I have a recipe for a timballo (a pasta-filled casserole entirely encased in a layer of meatloaf) that I want to make as 8 small, pot-pie-sized invidual portions instead of as a giant 8-serving behemoth.

No problem, you might think, just divide the pasta and meatloaf equally among enough small pans to use up all the ingredients. But you would be wrong. The original recipe timballo is baked in a springform pan, with meatloaf mix pressed into the bottom of the pan, up the sides of the pan and over the top of the casserole, forming a shell around the entire casserole. When you divide the mixture among smaller pans, you keep the same amount of pasta filling but wind up with a lot more "shell" surface to deal with.

In fact, according to my calculations the ratio of shell surface area to pasta volume is 1.4:1 for pot pies and only 1.1:1 for the springform pan. So the problem for this recipe is in figuring out by how much to increase the meatloaf part of the recipe to fully cover the smaller pans while simply dividing the pasta part of the recipe.

The trickiest part was figuring the volume of the slope-sided pot pie pans. I decided to conceptualize them as a slice of a cone. I used bamboo skewers taped to the side of the inverted pan to create the "invisible" portion of the cone and measure its height. I calculated the volume of the full cone and the volume of only the "invisible" part (using the top of the inverted pan, ergo the bottom) as the base of the "invisible" cone and subtracted the "invisible" volume from the full volume to find the volume of the slice remaining.

It's not exactly A Beautiful Mind, but I'm so excited to have such a tangible example of math in action. And so for any math geeks out there who want to play along, here is what I did:

__Formulas__:Area of circle = π*radius²

Circumference of circle = π * diameter

Area of rectangle (formed by the side of a circular pan) = Circumference * height

Volume of cylinder = area of circle * height

Volume of cone = 1/3 volume of cylinder

Area of trapezoid (formed by the side of a sloped circular pan) = (short circumference * height) + (1/2 difference of short and long circumferences * height)

Volume of slice of cone: Volume of estimated whole cone - volume of estimated "invisible" cone

__Springform pan__:Radius (r) =4.5"

Diameter (d) = 9"

Height (h) = 3"

**Volume = πr² * h = π * 4.5 * 4.5 * 3 = ~191 in^3**

**Surface Area = area of top and bottom disks (2πr²) + area of side layer (πd * h) = (2*π*4.5*4.5) + (π*9*3) = ~212 in²**

__Pot Pie pans__:Radius of top (rt) = 2.5"

Radius of bottom (rb) = 1.625"

Diameter of top (dt) = 5"

Diameter of bottom (db) = 3.25"

Height (h) = 1.5"

Estimated Full Cone Height (he) = 5"

**Volume = volume of full cone (1/3 * πrt²* he) - volume of invisible cone (1/3 * πrb² * (he - h)) = (1/3 * π * 2.5 * 2.5 * 5) - (1/3 * π *1.625 * 1.625 * (5 - 1.5)) = ~23 in^3**

**Surface Area = area of top disk (πrt²) + area of bottom disk (πrb²) + area of side layer ((πdb * h) + ((dt - db) * h * 1/2)) = (π*2.5*2.5) + (π*1.625*1.625) + ((π*3.25*1.5) + ((5-3.25) * 1.5 * .5)) = ~32 in²**

And here is where we move more fully into applied mathematics, wherein "close enough" is "good enough". I multiply the pot pie pan results by 8 since my recipe will make 8 servings. I verify that the volume (ie pasta amount) is unchanged, and calculate the ratio of springform to pot pie surface area (meatloaf mix) to determine by how much to scale the meatloaf mix up.

**8 * pot pie volume = (8 * 23 in^3) = 184 in^3**

**Springform volume = 191 in^3**

That's close enough for me.

**8 * pot pie surface area = 8 * 32in² = 256 in²**

**Ratio of springform surface area to 8x pot pie surface area = 212 in² : 256 in² OR ~ 1 : 1.25**

Which means that I have my answer...I'll scale up the meatloaf mix ingredients by about 1/4 to accommodate the smaller (read: greater surface area) pans. Pin It

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