Gravity Calculations in Metric

The Grim Beaker

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Hi
Sorry if this has been asked already, but after scouring brewing websites for days and getting multiple headaches, I'm still no nearer find the answer to the question;

How do I calculate gravity using metric rather than US units?

I've read plenty of things that use the US units, for example;

A rule of thumb is one pound of liquid extract per gallon of water for a light bodied beer. One and a half pounds per gallon produces a richer, full bodied beer. A pound of LME typically yields a gravity of 1.034 - 38, as measured by a hydrometer, when dissolved in one gallon of water. DME yields about 1.040 - 43. These yield values are referred to as Points per Pound per Gallon. If someone tells you that a certain extract or malt's yield is 36 points, it means that when 1 pound is dissolved into 1 gallon of water, the gravity is 1.036. If that 1 pound is dissolved into 3 gallons, its gravity would be 36/3 = 12 or 1.012. The gravity is how the strength of a beer is described. Most commercial beers have an Original Gravity (OG) of 1.035 - 1.050.

Example of Gravity Calculations
If you want to brew 5 gallons of 1.040 gravity beer, this would call for 5 lbs of DME having 40 pts/lb./gal, or 5.5 lbs of LME having 36 pts/lb./gal.

i.e. 1.040 = 40 pts/gal x 5 gal = 200 pts total

200 pts = 36 pts/lb. x (?) lbs => (?) lbs = 200 / 36 = 5.55 lbs.

5.55 lb. of 36 pts/lb./gal LME are needed to make the same 5 gallons of beer.

Note: The same concept can be used with the SI units of Liter Degrees per Kilogram, i.e., L°/kg or pts./kg/L. The conversion factor between ppg and L°/kg is 8.3454 x ppg = L°/kg.

The above makes sense, however I need to work in metric units. The underlined note above states the concept can work in metric and even gives the factor to convert between ppg and L°/kg (ppg x 8.3454), however I can't find anything that will help me work everything out in metric. i.e.;

1.040 = ??? pts/L x 20L = ??? pts total

??? pts = ??? pts/kg x (?) kg =. (?) kg = ??? / ??? = ??? kg

??? kg of ??? pts/kg/L LME are needed to make the same 20 litres of beer.​

Does any of this make sense??? I've been going round in circles with this for days and now my head is so scrambled I'm not sure I could even find my way out of the front door.
 
Specific gravity is not unit specific - it's a dimensionless ratio of the weight of a solution to the weight of pure water, in metric Kg/Kg. Brix/Plato are both the percentage of sugar (sucrose) of a solution, that is a 10 degree Plato solution is 10% by weight sucrose. Is it the specific gravity you're interested in, or the points? Homebrewers tend to work in gravity points, the digits after the decimal multiplied by 1,000. That is a 1.046 wort has 46 gravity points. I use that number because sucrose, the standard for sugar solutions because there is no water attached to the molecules, yields 46 points per pound per gallon.

LME is generally 38 ppg. In metric, multiply that by the 8.3454 to get 317.13 points/kilogram/liter. If you want a 1.040 SG beer, you are making 20 liters, how much LME do you need?

The total number of points you need is the gravity points desired (40) * the volume (20l), or 800 total points of gravity. Now divide that by the 317.13 points per kilogram to get 2.523 Kg of LME.

Does that make sense?
 
Let’s say we want to use DME with a ppg of 45. We need to compare that to Pure Sucrose to get the extract potential of the DME:

45 / 46.21 = ~ 0.973 or 97.3%

Let’s assume we want to use 5 kg of DME into 32 l of water.

We can then use the following formula:

Gravity (Plato) = (100 * 5 * 0.973) / (32 + (5 * 0.973)) = ~ 13.2 P or ~ 1.054 S.G.
 
Specific gravity is not unit specific - it's a dimensionless ratio of the weight of a solution to the weight of pure water, in metric Kg/Kg. Brix/Plato are both the percentage of sugar (sucrose) of a solution, that is a 10 degree Plato solution is 10% by weight sucrose. Is it the specific gravity you're interested in, or the points? Homebrewers tend to work in gravity points, the digits after the decimal multiplied by 1,000. That is a 1.046 wort has 46 gravity points. I use that number because sucrose, the standard for sugar solutions because there is no water attached to the molecules, yields 46 points per pound per gallon.

LME is generally 38 ppg. In metric, multiply that by the 8.3454 to get 317.13 points/kilogram/liter. If you want a 1.040 SG beer, you are making 20 liters, how much LME do you need?

The total number of points you need is the gravity points desired (40) * the volume (20l), or 800 total points of gravity. Now divide that by the 317.13 points per kilogram to get 2.523 Kg of LME.

Does that make sense?

Nice one, thanks.
I was sort of close to getting there at one point, but then I got it into my head that the gravity points (eg 40 for a SG of 1.040) was tied to gallons, probably because of this bit;

i.e. 1.040 = 40 pts/gal x 5 gal = 200 pts total

So I assumed it would be a different value for litres, and not just the same; 40.
And I then I started wondering how I would even know what the points/kg/litre for an ingredient is (eg LME) if I didn't already know what the PPG was to being with and then converting it. Ended up down a rabbit hole and a headache.
Thanks a lot for the help.
 
Let’s say we want to use DME with a ppg of 45. We need to compare that to Pure Sucrose to get the extract potential of the DME:

45 / 46.21 = ~ 0.973 or 97.3%

Let’s assume we want to use 5 kg of DME into 32 l of water.

We can then use the following formula:

Gravity (Plato) = (100 * 5 * 0.973) / (32 + (5 * 0.973)) = ~ 13.2 P or ~ 1.054 S.G.

Thanks, however not sure on something if wouldn't mind putting up with my idiocy for a bit longer...

Gravity (Plato) = (100 * 5 * 0.973) / (32 + (5 * 0.973)) = ~ 13.2 P or ~ 1.054 S.G.
So the above is;

Gravity (Plato) = (100 * the kg of DME * the extract potential (EP) of the DME) / (the litres of water + (the kg of DME * the EP)) = the Plato &/or the specific gravity.​

Is that right?
If so, what is the "100" in reference to in the below;

= (100 * 5 * 0.973)
 
Thanks, however not sure on something if wouldn't mind putting up with my idiocy for a bit longer...

Gravity (Plato) = (100 * 5 * 0.973) / (32 + (5 * 0.973)) = ~ 13.2 P or ~ 1.054 S.G.
So the above is;

Gravity (Plato) = (100 * the kg of DME * the extract potential (EP) of the DME) / (the litres of water + (the kg of DME * the EP)) = the Plato &/or the specific gravity.​

Is that right?
If so, what is the "100" in reference to in the below;

= (100 * 5 * 0.973)

Let me pull my actual calcs from my extract troubleshooter. I whipped this up late last night from Braukaiser but I know that my calcs have a simpler format. Give me a little bit and I’ll post up a better calc with more detail.
 
Ok. Let me make a quick distinction at the outset here by defining Maximum Extract Potential versus Dissolved Extract:

Maximum Extract Potential (kg) = Fermentable Mass (kg) * Fine Grind (As-Is) Extract (%)

where FGAI (%) for Malt = Fine Grind (Dry Basis) Extract (%) * (1 - Moisture Content (%))
where FGAI (%) for Malt Extract/Brewing Sugars = Potential (p/lb/gal) / 46.21

Dissolved Extract (kg) = Maximum Extract Potential (kg) * Efficiency (%)

We can see that the above equations can be used in a few ways:

1.) We can use it for DME/LME and brewing sugar contributions;
2.) We can use it to determine our thoeretical maximum extract for malt, which we can then apply a conversion or mash efficiency % to to get actual dissolved extract.

To answer your question from above, the factor of 100 simply comes from the calculations of Kai's I use for calculating °P/°Bx. I have always believed it was a consequence of having to convert the decimal percentages in the upstream parts of the equation.

Lastly, the calculation for converting from (°P/°Bx) to S.G.:

S.G. = 1 + (°P/°Bx / (258.6 - 227.1 * (°P/°Bx / 258.2)))

With that out of the way, we can proceed to the main calculation:

Estimated Extract/Gravity (°P/°Bx) = 100 * (Dissolved Extract (kg) / (Dissolved Extract (kg) * Volume (l))

Keep in mind that if you were concerned with First Wort Extract (kg), you'd use Conversion Efficiency (%) to get Dissolved Extract (kg) and Strike Volume (l) to get Estimated Extract/Gravity (°P/°Bx). Conversely, if you wanted Kettle Extract (kg), you'd use Mash Efficiency (%) to get Dissolved Extract (kg) and Pre-Boil Volume (l) to get Estimated Extract/Gravity (°P/°Bx).

So let me shut up and run the calculations. Let's look at 2 different scenarios and target First Wort Extract/Gravity (°P/°Bx).

Scenario 1: Extract Batch

We would like to know what the Extract/Gravity (°P/°Bx) would be if we used 5 kg of DME into 32 l of water. We assume 45 p/lb/gal for the DME:

FGAI (%) = 45/46.21 = 0.973 (97.3%)

Since there is no efficiency component in this calculation:

Dissolved Extract (kg) = Maximum Extract Potential (kg) = 5 * 0.973 = ~4.869

Estimated Extract/Gravity (°P/°Bx) = 100 * (4.869 / (4.869 * 32) = ~13.2 °P/°Bx or ~1.053 S.G.

Scenario 2: Grain Batch (First Wort Extract)

We would like to know what the Extract/Gravity (°P/°Bx) for a No-Sparge batch would be if we used 5.67 kg of Pilsner Malt with DBFG (%) of 80.5% and Moisture Content (%) of 4.20% into 32 l of water. We assume 93% Conversion Efficiency:

FGAI (%) = 0.805 * (1 - 0.042) = 0.771 (77.1%)

Maximum Extract Potential (kg) = 5.67 * 0.771 = ~4.371

Dissolved Extract (kg) = 4.371 * 0.93 = ~4.065

Estimated Extract/Gravity (°P/°Bx) = 100 * (4.065 / (4.065 * 32) = ~11.2 °P/°Bx or 1.045 S.G.

You can extend this basic understanding out to other scenarios by subbing the correct efficiency values and volumes.

Hope this helps.
 
Metric is nicer to work with regardless but yeah gravity isn't an area you need to worry about.
 
Ok. Let me make a quick distinction at the outset here by defining Maximum Extract Potential versus Dissolved Extract:

Maximum Extract Potential (kg) = Fermentable Mass (kg) * Fine Grind (As-Is) Extract (%)

where FGAI (%) for Malt = Fine Grind (Dry Basis) Extract (%) * (1 - Moisture Content (%))
where FGAI (%) for Malt Extract/Brewing Sugars = Potential (p/lb/gal) / 46.21

Dissolved Extract (kg) = Maximum Extract Potential (kg) * Efficiency (%)

We can see that the above equations can be used in a few ways:

1.) We can use it for DME/LME and brewing sugar contributions;
2.) We can use it to determine our thoeretical maximum extract for malt, which we can then apply a conversion or mash efficiency % to to get actual dissolved extract.

To answer your question from above, the factor of 100 simply comes from the calculations of Kai's I use for calculating °P/°Bx. I have always believed it was a consequence of having to convert the decimal percentages in the upstream parts of the equation.

Lastly, the calculation for converting from (°P/°Bx) to S.G.:

S.G. = 1 + (°P/°Bx / (258.6 - 227.1 * (°P/°Bx / 258.2)))

With that out of the way, we can proceed to the main calculation:

Estimated Extract/Gravity (°P/°Bx) = 100 * (Dissolved Extract (kg) / (Dissolved Extract (kg) * Volume (l))

Keep in mind that if you were concerned with First Wort Extract (kg), you'd use Conversion Efficiency (%) to get Dissolved Extract (kg) and Strike Volume (l) to get Estimated Extract/Gravity (°P/°Bx). Conversely, if you wanted Kettle Extract (kg), you'd use Mash Efficiency (%) to get Dissolved Extract (kg) and Pre-Boil Volume (l) to get Estimated Extract/Gravity (°P/°Bx).

So let me shut up and run the calculations. Let's look at 2 different scenarios and target First Wort Extract/Gravity (°P/°Bx).

Scenario 1: Extract Batch

We would like to know what the Extract/Gravity (°P/°Bx) would be if we used 5 kg of DME into 32 l of water. We assume 45 p/lb/gal for the DME:

FGAI (%) = 45/46.21 = 0.973 (97.3%)

Since there is no efficiency component in this calculation:

Dissolved Extract (kg) = Maximum Extract Potential (kg) = 5 * 0.973 = ~4.869

Estimated Extract/Gravity (°P/°Bx) = 100 * (4.869 / (4.869 * 32) = ~13.2 °P/°Bx or ~1.053 S.G.

Scenario 2: Grain Batch (First Wort Extract)

We would like to know what the Extract/Gravity (°P/°Bx) for a No-Sparge batch would be if we used 5.67 kg of Pilsner Malt with DBFG (%) of 80.5% and Moisture Content (%) of 4.20% into 32 l of water. We assume 93% Conversion Efficiency:

FGAI (%) = 0.805 * (1 - 0.042) = 0.771 (77.1%)

Maximum Extract Potential (kg) = 5.67 * 0.771 = ~4.371

Dissolved Extract (kg) = 4.371 * 0.93 = ~4.065

Estimated Extract/Gravity (°P/°Bx) = 100 * (4.065 / (4.065 * 32) = ~11.2 °P/°Bx or 1.045 S.G.

You can extend this basic understanding out to other scenarios by subbing the correct efficiency values and volumes.

Hope this helps.

I'd be lying if I said that some of that didn't go over my head at this point, but I'm starting to get there.

To answer your question from above, the factor of 100 simply comes from the calculations of Kai's I use for calculating °P/°Bx.
So, correct me if I'm wrong here but, that 100 is a constant within the formula.

So if I wanted to know what adding 0.1kg of malto-dextrin with a PPG of 39 to 20 litres of water would do to my gravity, I could use the formula and I'd see that that amount would have a value of ~0.3 °P/°Bx / S.G. ~1.002. That sound right?
 
I'd be lying if I said that some of that didn't go over my head at this point, but I'm starting to get there.

To answer your question from above, the factor of 100 simply comes from the calculations of Kai's I use for calculating °P/°Bx.
So, correct me if I'm wrong here but, that 100 is a constant within the formula.

So if I wanted to know what adding 0.1kg of malto-dextrin with a PPG of 39 to 20 litres of water would do to my gravity, I could use the formula and I'd see that that amount would have a value of ~0.3 °P/°Bx / S.G. ~1.002. That sound right?

I’m going to dig and find out what the factor of 100 is doing. I always thought it was a consequence of dealing with the percentages so I should be able to eliminate it.

First now:

°P/°Bx = 100 * ( ( 0.1 * ( 39 / 46.21 ) ) / ( ( 0.1 * ( 39 / 46.21 ) ) + 20 ) )
 
Nice one.

I'm seeing a slight difference in the S.G. and Plato figures once I convert everything from US units to metric and calculate based on those values, but I assume that would be due to that fact that 5 gallons doesn't equate to 20 litres but close to 19, and therefore the PPG & points per kg per litre will differ? Maybe I should stick with using PPG in my formulas.
 
Nice one.

I'm seeing a slight difference in the S.G. and Plato figures once I convert everything from US units to metric and calculate based on those values, but I assume that would be due to that fact that 5 gallons doesn't equate to 20 litres but close to 19, and therefore the PPG & points per kg per litre will differ? Maybe I should stick with using PPG in my formulas.
Sounds right to me. PPG, PKgL do not change, only the volume is different. S.G. and Plato are not a completely linear conversion so that might be part of your problem. Bottom line, if you're off by a point or two, relax, don't worry, have a homebrew.
 
Sounds right to me. PPG, PKgL do not change, only the volume is different. S.G. and Plato are not a completely linear conversion so that might be part of your problem. Bottom line, if you're off by a point or two, relax, don't worry, have a homebrew.
It definitely seems (to me anyhow) that it is caused by my converting the PPG.
In terms of getting technical with a brew it's very early days, but I've been doing bog standard kit brewing for years for my dad. Decided I was gonna elaborate on things for myself last year and have been tinkering ever since, and trying to build something in excel that once I add my ingredients, a lot of maths would be done automatically, which would help me play around with brews. But I kinda think you can't make changes without at least some understanding of what you're changing.

I think you're right though, time for a beverage or 2.
 

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