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Drag coefficient of a Missile

  1. Nov 14, 2007 #1
    Hi Does anyone know the drag coefficient of a missile. Would a missile have a lower drag coefficient then a torpedo?

    Thanks all
  2. jcsd
  3. Nov 14, 2007 #2
    Drag coefficients are about the cross section of the object to the media displaced.

    A missile has a lower drag coefficient because the air is less dense, not because there is something special about a missile. Remember a torpedo IS a missile, just meant to be deployed in a really dense media.

    To calculate drag think about the cross sectional area of the missile as compared to the medium it is in. Then factor in the speed. What you get is the mass of media the missile must move out of it's path per unit time to proceed forward. That is 90% of drag. The rest is surface factors that affect laminarity of flow of the media as it is ejected/shifted out of the way and replaced behind the object.
  4. Nov 14, 2007 #3
    What kind of missile? Depending, I would say .01 - .03 depending, since missiles are relatively streamlined. A torpedo would have a higher induction of drag, so a higher drag coefficient. I'm not terribly familiar with missile design, like I am with aircraft design. this might http://books.google.com/books?id=NV...ie&sig=EKrNBWmYqypU777RHMuWfP76RDA#PPA596,M1"
    Last edited by a moderator: Apr 23, 2017
  5. Nov 15, 2007 #4
    I am trying to make this object have the lowest drag coefficient as is possible. The object being a towfish. Is a soft point preferred over a solid point? Where the change occurs from the taper to the cylinder should that be rounded? on the back end where the fins are should I taper that end? Is there an optimal thickness vs length?

    Ultimately I will need to find out the drag coefficient to understand the depth of the towfish at given line leads.

    I appreciate all replies
  6. Nov 15, 2007 #5


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  7. Nov 15, 2007 #6
    the drag coefficients of both missile and a torpedo are same,at the end of the day, a torpedo is a missile.
    what is different is the drag, due to the different medium of use. a missile is used in air, whereas torpedo shoots in water.
    drag = [Cd*density*(velocity^2)*plan area]/2, so the density of the media affects the drag

    Vanselena, you only need to know the drag coefficient of the geometry, because the geometry(relative to flow) DEFINES the coefficient of drag & lift. only you need to take different densities for the calculation of drag.
  8. Nov 15, 2007 #7
    Drag coefficients are based on medium: Cd = D / (.5 * r * V^2 * A)

    Basic Algebra, everything in an equation is inter-related.

    A torpedo would have a higher drag coefficient. I don't know water dynamics though. So I couldn't say how much drag water induces, other than alot.

    Wedges are very streamlined, X-43 for example, a hypersonic aircraft.
  9. Nov 16, 2007 #8
    magik revolver, drag and drag coefficients are two different things. coefficient of drag entirely depends on the geometry relative to airflow, whereas drag depends on the flow conditions.
  10. Nov 16, 2007 #9
    d and cd are different, but in determining cd, you MUST in the equation supply for the fluid dynamic drag.
  11. Nov 16, 2007 #10


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    The drag coefficient of a body is purely a function of its geometry, and is independent of the medium in which it is measured. Calculate the drag coefficient of a sphere from measurements of drag taken in air and in helium and you'll get the same result.

    So, if a torpedo and a missile are the same shape, their drag coefficients will be the same. It's the drag that will be higher for a torpedo, since the torpedo will be travelling in water; not air.
    Last edited: Nov 16, 2007
  12. Nov 16, 2007 #11
    OK. I think I get it. brewnog you are saying that Cd is dimensionless and independant of the media, which is true, and MagikRevolver is pointing out that you cannot calculate Cd without actually measuring the object In some media. It's a chicken and egg problem.

    The bottom line for the original question is, check me if I'm wrong here, to use a geometry that yields as laminar a flow as possible around the object for the expected use. In this case a towed marine object. Is there a standard way to figure out the Reynolds number for a towed object, or do you just have to SWAG and test it?
  13. Nov 16, 2007 #12
    everything can be done analytically wysard. do you know any cfd applications ??
  14. Nov 16, 2007 #13
    Computational Fluid Dynamics programs? I'm not familiar with any. Don't they still require assumptions about the nature of the boundary layers though? Ie: Water to Metal, or Air on Enamel Painted Surface, Air on Latex Painted Surface, Dirt on Plywood, that kind of thing? If they do, and have fine enough resolution that would certainly qualify in my book as the Mother of All SWAG tools for what the OP needs. Are there any open source ones you could recommend to the OP? I'd like to know also, 'cause I've seen a number of these drag/friction type quetions and it would be helpfull if I knew of a tool to point others towards to help them solve thier questions.

  15. Nov 16, 2007 #14
    hi, i need some help (...im new here...:) )
    after reading this thread, i still don't understand how to calculate the drag coefficient. Our teacher gave us this formula:
    R = (1/2)DA(rho)v**2

    i know v, A, and im trying to find D, but how to i find R (i think its the resistant force?) and what is rho (i know its the density of air, but what is the value if they don't give it to you, is it constant?)
    thanks if you can help !! :D
  16. Nov 16, 2007 #15


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    In the same media, the torpedo would be more effiecient, it has a tapered tail, a missle doesnt, but a missle has a rocket engine at the back, elminating the need for a tapered tail (as long as the rocket engine is on).

    Tear drop shapes, rounded front, tapered tails, have the lowest drag coefficient.
  17. Nov 16, 2007 #16
    Jeff: Makes sense to me. In that vein, the OP asked about fins (I presume for stability, rather than control surfaces as he explained it would be towed) would it not follow that fins that are also rounded at the lead edge and tapered to a point at the rear, as in a neutral wing shape or very elongated tear shape would probably also present the lowest drag in water?
  18. Nov 16, 2007 #17
    Is there a particular reason for the tear drop shape? Would it not make more sense to have a missile shape but then taper the back, since the missile shape would have a lower frontal profile then the tear drop?
  19. Nov 18, 2007 #18


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    Think of it as a tear drop that is skinny and has been stretched out. Cambered airfoils are shaped similar to tear drops with a curved tail. It's just a general description, the specifics can vary quite a bit. There are a large number of registered air foil shapes, and these are modified by thickness to chord percentage and the amount of camber.
    Last edited: Nov 18, 2007
  20. Nov 18, 2007 #19
    I'm going to quote N.A.S.A: "The drag coefficient is a number that aerodynamicists use to model all of the complex dependencies of shape, inclination, and flow conditions on aircraft drag. The drag coefficient CD is equal to the drag D divided by the quantity: density r times half the velocity V squared times the reference area A." Flow conditions: Viscosity, Compressibility, Mass.
  21. Nov 18, 2007 #20


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    This formula is correct, you already stated it, and nobody is disputing that.

    However, this doesn't mean that the drag coefficient of a body changes depending on what medium it is placed in, since the actual value of drag will be different in both media!
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