[EMAIL PROTECTED]">Having seen the graph for the first time today, I can certainly see why.On Tue, 2 Jul 2002, Ian Woollard wrote:OTOH I found a water rocket web page that implies that the Cd for a nose
cone varies between 0.6 and 0.8 or so; and Randall was using 0.7 for his
simulations (as was I).Those are awfully high numbers. Several references show plots more like
0-15-0.2 at subsonic speed, rising sharply to a peak of 0.5-1.0 just after
Mach 1, declining smoothly to around the original at about Mach 6. (One
source adds a comment that an overall average Cd is *not* a good
substitute for a Mach-dependent one -- even a vague approximation to the
shape of the curve is much better than trying to pretend it's flat.)
[EMAIL PROTECTED]">Actually I meant arctan(1/4) which is 14 degrees, but it's close.One caution: there are a lot of variables here. Those plots probably
assume a traditional rocket shape, long and pointy. Even going from a
narrow cone to a wide cone on the nose of a cone-cylinder shape raises Cd
a *lot*. Anderson's "Fundamentals of Aerodynamics" shows experimental
data on a cone-cylinder shape with a 30deg half-angle in the cone:
subsonic Cd about 0.5, peak about 1.0, hypersonic limit about 0.6.
Whereas "Handbook of Astronautical Engineering" shows a 12.5deg half-angle
cone-cylinder at 0.2, 0.4, 0.1, and Anderson shows a 15deg one having a
hypersonic limit of about 0.15.There's an equation for a flat plate that gives Cd of about 0.15 or so
for a 1 in 4 angle to the flow...What do you mean by "1 in 4"? If that's an angle of arcsin(1/4), that's
about 15deg.
[EMAIL PROTECTED]">Neat, missed that [a mistake on my "first day" as a hypersonic aerodynamicist, how will I liveNewtonian Cd for a flat plate is (as seen in both that web page and
Anderson's "Fundamentals of Aerodynamics") is 2*sin^3(alpha). But note,
that Cd is based on the plate area, as is usual for wings, not on its
cross-section perpendicular to the flow, as is usual for body drag
coefficients! Cross-section is area*sin(alpha), so the "real" Cd ends
up being 2*sin^2(alpha). For alpha = arcsin(0.25), that's 0.125.
it down ;-) ]
[EMAIL PROTECTED]">Perhaps if I multiply by the mach related V2 graph finagle factor like fig 4-8 in Sutton it might startThat's at Mach infinity, mind you. Much of the drag loss a rocket
experiences (for orthodox trajectories, climbing out of the atmosphere
quickly) is at relatively low speeds where that won't be a good
approximation. And even at hypersonic speeds, various empirical modified
forms of Newtonian theory tend to be preferred over the original.
to be ok.
[EMAIL PROTECTED]">Fair comment in general. My educated guess was that it would work ok with a cone though.I think a cone should be the same Cd.Not a safe assumption! Two-dimensional and three-dimensional flow are
different animals.
The hypersonic air flow pretty much can't see the rest of the surface and all it sees is
an angled surface that is locally flat; that's basically what the Newtonian theory
assumes. Since none of the scattered flows intersect I thought it would work out. But
any nose cone shockwave would probably interfere some I guess.
[EMAIL PROTECTED]">(Actually I wondered about the edges with a flat plate, I would bet they'd doIn this particular case, Newtonian theory actually is
pretty close to the real hypersonic limit of the cone -- much closer than
it is to the real flat plate.
horrible things, you'd get a shockwave all the way along it, that may be why it's better with
a cone).
[EMAIL PROTECTED]">
[EMAIL PROTECTED]">Thanks Henry, much clearer.Conclusions? Nope. I must admit I'm a bit confused right now. I haven't
managed to explain why some sources give a Cd of 0.7..Check those references, and the shapes they use, carefully...
[EMAIL PROTECTED]">
Henry Spencer
[EMAIL PROTECTED]
