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Is it possible to get a useful starting gravity reading when not everything is dissolved or suspended, but you know it ultimately will be?

Some context: In my pre-hydrometer days, I made a few cysers / honey ciders that really only amounted to apple juice, honey, and yeast. I never was able to get the honey to fully dissolve, it just went straight to the bottom and sat there, silently mocking me. After initially being concerned, I RDWHAHB'd, figuring the yeast would find the honey eventually. And they did! A couple weeks later, the honey layer was gone, leaving the beginnings of a delicious beverage behind.

(In hindsight, I wonder if this accidental timed release mechanism meant the yeast weren't as stressed as they could have been, actually improving the final product. But I digress.)

Back to my question: if I understand the physics of a hydrometer, it measures the buoyancy due to displacement of the water+stuff (dissolved/suspended). So it seems to me that anything not dissolved or suspended (thus resting on the bottom) will NOT be reflected in the gravity reading, any more than if I were to throw in a handful of marbles. Am I correct in my thinking here?

=== EDIT: anyone who comes this way, the following wasn't quite right, see answers below ===

If so, then it seems like it should be possible to calculate an "effective" OG, since I know how much adjunct I'm adding. If:

starting gravity = (density of wort)/(density of water)
                 = ((mass of wort)/volume)/((mass of water)/volume)
                 = (mass of wort)/(mass of water)  # since volume is equal
                 = (mass of water + mass of stuff)/(mass of water)
                 = 1 + (mass of stuff)/(mass of water)

then for the purpose of illustration and intentionally making the math easy, would adding 50g of honey to 1L of 1.050 AJ lead to an "effective" OG of 1.100? (assuming 1L of water = 1000g)

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Yes it is still useful. You at least know where you are starting, as chthon states the second part of adding 50g to 1l of 1050 solution doesn't give 1100 solution.

Here is a great table that illustrates that dissolved sugar in g/l is linear with SG.

I converted it to a graph here

Don't confuse this with adding 57 g of sugar to 1 l of water to get 1020 solution, that is not what it says. It states that in 1 litre of 1020 solution there is 57g of dissolved sugar.

As chtlon correctly states in their answer adding sugar to 1 litre of water gives a non-linaer relationships due to the change in volume. I am really struggling to find a table to support this even though I know it is true.

==== EDIT ====

Found some grpahs from Fermclac and the maths that makes them.

Also, 50g Honey != 50g sugar due to the water content of honey => 50g of honey is about 41g of sugar.

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Your first idea is correct. Your gravity reading will be incorrect if not everything is dissolved.

However, the second part on adding the honey is incorrect.

The gravity reading means how many times heavier than water the solution is. So if 1 liter of water weighs 1 kg, 1 liter of solution weighs 1.050 kg in your example. But this does not mean that only 50 g of sugar is present in 1 l solution.

Some chemists made measurements to find the relationship between the amount of sugar dissolved and the gravity of the solution. Their names were Balling, Brix and Plato. Due to them, the amount of sugar dissolved can also be expressed as g sugar/100 g solution. As a first approximation of the relationship between Brix(or Plato, or Balling), divide the decimals of the gravity by 4. So 50/4 is 12.5, which means a sugar solution of 1.050 contains 12.5g sugar per 100 g of solution.

So in case you have 1l with a gravity of 1.050, you have actually 1050 g of solution, where each 100 g contains 12.5 g of sugar, calculated from this your liter of solution actually contains 131 g of sugars.

The problem is in the fact that adding sugar to water, increases the mass, but also increases the volume. This is not linear and can not be derived in any way. The only formulas for this are based upon the measuring work of Balling, Plato and Brix.

Using the above rough estimates (because the division by 4 actually starts to deviate a lot above a gravity of 1.070), a 1 liter solution of 1.100 would contain 25g sugar/100 g solution, and 25 * 11 = 275 g, so to reach an OG of 1100 you would need to add 144g of sugar.

  • Thanks for the great info! Silly assumption on my part that the volume of a solution would be unchanged. – tempest_col Mar 9 '18 at 19:18

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