michaeldr Posted 21 February Share Posted 21 February This specifically relates to naval shore bombardment, but I'm hoping that any artillery expert out there can help me; what would be the radius of the 'mortal danger zone' (for want of a better term) from fragments of a 9.2-inch shell (common or lyddite – not shrapnel)? Perhaps also relevant, the ground where the shell fell was rocky with very little soil depth Thanks in advance, Michael Link to comment Share on other sites More sharing options...
Spaceman Posted 21 February Share Posted 21 February If the ground was rocky and the shell did not penetrate very far into the ground before it detonated then the lethal radius could probably have been measured in hundreds of yards. You have to bear in mind that the detonation velocity of Lyddite was over 7,000 m/s imparting considerable velocity to any shell fragments. Link to comment Share on other sites More sharing options...
Chasemuseum Posted 21 February Share Posted 21 February The danger zone is based on how the steel casing of the shell shatters at detonation. As a naval shell, it is heat treated steel to achieve penetration into steel plate. It actually has a complicated structural design. The point is that the design is to optimize penetration against naval armour, with detonation miles clear of friendly forces. So having the casing shatter into large fragments that are potentially lethal a mile away from the detonation point is quite acceptable. At the same time the function is not anti-personnel, so failure to injure personnel near the detonation is also acceptable. In other words it all comes back to probability. You could be potentially standing 100 yards from the shell and not be hit by a fragment, only suffering from blast effect (noise, dust, pressure wave) at the same time another person a mile away could be decapitated by a fragment. Link to comment Share on other sites More sharing options...
michaeldr Posted 21 February Author Share Posted 21 February Spaceman & Chasemuseum, Thank you both very much for the explanations regards, Michael Link to comment Share on other sites More sharing options...
Chasemuseum Posted 22 February Share Posted 22 February Michael, I am just sorry I couldn't give you an answer withy the clarity you wanted. Link to comment Share on other sites More sharing options...
michaeldr Posted 22 February Author Share Posted 22 February No problem Chasemuseum, There's a lot of guess work here, eg: was it a Common or a Lyddite shell? etc, (probably only a visit to see the log at Kew will tell, and even then that's not certain) 22 hours ago, Chasemuseum said: In other words it all comes back to probability. You could be potentially standing 100 yards from the shell and not be hit by a fragment, only suffering from blast effect (noise, dust, pressure wave) at the same time another person a mile away could be decapitated by a fragment. For this layman your above paints a vivid picture, indicating a danger zone of at least some several hundreds of metres Thank you again for your help here Link to comment Share on other sites More sharing options...
MikB Posted 23 February Share Posted 23 February (edited) Here's my shell splinter, picked up 20 years ago near the Ulster Tower on the Somme. I'd thought it was a funny-looking stone until I felt its weight. It weighs 117 grams or about 4.12 oz. When I found it, there were still faint machining marks on a slightly convex surface. I think it came from the base area of the ogive on a shell of around 6" or 15 cm. calibre. Ejected from the site of explosion at, say, approximately the speed of sound and almost certainly spinning, it would have had about 4,800 ft.lb. of kinetic energy - a little over twice that of a .303 and about like an elephant rifle. Edited 23 February by MikB Link to comment Share on other sites More sharing options...
Spaceman Posted 23 February Share Posted 23 February The 9.2 inch naval gun fired a massive 380 lb shell with the common shell being filled with about 30 lb of Lyddite. When that detonated, you had 350 lb of cast steel broken into sizeable pieces and ejected at supersonic speeds. If the shell detonated on rocky ground, a considerable amount of the shrapnel would have travelled horizontally and would have been lethal to anyone standing above ground up to many hundreds of yards away. Even the armour piecing 9.2 inch shell contained about 20 lb of high explosive and would have been just as lethal against personnel. Link to comment Share on other sites More sharing options...
JMB1943 Posted 23 February Share Posted 23 February (edited) MikB, K. E. = 1/2 M[ V * V] where M = mass in lb; V = velocity in f/s; here, M = 4/16 = 0.25 lb and V = speed of sound, 720 mph = 12 miles/min = 1056 f/s So, K. E. = 0.5 [0.125 * (1056 * 1056)] = 69,700 ft-lb. SEE EDIT BELOW to 139,400 ft-lb Similarly, 0.303 bullet (174 grains) at 2440 f/s, (muzzle velocity), K. E. = 1/2[174/7000 (2440 * 2440] = 74,000 ft-lb. One/Both of us needs to check our calculator batteries.....? Regards, JMB EDIT. K E = 0.5 [0.25 * (1056) * (1056)] = 139,400 ft-lb First battery change! Edited 23 February by JMB1943 typo Link to comment Share on other sites More sharing options...
MikB Posted 23 February Share Posted 23 February (edited) JMB, your energy numbers are in ft. poundals, not ft.lb., which latter is the normal way of expressing them. To get from ft. poundals to ft.lb., it's necessary to divide by g, in ft.sec^2, ie 32.174. A well-established formula for calculating the KE of a ballistic projectile in Imperial is: (V^2 in ft./sec x m in grains) / 450436.68 All the sources I've seen that give the muzzle KE of Mk.VII 303 ball come out at 2,300 ft.lb. to the nearest whole number, pretty much on the button for your 74,000/g. Edited 23 February by MikB Link to comment Share on other sites More sharing options...
JMB1943 Posted 23 February Share Posted 23 February Aaaaaargh…… Blame it on the hot Florida sun! I did think that even my numbers were very high, but did not twig it. Sorry for my misplaced “correction.” Regards, JMB Link to comment Share on other sites More sharing options...
MikB Posted 23 February Share Posted 23 February (edited) No prob. Didn't mean to be stuffy with the reply... There'll be a good deal more to trying to understand the residual energy at various distances from ground zero. Weighty projectiles like the splinter I've got will have big angular momentum and sectional densities in the direction of travel that will vary hugely, many times a second. Plus the initial velocity might, as Spaceman says, be much higher. Nevertheless I'd expect velocity loss with distance to be very rapid. Even some firearms rounds that use bullets with poor aerodynamic shapes can lose a big percentage of their energy very quickly - one example that caught my eye is .444 Marlin, which manages to shed 1,000 ft.lb in the first 100 yards. Edited 23 February by MikB Link to comment Share on other sites More sharing options...
JMB1943 Posted 23 February Share Posted 23 February It’s all good! I’m on that road to hell paved with ……misplaced corrections. Best to Just quietly slink away and fall on my sword! Regards, JMB Link to comment Share on other sites More sharing options...
michaeldr Posted 6 March Author Share Posted 6 March (edited) Further research has turned up an interesting film clip. A film maker accompanied the Austro-Hungarian artillery which arrived on Gallipoli in December 1915 and this is a clip from a film which they shot https://filmmirasim.ktb.gov.tr/tr/film/anafartalar-da-itilaf-ordularinin-puskurtulmesi At minute 6.57, you will see the crater which they were shown as the site of the impact of the 9.2-inch shell, a large fragment from which killed their fellow country-woman who was nursing there. Edited 6 March by michaeldr Link to comment Share on other sites More sharing options...
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