Where would the bullet land?

[Golfer]

Lets say:

1.) You're flying along in level flight at 25,000' and humming along at say chewing up about 6 miles a minute.
2.) In level flight your guns are aligned to be level with the flight path of the airplane
3.) You pull the trigger and fire one .50 caliber round.
4.) There is no wind or forces other than a standard still atmosphere acting on the bullet.

Where's the bullet land?  How far did it go?  How do you figure it out?  I'm trying to figure out how to calculate it rather than simply asking someone else to do it for me.

I came up with some rough calculations while flying along the other day trying to figure out using known .50BMG ballistics information I had but I'm at a loss on how to computer out that far to decide what the drop would be like when we're talking miles.

I was trying to figure out 43,000', Mach .80 (458kts, 528mph, 774.4fps) and deciding if we would have flown past the point the bullet finally hits the ground.  I don't think we would by a fair margin but I have no way to prove it or even make an educated guess.

Help?

The airplane is NOT on a treadmill 

[IrishOne]

 

[Tupac]

He's asking if the airplane would pass the bullet in flight before the bullet hits the ground.

[Chalenge]

The bullet will hit the ground in 51.7 seconds ( if you meant 43k alt and ignoring the possibility of uneven ground or bullet rise or terminal velocity). Does that help any?  

(39.4 seconds from 25k)

[Golfer]

Yeah sort of.  I can figure out how fast a bullet will fall from 43,000' but I can't figure out how far it travels in that time.

[mtnman]

... but I can't figure out how far it travels in that time.

I imagine it'll travel all the way to the ground.

[Golfer]

I imagine it'll travel all the way to the ground.

 

I don't know the linear distance covered from the point the projectile was fired to the point it hits sea level.

[FLS]

4 miles.

I don't do math, I just guessed.   

And you wouldn't catch it.

[hitech]

The bullet will hit the ground in 51.7 seconds ( if you meant 43k alt and ignoring the possibility of uneven ground or bullet rise or terminal velocity). Does that help any?  

(39.4 seconds from 25k)

Are you accounting for drag Challenge?

I,ll run a sim on it tomorrow.

HiTech

[E25280]

A long time ago, in a thread far, far away . . .

I watch that show also, anyway what your missing is deceleration (or more properly drag )  varies with the sqaure of vel.

The baisic equations forms are (combining constants all into BulletDragCO ).

Drag = Vel^2 * AirDensity * BulletDragCO.
and
Acceleration = Drag / BulletMass

<Snip>

HiTech

Without a lot of time to browse, I quickly found one site (wikepedia, I believe) that listed the muzzle velocity of a WW2 Browning .50 Cal MG to be 3,050 feet per second, or about 2,080 mph.

E25280: Intial deacceration on the 50cal would be in the 1400 FPSPS Range

HiTech

Of course, while digging this up to help anyone start the calculation, I see that the big man himself is on the case.   

[Golfer]

Are you accounting for drag Challenge?

I,ll run a sim on it tomorrow.

HiTech

Thanks Dale.

[Chalenge]

Are you accounting for drag Challenge?

I,ll run a sim on it tomorrow.

HiTech

No... drag either... I was going to do the same thing.  

EDIT: I dont believe the bullet would ever slow down to its terminal velocity (max speed in free fall) and since gravity pulls on it the entire time the times I gave should be loosely accurate. This is a fascinating application of math though and more complex than it appears at first (owing to the increase of fluid drag closer to the Earths surface).

[tf15pin]

You will start by writing a pair of force equation (horizontal component and vertical component) describing the external forces on the bullet (Drag, gravity, Coriolis force if you want to get fancy). These will be second order linear differential equations that when solved will give you the position as a function of time. You can then set the vertical position function equal to zero and solve for time. Then you use that time in the horizontal component equation to get the horizontal range. The consideration of air density changing with altitude will make the math a little harder but it is the same process. Years ago I took a dynamics class where we did ballistic calculations for battleship guns so if I can find the solution to that it should be useful.

My recommendation, find a java application that does something close to what you are looking for.

[Ghosth]

I disagree Challenge

The bullet is falling at its terminal velocity as soon as it leaves the gun. Falling for that long, the drag is going to eventually stop its forward motion, or nearly so.  Leaving you with only its downward terminal velocity.

I think a .50 is going to lose half its muzzle velocity in the first second, or 800 yards, beyond that it will be slowing rapidly.

Falling for over 30 seconds gives the bullet lots of time to shed minor forward movement via drag.

Interesting exercise!

[EskimoJoe]

I'm wondering if you guys are taking into account the bullet beginning to tumble
after a certain distance? I'm unsure if it would, but my gut tells me it might.
< not a math/science genius... yet.