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Calculating terminal velocity

  1. May 28, 2009 #1
    good day all, out of sheer curiosity i tried to calculate what could be my terminal velocity in a skydive, since i will be diving in soon.

    using the terminal velocity formula: Vt = sqrt[ (2mg) / (pACd)

    where:
    Vt = terminal velocity
    m= mass
    g = acceleration due to gravity
    p = density of fluid [air]
    A = relative area
    Cd = coefficient of drag

    m and g are easy to fill, but for the remaining variables i made somewhat reasonable assumptions.

    the density of air [considering it changes between higher and lower altitudes i took to be 1.22 kg/m3 @ 15 degrees C.

    for Cd, i looked up charts of drag coeffs for typical shapes and objects. i could not find anything specific on a skydiver but i did find the Cd of a human standing upright. i assumed this meant a human standing in a stream of air as in a wind tunnel, the human standing vertically and the air moving aginst him horizontally.

    so if you rotate that entire system 90* so that the human is falling vertically thru air, i figured similar Cd values could be used. in this case, the Cd range is between 1.0 and 1.3 so i used the mean of 1.15.

    it seemed reasonable but i later found this is not the case.

    for the area, i generalized it into a 2x1 meter rectangle just for simplicity's sake.

    so crunching all those numbers i get a result of terminal velocity Vt = 22.14 m/s

    now, the situation is that several references i've seen all say that the average terminal velocity for a skydiver is about 55 m/s. which means im under half that of that!

    i figured some of my assumptions may be off so i worked backwards using the Vt = 55m/s to find the other variables.

    m and g really can't be argued with, the relative area is simple but reasonable. density of air could affect it, but i doubt significantly.. that only leaves drag coefficient Cd.

    solving for that i get Cd = .172 !!


    according to a table on wikipedia (http://en.wikipedia.org/wiki/Drag_coefficient), this is closer to a 'streamlined half body'.

    and according to the table at the bottom of this page (http://www.bookrags.com/wiki/Drag_coefficient [Broken]) it is closer to a smooth sphere, which i am not.

    what am i missing here.. why is there such a difference fin Cd or a human skydiver than a human in a wind tunnel??

    it's probably a very simple and obvious answer but it's beyond me at the moment. can anyone explain this to me.. i'm stumped.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 28, 2009 #2
    One experiment is worth a hundred calculations, lots of "real world" data available for this question.
     
  4. May 28, 2009 #3
    Cd can vary for example with speed .Wiki gives the example of a sphere where Cd can vary between about 0.5 for laminar flow and 0.1 for turbulent flow.
     
  5. May 28, 2009 #4
    i've seen several sites with tables of figures for skydivers. they range from 50 - 56 m/s terminal velocity.

    none explain their calculatoins or what Cd value was used, nor the mass or area used in the calculation. you'd think it would be easy to find these numbers online but i've had no luck so far.

    still it strikes me odd that such a low Cd is used, considering a human in the spread-eagle position wearing gear isn't very 'streamlined'..
     
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