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Integrating the Lift Equation

  1. Nov 5, 2017 #1
    I am trying to calculate the lift generate by a helicopter rotor using the lift equation, which is
    [tex]
    L = \frac{1}{2} \rho V^2 * C_L * S \\
    where \\
    \rho\mbox{ = density} \\
    V\mbox{ = velocity of a point on the rotor} \\
    C_L\mbox{ = lift coefficient} \\
    S\mbox{ = surface area swept out by the rotor} \\
    [/tex]
    So I'm integrating across the length of the rotor. Since [tex]S=\pi r^2 \mbox{ and } V=\omega r[/tex] I get:
    [tex]
    L= \int_0^r \frac{1}{2}\rho (\omega r)^2C_L\pi r^2dr \\
    L= (\frac{1}{2})\rho \omega^2 C_L\pi \int_0^rr^4dr \\

    L= \frac{1}{2}\rho \omega^2 C_L\pi \frac{r^5}{5} \\
    [/tex]

    I'm thinking I must have missed something by now because the units do not make sense. If I solve just for units, I get:
    [tex]
    \rho = lb/ft^3 \\
    \omega = V/R = \frac{1}{min} \\
    r = ft \\
    [/tex]
    The rest have no units.
    However this leaves me with L = lb*ft^2/s^2. There is an extra ft in the numerator. Any idea where I went wrong or if this is even the correct approach to calculating rotor lift?

    Thanks in advance.
     
  2. jcsd
  3. Nov 6, 2017 #2

    Nidum

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    S is the effective surface area of one rotor blade and not the area swept out by that blade .
     
  4. Nov 6, 2017 #3
    According to J. Seddon, Basic Helicopter Aerodynamics, (link) "S is the disc area". Not saying you're wrong but I have compelling evidence that says otherwise.

    In any case, that still doesn't solve my unit problem.
     
  5. Nov 6, 2017 #4

    Nidum

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    Which chapter and which paragraph ?
     
  6. Nov 6, 2017 #5
    This is noted in the Notation List on page xii where he says.
    A = area of rotor disc
    and blade area is given on the next line as
    Ab = total blade area (N blades)

    Then in chapter 2, page 15 about half way down, he states the equation for drag in autorotation as

    D=(1/2) p Vc2 A CD
     
  7. Nov 7, 2017 #6

    boneh3ad

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    You aren't setting up your integrand correctly. You can't just slap a ##dr## on the end of an equation and then integrate it and expect it to work out. That ##dr## you stuck to the end is the source of your extra unit of length. If you start with
    [tex]L = \dfrac{1}{2}\rho v(r)^2 A C_L,[/tex]
    you note that ##v(r) = \omega r## is a function of ##r## (and, I suppose, so might ##C_L## and ##\rho##). You can't just use ##A## here, though, because ##v## isn't constant, so you need to integrate over all of the infinitesimal area elements ##dA## to come up with ##L##. In other words,
    [tex]dL = \dfrac{1}{2}\rho v(r)^2 C_L\;dA.[/tex]
    Since the only term varying there is ##v(r)##, you know you need to recast ##dA## in terms of ##dr##. I can't draw a picture here, but if you imagine a tiny area element here at any random point in the disc that spans ##dr## in radius and ##d\theta## in angle, then ##dA=r\;d\theta\;dr## since the area is small and it is essentially a rectangle. The total lift is then a double integral over those two variables,
    [tex]L = \dfrac{1}{2}\int\int\rho (\omega r)^2 C_L r\;dr\;d\theta.[/tex]
    Lucky for you, nothing varies in ##\theta##, so one of those integrals is trivially over ##0\leq\theta\leq 2\pi##. Your final integral should be
    [tex]L = \pi\omega^2\int_0^R \rho r^3 C_L\;dr.[/tex]
    I believe you will find those units satisfactory.
     
  8. Nov 7, 2017 #7
    That's awesome, thank you so much! One thing I was bad at college was setting up the equations. I could usually solve them when they were given but I had trouble setting them up for things like this. Your answer makes a lot of sense.

    Thanks again!
     
  9. Nov 9, 2017 #8

    JBA

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    By looking at the original basic equation, since the total disc area is a part of the equation, I am a bit confused as to the need for integration to obtain the lift for a rotor with a tip radius of r for the selected reference point for V.
     
  10. Nov 9, 2017 #9
    JBA, that's a really good point. I guess I wasn't sure what to use for velocity since the velocity changes depending on where you're measuring on the rotor. The original equation is taken from the lift equation for an airplane with a fixed wing so the velocity at each point along the wing is constant. Do you have a suggestion how I would better solve the equation for a rotor? Perhaps substitute angular velocity in instead.
     
  11. Nov 9, 2017 #10

    jack action

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    By definition, for any drag (or lift) force, the drag coefficient is:
    $$C_D = \frac{D}{\frac{1}{2}\rho A v^2}$$
    The area ##A## is a reference area. The reference area ##A## is typically defined as the area of the orthographic projection of the object on a plane perpendicular to the direction of motion (source). It represents the relative size of the object studied and could be defined in any way one wishes. This is why it is important to know how drag (or lift) coefficients were initially obtained to apply them to your model, such that you use the same reference area.

    So you wouldn't need to integrate anything if you already know the lift coefficient. All you need to know is the definition of the reference area - which is probably the swept area of the rotor blades in your case - and use the same one for your model.

    All of this is based on dimensional analysis to find ratios (dimensionless quantity) with the help of the Buckingham ∏ theorem.
     
  12. Nov 10, 2017 #11

    CWatters

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    I'm no expert but I have a feeling it's more complicated than this. I think the pitch and aerofoil (and possibly other things) will change along the length of the blade? In which case CL isn't actually a constant but will vary with r ??
     
  13. Nov 10, 2017 #12

    jack action

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    The complications doesn't matter.

    That is what the coefficient does: It encompasses all of those details for a given system. The area represents a significant area for the object to specify its size.

    For example, for car's lift coefficient, some use the car's area as seen from the bottom of the car, which is logic. But most use the frontal area because it is already used for the drag coefficient. The lift coefficient will be different in both cases, but it doesn't matter as ##C_L A## will be the same with both methods and both areas represent more or less the same relative size of the vehicle. Heck, sometimes they even use a volume to represent the area:

     
    Last edited: Nov 10, 2017
  14. Nov 10, 2017 #13
    From my understanding, the Coefficient of Lift is a function of angle of attack and camber of the wing. It is not dependent on velocity. CL probably does change along the length of the rotor but I think I can assume it is constant for a fairly decent calculation.

    jack action:
    This is a site that describes CL, CD, etc. for different airfoil shapes.
    http://airfoiltools.com/airfoil/details?airfoil=naca0012h-sa

    So now I know the lift coefficient but I'm still left with two unknowns, Lift and Velocity. How do I now solve this problem if not by integrating?
     
  15. Nov 10, 2017 #14

    jack action

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    Now, in that case you have the ##C_L## defined for a portion of the airfoil and not for the complete rotor. So you have to consider only the area used for the portion. As @boneh3ad specified:
    $$dL = \dfrac{1}{2}\rho v(r)^2 C_L\;dA$$
    In this case, most probably, ##dA = l_c dr##, where ##l_c## is the length of the chord. Again, you have to verify what was used as a reference area to determine the lift coefficient.

    Knowing that, you now can integrate, knowing that ##C_L## - and maybe even ##l_c## - is a function of ##r## (because the shape and the angle of attack can change along ##r##). So:
    $$dL = \dfrac{1}{2}\rho v(r)^2 C_L(r) l_c(r)\;dr$$
    The important data is not ##C_L##, it is the combined ##C_L A##, where ##A## is always, more or less, arbitrarily chosen.
     
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