Bypassing Points

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ljm

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I have been redoing my spreadsheet attempting (among other things) to eliminate gravity points from recipe calculations.

Starting from m = V*SG*d*P
m = mass (kg) extract
V = recipe volume
SG = specific gravity of wort
d = density of water at 20C (kg/L) = .998203
P = degrees Plato

Here are the results with 4.5359kg (10 lbs) of 2-row, in a batch volume of V=18.927L (5 gal),
m = 3.55208kg.

Method 1:
Nothing fancy, just 72.3804 maximum gravity points with SG = 1.0724, P = 17.601 (ASBC polynomial).

Method 2:
Substitute SG = 1 + P/(258.6-P/258.2*227.1)
This gives a quadratic (nice) in Plato = 18.787 -> S = 1.0776

Method 3:
Substitute for P with the ASBC poly: -616.868+1111.14SG -630.272SG^2+135.997SG^3
The subsequent quartic equation is not nice. So I used an online solver to get S = 1.0011 which is even worse. An error on my part is possible, but I tried it a few times with and without consolidating the constants.

Well, says I, method 2 is a bit off from the point system, but as much as I respect empirical work, the 42.xxx pppg (38x.xxx LDK) always seemed a bit squiffy.

So I tried a grain amount more in line with my typical starting max gravity using 5.443kg (12 lbs) 2-row. Here are the max values.
Method 1: S = 1.0869 -> P = 20.869
Method 2: P = 22.541 -> S = 1.0944
Not very good agreement at all. Things are getting worse.

Even at 5 lbs of grain, method 2 gave 37.5 pts versus method 1 at 36.2. So what gives? I expected ASBC would be an improvement. I realize these max gravities might be pushing the ASBC polynomial, but Geez...

So which should I trust more? Method 1 or 2 or none of them? (I really don't want Method 3 anyway 'cause it doesn't fit into a spreadsheet.)

Thanks in advance for some input.
LJM
 

bitteritdown

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So you're trying to eliminate the standard nomenclature of (pppg = points per pound per gallon = points):

Fine Grind Extract (points) * Pounds of Grain / Final Batch Volume = Max Extract (points)

37 points * 12lbs / 5g = 88.8 points

With an efficiency of 75%:

88.8 points * 0.75 = 66.6

Converting back to SG:

66.6 points / 1000 = 1.0666 = 1.067 OG

Even at 5 lbs of grain, method 2 gave 37.5 pts versus method 1 at 36.2. So what gives? I expected ASBC would be an improvement. I realize these max gravities might be pushing the ASBC polynomial, but Geez...
It appears that you're attempting to calculate the Fine Grind Extract which is a laboratory measured value that cannot be computed.
 
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ljm

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This is what I am trying to do without using gravity points: Given an extract percent (FG, CG, CGAI, whatever), calculate maximum expected SG or Plato. Then I can apply whatever efficiency my system has. No assumption of gravity points (eg 37 for 2-row) referenced to sucrose (method 1) need be used. Yes, it is more complicated that using 37 for 2-row, but inquiring minds want to know. And so does my spreadsheet.

Using extract percent of a grain, I can find the maximum mass (or weight) of extract obtained from that grain. My thinking is that this together with the batch volume and plato/SG math relationships should yield a plato (method 2) or SG (method 3) value. (No clue what happened to method 3. Oh well. Too mathematically complex anyway). But the results I presented above indicates that method 2 is in the ball park. The questions are: 1) is my methodology totally off base, 2)can this method be improved and 3) could it be more trustworthy than using sucrose-based gravity points?
 

bitteritdown

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Thanks for explaining.

In my example above the fine grind extract percent is:

37 (pppg of malt) / 46 (pppg of sugar) = 0.80435 = ~80%

12 lbs of malt * 0.80435 = 9.6522 lbs of fermentable and unfermentable starches

9.6522 lbs / 5 gallons of water = 1.23067016 SG -or- 49.5 Plato

Is that what you're attempting?
 
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ljm

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Thanks for explaining.

In my example above the fine grind extract percent is:

37 (pppg of malt) / 46 (pppg of sugar) = 0.80435 = ~80%

12 lbs of malt * 0.80435 = 9.6522 lbs of fermentable and unfermentable starches

9.6522 lbs / 5 gallons of water = 1.23067016 SG -or- 49.5 Plato

Is that what you're attempting?
Not really, but getting closer. Thanks for replying. I guess I am not very good at explaining.

I am starting with a percent extract from whoever makes the grain. Eventually CGAI, but for now I just need a working percent. This is the same percent from which your example of 37pppg is derived. (0.80 * 46 = 37)

For example Briess has Brewers Malt with CG extract = 80.0% and moisture 4.2%. That makes CGAI = 76.6% which would be the maximum extract I could expect if I had 100% efficiency.

With 5kg (working with the metric system makes some of the math more transparent - my preference), I could expect a maximum extract of 3.83kg = m in the equation of the first post. With my efficiency, this mass would be multiplied by 75%. It is not necessary at this point to show the method, but lets do it anyway: .75 * 3.83 = 2.8725kg. I won't round until the bitter end to minimize significant figure errors.

For a batch volume of 20L (wort, not water), this maximum extract could also be found by (I think) the formula presented in the first post:

m = V*SG*d*P
m = mass (kg) extract = 2.8725kg
V = recipe volume = 20L
SG = specific gravity of wort
d = density of water at 20C (kg/L) = .998203
P = degrees Plato

But I know the extract mass and want to use this formula to find SG or P. The trouble is that both SG and P are unknowns. The good news is that there is a pretty accurate equation giving SG as a function of Plato:

SG = 1 + P/(258.6-P/258.2*227.1)

The resulting quadratic equation in Plato gives 14.38 P from which I get 1.0585 SG.

By comparison, sucrose pppg is given somewhere around 46. I have been using 46.214 for no good reason. This translates to 385.68 LDK - points per kg per liter.(LDK = Liter Degree per Kilogram - the metric equivalent.) Then using traditional sucrose-based gravity points, we should get

385.68*2.8725kg/20L = 55.39 gravity points for SG = 1.0534 or P = 13.66

You could do this in the Queens Measurements, but my head hurts. And Britain mostly gave up those silly units, anyway.

Assuming my math was OK it's not a huge difference, but as gravity goes up, so does the gap. And that is my puzzle. Why aren't these SG and P values closer to each other? What am I missing? Did I do something bad? Is one better than the other? Are they both rough estimates and you can flip a coin.

I used to brew by knowing how many total pounds of grain my system could handle and how many ounces of hops would be tasty. If I wanted a dark beer, I would add some darker malt and less hops; light beer? maybe vary the light grains and use more hops. Mash at high or low T, depending. I never tried to estimate the OG although I would always measure it and it was always in the happy range of 1.05-1.07. Good enough for me. Hey, I even won some local competitions.

Now I have more time on my hands and am getting deeper into the weeds to learn some of the quantitative aspects of this dear hobby. Just because it is fun.
 
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ljm

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Thanks Gypsy,
Well, that formula is a different way of thinking about it: just a simple Plato definition applied to the mash, assuming 1kg/L for water. Gonna have to scratch my head for a while. There has got to be a catch. Just too obvious. I'll report back after trying it out. Am I trying to make it too complicated? It wouldn't be the first time.
 

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Max possible gravity would be a no-sparge, first-runnings only brew, so max strength is just a matter of your liquor to grist ratio. Generally though, you're looking at 20-24P, depending on your grist
 
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ljm

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Hmmm. The P value depends entirely on mash thickness = how much strike water. I could adjust the thickness to make it agree with whatever number I want. Seems like the formula is really intended to compare experimental with predicted gravity of first wort to obtain mash efficiency for a particular batch.
 

bitteritdown

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Would you mind sharing the equations you came up with for both SG and P?
 

ajdelange

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By some means you have concluded that you have 5 gallons of wort containing 3.55208kg extract implying 0.187673 kg extract per liter. I don't know how you got "maximum gravity points" from that (nor do I really know what that term means) but the specific gravity of a solution with that amount of extract in it is
1.0721000 so 1.0724 isn't a bad guess. In method 2 you used the inverse Lincoln equation to get an estimate of 1.0776 which is a fairly good estimate and I think you should probably just stick with that. If you want a very precise answer you will have to solve the quartic equation. If you can write code, root bisection is a great method for doing this but you can do Newton's method pretty easily in a spreadsheet (for a handful of iterations at least). The mass of extract in a liter of wort of specific gravity S is S*dens_water*P(S)/100. Make the new polynomial P_ in which the coefficients are those of the ASBC polynomial divided by 100 and multiplied by the density of water. Then the task is to find S which satisfies S*P_(S) = w where w is the extract in kg/liter. Start off by computing S0 = w/2.6055 + 1. That yields 1.0720 as an initial estimate (as good as your first guess). Now inert that into S0*P_(S0) to calculate the extract at S0. Subtract this from w to get the error. Calculate the slope of the extract vs SG curve from the polynomial slope = ((4*a3*S0 +3*a2)S0 + 2*a1)*S0 + a0 in which the a's are the modified ASBC coefficients with a3 being the one multiplied by S^3, etc. Divide the error by the slope to get a correction. Apply the correction to get S1 = S0 + correction. Now repeat using S1. Etc. If you put this into your spreadsheet cleverly using the $A$1 feature carefully you should be able to just copy and past as many stages as you like (depending on how accurate you want to be).
 
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ljm

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Thanks for the very helpful post AJ. I was hoping you might jump in. Although I have a habit of not contributing (not knowing if anyone would be interested) I have long followed your comments and benefited from your insight.

It is comforting to know that Method 2 is valid. I will also try Method 3 again (same as Method 2, but using the ASBC polynomial) just for fun to compare and see how it incorporates into the recipe formulation process.

By some means you have concluded that you have 5 gallons of wort containing 3.55208kg extract implying 0.187673 kg extract per liter.
My objective is recipe formulation without ppppg although it is included in Method 1 for comparison. I haven't set an SG goal. I also haven't bothered with system efficiency in any of the calculations because I assume I can tack that on when things start to get serious. E.g. multiply extract mass by efficiency and continue with Method 2 or 3 calculations.

Preparatory to Methods 2 & 3 I used as a starting point the FGAI (unrealistic, but this is a proof of concept exercise), then FGAI * kg grain = kg of extract in a 5 gallon (18.93 L) batch.

I don't know how you got "maximum gravity points" from that (nor do I really know what that term means)
Sorry about sloppy terminology. "Maximum gravity points" refers to the traditional Method 1 of mass grain*ppppg where I assumed the full 36.2 ppppg for the grain calculated from extract % * 46.214. (Neither extract mass nor efficiency were used.) Elsewhere I referred to "maximum" with reference to SG or Plato as values without accounting for system efficiency.

I generally like to know the best (or nearly so) way(s) of working a problem before introducing more approximations. It is good to know I am on the right track - that Methods 2 & 3 are valid and may be improvements on Method 1. (?) I realize I am splitting hairs on the calculations because my brew day is pretty sloppy anyway. But I would like to have a good recipe formulation process that might be used as a goal to improve the brewing process.

And besides, I am one of the many Americans who can't understand why we cling to 4 qts/gal, 32 fl oz/qt, 16oz/lb, 5280 ft/mile with water boiling at 212F. And ppppg. Even the Brits who started this business have given it up for the most part. Yes, I could use LDK instead of ppppg, but throwing a sucrose-related number (46 or 385) that seems to float up and down a bit adds another wiggle in the calculation.
 

ajdelange

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It is comforting to know that Method 2 is valid. I will also try Method 3 again (same as Method 2, but using the ASBC polynomial) just for fun to compare and see how it incorporates into the recipe formulation process.
Since you reported having a problem with the 'online solver' I suggest you give Excel's solver a shot. It requires setup for each problem which is a bit of a PITA. That's why I suggested the Newton' method approach. You type a value in a cell and the answer appears in another cell. If you hide the individual calculation on a second sheet this keeps the page with which the user interacts neater.
 
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ljm

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Both, if it's not too much trouble.
Keep in mind this is all metric units (kilograms and liters) and I have accumulated constants where possible to keep it simpler. The quartic is easiest to set up but hardest to solve.

Both of these equations start from the first equation at the top of this thread for finding mass of extract from other info. They differ only in the substitution: 1) Lincoln equation inverse expressing SG as a function of Plato and 2) ASBC equation expressing Plato as a function of SG.

AP^2 + BP + C = 0
A = 0.120449
B = 258.6+0.879551 * m/(d*V)
C = -258.6*m/(d*V)
d = 0.998203 kg/L for density of water at 20C
m = kg of extract = % extract * kg grain
V = batch volume in Liters

The Plato result is a fraction in decimal form - multiply by 100 to get Plato degrees.

Here is the quartic to solve for SG that I haven't yet succeeded in conquering, but hopefully will with AJ's advice. The constants come directly from the ASBC polynomial. m, V and d are defined above.

pSG^4 + qSG^3 + rSG^2 + sSG - m/(Vd) = 0
p = 135.997
q = -630.272
r = 1111.14
s = -616.868

With luck I have transcribed everything correctly.
 
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ljm

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you have 5 gallons of wort containing 3.55208kg extract implying 0.187673 kg extract per liter
the specific gravity of a solution with that amount of extract in it is 1.077656
The quartic was so easy to set up - much easier than the messy quadratic. What could go wrong?

Started with the ASBC equation: 135.997 SG^3 - 630.272 SG^2 + 1111.14 SG -616.868

Multiplying by SG and .998203/100 and subtracting the kg extract/L gave the quartic

1.357526 SG^4 - 6.291394 SG^3 + 11.0914328 SG^2 - 6.157595 SG - 0.187672 = 0

Using the Newton-Raphson method for convergence to the kg/L extract value (.187672) I got convergence to SG = 1.072100 - virtually the ppppg result. Suspecting I screwed something up, I put the equation into the online solver I couldn't previously get to work and got exactly 1.072100 for the full polynomial converging to zero. Whoopee, sort of.

Recall the inverse Lincoln-produced quadratic gave 1.0776.

So at least in this case, ppppg is the winner. Seems counter intuitive.

Is this the end, or am I still missing something?

Thanks.
 

ajdelange

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The correct solution, to 9 decimal places (not that we would ever need that many but the Plato tables go to 6) is 1.072100038 not the value I gave in N0. 13 (subsequently edited) which makes the simple S = w/2.6055 + 1 ( w is kg/L) quite appealing - better than the Lincoln Equation. The reason it works this way is that the actual relationship is very linear so that Newton goes to the right answer in a step or two at most. In this particular case the first correction to S = w/2.6055 + 1 is 7E-5 and the second -4E-10. IOW you don't need a second step. Checking at a couple of other sugar levels condfirms that a second step is not needed but the first step does refine the answer to at least 6 decimal places.

Nonetheless it is a bit of a pain to have to evaluate the 4th order polynomial to get w from S and the 3rd order one to get the slope. But, as we just noted, this is a very linear problem which means the slope is almost a constant. You can, therefore, not bother to compute it but rather just use 2.6055.
 
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ljm

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This is very helpful, but also raises more questions. (That's good, right?)

Observation: When I used Newton on the quartic and came up with 1.072100 in one step, it seemed prudent to explore the unlikely possibility that there was another root in the near neighborhood. S0 = 1 arrived at the same answer in 2 iterations; S0=2 got there in 4 steps. Very efficient method.

Very interesting and still baffling to me:

The reason it works this way is that the actual relationship is very linear so that Newton goes to the right answer is a step or two at most.
I assume "actual relationship" refers to experimental SG vs w data - the stuff ASBC keeps under wraps. If so, does the equation (S=w/2.6055+1) come from data just in the neighborhood of 1.07? What are the limitations? Would further refinements (e.g. addition of other parameters or constants and/or further application of Newton) ever yield significant improvement over a wide range of wort gravities? In other words, can it be kept very simple?

Going back to post no. 1, my thought would be to use this (or similar) equation or the inverse Lincoln (Method 2) to arrive at S0 and further refine it to at least the 4th decimal point with the ASBC equation (Method 3) solved by Newton. I suppose I could even arrive at a guesstimate by finding Plato by equating wort volume and water volume ~= kg water to go with my kg extract, then -> S0. Newton would massage the end result of each of these methods to a good guess at SG.
 

ajdelange

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This is very helpful, but also raises more questions. (That's good, right?)
Yes, absolutely!

Observation: When I used Newton on the quartic and came up with 1.072100 in one step, it seemed prudent to explore the unlikely possibility that there was another root in the near neighborhood. S0 = 1 arrived at the same answer in 2 iterations; S0=2 got there in 4 steps. Very efficient method.
Caveats here. The "actual relationship" is S*P_(S) with P_(S) being the ASBC polynomial with its coefficients multiplied by the density of water divided by 100. Therefore, it is only valid for 1.000 < S < 1.080 which is, IIRC, the domain over which the ASBC polynomial is valid. This means that any solution giving an answer outside the range 1 to 1.080 is invalid and this implies that 0 < w < 0.2 kg/L. If you want to figure worts of > 20 °P you will have to find a polynomial which has a wider range than the ASBC polynomial and if asked to pick one I'g go with the ICUMSA (International Committee for Uniform Methods In Sugar Analysis) polynomial for 20 °C as it and the ASBC polynomial seem to agree very closely over the range for which the ASBC polynomial is valid.



I assume "actual relationship" refers to experimental SG vs w data - the stuff ASBC keeps under wraps.
Yes but it hardly kept under wraps. It is just the Plato Commission data tweaked for 20/20 apparent SG. It is published in the ASBC MOA's, Analytica etc.

If so, does the equation (S=w/2.6055+1) come from data just in the neighborhood of 1.07? What are the limitations?
No, it is derived from a plot of S*P_(S) vs S -1 The best straight line fit to that data has a slope of 2.6055.

Would further refinements (e.g. addition of other parameters or constants and/or further application of Newton) ever yield significant improvement over a wide range of wort gravities? In other words, can it be kept very simple?
As noted earlier it is possible to expand to a wider concentration range (and even different temperature if you want but that's getting hairy) by using the ICUMSA data rather than the Plato Commission data.
 

ajdelange

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I looked into expanding the range of this methodology to stronger solutions. To do that all I have to do is extend Dr. Plato's work to higher Plato worts. Fortunately the folks at ICUMSA (I picture blokes in khakhi shorts and topees) have already done this for us. To start we want some assurance that the ICUMSA results and the ASBC results agree over the domain of the ASBC polynomial (1.00 < S < 1.083). Comparing what ICUMSA gives us for the SG of a solution of given strength to what the Plato data says for that same solution we find, in the domain of interest, that the peak discrepancies are +.017 SG points (thousandths of an SG unit) at about 3 °P and -0.45 at 20 °P, and - 0.005 point at 10 °P. The differences lie on a smooth curve. Thus it seems that the ICUMSA data is acceptably close to the ASBC data where they overlap. We already have P(S) = ((135.997*S - 630.272)*S + 1111.14)S -616.868) which gives us strength w/w from S < 1.083. We now just need a polynomial, P_X(S) which gives us the same information from the ICUMSA data for stronger solutions. This polynomial is P_X(S) = ((((((50.3643*S -491.324)*S + 2104.96)*S -5157.12)*S +7897.64)*S -7723.45)*S+4711.92)*S -1392.99 valid whenever 1 < S < 1.51923 (95% w/w). We had to go to additional powers as there is a lot more information in the data spanning 0 to 95 °P than there is in the restricted range of 0 - 20 °P to which the ASBC polynomial is limited. You can use P_X(S) , where X stands for 'extended' exactly as you used P(S) previously.
 
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ljm

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Great work!! I had been fruitlessly looking on the internet for some ICUMSA-ish polynomials that might be useful for a shotgun wedding with ASBC. I don't know how you came up with this (afraid to ask or more properly afraid to get an answer), but I will gratefully test it. We'll see how efficiently Newton works with a 7th order polynomial over the range. I expect a need to limit the initial guess to something narrower than 1 < S0 < 1.5 for high and low gravity worts. But you never know. Exciting stuff!

If you like I can come back here to post findings. Or you probably already know how well it works... Anyway I am about to find out.

Much thanks again for another solid contribution. Lovin' it.
 

ajdelange

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When the Plato tables were made, they measured specific gravity of solutions of known strength w/w. ASBC plotted °P vs SG and fit a curve through the plotted points. I did exactly the same thing except that I got my specific gravity data from the ICUMSA formula. Otherewise, the process is exactly the same.

You will have to come up with a method of initializing Newto Raphson. This will, again, be exactly what we did for the ASBC data. You will need to plot water_dens*S*P_X(S)/100 and find the best linear or perhaps quadratic fit to it. Then, for a given kg/L insert that into the fit and solve for S0.
 
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ljm

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I have spent a little time looking at the expanded P_X polynomial and it is remarkable how much agreement there is between the ASBC and your P_X construction - even though ASBC is only advised up to 1.083. This should be of special interest to a homebrewer with typical low precision equipment. That includes me, at least for now. I think there may be a trio of Plato/Brix hydrometers in my future to add at least another decimal.

For example, I input SG values up to SG = 1.1 into both equations and looked at the resulting w = kg/L and Plato values:

SG / w (ASBC) / w (P_X) / ASBC / P_X
1.010 0.025817 0.025867 2.5608 2.5656
1.040 0.103746 0.103744 9.9935 9.9933
1.080 0.208400 0.209297 19.3310 19.3215
1.100 0.260988 0.260906 23.7689 23.7614

This suggests to me that one needn't be too anal about finding a starting specific gravity for convergence of (water dens)*S*P(S) to the expected recipe kg/L extract. As to the Newton-Raphson method, I tried some reasonable guesses (e.g. recipe target SG) and unreasonable ones with both polynomials and they converged speedily. Generally not more than 4 iterations for P_X. Certainly your linear approximation S = w/2.6055 + 1 is more than adequate for even the highest gravities a homebrewer is likely to encounter. (Anything greater than SG 1.1 I am not likely to prepare or drink, anyway.)

Thanks again, AJ.
 
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