Brett: "Foul, my friend!" I'm still with you, but the Lake Placid
track is not instrumented to do what you describe and those values
can't be derived from the givens. Besides, if you are taking us down
that road, you didn't go nearly far enough! <grin>

Okay, I propose we take this off-line and into Professor West's
virtual classroom. We can push around the problem a bit there and if
things get out of hand, then we'll simply have to move it into Lisa
G's classroom to get it solved. If you are interested in following
this further, send me a DIRECT e-mail saying you want in on the
discussion and I'll start a new thread. This way we'll keep the
ADK-Luge discussion board about the sport and not Newtonian physics
which is where we might be headed. --Jim D.


--- In adk-luge@yahoogroups.com, "csource_brett" <bwest@...> wrote:
>
> Jim D:
>
> Only because this is interesting math and a good riddle, I think I
need to challenge your
> formula again. However, you are getting much closer.
>
> You have derived the "theoretical distance," but not the "actual
distance." (There is such a
> thing, and they are different.) In order to derive the actual, you
would need the speed of
> the losing Luger at the exact moment their running time equals the
final time of the
> winning luger. You also need the speed of the losing Luger exactly
when they cross the
> finish line. You then calculate the average speed of the losing
luger during the period they
> were sliding for a longer time than the winning Luger. With this
average speed, and the
> known time delta, you then can derive "actual distance" between
winner and loser.
> Amazingly, the winning Lugers speed (at any point on the track) is
completely irrelevant to
> this equation. Further, only the speed of the losing Luger for their
extended time on the
> track (relative to the winner) matters. Lastly, do do it completely
accurately, you would
> need to do the formula twice, once for each run, then add the
distance values together to
> calculate the actual over the 2 runs. What is cool about this, it
could conceivably be a negative distance if the losing luger won one
of the 2 race runs. If this was the case, then
> the winning lugers time would be relevant to the overall equation-
but only if they lost
> one of the 2 runs. Still with me?
>
> I suspect only you and I care about this Jim.
>
>
> Brett West
>