Bunny Living Space

For more than a year now, two indoor bunnies – Nutmeg (left) and Tila (right) – have been part of our family. 

Our bunnies are very inquisitive and get along with each other only MOST of the time.  As a result, they spend most of their time in separate, adjacent pens that take up most of our living room, each with her own custom-built 3-story "condo" (the white, wire-mesh, structure at the back):

I wondered how large a space does this "feel" like to them, compared to the humans in the house?  To put it another way, if you were to shrink a human down to the size of the bunnies, how large of a house would this feel like?

Measuring the Pens

First, I needed to figure out how many square feet of space the pens actually take up.  The walls of the pens are flexible, so I choose one of the simplest configurations to compute:

The condos take up 14×42" of space per level.  6 levels result in 3,528in² of space.

Each condo has a front porch, which allows the bunnies to climb up or go under, doubling the floorspace provided.  Each porch has an additional 196in² of space.

I’ve broken the pen itself into three blocks – A, B, and C.

A: 140 x 33 = 4620in²
B: 76 x 32 = 2432in²
C: 0.5 x 32 x 32 = 512in²

So, the total space is

Floor Space	Area
3528	Condos
196	Condo Porch 1
196	Condo Porch 2
4620	Pen Area A
2432	Pen Area B
512	Pen Area C
11484	Total

11,484in² works out to 79.8ft².

Bunny:Human Ratio

Next I needed a bunny-to-human ratio.  I came up with a few possible approaches here.

The first was simply measuring the bunnies’ feet compared to my own.  With my shoe on, my foot is exactly 12 inches long.  The bunnies REALLY don’t like me touching their feet, let alone measuring them, so I decided to estimate them at 1in long, which gives a nice ratio of 12:1.  Converting "human feet" to "bunny feet", then:

12² / 1² = x / 79.8

x = 79.8ft² * 144

x = 11,491ft²

So, if we shrank a human to be 5.9" tall (1/12 my current height), would the pens feel like an 11,491ft² home?  Possibly, but to me that seems too large.

A second is to compare our relative heights.  I’ll use Tila’s dimensions here since she was the more accessible one when I took this photo:

This is probably their most common snoozing positions – the "loaf".  When loafed, Tila is 6" tall at her highest point.

I stand 71" tall, which means there is a ratio of 11.8:1 between us.  Using the second formula above, we get:

    x = 79.8ft² * (11.8)²
    x = 11,111ft²

A pleasantly repeating solution, but subjectively it still seems too large.

A third possible approach is to compare the bunny’s length to my height.  I walk upright, but bunnies spend virtually all of their time with "four on the floor".  When loafed, Tila is 10" long.  Compared to my 71" of height, that’s a ratio of 7.1:1.  Using the formula above:

    x = 79.8ft² * (7.1)²
    x = 4,023ft²

So, at the bunny’s scale, their pens are the equivalent to a 4,000ft² home with each bunny getting roughly half of that.  Would each pen feel like a 2,000ft² home to a 10" tall person?  I think that’s closer to what I was expecting.

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Squares Squared

Let’s say we’re going to calculate the product, P, of two integers, X and Y, each of which are squares.  Their product will also be a square:

P = X • Y

But since X and Y are squares, they can each be broken down into their roots, X_r and Yr, and the problem rewritten as follows:

P = X_r • X_r • Y_r • Y_r

Next, we can rearrange the terms like so:

P = X_r • Y_r • X_r • Y_r

P = (X_r • Y_r) • (X_r • Y_r)

P = (X_r • Y_r)²

Therefore, P will be a square.

More Positive

Here’s a numerical curiosity I came across.  I checked out a book from the library, and while the librarian correctly recorded the due date as “6/10/22”, the latter “2” looked like a 0 from a distance, so I read it as “6/10/20”.  I wondered if there was a clever way to relate the three numbers together.

There are probably many ways, but what I came up with was “If you add 6 and 10, and then add the difference between those two, you’ll get 20”.  Then I started playing with other pairs of numbers – 3 and 5, 1 and 4, and so on.  All of the results came out even.  Was that a general rule?  Given 2 integers a and b, will this always produce an even result?

Let a and b be positive integers, and r the result of our arithmetic machinations.  Our formula becomes:

r = a + b + |a – b|

Now, let’s consider the possibilities:

• If a is larger, then the absolute value portion reduces to simply “a – b”, the b’s cancel, and what’s left is 2a
• If b is larger, then the absolute value portion reduces to simply “b – a”, the a’s cancel, and what’s left is 2b
• If a and b are the same, then the absolute value portion is 0, leaving 2a (or alternatively, 2b)

Interestingly, not only will this always produce a positive result, it will always be twice the larger integer.

That works if we start with positive integers.  What happens if we introduce other combinations?

• Two negative integers: the result ends up being twice the smaller of the two negatives.  For example –6 and –10 produce a result of –12.  You can also define “the result will be twice the more positive value”.
• One positive and one negative: The result is still twice the more positive.
• One positive and 0?  Still twice the more positive number.

What about one negative and 0?  That interestingly follows the same rule – the result will continue to be the more positive of the two, which is 0 in this case.  If we let a be the negative number and b be 0, the formula reduces to

r = a + |a|

With a being negative, this will always cancel out, leaving 0.

Is there a name for any of this?

* Edited to correct the origin story of this proof.