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We've come up with a recipe for a cake that has four different layers (a dough base plus fillings) and is then covered in chocolate mousse. We tested that recipe as a small sample and concluded that the ideal proportion was to have 70 grams of base layers for 80 grams of chocolate mousse, creating a tight cylindric 150 gram cake. So far, so good.

We then proceeded to make a bigger version, with a wider mold, which would weight 1.88 kilograms, so we scaled our recipe and found out that 875 grams of base layers for 1kg of chocolate mousse. But the problem is, the mousse layer becomes so large that it collapses.

Fixing this is fairly obvious: use less mousse. Maybe, since the mousse is less dense than all the other layers, increasing its total weight has a much bigger impact in terms of volume. But how can we define our recipe in terms of volume rather than weight, then?

I feel like we're missing something very obvious here, but if the proportions of the recipe are the same as the small, initial version, why do we get such different results? How can we scale the recipe in order to achieve a consistent amount of mousse, regardless of the size we attempt to make (i.e. what kind of math is required to make a 1kg, 2kg and 3kg cake from one single recipe / set of proportions, considering the form factor stays the same -- still round, same height, just different diameter -- if it's even possible to do so)?

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This sounds less cooking and more physics. ^^ –  Jay Jan 17 '12 at 2:11
    
There's a few confusing things to me in your question. You talk about "then covered in chocolate mousse" and also "the mousse layer" - are you covering the cake in mousse or is there a whole layer of mousse? Are you asking how to scale the amount of mousse in terms of the math or are you asking how to make a large batch of mousse logistically? –  rfusca Jan 17 '12 at 2:29
    
@rfusca I've edited the question to attempt to clarify those confusing bits, but in effect 1) I'm covering the cake in mousse, and 2) I'm asking how to scale in terms of the math. –  tbuteler Jan 17 '12 at 10:36

1 Answer 1

up vote 2 down vote accepted

Density = mass / volume, so if you double the mass of mousse, the volume should also double. The only way you can get more than double volume is if the density decreases, which I doubt is happening.

If you're covering the sides and top with mousse, than mass--which is directly related to volume--isn't going to help, you need to think about surface area. For a cylindrical cake frosted on the sides and top, the surface area and volume are

SA = pi d^2 / 4 + pi d h
 V = pi h d^2 / 4

with d the diameter and h the height. You could imagine two cakes of the same mass, one thin and very broad with a very big surface area, the other a bit tall and small around with much smaller surface area. So you'll need to take both height and diameter into account when figuring how much more mousse you need.

Edit

We can simplify a bit. The ratio of new mousse to old mousse in volume or mass should be the same as the ratio of new SA to old SA. Using your edit that the height stays the same:

Ratio = SA_new / SA_old = d_new / d_old * (d_new  + 4h) / (d_old + 4h)

I'm going to assume that your height is pretty big, and that four times the height is maybe getting close to the bigger diameter. Then we could approximate the ratio above (there is some height h such that 4h is between d_old and d_new such that)

Ratio = d_new^2 / d_old^2

So if you're going from an 8-inch diameter to a 12-inch diameter you'd want to make 144 / 64 = 2.25 times the mousse. And, as this is pretty rough, I'd add a little to that to be safe and make 2.5 times the mousse. (And having a little extra mousse doesn't seem like the worst thing in world.) If my guess about the height is wrong, just measure and use the actual formula, you'll get better results! Always make a little more than the you expect to need though, because nothing's worse than having not quite enough.

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You are right. The top layer cannot be scaled like the other layers can, and the shape and size of our mold cannot be ignored. I just wish there was a more straightforward way to scale it without distorting the final result so much. A rule of thumb, perhaps? Anyway, thanks for the answer! –  tbuteler Jan 17 '12 at 9:37
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The rough approximation is that surface area grows with the square, volume with the cube. So if you double something (measuring diameter or radius) you get 2²=4 times the surface area, but 2³=8 times the volume. If you measure weight, which is more or less volume, 8 times the cake = 4 times the mousse (0.5). But 3³=27 times the cake = 3²=9 times the mousse (0.33). So there just isn't a linear scaling that works. –  derobert Jan 17 '12 at 22:14
    
@derobert right (and +1), and when the height is held constant (as in this case) even a pure power rule like (volume ratio)^2/3 won't be good, even though it would make a good rule of thumb in a spherical case. –  Gregor Jan 18 '12 at 0:00
    
Linear scaling will not work, indeed. What really seems more straightforward and safe right now is to figure out the scaled recipes empirically, then (if it's still necessary or worth the trouble) build an alternative, non-linear scaling model based on those experiments. Thanks a lot for your help! –  tbuteler Jan 18 '12 at 10:46

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