When do I take off?
You made millions in GOOG shares and is now the proud owner of a light aircraft to make your lunch trips from office to home (of course, home is situated on a hill overlooking the oceans in the distance and near a forest with ferns). While you are rich and smart, you still carry a peculiar grad student attribute much to your wife's annoyance: you dont find the necessity to lunch at any fixed time in the afternoon.
It takes 15 minutes one way on a calm day without any wind and you spend 30 minutes at home to eat. Your house is due so
uth of your airport (which adjoins your office). You notice that the wind always blows along the north-south axis throughout the year.
All things being the same, the wind being the only variable, what should your strategy be to minimize the time of flight?
ps: Wild shots are now accepted!
It takes 15 minutes one way on a calm day without any wind and you spend 30 minutes at home to eat. Your house is due so
uth of your airport (which adjoins your office). You notice that the wind always blows along the north-south axis throughout the year.
All things being the same, the wind being the only variable, what should your strategy be to minimize the time of flight?
ps: Wild shots are now accepted!
posted by littlecow at 11/12/2005 04:53:00 PM
9 Comments:
It should be at the time when the wind speed is the least.
And why is that? Addionally, when does the windspeed become the least?
You cannot minimize the 30 min you spend with your wife(rather you would not like to!!) so the only thing that can be minimised is the travel time.
Now the wind speeds have a diurnal variation (for a graph look here)
http://www.windpower.org/en/tour/wres/variab.htm
For minimizing travel time, its best to have weakest head wind(opposite to direction of travel) and strongest tail wind(vice versa).
Therefore I think, going a little early (before noon) and coming back after 12(but before 3) should minimize the travel time.
@
PS: I have ignored the topographical differences as there is an effect of hill...wind speeds increase as we go uphill on windy side..again:
http://www.windpower.org/en/tour/wres/hill.htm
bal, @, the little bomb: you are all right. what i was looking for was proof for this: "the total time of flight is minimal when no wind blows". since all of you have flexed your muscles of creativity and come up with decent answers that i cannot wrong, all of you get 1/2 points. but i guess the problem was too easy and hence i am not updating the scores. bal still leads but I suspect he is going to face fierce competition in the upcoming days!
Dude, what you were looking for was this:
Let the distance from office to house be "d"
Let the avg speed of jet be "u"
Let the avg speed of wind be "v"
Round trip time = 2d/u (no wind) - (1)
Round trip time = d/(u+v) + d/(u-v) (with wind)
= 2du/(u^2-v^2) - (2)
Ratio (1)/(2) = (u^2 - v^2)/(u^2)
which increases as v gets smaller....
Hence, go at the time the breeze is least. Came here too late I guess but had worked this before giving the answer :(
Grenade, your answer depends on how much "v" would lessen the time for travel I guess... I mean at the end of the day you might have so fierce breeze against direction of travel that it might have been better to go at noon when wind blows least...
@bal: this problem was too easy - i posted it here as it was interesting. better luck with the next problem!
Wokay... your the boss... I guess..
@grenade: if you so insist to look at devious ways of answering this problem, one can always look for a DSL/cable connection to work remotely - that will also satisfy @'s desire to stay close to the wife! :-D
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