## The Distance and Power of Mr Rogers Pass

The Green Bay Packers and Detroit Lions played their second game of the 2015 season on December 3rd. It ended with a highly unlikely Hail Mary catch by Richard Rodgers to score the winning touchdown. In the days following the game I heard many, many different people expressing excitement and admiration for the play, with one colleague even declaring Aaron Rodgers “The best quarterback in the NFL”.

The league posted a video clip asking “How far exactly did Rodgers’ Hail Mary Travel?”. The official marking of 61 yards is shown at the bottom of the screen. This is the distance from the line of scrimmage at the start of the play to the front of the end zone line. Ben Rohrbach expressed his belief that the total travel distance of the ball was close to 100 yards, including the vertical and horizontal elements.

When a ball is thrown, kicked, or otherwise sent with force into the air, the vertical profile will be shaped like an upside down parabola. This is because gravity acts with constant downward acceleration against the object. The total distance along this segment of a parabola is known as the arc length. It can be calculated by integrating the changing height across the horizontal distance. Wolfram’s discussion is here, and Had to Know’s is here.

With that out of the way, the objective is to calculate the actual arc length of Aaron Rogers December 3rd pass. The same method can be used to calculate golf drives, log throws, rock catapultations, etc…. Note: Air friction is neglected in the following discussion.

The horizontal distance of the pass is calculated via the Pythagorean Theorem using length and width coordinate estimates from the back end zone and visual top of the field. The image below shows how this was done for Aaron Rodgers throw. The throw starts an estimated 39 yards from the back end zone, and ends 103 yards from the same end zone. At the start it’s an estimated 13 yards from the top sideline, and it ends about 18 yards from the same sideline.

The second step is to estimate the total height attained by the ball, minus the height of the passer. Here, I assume that the ball was caught at the same height from which it was thrown. The vertical displacement in yards can be calculated by estimating the hang time, squaring it, and multiplying by 1.34058.

Once the vertical and horizontal displacements are measured, the rest can be calculated in a straightforward manner. Using my estimates of pass start at (13,39), catch at (18,103), and hang time of 5.6 seconds I got the following results:

- Horizontal Displacement: 64.195 yards
- Vertical Displacement: 42.041 yards
- Total Travel Distance: 110.716 yards
- Velocity of Pass: 12.706 yards per second
- Energy Imparted to Football: 21.405 foot pounds
- Power Output of Aaron Rodgers during pass: 0.216 horsepower

This admittedly armchair analysis vindicates Rohrbach, and calculates Aaron Rodgers throwing technique to be as strong as just over a fifth of a horse. The image of my Excel output is below, with the following cell formulas:

- A6 =SQRT(C2^2 + C4^2)
- A11 =SQRT((A6^2/4) + (4*B9^2)) + (A6^2/(8*B9))*ASINH(4*B9/A6)
- A13 =(3*B11)^2*(0.94799/(2*32.174049))
- B9 =1.34058*A9^2
- B11 =SQRT((A6/A9)^2 + 10.72467*(A9/2))
- B13 =(A13/B6)/550
- C2 =ABS(B2-A2)
- C4 =ABS(B4-A4)

## Discussing Unemployment in 1913

Pigou A C. “Unemployment”, pages 29-35. Published 1913 by William and Norgate, London. Lots of really good stuff in this book–lessons any person in the developed world can see and understand, many of which have been better quantified by recent research. You can get the whole thing on The Internet Archive.

Pigou points out the long term effects of even relatively short durations of unemployment on individuals beginning halfway down page 32. The same effects are being rediscovered (and measured) today.

## Solve If You Are A Genius — Of Ambiguity

The sun also ariseth, and the sun goeth down, and hasteth to his place where he arose. The wind goeth toward the south, and turneth about unto the north; it whirleth about continually, and the wind returneth again according to his circuits. All the rivers run into the sea; yet the sea is not full; unto the place from whence the rivers come, thither they return again. All things are full of labour; man cannot utter it: the eye is not satisfied with seeing, nor the ear filled with hearing.—Ecclesiastes 1:5-9^{9}The thing that hath been, it is that which shall be; and that which is done is that which shall be done: and there is no new thing under the sun.

Among other things, Ecclesiastes deals with ambiguity, assumptions, and the ways we react emotionally when our assumptions turn out to be inaccurate. Anyone who does the social media/blogosphere thing has seen at least one “Solve If You Are a Genius” meme. They’re great discussion starters, really good for getting folks to take a position on something that shouldn’t be controversial. I found the one I’m about to discuss on LinkedIn. It generated 51 answers, many of them with reasoned explanations, and some folks prompting others to justify their responses.

As far as I can tell, these memes are good at creating discussion for three reasons:

- They appear mathematical. It’s close enough to get everyone thinking mathematically, or at least arithmetically. The notation isn’t quite right, but it’s enough to get our brains to infer that something mathematical is going on.
- There’s just enough information to make the numbers look like they have a linear pattern, and it’s good enough for just about everyone to be able to discern one. Actually, it activates the human instinct to seek patterns, even in random noise. Bud Light’s football season commercials satirize the same tendency.
- Presentation—the words at the beginning take advantage of the math classes of our early school years, when most of us learned that math problems have only one solution.

The story on this meme, however, is that there are many reasonable answers—an infinite number, in fact. But in order to choose an answer, our brain has to make at least one assumption. Some assumptions may be more reasonable or easily explained than others. I’ll run two cases right quick and state my assumptions to show how they impact response to the meme.

**Case One**

**Assumptions:** ***Values on the left side produce the values on the right side via arithmetic operations. ***Not all strategic statements, x…. –> y, are included on the list.

**Corollary:** Completing the list is imperative to answering the question

**Strategy: ** Rewrite the list with gaps included. Number the statements. Relate the right and left sides arithmetic operations. Use the algorithm to fill in the gaps.

At this point, it outta be easy to fill in all the gaps, right? Sure: we can just accept our prior assumptions, input the results, and walk away confident that we got the right answer. Except maybe not: even if the presence of row 3 is reasonable, can the same be said about row 1? And if row 1 doesn’t exist, how do we deal with the non-existent value of 2 on the left side of the equation? Probably the best thing is to treat it as a 0—but maybe 1 is a better choice. On the other hand, maybe we just say it’s undefined. Now there are four reasonable possibilities:

Using the set of assumptions selected in Case One, the solution set is {0, 3, 6, DNE}. Here’s another way to look at it:

** Case Two**

**Assumptions: ** Values on the left side and right side appear correlated, but they are not

**Strategy: ** Fill in the value in question following the pattern on the right side only

By working through these two cases five possible answers have been found—all based on arithmetic patterns and reasonable assumptions. There are probably more answers to be found by following these same lines of reasoning, but for now I’d like to abandon it. I’d like to engage instead in some polynomial interpolation and extrapolate from there. In short, I will cheat the data. (Yes, I really said that!) In this case we just use it to cram together a non-singular matrix using an equation of the form:

p_{n}(x) = α_{n}x^{n} + α_{n-1}x^{n-1} … + α_{0}x^{0 }

For this exercise y = p(x) = right hand values, X = left hand values taken to the appropriate exponent, α-values will be represented by letters, and n starts at the top and ends at the bottom. Therefore the equation will take the form:

p(x) = ax^{3} + bx^{2} + cx + d and p(8) = 56, p(7) = 42, p(6) =30, p(5) = 20 . These assignments yield the following matrix:

This gives a value of p(3) = 6. The same answer pops up when p(x) = ax^{4} + bx^{3} + cx^{2} + dx. Using a standard, perfectly sensible linear interpolation, we find the pattern to be p(x) = x^{2} – x. But if p instead takes the form:

p(x) = ap^{n} + bp^{n-1} + cp^{n-2} + dp^{n-3} where n > 4

values of p(3) other than 6 begin to show up. For example, when n = 5, p(3) = 6.214285714285708…, and when n = 6, p(3) = 1.993112244897960…

Based on this we can safely say that there are many, many answers to the “Solve If You Are a Genius” meme. The set is almost as big as the set of polynomial functions, which just happens to be infinity. But this answer brings in two new questions:

1. How are the real answers bound for different polynomial forms?

2. Are answers limited to real numbers?

I’m not sure how to prove the first one analytically, but I will attempt to address it numerically in a future post. With the second question, my instinct is that under some conditions they are. When I solved the polynomial equation, I assumed the constants a, b, c, and d had only real parts. But it’s not difficult to imagine they take the complex form of q = y + zi. My sense is that under an even set of knowns consisting only of real numbers it is possible to have a set of z-constants that sum to zero for each known in the set. That is to say, the imaginary component is trapped within the null space. When the unknown gives the set an odd number of equations, then I think it’s possible for the imaginary component to show up in the answer. However, if it is an odd set of known consisting only of real numbers, the unknown must be a real number also. Confused yet? Me, too!

Anyway, that’s enough for this post. What do you think of the meme?

**Update (12/15/2013)**

I spent about a week chewing over my last paragraph and another week considering the best format to answer the questions I posed at the end. Here’s a couple comments addressing them….

- First Question–How are real answers bound for different polynomial forms: It’s not necessary to address this numerically because it can be shown analytically that they can be any real number. With polynomial interpolation it was possible to produce four independent equations with which to solve for the four independent variables–a, b, c, and d. I chose to unbind the highest order exponent as a variable j. The rest of the exponents were set to j-1, j-2, and j-3. In doing this, I actually was creating a fifth independent equation. In doing this, the system becomes specifiable to five values in the right hand column. By having five independent equations, any desired value can be specified for 3 = ?. There’s also no rule that says the exponent-values must be bound to the pattern j, j-1, j-2,….j-n. If exponents can be selected at will, then a total of 8 independent equations can be fabricated, allowing one to solve a system of eight question marks from the four known relations.
- Second Question–Is the range of reasonable answers bound only to real numbers:

The answer is no, answers can be real or complex. The imaginary components in the four known relations can be designated as b1 = b2 = b3 = b4 = 0. But if there are five independent equations, b5 is not constrained to zero. The same principle above applies, and it can be any positive or negative number.

The number of ways to consider the Solve If You Are Genius meme are as boundless as imagination, but there’s one more method I want to address here. Using the same polynomial interpolation p(x) can take the form:

p(x) = ap^{n} + bp^{n-1} + cp^{n-2} + dp^{n-3 }+…+ e

In this case the unknown can take the value of any number. What’s really happening is the unknown is being constrained to a one-dimensional line in R^{n}. Personally, I’d rather have a plane. But a line can be just as good. Cheers!