Canister in CMBB: Realistic or Hollywood?

[Wol]

In terms of danger to personnel, It is worth noting that a Hazardous fragment in current US thinking is one with only 58 ft-lbs of enegy

[flamingknives]

Ok gents, here are the images rexford has scanned and sent me.

They're moderately sized but there are quite a few of them.

The whole list is at this web page

[flamingknives]

Whups, they're huge!

I might fix that later, but then again...

[John Kettler]

flamingknives,

Please do! They look great and technically meaty, but are badly oriented, thus hard to view, and are overtaxing my barely running machine.

Troops,

Am surprised no one has even tried to discuss the

enhanced kinetic energy issue I raised, though I do appreciate the T-34 cannon inputs. The effectiveness modeling continues to intrigue me, but BFC has released 1.03 without our latest discoveries.

Regards,

John Kettler

[Kiff01]

X

No joke intended. I was trying to get the thread to jump to page 7. The images above are too dam large. My machine is having a tough time opening page 6 of this thread.

[ June 06, 2003, 10:05 AM: Message edited by: Kiff01 ]

[rexford]

Originally posted by flamingknives:

Whups, they're huge!

I might fix that later, but then again...

Thanks for the CMBB posting and for sharing the web site.

Velocity at range can be estimated by taking the range and flight time and using this simple equation:

final velocity at range = (2 x range - flight time x muzzle velocity)/flight time

The equation assumes a linear drop in velocity with time and range, which may not be too bad compared to the work involved with a more exact solution.

[Kiff01]

Great images, but the size is far too large for a discussion forum. Reading additional posts requires scrolling across the screen. Even with my 21" screen. What does it take to get this thread onto a new page, or can the size of the posted images be reduced? Thanks.

[rexford]

Using a ballistic trajectory program developed by William Jurens and Nathan Okun, I estimated the down range velocity for 76.2mm shrapnel balls fired as cannister. Assumptions are 626 m/s muzzle velocity, an added 60 m/s for shrapnel powder charge (per John Kettler's figure, which may not be 100% correct but is a start), and balls act independently of one another with little influence on aerodynamics of other balls (simplifying assumption that is not technically true).

Balls are 10mm diameter with 0.011 kg weight (first run assumption).

Detonation takes place 10m to 20m from gun, as per German report info.

Following figures are estimated down range velocity of shrapnel balls:

0m from projectile detonation: 686 m/s

85m, 475 m/s

165m, 346 m/s

198m, 311 m/s

255m, 265 m/s

299m, 237 m/s

399m, 188 m/s

While shell casing would add fragmentation, many casing fragments from air burst would be to sides (if it parallels HE air burst shape), and down range density of casing fragments at 100m might be very low. But this is just a guess.

[John Kettler]

Kiff01,

Your huge post may be some sort of protest or visual wisecrack, but it's only made an already bad problem worse, at least on my rig. Please prune it down to some token size so that it doesn't take an hour to scroll through it.

rexford,

Thanks for taking up the analytical cudgel! If the

delta vee from the shrapnel's bursting is off, the

honor is Jeff Duquette's, since I used his data as my point of departure. I think some digging in my copy of TRAINING FIELD ARTILLERY DETAILS may allow me to crosscheck his data. I'd bet my KE numbers are at least roughly in the ballpark, though. Am

positive that a 10mm lead ball at 457m/s is much worse than .45 ACP!

Regards,

John Kettler

[flamingknives]

OK, I'm on it. Give me a bit of time though

{Edit: Hah! Job's a good 'un}

[ June 06, 2003, 07:10 AM: Message edited by: flamingknives ]

[Lady Roxanne]

From the 1.03 patch Readme file:

"Soviet 45mm antitank gun and 76.2mm ZiS-3 field gun may use canister ammunition."

This little change may have a significant effect on the CMBB battlefield. It's time to experiment a bit. I wonder if the costs of these guns has gone up?

EDIT: These guns will not be shy about using their cannister rounds on infantry targets, but the 45mm AT must be manually targetted. I tested at 170 meters. At this range there were no cannister inflicted casualties; however, the troops in the open were instantly pinned by the cannister. Once that happened, the AI proceeded to pummel the immobile troops with regular HE. All troops were in-command regulars.

If you spot a 45mm AT gun, don't get too close with infantry unless you can suppress the heck out of it. I'm sure it's a devastating anti-personnel weapon now. Players may even purchase them for this reason. They're cheap, and one shot does a squad. Cannister rounds don't miss easily either.

[ June 06, 2003, 09:45 AM: Message edited by: CrankyKris ]

[Lady Roxanne]

I may be wrong about lethality. Even at 65-70 meters the 45mm cannister does not inflict casualties on defenders IN WOODS. The stuff will break an in-command regular squad very quickly however.

Don't be surprised to see the 45mm fire 3 cannister rounds in the first 4 shots at infantry. I loaded my test guns with 5 rounds of cannister along with all the other ammo.

[ June 06, 2003, 10:08 AM: Message edited by: CrankyKris ]

[c3k]

A thought occurred as I read the mathematical models being presented. (If my following assessment is wrong, my apologies.)

It seems as if there's a basic flaw in the "to hit" probabilities being calculated.

Assume the muzzle of the firing weapon is 2 meters above a level field. If the shrapnel round is fired on it's "cannister" setting, it will explode approximately 10 to 20 meters away from the muzzle. From that point, all the splinters and shrapnel balls will expand in a cone (whether weighted to the center, annular, or equally doesn't matter at this point).

Let's assume the angle of the cone is such that for every 10 meters forward, the base of the cone expands by 2 meters. (Obviously this could be changed, but it allows for a concrete example.)

So, 10 meters after the charge detonates (let's say from 10 meters away from the muzzle), so at a muzzle distance of 20 meters, the cone has a radius of 2 meters. The bottom-most set of "balls" will impact the ground at this point.

I'll use a range of 110 meters for my example.

At 110 meters from the muzzle, our expanding cone will have a radius of 20 meters. Since the muzzle is only 2 meters above the ground, the bottom 18 meters of the cone will have been absorbed by the ground. (This leads to a very rough approximation that about half the "balls" have impacted the ground by this point.)

Now that we've looked at the bottom of the cone, let's look at the top: at 110 meters from the muzzle, the cone has a radius of 20 meters, measured from 2 meters above the ground (muzzle height). Hence, the top of the cone is 22 meters above ground level. I know of very few personnel targets which could be situated at that height above ground.

In essence, all targets would be located from ground level to a height of 2 meters. This is just a sliver of the total projected area. In fact, it could be approximated as a rectangle measuring 2 meters by 40 meters (diameter of the cone at 110 meters from muzzle).

Given the area of a circle with radius, r, of 20 meters is approximately 1250 square meters, and that the rectangular portion of the cone has an area of about 80 square meters, or as a percentage of the total area, 80/1250 = 6.4%.

So, take the .5 square meter target area given for a human target, and place it in a region with only 6.4% of the shrapnel "balls".

I believe all the models assumed the targets were randomly located _anywhere_ within the cone.

I hope I've gotten my point across. If I'm totally wrong, feel free to let me know.

Thanks,

Ken

[MikeyD]

So by your calculations at a 110m distance a .5m square target would be expected to be struck by 3.5 ball out of 549 in the case shot round? At 10 grams per ball that's 35 grams of lead suddenly entering your body. Ouch!

[Redwolf]

So did anybody find something about the penetration the canister pellets may have?

Will they go through a steel helmet?

How much are they stopped by earth relative to 7.62mm?

[MikeyD]

The posts above stating the balls were just 10-11 grams each came as a surprise to me, I was expecting something more robust (how much did a Civil War rifle miniball weigh?). But that would make them about the size of a revolver slug travelling at (more-or-less) handgun slug velocities. I'd imagine that would make close-range helmet penetrations rather likely, with the penetration curve falling steeply due to rapid decrease in projectile velocity.

[flamingknives]

c3k: a few points, if you'll oblige me.

Let's assume the angle of the cone is such that for every 10 meters forward, the base of the cone expands by 2 meters. (Obviously this could be changed, but it allows for a concrete example.)

No need to assume an angle - it's given in the above documents and doesn't really need much German to understand.

The angle is 15degrees so at 80m from the detonation point, the width of the pattern is 20m. The length depends of the angle of incidence with the ground.

From that point, all the splinters and shrapnel balls will expand in a cone (whether weighted to the center, annular, or equally doesn't matter at this point).

Perhaps not right here, but you later make a conclusion based on the assumption that the shot is evenly distributed across the area.

As the distribution is most likely not even, but concentrated towards the centre, the 6.4% of the area will have more than 6.4% of the shot.

In addition, you have some of the higher trajectory shot coming back down (although this is probably insignificant at this range)

I believe the examples assumed that the .5m target could be within any of the cones given - i.e that at the target range it would fall into one of the concentric circles subtended by the cones described (0.1, 0.2, 0.3 and 0.4 radians, IIRC) The figures then given were the liklihood that the target would be struck with a ball (or many balls)

In effect, you can imagine a set of concentric circles (like a target) that describe a cross-section of the shot at that range. The circle that the target falls under determines the probability of being hit by a ball at that location. You could draw a line through it at any point to simulate ground level relative to the direction fo the shot, but the probability of being hit depends on the circle the target fall under.

Basically the circles described cover the density of balls.

If that makes absolutely no sense, I apologise - I can see it in my head, but can't really explain it that well.

For the German speakers out there, would I be right to assume that 'Mundungswaagerechte' means horizontal?

[tar]

For the German speakers out there, would I be right to assume that 'Mundungswaagerechte' means horizontal?

Yes. More precisely, a horizontally along the line of fire.

Thinking a bit more about the instructions, it appears that in order to get a good pattern, the gunner was supposed to fire so that the shell would burst only a couple meters above the ground. It noted that this would result in some shots hitting the ground (presumably before bursting?) and thus being ineffective. In other words, it looked like one had to reckon with a certain percentage of wasted shots.

[Chad Harrison]

Wow, you guys are amazing. How you guys find this stuff on the web and more importantly how you realize that you actually found something amazes me! Great work.

So okay, I am sold on the fact that it was used and I can live with the quantity of it. But have we come to a consensus on the actual effect of it in CM?

Chad

[rexford]

The following sites have some interesting info on shrapnel:

http://www.army.mil/cmh-pg/faq/shrapnel.htm

http://www.riv.co.nz/rnza/hist/shrap/

1.lethal velocity for shrapnel is 137 m/s in one case (18 pdr shrapnel with 13mm lead balls)

2.WW I resulted in creation of shrapnel helmets

3.shrapnel has bigger effective area than HE (verified by comparison of shrapnel area to U.S. curves for HE ground or air burst effect)

4.shrapnel case not meant to fracture or fragment so does not add to effective pieces in air (would apply to Russian 76.2mm shrapnel if same design theory was applied)

[ June 06, 2003, 10:58 PM: Message edited by: rexford ]

[rexford]

3ck:

The concept you described is what would occur if the muzzle was horizontal. Time would also enter into the equation and gravity would pull the balls down, but not by much compared to the ball scatter and probably could be ignored.

The hit probability theory that John Salt used appears to assume that each ball has a chance to be anywhere within a ring, so if there are 300 balls in Ring A each one has a chance of being on the target area within the ring (probability is function of target area divided by area of ring). And each one has a chance of being towards the bottom of the ring and smashing into the ground.

So if one half of a ring is in the ground this does not change the estimates cause balls can be anywhere within a ring.

==========================================

At 115m distance from gun with 15m detonation distance, cone of scatter is 25m wide and center is 2m above ground (assuming T34 gun height is 2m).

If flight path of balls is not significantly altered by gravity to 115m and it doesn't seem so (see below analysis), a standing 2m tall man at 115m would be fully within Ring A if directly on centerline. Center of cone is 2m above ground, Ring A area is 1/4th of total diameter at 115m or 6.25m across and 3.13m above or below cone center along gun aim line, which catches everything down to ground level on flat terrain.

Because Ring A area is circular shaped at any given range, a standing man at 3m distance from gun aim line would be partially within Ring A and mostly in Ring B.

=================================================

Flight Time Considerations (not too relevant)

Assume shrapnel detonates 15m from muzzle and target is at 115m range. 626 muzzle velocity and 686 m/s initial shrapnel ball velocity.

Shrapnel cone expands 1m from line of fire for every 4m range, so at 100m from gun is 25m diameter cone.

Flight time to 15m detonation: .024 seconds

Flight time from detonation to 115m range: 0.18 seconds

Gravity drop to 115m range is negligible. -0.20m.

At 315m target range with horizontal barrel flight time would be 0.81 seconds with -3.2m gravity drop.

Gravity drop is small compared to scatter cone.

[tar]

At 315m target range with horizontal barrel flight time would be 0.81 seconds with -3.2m gravity drop.

Gravity drop is small compared to scatter cone

I'm not so sure about that. With a burst height of around 2m or so, the gravity drop will have rendered a lot of the balls ineffective, thus limiting the useful length of the cone.

Some of the other diagrams seemed to indicate an area of effect of around 80 x 20 meters.

[Mr. Tittles]

The americans developed the POZIT proximity fuze with the use of an HE payload?

Why not the grape shot-like Shrapneler? It could blow the load and the electronics into a human's body well.

It tittlates me so to think of these things!

It would be a good enough disaster to have a white phosh type shelly that would sploog after a POZIT initiation and hose nasty hynies from above. It would just gush the flaming chemical spray down into the defenders holes. I like that.

[ June 09, 2003, 02:55 AM: Message edited by: Mr. Tittles ]

[ June 07, 2003, 02:57 AM: Message edited by: Mr. Tittles ]

[John Kettler]

Things continue to develop, I see.

In no particular order:

If we're going to analyze the 76mm field gun case or similar for near muzzle fuzing, then I'd argue that a 1 meter barrel height would be a better choice than the 2 meters I proposed for the T-34 case.

Addition of 45mm and 76mm canister to 1.03 is both good and bad. It's good in that it restores a missing capability; it's bad in that it denies some of the tactical flexibility the multifunction shrapnel round had that a straight canister round lacked. Since canister is modeled in CMBB as a muzzle action projectile, there is thus a real loss

in in game capabilities as compared to real world ones.

I think we're well on our way to developing some sort of reasonable criteria for both helmet defeat and lethal velocity against uniformed bodies without helmets. To a first order, I suggest that

a shrapnel/canister ball of mass greater than or equal to the bullet from .45 ACP and traveling at 253 m/s (830 fps) muzzle velocity be deemed capable of penetrating a typical steel helmet. This is based on U.S. Army WW II firing demonstrations in which the M-1911-A1 fired .45 ACP completely pierced one side of the Stahlhelm at ranges of 15-20 meters, knocking it clean off the head of a uniformed mannequin. Likewise, if the ball is smaller but the KE the same or greater than the baseline case, we ought also to consider such a hit as being enough to defeat the helmet. Somewhere around here I have the Fackler (U.S. Army specialist in terminal ballistics on biological targets) criteria for combat wounding. These should provide some insight into the validity of the numbers cited for a 10mm ball.

Onward!

Regards,

John Kettler

P.S.

This would be so much easier if we had the right volumes from the JMEM (Joint Munitions Effectiveness Manual) series. Most of the ones I used a decade plus ago were classified, though.

[ June 07, 2003, 03:35 AM: Message edited by: John Kettler ]

[rexford]

Originally posted by tar:

</font><blockquote>quote:</font><hr /> At 315m target range with horizontal barrel flight time would be 0.81 seconds with -3.2m gravity drop.

Gravity drop is small compared to scatter cone

I'm not so sure about that. With a burst height of around 2m or so, the gravity drop will have rendered a lot of the balls ineffective, thus limiting the useful length of the cone.

Some of the other diagrams seemed to indicate an area of effect of around 80 x 20 meters. </font>