﻿ True Story - How to Track a U.F.O.  Page - kingkonglomerate.com How to Track a U.F.O.

Here is an easy way to track a U.F.O that isn't too expensive.   Take two camcorders and seperate them by 3 miles.   Place them about 5 miles away from the city center of, say, Phoenix, Arizona.   Aim them toward the city but parallel to each other.  Turn them on and record video continuously.

If a small airplane were to take off from the Phoenix Sky Harbor International Airport at night leaving its landing lights on and flys over Phoenix, all you will see of the airplane on your videos will probably be the airplane's landing lights and perhaps the red and green navigation lights on the tips of its left and right wings.   But you will have recorded its entire flight path over the the city.

Now imagine a flat planar triangle with a base 3 miles across.   The left camcorder is at the left end of this base.   And the right camcorder is at the right end of this base.   And the airplane is at some point over Phoenix.   The angle formed at the left camcorder between the base and the airplane is called, angle L.   The angle formed at the right camcorder between the base and the airplane is called, angle R.   And the angle formed at the airplane is called, angle P.

You only need a good military type compass to shoot bearings from each camcorder position to get several fixed angles.   With known references along the Phoenix skyline, you essentially have a calibrated angular vertical grid across each video frame.  If the airplane appears directly in front of a camcorder it will form an angle of 90 degrees relative to the base.   To keep our example simple, we will assume that the airplane only flys between the two 90 degree reference angles from each camcorder.

The side opposite angle P is the triangle's base called, p, with length 3 miles.   The side opposite angle L is called, l, with unknown length.   The side opposite angle R is called, r, with unknown length.

Let's say that in one video frame angle L is 63.435 degrees and angle R is 75.964 degrees.   If we construct a triangle with these characteristics and knowing that the sum of the angles of a triangle is always 180 degrees, we know that the angle made at the airplane, angle P, is 180 - (63.435 + 75.964) = 40.601 degrees.

The Law of Sines states: (sin P) / p = (sin L) / l = (sin R) / r.   We know all three angles in degrees and we know that the length of p = 3.   This is enough information to calculate the lengths of the other two sides.   So l = 4.123 and r = 4.472.   Therefore, the airplane is directly overhead at 4.123 miles from R and the airplane is directly overhead at 4.472 miles from L.   (see below)

L = 63.435 degrees -> (sin L) / l = 0.21692916 => l = 4.1231322
R = 75.964 degrees -> (sin R) / r = 0.21692916 => r = 4.4721675
P = 40.601 degrees -> (sin P) / 3 = 0.21692916

You can establish a calibrated angular horizontal grid across each video frame just as was done vertically.   So how high is the airplane?  Suppose the inclination angle at L is 33.855 degrees and the inclination angle at R is 36.040 degrees.   The slant range from L to P is srLP and the slant range from R to P is srRP.

We can construct a vertical planar right triangle with base = 4.472 and with inclination angle at L = 63.435 degrees at one end and a 90 degree angle at the other end.   The hypotenuse of this right triangle is srLP.   We know that 4.472 = srLP * (cos 33.855) -> srLP = 5.385.   Therefore, the airplane's height = 5.385 * (sin 33.855) = 2.999 miles or about 3 miles high.

We can construct another vertical planar right triangle with base = 4.123 and with inclination angle at R = 36.040 degrees at one end and a 90 degree angle at the other end.   The hypotenuse of this right triangle is srRP.   We know that 4.123 = srRP * (cos 36.040) -> srRP = 5.099.   Therefore, the airplane's height = 5.099 * (sin 36.040) = 2.999 miles or about 3 miles high, again.

We've come a long way.   We now know the relative position of the airplane and its height.   But we still need to know the airplane's heading and speed.   So we will now establish a 3-dimensional grid in space with three axis:  X, Y, Z.

For arguments sake the positive direction of X points toward East and the positive direction of Y points toward North and the positive direction of Z points Up.   Let's say that L is a fixed point at the origin of this three dimensional grid at (0,0.0).   Let's say that R is a fixed point three miles to the right along the X axis at (3,0,0).   Using the information we've already calculated we know that the airplane is at P1 (2,4,3).

30 seconds later, using the same methods above, we find the airplane at P2 (1,3,3).   Therefore the airplane has traveled 1 mile along the X axis going west and 1 mile along the Y axis going south.   Constructing a right triangle 1 mile on each side of a 90 degree angle we get 1.414 miles for the distance traveled in 30 seconds.   Thus the speed is 170 miles per hour.

Using the Law of Sines again we get (sin 90) / 1.414 = 0.70721.   So (sin x) / 1 = 0.70721 -> x = 45 degrees.   And this is 45 degrees beyond a line parallel with the Y axis in the negative direction therefore 180 + 45 = 225 degrees is the airplane's heading.

Have you noticed anything remarkable?  This can all be done passively with no radar, lasers, GPS, etc.   Just set up two or more video, infrared, night vision, etc.  camcorders.   If you had done this on the night of March 13, 1997, you could have determined with good accuracy the position, heading, speed, and altitude over the course of the entire flight of the Phoenix Lights.   With this flight data along with wind pattern data over Phoenix at the time, you could greatly refine and advance our understanding of exactly what happened that night.

If you are inclined to make a name for yourself there should be enough video of this event still available.   If you decide to seriously pursue this goal, enhance your methods by subtracting one frame from another to eliminate all stationary background to highlight only moving objects.   And expand the slant range lines to three dimensional tubes of appropriate radius in space.   Then solve for their intersection to locate actual moving objects in space.   And good luck.   I look forward to seeing your work.

Latest page update: 21 Feb '12
Page first published: 21 Feb '12