Drive band dimensions

[mancpal]

Hopefully a quick question.

Could anybody tell me the dimensions of the drive band on an eighteen pounder please?

 

Simon

[Yves]

Bonjour,

Hope it can help.

Don't speak very well to help more.

Sorry

Yves

Edited 4 June , 2020 by Yves
Drive band for shell star 18-pr

[MikB]

Width of the driving band : 0.885 - 0.915" (~22,48 - 23,24 mm). Not clear whether this includes 0.025" x 2 (0,635mm x2) retaining undercuts.

[mancpal]

Thank you both for your replies, just what I was after.

 

Simon

[mancpal]

Rather than start a new thread could I ask if the drive bands were copper or an an alloy? I also wonder if this changed during the course of the war as materials ran short?

 

Simon

[MikB]

There may be exceptions, but I think plain copper was favourite for good function, easy working and availability. Most alloys are harder, brittler and require metals that are less abundant and harder to run through a blockade. Germany's copper supply got difficult, but I believe they put more effort into substitution on small arms ammunition than artillery.

[mancpal]

Thanks for your reply. It confirms my suspicions as I'm sure I've read they were copper but I wondered  in the latter years if there was still sufficient material available.

 

Simon

Edited 11 June , 2020 by mancpal

[Rod Burgess]

The drawings of the driving band groove that Yves posted come from the 1915 Treatise on Ammunition (available online  - just google it). They are generic, rather than specificto the 18-Pdr. The groove for the 18-Pdr shells was the same for all types (HE, shrapnel, chemical, star etc) and had only two waved ribs, not four.

MikB is correct: Almost all British driving bands in WW1 were copper. It wasn't perfect but it was the best compromise available.

 

The following comes from a write-up I've done for an 18-Pdr shrapnel round in my brother-in-law's possession.

 

A groove to accommodate the driving band was machined into the shell body near its base, which position gave the greatest accuracy of shooting. However, it had to be far enough forward that sufficient metal was behind the band to withstand the immense force imparted when the band took the rifling and which tended to tear the base off the shell. The driving band must be firmly attached to the shell body so that it could not be torn out of its groove when it took the rifling and this was achieved by undercutting the groove.

Two waved ribs were machined into the bottom of the groove to secure the driving band both fore-and-aft and to prevent it rotating on the shell. Chisel cuts across the ribs allowed the air trapped between them to escape when the band was pressed into place.

Into the groove was fitted a driving band, larger in external diameter than the bore of the gun. When the gun is fired, the soft metal of the driving band is compressed into the grooves of the rifling and cut into by the lands, overflowing to some extent towards the projectile’s base and transferring the turn of the rifling to the body of the shell, causing it to spin. The metal of the driving band should ideally be soft enough not to throw an excessive strain on the base of the projectile when it takes the rifling and to take the rifling readily, but not so soft as to strip; it should have a high melting point and should neither cause smoke nor leave a deposit in the bore. At the time this shell was made, the best available compromise was the use of copper, even though it did vaporise to some extent, causing brown smoke and fouling the bore. Experiments were being conducted at the time with cupro-nickel driving bands but these were not used on 18-Pdr ammunition. The driving band was manufactured by taking a slice from the end of a drawn copper tube of the right diameter which was then annealed and forced into the groove round the shell body by a hydraulic press before being machined to the required shape.

The profile of the driving band was critical to the correct functioning of the projectile. The band was designed to ensure that it completely sealed the windage between the sides of the shell and the inside of the gun’s bore. Its leading edge was so shaped that it prevented over-ramming of the shell into the chamber of the gun. The rear edge was shaped to prevent “fringing” or “fanning”, which is where surplus copper is dragged back by the lands of the rifling as the projectile passes down the bore of the gun and forms a sort of fringe behind the driving band. When the shell leaves the muzzle, this fringe is no longer supported and the pressure of the gas behind it turns it up irregularly at various angles creating asymmetric air drag and deflecting the shell from its true course of flight. There were some fifteen or more different profiles of driving band in use on British projectiles during World War 1.

• 
  Mark I and II 18-Pdr and 13-Pdr shells used “No 13 Special Narrow” bands specifically designed for them. The band had a short, steep leading edge, a parallel mid-section and a rear portion sloped to prevent fringing. At the extreme rear edge was a narrow lip which prevented the shell from being inserted too far into the cartridge case. Behind the band, the shell had a semi-circular cannelure (groove) into which the sides of the case were pressed to hold the two components together. The external diameter of the band was 3.39 inches (86.106 mm).
• Mark III 18-Pdr and 13-Pdr projectiles used an improved “No 14 New-style Special Narrow” band, again specially designed for them. Instead of pressing the sides of the cartridge case into a cannelure behind the driving band, the rear edge of the band was now extended and so shaped that the mouth of the case could be “coned” tightly around it, securing the shell to the case. The external diameter was unchanged at 3.39 inches (86.106 mm).

As soon as the driving band was finished it was protected from damage during the rest of the manufacturing process by wrapping a temporary webbing or fabric “grumett” around it.

 

Close-up of the 18-Pdr shrapnel round in the Canadian War Museum. The Mk III shell is fitted with a No14 driving band and the rim of the Mk II cartridge case is ”coned” tightly onto the tapered rear section of the band.

 

 

Edited 20 September , 2020 by Rod Burgess
Formatting error

[MikB]

That's a good explanation, but I'm a bit unsure of: "...sufficient metal was behind the band to withstand the immense force imparted when the band took the rifling and which tended to tear the base off the shell. " 

 

Since the propellant gas pressure would be acting across the whole base of the shell and the exposed bits of the rear face of the driving band, the only way I can visualise the base tearing off is through largely inertial plus possibly frictive resistance to the sudden torque imposed by entering the rifling, perhaps causing failure at the change of section at the pusher disc seating.

 

I'd think that over-ramming of the shell in fixed ammunition would be prevented by ensuring there was freebore ahead of the driving band before entering the leed, with allowance for the greatest permissible accumulation of tolerances over the multiple components involved. Certainly it looks that way in footage of such guns in action - there's no obvious extra force required to close breeches and compress drive bands.

[Rod Burgess]

From the 1915 Treatise (p150):

 

 

The driving band should be as near the base of the projectile as possible, it being generally found that the more rearward position of the band gives the most accurate shooting. In practice, however, this is limited by the thickness of metal behind the band which is necessary to support the great strain thrown upon the shell when the band is forced through the grooves tending to tear off the base.

 

The gas pressure acting on the larger area of the projectile base exerts several orders of magnitude greater force on the projectile than is applied to the smaller area of the driving band. The inner edge of the driving band is very definitely being pushed forward by the gasses acting on the projectile base but the outer edge is being retarded by its contact with the bore and more so by the action of the rifling (hence the tendency to fringing). There is therefore a huge shear force within the thickness of the driving band and this is sustained by the stiffness of the metal and by the shape of the undercut groove which prevents the band just ripping off the tail of the projectile. That force is transferred by the ribs in the groove and by the rear undercut to the outer layers of the shell wall. Thus, there is a compressive longitudinal load on the shell wall behind the band and a longitudinal tensile load ahead of it. The effect of these loads is dramatically worded as a tendency to try to tear the base off the shell, despite the pressure of many tons per square inch of gas trying to push it up the bore.

 

On over-ramming...

The shell is rammed as far as it’s going to be before closing the breech. The breech block is not designed to impart any forward force on the base of a round when being closed. With a QF round, the case cannot be over-rammed because of its rim which fits exactly into a rebate in the face of the breech. In the case of a fixed round, that positions the projectile precisely fore-and-aft; with separate loading QF the last thing you want is to damage the leading edge of the case by smacking it into the base of a projectile which has not been properly rammed first. Separate-loading QF and BL projectiles are forcibly rammed to the point where the driving band just bites on the rifling before loading the cartridge. (On the M109 155mm, which I know best, one man could easily push the 98 lb shell off the loading tray and into the breech, but it took two or three, giving a mighty, co-ordinated heave to hand ram it into the correct position for firing.) Again, the Treatise has words on the subject on p152:

 

 

The first portion of the band should be so shaped as to prevent “over ramming” of the projectile.

 

Clearly, the leading edge of the driving band is only really a factor in preventing over-ramming in separate-loading ammunition, because the case rim does the job for fixed rounds.

Edited 20 September , 2020 by Rod Burgess
Another flipping formatting error!

[MikB]

I'm struggling to understand the longitudinal tensile load ahead of the driving band. There are of course local tensile loads within the band as about half of the circumference is engraved by the rifling lands, and the copper flashing from this is presumably pushed into the reverse taper behind the max diameter and ahead of the case mouth stop flange.

But the shell is accelerating up the barrel at an average of a bit over 6000g - it'll be much more at the start and much less near the muzzle - so it's hard to see how there could be  tensile loads within the shell that could bear significant comparison with the compressive. 

[Rod Burgess]

You're right. And I suspect you're probably far more of an engineer than I am.

Please correct me if I'm getting my physics and engineering wrong but, the way I reason it out is as follows:

 

a) The only reason there's any force, load stress or strain is because the base of the shell (and all rearward-facing areas) are exposed to the pressure of the propellant gasses, causing forward acceleration of the projectile;

b) At the outer periphery of the driving band, there is friction and resistance to that forward movement caused by contact with the bore and engagement in the rifling;

c) Taken together, these opposing forces create a shear force in the thickness of the driving band;

d) Thus, while the disc of the projectile base is being forced forwards, the waved ribs and rear undercut of the driving band groove experience a force towards the rear of the projectile. These opposing forces must, presumably create a compressive load in the wall of the projectile aft of the driving band;

e) In fact and more precisely, the force in the rear part of the projectile wall must also be a shear force, with the inner part of the wall thickness being forced forwards by the base and the outer part being forced rearwards by the load transmitted through the ribs and undercut;

f) If I've got that right, you're correct and I was wrong: there's not a simple tensile load in front of the driving band - there's a shear force within the rear wall which is trying to rip the outer layers of the aft portion of the wall clean off the back of the accelerating projectile, while the base, the inner layers of the wall and everything forward of the driving band continue up the bore.

 

If my reasoning is sound, then the Treatise is over-simplifying, though the net effect is still effectively the same - if insufficient thickness and strength exists in the portion of the projectile wall behind the driving band groove, the tendency would be to rip the back end of the shell apart.

Does that make sense, or am I getting things wrong?

 

Incidentally, I've never actually calculated the mean acceleration of a typical projectile. 9000g would certainly make your eyes water a little, wouldn't it? I've also never really worked out the deceleration in flight, have you? What sort of velocity would, say, and 18-Pdr shrapnel shell have at the point of burst at a typical range of say 4000 or 5000 yards? I ask because I know that the balls were considered to need a striking energy of around 60 ft.lbf which, at 41 balls to the lb, I calculate to equate a striking velocity of around 281 ft/s even 300 yards or so from the point of burst. (Don't you just love imperial units?) I've probably got my sums wrong, anyway. It's probably a good thing I wasn't an armaments designer!

Edited 21 September , 2020 by Rod Burgess
Typo

[Rod Burgess]

We've wandered a little from topic ("measurements of the driving band") but I've just done the calculation of mean acceleration in the bore for an 18-Pdr. I get 6055.85g. Double-check my working for me, please.

 

mv = 492 m/s (1615 ft/s)

Thus mean velocity in the bore = 492/2 = 246 m/s

Length of rifling = 2.038 m (24.32 calibres or 80.232 in)

Time in the bore = Length / velocity = 2.038 / 246 = 8.28 x 10^-3 s

Acceleration = Change in velocity / time = 492 / 8.28 x 10^-3 = 59387.63 m/s²

1 g = 9.80665 m/s²

Acceleration = 59387.63 / 9.80665 = 6055.85 g

 

I've also done "rate of spin" at the muzzle.

Calibre = 83.8 mm (3.3 in)

Rifling is 1 turn in 30 calibres (ie in 0.083 x 30 = 2.49 m) (uniform turn, rather than increasing as in some equipments)

Length of rifling = 2.038 m

Turns in the bore = 2.038 / 2.49 = 0.82 turns

Spin = 0.82 turns in 8.28 x10^-3 s = 99 r/s = 5942 r/min  (NB this value is wrong: it should be twice this - see below)

 

If I've got that spin correct, that's some torque applied through the driving band to the waved ribs in the projectile groove. The total loads applied in that area of the shell wall, both longitudinal shear and torsional, are pretty impressive. No wonder they didn't make shells out of plasticene.

Edited 21 September , 2020 by Rod Burgess

[MikB]

I got 6077.68g in Imperial with g = 32.17 ft./sec.^2.

 

Rifling twist is 1 turn in 99 in. or 8.25 ft.

MV = 1615 ft./sec. so 1615/8.25 = 195.75r turns/sec, or 11745.45 rpm.

 

Peripheral velocity will be (3.142 x 3.3)/12 x 195.76 = ~169 ft./sec., also achieved in 0.00827 sec. (time in bore), so average torsional acceleration will be about 636g.

 

It makes my brain hurt to think about how to calculate the torsional stress on the shell body and compare it with the strength of the material, but there's the change in section in the body where the shrapnel pusher disc sits in your sectional photo, and my guess would be that's where it might be vulnerable. Don't know whether there'd been history of broken shells in early trials, but it's probably academic whether you decide to talk about bases being torn off shells, or shell bodies being torn off bases...

[Rod Burgess]

Funnily enough, I got 12,000 rpm in an earlier calculation but thought I'd done something wrong, so I started again using metric units with a different calculation method. I can see where I went wrong above. You're right: spin is 11,884 rpm using my metric values as rounded above. I'd say our two sets of calculations agree pretty well.

Now for the shell's residual velocity at say 4500 yards...

Edited 21 September , 2020 by Rod Burgess
I must learn to spell.