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Sky diver project

  1. Dec 16, 2007 #1
    1. The problem statement, all variables and given/known data
    My physics teacher has given me a project to design some kind of device to allow landing from 5,000 feet without a parachute. So far we have been taught nothing but forces, velocity, acceleration, displacement, and vectors so I'm kind of at a loss. First step of the project is to find out how much force a person will experience hitting the ground when he jumps off a plane with just a wingsuit ( http://en.wikipedia.org/wiki/Wingsuit_flying ), so naturally I used force = mass x acceleration, but then then I noticed that the acceleration for gravity is always 9.81 m/s^2, which means that the altitude doesn't even matter! So obviously I am missing something, but what is it?

    To make things even more complicated, the wingsuit gives the sky diver a lot more drag, I know the drag equation which seems easy enough to calculate, but I'm not sure what the drag coefficient or reference area is. Also, does the wing suit actually create lift even though the sky diver is going straight down? Or does lift need some kind of horizontal movement?

    Oh and one more thing, does anyone know the maximum amount of force a human body can take and still survive?

    2. Relevant equations

    F = MA



    3. The attempt at a solution

    Tried to use F = MA, but got stuck on the acceleration part.
     
  2. jcsd
  3. Dec 16, 2007 #2

    Shooting Star

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    The altitude doesn't matter? What do you exactly mean by that? Jump from six inches high and then jump from, say, four ft, which is very safe, but you'll understand that altitude matters.

    Anything with a drag will reach terminal velocity. The simplest approximation is that the drag is proportional to the speed. I don't know much about wing suits, so I can't help you there. But you already know how to calculate drag. Think of something which will make the terminal velo tolerable for a human to impact at. The speed of impact is important. That you can find out from the net.

    Lift will need horizontal movement, like paragliding etc. Your poblem is something falling vertically.
     
  4. Dec 16, 2007 #3
    Exactly, which is why I'm saying I don't understand how the force of impact is calculated with F = MA if the A = 9.81 in freefall, because if A always equaled 9.81, then how does the altitude get factored in? I'm missing something here!
     
  5. Dec 16, 2007 #4

    Hootenanny

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    You are indeed correct that the force acting on the sky diver under free fall is constant. However, to examine the force of impact it my be more useful to to rewrite Newton's second law thus;

    [tex]\underbrace{Ft}_{\text{Impulse}} = m\left(v_1 - v_0\right)[/tex]

    Where t is the contact time and v1 and v0 are the final and initial (impact) velocity; in our case after the impact our sky diver will be stationary and hence v1=0. Therefore, the impulse is proportional to the impact speed. And therefore;

    [tex]F = -m\frac{v_0}{t}[/tex]

    So qualitatively, the greater the impact speed and the shorter the contact time the greater the impact force. Does that make a little more sense?
     
    Last edited: Dec 16, 2007
  6. Dec 16, 2007 #5

    Shooting Star

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    Hootenanny has given you the math. F=ma is not the right eqn.

    Simply put, if you fall in vacuum, then the speed of impact is proportional to the sqrt of height from where the object is dropped. The higher you fall from, the more the injury. That does not require Physics to understand. In air, there is a drag, so that the speed won’t go up indefinitely. It will approach the terminal velocity.

    What you have to find out is the highest tolerable impact speed for humans. Then adjust the drag so that the terminal velocity is that particular impact speed.
     
  7. Dec 16, 2007 #6

    Hootenanny

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    Although technically correct, a body under free fall experiences no other force other than it's own weight. Therefore, we can ignore drag.
     
  8. Dec 16, 2007 #7

    Shooting Star

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    I'm confused. This whole problem is about drag, and how to utilize it in landing from 5000 ft. At least, I presumed that's what the OP meant when he was talking about jumping without a parachute but only a wingsuit. I may be wrong, but the OP has to clarify more in that case.

    Actually, a wingsuit provides lift and steerage, so it’ll be better than a parachute. So, something else is required. Judging by the OP’s post, I am assuming that the problem given relates to drag.
     
  9. Dec 16, 2007 #8

    Hootenanny

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    I apologise, I just read 'free fall' in the OP's previous post and assumed we were dealing with a free fall problem. Reading the full thread, I realise I was mistaken :redface:

    Just to clarify for skatj, in this case we do not have free fall since the skydiver's weight is not the only force acting, we also have the drag of the wingsuit and therefore the acceleration (whilst falling) will not be equal to 9.81 m/s/s.
     
    Last edited: Dec 16, 2007
  10. Dec 16, 2007 #9
    Yea that makes a lot of sense! Thanks. How would I know the impact time though? Would I just generalize it to something very small? (Btw, since a skydiver will reach terminal velocity, then doesnt that mean after a certain altitude the force won't increase because he's already hit a max speed?)

    Well, I took the problem in steps. First, I would see how see how much force the sky diver would take without drag, then I would see how much force he would experience with the wingsuit on. Any hints on how to do that? :(

    So the drag force vector will just be acting in the opposite direction right? But how does that relate to the velocities.
     
    Last edited: Dec 16, 2007
  11. Dec 16, 2007 #10

    Shooting Star

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    Hi skatz,

    In fact, there is no accn during the later stages of fall, because the body is traveling (almost) at the terminal velocity, provided that it can reach terminal velo within 5000 ft. Also, if the terminal velo is not reached within 5000 ft, the impact speed would anyway be less than the terminal velo.

    Now tell us whether you know the eqn for motion with drag? Then an attempt can be made, provided you can find out the highest impact speed a man can tolerate. Or have you thought of something completely different for the project?

    EDIT: (I didn't see your previous post while typing this one.)
     
    Last edited: Dec 16, 2007
  12. Dec 16, 2007 #11
    I'm not familiar with that equation, sorry :(

    I couldn't find it on google either
     
  13. Dec 16, 2007 #12

    mgb_phys

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    Drag force is
    F = 1/2 rho v^2 C A
    rho is the density = 1.293 kg/m3 for air (at sea level and 0deg)
    V is the velocity
    C is the drag coefficent
    A is the area

    This makes sense if you think of each term:
    Density makes more drag - try running in treacle
    Going twice as fast means 4x as much wind resistance, look at the shape of race cars
    Larger area is more drag, this should be pretty obvious
    C is the tricky part there is a table estimating it for skydivers on http://en.wikipedia.org/wiki/Drag_coefficient

    You are correct that there is a terminal velocity - eventually drag matches your downward accelaration and you don't go any faster. You can derive it from the above equation -

    V = root ( 2mg / rho A c )
     
  14. Dec 16, 2007 #13
    So that's the equation for terminal velocity, and if the terminal velocity will always be the impact speed, then I need to find out how to make terminal velocity low enough so the force is under the maximum survivable force taken by a human body right?

    In that equation, m = mass, g = acceleration from gravity?, rho = ?, A = ?, and c = ?
     
  15. Dec 16, 2007 #14

    mgb_phys

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    The same as the equation above, rho is the density of the air
    A is cross section area and C is the drag.

    All a parachute does is make the terminal velocity low enough that you can survive.
    The other alternative is not to slow down the flight but make the landing accelearion low enough. Ie make the slowing down last long enough ( a big spring/air bag?)
     
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