Steve Kass

A few months ago, I posted a graph and parametric equations for conchiglie rigati. Today I’m sharing a graph and equations for cavatappi. As before, I started with equations from Chris Tiee’s 2006 class notes for vector calculus.

Coming soon, a graph and equations for Möbius pasta.

Last night, as a few of us not backing up Ray Davies in Philadelphia gathered for dinner in Hoboken, I spotted this holiday mispostrophe (ssp. dyspostrophe).

It probably wasn’t intentional on Trader Joe’s’s part, but the mispostrophe distracted me from the numbers on the box — especially 24 and 50. But only briefly; the pressing question quickly loomed.

If 24 Milk Chocolates weigh 50 grams altogether, aren’t they too small?

The appropriate comparison was obvious: M&M’s®. Little did I suspect it would be something of a challenge to find out the true weight of one regular M&M.

Disregarding outliers like “I think it is about 15g; 15 grams is perhaps the answer,” answers on the web (to the question of an M&M’s weight) generally fell into two camps. There was a handful of a-bit-less-than-a-gram answers, like “There are about 500 Plain M&M’s per pound,” and there was also a handful of around-2-grams answers, like “After an experiment, of weighing M&M’s, here were the results. 1) 2.208 g 2) 1.882 g 3) 1.904 g 4) 2.438 g.”

After considerable “research,” but no direct measurement, I’m swayed, not by any attestations of milligram precision, but by the preponderance of evidence [and 1] that one regular M&M weighs a bit less than a gram. Which conclusion is consistent with my personal experiences as a candy sorter (when I can find an uncluttered flat surface, which isn’t very often).

From the web’s many M&M Q&A (or should I say Q&“A”?) a few examples:

• Q: What is the weight of one M-and-M candy? [link]
  A: I think it is about 15g; 15 grams is perhaps the answer
   
• Q: How much does an M and M weigh? [link]
  A: When we counted the number of M&M’s in a 12.6oz bag, we got 404, which means there are 32.06 M&M’s/oz, which means that each M&M weighs 1.13 grams. [SK: If you divide backwardsly, perhaps. Otherwise, each M&M weighs (on average) about 0.88 grams.]
   
• Q: How many m&m’s do you reckon are in 7oz? I’m ordering custom m&ms, and they come in 7oz bags. I need about 1000 m&ms, total. how many bags should I order? [link]
  A1: [Best Answer] 10 bags, maybe around 75 or 100 in each bag. [SK: Better safe than sorry.]
  A2: 2 or 3.
   
• Q: How much does a single plain m&m weigh? [link]
  A: After an experiment, of weighing M&M’s, here were the results. 1) 2.208 g 2) 1.882 g 3) 1.904 g 4) 2.438 g.

As for the pressing question, I’ll cautiously answer it “No” and hope Toby and Theo agree. Two or three M&M’s-worth of chocolate every day for most of a month — for those endless days, those sacred days, believe me — is not so bad. Despite anyone’s opinion that one serving of M&Ms comprises 208 grams (and 1023 calories).

Chuck roast in the freezer,
Carrots in the fridge.
That spells string meat for dinner.

Looks real bad,
But tastes real good, 
We’re havin’ string meat for dinner!

Wandering the internet today, I stumbled upon pasta and mathematics. At the same time. Chris Tiee, a teaching assistant for one of UCSD’s vector calculus courses, had put into his class notes back in 2006 a short and very cute parametric equations quiz: match the parametric equations to the pasta shape. And he (or UCSD) conveniently left his notes on the web for posterity — or should I say pastarity?

His parametric equations were pretty basic — absolutely fine for a vector calculus quiz — and I thought I might be able to touch them up a bit. Here’s what I came up with for conchiglie rigati.

This exercise is also my excuse for finally getting MathJax up and running on my blog. [Update: I’ve disabled MathJax, because it mucks up non-LaTex posts that have $ characters. At some point I’ll figure out how to configure it amicably, but for now, the pastametric equations are provided as an image file.] You might find that this page loads slowly, and I don’t yet know if I can do anything about that. If  you don’t see any equations below the picture, however, please let me know.

When a sudden urge for Thai food strikes, I never have all the ingredients on hand. Today I had enough for a try, and the result was pretty good, if not pretty-pretty.

Soup (serves 1 glutton or 4 polite folk)

• Put into a 5-cup or larger appliance that slow cooks:
    1 lb. chicken tenders, cut in 1" pieces,
    2 small cans sliced mushrooms, drained,
    2-3 fresh shallots, peeled and quartered,
    3 cups water,
    1/2 cup marble-sized potatoes, and
    2 tablespoons tom yum paste.
• Slow cook (185-190°F) for 60-90 minutes.
• Stir in:
    1/2 can of coconut milk
    2 frozen kaffir lime leaves, cut into very thin strips, and
    up to 1 tablespoon palm sugar (to taste).
• Simmer for another 30 minutes.
• Salt liberally to taste.

Notes:  
  If you have fresh cilantro, chop and add a small handful right before serving. 
  Use whole straw mushrooms instead of button mushrooms, if you have them.
  Boneless breast might do, but my Costco chicken tenders were very tender.
  I’ll leave out the potatoes next time. They didn’t help any.
  Regular or light brown sugar is fine.
  If you want it spicy-hotter, add Vietnamese or Thai chili paste or crushed red peppers. 
  There’s no good substitute for fresh shallots. Onion might do; avoid boxed shallots.
  Don’t use dried kaffir lime leaves, only fresh or frozen, or substitute half a lime’s juice.
  For a low-fat version, leave out the coconut milk and use broth for half the water.
  If you don’t have a slow-cooking appliance, use the stove, but avoid boiling the soup.

Maybe I got lucky with this tom yum paste.

Today’s eLetter from the folks at Fine Cooking began “Baked Pasta 259,200 Ways. We did the math.” As you can imagine, I did the Baked Pasta Recipe Maker math, too. I figure it’s 16,128,000 ways, or about 60 times the number Fine Cooking found when they did the math. Here’s the calculation. The recipe maker walked me through the following steps:

• Choose one or two of four Flavor Bases
• Chose one of three Sauces
• Choose two or three of nine Sauce Enhancers
• Choose one of eight Pastas
• Choose zero, one, or two of five Vegetables
• Choose two or three of six Cheeses

Assuming no choice combinations are forbidden (the recipe maker doesn’t appear to prevent you from adding olives and sherry vinegar to sausage and chicken in pink sauce, for example), you find total number of different ways to make a choice at every step by multiplying together the numbers of choices at each step.

It’s easy to count the number of ways to “choose this many of those Things.” If this many is k, and those Things are n in number, the number of ways to choose k of the n things is “n choose k,” sometimes written as C(n,k). These numbers can all be found in Pascal’s triangle. As it’s shown here, C(n,k) is in the row labeled with the n value, under the column labeled with the k value. Here’s how to use the triangle to find the value of C(9,3):

• To choose one or two of the four Flavor Bases, there are C(4,1) = 4 ways to choose one plus C(4,2) = 6 ways to choose two, for a total of 10 ways to choose this item.
• To choose one of the three Sauces, there are C(3,1) = 3 ways.
• To choose two or three of the nine Sauce Enhancers, there are C(9,2) = 36 ways to choose two plus C(9,3) = 84 ways to choose three, for a total of 120 ways.
• There are C(8,1) = 8 ways to choose a pasta.
• There are C(5,0) + C(5,1) + C(5,2) ways to choose up to two vegetables, or 1 + 5 + 10 = 16 ways.
• There are C(6,2) + C(6,3) to choose the Cheeses, or 15 + 20 = 35 ways

Multiplying these numbers of choices for each step yields 10·3·120·8·16·35 = 16,128,000 ways, about 60 times as many as Fine Cooking found when they did the math. Counting ways isn’t standard recipe math, and I’d like to note that Fine Cooking’s math is generally fine when it comes to ounces, grams, cups, servings, and calories.