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Resultant wind speed/direction .

  1. Sep 21, 2008 #1
    Resultant wind speed/direction.....

    Hi, new here and am hoping this is in the correct forum and if so that someone can help with this little problem.

    Basically I would like to know if there is a "quick" way of calculating an average or resultant wind speed and direction given a selection of 5 or so winds. My aim is to establish the mean wind that acts on a parachutist dropped from, say, 12000ft. If I take off in my plane and note the wind speeds and directions at 2000ft intervals as I climb up I can then manually plot them on paper, draw the resultant and measure the necessary angle; the length of the resultant divided by the number of samples gives me a wind strength. This average wind speed and direction can then be used to calculate the release point for the parachutist so that he lands on the drop zone.

    What I was wondering was if there was a way of calculating it accurately and, importantly, quickly using a calculator, rather than drawing it out on graph paper. I can obviously calculate a resultant from 2 winds by dredging up some schoolboy math but anymore than 2 and I'm stumped (other than repeatedly resolving winds in pairs until left with one resultant!)

    I'm no math/physics genius so please be gentle!
  2. jcsd
  3. Sep 21, 2008 #2


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    Re: Resultant wind speed/direction.....

    Real problem or homework problem (I won't give my guesses on a real problem, since I don't want to kill a real parachutist). Do you mean 5 simultaneous winds?

    LHS of this formula adds the individual vectors, RHS is the resultant (only works with cartesian coordinates):
    (v1x,v1y,v1z)+(v2x,v2y,v2z)+...(v5x,v5y,v5z)=(v1x+v2x+v3x+v4x+v5x, v1y+v2y+...v5y, v1z+v2z+ ...v5z).

    Again, just my guess.

    Edit: I just re-read your post and understand the winds are not simultaneous, so the resultant gets divided by 5 as you said. Also, I realised I don't know how the windspeed is specified, ie. is it already in cartesian form, or is it specified by speed and angle?
    Last edited: Sep 21, 2008
  4. Sep 22, 2008 #3
    Re: Resultant wind speed/direction.....

    Hi and thanks for the reply. This is based on a "real" problem; I'm a pilot and we normally calculate this using a pda with an excel spreadsheet on it. I've "reverse engineered" a lot of the formulae in the spreadsheet (I'm nosy like that) but was a little stumped with the wind vector thing. If the pda goes the way of most things electronic when they're needed most (!) then we revert to the manually plotted vector diagram to resolve an average wind, which works fine.

    The winds are recorded every thousand feet as we climb up to drop altitude. They are recorded in degrees magnetic and nautical miles per hour (kts). Thus a wind out of the north east at 10kts would be recorded in pilot-speak as "045 at 10". I guess I would need to resolve that into cartesian vectors then?

    Whilst the winds are not recorded simultaneously they are treated as being such for the purpose of the exercise. Put them all into the spreadsheet and it miraculously spits out an average wind for us. :smile:

    Don't worry though - this is purely a personal interest/idle curiosity thing by the way - I shan't be scattering parachutists across the countryside if I can't work this out! :eek:
  5. Sep 22, 2008 #4


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    Re: Resultant wind speed/direction.....

    Whew! :rofl: I shall continue with my guessing then. :smile:

    Yes, I would try resolving into cartesian components.

    For a given speed v, and direction θ:
    vx=vsinθ, vy=vcosθ

    Then adding:
    (v1x,v2y)+(v2x,v2y)+...(v5x,v5y)=(v1x+v2x+v3x+v4x+v5x, v1y+v2y+...v5y)=(vRx,vRy)

    From the resultant (vRx,vRy),
    get the magnitude: vR=sqrt(vRx2+vRy2)
    and direction: θR=arctan(vRy/vRx)
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