaze = weight-of-water +  weight-of-glaze-powder

volume-of-glaze = weight-of-water/SG-of water +  weight-of-glaze-powder/SG-of-glaze-powder

Once you can agree with that, the rest is just arithmetic.
... Or find a calculator to do it for you. Thank @Minfor the reference.

PS Google makes quite a good calculator: enter 1371.1/1.0 + 1268.9/2.6 and you get

... I expect other browser do the same.

 

Edited November 16, 2022 by PeterH
typo

[PeterH]

And if all else fails there are "equation solvers" about which can perform some of the legwork, such as:
Equation Solver https://www.dcode.fr/equation-solver

Input
60*1000/1.7  = 4/3*pi*(50/2)^2*55 - 4/3*pi*(50/2-t)^2*(55-t)
v = 4/3*pi*(50/2)^2*55
c = 4/3*pi*(50/2-t)^2*(55-t)

... and ask it to solve for t  gives you

Only time I've had to use it so far was calculating the likely wall thickness of a  Qvevra!

... and how sweet it  still remembers the calculation for me.

Edited November 16, 2022 by PeterH

[Pir]

5 hours ago, PeterH said:

SG-of-glaze = weight-of-glaze/volume-of-glaze

weight-of-glaze = weight-of-water +  weight-of-glaze-powder

volume-of-glaze = weight-of-water/SG-of water +  weight-of-glaze-powder/SG-of-glaze-powder

Once you can agree with that, the rest is just arithmetic.

This I understand. Although "weight-of-glaze-powder" is the unknown, in my case. Brongniart’s formula, as it's written, with the -1 and the 1-1 etc., still doesn't compute for me, but I'm happy to use the calculator. I wonder if you'd like G. Spencer Brown's book, The Laws of Form, which I always think of when confronted with impossible looking mathematics. I can't get past the first five pages, myself.

[PeterH]

20 hours ago, Pir said:

Brongniart’s formula, as it's written, with the -1 and the 1-1 etc., still doesn't compute for me

I assume you mean

Can you elaboration on your problems? After an initial abortive attempt, I re-read it's documentation* and then it seemed to work fine for me.

Mwet = your weight of glaze in bucket = 2640
Swet = your measured specific gravity = 1.42
Sdry = Brongniart’s recommended value = 2.6

giving Mdry = weight of glaze powder in the bucket = 2640*(1-1/1.42)/(1-1/2.6)

Plugging this into google gives 1268.9 again

* As they say in software engineering, if all else fails RTFM (read the f-ing manual). A tribute to the the fact that much of the available documentation is often inappropriate or of poor quality.

Edited November 17, 2022 by PeterH
typo

[Pir]

By golly, I did it. I mean, manually, pen-and-paper. It worked out same as the calculator. Neat! And I see you're using the second method because I didn't offer the volume of the glaze... and I suppose finding the volume of the dry materials wouldn't be helpful, as dry glaze powders aren't measured by volume (at least not by me).

Tonight I'll do a real-world test and see if it corresponds.

Thanks, Peter.

 

[PeterH]

1 hour ago, Pir said:

and I suppose finding the volume of the dry materials wouldn't be helpful, as dry glaze powders aren't measured by volume (at least not by me).

Don't go there, these formula use the "particle density" rather than the "bulk density".

See wiki for the full story:
https://en.wikipedia.org/wiki/Bulk_density
https://en.wikipedia.org/wiki/Particle_density_(packed_density)

But imagine a quantity of sand. It's weight is easy, but what is it's volume? You can measure it in a measuring cylinder, getting a value for sand plus inter-grain air. Or you can pour it into a measuring cylinder already containing some water, and see how much the volume increases (avoiding measuring any trapped air).

[Callie Beller Diesel]

In the documentation for that calculator, it does state that the “constant” value of 2.6 for Sdry can be inaccurate if your glaze contains a lot of a dense material like tin. But outside of that, on a practical level, it does work. I have used that particular one to fix glaze batches before, when I found out the hard way I’d inadvertently left out a material. It might not work well for fixing something with minuscule amounts of a glaze ingredient that has a very strong effect on the glaze, but for most things, it should.  I’m thinking of chrome tin pinks, or something with fractional amounts of cobalt or silicon carbide. 

[Pir]

That damn inter-grain air--I can imagine that factor, as though in a movie, being neglected and then throwing everything wildly off course. 

[Pir]

35 minutes ago, Callie Beller Diesel said:

It might not work well for fixing something with minuscule amounts of a glaze ingredient that has a very strong effect on the glaze, but for most things, it should.  I’m thinking of chrome tin pinks, or something with fractional amounts of cobalt or silicon carbide. 

For me it was black stain. Would that be such material, in your opinion? Since I added less, based on my incomplete calculation, than the amount suggested by old Brongniart, I can always add more if the test comes out gray.

[Callie Beller Diesel]

According to the MSDS sheet, mason stain 6600 has a specific gravity of 5.2. That’s higher than the values listed in the wiki for things like alumina and silica, but it’s a bit lower than, say, tin or cobalt. That higher density of the stain *MIGHT* effect that 2.6 number if it was present in the recipe in larger quantities. But given that you’re only adding 3% or so to the glaze, I don’t think it’ll change the combined density of the dry ingredients very much. You’d have to math it out to verify though.

Pieter mentioned lead and tin because it’s not uncommon to use 10% tin (sg 6.85-6.95) in a recipe to opacity it, or use lead in double digit percentages because it’s a flux. He did not mention cobalt oxide though, and it has an SG of 6.07-6.66 according to his numbers. Cobalt oxide is typically only needed in tiny, single percent quantities, so it’s effect on that combined density value would be less. 

 

[Pir]

1 hour ago, Callie Beller Diesel said:

According to the MSDS sheet, mason stain 6600 has a specific gravity of 5.2.

Ah! Okay. I just took 700 grams of a base glaze and I want to add 5% black stain.

At the 2.6 average SG-dry, the calculator determines there are 389.1 grams dry = add 19.45 grams black.

At 5.2 for Black 6600, we get 296.5 grams dry = add 14.8 grams black.

So... some difference... I'll go with the 14.8 grams.

Where did you find the MSDS sheet?

Thanks!

 

[Callie Beller Diesel]

The top google search for sds mason stain 6600 leads to the one on the Mason website. SDS or MSDS sheets by law have to be available for free upon request, so most suppliers just put them on their websites.

[Pir]

Thanks Callie

[JohnnyK]

How about you take 100g of the wet glaze and dry it out? You could spread the glaze on a cookie sheet and air dry it first and then maybe put it in an oven at about 150 degrees for a while to dry it out further. Then measure the weight of the dry materials. Then you can easily calculate the weight of the additive...   As I see it, you could have done this in less time than you're spending trying to make the above calculations and the results would be definitive...I think...

[Pir]

1 hour ago, JohnnyK said:

How about you take 100g of the wet glaze and dry it out? You could spread the glaze on a cookie sheet and air dry it first and then maybe put it in an oven at about 150 degrees for a while to dry it out further. Then measure the weight of the dry materials. Then you can easily calculate the weight of the additive...   As I see it, you could have done this in less time than you're spending trying to make the above calculations and the results would be definitive...I think...

Very likely! Will keep that in mind for my next blunder.

[elaine clapper]

Oh my, how I wish I had paid more attention to math and chemistry when I was in high school!