CORRECTIONS TO “FROZEN DESSERTS” FORMULA PAGE 62 AND 63 Mathematical Formulation
1. 2. 3. 4. 5. 6. 7. 8.
Determine the desired yield for the product base. Calculate the amounts for each percentage from the chart on page 61. Calculate the amount of nonfat dry milk needed. Determine the total weight of nonfat solids and milk. Calculate the weight of the liquid. Calculate the weight of the milk. Calculate the dairy amount distribution Finalize amounts in the recipe.
To obtain 8000 gr. of custard-base vanilla ice cream base, we will start off with a desired percentage, based on the numbers in the chart above. They will be increased, decreased or left as they are based on the obtained results: do you like or dislike the product? Does it need more fat? Does it need to be sweeter? Is it icy? Remember, this is only a mathematical exercise, not a recipe per se, and it is merely the starting point for a finalized recipe:
Total fat: 9% of 8000 gr. = 720 gr. Nonfat solids: 8% of 8000 gr. = 640 gr. Sugars: 17% of 8000 gr. = 1360 gr. Egg yolks: 5% of 8000 gr. = 400 gr. Stabilizer: .4% of 8000 gr. = 32 gr. Total: 39.4% of 8000 gr. = 3152 gr.
(Vanilla bean: 2% of 8000 gr. = 160 gr. Not considered part of the total solids.)
The next step is to calculate how much milk powder the recipe requires. Remember, milk powder is largely responsible for the appropriate structure of the base. Based on the above numbers, we first need to determine the total weight of nonfat solids and milk. To figure this out, we need to
calculate the weight of the liquid (without fats, proteins, etc., just the liquid, also known as “serum”). For this, we add the weight of all the known solid ingredients except the non fat solids:
720 g (total fat) + 1360 g (sugars) + 400 g (egg yolk) + 32 g (stabilizer) = 2512 g (weight of total solids)
8000 g of total base – 2512 g of total solids = 5488 g of the liquid or “serum”
Then use the following formula in order to calculate the amount of milk powder needed for a formula:
(Required weight of nonfat solids in kilograms) – (weight of the liquid in kilograms x .092**) / (Weight of nonfat solids in 1 kg. skimmed milk powder***) - (.092**)
*These amounts need to be converted to kilograms for precision. ** The .092 factor represents the amount of non-fat dairy solids in liquid skim milk (9.2%); the water content is 91% and the fat content is 0.1%. ***The weight of nonfat solids in 1 kg. of skimmed milk powder is always 970 gr. (or .970 kg). There will always be a minimal trace of fat in skim milk (see dairy chart page xxx ingredients) (Do you want to call for this as “Weight of nonfat solids in 1 kg. of skimmed milk powder in kilograms” as well? Yes
Based on the formula and our numbers, we have the following:
.640 – (5.488 x .092**) = .640 - .504 = .136 = .155 kg .970*** – .092**
.878
.878
Therefore, the required amount of powdered milk to add is 155 gr.
Next you need to calculate the amount of milk required. Add all the ingredients and then subtract the total weight, in this case 8000gr.
1360 g (sugars) + 400 g (egg yolk) + 32 g (stabilizer) + 155 g (powdered milk) = 1947 g 8000 g of total base – 1947 = 6053 gr. of milk
The recipe so far: 6053 gr. Milk 155 gr. powdered milk (160 gr. vanilla beans) 1360 gr. sugars 32 gr. stabilizer 400 gr. egg yolks 8000 gr. (without vanilla beans)
At this point, the recipe is not quite yet finished. If you calculate the total fat percentage of this recipe vs. the original desired fat percentage (9%), it is way off. Presently, if we add the amount of fat found in the milk: (6053 x 36% = 217.9 g) and the amount of fat found in the yolks (400g x 33% = 132 g), we have 349.9 g, which is only 4.37% of the total weight of the recipe. 9% of 8000 g is 720 g, so that means that we are 370.1 g (or 4.63%) of fat short. When you formulate with lower desired fat percentages, for example 4 or 5%, the final actual fat percentage is very close to the desired percentage by fractions. The trouble starts when you formulate for larger amounts of fat, such as in this example. What you need to do now in order to achieve the desired percentage of fat is to refresh your memory and try to remember some very complicated (at least for me) algebra principles. Why is this? Because the only way to obtain the desired amount of fat is to add heavy cream and reduce the amount of milk. You cannot change the amount of yolks, because that would throw the recipe off. You cannot keep adding milk to reach the desired amount of fat because that would also throw the recipe off. The only option is to add heavy cream and reduce the amount of milk, because the chemical makeup of both is essentially the same (except for the amount of fat) and it won’t alter the recipe negatively. The recipes in this book on page 371 are all formulated with low (under 6%) percentages of fat and therefore did not require the addition of heavy cream to obtain the desired results. However, you may want to have recipes with higher fat contents and in this case you will need to take the recipe formulation further. The big question is: how much milk do you have to reduce and how much heavy cream do you need to add of the 6053 g? The first step is to not consider the 6053 g as milk, but as a mix of milk and heavy cream. Next, consider the only fat presently to be that of the egg yolks: 132 g (400 g x 33% = 132
g). This means that the recipe needs 588 more grams of heavy cream to reach the 720 g of total desired fat. Now, we will need to use algebra by utilizing the substitution method:
x = g of milk (the unknown amount that we are trying to calculate) y = g of heavy cream (the unknown amount that we are trying to calculate)
therefore:
Part a)
x + y = 6053
What do we know of each ingredient? We know that milk contains 3.6% fat and that heavy cream contains 40% fat (if using other fat percentage heavy creams, simply substitute this amount). Convert these percentages into decimal numbers and continue with the formula this way:
Part b) 0.036x + 0.40y = 588 (this is the amount of fat you need to obtain the total desired amount of 720 g). Think of it this way: the percentage of fat of each ingredient (milk and heavy cream), when multiplied by its actual amount will produce a number for each, and when you add these numbers they will add up to 588.
Next, move the x over (not to sound confusing, but you could do the same with the y instead.) from Part a). y = 6053 – x (when you move sides, the sign has to change)
Now perform the substitution on Part b)
.036x + .40(6053 – x) = 588 (You have literally substituted the value of “y” in Part b))
.036x + 2421.2 - .40x = 588 (here, you multiplied .40 x 6053, and the result was 2421.2, and then you multiplied .40 x (-x), and the result was -.40x). Now we need to combine like-items.
.036x + (- .40x) = -.364x
588 – 2421.2 = -1833.2
With these numbers we can now continue with the equation:
-.364x = -1833.2 .364
-.364
You divide the -.364 x .364 to eliminate x values; to keep the equation in balance, you need to perform the same operation on both sides, which is why you needed to divide -1833.2 by -.364. In simpler terms, because on the left side of the equation you were obligated to divide by .364, you are also obligated to do the same on the right side of the equation. At this point, by dividing -.364x by .364, you have only “x” left. Therefore
x = -1833.2 -.364
x = 5036.26 This is the amount of milk the recipe will require.
Therefore:
y = 6053 – x , or, 6053 – 5036 (rounded out) = 1017 this is the amount of heavy cream the recipe will require.
Final recipe: 5036 g milk 1017 g heavy cream 155 gr. powdered milk (160 gr. vanilla beans) 1360 gr. sugars 32 gr. stabilizer
400 gr. egg yolks 8000 gr. (without vanilla beans)
