Fun Fact of the Week: Hum Reading Length, or: An Exercise in Dimensional Analysis

Sometimes it feels like these HUM 110 readings are sooo long. But how long are they, actually? 

I would like to find, using dimensional analysis and some experimentation, how long (in distance) the Hum readings are. 

To begin with, let’s consider this equation: Length of reading= (Length / Word)# Words. For the (Length / Word) term, we could go through all the work of determining an experimental average, or we could also just Google some stuff and multiply it together. So! Consider: (Length / Word)=(Length / Letter)(Length / Word). 

The average number of letters per word can be calculated pretty easily by pasting a few pages worth of reading into a doc and then Control-Shift-C-ing the number of characters (excluding spaces) and the number of words. For example, the first three pages of Their Eyes Were Watching God have an average word length of 4.28, pretty small due to all of the dialect dialogue, whereas an academic paper by W.E.B. DuBois would have a higher ratio, closer to 5.90. I think that a fair average for all of these papers would be 5.2, so let’s go with that. I’ll add 0.5 to this to account for a space following each word, to get 5.7. 

Now, the length of a letter would be determined by font, font size, spacing, view size, etc, but for today’s challenge, let’s look at 12pt. Times New Roman font. So how big is each of the letters, length-wise? According to my online research (Gregory MacNaughton didn’t respond before my deadline), a Times New Roman ‘a’ has a height-to-width ratio of 1:0.44. A fun fact is that ‘pt,’ as in ‘10pt, 12pt, 23pt font’ is an actual unit of measurement, equal to 1/72 of an inch. So, using this information, we say, 

12 Height (pt)(0.44 Width(pt) / 1 Height(pt))(1 in / 72 pt)(2.54 cm / 1 in)=0.186cm is the width of an ‘a.’ For the sake of simplicity, we’re going to say that, on average, the letters are all this width, as the number of thin ‘i’s and ‘l’s should cancel out the wider ‘m’s and ‘w’s, particularly when taking punctuation into account. 

So, now we can say (0.186 cm / Letter)(5.7 Letter / 1 word)=(1.06 cm / Letter) for our length per word. Now all we need to do is count the number of words, and we’ll be golden. I couldn’t think of a better way to do this than experimentally, but I don’t have all day, so I’m going to make some approximations along the way. First, I’ll divide up the readings into three groups, Article, Poetry, and Prose, then count the number of pages for each, and multiply by the word density (wordspage). This must be done as Gilgamesh, for example, has a density of around 350 words per page, which is much lower than James Weldon Johnson’s “The Making of Harlem,” at ~800 words per page. So (I did this instead of my Hum reading), the count is (Fall+Spring) Article: 295+310, Poetry: 590+75, Prose: 395+560, plus ~365,000 words for Genesis, Exodus, the Iliad, Their Eyes Were Watching God, and Invisible Man combined. With some light testing, I found the average word density of each genera were Article; ~425, Poetry: ~300 words per page, and Prose: ~450 words per page. So, multiplying this all out gives a grand total of Article: 257,100, Poetry: 199,500, Prose: 429,700, plus the additional books, gives a grand total word count of 1.25 Million Words! Therefore, given that you have done all of the reading in Hum this year, you will have read for a total length of:

1.25 106words(1.06 cm / 1 word)(1 km / 100,000 cm)=13.2 kmGiven that Eliot Hall has a perimeter of ~220 meters, you could wrap around the building 60 times with your string of 12pt. letters. Stretching into the sky, that’s well over the height of Mt. Everest, and a little over the cruising altitude of a jet. So, be proud of yourself, and get ready for Invisible Man, which accounts for 1.5 km all on its own.