C /|\ / | \ / | \ / O \ / . . \ /x . .z\ /. w y .\ A-----------------B D
Two observers, A & B, are a distance D apart on flat ground and are both
looking at the same object C above them (just for the sake of argument,
let's say it's the point of maximum height of a water rocket.)
With the angles they measure,
they want to determine the height of C above the ground.
Assume that point O on the ground is directly below C
(AOC and BOC are right angles).
Find OC in terms of D and the necessary angles (I don't think all of the angles will be needed.)
Source: Original (personal necessity...) A good starting place for water rocketry is: www.h2orocket.com
OC = D * tan(x) * sin(y) / sin(w+y)
This is found by first using the sine law to find AO, then multiplying by tan(x) to get OC.
If either w or y is zero, then this formula won't work.
Find a formula for OC in this special case.
Wilfred Theunissen sent the following solution to this possibility:
If w=0 then y=0 too, so we have the following three situations: Case1: O is between A and B (so x and z < 90 degrees) Case2: O isn't between A and B (say x>90 degrees and z<90 degrees Case3: O=A or O=B : this is trivial. Let AB=D and Case1: Cosine-rule: 1) D^2 = AC^2 + BC^2 -2*AC*BC*cos(180-(x+z)) 2) sin(x) = OC/AC so AC = OC/sin(x) 3) sin(z) = OC/BC so BC = OC/sin(z) Substituting 2) and 3) in 1) and cos(180-alpha) = -cos(alpha) 4) D^2 = (OC/sin(x))^2+(OC/sin(z))^2+2*OC^2*cos(x+z)/(sin(x)*sin(z)) multiplying with sin(x)^2*sin(z)^2 we get 5) sin(x)^2*sin(z)^2*D^2 = OC^2 * (sin(x)^2+sin(z)^2+2*sin(x)*sin(z)*cos(x+z)) Rewriting de righthand-side we get: 6) sin(x)^2*sin(z)^2*D^2 = OC^2 *((sin(z)+cos(x+z)*sin(x))^2+sin(x)^2*sin(x+z)^2) so OC can be calculated from 6) Case2: Cosine-rule: 1) D^2 = AC^2 + BC^2 -2*AC*BC*cos(180-(x+z)) 2) sin(180-x) = OC/AC so AC = OC/sin(180-x) 3) sin(z) = OC/BC so BC = OC/sin(z) But because sin(180-x)=sin(x) we get the same formula as in Case1 !!!!