What Is the Drag Coefficient of a Missile Compared to a Torpedo?

[Vanselena]

Hi Does anyone know the drag coefficient of a missile. Would a missile have a lower drag coefficient then a torpedo?

Thanks all

 

[wysard]

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.

 

[MagikRevolver]

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"

 

[Vanselena]

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

 

[stewartcs]

This may help...

http://en.wikipedia.org/wiki/Drag_(physics)_derivations

 

[ank_gl]

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.

 

[MagikRevolver]

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.

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.

 

[ank_gl]

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.

 

[MagikRevolver]

d and cd are different, but in determining cd, you MUST in the equation supply for the fluid dynamic drag.

 

[brewnog]

Drag coefficients are based on medium: Cd = D / (.5 * r * V^2 * A)

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

d and cd are different, but in determining cd, you MUST in the equation supply for the fluid dynamic drag.

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 traveling in water; not air.

 

[wysard]

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?

 

[ank_gl]

everything can be done analytically wysard. do you know any cfd applications ??

 

[wysard]

everything can be done analytically wysard. do you know any cfd applications ??

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 their questions.

Cheers!

 

[i<3math]

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 I am 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

 

[rcgldr]

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.

 

[wysard]

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?

 

[Vanselena]

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.

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?

 

[rcgldr]

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?

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.

 

[MagikRevolver]

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.

 

[brewnog]

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.

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!

 

[caslav.ilic]

It's not very good to look at airborne missile shapes when thinking about a towfish. Airborne missiles fly supersonic or transonic, and the optimal shape there -- in particular length-to-diameter ratio -- is different than what would be for the low subsonic, incompressible (where towfish operates). Torpedo shape may also not be the best model, as torpedos must fit into standard tube diameters.

I'd look into submarine hulls instead. A more teardrop shape than that of the submarine, without the distinct cylindrical midsection, would have slightly less drag, but the tradeoff with less convenient internal arrangement is probably not worth it.

Which brings me to the quite important point, that "the best shape" depends strongly on the internal subsystems that need to fit into the object. At the very least, one should talk about "best shape for constant internal volume", but that's assuming absolutely flexible or fine grained internals.

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.

This is not really correct. If the fluid were inviscid, the object would still displace fluid around itself, but the drag would be zero. In this application, everything is in viscosity and boundary layer (lest cavitation occurs). For example, although a teardrop shape of a given internal volume would have greater frontal profile than the cylindrical shape of that same volume, it would still have less drag.

Is there a particular reason for the tear drop shape?

In the Stokes flow, which would be like "normal" flow but at very low Reynolds number (~1), the optimal shape is a ball. For hypersonic flows, an extremely elongated (factor ~15 or more) spindle. I.e. optimality of the shape depends strongly on the flow type.

For the high Reynolds number incompressible flow, where submerged objects typically operate, it's a tradeoff between viscous and pressure drag. Viscous drag depends on the wetted area; for given internal volume, the ball has the least wetted area. However, any separation of the boundary layer will cause pressure drag, which outweighs viscous drag up to some point. So, the ball has low viscous, but high pressure drag, and then you start to "strech" it in direction of the flow -- the viscous drag increases (more wetted area for constant internal volume), but the pressure drag decreases (less separation of boundary layer). Thus the minimum sum point is reached at length-to-diameter ratio of about 7. The optimal teardrop is designed such as to provide the least possible pressure drag for the given length-to-diameter ratio.

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?

I'd say just pick out any submarine hull, can't go very wrong with that :) Lenth-to-diameter ratio of subs is 7 to 10. The bow shape is not too important (unlike for an airfoil), but any abrupt cutoff of the tappered fin will add a lot to the drag.

If at hand, you could pick a shape out of some experimental tables, for the similar Reynolds number (diameter based probably), and use the drag coefficient given there. One reference would be Hoerner, "Fluid-Dynamic Drag", which has some experimental values as well as empirical formulas for drag coefficients; ballpark coefficient 0.06-0.09 for the clean shape, with reference area that of the max cross-section. And then you need to factor in the fins, and possibly any small protuberances, which may add a lot to the drag, say twice of the clean shape.

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

Argh, for this it seems to me you would need quite a precise drag estimate :) What application are we talking about here -- a theoretical problem, simulation, or the real thing? In the real thing scenario, I don't think any of the textbook estimates or general experimental data will yield accurate results; probably would need calibrating at site.

 

[V Mad]

OK. I think I get it. brewnog you are saying that Cd is dimensionless ...

Is that not the definition of coefficient, a dimensionless factor?

Having said that, there may be different Cd applied to different media as their values are obtained experimentally.

I thought that the drag coefficient (as applied to motor vehicle technology) was simply the measured drag of the subject, divided by the measured drag of a rectangular block of the same frontal cross section. Sounds a bit too simplified though doesn't it?

The shape of rockets, missiles etc is not all about minimising drag though, it is about stability (and if guided, response to guidance controls).

 

[wysard]

Gods how I love being quoted out of context.

My point was, as clearly stated as I was able, that the equations given for Cd rely at some point on someone actually taking some object and measuring it in a real scenario. At the end of the day, that is exactly what the tenor of this thread implies. That all the SWAG math in the world must ultimately come down to someone somewhere measuring the darn thing so it can be used by others. And for the record, I did point out that the whole problem was one of the boudary layer, hence my question about Reynolds numbers for turbulence calculation at the boundary.

A sphere may be the "optimal" shape, but even the slowest cretin to wield a slipstick can tell that in anything less than a perfect vacuum it can create a significant amount of turbulence induced drag. The whole point of a teardrop or other enlongated shape is to "fill in" the gross turbulence areas with solid matter and preserve the laminarity of flow about the object, or at least limit the turbulence to minor boundary layer effects.

I still think the best idea on the thread is the person that suggested a decent computer fluid dynamics program to give the OP a good modelling environment to get a "best guess" before buiding a prototype to actually do testing with.

 

[pmbasa]

I have searched a lot in the internet for the coefficient of drag, but to no avail.. All i found out is that u can achieve the coefficient by experimentation... the drag formula (Fd = 1/2 * Cd * rho * A * v^2) is correct. but the equation used in solving the drag coeficient is somewhat stupid to think that it is Cd = 2 * Fd / (rho * A * v^2) because u just derived it form the drag formula, and if u insert it into the drag formula, ull see that it is just stupid.. here's a little info:
Object Cd
Streamline body 0.04
Solid Hemisphere 0.42
Thin Disk 1.1
cube 0.8
Person (upright position) 1.0 - 1.3

hope this helps

 

[ank_gl]

you see it stupid because you are thinking stupidly. since it is determined experimently, it has to be in above said form, or else how ll you find it genius?? had it been an analytical solution, you would have found something similar to the first equation you quoted
errr... stupid

 

[Vanselena]

Gentlemen, all information is helpful, please let's keep this civil

 

[pmbasa]

you see it stupid because you are thinking stupidly. since it is determined experimently, it has to be in above said form, or else how ll you find it genius?? had it been an analytical solution, you would have found something similar to the first equation you quoted
errr... stupid

I am just pointing out that the equation used for the coefficient of drag is not helpful unless u have the drag force, but u also need to find the drag force using the coefficient.. it's stupidly circular.. and I am really sorry if i keep using the word stupid and offending someone.. anyway, what I am really pointing out is u can not use that drag coefficient equation.. but here's another info.. the drag coefficient is quite relative to the reynolds number.. i forgot the reynolds equation.. it is used to find the turbulence, i think.. and so the higher the value, the more turbulent, and i think the higher the coefficient of drag too.. the reynolds number by the way also rely on 'nu' which is the viscosity of the liquid.. 'rho' is the density, so 'nu' is the viscosity..

Is that not the definition of coefficient, a dimensionless factor?

Having said that, there may be different Cd applied to different media as their values are obtained experimentally.

I thought that the drag coefficient (as applied to motor vehicle technology) was simply the measured drag of the subject, divided by the measured drag of a rectangular block of the same frontal cross section. Sounds a bit too simplified though doesn't it?

The shape of rockets, missiles etc is not all about minimising drag though, it is about stability (and if guided, response to guidance controls).

You got a point there. Objects that are made to move swiftly and smoothly are streamlined to minimize drag.. and also to avoid turbulence.. turbulence affects the path of the object as it passes through a fluid.. try reading information about lift or the magnus effect to understand the usefulness of turbulence... or otherwise, the bad effects.

 

[patrickwilson]

A lot of work was published on cd of bullets in the sixties, which might be a similar problem. It might help to look up "ogive".

 

[brewnog]

I forgot about this thread.

I am just pointing out that the equation used for the coefficient of drag is not helpful unless u have the drag force, but u also need to find the drag force using the coefficient.. it's stupidly circular..

This is not the case. The drag force can be directly measured. You don't need to know the drag coefficient; that's the point of the test. Put a body in a flow stream and measure the force acting upon it.

 

[pmbasa]

Well I guess, that is somehow quite problematic(to me). How do you mesure it? with a ruler? XD.. My point is, and I am sorry if i ddint point it out, that the coefficients (magnus, drag, lift, whatever!) can not be exactly determined by that formula. It is because the coefficients change along with the velocity. Funny, I am not sure. It depends on the reynolds anyway.

 

[ank_gl]

ok dude, you have gone far enough being stupid.
drag can be calculated experimentally.
and the drag coefficient can be obtained from von karmen integral if you have the velocity profile, which can be accurately assumed from the flow conditions.
and the drag coefficient can also be calculated using the drag formula, provided you have the value of drag.

and phusllllesssssee, DO NOT say that drag coefficient changes with velocity, infact no coefficient does so, drag coeff entirely depends on geometry relative to flow.
think yourself dude, if the drag coefficient changes, what's its use??

next time you comment anything, make sure you are fully prepared

and sorry for last time, i didnt mean that stupid thing in any offensive manner and neither now

 

[sharantillu]

Dear i<3math :

In fluid dynamics drag is the resistance offered by the fluid to the motion of a body. Drag is a force that depends on velocity, coefficient of drag(which is different for different geometries), density of the medium(fluid) in which the body is moving and the maximum cross-sectional area of the body exposed to the fluid. Drag can be calculated using standard formula

compare the formula given to you -->R = (1/2)DA(rho)v**2

with the standard formula -->drag = [Cd*density*(velocity^2)*plan area]/2

you can find that the resistance force R = Drag(D) [in Newton],

Cd = Coefficient of drag = D

rho = density (for air it is 1.225 Kg/m**3) & V = velocity (m/s)

velocity here is the relative velocity between the body and the fluid

I think now you can calculate the drag force on the given body

 

[nekkanti26]

howw to caliculate drag on a javelin missile..?

 

[nekkanti26]

how to caliculate drag on a javelin missile.?