# A bullet fired from a moving platform.
**URL:** https://boards.straightdope.com/t/a-bullet-fired-from-a-moving-platform/226686
**Category:** Factual Questions
**Created:** [January 30, 2004, 2:29am UTC](https://boards.straightdope.com/t/a-bullet-fired-from-a-moving-platform/226686 "2004-01-30T02:29:30Z")
**Posts on this page:** 20
**Page:** 1
### Author:  [@jimpeel](https://boards.straightdope.com/u/jimpeel)
#### Post date: [January 30, 2004, 2:29am UTC](https://boards.straightdope.com/t/a-bullet-fired-from-a-moving-platform/226686/1 "2004-01-30T02:29:30Z")
We are having a discussion at [http://www.thehighroad.org/showthread.php?s=&threadid=61836](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.
---
### Author:  [@Eleusis](https://boards.straightdope.com/u/Eleusis)
#### Post date: [January 30, 2004, 2:41am UTC](https://boards.straightdope.com/t/a-bullet-fired-from-a-moving-platform/226686/2 "2004-01-30T02:41:48Z")
> [@](#):
>
> 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.
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.
---
### Author:  [@Princhester](https://boards.straightdope.com/u/Princhester)
#### Post date: [January 30, 2004, 3:17am UTC](https://boards.straightdope.com/t/a-bullet-fired-from-a-moving-platform/226686/3 "2004-01-30T03:17:27Z")
> [@jimpeel](#):
>
> The barrel of the firearm, relative to the bullet, is traveling ZERO ft/sec regardless of the velocity of the firearm containing the bullet.
Why would the barrel be travelling at a speed any different to the firearm to which it is attached?
> [@](#):
>
> 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.
You seem to be saying in the first sentence the bullet and firearm have a relative velocity of 900ft/sec and in the second sentence 1800 ft/sec.
The former is correct. How you figure the latter is somewhat obscure.
---
### Author:  [@kniz](https://boards.straightdope.com/u/kniz)
#### Post date: [January 30, 2004, 4:02am UTC](https://boards.straightdope.com/t/a-bullet-fired-from-a-moving-platform/226686/4 "2004-01-30T04:02:29Z")
[indent][indent][indent][indent][indent][indent]🆒[/indent][/indent][/indent][/indent][/indent][/indent]
[indent][indent][indent][indent][indent] **It’s all relative!** [/indent][/indent][/indent][/indent]
---
### Author:  [@Quint\_Essence](https://boards.straightdope.com/u/Quint_Essence)
#### Post date: [January 30, 2004, 4:29am UTC](https://boards.straightdope.com/t/a-bullet-fired-from-a-moving-platform/226686/5 "2004-01-30T04:29:24Z")
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.
---
### Author:  [@jimpeel](https://boards.straightdope.com/u/jimpeel)
#### Post date: [January 30, 2004, 4:42am UTC](https://boards.straightdope.com/t/a-bullet-fired-from-a-moving-platform/226686/6 "2004-01-30T04:42:36Z")
This is what one of our members said also.
---
### Author:  [@Xema](https://boards.straightdope.com/u/Xema)
#### Post date: [January 30, 2004, 4:52am UTC](https://boards.straightdope.com/t/a-bullet-fired-from-a-moving-platform/226686/7 "2004-01-30T04:52:48Z")
> [@jimpeel](#):
>
> 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.
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.
---
### Author:  [@jimpeel](https://boards.straightdope.com/u/jimpeel)
#### Post date: [January 30, 2004, 5:05am UTC](https://boards.straightdope.com/t/a-bullet-fired-from-a-moving-platform/226686/8 "2004-01-30T05:05:06Z")
> [@Princhester](#):
>
> Why would the barrel be travelling at a speed any different to the firearm to which it is attached?
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.
> [@](#):
>
> You seem to be saying in the first sentence the bullet and firearm have a relative velocity of 900ft/sec and in the second sentence 1800 ft/sec.
>
> The former is correct. How you figure the latter is somewhat obscure.
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.**
---
### Author:  [@jimpeel](https://boards.straightdope.com/u/jimpeel)
#### Post date: [January 30, 2004, 5:14am UTC](https://boards.straightdope.com/t/a-bullet-fired-from-a-moving-platform/226686/9 "2004-01-30T05:14:19Z")
> [@Xema](#):
>
> 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 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.
---
### Author:  [@Gorsnak](https://boards.straightdope.com/u/Gorsnak)
#### Post date: [January 30, 2004, 5:21am UTC](https://boards.straightdope.com/t/a-bullet-fired-from-a-moving-platform/226686/10 "2004-01-30T05:21:08Z")
> [@jimpeel](#):
>
> … 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.
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.
---
### Author:  [@jimpeel](https://boards.straightdope.com/u/jimpeel)
#### Post date: [January 30, 2004, 6:02am UTC](https://boards.straightdope.com/t/a-bullet-fired-from-a-moving-platform/226686/11 "2004-01-30T06:02:07Z")
> [@Gorsnak](#):
>
> 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.
I had already looked that event up and posted it at the referenced site.
> [@](#):
>
> I believe that there is only one instance of a self-shootdown. Here are a couple of writeups the self-shootdown Sep 21, 1956 by an F11F-1.
>
> [F11F-1 Shoots Itself Down](http://www.aerofiles.com/tiger-tail.html)
>
> > [@](#):
> >
> > A Tiger Bites Its Tail
> > On Sep 21, 1956, Grumman test pilot Tom Attridge shot himself down in a graphic demonstration of two objects occupying the wrong place at the same time — one being a Grumman F11F-1 Tiger [138260], the other a gaggle of its own bullets…
> >
> > It happened on the second run of test-firing four 20mm cannon at Mach 1.0 speeds. At 20,000’ Attridge entered a shallow dive of 20°, accelerating in afterburner, and at 13,000’ pulled the trigger for a four-second burst, then another to empty the belts. During the firing run the F11F continued its descent, and upon arriving at 7,000’, the armor-glass windshield was struck, but not penetrated, by an object…
> >
> > Attridge throttled back to slow down and prevent cave-in of the windshield, flying back to Grumman’s Long Island field at 230 mph. He radioed that a gash in the outboard side of the right engine’s intake lip was the only apparent sign of damage other than for the glass, but that 78 percent was maximum available power without engine roughness occurring…
> >
> > Two miles from base, at 1,200’ with flaps and wheels down, it became evident from the sink rate that the runway could not be gained on 78 percent power. Attridge applied power and said “the engine sounded like it was tearing up.” It then lost power completely. He pulled up the gear and settled into trees less than a mile short of the runway, traveling 300 feet and losing a right wing and stabilizer in the process. Fire broke out, but, despite injuries, Attridge managed to exit the plane and get away safely, to be picked up by Grumman’s rescue helicopter.
> >
> > Examination of the F11F established there were three hits — in the windshield, the right engine intake, and the nose cone. The engine’s inlet guide vanes were struck, and a battered 20mm projectile was found in the first compressor stage…
> >
> > See image of bullet at [http://www.aerofiles.com/20mm-culprit.jpg](http://www.aerofiles.com/20mm-culprit.jpg)
> >
> > How did this happen? The combination of conditions reponsible for the event was (1) the decay in projectile velocity and trajectory drop; (2) the approximate 0.5-G descent of the F11F, due in part to its nose pitching down from firing low-mounted guns; (3) alignment of the boresight line of 0° to the line of flight. With that 0.5-G dive, Attridge had flown below the trajectory of his bullets and, 11 seconds later, flew through them as their flight paths met…
>
> [http://www.f-16.net/library/stories/owngoal.html](http://www.f-16.net/library/stories/owngoal.html)
>
> > [@](#):
> >
> > The Grumann-made F-11 Tiger achieved some notoriety as being the first aircraft in history to “shoot itself down”. Two F-11s were experimentally fitted with massive J-79 engines, giving them thrust-to-weight ratios of nearly 1 to 1. Although their increased weight and limited fuel capacity made them unsuitable for carrier duty and impractical as fighters, the two aircraft were nonetheless tested extensively.
> >
> > On one such test flight, an F-11 was making passes on a gunnery target. The pilot fired a short burst to clear his guns and began a shallow dive at full power. After several seconds, the pilot pilot pulled the aircraft into a shallow climb and simultaneously began a rapid deceleration maneuver. Almost instantly, the F-11 was racked by a series of explosions, forcing the pilot to abandon his stricken aircraft. The accident investigation board concluded that the F-11 had outrun its own cannon fire, only to have the cannon fire catch up with the aircraft when it decelerated!
---
### Author:  [@Valgard](https://boards.straightdope.com/u/Valgard)
#### Post date: [January 30, 2004, 6:03am UTC](https://boards.straightdope.com/t/a-bullet-fired-from-a-moving-platform/226686/12 "2004-01-30T06:03:27Z")
> [@jimpeel](#):
>
> **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.**
Hiya Jim,
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.
---
### Author:  [@Valgard](https://boards.straightdope.com/u/Valgard)
#### Post date: [January 30, 2004, 6:10am UTC](https://boards.straightdope.com/t/a-bullet-fired-from-a-moving-platform/226686/13 "2004-01-30T06:10:16Z")
…and I see you’ve already found the real examples.
Boy, that’d be a sucky “kill” marker to have to paint on what’s left of your plane, huh?
---
### Author:  [@jimpeel](https://boards.straightdope.com/u/jimpeel)
#### Post date: [January 30, 2004, 6:17am UTC](https://boards.straightdope.com/t/a-bullet-fired-from-a-moving-platform/226686/14 "2004-01-30T06:17:26Z")
It would really suck to realize that you are one-fifth of an ace and five-fifths of an ass at the same time.
---
### Author:  [@kanicbird](https://boards.straightdope.com/u/kanicbird)
#### Post date: [January 30, 2004, 11:35am UTC](https://boards.straightdope.com/t/a-bullet-fired-from-a-moving-platform/226686/15 "2004-01-30T11:35:25Z")
Unless the bullet is traveling at orbital velocities, It’s going to fall at about the same rate and is almost independant of forward velocity.
---
### Author:  [@Mangetout](https://boards.straightdope.com/u/Mangetout)
#### Post date: [January 30, 2004, 11:56am UTC](https://boards.straightdope.com/t/a-bullet-fired-from-a-moving-platform/226686/16 "2004-01-30T11:56:22Z")
The ground is irrelevant to the speed of the bullet within the moving plane - it’s as simple as that, otherwise…
> [@](#):
>
> Just remember that you’re standing on a planet that’s evolving
> And revolving at nine hundred miles an hour,
> That’s orbiting at nineteen miles a second, so it’s reckoned,
> A sun that is the source of all our power.
> The sun and you and me and all the stars that we can see
> Are moving at a million miles a day
> In an outer spiral arm, at forty thousand miles an hour,
> Of the galaxy we call the ‘Milky Way’…
In other words, there are no privileged reference frames.
---
### Author:  [@alterego](https://boards.straightdope.com/u/alterego)
#### Post date: [January 30, 2004, 12:12pm UTC](https://boards.straightdope.com/t/a-bullet-fired-from-a-moving-platform/226686/17 "2004-01-30T12:12:35Z")
[There once was a racer named Fisk
Who took a considerable risk
When his dragster got traction
The Fitzgerald contraction
Reduced his wazoo to a disc](http://www.straightdope.com/classics/a2_095.html)
> [@](#):
>
> You probably have the idea that if you are standing in a bus moving at speed u, and you walk forward at speed v, your total forward speed w is expressed by the straightforward sum u + v. Alas, this is a cruel illusion.
>
> In reality, what we might call “addition of velocities” is governed by the awe-inspiring equation
>
> w = (u + v)/(1 + uv/c^2)
>
> where c^2 is the speed of light squared. (This may give you pause next time you hike to the can on a Greyhound.)
>
> At so-called Newtonian (i.e., slow) speeds, the term uv/c^2 is pretty close to 0, and the equation reduces down to the familiar w = u + v.
>
> However, if we are traveling at, say, 0.9c (nine-tenths the speed of light), and we shoot a bullet forward also at 0.9c, we discover via the above formula that the slug does not attain an overall speed of 1.8c (i.e., more than the speed of light), but rather a modest
>
> (0.9c + 0.9c)/(1 + [0.9]^2) = 0.994c
---
### Author:  [@Desmostylus](https://boards.straightdope.com/u/Desmostylus)
#### Post date: [January 30, 2004, 12:47pm UTC](https://boards.straightdope.com/t/a-bullet-fired-from-a-moving-platform/226686/18 "2004-01-30T12:47:27Z")
> [@alterego](#):
>
> There once was a racer named Fisk
> Who took a considerable risk
> When his dragster got traction
> The Fitzgerald contraction
> Reduced his wazoo to a disc
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.
---
### Author:  [@jimpeel](https://boards.straightdope.com/u/jimpeel)
#### Post date: [January 31, 2004, 5:32am UTC](https://boards.straightdope.com/t/a-bullet-fired-from-a-moving-platform/226686/19 "2004-01-31T05:32:01Z")
[http://www.archlug.org/pinguin.html](http://www.archlug.org/pinguin.html)
First mouse click launches bird
Second mouse click swings bat
---
### Author:  [@Princhester](https://boards.straightdope.com/u/Princhester)
#### Post date: [January 31, 2004, 5:59am UTC](https://boards.straightdope.com/t/a-bullet-fired-from-a-moving-platform/226686/20 "2004-01-31T05:59:05Z")
OK, I’m on 492.2. Anyone beat that?
[Next page](https://boards.straightdope.com/t/a-bullet-fired-from-a-moving-platform/226686.md?page=2)