Mr. Tittles Posted November 12, 2003 Share Posted November 12, 2003 Lets analyze the US 76mm HE round. I will leave it to others to get out their scientific calculators and check me but I get 673 M/s static fragment velocity. I used 12.9 # total shell weight and 0.9 # HE weight. So little m is 12/2.2 and big M is 0.9/2.2. use D=7507 to make everyone happy. This was a quick round so lets say the forward velocity is 800 M/s. This gives 1045 M/s fragment velocity. I am disregarding shell spin because I dont know it for this weapon. It may add another 100 M/s to the static velocity. and the arctan function gives a forward angle of 40 degrees (!). Lets compare the US 75mm HE vs the 76mm HE against a building. Germans are inside and its a moderatedly strong brick building and the Germans have hastily thrown tables, furniture up against the walls. The 75mm HE strikes the wall at 90 degrees and it has a point detonating fuze. In 50 microseconds or so, it has blown up and spent itself. Where did the fragments go? Mostly sideways at an 80 degree angle. Those fragments that strike the outer walls are doing so at an extreme angle. The main effect it would have on the building would be its blast effect from the HE and the fuze and forward elements of the shell projecting forward. Now the 76mm shell. It also strikes the building and detonates with a PD fuze. Where does its fragments go? Forward into the structure. It literally will spray the interior with a cone of fragments like a directed energy shell of sorts. What can we safely assume (if thats ever a good idea) about the size of the fragments of each shell? The 75mm HE shell would break up into smaller fragments. The greater quantity of metal in the 75mm HE shell would make overall more fragments. Nice small ones that easily kill/maim WWII era men with limited/no body armor. But in this situation, the 76mm HE shell effectiveness would put more mass of fragments, with a better size for piercing buildings, at the intended target. This example is just meant to demonstrate that velocity, angle of attack and fuze can make or break an attack. The much maligned US 76mm HE shell was probably poor at killing a wave of infantry in the open. Even using a delay setting on its fuze to achieve a ricochet would still have an elongated/narrow kill zone. The US TDs with the 76mm were very useful when attacking fortified buildings. [ November 12, 2003, 04:37 PM: Message edited by: Mr. Tittles ] 0 Quote Link to comment Share on other sites More sharing options...
Redwolf Posted November 12, 2003 Share Posted November 12, 2003 Why do you assume the fragments play any signficant role in getting a wall down? 0 Quote Link to comment Share on other sites More sharing options...
Mr. Tittles Posted November 13, 2003 Share Posted November 13, 2003 They play a significant role in penetrating the wall as high velocity fragments with a nice quality. They are hard metal. The fragments and wall debris would fan out into the room in a lethal cone. The combined effect of the HE blast and the narrow focusing of the fragments would tear through a brick wall. They would make a nice hole about a foot across or more though. Several rounds fired across a wall of a house would bring down the wall. The effect is more akin to a high velocity cannister. I recently read where 17 pr. HE had a similar effect. The tank commander remarking how it always 'blew out the back wall of the house'. 0 Quote Link to comment Share on other sites More sharing options...
Redwolf Posted November 13, 2003 Share Posted November 13, 2003 I don't buy that. The tiny fragments hitting the wall have an overall kinetic energy that is insignificant compared to the blast and goes directly off the blast. Or in other words, the effect is the same as if the blast would be alone. That 17pdr observation is definitivly not to be interpreted as shell fragments penetrating both walls at once. With a wooden wall things might be a little different because the structure of the wood may be affected by cutting of sharp splinters, but that is very hard to tell. 0 Quote Link to comment Share on other sites More sharing options...
Mr. Tittles Posted November 13, 2003 Share Posted November 13, 2003 The shell weighed 12.9 pounds. How did you calculate the transformation of 12 pounds of hard metal into 'tiny' fragments? If the 'tiny' fragments are going 1045 M/s ( uh, thats bullet fast?) how tiny are they that they become insignificant? Lets say the shell broke into 100 pieces. Thats 0.12 pounds each. Thats about 1/8 pound of hard metal going 3000 fps? Its a good question and the next challenge. Is there a formula that gives fragment size? In reality, shells will break up into an array of different sized fragments. I will try to find a book that shows a US 75mm that detonated in a sand trap. Sort of an igloo shape. The sand is then sifted and 90+ percent of the metal is recovered. In any case, redwolf is wrong. The metal does not disappear and the conservation laws must be obeyed. Brick walls are not that tough either. they are very susceptible to shock/fracture. An AP round will make a hole much larger than its own diameter in a typical brick wall. I think redwolf is falling into another one of those mental pitfalls. He is basically saying that if we took a 76mm shell and rigged it so it would explode against a wall on command, at zero velocity, the effects would be the same as firing it at the wall with a 76mm gun. This, is clearly, wrong. Or is he saying that we can slip the metal shell off the explosive and its just as effective? [ November 13, 2003, 12:18 AM: Message edited by: Mr. Tittles ] 0 Quote Link to comment Share on other sites More sharing options...
Hurryin Heinz Posted November 13, 2003 Share Posted November 13, 2003 This is my first post, by nature a quiet person, so pleeeeaaaaaaaasssseeee dont kill me. I just want to say that I love this thread and it has been entertaining and wonderfully informative, and that I would love to see it taken as far as it can go. To this end, lately in this thread, just a lil concerned about something that seems to be killing it, mr tittles seems to be overly unfriendly about a rather complex matter, bogging the thread down instead of helping it to move forward. Isn't so much a matter of who is right or who is not, more of a matter of a spectator enjoying a throughly engaging effort by both parties involved, wishing the topic to not suffer a premature death. (just to be clear, love pretty much every post in this, some wonderful efforts and links by all, very much including mr tiddles ) 0 Quote Link to comment Share on other sites More sharing options...
Wol Posted November 13, 2003 Share Posted November 13, 2003 The point about naval shells was simply that that they demonstrate the importance of shell geometry more clearly than any other rounds becuse of the very large range of bursters and shell shape. In the exampoles cites, not only does the british shell have a larger and more powerful burster, but that as it has a dis-proportionally shorter head, the effect on fragmentation is even more pronounced than a simple comparision of charge to weight ratios. Second, that much of the interst in fragmentation effect is driven by the naval ordnance concerns with fragment effects on armour. Fragment effects are also very important for consideration of penetration of light armour. The US spent years working on this starting with Ronald Gurney in 1943, and through project THOR, to the current US FATPEN, and HAZFRAG and Mott-CALE models. Even small 1-20 grain fragments can do a lot of damage if going fast enough. 0 Quote Link to comment Share on other sites More sharing options...
Mr. Tittles Posted November 13, 2003 Share Posted November 13, 2003 http://members.aol.com/Andrewvict/WRHDPRFM.pdf heres good stuff. i save the pdfs to my hard drive myself 0 Quote Link to comment Share on other sites More sharing options...
Mr. Tittles Posted November 13, 2003 Share Posted November 13, 2003 Originally posted by Hurryin Heinz: This is my first post, by nature a quiet person, so pleeeeaaaaaaaasssseeee dont kill me. Weeeeeeeeeeeeeeellllllllllllllllllll, since you did say please, we won't kill you. This time anyway. [ November 13, 2003, 12:15 AM: Message edited by: Mr. Tittles ] 0 Quote Link to comment Share on other sites More sharing options...
Mr. Tittles Posted November 13, 2003 Share Posted November 13, 2003 Mave = 2 B^2 (to/di )^2(to + di)^3 (1+0.5 M/C) = average fragment mass (2) where: C = mass of explosive charge M = total mass of cylindrical section of case B = a function of the explosive and the case material in equation (2), the Gurney-Sarmousakis equation. As a rough approximation, for a mild steel case, B = 338.1/PCJ, with the detonation pressure (PCJ) in kbar. to = case thickness in inches di = case internal diameter (i.e., explosive diameter), inches I think I have enough of this data to do the 75mm HE sherman m48 round. [ November 13, 2003, 12:12 AM: Message edited by: Mr. Tittles ] 0 Quote Link to comment Share on other sites More sharing options...
Mr. Tittles Posted November 13, 2003 Share Posted November 13, 2003 Fragment Velocity The "initial velocity" of fragments released in the detonation of a cased explosive is approximated by the Gurney formulas. Actually, the velocity found in this way is the maximum velocity achieved by the fragments during the acceleration phase, and applies to the expanding warhead case fragments only at distances from the warhead center greater than about twice the warhead initial radius. An initial step increase in the case velocity is imparted by passage of the detonation shock wave through the metal. For cylindrical steel warheads, initial elasticplastic expansion of the case occurs as it expands from the original radius to about 1.2 times the radius. At the end of this phase the case radial velocity is about 60% of the calculated "Gurney velocity." The maximum velocity (95 to 100% of the Gurney velocity) is that achieved at the end of fragment acceleration with the fragments at a radius of about 1.6 to 1.8 times the initial warhead radius. At this time, the detonation products appear in the openings in the fracture and subsequently develop an expanding cloud beyond the fractured warhead case. In the final phase of terminal flight, beyond about 20 times the initial warhead radius, unhindered fragments emerge from the detonationproducts cloud (typically reduced in speed by drag to about 90% of the Gurney velocity). Subsequent fragment velocities are subject to deceleration by continuing drag forces in the surrounding medium. The simplest expression of the Gurney formula for symmetrical configurations is VGurney = {2E / (µ + n/[n + 2])^1/2} (15) where µ = M/C, and M= mass of metal in “warhead case” and C= mass of explosive charge. Ö2E = “Gurney constant” in units of m/s or ft/s. Values of n are 1 for a flat sandwich of explosive between two equivalent flat metal plates, 2 for a cylinder, and 3 for a sphere. and pictures too http://www.asc2002.com/summaries/h/HP-15.pdf Basically, a shell will stretch out initially under the incredible rise in pressure. The walls get thinner. It then cracks along the length of the shell. These long 'sabers' can be further cracked if the explosive is strong enough. The explosive wave will precede the fragments initially. So in the case of redfords brick wall, the blast attacks the structure first. The velocity of the blast is quickly dropped and the velocity of the fragments overtake it. 0 Quote Link to comment Share on other sites More sharing options...
Mr. Tittles Posted November 14, 2003 Share Posted November 14, 2003 Thought experiment: Our two shermans, one with a 75mm and the other with a 76mm, decide to shoot at the adjacent wooden farm building. Its about 60 feet by 60 feet in area. This time the 75mm sherman shoots first but decides to use a delay setting on his HE shell. He fires at the building, the shell strikes the outer wall, the detonation train is started within the shell, the shell was traveling at 1800 fps when it strikes the outer wall (at a 90 degree angle lets say). The shell detonates beyond the building's back wall and kills a bunch of chickens. The 76mm tank commander shakes his head and tells the loader to use superquick again. They fire at the wall and see that the building has an internal flash and smoke/debris flys out the windows. Is it possible for the delay triggered 75mm HE shell to penetrate both walls and explode outside the back of the building? What could possibly be going on with the 76mm SQ HE shell? How could it go off inside the building? 0 Quote Link to comment Share on other sites More sharing options...
Wol Posted November 14, 2003 Share Posted November 14, 2003 How long is a piece of string It depend on the delay, and whether it works as designed. I think most US delays are about .03 seconds, but there is some 25% variation I think, and it depends on how resistent the wall is (probably not very much). 0 Quote Link to comment Share on other sites More sharing options...
Guest Mike Posted November 14, 2003 Share Posted November 14, 2003 0.03 seconds is 54 feet at 1800 fps - of course the velocity would be less once the round hit the wall, but it seems to me that if he's using a delay firing at het wall of a house there's quite a good chance the shell will explode on the far side! 0 Quote Link to comment Share on other sites More sharing options...
Mr. Tittles Posted November 14, 2003 Share Posted November 14, 2003 Shell delay is usually 0.05 seconds or 50 milliseconds. But Wol raises the excellent real-world point about variance. I think 25% would be excessive though. But just how fast is 'Superquick'? Obviously less than 0.05s but obviously nothing is instantaneous either. Superquick would be the sum total of time for mechanical initiation, that is, the time it takes before the actual physical act of touching an object to start the detonation train', to the actual completion of the explosion. The x ray photos to the link above show that the detonation itself only reaches so far in so much time. The process is probably a mechanical system (spring/plunger/etc) followed by a chemical system (or series of chemical systems) to initiate the detonation of the explosive material. Some explosives need more than one booster so the 'superquick' would be longer. If the shell is moving at 2600 fps (800 M/s) and is about a foot long, it would need about 1/2 millisecond to get inside the building. Note that the explosive process could be going on as this forward motion is happening. Note also that detonators are typically on the front of the shell and that is where the detonation process starts from. My gut feeling about the explosion is that it, like most things, takes the path of least resistance. The front of the shell, if it is the weakest, will allow most of the blast (once it is beyond the stretching/cracking phase to escape forwards. Look at the pic of the 45mm HE shell that I had a link to earlier. It has a curious internal shape. Is the thinning of the shell wall towards the front of the shell a designed feature? Just food for thought. [ November 14, 2003, 11:15 AM: Message edited by: Mr. Tittles ] 0 Quote Link to comment Share on other sites More sharing options...
Mr. Tittles Posted November 14, 2003 Share Posted November 14, 2003 Originally posted by Wol: The point about naval shells was simply that that they demonstrate the importance of shell geometry more clearly than any other rounds becuse of the very large range of bursters and shell shape. In the exampoles cites, not only does the british shell have a larger and more powerful burster, but that as it has a dis-proportionally shorter head, the effect on fragmentation is even more pronounced than a simple comparision of charge to weight ratios. Second, that much of the interst in fragmentation effect is driven by the naval ordnance concerns with fragment effects on armour. Fragment effects are also very important for consideration of penetration of light armour. The US spent years working on this starting with Ronald Gurney in 1943, and through project THOR, to the current US FATPEN, and HAZFRAG and Mott-CALE models. Even small 1-20 grain fragments can do a lot of damage if going fast enough. I believe most naval shells are like AP rounds with strong noses and rear fuzes. I could be wrong. I think the use of Kevlar, body armor, IFV has driven the size of modern useful fragment size upwards. In WWII, the average ground pounder was pretty much 'naked' to fragments of most size. Fragments that are very small can still be deadly but they 'scrub' their velocity quickly. These micro-frags make small entry holes and then go on erratic paths through tissue in the body. 0 Quote Link to comment Share on other sites More sharing options...
Redwolf Posted November 14, 2003 Share Posted November 14, 2003 Before you go on a total rampage you should take a little breath and maybe think over some issues. With regards to your reply to my assumptions: A high load of explosives exploding on contact with a wall will inject a given impulse into the wall and it will transfer a certain amount of energy. It doesn't really matter whether impulse and energy are transmitted directly or with a layer of metal between them, the impulse and energy are the same. Except that the energy required to burst the casing into fragments takes energy which goes off your destructive effect on the wall. In the case of a wooden wall I could give you the benefit of the doubt that an explosion driving metal splinters may be more effective from cutting effects, but for a brick wall it certainly doesn't matter. 0 Quote Link to comment Share on other sites More sharing options...
Mr. Tittles Posted November 14, 2003 Share Posted November 14, 2003 http://www.bravecannons.org/the_gun/munitions.html Heres some good fuze info. Even though its Vietnam, its very topical. 0 Quote Link to comment Share on other sites More sharing options...
Mr. Tittles Posted November 14, 2003 Share Posted November 14, 2003 KE=0.5M*V^2 Thats the KE of the unexploded HE round. It does not disappear or become negligible. Its 0.5*5.86*800^2=1,875,200 kg M/s^2. Its stupendous. Its like a bowling ball hitting the wall at supersonic speeds. It does not matter that it may be breaking apart against the wall. A similar effect is comets/asteroids zooming towards the earth. If you break them up, they will continue onwards as a collection of very fast smaller projectiles. Energy does not disappear. 0 Quote Link to comment Share on other sites More sharing options...
Redwolf Posted November 14, 2003 Share Posted November 14, 2003 Err, is that supposed to be an answer to my posting? If so, of course the kinetic energy that the metal had before the shell burst will count against the walls. 0 Quote Link to comment Share on other sites More sharing options...
Mr. Tittles Posted November 14, 2003 Share Posted November 14, 2003 Originally posted by redwolf: I don't buy that. The tiny fragments hitting the wall have an overall kinetic energy that is insignificant compared to the blast and goes directly off the blast. Or in other words, the effect is the same as if the blast would be alone. Its in response to this mostly. 0 Quote Link to comment Share on other sites More sharing options...
Jim Boggs Posted November 14, 2003 Share Posted November 14, 2003 Thought Experiment: I have some valuable and interesting data that I can: A. Share with other people and respond to any questions or comments in a friendly and open interchange. B. Hold over your head while commenting in a condescending and caustic manner, as I gloat over my obvious intellectual superiority. 0 Quote Link to comment Share on other sites More sharing options...
Mr. Tittles Posted November 14, 2003 Share Posted November 14, 2003 The thread is meant to challenge misconceptions, mental pitfalls and sometimes psuedo-science and other fun stuff. Its meant to make people respond and reveal what they think is going on. Perhaps its not for Mr. and Mrs. John Q. Milktoast. I think the way the thread has developed has actually made me think and investigate new ideas/information. I am sure that many people have read the links and feel the same way. I think you are saying that I think I am right about everything or am in some silly pissing match. I may certainly be wrong for all I or you know. You may have a problem with people that actually try to back up what they say. I don't know. It probably doesn't really matter. 0 Quote Link to comment Share on other sites More sharing options...
Jim Boggs Posted November 14, 2003 Share Posted November 14, 2003 Originally posted by Mr. Tittles: Post note: I was really hoping Emrys would wander in and put one of his own big boots in his mouth but JasonC has beat him to it. Gee I don't know what would make me think that you weren't in it strictly for the scientific knowledge. :confused: :confused: 0 Quote Link to comment Share on other sites More sharing options...
Mr. Tittles Posted November 14, 2003 Share Posted November 14, 2003 Actually was hoping JonS would come in and accuse me of being a propagandistical ogre. But lets not let the thread 'bogg' down. 0 Quote Link to comment Share on other sites More sharing options...
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