Wednesday, June 27, 2012

Perspective in astronomy pictures

One movie that had a profound effect on me was E.T. Besides opening my eyes to a much larger world besides the earth we live in, I remember being amazed at Spielberg's beautiful filmwork and storytelling. Especially striking however, an image that often pops in my head is the picture of Eliot and the alien riding their bike transiting the moon. I must admit being tricked by the picture for many years and longing to see a moon that was that big! I always loved seeing the moon on the horizon because of the optical illusion it provides, but the illusion was never THAT big!  I want the moon to fill up more than half the sky like that!

Of course, I learned later the moon itself was zoomed in on, the biker superimposed on top, but still, what a great picture!

The thought for today is to calculate how far away are Elliot and the alien, given what we know about the true size of the moon? And we'll look at another "true" photograph from yesterday to calculate how far away the fisherman were.

One important fact that we need to know is the angular width of the moon. If you've been following the Summer Math Series so far, you'll remember I told you the moon and sun both occupy approximately 1/2 degree in the sky. This coincidence is a gift of God allowing for eclipses and measuring the solar system, but more about that later.

I "photoshopped" the image of E.T. to see how wide the bike and boy are. It looks like 4.25 "bikers" fit across the moon, so that means the angular size of a biker is:
$\frac {\tfrac{1}{2}^{\circ}}{moon } \frac{1 moon}{4.25 bikers} \approx 0.1^{\circ} / biker$

Using some trigonometry we can calculate the distance away the biker is. First, a triangle picture.  I am assuming that the biker in the picture is pretty much perpendicular to me, and so I draw a right triangle as shown:
The angle, height, and distance are all related by the tangent function:
$tan(\Theta )= \frac {opposite}{adjacent} = \frac{h}{distance}$
That means that if we know two out of those three things, we can calculate the third. Earlier we measured the angle, and if I assume the height of the biker to be approx 5 feet, we can calculate the distance by rearranging the equation:
$distance = \frac {h}{tan(\Theta )} = \frac {5 ft}{tan(.1^{\circ})} \approx 3000 ft$
Notice, if you type this into your calculator, you might get a more precise number -- I actually got 2864.786067 feet. That amount of precision is misleading, because I made two pretty uncertain measurements in the height of the biker (how do i know he isn't 5' 2" or 4' because he's sitting down?) and in the angular width of the biker (maybe its more like 4.5 bikers per moon, or only 4?)  When this situation arises, one rule of thumb is to round to as many numbers as you had when you made your assumptions. Since 5' had only one digit, I'm going to round to the nearest 1st digit.

So, if this picture was real -- the bikers would have had to be about a half mile away, giving you some idea of the telephoto lens required to make the shot.

Using similar analysis, we can return to the real picture that we looked at yesterday and calculate how far away the fisherman are:
The fisherman are approximately 2 suns tall, or 1 degree.  Assuming that they are adults of approximately 6 foot in height, I calculate:
$distance = \frac {h}{tan(\Theta )} = \frac {6 ft}{tan(1^{\circ})} \approx 343 ft$
To me that looks about normal for a foot ball field away, so this picture was probably taken with very little zoom.