Quote:
Originally posted by Deejax:
Aren't you forgetting the distance you travel while picking up the passengers? The triangles you describe assume an instantaneous pickup of passengers.
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No I'm not. If you read my post carefully:
all five triangles are seperated by the distance 5 which is what you and the train cover while loading in the passengers
Quote:
The helicopter travels 9 different path from base to train and back. And I don't think they will make 'nice' triangles to calculate.
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That depends on what "nice" triangles are but:
the first path TO the train is not part of one of my triangles
the first part BACK builds a triangle with the second part TO the train and so do the other pairs.
I'm trying to do an ASCII sketch
-------------------------B(ase)-----------------------------
Chicago______A___________S______________C_________ _Milwaukee
S is the point the train reaches when the helicopter reaches base (and I assume instantaneous unloading)
Now assuming a Blackhawk speed of 225mph (=2.5*90)
we get AB = 2.5*AS
and BC = 2.5*CS
AB/sinASB = AS/sinABS => 2.5*AS/sinASB = AS/sinABS => sinABS/2.5 = sinASB
same for the triangle BSC
BAS=180-ABS-arcsin(sinABS/2.5)
same for the triangle BSC
if you now introduce a height (orthogonal distance between B and AC) it
takes a little bit more trigonometry and a system of equations to figure it out.
It's not that difficult BUT it's certainly a LOT of work
[ 08-22-2003, 12:07 PM: Message edited by: Faceman ]