When we last left off in part 1, we were about to look “head on” at the three dimensional graph of the Tinseth Formula. If you have not yet read part 1, I suggest you do before continuing.

Now our 3-D graph looks 2-D. The upper edge of the graph (that extends into the purple) represents how the IBUs of 1.000 specific gravity wort (essentially water) changes over time. The bottom edge represents how the IBUs of 1.100 specific gravity wort changes over time. In between we have all of the possible specific gravities between 1.000 and 1.100 represented and how the respective IBUs change with time. The shape of the curve becomes compressed, but it remains the general shape of our familiar graph of IBUs versus time.

For clarity I have included the same graph in 2-D demonstrating worts of specific gravity 1.000, 1.025, 1.050, 1.075, and 1.100. The darker the line is, the higher the specific gravity is.

Now what happens if we look at the graph “head on” in the other direction? Look at the 3-D graph of IBUs versus Specific gravity and time. We will now be looking across that surface from left to right so we can see what happens to the possible IBU values as gravity increases.

In this view we are seeing how the essentially how the potential IBUs decrease for a 90 minute boil as the gravity increases. The top edge of the graph represents what happens if you vary the gravity from 1.000 (far left) to 1.100 (far right) while the hop boil time remains fixed at 90 minutes. Everything below that represents the possible IBU outputs for every gravity when we reduce the hop boil time.

We can see that how this makes sense by turning to theoretical brewing for an example. If I make six gallons of 1.100 gravity wort and boil one ounce of 7.8% Palisade hop pellets for 90 minutes I will have 15 IBUs. Therefor I know that if I decrease my hop boil time I can achieve a range of IBUs between 0 and 15.

**Important ramifications of specific gravity: **

Extract brewers should be paying attention here, but I am also speaking to anyone who uses extract in partial mashes. Consider the following scenario.

Cortney is making 6 gallons of strong blonde ale from extract and one ounce of Palisade hops (7.8% alpha acids) for bittering. A gravity of 1.060 seems appropriate to her. She decides that she wants to boil the hops for 90 minutes, since that would give her a reasonable 22 IBUs. She also read on an extremely helpful online forum that she can add the extract late in the boil to help her achieve the appropriate beer color and flavor. She adds the hops at the beginning of the boil (90 minutes before completion of the boil) and then adds the extract at 15 minutes (before completion of the boil). Two months later she cracks her first blonde ale open and reports that “It’s pretty good ale, but it’s much more bitter than I wanted”.

So what happened here to cause her beer to be too bitter? She started off by boiling her hops in water, which we can assume to be near 1.000, and then much later added extract to achieve her desired 1.060. Here is the graph of how gravity affects a 90 minute boil of an ounce of 7.8% alpha acid Palisade hop pellets.

The green marker indicates the gravity Cortney used for her calculations. The purple marker indicates the theoretical final gravity of the water had she not added extract within the last 15 minutes. Reason indicates that the IBUs Cortney actually achieved falls somewhere between the purple marker and the green marker (a good guess would be much closer to the purple). I will now take this a step further by trying to make a good estimate at the actual IBUs Cortney achieved.

We know that there must be a correlation between IBUs versus time and the rate of change of IBUs versus time. Our exponential function of IBUs versus time has a very fast rate of change between flameout and 30 minutes and a slow **rate of change** between 30 and 90 minutes. I can demonstrate this without much math just by drawing slopes (lines connecting two points) on the graph.

On the left we see the function of IBUs depending on time. On the right we see two average rates of change applied to the function. These are slope lines (known to some as secant lines) that describe, on average, how fast the quantity of IBUs are changing. The general idea: the steeper the slope, the greater the rate of change.

Using calculus we can “take the **derivative**” of a function and find the exact rate of change for any point in the function. Again I will spare you the details, but I encourage you to look into calculus if it interests you. I will provide the derivative followed by a graph of the derivative. Also note that the units for the rate of change of IBUs are “IBUs per minute”.

Remember that this graph is of the rate of change, meaning a higher the value in the graph represents a greater rate of change. This graph keeps in line with what our ideas are concerning the exponential growth of IBUs; they begin with fast increases then slowly taper off with slower increases. Now I will compare the values for the rate of change of IBUs between water (1.000) and Cortney’s ideal wort (1.060).

I hope you notice the substantial differences in the respective rates of change of IBUs. This is where Cortney’s beer took the unexpected turn for the bitter. Boiling her hops in water greatly changed the utilization. This will have a profound effect on the bitterness.

I will take a mathematical stab at the IBU quantity Cortney achieved. I can **integrate** the rate of change of each component of the boil: from 0 to 75 (prior to the addition of extract) plus 75 to 90 (when the wort was 1.060). This will return to us a value for IBUs achieved. That long vertical S-shaped symbol is an **integrand**. You don’t need to know how to do the math, just realize that there are values and variables above and below the integrand. The value zero is the beginning of the hop boil, a is the time since the hop boil began that the extract was added, and F is the total amount of time the hop was boiled.

After plugging in the values for this scenario, the number of IBUs Cortney likely achieved was 37.3. As I thought, her value fell between the value for wort of specific gravity 1.000 (37.7 IBUs) and wort of specific gravity 1.060 (22.0 IBUs).

The fact that Cortney added the extract with 15 minutes to go probably helped the overall appearance and possibly flavor of her beer, but it also threw off her calculations for bitterness by roughly 15 IBUs. Cortney’s professional palette was able to detect this difference and I am sure that many of us would also notice an additional 15 IBUs. There was a 68% difference when compared to her calculated value.Imagine what might have happened if this had been an IPA recipe with a larger bittering addition or higher alpha acid hop.

We can also look at how the IBUs vary during the 90 minute boil based on when the malt extract was added. We know that the boil time F is fixed at 90, but we can allow a to vary from 0 to 90.

Here are all of the possible times Cortney could have added the malt extract and the resulting effect on the IBU output. On the far left we see that she started with the malt extract added when the hops were added. This would be the equivalent of her 1.060 post boil wort. On the far right is the case where she adds the malt extract at flame out, making the effective post boil wort 1.000 (as far as the Tinseth formula is concerned).

This is also useful information because extract or partial mash brewers can have more control over the bittering additions for their beers. Higher utilization means less hops are required which in turn means more hops for something else. By bolstering your barleywine with extract at the end of the boil you can use fewer hops to achieve the same IBU goal. An imperial stout might only require two ounces of high alpha acid hops to balance its roast-y goodness merely by delaying the addition of extract.

**Side Note: The Partial Boil**

Some extract brewers are forced to use a partial boil. This does have an effect on the hop utilization and the Tinseth formula must be tailored to this. For the value of g you must use the post boil gravity **before** you add water. The value for Vf that you must use is the volume of wort in the fermenter. To be clear this means **after** you add the water. You may want to consider adding all or part of the extract late in the partial boil to help mitigate the loss of utilization due to a partial boil.

**Concluding Remarks:**

I hope you find what I have written here interesting and, above all, useful. In this article, I employed concepts from basic multiplication all the way to integral calculus in an effort to achieve a better understanding of brew science (or art, whichever you feel best describes what you do). Technology and software have come a long way in making these calculations instantaneous and effortless, but they rob us of an opportunity to immerse ourselves in the math that helps us analyze and better understand the choices we make when we brew.

(There is a PDF of the original word document available for those who are interested.)

Interesting article.

Did my own maths on the formula but you gave me more insight.

Never thought about the IBU potential, and how it could be used.

Can Cortney calculate the IBU without doing the integral?

Cortney used for the first 75 minutes of the boil 95.02% of the IBU potential as showed in part 1.

So 4.98% is left for the rest of the boil.

This gives a virtual alpha acids of 7.8% * 4.98/100 = 0.39%

Boiling this virtual hop for 15 minutes with a gravity of 1.060 gets 0.5 IBU.

Together with the 36.8 IBU from the 75 minutes boil she gets 37.3 IBU.

With less rounding its 37.3429 IBU.

What do you think about this calculation?

My own maths resulted in a formula like this:

0 = c + f(t) + f(g) + f(w) + f(a) + f(V) + f(I)

That’s not very impressive, but the it can be used for things like here:

http://en.wikipedia.org/wiki/Nomogram

and here:

http://en.wikipedia.org/wiki/Circular_slide_rule

For short, this is this is the outcome:

https://drive.google.com/folderview?id=0B_o-Wm3o8HPdVk5kb3F1Wjgwb1E&usp=sharing

Now I wonder, where should I put the IBU potential on those graphs?

@saleb

Wow, excellent thoughts on the calculation using the U% figures!!! This is something I had not considered. It’s 11:12 PM here at the moment, so I’m not exactly on the ball here… but if we look closely:

Already I see one problem: the total percentages of IBU potential utilized add up to 100%, which is impossible. We can never achieve 100% of the IBU Potential.

I do see some merit in this still, however, so lets carry on.

The total U% available to us is 97.3%. If, however, we subtract the U% value at 75 (95.0%), we can describe the U% from 75 to 90 minutes. This works out (rounded) to 2.3%.

Ipot for 1.000 is 38.7.

Ipot for 1.060 is 22.6.

(38.7*.95)+(22.6*.023)= 37.3

The numbers do check out, and it seems that it works. Without closely looking at the products involved in my integration, I can bet that the subtraction portions involving e will look very similar to the case where we subtract the formulas for U% evaluated at the given times.

Very clever and thank you for your comments!

I will have to look into your own formula, since I am not familiar with that material or nomenclature, but I will send you a private message with what I do find out.

I use the method described by @saleb. Regardless, great work here! I’ll PM you and send you my spreadsheet.

What’s needed to get a look at the original PDF of the ‘Word’ document?