BBQ Done When?

2 points by graycat 2y ago ↗ HN

Question:

Roasting cooking pork shoulder for BBQ and want to know when it will be done.

Why Care:

The usual approach to cooking BBQ is "low and slow". So, with oven temperature of 210 F (which is "low") the cooking time to reach a BBQ recommended internal meat temperature of 180 F might be 20 hours ("slow").

So, after a few hours of cooking noting, say, two times and internal meat temperatures and considering other activities, sleeping, etc., can want to know how many more hours will be needed to reach the 180 F.

So, made a physics assumption, did some math derivations, and wrote the simple software for the arithmetic. The physics assumption ignores the temperature stall sometimes seen in BBQ cooking.

Details:

At what we can call time 0, the internal temperature was 155 F.

At time 157 minutes, the internal temperature was 170 F.

Answer: At 300 minutes the temperature should be 180.071 F.

So, we start with the oven temperature and the internal meat temperature at two times so far, and from that data we can find the internal meat temperature at any time.

The physics and math:

     Y -- oven temperature (F)

     t -- time (minutes)

     y(t)  -- internal meat temperature
              (F) at time t

Physics:

At time t (minutes) the rate of temperature increase is directly proportional to the difference between the oven temperature and the internal meat temperature. So, with constant of proportionality k, we have

     y'(t) = k*(Y - y(t))

That is a first order, linear, ordinary differential equation.

Math Solution:

For some constants k and c, the solution is:

     y(t) = Y - e^(-k*t + c)

To simplify the notation, we have quantities a1 and a2 defined below. The two times are t1 and t2. Then we have the values of k and c:

     k = (a2 - a1) / (t1 - t2)

     c = a1 + t1*k

The arithmetic in simple (Rexx programming) syntax:

     The oven temperature:

          Y = 210

     Time in minutes of the first meat
     internal temperature measurement:

          t1 = 0

     The first temperature (F):

          y1 = 155

     Taking the natural logarithm, that
     is, base e = ~2.71828

          a1 = RxCalcLog(Y - y1)

     Second time (minutes):

          t2 = 157

     The second temperature (F):

          y2 = 170

          a2 = RxCalcLog(Y - y2)

     So using the t1, t2, a1, a2 from
     above we have k and c:

          k = (a2 - a1) / (t1 - t2)

          c = a1 + t1 * k

     For a Do-End loop to find many
     time-temperature pairs, our time
     step (minutes):

          dt = 60

          s = '    Time (minutes)    Temperature (F)'

          Say s

          Do j = 0 to 22

               t3 = j * dt

     Taking e^(-k t3 + c) we get our
     desired temperature y3 at each of the
     times t3:

               y3 = Y - RxCalcExp(-k*t3 + c)

               Say Format( t3, 18) Format( y3, 14, 3)

          End

From the output of the

          Do j = 0 to 22

we get at 300 minutes the temperature will be 180.071 F.

brodouevencode 2y ago ↗

A few notes:

- is this a question or a statement?

- this looks like a partial diff-eq textbook problem - if so no one is going to check your homework for you

- you will always see a stall when cooking meat

- 180F is not the ideal target temperature for tougher meats - it should be 200-205F for full collagen/fat breakdown

- pedantry: barbecuing requires indirect heat and usually requires smoke (depending on who you ask), otherwise you're just roasting

graycat 2y ago ↗

> - this looks like a partial diff-eq textbook problem - if so no one is going to check your homework for you

In conflict, my post had

> That is a first order, linear, ordinary differential equation.

So, it's "ordinary", not "partial". Big difference.

It's an especially simple differential equation and can be solved with just a first course in calculus, even at the high school level. So, a secondary objective of the post was to give a simple calculus solution of a kitchen question otherwise not so easy to answer. To encourage readers to do the derivation themselves, I deliberately omitted the details. I explained why there is a reason for getting the solution.

> check your homework

It's been a long time since I assigned "homework" and longer since I did "homework".

My post has nothing to do with "homework". Instead, as I explained:

> ... can want to know how many more hours will be needed to reach the 180 F.

I wanted to know so did the work. Since others cooking "low and slow" could easily want to "know" I did the post.

For BBQ from a picnic pork shoulder, that's just "roasted" pork as I described plus some sauce added just before eating.

For a piece of meat as large a picnic pork shoulder, smoke during the cooking has no chance of adding any flavor inside the meat. If you want some smoke flavor, add some smoke from a bottle to the sauce.

With such BBQ, the meat is supposed to separate easily into fibers that are soft, flexible, and moist, i.e., succulent. A flat statement is that the fibers are always tender and meat is tough only because of collagen not yet melted.

But to melt the collagen, cooking slowly to 180 F is fine. I've also done well with 175 F. Google confirms that collagen melts at 160 F. Sooo, for melting collagen, some hours over 160 F up to 180 F works fine.

Temperatures higher than 180 F risk having the meat proteins unwind, expel their water, and become hard, dry, and brittle, e.g., like some of the crust on the outside surface, even in a 210 F oven.

Cook your meat to 200+ F with a lot of smoke if you want as you read partial differential equations -- for that subject there are the volumes of Courant and Hilbert now available in PDF.

For BBQ, consult the large volumes of physicist Nathan Myhrvold, also winner of the BBQ contest "Memphis in May".

Memphis is supposed to be a good town for BBQ. I grew up there, e.g., classmate of one of the sons of the founder of Holiday Inn, went to the same college as SCOTUS Justice Amy Coney Barrett, and returned later to save FedEx twice, the second time with a differential equation. My office was next founder F. Smith's. I gave a lecture with one problem that became my Ph.D. dissertation and another problem that became the dissertation of another student. Sooo, I have a background in Memphis. Again, ate a LOT of BBQ in Memphis.

Currently I'm in Knoxville, home of Buddy's BBQ that got its start from getting famous at the 1982 World's Fair. I've eaten a lot of pork BBQ at Buddy's, but what I cooked while writing my post tastes the same. One difference is that I follow west TN that chops the cooked meat while east TN pulls it. One more difference is, sometimes as in Memphis style, I put slaw on the sandwich.

One reason for the home cooking is that can have a BBQ meal quickly without driving. Another reason is cost: The BBQ I just did costs 31 cents per ounce of the meat ready to eat, and 3 ounces makes a good sandwich. Via partial differential equations you may be able to show that 3 ounces would cost 93 cents. Buddy's is good, but as I recall they charge $5+ per sandwich.