While on the beach one day, I found a triangular rock. So I started on some beach relief.
Then I iterated.
And I iterated again.
And I continued to iterate, until the whole beach was one big Sierpinski Triangle!
Appreciation
More Insignificant Digits
Are there any numbers we regularly interact with that are less significant than this 
This fraction of a penny is even more pointless than these 2-decimal-place accuracy mile markers.
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Appreciation Geometry
CDs, Prisms, and Parallelepipeds
I’ve had some fun playing around with old CDs and CD cases recently.
In addition to demonstrating Cavalieri’s principle both with discs and their cases, I’ve found some other mathematical uses for these objects.
Here I’ve used a stack of cases to demonstrate the difference between some geometric solids.
On the left, we have a right rectangular prism. Give that prism a slight push in a direction perpendicular to a side and you get an oblique rectangular prism.
And if you give that original prism a push along a diagonal of the base, you’ll get one of my favorite geometric objects, a parallelepiped. It may not be the perfect parallelepiped, but I’ll take it!
Related Posts
- The Perfect Parallelepiped
- CD Packing Problems
- CDs and Cavalieri’s Principle
- CDs and Cavalieri’s Principle: Part 2
Appreciation Representation
Dancing Bubble Sort
This is a fun and whimsical demonstration of bubble sorting through dance!
http://www.youtube.com/watch?v=lyZQPjUT5B4
The dancers arrange themselves in numerical order in the same manner one would bubble sort an unordered list. One by one, each number “compares” himself with the number on his left; if they are out of order, they switch places. Make you’re way down the list, and start again at the front. Repeat until no one switches places and voila! everyone’s in order!
And just to be thorough, the troupe does dance-representations of Insert-Sort, Shell-Sort, and Select-Sort algorithms as well!
Appreciation Geometry
CDs and Cavalieri’s Principle: Part 2
After demonstrating Cavalieri’s Principle with empty CD cases, I thought I’d do the same with the actual CDs.
Here we see a bunch of discs stacked up to make a right cylinder.
To compute the volume of this cylinder, it would be sufficient to know (a) the volume of one CD, and (b) the number of CDs in the stack. We would simply multiply the two together to get the volume.
The argument is less obvious, but essentially the same, regardless of how the CDs are stacked! So this “prism”
has the same volume as the original cylinder. Now, this object should also have the same volume
however some center-of-mass issues may foil our elegant mathematical demonstration.
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