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| || ||Mad City|
|Title:||"Mathematician's Corner" : Lesson 2|
|Date:||Thursday March 24th, 2011|
|In our previous episode of “Mathematician’s Corner”, we determined that Usain Bolt would have finished the Mad City 100k in 2 hours, 39 minutes, 40 seconds if the blazing pace of 23.35 miles per hour that he ran during his world record 100-meter run could be maintained indefinitely. In this episode we’ll bring the opposite end of the speed spectrum into our discussion, and do some fun comparisons. As fate would have it, a recent news story provides us with a perfect model for these comparisons.|
Last Saturday, a 400-pound man named Kelly Gneiting finished the Los Angeles Marathon in 9 hours, 48 minutes, 52 seconds. In decimal terms, this is approximately 9.81444 hours, and since a marathon is 26.21875 miles, Gneiting’s average speed was:
| = ||2.67 miles/hour|
Now, how long would Kneiting have taken to finish the Mad City 100k at this speed? Since the race is measured in kilometers, we’ll first convert his speed to kilometers per hour:
| = ||4.30 kilometers/hour|
By dividing this number into the race distance of 100 kilometers, we discover that Gneiting would have taken about 23.26 hours, or slightly over 23 hours, 15 minutes to finish. Now remember that the Mad City 100k race consists of 10 laps of a 10-kilometer loop course through the scenic University of Wisconsin Arboretum. In his 2 hour, 39 minute sprint, Usain Bolt would have “lapped” Kelly Gneiting 8 times by the time Bolt crossed the finish line.
This raises an interesting question (and brings us to our more challenging mathematical computation for today): Is there a simple formula for determining how many times a faster runner will “lap” a slower runner in a multi-loop course, presuming each runner’s speed is known? Let me explain how I determined that Bolt would have “lapped” Gneiting 8 times.
First, realize that distance is directly proportional to speed (e.g. if you run twice as fast as someone else, you’ll travel twice as far in the same amount of time). Thus, if each man runs for the same amount of time, the ratio of Gneiting’s speed to Bolt’s speed should be equal to the ratio of the number of loops each man has completed at any given moment in time. We therefore have the following proportion comparing the two runners at the moment Bolt crosses the finish line:
| = ||x loops|
To solve a proportion such as this for the unknown x, we “cross-multiply” the 2.67 and the 10, then divide by 23.35:
| x || = ||(2.67)(10)|
| = ||1.14|
In other words, Gneiting will have run only about 1.14 loops by the time Bolt finishes his 10th and final loop. Why does this imply that Bolt “laps” Gneiting 8 times? Well, since 10 minus 1.14 is equal to 8.86, Bolt will be 8.86 laps ahead of Gneiting by the time Bolt finishes. Now think about it: When Bolt is exactly one lap ahead, he’s right beside Gneiting, “lapping” him. When he’s exactly two laps ahead, he’s right beside him again, “lapping” him a second time. And so on. Therefore a lead of 8.86 laps means that Bolt has “lapped” Gneiting 8 times, and fallen just short of doing so a 9th time.
Now let’s generalize: if you (the faster runner, of course) run at a speed of r1, and your friend (the slower runner) runs at a speed of r2, and the race consists of a total of L laps, then the quantity gives the number of laps that your friend will have completed by the time you finish all L laps, and the quantity
tells you how many laps you’re ahead of your friend at the moment you cross the finish line. The whole number portion of this quantity represents the number of times you’ve “lapped” your friend.
Interestingly, this calculation is completely independent of the lap distance! It can be 10 kilometers, 1 mile, 100 miles, or whatever! All that matters is the number of laps, regardless of distance. The calculation is also independent of the units used to measure the runners’ speeds. They can be measured in miles per hour, kilometers per hour, feet per second, or even millimeters per century, as long as both runners’ speeds are expressed using the same units.
Of course, the rate that most runners are familiar with is their “per mile pace”, e.g. 8 minutes/mile. However, you should not use a per mile pace (or a per kilometer pace) in the above calculations, since it does not actually represent a “speed”. That’s because speed is always equal to a distance unit divided by a time unit, not the reverse. Fortunately, you can get around this issue by using the reciprocal of the per mile pace as the runner’s speed.
Brenda can run the Mad City 100k at a pace of 10 minutes per mile, while her friend Kevin can only manage 12:45 miles. How many times will Brenda “lap” Kevin during the race?
|Brenda’s pace of || 10 minutes |
| corresponds to a speed of || 1 mile |
| = 0.1 mi/min|
|Kevin’s pace of || 12.75 minutes |
| corresponds to a speed of || 1 mile |
| = 0.0784 mi/min|
We therefore take r1=0.1, r2=0.0784, and L = 10 laps, and apply the above formula:
Brenda will therefore “lap” Kevin twice during the race.
| L || - ||r2 * L|
| = 10 - || 0.0784 * 10 |
| = 2.16|
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|Posted:||March 24th, 2011 1:30 pm|
|Last View:||August 9th, 2016 7:21 pm|