Posts tagged ‘formula’

Fowl Formulae for Thanksgiving

Okay, I know you’re going to find this hard to believe, but there is disagreement on the internet. And I don’t mean about some insignificant topic like gun control or taxes or health care or the value of 6 ÷ 2(1 + 2). This is big. This is important.

We’re talking turkey. Literally.

According to the British Turkey Information Service — yes, there really is such an agency — the amount of time you should cook your turkey at 375° F can be found with the following formula:

t =    \begin{cases}    20w + 70 & w < 4; \\    20w + 90 & w \geq 4,    \end{cases}

where t is the cooking time in minutes and w is the weight of the turkey in kilograms.

If you’d rather not do the math yourself, try the British Turkey Cooking Calculator, which will not only give you the cooking time but also the defrosting time and the size of turkey to buy for a given number of servings.

By comparison, the Meat Chart provided by FoodSafety.org says that turkey should be cooked at 325° F for 30 minutes per pound.

But the cooking website allrecipes.com says that only 20 minutes per pound is sufficient if you bake the bird at 350° F.

Whereas the good folks at delish offer the following guidelines:

Delish - Turkey Cooking Times

Cooking Times at 325° F from delish.com

which translates to the lovely formula

t =    \begin{cases}    5w + 125 & w \leq 10; \\    15w & w = 12; \\    7.5w + 120 & w \geq 14,    \end{cases}

but requires that you interpolate if your bird weighs an odd number of pounds. (Like 86 pounds, the world record for heaviest turkey ever raised. Even though the units digit is 6, you’d agree that 86 is an odd number of pounds for the weight of a turkey, no?)

As you might suspect, Wolfram Alpha has a more mathematically sophisticated formula:

\displaystyle t = T \times \left( \frac{w}{20} \right)^\frac{2}{3}

where t is the cooking time in hours, w is the weight in pounds, and T is a coefficient to account for cooking environment. For normal conditions, T = 4.5, and the equation reduces to

\displaystyle t = 36.64w^\frac{2}{3}

if you use minutes instead of hours for the unit of time.

But this feels a little like a math joke; below the formula, Wolfram offers the following:

using the heat equation for a spherical turkey in a 325° F oven

Falling into the wrong hands, that idea could lead to an horrendous modification of the spherical cow joke…

The turkeys at a farm were not gaining sufficient weight in the weeks leading up to Thanksgiving, so the farmer approached a local university to ask for help. A theoretical physicist was intrigued by the problem and offered his assistance. He spent several weeks at the farm, examining the turkeys and filling his notebook with equation after equation. Finally, he approached the farmer and said, “I have found a solution.”

“Oh, that’s excellent!” said the farmer.

“Yes,” said the physicist. “Unfortunately, it only works for spherical turkeys in a vacuum.”

The Wolfram formula is very similar to one suggested by physicist Pief Palofsky, who apparently dabbled in poultry when not winning the National Medal of Science.

\displaystyle t = \frac{2}{3} w^\frac{2}{3}

and when converted to minutes instead of hours, this becomes

\displaystyle t = 40w^\frac{2}{3}.

According to Turkey for the Holidays, the average weight of a turkey purchased at Thanksgiving is 15 pounds. The cooking times for a 15-pound bird, based on the formulae above, appear to have been chosen by a random number generator.

Recommender Time (min) Temp (° F)
British Turkey Information Service 226 minutes 375
Foodsafety.org 450 minutes 325
allrecipes.com 300 minutes 350
delish.com 233 minutes 325
Wolfram Alpha 223 minutes 325
Pief Palofsky 243 minutes 325

Even if you limit consideration to those who suggest a cooking temp of 325° F, the range of times still varies from just under 2¾ hours to a staggering 7½ hours. Wow.

With Thanksgiving just around the corner, where does all of this contradictory information leave us?

A number of sites on the internet claim that the only way to adequately check the doneness of a turkey is with a meat thermometer.

turkeyThe folks at recipetips.com claim that a turkey can be removed when the temperature is at least 170° F for the breast and 180° F for the thigh. Yet on the very same page, they claim, “Turkey must reach an internal temperature of 185° F.”

On the other hand, the folks at the Food Lab claim a turkey can be safely removed when the breast temperature reaches 150° F, because after resting 15‑20 minutes before carving, the amount of remaining bacteria will be minimal. They explain, “What the USDA is really looking for is a 7.0 log10 relative reduction in bacteria,” particularly Salmonella, which means that only 1 out of every 10,000,000 bacteria that were on the turkey to start with will survive the cooking process. And according to the USDA guidelines, a turkey that maintains a temperature above 150° F for 3.8 minutes or longer will reach that threshold for safety.

Which has to make you wonder — if 3.8 minutes at 150° F is supposedly adequate, why then does the USDA Food Safety and Inspection Service recommend that the minimum internal temperature of the turkey in the thigh, wing, and breast should be at least 165° F? Who knows. I suspect it’s typical government over-engineering to remove all doubt.

So, how long should you cook your turkey? Hard to say. But if you put your turkey in the oven right now, it should be done by November 23.

When the turkey is finally ready, here are a few math jokes you can tell around the Thanksgiving table.

What do math teachers do on Thanksgiving?
Count their blessings!

What does a math teacher serve for dessert on Thanksgiving?
Pumpkin Pi.

How do you keep private messages secure on Thanksgiving?
Public turkey cryptography.

Thanksgiving dinners take 18 hours to prepare. They are consumed in 12 minutes. Halftimes take 12 minutes. This is not coincidence. 
~ Erma Bombeck

Gobble, gobble!

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November 14, 2017 at 12:32 pm Leave a comment

Hiking Routes, Square Roots, and Trail Ratings

Not all those who wander are lost.

You may have read that line in Gandalf’s letter to Frodo Baggins in The Fellowship of the Ring, but nowadays you’re more likely to see it on the t-shirts and bumper stickers of hikers.

Hiking is a popular sport, and the 47 million Americans who reported that they’ve taken a hike in the past 12 months (Statista) had a lot of different trails to choose from: American Trails maintains a database of over 1,100 trails, and Backpacker‘s list of America’s Best Long Trails offers an impressive 39,000 combined miles. Plus, there are thousands of miles of trail not on either of those lists. With so many options, how are you supposed to choose?

Tour Du Mont Blanc

A wealth of information is provided for most hiking trails. But while some information — like distance and elevation gain — is absolute, other information leaves room for interpretation. What does it mean when the Craggy Pinnacle Trail just outside Asheville, NC, is described as a “moderate” hike? The Explore Asheville website says,

Moderate hikes could range anywhere from a few to ten miles with an elevation gain up to 2,000 feet.

By those standards, a three-mile hike with a 10% grade would be considered moderate. No, thank you.

Unfortunately, there is no standardized system for determining trail difficulty. Most of the time, the trail rating is a nebulous qualitative combination based on an examination of the terrain, trail conditions, length, elevation gain, and the rater’s disposition.

But I tip my hat to the good folks in Shenandoah National Park who have attempted to quantify this process. Their solution? The simple formula

r = \sqrt{2gd}

where g is the elevation gain (in feet) and d is the distance (in miles). The value of r then corresponds to a trail rating from the following table:

Numerical Rating Level of Difficulty Estimated Average Pace (miles per hour)
< 50 Easiest 1.5
50-100 Moderate 1.4
100-150 Moderately Strenuous 1.3
150-200 Strenuous 1.2
> 200 Very Strenuous 1.2

Elevation gain is defined as the cumulative elevation gain over the entire hike. So if the hike climbs 300 feet over the first mile, then descends 500 feet over the next 2 miles, then goes back up 200 feet to return to the start, the elevation gain is reported as 300 + 200 = 500 feet.

Old Rag is one of the most popular hikes in northern Virginia. Known for the half-mile rock scramble near the top, this trail boasts an impressive 2,415 feet of elevation gain over 9.1 miles. Applying the formula,

r = \sqrt{2 \cdot 2415 \cdot 9.1} \approx 209.6

which means Old Rag’s level of difficulty would be “very strenuous.”

This formula could lead to several activities for a middle or high school classroom:

  • Draw an elevation map depicting a trail on which any type of hike (from easiest to very strenuous) would be possible, depending on how far a person hiked.
  • With distance on the horizontal axis and elevation gain on the vertical axis, create a graph that shows the functions for easiest to very strenuous hikes. (See Figure 1.)
  • If you were on a trail with an average elevation gain of 300 feet per mile, how long would you have to hike for it to be considered a moderately strenuous hike?
  • If one 5-mile hike is rated “easiest” and another 5-mile hike is rated “strenuous,” what’s the minimum possible difference in elevation gains for the two trails?

Students could also do a comparison between this trail rating formula and the geometric mean, if you wanted to go really crazy.

Feel free to drop some of your ideas into the comments.

Figure 1. The graphs for various difficulty ratings, using the Shenandoah trail rating formula.

Just as every good hike comes to an end, so must this blog post. But not before we laugh a little.

As it turns out, there’s a math joke about hiking…

An actuary has been walking for several hours when the trail ends at the edge of a river. Having no idea how to cross, she sees another hiker on the opposite bank, and she yells, “Hey, how do I get to the other side?”

The man across the river — a math professor — looks upstream, then downstream, then thinks a bit and finally says, “But you are on the other side!”

It’s a math joke about hiking as much as any joke about any topic is a math joke, if you insert the correct professions.

There’s a great non-math joke about hiking, too…

A fish is hiking through a reservoir when he walks into a wall. “Dam!” he says.

And there is a very mathematical list about hiking, which might be considered a joke if so many of the observations weren’t true…

Eight Mathematical Lessons from the Trail

  1. A pebble in a hiking boot will migrate to the point of maximum irritation.
  2. The distance to the trailhead where you parked remains constant as twilight approaches.
  3. The sun sets at two-and-a-half times its normal rate when you’ re trying to reach the trailhead before dark.
  4. The mosquito population at any given location is inversely proportional to the effectiveness of your repellent.
  5. Waterproof rainwear isn’t. But, it is 100% effective at containing sweat.
  6. The width of backpack straps decreases with the distance hiked. To compensate, the weight of the backpack increases.
  7. The ambient temperature increases proportionally to the amount of extra clothing in your backpack.
  8. The weight in a backpack can never remain uniformly distributed.

Go take a hike!

September 23, 2017 at 5:01 pm Leave a comment

Lighthouses and Math in Oregon

My family and I recently spent a week along the Oregon coast, where we hiked, biked, and — of course — did some math.

While on a hike through Ecola State Park, we had this view of the Tillamook Rock Lighthouse, a structure 62 feet tall that stands atop an offshore island:

Tillamook Lighthouse

Never one to resist the opportunity to pose a math question, I asked my sons, “How far do you think it is from here to the lighthouse?” My sons — never ones to resist reading the informational placard at a trailhead — already knew the answer and quickly responded, “About 1.2 miles.”

Ah, hell no. I’m itching for a mathy conversation, damn it, and you’re not getting off that easy! So I then asked, “How far do you think it is from here to the horizon?” Luckily, the answer to that question hadn’t been included in the display at the bottom of the hill.

Eli used proportional reasoning. He extended his arm and measured the distance from the shore to the lighthouse between his thumb and index finger; then, he used that same unit to measure the distance from the lighthouse to the horizon. “I’d guess about 2 miles,” he said. “It’s farther from here to the lighthouse than from the lighthouse to the horizon.”

Alex used intuition. “I think it’s a little farther,” he said. “Maybe three miles?”

“Actually,” I said, “it’s much farther than that. I won’t tell you how much farther, but I will help you figure it out.” (Typing that last sentence makes me realize that having a constructivist father is not an easy burden to bear. My answer to most questions is usually another question.)

After much discussion, we finally concluded that we were about 90 feet above sea level. Estimating vertical distance, it turns out, is not an easy task, especially for eight-year-olds. But once we agreed, the rest of the math was more or less straightforward. The image below provides a reasonable model for the situation:

Distance to HorizonIn that model,

  • h = height above sea level of the observer
  • R = radius of the Earth
  • x = the straight-line distance to the horizon from the observer

The sight line of the observer is a tangent line to the circle, and the point of tangency is the horizon that the observer sees. Consequently, a right angle is formed by the tangent line and radius. We now have a right triangle with which to work, so we can lean on the Pythagorean theorem. But first we’ll need some numbers.

At the Tillamook Rock Lighthouse, we had the following numbers:

  • h = 90 feet (estimated)
  • R = 4,000 miles (give or take)

And x, of course, is the unknown distance to the horizon that we are trying to find.

This leads to the following:

\begin{array}{rcl}    x^2 + R^2 & = &(R + h)^2 \\    x^2 + 20\,000\,000^2 & = &(20\,000\,000 + 100)^2 \\    x^2 & = &40\,000\,010\,000 \\    x & \approx &62\,000 \, \textrm{feet} \end{array}

A rough estimate of the Earth’s radius in feet is 4,000 × 5,000 = 20,000,000, which is why that number appears in the equations above. Two things should be noted here. First, using 5,000 for the number of feet in a mile instead of the more accurate but less friendly 5,280 allows for easier calculations throughout. Second, we’re only looking for an estimate of the distance to the horizon, so we should feel free to take some liberties with the numbers if it’ll help with computation.

Using 20,000,000 instead of 4,000 allows feet to be used as the units throughout, so the answer obtained at the end (x = 62,000) is also in feet. To find the distance to the horizon in miles, we’ll need to divide by 5,000, which gives:

62,000 ÷ 5,000 = 12.4 miles

Now that’s all well and good, and the boys were amazed that the horizon is so far away. But what if you want to generalize to find the distance to the horizon from any height above sea level?

The answer depends on the math, boating, or physics book you choose to reference.

One possibility is to use this graph, which we discovered a few days later while visiting the Yaquina Head Lighthouse:

Graph - Distance to Horizon

Graph at the Yaquina Head Lighthouse, Newport, OR

It shows the distance to the horizon on the horizontal axis and the feet above sea level on the vertical axis, which I suppose makes sense from a visual perspective; but it’s more conventional to put the independent variable (feet) on the horizontal axis and the dependent variable (miles) on the vertical axis. Nonetheless, the information provided by the graph is still useful, and it nicely shows that the relationship between height and distance to the horizon is clearly not linear.

On the other hand, if you search for “distance to horizon formula” online, you’ll find any of the following formulas:

\begin{array}{rcl}    d & = &\sqrt{1.5h} \\    d & = &\sqrt{\frac{7h}{4}} \\    d & = & 1.17\sqrt{h} \\    d & = &R\times\arccos\bigl({\frac{R}{R+h}\bigr)}    \end{array}

The differences are due to whether miles or nautical miles are used, as well as whether you’d like to account for the refraction of light, which greatly influences the distance an observer would be able to see. Personally, I prefer the first formula for its simplicity, but you’re free to use whichever one you like.

Sadly, I don’t know a single joke involving the horizon. (Do you? Post in the comments.) So I’ll leave you instead with a quote about the horizon:

Make the horizon your goal. It will always be ahead of you.
— William Makepeace Thackeray

September 9, 2015 at 7:42 am 3 comments

Good Luck on Friday the 13th

A friend is taking a business trip today, and I asked if he was worried about flying on Friday the 13th. “I think it’s unlucky to have superstitions,” he replied.

If you’re a baseball player and you see a cross-eyed woman today, you might want to spit in your hat to avoid bad luck. (Or so they say.)

As it turns out, the first Friday the 13th of each year is “Blame Someone Else Day,” so if you don’t like this post, you should defintely let my wife know about it.

To ensure that you point a finger at others on the correct day, wouldn’t it be helpful to know when the first Friday the 13th of the year will occur? Consider yourself lucky, because you’ve stumbled across this post. I have two methods you can use for determining which months contain a Friday the 13th.

Method 1: Look-Up Table

Take the last two digits of the year, yy.

  1. Calculate the sum yy + ⌊yy/4⌋. (The notation ⌊x⌋ indicates the floor function, which is the greatest integer less than x, so ⌊π⌋ = 3 and ⌊7.28⌋ = 7, for example.)
  2. Determine the remainder when that sum is divided by 7.
  3. Look up the remainder in the table below.

Friday the 13th Table

For example, in 2013, the calculations would give 13 + ⌊13/4⌋ = 13 + 3 = 16, which has a remainder of 2 when divided by 7. Since 2013 is a non-leap year, the table tells us that Friday the 13th will occur in September and December this year.

Note that the table only works for dates in the 2000’s. To modify the process for dates in the 1900’s, you need to add 1 in the first step; that is, find the sum yy + ⌊yy/4⌋ + 1. Then proceed as described above.

Also note that four lines in the table are highlighted in pink and yellow. Leap years always cause problems. The yellow lines in the table indicate years in which an extra Friday the 13th occurs in January or Feburary because of leap year, and the pink lines indicate years in which a Friday the 13th in January or February does not occur because of leap year.

Method 2: Internet

Go to http://www.timeanddate.com/calendar, enter the year in question, and examine all 12  months to see which contain a Friday the 13th.

[Ed. Note: Though perhaps more efficient, the use of Method 2 is highly discouraged and less fun than Method 1. Using Method 2 instead of Method 1 will result in the automatic revocation of your Geek Card. Plus, there’s a practical issue: What will you do when you have a time-sensitive need to know the month of the first Friday the 13th in the year 2044, say, and you find yourself in a location without Internet access? Shudder. Consequently, learning Method 1 is just as critical to you as learning to calculate change mentally is to a grocery store cashier, who lives in perpetual fear of power outages.]

Incidentally, there is at least one Friday the 13th every year. In addition, every month has four Fridays the 13th in a 28-year period, which means there are an average of 1.71 Fridays the 13th each year. (There is a slight snafu regarding this last fact, because years ending in 00 aren’t leap years if the year is not a multiple of 400, but whatever. It’s true most of the time.)

September 13, 2013 at 1:45 am 2 comments

Flatulence Funniness Formula

Little kids love flatulence.

I was reminded of this when my two five-year-old sons Alex and Eli, my wife Nadine, and I were playing Blokus at the dining room table, and one of us (who shall remain nameless) brought forth a sound slightly louder but a little less musical than a kazoo.

“Excuse me,” said the offending party. And then both boys giggled uncontrollably until they were out of breath and red in the face.

“Are farts funny?” Nadine asked, after they collected themselves.

“Well, they are,” Eli began, “but the older a person is, the funnier their farts are.”

“That’s true,” Alex agreed, “but it also matters how much they fart. The more often someone farts, the funnier it is.”

Intrigued, Nadine had a series of follow-up questions. “Are grandma’s farts funny?” she asked.

Both boys answered quickly and unequivocally in the negative.

“Are daddy’s farts funny because he farts a lot?” she then asked.

“Daddy farts a lot, and he’s also old — that’s why his farts are so funny,” Eli explained.

“But his stinky farts are not funny!” Alex insisted.

All of this pontification got me to thinking: someone ought to devise a formula to estimate how funny a fart is. Given my expertise, I decided that that someone should be me.

Alex and Eli had already identified three factors — frequency, age of offending party, and odor. In addition, I believe that sound quality, attractiveness of the offending party, and inappropriateness of the situation are also important criteria.

But I had no idea how these components should be incorporated. To do this justice was going to require some research.

According to the National Institutes of Health, an average person breaks wind between 14 and 23 times per day. (Wow!) So based on Eli’s belief that more often is funnier, but also based on my experience that there can definitely be too much of a good thing, it seems that a flatulence frequency slightly above average — maybe one fart every 40 minutes — would be funny, but significantly above average — such as one fart every 2 minutes — would be devoid of humor and, quite frankly, disgusting. Consequently, the frequency (F) in the formula below takes into account the number of rectal tremors released within earshot of another human being during the past hour (not counting the current infraction).

Age (A) is a delicate issue in this analysis. Farts from very young kids can be funny, and I personally think that farts from really old people are hysterical. For people in between, age isn’t much of a factor — unless you’re a teenage girl, in which case the embarassment of a public toot would be too much to bear.

The loudness (L) of a fart is measured in decibels, as is any sound. For a fart to be funny, it has to be loud. Silent farts are not funny — they can’t evoke laughter if no one hears them, and they most definitely aren’t funny if they fall into the silent-but-deadly category. But any sound over 130 decibels immediately causes ear damage, so the formula below assumes that no fart will be greater than 130 dB. It’s difficult for me to fathom that a fart could exceed that level — that would put it on par with a jackhammer or gunshot. (Great googly-moogly! If you’re reaching that standard, change your diet!)

The beauty (B) of the offending party, measured on a scale of 1‑10, also has an impact. No one wants to imagine — or worse, actually witness — a supermodel letting one rip. But on the flip side, most of us are equally mortified by the thought of a repulsive person cutting the cheese. The formula therefore uses a quadratic relation, so that a person with middle-of-the-road looks receives higher marks.

As Alex said, stinky farts are not funny, so smell (S) is a major concern. As it turns out, odor has four properties: odor concentration, odor intensity, odor quality, and hedonic tone. Biochemists use very detailed methods to measure each of these properties; for the purpose of this analysis, however, only odor intensity will be considered. A complex formula called the Weber-Fechner law ranks the odor intensity of a smell on a scale from 0 (no odor) to 6 (intolerable). You don’t need to use some complex formula; let your nose be your guide. If S = 6, the formula below automatically zeroes out.1

Finally, the inappropriateness (I) of the situation has to be considered. The formula below accepts values from 1 (“whatever, no big deal”) to 10 (“oh, no, you didn’t!”). Of course, this is a highly subjective factor. While it is certainly inappropriate for a groom to pass gas while reciting his vows on the altar, it’s a whole lot funnier than releasing an air biscuit while delivering the eulogy at your father’s funeral. So I = 4.2 for the gassy groom, but I = 9.9 for the guy who shows no respect for the deceased.

Put this all together, and what do ya get?

H = \frac{4}{{4 + F}} \cdot \frac{{30 - 5S}}{{9I}} \cdot \left( {\frac{L}{{13}} + \frac{1}{{1000}}{{\left( {A - 15} \right)}^2} - \frac{2}{{5}}{{\left( {B - 5} \right)}^2} + 10} \right)

For the most part, the formula will return values in the range 0‑100, so it works sort of like a song rating on American Bandstand.

Yeah, it’s not perfect, and I’m happy to hear suggestions for improvement. But it works pretty well. A middle-aged (A = 45) woman with average looks (B = 6) who kills the canary for the first time that evening (F = 0) with a loud (L = 95) but relatively mild (S = 1) honk while playing Scrabble® with her sister (I = 1) would score a 49.5; and having witnessed my mother and aunt bust a gut at this very scenario on numerous occasions, my instinct is that such a score is about right.

On the other hand, a man of any age who releases a silent-but-deadly (L = 0, S = 6) would score a 0 regardless of his looks, his age, how long since his last incident, or the situation. And that feels right to me, too. That sick SOB! How dare he?

And lest you think you’ve read all the way to the bottom with not a single joke, don’t dismay! Here’s a joke that’s both mathy and relevant to the rest of this post. (Be afraid… be very afraid.)

What do you call a math teacher who never farts in public?
A private tooter.


1 It occurs to me that (a) I’ve spent entirely too much time doing the research for this post and (b) I now know more about flatulence than I ever wanted or needed to.

August 20, 2013 at 9:55 pm Leave a comment

Mathematical Finances

Got a bead of sweat running down your forehead as you frantically race to complete your 1040? Here are a few math finance jokes to relieve the stress.

Financial Trigonometry: If someone asks you to cosine, don’t sine! Instead, go off on a tangent! That’ll save you $40,000!

Financial Algebra: My wife leaves Houston at 8:39 a.m. on a plane bound for Albuquerque. She arrives at 9:42 a.m. and spends the next three days at a hotel with my best man. If she then decides to leave me for him, how long will it take me to pay off the Visa bill from this trip of infidelity, assuming an annual percentage rate of 18.5%? 

Financial Formula: Easiest way to determine your cost of living? Take your income, then add 10%.

And just in case you needed another reason to never trust a financial mathematician…

A pure mathematician asks, “Would $30,000 be too much?”

An applied mathematician asks, “How about $60,000?”

And a financial mathematician says, “How about $300,000? That’d be $135,000 for me, $135,000 for you, and $30,000 for a pure mathematician to do the work.”

April 15, 2013 at 7:00 am Leave a comment


About MJ4MF

The Math Jokes 4 Mathy Folks blog is an online extension to the book Math Jokes 4 Mathy Folks. The blog contains jokes submitted by readers, new jokes discovered by the author, details about speaking appearances and workshops, and other random bits of information that might be interesting to the strange folks who like math jokes.

MJ4MF (offline version)

Math Jokes 4 Mathy Folks is available from Amazon, Borders, Barnes & Noble, NCTM, Robert D. Reed Publishers, and other purveyors of exceptional literature.

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