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An educated guess--a significant source of sound for some firearms is actually the sonic boom created by the bullet. If you made a bullet that could maintain a Busemann Biplane form in flight, however, there would be no telltale crack. Coupled with a good suppressor, I can definitely see how the military would be interested in this sort of thing. --Lode Runner (talk) 09:30, 15 December 2007 (UTC)Reply
There is a paper kicking around on the net somewhere somebody tested biplane based ammunition and tried to persuade the military to buy it. It had maybe 20% extra range or something because it wasn't wasting energy on the shock; but range isn't as useful as you would expect, since it's even harder to aim when you're going further and stuff like that. I'm not aware, and can't recall any mention of it being extra quiet, but it would be expected to be I suppose.- (User) WolfKeeper (Talk) 10:11, 15 December 2007 (UTC)Reply
Of course, the lack of a sonic boom would be good for range too. I forgot. Given the push towards unmanned and computer-assisted attack vehicles (I have it on good authority that in the latest attack helicopter systems, it's pretty much "point 'n' click" and the computer does the shooting for you), I bet that 20% will come in handy some day. Also, in addition to added range, the bullets should be going faster when they hit their target (at *any* range), thus resulting in more penetration. Faster bullets should mean a flatter trajectory, too, which should make aiming easier at medium ranges. --22:29, 15 December 2007 (UTC)
No, expansion fans are not discontinuities (though often depicted as a fan of discrete lines) and isentropic, unlike shock waves. Across shock waves, static pressure rises and total pressure decreases (thus increase in entropy), whereas across expansion waves, static pressure decreases and total pressure remains constant.--Butakun00:52, 1 January 2007 (UTC)Reply
Some comments,
The waves emanating from the edges of the vehicle are oblique shock waves and not expansion waves. They are formed because the flow is being turned into itself. Other way to look at is, since the area is decreasing from front to middle, there has to be a rise in pressure, which is accomplished by shock waves. If the vehicle is moving too fast slow, the shock waves will become more and more normal to the flow till they form a normal shock.
How the waves get reflected is a different matter. If the vehicle is flying at design speed (as indicated in the caption of the figure), and the shock waves are incident precisely on the throat, they will get reflected as expansion waves and one will see an expansion fan. The figure needs to be corrected for this. At other speeds depending on where the shocks waves hit the vehicle, they may get reflected as shock waves or expansion fan.
One may look at an expansion fan as a collection of infinite number of shock waves. They are very weak shock waves and are referred to as Mach waves. Thinking of these waves as shock waves can lead to confusion and should be avoided, but is not incorrect. Across a Mach wave the stagnation properties do not change, the static properties change by small extent and may increase or decrease. A collection of many Mach wave at one point will lead to a normal or oblique shock (this will happen is the Mach wave converge). If the Mach waves diverge they will form an expansion fan.
1) The first part I agree with. What I was referring to was not the leading edge shocks. However, as for the second part, I do not agree completely. If the throat area is sufficient enough to swallow the mass flow, the oblique L.E. shocks will be weak shocks, meaning that the shock angle will decrease as Mach number increases. 3) I agree also, but I used the word "shock" simply to mean compression discontinuity, but yes, Mach wave is an infinitesimally weak shock of which Prandtl-Meyer fan is made.--Butakun11:37, 30 January 2007 (UTC)Reply
I agree, I meant to say slow for as Mach number decreases, the shock will become more normal to the flow. I have made the appropriate corrections. Thank you. Myth (Talk) 12:05, 30 January 2007 (UTC)Reply
I've updated the file slightly, it's possibly an improvement. Ideally we would replace it with a still or something from the Japanese groups simulation; but we would need to negotiate a license- I haven't tried that as yet. Mostly my aims for this picture are so that people get atleast a vague idea what's happening, how the cancelation occurs, but it's still very half-arsed.WolfKeeper08:59, 30 January 2007 (UTC)Reply
Latest comment: 14 years ago10 comments3 people in discussion
I am not sure if it is correct to say that the vehicle produces no wave drag. Any vehicle traveling at supersonic speed will produce a wave drag (unless it is straight line, which aligns itself with the streamlines). The reason we observe wave drag is because across a shock wave (both normal and oblique shock) there is drop in the stagnation pressure. This loss can be related to the wave drag that a vehicle produces. Myth (Talk) 10:07, 30 January 2007 (UTC)Reply
Busemann's biplane only works if you align it with the stream. The theory says that there is no wave drag, and the theory and the biplane have been successfully tested in a wind tunnel.WolfKeeper16:25, 30 January 2007 (UTC)Reply
The external link provided in the article mentions that the configurations they tested had lower wave drag. They also say that the wave drag is minimized for particular operating conditions. These claims I can agree with. Note that it never says that the wave drag is reduced to zero. The presence of oblique shock is enough to ensure that there will always be some amount of wave drag (maybe small but not zero). Myth (Talk) 21:34, 30 January 2007 (UTC)Reply
You may well be right, but we need a reference I think. Zero is a good minimum(!) They can't show that it does or does not zero out in a simulation anyway.WolfKeeper21:46, 30 January 2007 (UTC)Reply
The pressure plot shows the variation in pressure on the lower "wing" if we follow the dashed line. I will see if I can find any references. I didn't know the concept of Busemann's biplane till yesterday. btw even if they had infinite-precision computations, they won't be able to show that the wave drag is zero.
Here are few google results that I found, this and this go to page 59 of the pdf file. Both these references say that there will be no wave drag, however I am not able to understand why. I made the following sketch to show what I understand is happening. Clearly the pressure is higher on the surface BC than on surface CE. If we add up the pressure on the top and bottom "wings", the net component of the pressure will be in the flow direction (i.e. a drag on the vehicle).
The reason as to why the other references are not correct is because they are representing the expansion fan by a single wave, which, IMO, is not correct (unless the "wing" is thin and there is no significant spread in the fan before it reaches the other side). I may be wrong here or maybe making too much fuss about concepts like . Sorry about that, but just a doubt I wanted to get clarified. Thanks. Myth (Talk) 23:30, 30 January 2007 (UTC)Reply
Well, if you shape it as the biplane then you could. But the shape is highly restricting and gives no lift. Rumour has it that this shape has been tested as a way to silence weird-shaped bullets though.WolfKeeper16:25, 30 January 2007 (UTC)Reply
I agree, I hadn't though long enough about that earlier. I can think of geometries where you do not have shock waves. But they will require isentropic compression and expansion using Mach waves and also that the compression waves do not converge to a point to form a shock wave. But this is not the case here, where we have oblique shocks from the leading edges of the vehicle, getting reflected as expansion waves. Myth (Talk) 21:34, 30 January 2007 (UTC)Reply
I think in practice there's still a shockwave, and so the sound is lower, but still present, at least with the ammunition that was tested- there's a link in the article if you're that interested.- (User) WolfKeeper (Talk) 22:42, 15 December 2007 (UTC)Reply
Myth is entirely correct on this. There will still be a stagnation pressure loss through the oblique shocks that cannot be recovered in the isentropic expansion. Note that shocks (oblique or otherwise) are ALWAYS non-isentropic and will ALWAYS have an associated stagnation pressure loss. This stagnation pressure loss directly leads to wave drag. 128.61.191.66 (talk) 20:01, 27 April 2010 (UTC)Reply
Latest comment: 7 years ago1 comment1 person in discussion
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