| The situation looks something like this:
|\ /| Wa + Wb = W
| \ / |
| \ / | W� + Ya� = A�
Yb ! X | Ya W� + Yb� = B�
| /|\ |
| / | \ | Wa/C = W/Ya (by similar triangles)
|/ | \| Wb/C = W/Yb (by similar triangles)
+===+===+
Wa Wb
We want to solve for W, the distance between the two walls.
We note that:
W Wa Wb W W
- = -- + -- = -- + --
C C C Ya Yb
or
1/Ya + 1/Yb = 1/C
Realizing that we can't readily deal with Ya and Yb in our equations,
but that we can deal with Ya� and Yb�, we manipulate this to:
(Ya+Yb)C = YaYb
C�(Ya�+2YaYb+Yb�) = Ya�Yb�
2C�YaYb = Ya�Yb� - C�(Ya�+Yb�)
Squaring both sides again, ...
4C^4(A�-W�)(B�-W�) = ((A�-W�)(B�-W�) - C�(A�+B�-2W�))�
Notice that, by scaling, we may choose C = 1. Then, the above is an
equation in A�, B�, and W�, and it is a fourth degree equation in W�.
4(A�-W�)(B�-W�) = ((A�-W�)(B�-W�) - (A�+B�-2W�))�
Changing terms:
4(A-W)(B-W) = ((A-W)(B-W) - (A+B-2W))�
Because this is symmetric with A and B, we are able to rewrite it as
an equation involving W, and two independent symmetric functions of
A and B, such as (A�+B�) and (A+B). A simplifying substitution is
D = A+B-2, so that the W� term becomes -2DW�. Then, the equation
can be solved by standard techniques for solving 4th-degree polynomials.
|