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Trekkie Trig

Date: 08/13/2013 at 15:05:22
From: Keith Harvey
Subject: Star Trek The Motion Picture Math

Question:

How do you use trigonometry to find the distance between the deadly
asteroid and the refitted Enterprise 1701 in the wormhole?

How do you do the calculation?



Date: 08/26/2013 at 22:19:58
From: Doctor Jeremiah
Subject: Re: Star Trek The Motion Picture Math

Hi Keith,

There are lots of possible ways. 

One interesting way is stereo photogrammetry. Basically, you take two
photos of an object from two cameras on either side of the ship. If both
cameras point straight ahead, and you know the angle of view from the
cameras and how far apart they are (the beam width of the ship), then you
can measure the distance to any point shown in the photos.

If you're not already familiar with angles of view, check out the
Wikipedia article

  Angle of View
    http://en.wikipedia.org/wiki/Angle_of_view 

Let's say that the angle of view from the cameras is 50 degrees -- just a
guess, but a reasonable one. The photos could look like this:

        left camera              right camera
   +---------+---------+     +---------+---------+
   |         |         |     |         |         |
   |         |         |     |         |         |
   |         |    o    |     |    o    |         |
   |         |         |     |         |         |
   +---------+----+----+     +----+----+---------+
   |         |--a-|    |     |    |-b--|         |
   |         |---50/2--|     |--50/2---|         |
   |         |         |     |         |         |
   |         |         |     |         |         |
   +---------+---------+     +---------+---------+

If the width from the center of the photo to the edge is 25 degrees (half
of 50), then you can use ratios to calculate how many degrees the asteroid
is from the center in both photos. Call those angles a and b.

For example, if the object is halfway between the center and the edge,
then the angle is 22.5 degrees. Now you can draw a picture like this:

     left
    camera
   --- +-----------------
    |  |   +   a
    |  | 90-a  +
    |  |           +
    |  m               +
    d  |                   +
    |  +--------D-------------- o
    |  |                   +
    |  n              +
    |  |  90-b   +
    |  |    +    b
   --- +-----------------
     right
    camera

Now you have two angles (90 - a and 90 - b) and one side length 
(d = distance between cameras), so you can use trigonometry to solve for
the distance to the object, a distance I have labeled as D.

If you do the algebra, you get

        d tan(90 - b) tan(90 - a) 
   D = ---------------------------
        tan(90 - b) + tan(90 - a)

To figure that out, I started with these three relations, taken directly
from the picture:

   d = m + n
   D = m tan(90 - a)
   D = n tan(90 - b)

I changed them up a bit:

   m = D / tan(90 - a)
   n = D / tan(90 - b)

Then I eliminated m and n:

   d = m + n
     = D / tan(90 - a) + D / tan(90 - b)

Finally, I solved for D:

   d tan(90 - a) tan(90 - b) = D tan(90 - a) tan(90 - b) / tan(90 - a)
                             + D tan(90 - a) tan(90 - b) / tan(90 - b)
                             = D tan(90 - b) + D tan(90 - a)
                             = D [tan(90 - b) + tan(90 - a)]

So,

   D = d tan(90 - a) tan(90 - b) / [tan(90 - b) + tan(90 - a)]

Let's say that a = 12 and b = 10, and the distance between the cameras is
131m (I am pretty sure that is the right value for a MkII Constitution
class Enterprise).

Then:

   D = d tan(90 - b) tan(90 - a) / [tan(90 - b) + tan(90 - a)]
     = 131 tan(90 - 10) tan(90 - 12) / [tan(90 - 10) + tan(90 - 12)]
     = 131 tan(80) tan(78) / [tan(80) + tan(78)]
     = 131 * 5.67 * 4.70 / (5.67 + 4.70)
     = 131 * 26.68 / 10.38
     = 131 * 2.57
     = 336.86 meters

And of course if the object appears in a different place in the photo, 
then the angles will be different and so will the distance.

Warning: The closer the image appears to the center of one of the photos,
the harder it is to calculate accurately with this formula. If it is near
the center of a camera, then the picture becomes almost a right triangle,
and you should use the trig formula for one right triangle.

Let's say for example that b = 0 and a = 45:

     left
    camera
   --- +-----------------
    |  | +   a
    |  |   +
    |  |     +
    |  | 90-a  +
    d  |         +
    |  |           +
    |  |             +
    |  |               +
    |  |                 +
    |  |                   +
   --- +--------D----------- o
     right
    camera
  
   D = d tan(90 - a)
     = 131 tan(90 - 45)
     = 131 tan(45)
     = 131 (1)
     = 131 meters
  
Let me know if you need a bit more explanation.

- Doctor Jeremiah, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Trigonometry

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