
#1
Feb513, 09:31 AM

P: 2

Back in World War Two, the Russians developed an experimental weapon called so eloquently "Fire Hedgehog" that fired 88 ppsh submachine guns straight down as a volume strafing device. Knowing quite a bit about straight line ballistics, i was curious to know more about the effectiveness of this device. So here is the question that I could not figure out.
Question: At what height would would the aircraft have to obtain for this device to be effective, and subsequently at what height would this weapon pose no threat? At some point the initial velocity of the bullet being fired will be lost due to wind resistance which is more of a fluid dynamics question and will need to be figured out in order to get this to work correctly. Below I will set up the experiment with as many variables as I can think of: Experiment: Aircraft: Russian Tu2 Traveling at 379mph or 555fps. Weapon: Russian PPSH Submachine gun statistics: Muzzle Velocity 488 m/s (1,600.6 ft/s) Fires the 7.62x25 Tokarev Bullet Statistics: Bullet Diameter .310in Bullet Weight 5.5g Velocity 525 m/s (1,720 ft/s) Energy: 760 J (560 ft·lbf) Atmospheric Conditions: Altitude: 0 MCL Lateral Wind: 0 Temperature at 0 MCL: 32F (0c) If more variables are needed let me know and i will attempt to supply the rest. 



#2
Feb513, 09:47 AM

P: 3,545





#3
Feb513, 10:14 AM

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P: 38,886

It appears that Recurve is asking about the horizontal motion. And he is asking about a maximum effective height, not minimum.
However, in order to answer that we would need more information. What is the area of the region we wish to fire at? What coefficents for air resistance are we to use? Are we to assume that resistance is proportional to speed or speed squared? 



#4
Feb513, 10:44 AM

P: 2,861

Shooting a bullet straight down?
You also need to decide how slow a bullet can go and still be effective. On Myth Busters their experiments appears to show that the terminal velocity of a bullet wasn't lethal, however they interviewed medical experts that had seen cases where falling bullets had killed or injured somone.
I think most bullets are supersonic so that might complicate the equations needed to calculate the speed. 



#5
Feb513, 10:53 AM

P: 2,861

http://www.slate.com/articles/news_a...g_bullets.html
http://hypertextbook.com/facts/2002/...toniadis.shtml For example.. 9mm Luger (Parabellum) (124 gr RN) Muzzle Velocity 1120 ft/s Maximum Range 2400 yard, 1.36 mile Impact Velocity 350 ft/s So it would seem reasonable that a plane firing a bullet from 12 miles up might be harmful as the velocity could still be over 300 ft/s. 



#6
Feb513, 11:05 AM

P: 2,861

If the idea is to cover the ground with a certain density of bullets (eg one per square foot?) you might like to work out the accuracy with which each gun must be set in relative to it's neighbours to achieve even coverage from a given height. For example 1 foot accuracy from 8000 feet is about ..
Tanθ= 1/8000 θ = tan^{1}(1/8000) = 0.007 degrees 



#7
Feb513, 01:05 PM

P: 2

Well the effectiveness of the weapon its self is not really what I am concerned with. I figured out the the basic effectiveness of the weapon its self at 100 feet off the ground is a maximum burst of 4.6 seconds due to ammo capacity of each weapon which can cover 2,000 feet laterally. This assuming the aircraft is moving at 379mph.
My question here is specifically to the bullet its self. Since there is no minimum effective range for a bullet, I am trying to figure out the maximum effective range of the bullet fired straight down. So at what point does the bullet slow from a supersonic state to terminal velocity. (While it can be argued that some bullets even at terminal velocity can be deadly, in this instance we are going to assume that 5.5 grams is not deadly even at terminal velocity) As i stated before, this is more difficult for me to figure out due to the fact that it is not being affected by the downward pull of gravity like a bullet fired on a level plain is. While what CWatters mentioned about the ballistics of the 9mm hold true when the bullet is fired latterly, the maximum effective range will be increased due to the lack of gravitational pull acting upon it in a negative way. As far as the air resistance coefficient, I am trying to keep everything as simple as possible in order to get a more more definitive answer and have a baseline. Obviously the more factors you throw in. Horizontal windspeed, temperature, etc... the more complicated it can get. So for now the most simple is what I am trying to figure out. 



#8
Feb613, 10:33 AM

P: 2,861

The forces acting on the bullet will be drag (in the upward direction) and gravity (in the downward direction). These forces can be added together to find the net force on the bullet. 



#9
Feb613, 11:02 AM

P: 2,861

I know the following is incorrect but for the moment...
Lets assume the bullet was is subject to constant deceleration. If you know the initial velocity (U), the final velocity (V) and the distance (S) from that table I posted you could use.. V^{2} = U^{2} +2as .......................(1) to work out the deceleration (a) due to air resistance when fired horizontally.. For example for the 9mm round Initial Velocity 1120 ft/s = 341m/s Distance 1.36 mile = 2189m Impact Velocity 350 ft/s = 107m/s From equation 1... a = (V^{2}  U^{2})/2s a = 24m/s^{2} So that's the typical deceleration when fired horizontally. So the net acceleration when fired vertically downwards is going to be something like.. gravity  drag... 9.8  24 = 14m/s^{2} You can then put that back into eqn 1 to work out the corresponding height for any final velocity you want. You can also use it to find the approx height at which it reaches terminal velocity. HOWEVER this approach is unlikely to give accurate results because air resistance is not constant, it's typically proportional to velocity squared so the deceleration won't be constant and equation 1 is invalid. In short you need someone who knows the aerodynamics of bullets to write you the equivalent equation. This might be complicated if the initial valocity is close to or above the speed of sound? 



#10
Feb713, 06:55 AM

P: 1,440

It would be effective at a greater range fired vertically down than it would fired horizontally. The air resistance wouldn't change between the two cases but the direction of the gravitational force would.
At any rate, to determine this, you will need to take into account the muzzle velocity of the bullet (and at what altitude the firing occurred) and the shape of the bullet. Assuming the bullet is never above about Mach 0.8 it would be fairly straightforward after that assuming that you could make a reasonable approximation of the drag coefficient. If you get much faster than that you end up in the transonic and possibly supersonic regimes during flight, adding quite a bit of complexity to the problem since the drag coefficient is no longer constant. 


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