Reducing Turkey Cooking Time

2011 November 11
by Daniel Lakeland

My wife sent me a link to a NY Times Video about cooking a Turkey in 45 minutes (actually they take 35 mins in the video).

So of course, given my obsession with modeling the cooking time of Turkeys, and the inevitable need to come up with the physical part of the model, now is as good a time as ever to discuss why this method works. Mark Bittman implicitly implies that its by increasing the exposed surface area. Although this explanation is intuitive, the actual change in the surface area is quite small, only related to the new surfaces he makes with the knife. There seems to be something wrong.

To see what is going on, let’s use the Feynman-Kac particle diffusion representation of the heat equation.  The temperature at a point x inside the Turkey is related through the heat capacity to the concentration of thermal energy. The thermal energy comes from a large number of packets of heat which diffuse in from the boundary of the Turkey, the surface area that Bittman mentions. No other sources of heat (such as chemical reactions within the Turkey) are considered, and we’ll ignore the flow and evaporation of fluids for the moment, an assumption which is not valid, but probably sufficiently accurate for our purposes.

Now a gaussian random walk in 3 dimensions takes some time to go a certain distance. In particular, at a time $$t$$, the distance between the origin point and some point where a particle of heat has gone to (in backwards time) is $$\sqrt{\Delta x^2+\Delta y^2+\Delta z^2}$$. Now consider the value of $$\Delta x$$. Using a nonstandard analysis approach, $$\Delta x$$ is the sum of N iid random variables whose standard deviation is related to the thermal conductivity of the material, and $$\sqrt{dt}$$ where $$Ndt = t$$ the current time.

Why the $$\sqrt{dt}$$? The answer lies in the fact that the sum of the variances of N iid random variables is the variance of the sum of the random variables $${\rm var}(\Sigma x_i) = \Sigma {\rm var}({x_i})$$. This follows from the linearity of expectation, the independence of the random variables, and the fact that the variance is the expected value of the square of the random variable. Basically if we square $$\Sigma x_i$$ we wind up with $$N$$ terms $$x_i^2$$, and $$O(N^2)$$ terms that are the expectation of the product of two independent random variables, which go to zero due to independence.

We are at time $$t = N dt$$ and we want the variance of the sum of all our individual infinitesimal random variables after this amount of time to be independent of the size of our infinitesimal $$dt$$. This means if we use $$N/k$$ random steps so that $$t = N/k (k dt)$$ , we should scale the variance of the individual random variables so that we do not change the variance in our sum. If we are going to do this, and keep the total variance after a certain time constant (independent of the choice of dt), we need the variance in the random variable to be linear with time, since the standard deviation is the square root of the variance, the standard deviation should scale like $$\sqrt{dt}$$. We’ve got three directions, x, y, and z, so the total 3D Euclidean distance is the square root of a scaled chi-squared variable with 3 degrees of freedom.

How does this help us? It tells us that the time it takes on average for a packet of energy to come from the boundary and get to some point deep within the turkey goes like $$R^2$$ where R is the distance of the point from the boundary. Now the heat doesn’t just come from the closest point on the boundary, it comes from a mixture of points on the boundary, and in general what we want is some kind of average. But the point is that when we can make the distance between the deepest part of the Turkey and the surface of the Turkey smaller, we should expect to decrease the time it takes to cook the inner part of the Turkey quadratically.

Let’s see how this should work, supposing that Bittman’s Turkey would normally take say 3 hours to cook, and now takes 1/2 hour to cook. How much did he decrease the distance from the inner part of the Turkey to the boundary? Let $$x’$$ be the distance after splaying the Turkey, and $$x_0$$ be the distance in an un-modified turkey, then $$(x’/x_0)^2 \sim (1/2)/3 \rightarrow x’ = \sqrt{1/6} x_0 \approx 0.41 x_0$$. It seems very reasonable to me based on just watching him splay the Turkey that after breaking it up, the distance between the deep parts of the Turkey and its surface is cut to say 40% of the original value, and this is the type of change we predict we would need to cut our time to 1/2 an hour. In particular by breaking the bird in a few thick places he puts new surface area near the thickest portions of the breast, and he eliminates the interior cavity which brings the high oven heat directly to the interior surface of the Turkey, making the “farthest part” of the breast more like 1/2 the thickness of the breast, not the full thickness of the breast (which it would be if you ignore heating from the inside of the cavity).

Voila, magic, or science?


3 Responses
  1. November 12, 2011

    Interesting, Dan. I can’t really follow the math, but I’ve been cooking fully stuffed turkey’s in 2 1/2 hours for years. You put the turkey on a rack over about a quart of water in a tightly covered roasting pan. Put it into the oven at 500 degrees for 2 hours, uncover and cook at 450 for 30 min. Delicious. I notice neither Mark Bittman nor you propose a temperature–a pretty important part of the equation.

    • Daniel Lakeland
      November 12, 2011

      The recipe says 450F for 20 mins, then 400F or 350F for the remainder depending on how quickly it’s browning.

      The temperature is of course important, but it’s not as important as the distance that the thermal energy has to penetrate into the meat. The reason he could drop a 2-3hr cooking time to 35 to 45 minutes is basically because he’s getting the high oven temperature “closer” to the meat, I think mostly by eliminating the interior cavity and letting the high heat circulate on both sides of the meat.

      • Daniel Lakeland
        November 12, 2011

        Another question is how dry and potentially over-cooked is the outermost portion of the meat? Perhaps with less time he also loses less moisture. My main method has been to start at high heat and steadily reduce the heat in an effort to prevent the outer portions from drying out while ensuring the inner portions are fully cooked. My final temperature is sometimes in the vicinity of 325, and if I had the oven free I would then set it down to 175 and let it sit for a half an hour right before serving. Usually we need the oven for something else though.

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