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Free-fall formula, adding air resistance

  1. Jan 6, 2009 #1

    Nim

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    How do you add air resistance to these 2 equations?

    #1 sqrt(2 * d / g) = time until impact
    #2 sqrt(2 * d * g) = velocity on impact

    The equation for air resistance is:

    Fd = 0.5 * Air Density * Velocity ^2 * Frontal Area * Drag Coefficient

    How would you combine this equation with the above two?

    edit: I thought I figured it out doing:

    sqrt((2*d/g)/(p*A*Cd))
    sqrt(2*d*g)*(p*A*Cd)

    I got the right answer it seemed, but I must have been using just the right variables. When I changed the fluid density from 1.2929 (Earth's Atmosphere) to 1000 (water), the answer was definitely wrong.
     
    Last edited: Jan 7, 2009
  2. jcsd
  3. Jan 7, 2009 #2

    clem

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    It's not that simple. You have to solve a non-linear differential equation.
    You can make some approximations for short and for long times.
     
  4. Jan 7, 2009 #3

    Nim

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    I finally found something close at this website: http://keisan.casio.com/has10/SpecExec.cgi

    #1 Sqrt(m*g/k)*Sqrt(m/(g*k))*Log(Cosh(t/Sqrt(m/(g*k))))
    #2 Sqrt(2*m*g/k)*Tanh(t*Sqrt(g*k/(2*m)))

    The problem is that you have to enter TIME instead of DISTANCE... The same website has the ones that use distance, but they don't take air resistance into account, they are exactly the same as the ones I posted.

    I actually had an easier time creating a real-time simulation in JavaScript of a falling object using Euler's integration method. I wanted the equations though so I could see how accurate my simulation was. So far it looks accurate enough, but I can only test it in a vacuum with the 1st two equations.
     
  5. Jan 7, 2009 #4

    HallsofIvy

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    The formula is [itex]mdv/dt= -g+ Kv^2[/itex] where I have incorporated all the constants, Air Density, Frontal Area, and Drag Coefficient in to "K". That is, as clem said, a non-linear differential equation but a relatively easy separable first order equation. It can be "separated" as
    [tex]m\frac{dv}{Kv^2- g}= dt[/tex]
    The denominator factors as [itex]\left(\sqrt{K}v- \sqrt{g}\right)\left(\sqrt{K}v+ \sqrt{g}\right)[/itex] so that can be integrated using "partial fractions".
     
  6. Jan 7, 2009 #5

    Nim

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    Thank you. Unfortunately, I am math illiterate. I don't know what any of this means, or how to use it. I see 4 equations? I need to arrange them somehow? Like enigmatic Lego blocks? The language is alien to me as well, such as "partial fractions", "separable first order equation", and "non-linear differential equations".
     
  7. Jan 7, 2009 #6

    clem

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    With a name like Nim, I thought you were a brilliant young mathematician.
    You need more than PF. You need someone to do all the work.
    Find a friend who knows calculus.
     
  8. Jan 7, 2009 #7

    Nim

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    I just made it up. I found out later that it's also the name of a math game. I didn't realize it would take so much work. I have a friend who knows calculus. But... I think I need a friend that knows advanced calc? and classical physics? I had them look at it just in case, and it was definitely a no go. They were able to explain a bit of what HallsofIvy was doing, after I explained a little physics to them, but they didn't know what to do from there.
     
  9. Jan 8, 2009 #8
    HallsofIvy has the correct equation but it can be solve using trig substition which is fairly easy. First, multiply both sides by -1 to reverse the signs in the denominator. Then divide both sides by m. This leaves an integral that can be solved using the substitution Kv^2 = gsin(theta). Using this equation, solve for v^2, take the square root of both sides to solve for v, then take the derivative of both sides. Substitue the expression for dv and Kv^2 into the original integral, factor out the g in the denominator obtaining g(1-sine squared theta). The expression (1-sine squared theta) equals cosine squared theta. Then the integrand becomes a constant times secant(theta) which is easily solved.
     
  10. Aug 27, 2009 #9
    Everything you guys said makes sense, but what are the limits for the separable integrals? If you want to find the time if takes after 100 m of free fall, what do you for for the limits for the left side dv and right side dt?
     
  11. Aug 27, 2009 #10
    I think the solution if you want to find time it took to travel say 100 meters will involve the chain rule (dx/dt)(dt/dv) or something like that, but i'm not sure how to incorporate that.
     
  12. Aug 27, 2009 #11
  13. Aug 27, 2009 #12
  14. Aug 27, 2009 #13
    You might also study the physics of air drag and terminal velocity. There are two types of air drag: Stokes Law (low turbulence) Force linear with velocity, and turbulent drag (force quadratic in velocity). See
    http://en.wikipedia.org/wiki/Drag_(physics)
     
  15. Aug 28, 2009 #14
    That problem is close but different. It looks like that if for something being thrown up with air resistence and they are solving for max height it will reach.

    My problem is as follows:
    A sky diver with a mass of 75 kg jumps from an aircraft. The aerodynamic drag force acting on the sky diver is known to be Fd=kv^2 where k=0.228 N sec^2/m^2. Determine the maximum speed of the freefall for the sky diver and the speed reached after 100 m of fall.

    I've already done some searching on the net and have found what equations I need. They are these:

    http://keisan.casio.com/keisan/lib/...ime and velocity)/rakka3.GIF?51069941285.4036

    All I want to know is how do you get from the first equation to the second equation.
    I know you have to solve the differential equation using seperation method and the chain rule. IE replace dv/dt with (dv*dx)(dx/dt)=(dv/dx)v
    But how exactly do you solve that differential equation? What does it look like? What are the limits for the integrals?
     
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