- 1:04 pm - Tue, Oct 22, 2013

This is the reference photo for my album cover by @michaelramstead

- 3:47 am

tavlo:

rosaparking:

entropiaorganizada:

hookteeth:

… Y’see, now, y’see, I’m looking at this, thinking, squares fit together better than circles, so, say, if you wanted a box of donuts, a full box, you could probably fit more square donuts in than circle donuts if the circumference of the circle touched the each of the corners of the square donut.

So you might end up with more donuts.

But then I also think… Does the square or round donut have a greater donut volume? Is the number of donuts better than the entire donut mass as a whole?

Hrm.

HRM.

A round donut with radius R_{1} occupies the same space as a square donut with side 2R_{1}. If the center circle of a round donut has a radius R_{2} and the hole of a square donut has a side 2R_{2}, then the area of a round donut is πR_{1}^{2} - πr_{2}^{2}. The area of a square donut would be then 4R_{1}^{2} - 4R_{2}^{2}. This doesn’t say much, but in general and throwing numbers, a full box of square donuts has more donut per donut than a full box of round donuts.

The interesting thing is knowing exactly how much more donut per donut we have. Assuming first a small center hole (R_{2} = R_{1}/4) and replacing in the proper expressions, we have a 27,6% more donut in the square one (Round: 15πR_{1}^{2}/16 ≃ 2,94R_{1}^{2}, square: 15R_{1}^{2}/4 = 3,75R_{1}^{2}). Now, assuming a large center hole (R_{2} = 3R_{1}/4) we have a 27,7% more donut in the square one (Round: 7πR_{1}^{2}/16 ≃ 1,37R_{1}^{2}, square: 7R_{1}^{2}/4 = 1,75R_{1}^{2}). This tells us that, approximately, we’ll have a 27% bigger donut if it’s square than if it’s round.

tl;dr: Square donuts have a 27% more donut per donut in the same space as a round one.

did u major in donut math for nerds

Sweet Neptune. I’m changing my major to doughnut math for nerds. Can you imagine the labs?!

(Source: nimstrz, via screamingdonkey)