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Question about Firing tank shell in freefall 
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#1
Feb2212, 11:52 AM

P: 10

Im new to these forums. Hope i posted this in the right section.
I watched the movie "The ateam". There is a scene where there is tank under free fall, and they fire the canon sideways to move the tank whilst free fall. I wish to calculate if this is possible or if possible, how much exact distance will it move. Ive tried a lot of methods, Velocity of the tank falling in air, velocity of canon and there for the impact force.. Ive tried Getting the force of the tank falling and force applied by canon on tank.. Or do i just need to calculate the recoil force? Pls help me, im all confused. I want to know if a tank falling in air moves more due to canon recoil than in air and what formulae can i use. Thanks a lot :) 


#2
Feb2212, 11:58 AM

P: 2,992

You need the mass of the tank, the mass of the shell and the velocity of the shell but I think you're going to find that this is a problem for the show mythbusters. I'd look at it more as a momentum problem where horizontally you're conserving momentum so if the shell has a momentum of mshell * vshell then it should be equals to mtank * vtank. then solve for vtank and then use time to determine how far it moved horizontally



#3
Feb2212, 12:14 PM

P: 2,992

tank info:
http://www.militarytoday.com/tanks/m1a1_abrams.htm tank mass: 57 tons = 51 709.5302 kilograms http://www.inetres.com/gp/military/cv/weapon/M256.html shell mass: 41.2 lb =18.7 kg Nominal Velocity: 5,510 ft/sec (1,679 m/s) So I get 0.607185946 m/s for the tank : (((18.7 * 1 679) / 51 709.5302) * 3 600) / 1.609344E3 = 1.35823628 miles/hour a person can walk about 3 mi/hour then you have the air resistance and the fact that they are floating down by parachute not really free fall. 


#4
Feb2212, 12:27 PM

P: 10

Question about Firing tank shell in freefall
Alright I see I've to use conservation of momentum. So now the velocity obtained is the velocity the tank should start moving at?
If I consider air resistance, howwould I go about calculating? I can assume a coefficient of drag around 1.1 appx rt.. Another question I have is, when the canon is fired Nd the tank moves, doesn't it move in the direction of the two resultant forces, one the direction of falling of the tank wih f=m*g and the other the force applied by the canon? 


#5
Feb2212, 01:05 PM

P: 2,992

think of this as a spaceman floating in space and he throws a ball. Conservation of momentum says that the momentum before must equal the momentum afterwards.
so tank and shell are falling together with zero horizontal movement then the shell is thrown in one direction and the tank moves in the opposite direction. For this kind of problem you definitely need to consider air resistance especially since the tank is so massive. You also need to consider the parachute too which provides tremendous drag and makes the system kind of like a pendulum which means the tank will swing back and forth after the shell is fired. 


#6
Feb2212, 08:16 PM

P: 10

Alright I'm getting the concepts. So if i calculate the drag force, it should be f=Cd * .5 * rho * A * V^2 rt.
If i were to Ssume the tank in free fall, can u help me calculate the first part, the tank firing the shell sideways in x direction while falling in y direction? Thanks a lot 


#7
Feb2212, 09:24 PM

P: 2,992

Is this for a course?



#8
Feb2312, 01:30 AM

P: 10

Nope. Personal pleasure :)



#9
Feb2312, 03:31 AM

P: 10

I want to ignore the parachute. Just a tank falling in free air



#10
Feb2312, 03:50 AM

P: 3

Use conservation of linear momentum and keep in mind that the centre of mass of the system still falls along the same path before the tank fires the shell and after it fires the shell. Once you find the velocity of the tank (or the shell, whichever one you want), it becomes a fairly easy projectile motion problem for the tank or the shell.... this is all assuming no air resistance though.



#11
Feb2312, 04:10 AM

HW Helper
PF Gold
P: 1,961

If the tank is at a height h and falling with speed s when the cannon is fired, then it will touch the floor after t seconds.
st + (1/2)gt^{2} = h Calling the mass of the cannon ball m and the mass of the tank is M, we can find the sideways speed of the tank v (after cannon is fired, speed V) from conservation of momentum v = (M/m)V The sideways distance it will travel by the time it touches the floor is vt. 


#12
Feb2312, 11:34 AM

P: 10

ive done all those calculations. i got the idea of projectile motion today while sitting in my college bus :)
I have done some calculations of the projectile motion assuming a horizontal launch with height = height of tank from the ground. (appx 304mt or 10,000ft) Now i want to add air resistance to this projectile motion. Can you explain how i can use airresistance? Do i have to split my motion in each component and calculate air resistance in each direction, (x and y) and then subtract and then find the resultant? OR is there any other way? And is drag force = C * V^2? 


#13
Feb2312, 11:49 AM

P: 2,992

yes you'd have to consider x and y direction as wind resistance is against the direction of motion. It usually related to the square of the velocity. Normally objects have a terminal velocity of 400mph but for a tank I dont think that would apply.
In the end at 10,000ft, it would take 25 secs to hit the ground: 10,000 = 1/2 g t^2 = (0.5)*(32) t^2 and solving for t ==> 25 secs and the horizontal motion is estimated to be about 2ft/sec or 50ft and here's a NASA reference with more details on calculating drag: http://www.grc.nasa.gov/WWW/k12/airplane/termv.html 


#14
Feb2612, 07:00 PM

P: 14

hmm this is interesting..in my reasoning ignoring air resistance the shell is also in free fall before its shot. then when shot the bullet causes a reaction and pushes the tank back. the reachtion is equal to the action of shooting the bullet..if you know the initial velocity of the bullet when it travel a distance d to exit the tanks cannon youll knoe the acceleration a=(vf^2vi^2)/2d.....vi=0. and with the mass of the bullet youll the the force on it f=ma and thus the reaction and apply that force to the mass of the tankk and youll get he acceleration of the tank in opposite direction probably small number.
now my question .. if the canon is horizontal and we ignore earths curvature the bullet and tank will land at the sae time..right? projectile motion and freefall..but if the tank shots hard enough the projectile will become a satellite and the tank will fall right? but too hard and the bullet will escape earths gravity and wonder in space forever..can the tank and projectile become satellites at the same time? i think that if the tank shoots hard enough to become asatellite then the bullet (which i assume is much lighter than the tank) must reach escape speed and wonder forever..but hey noe theres a tank satellite !!!!hahahahahaha! imagine looking up and seeing a tank flying uncontrollably across the sky hahaha. then my question would be "dude is that a tank?" .....good movie 


#15
Mar2012, 12:41 PM

P: 10

Ok ive made a lil progress in a similar situation.. ive been trying to study projectile motion with air drag..So far i have my equations with a linear model of drag. Now i need some help.
If i take a quadratic model of drag, ie f [itex]\propto[/itex] v^{2} and in two dimensions, i get the equations from various sources as f_{x} = Cd x [itex]\leftV\right[/itex]V_{x} where [itex]\leftV\right[/itex] = √(V_{x})^{2}+(V_{y})^{2} now how do i solve the equations for each direction as there is a Vy term in the x direction and Vx terms in y directions.. 


#16
Mar2012, 12:56 PM

P: 477

There is another aspect to this, unless the line of the barrel goes through the centre of mass of the tank you are going to produce torque about one or possibly all three axis of rotation causing the tank to start to spin, rotate or tumble.



#17
Mar2312, 11:25 AM

P: 10




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