| |
d . |
|. |
| . |
| .30 |
| . - c
| . |
| .- |
| . |
| 20- . |
| .|
0|-________.___>
0 a
I have been asked by several how I arrived at the answer to the SLP
(silly ladder problem) of an alley width of 12.3119 feet. It takes one
back to junior high school, which is always a fun thing to do if you
are masocistic.
Consider the above diagram (which requires mono spaced type to look
correct). Let the origin of a Cartesian coordinate system be the lower
left-hand corner. We have a 20 foot ladder (the - line) leaning up c
feet against one wall, and a 30 foot ladder (the . line) leaning up d
feet against the opposite wall. Let the alley width be "a". Recalling
your junior high math stuff (Pythagorus' theorem, equations of straight
lines, solving two linear simultaneous equations) we have by Pythagorus
that
c = Sqrt(20^2 - a^2), and d = Sqrt(30^2 - a^2).
Thus the equation (y = mx + b) for the 20-foot ladder line is
y = (c/a)x, and that for the 30-foot ladder line is
y = -(d/a)x + d.
Solving these two equations simultaneosly for x and y (set them equal
to solve for x, then substitute back into the first to get y) gives the
coordinates at which they cross. The y value (as a function of a) for
this solultion is
y = cd/(c+d),
and we require the value of "a" that makes this 10 feet, i.e., the a
such that
cd/(c+d) = 10. This is 12.3119 feet. Note that a = 0 also solves this
problem as it was stated, although it is not the "interesting" answer.
best wishes,
dave belsley, w1euy
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