Документ взят из кэша поисковой машины. Адрес
оригинального документа
: http://www.badastronomy.com/bad/movies/mummyreturnsmath.html
Дата изменения: Unknown Дата индексирования: Sun Apr 10 00:07:16 2016 Кодировка: Поисковые слова: п п п п п п |
|
![]() |
|||||||||||||||||||||||||||||||||||
Buy My Stuff |
The Math of The Mummy ReturnsSo, how did I do that fancy math to figure out how fast the shadow moves? Well, remember in high school when your math teacher told you that someday math might save your life? Well, here is where Rick O'Connell should have paid attention in school (with apologies to the character Val Kilmer plays in Red Planet).
Here's where the trig comes in. We can calculate the length of the shadow if we know the Sun's angle above the horizon and the object's height. In this case, the length equals the height divided by the tangent of the Sun's angle:
We have to be careful, because the Sun has to clear the mountains first. O'Connell won't see the Sun until it rises over the mountains. At what angle does the Sun clear the mountains? We can use the above equation to solve for the angle:
Assuming the mountains are one kilometer high and 10 away, that means that
or, the Sun must be 5.7 degrees above the horizon to clear those mountains. So where does that leave us? Well, we know how high up the Sun is, and the length of the mountains' shadow at that point (10 kilometers). So the next question is, how fast is the shadow moving? There are two ways to do this. You could take a calculator, plug in different angles (say, increasing by one degree each time) and get the shadow length. The Earth spins 360 degrees in 24 hours, which means it spins 1 degree in 4 minutes; that in turn means the Sun moves 1 degree every 4 minutes in the sky. You can then convert your numbers into a velocity. Or, you can do a bit more clever math. The shadow length depends on the Sun angle, so the change in the length depends on the change in that angle. That means you can take a derivative! If you figure out how the length changes with time, you get
where, yes, that's the co-secant of the angle. Luckily, I can program a computer to calculate this for me, and let the angle start at 5.7 degrees and run it up to where the Sun is way up the sky. If I do that, I get the following graph.
|
|