Scourge of War Boards

I was very impressed with the previous debates on firing angles etc, and have a similar question to pose to our resident experts.

Yes it's a bit morbid, but purely academic!

Assuming a 12lb'er hits a marching line of men - perfectly parallel, i.e. it will contact each soldier, it isnt dropping or rising, and hasn't hit the ground yet, how many men would it take out before slowing to a non-lethal velocity?

I dont mind what size soldier is chosen, just mention it in the reply.

You can ignore any air resistance, or deceleration as the shot passes trough the gap between marching troops.

Con20or,

I think it would depend on the relative density of each individual as the shot passed through him, what he was wearing, equipment, etc...

I don't think you could ignore deceleration as it passed through the individual.

More importantly, why in the hell did you post this question knowing we'll get treatises on experimental and theoretical physics as well as empirical data?

B

Con20or,

I think it would depend on the relative density of each individual as the shot passed through him, what he was wearing, equipment, etc...

I don't think you could ignore deceleration as it passed through the individual.

More importantly, why in the hell did you post this question knowing we'll get treatises on experimental and theoretical physics as well as empirical data?

B

Lol - looking forward to seeing the empirical data - MP does need spicing up some times but thats going a bit far

I'll clarify this - deceleration between troops is as the shot passes from one soldier to the next... i.e the momentum is the same when it leaves one soldier as when it reaches the next...

Where are MTG and HTS when you need them?


B

born2see wrote:

More importantly, why in the hell did you post this question knowing we'll get treatises on experimental and theoretical physics as well as empirical data?

And indeed you shall. This is actually a straightforward calculation that yields an interesting result. The result is happily confirmed by the historic record.

This thread should actually be titled "Energy Dissipation Of A Sphere Passing Through a Medium." Please change it Con20or. I am quite sure it will garner 10K hits within a week if you do. :laugh:

We'll use the law of conservation of energy to solve this.
The kinetic energy of the sphere is: E = 1/2mv**2
m=mass of sphere, v=velocity of sphere

The cannonball will lose energy as it passes through the medium, (people). That is a messy calculation, no pun intended. So we'll use an approximate solution that is valid for spheres which move slowly through a medium.
The drag force on such a sphere is: f = 1/2Cpav**2
C=drag coefficient, p=density of medium, a=cross sectional area of the sphere

The energy loss due to drag is just: E = fd d=distance sphere travels in the medium
So 1/2Cpadv**2 = 1/2mv**2
Canceling out like terms, we end up with: d = m/Cpa

That's rather remarkable when you think about it. The distance the cannonball travels is independent of the initial velocity. Extra credit will be awarded if you can explain why this is so.

Time for numbers:
m = 5.6kg (12.3 lbs ball)
a = 0.01 M**2 (4.5 in. dia.)
C = 0.1-0.47 depending on the smoothness of the sphere
Human are not much more than bags of salty water.
p = 1027 kg/m**3

Plugging in the numbers we get
d = 0.55/C or d = 1.16 - 5.5 M

Human torsos are about 34cm wide and 26cm thick, (if you significantly exceed these measurements you are part of the obesity epidemic)

So 4 - 21 humans would be killed if the ball strikes head-on and 3 - 16 in enfilade.

The largest number of casualties from solid shot I have seen recorded is 14 at Shiloh and 12 at Gettysburg. Both figures fit nicely into the calculated ranges.

More commonly, the cannonball was descending at some significant rate depending on the range and elevation of the gun relative to the infantry. There was an example cited from GB IIRC where a single cannonball killed four men, the first was hit in the head, the 2nd in the chest, the 3rd in the hips and the fourth lost both legs at the knees. The reaction of the 5th man in that line was not recorded. :sick:

-Jim

very interesting! good work MTG.

That's rather remarkable when you think about it. The distance the cannonball travels is independent of the initial velocity. Extra credit will be awarded if you can explain why this is so.

That is quite remarkable. I am going to venture that it is because both KE and the retardive force vary with the square of velocity.

Even so, I would prefer to get hit by a 12 lbr rolling along the ground 2 mph

Group Captain Mandrake:

I am going to venture that it is because both KE and the retardive force vary with the square of velocity.
Even so, I would prefer to get hit by a 12 lbr rolling along the ground 2 mph

Gutsy, your first post and you venture into a very deep pool. Yes, that's the correct answer. Both the kinetic and the dissipation of that energy are dependent on the same power of the velocity. So if the sphere is moving 1cm/sec or 500M/sec it will travel exactly the same distance when the kinetic energy will drop to 0. The reason that doesn't happen on the battlefield, is that we bags of salty water have a rather thin, but tough membrane surrounding us.

Welcome to the forum.

Huzzah!