Yeah...

Must I think in Russian?

Paging Old_Painless! He needs to replicate this experiment ( http://www.theboxotruth.com/the-box-o-truth-55-slingshots-and-the-box-o-truth/ ), now being fired from the bed of a pickup truck doing 122 mph.

I wonder how fast would a plane have to go so that the pressure behind the bullet caused by the expanding gases would be equal to the air pressure at the stagnation point on the end of the barrel?

At that point would the bullet velocity would be much lower?

Well, I never claimed to be an Aerospace Engineer. I am an ME that does a sometimes poor impression of being an Aeronautical Engineer in my day to day job. But you are correct, I did mis-speak earlier, and it makes sense based on the kinds of aero stuff I do on a near daily basis, but let's not get into that.
I think most of us are in agreement on most points. I think my first post that I wrote all the way back on page 1 still is correct.

I guess to fully answer the OPs question you must ask a few more pertinant questions to have a definitive answer.
1) What is the angle between the bore axis and the direction of travel ?
2) What is the angle of attack for the aircraft?
3) Will the aircraft maintain a constant airspeed and constant altitude?
If the answer to these questions are 0, 0, and yes; than I can say with a high degree of certainty what will happen. If they are not, the situation becomes more complicated quickly.

I would agree.

There are a few variables that aren't clear with most of us.  In the case of the airplane that shot itself down, the bullet impact the plane well after it was fired.  The picture showed the impact happened a few miles ahead and down.  The bullet will not hit the plane immediately after it is fired, but it is possible to out run the bullet before it hits the ground.

10,000 psi would be a conservative estimate of pressure as the projectile leaves the barrel.  An object experiencing 10,000 psi of dynamic pressure in a 1.25 kg/m^3 density fluid (Earth's atmosphere at sea level) would be moving fast enough to escape earth's gravity.  Such an aircraft would cruise with the wings generating lift DOWNWARDS, assuming it is made of unobtanium and doesn't melt and/or ignite.

A 10000 psi Q is something like Mach 32 at sea level! 106,000ºF stagnation temp!  Holy plasma!
[not for official use, because that's outside the bounds of my spreadsheet, way out of bounds]

Mach 1.2, 0.5º0.13º AOA, SSL conditions, 5.56x45mm 77SMK ... you will fly into the bullet in level flight in 1.540.51 seconds.

Lesson learned: don't equip your fighter with a AR-15 Mk12, spend the extra money for the M61.

ETA:
4067 fps muzzle exit, 1380 fps aircraft velocity, 0.345 fixed-value drag coefficient.

ETA2:
Had a bad cell reference and wasn't looking at the right time step.  Updated values now shown.

our galaxy moves at something like 1.3 million mph, our sun orbits in that galaxy at 483,000 mph it also has some smaller movement of around 43,000 mph the earth orbits the sun at 66,000 miles an hour and we rotate on that earth at about 1,000 miles an hour our plane is traveling at mach 1 which is 761 mph the gau 8 it is housing has a muzzle velocity of 3500 ft/s or 2386.36 mph


so our bullet is traveling at 1,896,147 mph roughly as observed by the constant background radiation if all of those vectors align

So, say there is a treadmill in the plane that turns toward the rear and always exactly matches the speed of the plane. Also, there is a gun that produces a muzzle velocity on the ground that exactly matches the speed of the plane. What will the velocity of the projectile be relative to the ground if fired from the treadmill toward the rear of the plane, and will it leave the plane? How about firing it forward?

What angle for the y component so we end up with stupid redneck thrown beer bottle missing me at 87 mph?
 

If you designed bomb sights for bombers, all your bombs would miss because they would all overshoot the target since you think that the bombs would somehow lose the forward speed of the aircraft.



Go drive down the road at 10 mph and drop a ball bearing out the window when you pass a light post.  The ball bearing will not land at the base of the light post since it will have forward speed.



At one time, kids would ride their bikes and throw newspaper's in people's yards.  Pretty sure they would get the concept.
 

That's why tracers were so important.



 

Gravity will determine how long it takes for the ball to hit the ground.



The forward speed will determine the point on the ground that is hit relative to the release point.



 

Updated for (slightly) more realistic conditions:

M61 w/PGU-28/B 3450 fps nominal muzzle velocity

Mach 1.2, SL, 6º AOA, 4830 fps equivalent muzzle velocity, 1380 fps aircraft velocity, ~1.03 G1 BC (0.439 G6 BC) ...
... aircraft flies into the bullet roughly 14.5 seconds later, some 20,100 feet down range

6º AOA isn't realistic for 1g flight at sea level, but neither is 1.2M.  But, the model is now robust enough to handle different altitudes, and if I give the aircraft some maneuver after flying, I image there would be innumerable intercepts.

I'm also outside the bounds of my G1 reference drag for the first 0.28 seconds, and haven't done anything to account for Magnus forces, but I don't care.  If this bothers you, write a poem about it, and I'll consider updating the model.

That is an absolutely perfect illustration of the conservation of momentum.



In the world according to waterglass, we should see those bombs trailing away from the aircraft which is probably moving between 500 and 600 mph.



 

If that were the case then it should impact the ground at point you released it instead of where it will actually hit which will be well forward of that point.



Drag will slow down the forward speed of the ball bearing, but that takes time.  The ball bearing will likely still have forward momentum although it will have a velocity of less than 10 mph due to the drag.  However, when it bounces, it will probably still have enough momentum left that it will "follow" the car.





 

I believe we may have identified why they're having so much trouble getting the gun on the F-35 to work.

I dropped in for the physics fail and was not disappointed.

 

Well a few people have thoroughly discredited themselves on scientific matters.

Well, to be fair, waist gunners generally didn't actually hit anything anyway.

This thread entertains.

has anyone brought up if the craft was traveling at the speed of light?

LOL - wow that is obscure but I got it Major Gant.

what in the hell?

ps
did you stop and think about what happens if, instead of dropping the ball bearing out the window as you are driving down the road at 10MPH, you drop the ball bearing into your lap?  what is the velocity of the ball bearing after it comes to rest in your lap?  

so, if you drop the ball bearing out the window, it doesn't get velocity imparted on it by the car, but if you drop it into your lap it does get velocity imparted by the car.  

ok, got it, makes perfect sense...



ar-jedi

True

YES. . .

So for this answer I will assume the aircraft is flying in steady level flight at sea level with the following conditions:

P=101 kPa
T= 293 K
gamma =1.4
a=(gamma R T)^.5 = 343.1 m/s

I will also assume the aircraft is firing a 50 BMG which wiki states has an average chamber pressure of 378,680 kPa (quoting TM43-0001-27, that I didn't care to read). Assuming that is a gauge pressure (lol because that matters) giving an average chamber pressure of  ~ 378800 kPa.

So, deferring to our good friend Bernoulli and his equation (google it) we know that given a static pressure, like we have here, the pressure at a stagnation point in a flow increases with velocity. That is that as the flow over the aircraft (yes the aircraft is stationary here and the airflow is moving over it, just a reference frame change don't freak out) will hit the end of the barrel and stop. The air at this point of stopped flow will be at a higher pressure than the free stream fluid. So the question here is how fast do we have to travel before the pressure at the end of the barrel is so great it prevents the bullet from exiting.

Well first forget Bernoulli's equation, that's for incompressible flow and we are dealing with Mach number in excess of 0.3 where compressible effects become significant. So we must now consult isentropic law:

Total Pressure: Pt = P [1 + ((gamma-1)/2)M^2 ]^(gamma/(gamma-1))

since gamma =1.4:   Pt=P[1+ 0.2*M^2]^3.5

So now assume we must overcome the quoted average chamber pressure of 378800 kPa at the muzzle, our static pressure is 101 kPa, solving for Mach number yields: 6.892

That is at Mach 6.9  at sea level (~1.5 Miles / second or 5300 MPH) the stagnation pressure would exceed the average chamber pressure of the 50bmg round preventing it from accelerating within the barrel and I assume explode.

Unfortunately at the same conditions the stagnation temperature is given as Tt=T[1+ 0.2M^2], giving 3082 K. At 3082 K I assume there isn't a gun or aircraft left anyway and the question was pointless.