jwduquette1 Posted June 20, 2005 Posted June 20, 2005 (edited) Any guesses on the penetrator weight? ATK indicates 10kg (sabot + penetrator). Assuming the same sort of weight ratio of sabot weight to projectile weight as we have seen for M829, M829A1 and M829A2, I would guess the penetrator to be in the realm of 5.1Kg. However if the sabot has had a substantial redesign in terms of lighter materials employed, perhaps the sabot weight to projectile weight ratio is lower than previous iterations of M829. I'd guess the rod length to be about 770mm, although a small bit of this on either end involves a slightly reduced rod diameter -- threaded (or is it epoxyed) on the rear for the fin assembly, and frustum under the windshield. Assuming the same sort of rod diameter as seen in M892A2, than L/D is perhaps 36.7. But is it safe to assume the rod is unsheathed/unjacketed? It's pretty bloody long after all. Is M829A2 sheathed -- does it have a steel jacket? Edited June 20, 2005 by jwduquette1
Paul Lakowski Posted June 21, 2005 Posted June 21, 2005 Any guesses on the penetrator weight? ATK indicates 10kg (sabot + penetrator). Assuming the same sort of weight ratio of sabot weight to projectile weight as we have seen for M829, M829A1 and M829A2, I would guess the penetrator to be in the realm of 5.1Kg. However if the sabot has had a substantial redesign in terms of lighter materials employed, perhaps the sabot weight to projectile weight ratio is lower than previous iterations of M829. I'd guess the rod length to be about 770mm, although a small bit of this on either end involves a slightly reduced rod diameter -- threaded (or is it epoxyed) on the rear for the fin assembly, and frustum under the windshield. Assuming the same sort of rod diameter as seen in M892A2, than L/D is perhaps 36.7. But is it safe to assume the rod is unsheathed/unjacketed? It's pretty bloody long after all. Is M829A2 sheathed -- does it have a steel jacket?186066[/snapback] Humm this puts a different spin on things. I also got 78cm with 37:1 L/d but assumed a much higher MV. With these figures and that adjusted MV ,its penetration is going to be in the same region as the German DM-53 APFSDS.
tanker_karl Posted June 22, 2005 Posted June 22, 2005 Thanks for the estimates, Jeff and Paul. Regards.
Andrew Jaremkow Posted June 24, 2005 Posted June 24, 2005 The M829A2 is a monoblock with an aluminum windshield and steel tip. The M829A3 is also monoblock, according to what's been said over the years. However, it's switched to a steel windshield, which is unusual. It certainly doesn't need one based on heating effects, given the low speed of the round, so there's something different going on with the front end. But whether it's as simple as wanting stronger material to allow a thinner windshield, or a more complex structural role with terminal ballistic implications, I simply don't know. The M829A3 sabot is carbon fibre composite, like the M829A2, but with a redesign for lower parasitic mass. It has to extend further, however, to cover the longer rod, so I'm not sure if there was a net weight savings. As far as the DM53 goes, the Germans were heavily involved in the later stages of the M829A3's propellant development, so I expect there are similarities in that regard.
Sebastian Balos Posted June 24, 2005 Posted June 24, 2005 The M829A2 is a monoblock with an aluminum windshield and steel tip. The M829A3 is also monoblock, according to what's been said over the years. However, it's switched to a steel windshield, which is unusual. It certainly doesn't need one based on heating effects, given the low speed of the round, so there's something different going on with the front end. But whether it's as simple as wanting stronger material to allow a thinner windshield, or a more complex structural role with terminal ballistic implications, I simply don't know. The M829A3 sabot is carbon fibre composite, like the M829A2, but with a redesign for lower parasitic mass. It has to extend further, however, to cover the longer rod, so I'm not sure if there was a net weight savings. As far as the DM53 goes, the Germans were heavily involved in the later stages of the M829A3's propellant development, so I expect there are similarities in that regard.187453[/snapback] Perhaps the projectile uses some mind of heat insulating ceramic-metal layer, which can be deposited on steel, so they used thinner steel windshield. This way, you'd perhaps save weight, as well as length. As far as I can see from the cross-sectioned round, the tip is sharp-if the model is accurate. This site quotes propellant weight of 8.1 kg and projectile weight of 10 kg. http://www.defense-update.com/products/digits/120ke.htm To me, lowering mv for a projectile with higher L/D ratio makes sense. Longer projectile might be less affected by ERA, since a smaller portion will be cut-off, i.e. a longer portion remains to penetrate main armor. Regards, Sebastian
arrow Posted June 24, 2005 Posted June 24, 2005 Is somebody interested in guessing muzzle velocities of APDSFS from the L44-barrel of the 120mm smooth bore canon? The following equation could be helpfully: vo=a*tanhyp(b*ml/mp+c) where isml = mass of propellantmp = accelerated mass of projectilea = 2600b = 0.35c = 0.39 the ratio of ml to mp have to be greater than 0.7 Examples: M829A3 ml = 8.1 kgmp = 10 kgvo = 2600*tanhyp(0.35*8.1/10 + 039) = 2600*tanhyp(0.6735)= 2600*0.5873 = 1527 m/sthe nominal velocity is 1555 m/s M829A1ml = 7.9 kgmp = 9 kgvo = 1567 (nominal velocity is 1575 m/s) DM33 Rheinmetallml = 7.5 kgmp = 7.2 kgvo = 1658 (nominal velocity is 1650 m/s)
Paul Lakowski Posted June 24, 2005 Posted June 24, 2005 Is somebody interested in guessing muzzle velocities of APDSFS from the L44-barrel of the 120mm smooth bore canon? The following equation could be helpfully: vo=a*tanhyp(b*ml/mp+c) where isml = mass of propellantmp = accelerated mass of projectilea = 2600b = 0.35c = 0.39 the ratio of ml to mp have to be greater than 0.7 Examples: Thanks for this Arrow! couple of question Whats "tanhyp"? Are the 'a b c' constants? If the mass to charge ratio falls below 0.7 what do you use then? What are the limits of the formula?
DB Posted June 24, 2005 Posted June 24, 2005 Thanks for this Arrow! couple of question Whats "tanhyp"? Are the 'a b c' constants? If the mass to charge ratio falls below 0.7 what do you use then? What are the limits of the formula?187692[/snapback] "hyperbolic tangent" function. On my Casio calculator, you punch the "hyp" button before you punch the tan button normally seen as "tanh". http://mathworld.wolfram.com/HyperbolicTangent.html David
tanker_karl Posted June 24, 2005 Posted June 24, 2005 (edited) Is somebody interested in guessing muzzle velocities of APDSFS from the L44-barrel of the 120mm smooth bore canon? The following equation could be helpfully: vo=a*tanhyp(b*ml/mp+c) where isml = mass of propellantmp = accelerated mass of projectilea = 2600b = 0.35c = 0.39 187569[/snapback] Interesting. Thanks for posting this, Arrow :-) Seems to me that you like the 'tanh' function very much these days ;-) More seriously, I would be interested in knowing how you came up with the idea of using the 'tanh' function in your best fit equation (I suppose the a, b and c are the best fit constants) ? Is it through trials and errors or is the 'tanh' function especially appropriate for describing specific phenomenons in physics ? Sorry if this question is slightly OT, but I wanted to ask when you first came up with your penetration equation and I forgot. So I thought I might take this opportunity (before I forget again). Best regards. Edited June 24, 2005 by tanker_karl
arrow Posted June 24, 2005 Posted June 24, 2005 Whats "tanhyp"?Hi PaulDavid already gave the answer.Are the 'a b c' constants? Yes, best fitted constants (derived from test results until 1997)If the mass to charge ratio falls below 0.7 what do you use then?For conventionel propellants and guns there is a good equation from E. Schmidt (US Army Research Laboratory, Aberdeen): Vo = 3050*sqrt(ml/mp/(ml/mp+3)) But for modern tank guns the values are too low - for the M829A3 vo=1406 m/s.What are the limits of the formula?As I posted: valid for 120mm guns as used in M1A1 and Leopard 2 and charge ratio greater then 0.7 (with decreasing charge ratio below 0.7 the results became worse; for example with ml=0 one gets for all mp a muzzle velocity of 966 m/s!) Arrow
arrow Posted June 24, 2005 Posted June 24, 2005 Seems to me that you like the 'tanh' function very much these days ;-) Is it through trials and errors or is the 'tanh' function especially appropriate for describing specific phenomenons in physics ?Hi Karl Oh I love tanh!!!No, there is a physical reason. With solid propellants the muzzle velocity is limited through the sound velocity of the combustion gases. This velocity is in the region of 2600 m/s. Tanh of a big value tends to 1 (ml gigantic and mp very small lead to a muzzle velocity of 2600 m/s). Arrow
tanker_karl Posted June 24, 2005 Posted June 24, 2005 (edited) Hi KarlOh I love tanh!!!No, there is a physical reason. With solid propellants the muzzle velocity is limited through the sound velocity of the combustion gases. This velocity is in the region of 2600 m/s. Tanh of a big value tends to 1 (ml gigantic and mp very small lead to a muzzle velocity of 2600 m/s).Arrow187764[/snapback] Hi Arrow, I just realized I did not ask my question properly. This is what I meant to ask you : Beyond the fact that tanh is a *convenient* hyperbolic function (with an asymptotic value of 1 in the positives and y values > or equal to 0 for positive x values), does it have specifics that makes it particularly suitable for describing physical phenomenons ? I am specifically wondering whether the *shape* of the 'tanh' function for x belonging to the [0;e] interval would provide a specially good *fit* to describe certain physical phenomenons (and if yes, why). I hope the question is clear. Regards. PS : just curiosity - are you using the NLREG software ? Edited September 26, 2005 by tanker_karl
jwduquette1 Posted June 25, 2005 Posted June 25, 2005 The M829A2 is a monoblock with an aluminum windshield and steel tip. The M829A3 is also monoblock, according to what's been said over the years. However, it's switched to a steel windshield, which is unusual. It certainly doesn't need one based on heating effects, given the low speed of the round, so there's something different going on with the front end. But whether it's as simple as wanting stronger material to allow a thinner windshield, or a more complex structural role with terminal ballistic implications, I simply don't know. The M829A3 sabot is carbon fibre composite, like the M829A2, but with a redesign for lower parasitic mass. It has to extend further, however, to cover the longer rod, so I'm not sure if there was a net weight savings. As far as the DM53 goes, the Germans were heavily involved in the later stages of the M829A3's propellant development, so I expect there are similarities in that regard.187453[/snapback] Thnx Andrew. Regarding M829A3, I came across at least two DoD or US Army online pdfs indicating that the propellant used in the A3 is a new blend of some sort. Apparently the new propellant used in M829A3 is much less susceptible cook-off or spontaneous explosion if the cartridge is perforated by high velocity fragments or spall. No idea if the implication is that there was a trade off in how energetic the new propellant is or not. However, I thought perhaps the lower muzzle velocity of the A3 might somehow be connected. Just idle speculation.
jwduquette1 Posted June 25, 2005 Posted June 25, 2005 Hello Willi: Thanks for posting the function for estimating smoothbore muzzle velocity. This should come in handy at times. Best RegardsJeff D.
RPG_Holder Posted July 19, 2005 Posted July 19, 2005 We knew about these Limits since 1983 after Israel sent us several Captured T-72 in Lebanon that where intact but the Syrian Crews left them on the field, Motors running and the crews panic.184964[/snapback] I don't know where you got these false?
EchoFiveMike Posted September 15, 2005 Posted September 15, 2005 Was pawing up the display at a defense expo today and the penetrator is very substantial and of larger diameter than I expected. I'd guess 25mm or larger, how does that effect performance? S/F....Ken M
jwduquette1 Posted September 16, 2005 Posted September 16, 2005 (edited) Did you get any photos? ;o) My two cents worth on perforation estimates... If the Diameter was 25mm, and running with my previous length estimate of 770mm, than L/D = 30.8. Perforation estimate using Lanz/Odermatt: Remaining Velocity ----------Perforation Monolithic RHA @ 0-degrees1.3 Km/s --------------------------------540mm1.5 Km/s ---------------------------------650mm I think Lanz/Odermatt was developed based upon an extensive tungsten rod penetrator data base, so add say 9 or 10% to account for DU’s adiabatic shear banding characteristics. Some might use (+)12 or 13% -- I like (+)9 or 10%. My final perforation guesstimate is: Velocity -------Perf. Monolithic RHA @ 0-degrees------Perf. RHA @ 60-degrees1.3 Km/s ---------------------------596mm--------------------------349mm1.5 Km/s ---------------------------717mm--------------------------419mm Using Paul’s estimate of rod length – 780mm -- L/D now equals 31.2, and the associated perforation magnitudes are: Velocity -------Perf. Monolithic RHA @ 0-degrees------Perf. RHA @ 60-degrees1.3 Km/s ---------------------------604mm--------------------------353mm1.5 Km/s ---------------------------726mm--------------------------425mm I'd round to the nearest 10mm to account for unknowns, but in this case it was prolly more interesting to show the subtle difference between a 770mm rod and a 780mm rod. Size does matter inspite of what your wife might tell you. This assumes target UTS = 898.5Mpa. My assumed penetrator density is based upon U-3/4Ti alloy test rods. Edited September 17, 2005 by jwduquette1
jwduquette1 Posted September 17, 2005 Posted September 17, 2005 (edited) This was interesting. I came across a reference indicating that the total projectile length for M829A3 is on the order of 924mm, & total cartridge length = 982mm. I scaled from the same sectional image of M892A3 using 924mm total projectile length. The rod length maybe more like 800mm. I’d guess the windshield/nose-cone length would extend an additional 37 to 40mm beyond the rod/nose cone attachment point, and the fin assembly probably extends something like 84mm to 87mm past the arse of the rod/fin coupling/threads. That would put the rod length at something on the order of 800mm to 805mm. 924mm - (37mm+84mm) = 803mm Using L=800mm and D=25mm, the aspect ratio is on the order of 32, and the monolithic RHA perforation would be something like: Velocity -------Perf. RHA @ 0-degrees------Perf. RHA @ 60-degrees1.3 Km/s ----------------620mm-----------------------362mm1.4 Km/s ----------------686mm-----------------------401mm1.5 Km/s ----------------745mm-----------------------436mm Edited February 5, 2008 by jwduquette1
gewing Posted September 17, 2005 Posted September 17, 2005 How do these numbers compare to the earlier generations? This was interesting. I came across a reference indicating that the total projectile length for M829A3 is on the order of 924mm, & total cartridge length = 982mm. I scaled from the same sectional image of M892A3 using 924mm total projectile length. The rod length maybe more like 800mm. I’d guess the windshield/nose-cone length would extend an additional 37 to 40mm beyond the rod/nose cone attachment point, and the fin assembly probably extends something like 84mm to 87mm past the arse of the rod/fin coupling/threads. That would put the rod length at something on the order of 800mm to 805mm. 924mm - (37mm+84mm) = 803mm Using L=800mm and D=25mm, the aspect ratio is on the order of 32, and the monolithic RHA perforation would be something like: Velocity -------Perf. RHA @ 0-degrees------Perf. RHA @ 60-degrees1.3 Km/s ----------------620mm-----------------------362mm1.4 Km/s ----------------686mm-----------------------401mm1.5 Km/s ----------------745mm-----------------------436mm 223352[/snapback]
jwduquette1 Posted September 17, 2005 Posted September 17, 2005 Hi Gewing: Janes Ammunition Handbook indicates 120mm M829 APFSDS is capable of perforating 540mm of armor at a range of 2000meters (other estimates I have seen suggest perforation of 590mm/2000m). Using the Lanz/Odermatt Equation indicates M829 will perforate about 521mm of UTS-898.5MPa (BHN-269) armor at this range. However, if the 540mm figure derived from Janes represents testing against plate of about BHN-250 RHA (UTS = 832.3MPa) than Lanz/Odermatt is actually quite accurate at predicting perforation magnitude. L/O results in a perforation magnitude of 539mm of BHN-250 Armor at a range of 2000m. A lower plate hardness level would seem likely if thicker test plates were employed for whatever trials resulted in the 540mm figure. Moreover, MIL-A specifications for armor production indicates that RHA plates of thickness 4 to 6 inches would be produced at hardness levels between BHN-241 to BHN-277. Armor hardness levels for plates thicker than 6” are even lower. BHN-269 armor is more appropriate to plate thickness of 2 or 3 inches. This level of plate hardness is often used in 1/4 and 1/2 scale testing of rod penetrators. I'm guessing full scale trials would use thicker plates -- and therefore plates of lower hardness. Of course the 540mm figure may simply be a function of multiple plate stacking. But let’s not go there for the moment. Let’s assume our target is monolithic RHA of BHN-250. I normalized my previous M829A3 perforation figures to monolithic RHA of UTS-832.3MPA. I have plotted these values against my estimate of M829 perforation figures for the same plate hardness on the attached figure. Regarding the M829 perforating 590mm @ 2000m; this is also conceivable. Very thick RHA – the 9 to 12-inches thick stuff – is produced at BHN-212 to 241. Assuming someone actually was shooting holes in a 12-inch thick chunk of steel with M829, it is conceivable that 590mm of perforation could be achieved. Lanz/Odermatt predicts M829 can perforate about 583mm of RHA (BHN-212) at a range of 2000meters. Best RegardsJD
gewing Posted September 17, 2005 Posted September 17, 2005 Hi Gewing: Janes Ammunition Handbook indicates 120mm M829 APFSDS is capable of perforating 540mm of armor at a range of 2000meters (other estimates I have seen suggest perforation of 590mm/2000m). Using the Lanz/Odermatt Equation indicates M829 will perforate about 521mm of UTS-898.5MPa (BHN-269) armor at this range. However, if the 540mm figure derived from Janes represents testing against plate of about BHN-250 RHA (UTS = 832.3MPa) than Lanz/Odermatt is actually quite accurate at predicting perforation magnitude. L/O results in a perforation magnitude of 539mm of BHN-250 Armor at a range of 2000m. A lower plate hardness level would seem likely if thicker test plates were employed for whatever trials resulted in the 540mm figure. Moreover, MIL-A specifications for armor production indicates that RHA plates of thickness 4 to 6 inches would be produced at hardness levels between BHN-241 to BHN-277. Armor hardness levels for plates thicker than 6” are even lower. BHN-269 armor is more appropriate to plate thickness of 2 or 3 inches. This level of plate hardness is often used in 1/4 and 1/2 scale testing of rod penetrators. I'm guessing full scale trials would use thicker plates -- and therefore plates of lower hardness. Of course the 540mm figure may simply be a function of multiple plate stacking. But let’s not go there for the moment. Let’s assume our target is monolithic RHA of BHN-250. I normalized my previous M829A3 perforation figures to monolithic RHA of UTS-832.3MPA. I have plotted these values against my estimate of M829 perforation figures for the same plate hardness on the attached figure. Regarding the M829 perforating 590mm @ 2000m; this is also conceivable. Very thick RHA – the 9 to 12-inches thick stuff – is produced at BHN-212 to 241. Assuming someone actually was shooting holes in a 12-inch thick chunk of steel with M829, it is conceivable that 590mm of perforation could be achieved. Lanz/Odermatt predicts M829 can perforate about 583mm of RHA (BHN-212) at a range of 2000meters. Best RegardsJD223613[/snapback] I'm just glad there seems to be a significant improvement! I wish the US could field a round that could be mass produced and penetrate 1000mm at 2000m, just because I AM Americacentric. Maybe when we get rail guns further developed. I don't want the US to ever be in a situation of NOT having an anti-armor overmatching capability.
JWB Posted September 18, 2005 Posted September 18, 2005 OK Gewing according to the Desert Storm Factbook 120mm RM gun mounted on M1A1 could defeat more than 650mm at 1km-2km range.
gewing Posted September 18, 2005 Posted September 18, 2005 OK Gewing according to the Desert Storm Factbook 120mm RM gun mounted on M1A1 could defeat more than 650mm at 1km-2km range.223725[/snapback] That is a good start...
jwduquette1 Posted September 18, 2005 Posted September 18, 2005 OK Gewing according to the Desert Storm Factbook 120mm RM gun mounted on M1A1 could defeat more than 650mm at 1km-2km range.223725[/snapback] Hi JB: Your reference is probably talking about M829A1. The famous “Silver Bullet”. Coors Light served up at 1575meters per second. It is indicated in Jane’s to have seen operational use during Desert Storm with the M1A1. The penetrating rod is much longer than M829, but not as long as M829A3. The performance of the A1 should fall somewhere between the earlier iteration of M829 and the latter design M829A3. Best RegardsJD
DKTanker Posted September 18, 2005 Posted September 18, 2005 (edited) Hi JB: Your reference is probably talking about M829A1. The famous “Silver Bullet”. Coors Light served up at 1575meters per second. It is indicated in Jane’s to have seen operational use during Desert Storm with the M1A1. The penetrating rod is much longer than M829, but not as long as M829A3. The performance of the A1 should fall somewhere between the earlier iteration of M829 and the latter design M829A3. Best RegardsJD223788[/snapback] It is indicated? Why don't you ask some TN members. I'm sure they/we/I can state catagorically, and from first hand experience, that M829A1 was used during ODS, if in limited supply. As a side note. The M829A1 received the moniker 'silver bullet' because the windscreen is in fact silver in color. That the cure for several VDs is penicillin, which is also known as a 'silver bullet', is not coincidental. Edited September 18, 2005 by DKTanker
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