We are having a discussion at http://www.thehighroad.org/showthread.php?s=&threadid=61836 which started on the subject of explosive decompression due to bullet penetration of an aircraft skin, then the discussion changed to the ability of an aircraft to shoot itself down, and finally to what would happen if a bullet were to be fired from a moving platform in the direction opposite to the line of travel.
I believe that if a plane is flying 900ft/sec and a bullet is fired which has a muzzle velocity of 900ft/sec the two will travel apart at 1,800ft/sec. Others believe that the bullet will attain 0 velocity relative to a fixed object on the ground and simply fall straight down. Still others believe that if the platform is traveling faster than the muzzle velocity of the bullet it will travel backward.
I say this:
The barrel of the firearm, relative to the bullet, is traveling ZERO ft/sec regardless of the velocity of the firearm containing the bullet. The bullet, when fired, achieves a velocity relative to the firearm of 900ft/sec. The two part company, in relative terms, at 1,800ft/sec.
What say you?
Feel free to drop in to the discussion board and lurk a while.
If the plane is flying 900ft/sec and the bullet has a muzzle velocity of 900ft/sec, and the gun is fired backward, the net velocity will be zero and the bullet will fall straight down. In practive, however it will still be going 900ft/sec relative to the plane, and shoot a hole out the back. It’s better to visualize shooting off the back of a (very fast) train, where a stationary person on the ground will simply see it fall straight down.
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[indent][indent][indent][indent][indent]It’s all relative![/indent][/indent][/indent][/indent]
no matter which direction you fire the gun it will accelerate away from you at 900 m/s.
Relative to an outside viewer, if you fired it in the same direction you are moving, since the bullet and gun are already moving at 900 m/s the bullet will have a relative speed of 1800 m/s. If you fired the gun backwards, to YOU it would still accelerate away at 900 m/s but to an outside observer it would go from 900 m/s to a stop.
I think we can all accept that the velocity of the bullet can be expressed relative to the gun. But you seem unclear here. In one sentence, their relative velocity is 900fps. In the next it’s 1800 fps. ???
Based on other clues, it seems that 900 fps is correct. This means that if the bullet is fired in the direction opposite to the plane’s motion, its horizontal velocity relative to the earth is zero - it falls straight down.
I didn’t think I had to specify that the firearm is an assembly which contains a barrel. All portions of the firearm assembly are going zero ft/sec relative to the unfired bullet are they not? It is only upon pulling the trigger that the bullet achieves velocity relative to the firearm.
I see the fired bullet traveling at 900ft/sec away from the firearm which was at zero ft/sec relative to the bullet prior to firing. In my mind’s eye, I saw the aircraft at +900ft/sec and the bullet at +900ft/sec or a combined speed away from each other of +1,800ft/sec.
I now see how the opposing velocities would cancel each other out. The plane would be going +900ft/sec and the bullet -900ft/sec. Thus they cancel each other out. My prior contention was incorrect.
This does bring up the interesting concept that an aircraft, firing at another aircraft, using the above formula would be fully dependent on the trailing aircraft actually running into the expended rounds which would not be traveling toward it at all but dropping toward the Earth. 900ft/sec is a little over 613 mph.
So what happens when a plane fires a firearm that has a 900ft/sec muzzle velocity but is traveling at, say, 1839 mph or exactly three times the speed of the bullet? The bullet is moving away at 900ft/sec but the aircraft is closing on it at 1,800ft/sec (2,700ft/sec - 900ft/sec). Although the combined speed of the bullet fired from the plane relative to a fixed object is 3,600ft/sec the speed of the bullet relative to the aircraft that fired it is still 900ft/sec.
I saw the firearm with the unfired bullet having a relative velocituy to one another of zero as the firearm containing the unexpended bullet is traveling at 900ft/sec. I also saw the bullet achieving a velocity of 900ft/sec when fired but the firearm is now traveling at 900ft/sec away from the bullet. 900 + 900 = 1,800. Of course, this was incorrect as it is actually (900 + (-900)) = 0 relative to a fixed object so, in actuality, the plane must be traveling away from the bullet at 900ft/sec.
A long time ago when I was a kid and fascinated by all things aeronautical, I read an account of a prototype fighter jet that shot itself down. (This means you should not expect a cite to be forthcoming. :)) This would have been during the early days of supersonic flight. (The shooting, that is, not the reading.) Because the bullets are unpropelled after firing, they slow rapidly due to air resistance, while the powered jet overtakes them. Oops.
Perhaps a more ambitious soul than myself can google up the story.
Perhaps it’ll help to picture this from the standpoint of a ground observer.
Right before the gunner pulls the trigger, both the plane and the bullets are moving with the exact same speed relative to the ground - 2700fps.
Gun fires and bullet exits the barrel. Plane is still moving 2700fps, however the bullet is moving about 3600fps. Again, all speeds relative to the ground.
The bullet is moving ahead of the plane at a net 900fps, just like you’d expect. If it didn’t slow down, in exactly one second the bullet would be exactly 900 feet ahead of the plane.
Realistically, the bullet starts to slow down as soon as it leaves the gun - air resistance. The plane doesn’t slow down because the engine keeps it moving. So, if you know how fast the bullet decellerates due to drag then you can figure out how long it’ll take for the plane to catch up.
For example suppose that the bullet slows down smoothly at 1000fps/sec (I’m making this number up completely at random by the way):
vb(t) = 3600-1000t
vp(t) = 2700
(Velocity of bullet at time “t” = 3600-1000t, in fps)
(Velocity of plane at time “t” = 2700fps; it’s constant)
Integrating (chortle) we get the positions of the bullet and the plane at time “t”, taking position = 0 as the moment that the bullet leaves the plane.
xb(t) = 3600t - 500t^2
xp(t) = 2700t
Note that this is only good up to the point where the bullet has lost all forward velocity (t = 3.6s), otherwise xb(t) gets funny and we’ve got Boomerang Bullets.
To figure out where the plane catches up to the bullet, set the two equations equal and solve for t:
3600t - 500t^2 = 2700t
Which gives t = 1.8 seconds.
In level flight, the bullet will have travelled 4860 feet forward in those 1.8 seconds, and it will have dropped almost 52 feet vertically due to the pull of gravity.
So what does our ground observer see? Plane whips overhead at 2700fps. Bullets fly out of the cannon at 3600fps. In 1.8 seconds, the plane and bullets are both almost 5000 feet downrange. The bullets have lost all forward speed and are falling straight down, while the plane is 52 feet directly above the bullets, and it’s still cruising along at 2700fps. Unless the plane was in a shallow dive it won’t actually run into its own shells.
Realistically the bullets decelleration will be more complicated (drag is a function of velocity squared and blahdeeblahdeeblah) and the plane won’t be flying flat and level but you get the idea.
Really? What did Fisk’s wazoo look like before? The limerick only makes sense if it specifies some body part of Fisk that wasn’t disk-shaped to start with.
Anyway:
There was a young lady called Bright
Who could travel much faster than light.
She set out one day
In a relative way
And returned on the previous night.
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