Altitude of a Model Rocket

Date: 10/26/1999 at 12:52:46
From: Jack L. Swayze Sr.
Subject: 3D trigonometry

I enjoy model rocketry. I especially like having youngsters 
participate. The model rockets (as can be expected) will fly to some 
unknown altitude over the rocket site. Of course, they will not go 
straight, much less go straight up, so the only reliable way to 
determine altitude is by observation. Barring such expensive things as 
altimeters in the rocket itself, this has to be done by ground 
observation. With two observers, each measuring angle from north, and 
angle of inclination, you can determine altitude.

However, asking very young children to make both observations at the 
same time is difficult. One should be able to have three children, 
whose location on the ground (the plane) is predetermined, make only 
an observation of inclination, then mathematically project the 'cone' 
up from each point of observation, and determine the intersection 
point of all three cones, and calculate the altitude above the plane 
(ground) from that point. But what are the calculations? I need to 
create a computer program that will allow us to 'plug in' the three 
observations of inclination and have it 'churn out' the altitude (and 
perhaps also the downrange distance, assuming the children form an 
equilateral triangle on the ground, with the launch pad in the exact 
center of the triangle).

If I can get some algebraic formulas that can determine the altitude, 
and the downrange location from the known angles of inclination above 
the plane, then I can do the programming to put it into a computer.  
But what would be the algebra equations?

Date: 10/26/1999 at 14:21:57
From: Doctor Rob
Subject: Re: 3D trigonometry

Thanks for writing to Ask Dr. Math, Jack.

This is an interesting problem. I enjoyed working it out.

Set up a coordinate system. Say the launch point is the origin, the 
xy-plane is the surface of the Earth around there, and the z-axis is 
vertical. Suppose the coordinates of the 3 observers are (x1,y1,0), 
(x2,y2,0) and (x3,y3,0), which must not be collinear. Suppose the 
observed angles of elevation of the rocket at its apogee are a1, a2, 
and a3. Then the equations of the three cones are

     (x-x1)^2 + (y-y1)^2 = cot(a1)^2*z^2
     (x-x2)^2 + (y-y2)^2 = cot(a2)^2*z^2
     (x-x3)^2 + (y-y3)^2 = cot(a3)^2*z^2

If you subtract the third equation from the first and also from the 
second, you will have a pair of linear equations in x and y, which you 
can solve. This will give you x and y in terms of z^2. Now substitute 
those expressions into the third equation, and you will have a single 
quadratic equation for z^2. In terms of the variables x1, x2, x3, y1, 
y2, y3, a1, a2, and a3, this equation is lengthy to write down 
explicitly, but for numerical values of these, it is quite simple. The 
roots of this equation will give you possible values of the height. 
Discard negative values of z^2, and take the positive square root of 
any positive values. What you have are the solutions.

There may, indeed, be 2 solutions.

For example, suppose that you have

     (x1,y1,0) = (100,0,0)
     (x2,y2,0) = (0,100,0)
 and (x3,y3,0) = (-100,-100,0)

These form an isosceles triangle, but not equilateral. Next suppose 
that a1 = a2 = 26.575 degrees, and a3 = 45 degrees. Then the 
cot(a1) = cot(a2) = 2, and cot(a3) = 1. So you get the equations

     x = -(3*z^2+10000)/600
     y = -(3*z^2+10000)/60,
     180000*z^4 - 9600000000*z^2 + 50000000000000 = 0
     20000*(9*z^4 - 480000*z^2 + 2500000000) = 0

whose positive real roots are z = 50*sqrt(2)*(sqrt[39]+-3)/3, or 
76.485 and 217.907, approximately. The three observations are 
insufficient to resolve the ambiguity. The intersection points of the 
three cones above ground are

     ( -45.917,  -45.917,  76.485) and
     (-254.083, -254.083, 217.907)

Perhaps the ambiguity can be resolved in other ways. In this case, one 
solution is inside the triangle formed by the observers, and the other 
is rather far outside. Thus dead reckoning should be enough for a 
fourth observer, or even the given three observers, to distinguish 
between the correct and extraneous solutions.

Presumably a fourth observer, not collinear with any two of the 
others, could resolve this. You could compute the angle of elevation 
of each of the solution points from the fourth point,

     a4 = arctan(sqrt[(x-x4)^2+(y-y4)^2]/z)

and pick the one which agreed best with the fourth observation.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/