GERMAN 5.9 SHELL - KILLING RANGE?

[Guest]

  Could any ordnance  expert advise? I have a local casualty  where one obituary (his school) says that he was killed by a German 5.9" shell while whistling the school hymn-and that the shell landed almost at his feet. It struck me this might be an exaggeration as -is this right-anyone likely to have been close enough to hear him whistling  would also have been killed by the blast.

 

(I also assume that when a report-as with others- says that a shell landed"at his feet"(or similar), that this is a polite way of saying the man was blown to bits.

 

       What was the effective killing and blast range of this shell?- I regret Mr. Wikipedia has failed me on this.

 

(The  man was Captain A.J..Waugh, RAMC, attached 1st Bn. North Staffords,)

Edited 23 January , 2018 by Guest

[Chasemuseum]

Artillery projectiles are funny things when it comes to lethality. There are two main types of shell in WW1; shrapnel and HE (I am choosing to ignore canister, smoke, gas, illumination and stars).

A shrapnel shell is like a glorified mobile shotgun. The shell is a steel canister with a time fuse and a small black powder charge in a can inside at the base. The bulk of the shell is filled with lead balls in resin. On firing from the gun, inertia fires a cap that starts the fuse train running. If range, elevation and time of flight have all been calculated correctly, the fuse burns through when the shell is  in the air about 10m to 30m above the target, it flashes down to the charge in the base which fires the shrapnel balls out of the case like a shotgun. The target area is then subject to a lethal spray of approx. 10mm diameter lead balls.

 

If mis-calculated the shell will crash into the ground before this happens. The percussion feature of the fuse will fire, there will be quite a small bang. Balls and the shell body will be thrown about.

 

In the first case, the combination of a trench and a steel helmet will give very good protection against injury. To be caught out in the open is very bad news.  For the latter, it is possible to be quite close to where the shell hits and suffer no injury other than an urgent need for clean underwear. For the 77mm “whizz bang”, being fired from a field gun (C96nA or M16) it is likely to have a relatively flat trajectory, so the air-burst is most likely. A fuse setting/ranging error is likely to see the shell firing before or after the target. A “5.9” (Jack Johnson) is a howitzer. A fuse setting/ranging error is likely to result in either the shell bursting so far above the ground that the shrapnel is harmless or the shell crashing into the ground.

 

A HE shell is just that. A thick steel bodied shell filled with high explosive. Usually fuzed for impact, or impact with a short time delay. On impact, the HE detonates and rips the steel body apart. If the shell fires on impact but has just buried itself, most of the blast affect is directed upwards. A man standing 10 to 15m may be clear and safe from serious injury. (likely to still be knocked over by blast affect and may be injured/killed by falling debris). If the fuse is set for delay, it is likely to penetrate from 2m to 6m before detonating (ground conditions are an important factor). Most of the blast effect is directed immediately upwards, again a man standing 10 to 15m away is relatively safe. As the blast effect moves through the ground, it collapses trenches and dugouts. Men taking shelter in these are liable to be buried alive. For deep dugouts, it may only collapse the access passages, but if all of the access passages to a deep dugout can be sealed this can lead to the deaths of the trapped soldiers. The “daisy cutter” fuse is the “super quick” instantaneous fuze for an HE shell. If this can burst the HE shell before it buries itself, anyone on the surface is at great risk of being hit by fragments of shell body. For a 5.9 shell a fragment can easily be lethal at 150m. The counterpoint is that WW1 HE shells fragmented very inconsistently. It was possible to be relatively close to the point of detonation and not be struck by any fragments while men at a considerable distance away were cut in half.

 

What does all this mean. The soldier in the story was probably killed by a 5.9inch howitzer shrapnel shell that burst in the ground near him. Other soldiers not far away are likely to have been uninjured and able to be good witnesses to the event. Had it have been an HE shell with a super quick fuze – a surviving witness who could have named the unfortunate whistler’s tune is improbable.

 

I should also say, that soldiers on the Western Front became very adept at identifying different types of shell by how the detonated. Later in the war, officers were actually required to report the numbers and types of shells landing in their sector. This data being used to evaluate enemy artillery capabilities so they could be targeted for counter battery fire.

Cheers

Ross

[MikB]

A 5.9 or 6" HE shell weighs around 100lb. If the witness was positioned somewhere close, but lower than the victim, it sounds just possible. But embroidery seems very likely - possibly originating in a letter to relatives (or even to the school?) from his superior. I think you'd need other evidence to be sure.

[Chasemuseum]

The black powder charge in the 5.9 is only about 200gm (measured by calibrated eyeball and compared to a tin of powder used for target shooting). Buried in the ground and causing the shrapnel shell to burst it is not that big a bang. It is still several times the black powder charge found in most WW1 hand grenades, but they are bursting on the ground, not buried.

 

For this scenario, the story is credible. For the shrapnel shell air bursting at the correct height. - the idea of the witness is difficult to believe but not impossible. For an HE shell, the story is difficult to believe and for HE with a daisy cutter (super quick) fuze. I cannot accept the story.

 

Witness stories are often embroidered. For assessment; who were the witnesses (one or several), what was their relationship to the victim, how long after the event was the statement made, who was the statement made to, what where their likely motivations in making the statement?

 

Attached is a witness casualty statement to a Red Cross official by a soldier in a rear area regarding a casualty that had occurred nearly a year previously where there had been confusion regarding the casualty's fate.  The witness was not related to the casualty, they did shared a common surname.  My point is that here is a witness statement where the witness records a wounded soldier asking for his comrades to kill him. Something that is easy to believe but is very rarely document in any reliable fashion, as suicide was such a controversial subject at that period. In this case, the witness is from the soldier's company, it is part of a statement confirming the casualties probable death (as a hopeless case), made effectively as a confidential statement to a Red Cross official assembling data for incomplete casualty files. The data was not released to the family and was not released until after 2000.  Accordingly I choose to have confidence in this statement even though it is not supported by other statements.

Cheers

RT 

 

[Guest]

Are you sure about the "shotgun" model for shrapnel? I think it is a misconception that has permeated a fair amount of literature. 

 

The charge is very small relative to the large mass of shrapnel. The shrapnel and fuse have a greater mass than the shell-case and base plate in most HE shells of the period. The charge effectively forces the two apart so the relative motion of the shrapnel + fuse to the baseplate + shell-case are determined by their relative masses (law of conservation of momentum). The key factor is the change in momentum (inertia). In simple terms the greater the mass the greater the resistance to change in momentum. The base plate + shell-case are forced backwards relative to the shrapnel+fuse. Almost all of the forward velocity of the shrapnel comes from the forward velocity of the shell in flight at the time of the detonation. The sideways dispersal of shrapnel comes from the spin of the shell. 

 

The beaten zone from a shrapnel shell is quite complex. MG

 

Edit. Treatise on Ammunition (1915) page 184

 

"ACTION - The explosion of the bursting charge slightly checks the velocity of the bullets and acting through them blows off the base of the shell. The body then slides over them like a glove, the bullets continuing on their course with a velocity slightly reduced."

 

Edited 24 January , 2018 by Guest

[Guest]

 

I would not claim to be an expert (is there such a thing?) but I have some experience of using explosives. It would depend on a host of other factors. The effects of HE in confined spaces (exploding in rooms or near walls or example) are very complex. I would consider myself to e reasonably well read in the war diaries - another source for corroborating evidence.....It is quite possible (and there are plenty of examples in the war diaries) for men to be extremely close to exploding HE shells and be unscathed while men close by are killed. I can think of an example of when an HE shell made a direct hit on a Headquarters (in a room in a derelict house) and not all the men were killed. There is also a larger example of the caves at Soupir when the OBLI and Guards took direct hits while exiting the caves. Many were killed but not all.

 

Separately, there are plenty of examples in war diaries of men killed by shells who had no outward physical evidence of trauma. 

[Chasemuseum]

1 hour ago, QGE said:

Are you sure about the "shotgun" model for shrapnel? I think it is a misconception. 

Hi MG

Very sure. The shotgun effect is to clear the shrapnel from the shell body.

 

You are quite correct, the majority of the destructive forward inertial energy of the shrapnel balls is already present. The charge only needs to accelerate the balls out of the shell body, once out, they already have a lethal velocity/kinetic energy. 

Cheers

 

 

 

[Guest]

1 hour ago, Chasemuseum said:

Hi MG

Very sure. The shotgun effect is to clear the shrapnel from the shell body.

 

You are quite correct, the majority of the destructive forward inertial energy of the shrapnel balls is already present. The charge only needs to accelerate the balls out of the shell body, once out, they already have a lethal velocity/kinetic energy. 

Cheers

 

I am afraid I will have to politely disagree. In the interests of accuracy there is no 'shotgun' effect. it is a myth, or to put it technically, a misunderstanding of some basic physics. The 'bursting charge' of does not shoot the shrapnel forward it simply strips the base plate and shell case backwards (relatively) whilst in flight.. Almost all the forward velocity comes form the velocity of the shell which starts to decelerate as soon as it leave the barrel of the gun. The further from the gun the slower the shell in flight.

 

If you need evidence for the above it is contained in the Treatise on Ammunition 1915 (available via Archive.org) which actually states that the forward velocity of the shrapnel balls is checked and the velocity of the shrapnel is slightly reduced. i.e decelerates (due to the small amount of friction between the casing stripping back and shrapnel) rather than your suggestion that it accelerates. It is the complete opposite of what you suggest. 

 

 See post #5 for the full quote. MG

Edited 24 January , 2018 by Guest

[Neill Gilhooley]

I suspect you are both right.

The action of most shrapnel is described as: ‘The bursting charge… is intended to blow off the head and drive out the bullets, which are carried forward with the remaining velocity of the shell... Certain shrapnel are now fitted with a compressed pellet of powder instead of loose F.G.; this tends to give the bullets additional velocity ’ p.173 Treatise on Ammunition 1915

However for these shells: ‘The Shell, B.L. shrapnel, 4-inch, marks IV, V and VI… have the bursting charge contained in the head.’ p.184

[Chasemuseum]

Sorry to keep disagreeing, but this is basic Newtonian physics. Consider the shell in flight as the frame of reference. When the charge detonates, the fuse and shrapnel bullets are accelerated forwards while the shell body is accelerated in the opposite direction. Energy is conserved. The mass of the fuse and shrapnel are much greater than the shell body and captive pusher plate. As a consequence the rate of acceleration of the body is greater than the fuse and shrapnel.

 

Then consider both groups of components in relation to the earth as the frame of reference. The forces acting on the shell in flight are gravity, wind resistance and the coriolis component of acceleration (as the earth is a translating frame of reference). These forces and the forces of the bursting are the net accelerations acting on the components. Energy is conserved.

Cheers

RT

 

[Dai Bach y Sowldiwr]

6 minutes ago, Chasemuseum said:

Sorry to keep disagreeing, but this is basic Newtonian physics.

 

Am I right in saying that (ignore air, wind, earth rotation for a moment) the trajectory of the centre of mass of the shell is unchanged by the explosion of the fuse and expulsion of the shrapnel,  even though all the components radiate out at varying velocity from a point at the moment of explosion?

[MikB]

29 minutes ago, Dai Bach y Sowldiwr said:

 

Am I right in saying that (ignore air, wind, earth rotation for a moment) the trajectory of the centre of mass of the shell is unchanged by the explosion of the fuse and expulsion of the shrapnel,  even though all the components radiate out at varying velocity from a point at the moment of explosion?

 

If you're right, it seems to me to be purely theoretical, because there'll be no object occupying that centre of mass!

 

It's the rearward momentum of the body relative to the point of shrapnel-charge explosion that is equal to the forward one of the nose and bullets - energy isn't what's conserved in that context, or firing a rifle would give the shooter serious shoulder wounds. Of course, the forward velocity of the bullets is bound to be increased because the discarded body is not massless, but such increase is small relative to the current residual velocity of the shell at the time.

Edited 24 January , 2018 by MikB

[Chasemuseum]

Yes,

If we can ignore the forces of gravity, air resistance, and the earth's rotation. Then conduct an energy balance of the sum of the kinetic energy of all the components of the exploding shell after the explosion in air this will be the same as the kinetic energy immediately prior to the explosion.

Cheers

RT

[Guest]

1 hour ago, Chasemuseum said:

Sorry to keep disagreeing, but this is basic Newtonian physics. Consider the shell in flight as the frame of reference. When the charge detonates, the fuse and shrapnel bullets are accelerated forwards while the shell body is accelerated in the opposite direction. Energy is conserved. The mass of the fuse and shrapnel are much greater than the shell body and captive pusher plate. As a consequence the rate of acceleration of the body is greater than the fuse and shrapnel.

 

Then consider both groups of components in relation to the earth as the frame of reference. The forces acting on the shell in flight are gravity, wind resistance and the coriolis component of acceleration (as the earth is a translating frame of reference). These forces and the forces of the bursting are the net accelerations acting on the components. Energy is conserved.

Cheers

RT

 

 

The bursting charge does exactly what is says on the tin. It's aim is to simply separate the components, not to add forward lethal velocity.  It follows that the amount of gunpowder in the bursting charge would be the minimum required as any excess simply increases the payload for no additional benefit.  Consequently the bursting charge is minuscule relative to the mass of fuze, shrapnel and casing and base plate. All it does is strip the two apart. The lethal force is imparted by the velocity of the shell not the bursting charge..

 

HE is different as the explosive (lyddite or similar) provides the massive amount of chemical energy that when detonated is converted into a similarly massive amount of kinetic energy and heat (and light) breaking the casing into small parts which become the lethal elements. In addition the pressure wave can be lethal at short distances. 

 

The 'shotgun model' for shrapnel shells implies the shrapnel is fired out of the shell at a high velocity  - the implication is that the bursting charge acts like an airborne shotgun  - and it is this that imparts the lethal force. It doesn't. The reason for this is that as soon as the case is stripped away (in a tiny fraction of a second) the residual (small) force of the bursting charge dissipates immediately. At that point everything starts to decelerate (gravity aside) primarily due to air resistance.

 

Contrast this with the barrel of a shotgun. The barrel contains the explosive for for longer - which is why long barrels fire projectiles further -  accelerating the pellets and ejecting them at high velocity. At the point the pellets leave the barrel they start to decelerate. 

 

On your list of accelerators wind resistance is not an accelerator; it is a decelerator. There are tables showing the velocity of a shell in flight and how this decelerates in flight primarily due to the air resistance. While a flat trajectory shell fired from a gun will pick up some acceleration from gravity as it passed the high point of its trajectory it is negligible. 

 

The image below comes from a US manual of its 3" Field Gun. The Muzzle Velocity (MV) was 1,700 ft/s...after 3,000 yards the shell is travelling at 906 ft/s (a 47% reduction from MV) and by 6,000 yards the shell is travelling at 740 ft/s  - or less than half the MV. All of this is due to atmospheric resistance.   Importantly the velocity at 6,000 yards is less than the velocity at 3,000  yards when the shell was at or very close to peak trajectory. Despite the accelerating effects of gravity on the shell, by the time it reaches 6,000 yards its velocity is 18% less. The point is that air resistance more than offsets the additional effects of a gravity on a low trajectory shell. 

 

MG

 

Edit: The deceleration is non-linear and is the reason why the war diaries often refer to shrapnel having little effect as it bursts too high. As you doubtless know kinetic energy of a shrapnel ball is a function of 0.5 x mass x velocity-squared; a shrapnel ball loses 75% of its energy if it loses half its velocity. ...and is why there the diaries have countells examples of men being hit by shrapnel and surviving as the kinetic energy had dissipated due to the deceleration. 

 

 

 

Edited 24 January , 2018 by Guest

[Chasemuseum]

All firearms or artillery, where the projectile is not rocket assisted, will commence decelerating due to air resistance once they leave the muzzle. The calculation of wind resistance is complicated, but can generally be consider as a square function of velocity - that is, as the projectile loses speed, the effect of air resistance becomes less acute. This is why going for ever higher muzzle velocities is of limited value.

 

Similarly all projectiles regardless of the angle of incidence, are falling under the force of gravity from the moment they leave the muzzle.

 

I suggest we terminate this discussion. We interpret models differently and will not agree. We are going way off the topic of the original thread and I doubt that this is of any assistance to guest.

Good night

RT

[Guest]

On 24/01/2018 at 13:23, Chasemuseum said:

All firearms or artillery, where the projectile is not rocket assisted, will commence decelerating due to air resistance once they leave the muzzle. The calculation of wind resistance is complicated, but can generally be consider as a square function of velocity - that is, as the projectile loses speed, the effect of air resistance becomes less acute. This is why going for ever higher muzzle velocities is of limited value.

 

Similarly all projectiles regardless of the angle of incidence, are falling under the force of gravity from the moment they leave the muzzle.

 

I suggest we terminate this discussion. We interpret models differently and will not agree. We are going way off the topic of the original thread and I doubt that this is of any assistance to guest.

Good night

RT

 

Fine. For the record I don't believe that shrapnel is fired out of its shell like a flying shotgun.

 

If as implied  the bursting charge created a 'shotgun effect', it fails to explain why many shrapnel shells had the bursting charge at the front of the shell rather than the rear. It would have the opposite of the desired effect if the theory was corect... . As we know from the Treatise the effect was to very slightly reduce the shrapnel velocity which suggests/confirms the bursting charge had negligible propellent effect and was designed to separate the main components.... just a thought. MG

[MikB]

I'd agree with Chase - I think we've been wandering away from anything that'll help guest. I would also guess that, if the information was that the subject of his research was killed by a five-nine, that it'd be HE, not shrapnel - a witness would be less likely to be able to tell what calibre of shell a shower of shrapnel bullets came from, unless other evidence was to hand.

[Guest]

On 24/01/2018 at 13:56, MikB said:

I'd agree with Chase - I think we've been wandering away from anything that'll help guest. I would also guess that, if the information was that the subject of his research was killed by a five-nine, that it'd be HE, not shrapnel - a witness would be less likely to be able to tell what calibre of shell a shower of shrapnel bullets came from, unless other evidence was to hand.

 

I think the OP question has been answered: it is possible and there is plenty of diary evidence that men could be unscathed having been in very close proximity to exploding HE shells. 

 

Other evidence could be in the shape of fuses which could provide valuable information on fuse settings ( as an indication of range setting) and potentially the type of gun or howitzer for counter-battery work. This was quite common. Separately there were composite ammunition types that mixed HE and shrapnel.  MG