Collision when a falling tank impacts the surface of a lake

[Distern]

I'm writing a paper on the movie realism of the A-team movie. The basic situation is that the guys are falling inside a tank with terminal velocity into a lake. I'm stumped on how to calculate the force that is created on impact, though I imagined it being done with some kind of pulse equation. I'm not too good at physics but have already come to some conclusions about the problem at hand.

The terminal velocity reached by the tank is: 278.3 m/s
Mass of the tank with passengers: 19’614 kg
Frontal projected surface: 6.76 m2
Density of air: 1,225 kg/m3
The density of water: 998 kg/m3
The drag coefficient: 0.6 (approximated by its wedgelike shape)

You don't have to solve it for me, I just would appreciate some guidance in which way I should tackle this problem. Many thanks.

 

[berkeman]

Welcome to PF.

What kind of tank? A cylindrical flat-bottom tank? Or more like a Sherman Tank falling vertically? How are the people supported inside? Are you trying to calculate if the team has a chance of surviving? Do you have a clip of this part of the movie to help us with visuals? Thanks.

 

[Distern]

Welcome to PF.

What kind of tank? A cylindrical flat-bottom tank? Or more like a Sherman Tank falling vertically? How are the people supported inside? Are you trying to calculate if the team has a chance of surviving? Do you have a clip of this part of the movie to help us with visuals? Thanks.

here is a link to the relevant scene. But my question purely bases on how much force the collision generates. With the inclusion of the A-Team, I simply assume that they all have taken a downward facing position, in which the most force toleratable is equal to 45 G's. And sorry for not specifying what kind of "tank" is in question (yeah it's a fighting tank, an XM8 Bruford to be exact).

 

[jrmichler]

When the tank contacts the water, there will be a deceleration force. That force peaks at the point where the tank is penetrated to its maximum cross sectional area (about half way). The drag equation (search the term) should give a good estimate of the deceleration force from the water. That force, divided by the mass, will give you a deceleration. After that point, you need a numerical solution because the force changes as the velocity decreases.

The tank enters the water in a vertical orientation. It will probably change to a horizontal orientation after the water rips the parachute off. The change in orientation will cause the drag area and drag coefficient to change.

All of this can be done using a spreadsheet.

 

[Distern]

Thank you very much for your response! I have done as you said and have come up with a good result. But I wish now to put this into context and compare it to a deceleration hitting the ground (concrete for example). Assuming the two surfaces are not elastic, how can I determine an impact deceleration if I don't have the time given in which the collision happens?

 

[jbriggs444]

how can I determine an impact deceleration if I don't have the time given in which the collision happens?

@jrmichler has given you the calculation. You compute force and divide by mass.

There is a short-cut, however. The viscosity of water is 50 times that of air. The density of water is 830 times that of air. Generally speaking, for high speeds and large objects, the density matters more while for low speeds and small objects the viscosity matters more. When you have turbulent flow, density matters. When you have laminar flow, viscosity dominates. Which kind of flow you have can be determined based on the Reynolds number.

We know that since the tank is at terminal velocity, air resistance is one g. Without calculating the Reynolds number we can immediately say that the water resistance will start somewhere between 50 g's and 830 g's.

Take the low figure: 50 g's. At 278 m/s initial and 9.8 m/s^2 times 50 g's deceleration, it would take about 1/2 of a second to come to a stop. (Actually longer than that since the force will be decreasing during the impact).

For a properly prepared human, this might be survivable. Check out John Stapp.

However, my guess is that for an object as big as a main battle tank and an impact velocity of 278 m/s that it is the density ratio that comes into play. 830 g's is not survivable. Water feels pretty darned hard when you hit it at speed.

 

[Distern]

@jrmichler has given you the calculation. You compute force and divide by mass.

There is a short-cut, however. The viscosity of water is 50 times that of air. The density of water is 830 times that of air. Generally speaking, for high speeds and large objects, the density matters more while for low speeds and small objects the viscosity matters more. When you have turbulent flow, density matters. When you have laminar flow, viscosity dominates. Which kind of flow you have can be determined based on the Reynolds number.

We know that since the tank is at terminal velocity, air resistance is one g. Without calculating the Reynolds number we can immediately say that the water resistance will start somewhere between 50 g's and 830 g's.

Take the low figure: 50 g's. At 278 m/s initial and 9.8 m/s^2 times 50 g's deceleration, it would take about 1/2 of a second to come to a stop. (Actually longer than that since the force will be decreasing during the impact).

For a properly prepared human, this might be survivable. Check out John Stapp.

However, my guess is that for an object as big as a main battle tank and an impact velocity of 278 m/s that it is the density ratio that comes into play. 830 g's is not survivable. Water feels pretty darned hard when you hit it at speed.

Yes, I have already realized these calculations through an Excel sheet. Though I have failed to mention that the parachute actually presents a huge factor to slowing down the tank to a speed of 43m/s.

Here the Graph of the collision is shown. The x-axis shows the time, while the y-axis is just fo reference. The yellow line is the velocity of the tank and orange the acceleration. Feel free to leave some feedback!
But my previous question was different. I want now to determine the deceleration of the tank colliding on a hard, inplastic ground. How can I tacle this without knowing the impact time, or at least estimate it. Thx!

 

[jrmichler]

For some ideas about falling objects, search earth penetrating weapon GBU-37. Look closely at the velocity, mass, frontal area, and penetration depth for both Earth and concrete for different bombs. Then make the simplifying assumption that penetration depth is proportional to sectional density (total mass divided by frontal area). Further assume that the deceleration is constant until it comes to a stop.

Now you have all information needed to estimate the deceleration, and thus the force. If the tank is strong enough to withstand that force, you are done. If not, the tank collapses into a wad of crushed metal.

This approach is crude, not very accurate, but can be done easily and should be good enough for your purposes.