Calculation question (relating corn sugar to gravity)

I am considering adding some corn sugar to a recipe, and was trying to figure out in my head how much I would need.

In my mind, corn sugar is basically 100% soluble sugars. SG as I understand is a density relative to water ... so 1.000 is water, which would weigh 1Kg/Litre.

From that, I figured I could calculate the gravity quite simply... if I added .05 kg of corn sugar to 1 L of water, that 1 L of water should weigh 1.05 kg and the SG should be 1.05.

I followed this math out that if I wanted to increase a 20 L batch (~5gal) by 0.02 gravity, that I would need .02 kg /L x 20 L = .4 Kg or about 0.9lb

The recipe calculators don't agree... they come in about half... why is that? Where did I go wrong?

1 pound of corn sugar adds about 0.009sg to a 5.25gallon(20L) batch.

If 2.25 pounds adds 0.020sg to a 5.25 gallon batch then 1 kilo of corn sugar should add 0.020 to a 20L batch.

I used Beersmith and my calculator to figure this so I think I am correct!

I hope this helps! Good luck!

If you add .05 kg sugar to 1 L (which weighs pretty close to 1 kg) you will have a 100*0.05/1.05 =4.76 °P solution. Using the simplest conversion to SG that would give you a specific gravity of a about 1.0000 + 4*4.76/1000 = 1.019. The exact conversion is 1.01872


To calculate the sugar addition necessary to increase the strength of a given wort by a given amount proceed as follows. As an example let us suppose that we want to increase the strength of 1.040 wort to 1.050.
1. If you have the original solution strength in terms of specific gravity convert to °P or, if you have it in °P, convert to specific gravity. Both are needed. Use the ASBC tables, the ASBC polynomial:
°P = ((135.997*sg -630.272)*sg +1111.14)*sg - 616.868
the ICUMSA tables or polynomial, the Lincoln equation or, if all else fails the fact that 1 °P is approximately equal to 0.004 SG to do the conversion. Example: wort of SG 1.040 corresponds to 9.994 °P.
2. Find density of original solution by multiplying the specific gravity by .998203. Example: 0.998203*1.040 = 1.03813
3. Find the weight of 1 L of the solution by multiplying the density by 1000. Example: 1 L of 1.040 wort weighs 1038.13 grams.
4. Find the amount of sugar in 1L of the solution by multiplying the weight of 1L by °P/100. Example: 0.09994*1028.13 = 102.75 grams/L
5. Subtract the weight of the sugar from the weight of 1 L of solution. This is the weight of the water in 1 L of solution. Example: 1038.13 - 102.75 = 935.38 grams water
6. Determine how much sugar is needed with that much water from
s = w*(°P/100)/(1 - °P/100)
where s is the grams of sugar, w is the weight of water from Step 5 and °P is the new strength. Example: 1.050 SG corresponds to 12.39 °P and
935.38*(.1239)/(1-.1239) = 132.284
7. Subtract s from the amount of sugar already in 1L as determined in Step 4. Example: 132.284 - 102.75 = 29.534
8. Add this many grams of sugar to each liter of wort.

I'm sure this looks horrible but if you put it all into an Excel spreadsheet it's available thereafter. The only tricky part is calculating sg from °P and that can be done accurately enough by using the inverse of the Lincoln equation. Or you can invert the ASBC polynomial using the Excel Solver.

[Edit] The Lincoln Equation: °P = (463 - 205*sg)*(sg-1)
Inverse: sg = (668 - sqrt(668^2 - 820*(463 + °P)))/410

Brain still... computing...

Thanks ajdelange

mrkeeg said:

... so 1.000 is water, which would weigh 1Kg/Litre.

Click to expand...

Not exactly. Specific gravity is measured in reference to a specific temperature. If your hydrometer says "20C/20C" on it, then if you read the numbers off the instrument when the solution is 20C, it will tell you the density relative to water density at 20C, which is about 0.9982071 kg/L. So, specific gravity measurements depend on the reference temperature.

mrkeeg said:

The recipe calculators don't agree... they come in about half... why is that? Where did I go wrong?

Click to expand...

You forgot to take into account that by adding the sugar to the water, you are changing the volume of solution.