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Time to speed question

  1. Jul 18, 2007 #1
    (this is for a hobby, not homework)

    I fully understand the drag equation and can work it forwards and backwards.
    http://en.wikipedia.org/wiki/Drag_equation

    How can I use the same data (plus mass if needed) to determine time to top speed or a given speed?

    I'm sure there is a simple formula related to the Drag Equation that gives time to speed. I am looking everywhere and I can't find what I need.

    Thanks a billion in advance for a shove in the right direction! :biggrin:
     
  2. jcsd
  3. Jul 18, 2007 #2

    xez

    User Avatar

    I don't think you can find the equation you seek just
    because that's not all the information that's required.

    The drag equation gives the FORCE of the drag given
    velocity, et. al. but that says nothing about the
    FORCE or power or accelleration your vechicle can
    produce given a certain present velocity and drag forces
    acting upon it.

    You could say you have a fixed force of propulsive power
    available and a fixed mass of vehicle then in that case
    the net force on the vehicle would be
    F_net = F_propulsive - F_drag, and then via F = m*a
    your accelleration at any moment would be F_net/m, and
    the integral of accelleration(time)=F_net(time)/mass
    would be velocity(time), and that'd be a curve that'd
    even out and stay level when F_net=0 due to top speed
    having been reached.

    If you had a vehicle that was capable of a certain constant
    power output you could similarly calculate a resultant
    accelleration due to the force balance and again get
    to a velocity and top speed.

    More realistically, though, the power available from
    most engines will depend on their RPM, and consequently
    the gear ratio and consequently on the present velocity,
    so you have several complications to determine the
    actual accelleration.
     
  4. Jul 19, 2007 #3
    xez was right...

    Unless you made a simple assumtion, i.e. the force provided by the vihicle is constant, the answer is anything but far from simple... :rofl:
     
  5. Jul 19, 2007 #4
    Thanks xez and chanvincent, you guys rock!!!

    No easy answer? Dang!

    Ok maybe if I was more specific this would make more sense.

    I am talking about a theoretical aircraft with constant thrust.

    From the drag equation you can determine top speed

    Velocity(top speed)=Sqrt[(2T)/(P*A*Cd)]

    Here is a simple example
    T(thrust of engine) = 10000 Newtons
    P(atmospheric density) = 1.25 Kg/M^3
    A(reference area of craft) = 10 square meters
    Cd(drag coefficient) = .1

    The answer is 455.4 Km/h (have to change from M to Km)

    Isn't there a companion equation that determines time to a given speed or top speed?

    My biggest problem is that acceleration doesn't operate in a straight line. It starts out accelerating rapidly and gradually slows to 0 acceleration. Is there no simple time to speed equation for simple models lke this? I believe it, I just can't believe it! :bugeye:
     
  6. Jul 19, 2007 #5
    Whooops...

    Hypothetically speaking the craft weighs 1000 Kg (if that is the right kind of units).

    If there is no obvious answer that is fine. If there is an available equation that is a decent enough test number.
     
  7. Jul 19, 2007 #6
    Well, as I said be4, the answer is not simple if the force is not constant.

    But now you are talking about the constant force.. then the solution IS very simple,

    You have the drag force
    [tex]F_d = -\frac{1}{2}\rho v^2 C_p A[/tex]
    and the force from the trust
    [tex] F_{thrust} = \mbox{constant} [/tex],

    now the total force

    [tex] \Sigma F = F_d + F_{trust} = Ma [/tex]
    [tex]-\frac{1}{2}\rho v^2 C_p A+ F_{trust} = Ma [/tex]
    [tex]-\frac{1}{2}\rho v^2 C_p A+ F_{trust} = M\frac{dv}{dt} [/tex]

    I will leave this first order non-linear DE here for you to work with...
    Notice, luckily, there is an exact solution for this DE... :biggrin:
     
    Last edited: Jul 19, 2007
  8. Jul 19, 2007 #7
    chanvincent

    YOU ROCK!!!

    This might sounds like a retarded question, but... I am not an expert.

    I don't totally understand the first line.
    Total Force = Force of drag + Force of thrust? = Mass? (Ma = Mass?)

    Second line I get because it is the drag equation + Force of thrust? = Mass?

    The third line what is M(dv/dt)?

    DOH!

    Thanks a billion!!!!
     
  9. Jul 19, 2007 #8
    a is accelaration... dv/dt is also accelaration...
    You havn't learned calculus yet, have you?
    otherwise you should be familiar with dv/dt....
    well... Since you claimed that you fully understand the drag equation, I thought you knew calculus...

    although you do not know how do derive the answer... maybe you want to take a look...
    http://en.wikipedia.org/wiki/Drag_(physics)

    Hopefully you could understand... Cheer...
     
  10. Jul 19, 2007 #9
    Thanks Chan

    Actually I understand basic calculus and derivatives. 3X^2 = 6x and all that stuff.

    dv/dt being acceleration makes sense, velocity over time.

    But what the heck is M in Ma? Mass, weight, something else completely?

    THANKS!!!!!
     
  11. Jul 19, 2007 #10
    m = mass, a = accelration

    F=ma, => Total Force equals to mass times accelration...

    understand?
     
  12. Jul 19, 2007 #11
    Ok, awesome, I didn't spot the F=ma in that equation.

    So now that I have velocity over time on one side I can solve for time? That I can certainly do.

    One last tiny question on something I always screw up... my units.

    If I'm using Newtons for thrust, I should use Kilograms for Mass? Velocity (dv) is obviously meters per second. Time (dt) obviously seconds.
     
  13. Jul 19, 2007 #12
    Yes....

    It is not as easy as you think... I doubt that you could really solve it... but, yes, it could be solved...
     
  14. Jul 19, 2007 #13
    So

    dt=(M*dv)/(Fd+Fthrust)

    Did I solve that right? :bugeye:
     
  15. Jul 19, 2007 #14
    Yes... you got it... the answer should look like some hyperbolic tangent... just go ahead...
     
  16. Jul 19, 2007 #15
    YOU ROCK!!!

    Ok, I gotta go to bed but I'm going to start plugging in practice data tomorrow afternoon.

    Thank you so much, I could have never EVER done it without your help. It doesn't look too bad once it is finished but I was clueless. :yuck:
     
  17. Jul 20, 2007 #16
    Chanvincent :cry:

    dt=(M*dv)/(Fd+Fthrust)

    Fd+Fthrust = 0 at top speed. That makes sense because once the force of Thurst and Drag even out there is no longer any acceleration. But in F=Ma if F = 0 then a = 0 too.

    dt=(M*dv)/(Fd+Fthrust) = zero no matter what data is plugged in.

    I must be 1 nanometer away from the correct solution but.... what the heck?

    I do understand the basic concept of derivitives meaning, 4X^2 = 8x, but I honestly don't know when or why they are used. I'm beggin' ya here, I must be close but what do I do with it? I'm going to use this thing a million times once I am done but.... Dag!
     
  18. Jul 22, 2007 #17
    I told you the integral is not easy to do (in your level)

    [tex] dt = \frac{Mdv}{F_{trust}-\frac{1}{2}\rho v^2 C_p A} [/tex]
    [tex] \int_0^Tdt=\int_0^v \frac{Mdv}{F_{trust}-\frac{1}{2}\rho v^2 C_p A } [/tex]

    Any idea how do solve this? Hint:

    [tex] \frac{d}{dx} tanh^{-1}x = \frac{1}{1 - x^2} [/tex]

    :tongue2:
     
    Last edited: Jul 23, 2007
  19. Jul 24, 2007 #18
    Thanks chan!

    Sadly I've been sworn off working on this thing until next week.

    That integral does look like a monster. By if there is an ordered set of steps I'm sure I can do it. I'm gonna save that and come back to it next week.

    Thanks!
     
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