Calculations for adjusting Specific Gravity

This post is about how to calculate the amount of water to add or remove from a glaze to achieve a desired specific gravity.

Suppose you have a glaze with specific gravity S1, but you want to change this to specific gravity S2. Let W be the weight of the glaze. Then the weight of the same volume of water is W / S1, since
SG = weight of glaze / weight of same volume of water.
(Note that if you’re measuring the weight in grams, the weight of water is the same as its volume in millilitres)

Suppose we add to the glaze a certain quantity of water, of weight X. The glaze now weighs W + X, and the weight of the same volume of water is now W / S1 + X. This means the specific gravity is now
(W + X) / (W / S1 + X).
If we set this equal to S2, we can solve for X. The result is:
X = W(S1 - S2) / (S1(S2 - 1)). . . . . . . . . . . . .(Eq 1)

If S2 is greater than S1, i.e. if you want to increase the specific gravity, this will make X negative. In that case, the absolute value of X is the weight of water you need to remove from the glaze.

This calculation has been implemented in the online calculator, where instead of having to work out the initial specific gravity S1, you just enter the weight of glaze drawn into a syringe (or poured into a graduated cylinder), and the weight of the same volume of water.

Notice that in (Eq 1), if we know X, S1 and S2, we can solve for W:
W = X S1(S2 - 1) / (S1 - S2). . . . . . . . . . . .(Eq 2)
This is useful in situations where it’s impractical to weigh the entire bucket of glaze. In this case, you can take two measurements of specific gravity; the initial measurement S1, and a measurement S2 obtained after adding a known quantity of water, and stirring thoroughly. You can then use (Eq 2) to work out the initial weight of the glaze, and then plug that into (Eq 1) to work out how much water you should have added in order to reach a target specific gravity S3. Subtract the weight of water you’ve already added from this, and you’ll know how much water you still need to add (or remove, if you added too much).

It will be convenient to express the weight of water you still need to add, X2, in terms of the weight of water already added, X1, and the specific gravities S1, S2 and S3. It turns out that the easiest way to derive this equation is by starting with the glaze at the intermediate step, after the first quantity of water has been added. If the weight of the glaze at this stage is W1, then from (Eq 2),
W1 = X2 S2(S3 - 1) / (S2 - S3). . . . . . . . . . . .(Eq 3)
This is because we change the specific gravity from S2 to S3 by adding water of weight X2.

Starting from the intermediate step again, we know that if we subtract water of weight X1, we’ll change the specific gravity from S2 back to S1. This is the same as adding water of weight -X1, so from (Eq 2) again:
W1 = - X1 S2(S1 - 1) / (S2 - S1). . . . . . . . . . . .(Eq 4)

All that’s left to do is equate (Eq 3) and (Eq 4), and solve for X2:
X2 = X1(S1 - 1)(S2 - S3) / [(S1 - S2)(S3 - 1)]

This calculation has been implemented in the online calculator Be aware that this method is more sensitive to measurement errors; if S2 is very close to S1, the relative error in S1 - S2 may be large, in which case the relative error in X2 will be large.

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