I was away from work for Monday and Tuesday of this week. I had signed myself up for a Cost Engineering course at a hotel in downtown Edmonton. The course outline looked like it had some relevance to certain aspects of my work.
I was wrong.
It turned out to be two days of mostly monotony wherein our two instructors stood at the front of the small seventh floor room, backlit by the omnipresent glow of the Powerpoint presentation, and droned on about such topics as Cost Containment, Best Value Assessment, Authorization for Expenditure and Fixed Asset Management. There were a few islands of interest, amidst this sea of ennui, when one of the instructors would regale us with personal experience pulled from one of the many projects in which he had been involved whose success had hinged on accurate cost engineering. These were few and far between.
There were fifteen male attendees in the class and both instructors were also men. Women, apparently, aren't drawn to the cost engineering field, or at the very least have better sense than to attend a two-day course on the topic.
Midway through the second morning, I began to doodle. I became more entranced by the random blue lines my pen was making around and through the words on the page than I had been to date with the words themselves. And given my wont for logical and organised thinking, my doodles began to take shape. I drew a cube. Inside that cube I circumscribed a circle, whereby one point of the circle touched each of the four sides of the square. I then thought to myself, you know, the ratio of the surface area of the larger square to that of the smaller circle is a geometric constant. Let's figure out what that ratio is.
The diameter of the circle was equal to the length of the side of the square, so this ended up being quite easy. The ratio is: 4/π
I was very pleased with initial success. So now that the steam train was going, I wanted to build up a little speed. I drew a circle and circumscribed a square inside of that. The diagonal of the square is equal to the diameter of the circle. So, given that the diagonal created two right triangles, Pythagorus came to my aid to determine the length of the sides of the square. The ratio of the area of the larger circle to that of the smaller square was then: π/2
This is where I started to get really crazy.
Not content with a 2-dimensional world, I decided to branch out and explore the realms of 3-D. I was nearly unstoppable. I sketched a cube. Inside of that I drew a circle that represented a sphere, touching the cube on each of its six sides. Similar to my first problem, the length of one side of the cube was equal to the diameter of the sphere. A couple of simple volume formulae resulted in the ratio of the volume of the cube to that of the smaller sphere to be: 6/π
This is where it started to get a little trickier.
I drew a sphere, and inside that transcribed a cube wherein all eight corners of the cube made contact with the surface of the sphere. Cost Engineering had gone by the wayside. Powerpoint was merely the glow by which I guided my pen on the page. The only voice I now heard, no longer the Scottish brogue of the instructor from Cuthbert & Smith Consulting Ltd., was the one inside my head, driving me on to complete this fourth geometric conundrum!
It was no longer so easy... The diameter of the sphere could not be related directly to the length of one side of the cube. The diagonal of the cube was the answer to that problem; but how to relate the length of the diagonal to the length of one side of the cube? For that, too, is a geometric constant. Pythagorus again to the rescue!! I needed to use his theorem twice this time instead of once. More circuitous was the path, and more rewarding was the fruit of this labour, let me tell you. The ratio of the volume of the sphere to that of the smaller cube became: (π√3)/2
I sat back, basking now in the glow of the screen at the front of the room as well as that of the success of my mathematical meanderings. I tuned back in to the lecture taking place in front of me, realizing that the rest of the world, or at least the rest of the room, had failed to take note of my prowess. No matter. I was content without the accolades.
I came back to work yesterday realizing that I had earned 16 Professional Development hours over the course of the first two days of the work week, several of which had been spent contemplating the geometric relationships of doodles on my paper.
What the hell was up with THAT?!
The circle-inside-a-square was used in one of my favourite classes ever at university.
-Take a random number generator that gives numbers between 0 and 1 inclusive with as much precision as you like (the more the better though)
-Ask for two random numbers. Assign them to an (X,Y) coordinate. Ex: (0.75488, 0.17340)
-Now imagine a square going from (0,0) to (1,1). All of your randomly generated coordinates must fall within this square.
-Now imagine a circle with a centre at (0.5,0.5) with a radius of 0.5. This circle is inscribed within the square. Not all of the random coords will fall within this circle.
-You can now establish that:
(# of points in circle)/(# of points in square) =
(area of circle)/(area of square)
-You know the ratio of Circle Area to Square Area = Pi /4
-Thus, 4* percent of points in the circle gives you a value of Pi.
Using only random numbers between 0 to 1, you can get a very good value for Pi! Crazy! The longer you let it run and/or the better the precision of your random numbers, the better your estimate of Pi.
Posted by: fv | Thursday, 18 November 2004 at 02:38 PM
Math geeks! I'ma skeered. I'ma haid unnerneatha this here table 'til youse guys er dun.
Posted by: Paul | Thursday, 18 November 2004 at 02:46 PM