Bofors

[Crannogman]

Running low on 4" shells (and out of fish), I put my last dozen shells into a 5000-ton freighter, which proved insufficient. I closed to under 2000y at decks awash and began to pour in 40mm Bofors rounds. 1500 HE rounds and another 200 AA (using OTC) did not make an appreciable difference, except for one lucky hit taking out the freighter's bow gun.

A 4" shell weighs 33lb, while a 40mm shell is about 2lbs. Is there any reason to expect 1.5 tons of 40mm shells to be incapable of sinking a 5000-ton ship?

[TorpX]

A 4" shell weighs 33lb, while a 40mm shell is about 2lbs. Is there any reason to expect 1.5 tons of 40mm shells to be incapable of sinking a 5000-ton ship?

The shell may weigh 33lb, but the bursting charge is less than 1½ lbs. The bursting charge for the 40mm rd. is 0.15 lbs. Most torpedo warheads were 500 lbs., or more. The bursting charge is, I think, a better index of their relative capability.

There are many examples of submarine gun attacks that failed to sink ships, or sank them with only a great deal of difficulty. Imo, ships over 1,000 tons, and certainly those over 2,000 tons, should not be that easy to sink (with shells I mean).

If you go by the USN's official records, the tonnage of Japanese merchant ships sunk by shell fire was almost trivial. Most of those sunk in this way, were smaller ships/craft.

You didn't say what mods you are using. It can make a big difference. In RFB you can sink ships of this size with the deck gun, but I think it is more difficult than with stock. Don't know about TMO.

[Crannogman]

My only mods of note are RSRD and OTC. For reference, I had sunk a couple 4500-ton Heito Marus a day earlier using a total of 40 4" shells

[TorpX]

So, I take it you are using the stock version of RSRDC?

In stock, it is usually pretty easy to sink stuff with DG. I don't know if RSRDC would change that, or not.

For reference, I had sunk a couple 4500-ton Heito Marus a day earlier using a total of 40 4" shells

That is more or less what I would expect in the game, generally.

However, you have to realize that a 40mm gun is designed to tear up aluminum aircraft, not sink ships.

In RFB, you can sink ships fairly well, but you usually have to aim near the waterline, and be patient. The damage zones of the merchants are mostly below WL. In stock, iirc, you can hit them anywhere, and blow them up.

[Sniper297]

Silent 3ditor, open \Data\Library\shells.zon.

40mm has three types of ammo, ball for the flying machines, high explosive and armor piercing for ships.
17 is the AP, 19 is the HE.

I increased the explosive power and damage radius for those two to make it possible to sink small freighters with 10 clips or so of HE if I fire at the waterline. Got a destroyer chasing me and I can't dive for some reason I use armor piercing, but unless he's already damaged that takes 20-30 hits by which time my sub is a junkyard. Unless I get really lucky.

The modern AH-64 Apache helicopter has a single barrel 30mm gun, I've seen several videos of that gun tearing up armored vehicles and light tanks. My opinion the 40 SHOULD do AT LEAST that much damage if not more, so to me the BOFORS in game seems underpowered.

[Crannogman]

Yes, but even a heavy tank is only 70t. The Bofors as-is does seem to be decently effective for knocking out defensive weapons on the enemy ship - under 2000y, your crew should be able to score hits on occasion

[Crannogman]

The shell may weigh 33lb, but the bursting charge is less than 1½ lbs. The bursting charge for the 40mm rd. is 0.15 lbs. Most torpedo warheads were 500 lbs., or more. The bursting charge is, I think, a better index of their relative capability.

There are many examples of submarine gun attacks that failed to sink ships, or sank them with only a great deal of difficulty. Imo, ships over 1,000 tons, and certainly those over 2,000 tons, should not be that easy to sink (with shells I mean).

If you go by the USN's official records, the tonnage of Japanese merchant ships sunk by shell fire was almost trivial. Most of those sunk in this way, were smaller ships/craft.

You didn't say what mods you are using. It can make a big difference. In RFB you can sink ships of this size with the deck gun, but I think it is more difficult than with stock. Don't know about TMO.


Interesting that the 4" shell weighs 16x the 40mm shell, but its bursting charge is not quite 10x heavier (630g vs 68). Also, if http://www.navweaps.com/Weapons/Gun_Data_p2.htm is to be believed, it's only ~3x more effective (25 vs 8.3).

[Sniper297]

"Yes, but even a heavy tank is only 70t."

Main difference is it would take zero hits with a 40mm shell to sink a tank, put it in the water and it would go down by itself. :woot:

"The Bofors as-is does seem to be decently effective for knocking out defensive weapons on the enemy ship - under 2000y, your crew should be able to score hits on occasion"

Ridiculous part is the crew AI is actually pretty good at shooting down enemy aircraft, in fact a lot better than I am doing it manually. However, AFAIK there's no way to get them to load the BOFORS with HE or AP and fire at a ship, you either do it yourself or it don't happen. 4"50 deck gun is a different story, there's a mod for AA shells for the deck gun but stock game doesn't have it. I never tried the mod since the crew AI for the deck gun gets about 1 hit for 20 rounds fired even at close range when firing at ships, I can do better than that with a hangover.

There is a balance involved in hacking the shells.zon file - on the one hand it's tedious trying to finish off an unescorted merchie with hundreds of rounds, on the other hand you don't want to make the deck gun and BOFORS so powerful that torpedoes become superfluous. What I did was double the explosive power, radius, and armor piercing, then reduced the number of rounds carried (in the Data\Library\SubParts\*.sim file for the weapon). The idea is when you're out of fish and gotta sink something you can do it without wasting a lot of time, but you can't do it often because of limited ammo.

[Crannogman]

Actually, I meant have the AI use the deck gun against the waterline, while you man the AA gun and go after the merchie's gun.

I just began a career with Narwhal - 4 shells sank a 1800t freighter off the Mariannas, but it took 40% of my fuel to get there at 10kts

[TorpX]

Interesting that the 4" shell weighs 16x the 40mm shell, but its bursting charge is not quite 10x heavier (630g vs 68). Also, if http://www.navweaps.com/Weapons/Gun_Data_p2.htm is to be believed, it's only ~3x more effective (25 vs 8.3).

You are referring to this:

Two rules of thumb about Burster Power
1) The effect of the burster may be taken as being proportional to the square root of the weight of the bursting charge.
2) ...

I've read that, and I would really like to know where they get this idea.

It doesn't seem to be supported by the laws of physics. Moreover, it seems to fly in the face of documented battle reports, both of sub attacks on ships, and warship vs. warship battles.

And, if it were really true, why would navies even bother with torpedoes and bombs; just use small or medium size guns, and shower the enemy with high volume fire, to do the job. Reported instances of successful sub DG attacks tend to stand out because they were uncommon.

Someone, a while ago, was complaining that they couldn't sink a 1,000 ton ship with 20x 4 in. shells. I posted a short summery of an attack made by the Tambor:

There are many examples of submarines requiring many shells to sink even small ships, or not being able to sink them at all. For example, it was the Tambor, iirc, that happened to see a small 'ship' that was taking supplies to the stranded garrison on Wake Is. The vessel was something less than 500 t, and clearly not a good torpedo target. They surfaced and engaged with the 5 inch gun. They exhausted their ammunition, and ended up taking out the defenders with small caliber fire, before boarding the craft, capturing the First Officer, and finally setting the craft on fire. Nor was this extraordinary. Shells aren't nearly as effective as torpedoes, and this is why the torpedo malfunctions were so disabling( and frustrating). There was really little alternative to torpedo attack, unless the targets were very small. BTW, when I use the word small here, I mean something less than 500 tons. That was the criteria used by JANAC, and they made no attempt to track or verify sinking of anything less.

[Sniper297]

I think you're looking at it bass ackwards. I'm miles away from being a math whiz, but the square root is the number which multiplied by itself is the original number. Example bursting charge 16 pounds, square root of 16 is 4, so the "proportional effect" of a 16 pound charge would be 4 pounds, in other words weaker, not stronger. Dunno if that's true or not, I gave up trying to figure out algebra and women in 1968.

One factor left out - fire a 40mm HE shell into a merchant sailing in ballast and it will make a hole in the side, probably damage some internal bulkheads. Need a lot of small holes near the waterline to sink it.

Fire a 40mm HE shell into a merchant loaded with bombs torpedoes artillery shells or all the above, all it takes is one hit to set off the chain reaction that will blow pieces of the target into another ocean. Same with tankers, got one carrying crude or empty you're gonna need a dozen torpedoes to actually sink it, deck guns and BOFORS would be a complete waste of time. OTOH if the tanker is full of avgas, one 40mm HE sparks 10,000 tons of high octane, kaboom. Rather than proportional square root math you get the big bang theory with a secondary explosion.

:fff:

[Crannogman]

Indeed, the square root claim is entirely un-referenced. But that was all I could find. Another section on that website is dedicated to calculations of damage from shellfire - it is hampered by referring mostly to armored targets, but suggests that the lion's share of destruction is kinetic rather than explosive.

[Longknife]

Being pragmatic you can lay waste to the superstructure reducing it to twisted iron but if there are no holes below the WL the ship will not sink.

FWIW I gave up using the deck gun & I run from aircraft like a little girl from a spider only manning the AA as I am crash diving. While the secondary armament is no doubt fun to play with, in sim as in real life the only way to reliably put a ship down is to drive a torp into her side.

[Crannogman]

Being pragmatic you can lay waste to the superstructure reducing it to twisted iron but if there are no holes below the WL the ship will not sink.

FWIW I gave up using the deck gun & I run from aircraft like a little girl from a spider only manning the AA as I am crash diving. While the secondary armament is no doubt fun to play with, in sim as in real life the only way to reliably put a ship down is to drive a torp into her side.

I generally agree. When I first got a pair of twin 20mm guns I thought to sail with impunity. It was effective against the Zeroes for a day, but then I was out of ammunition. I generally save the deck gun for ships at anchor inside atolls, or else to finish off merchants that I have already holed. I do find the deck gun to be useful against escorts - when there's one minesweeper guarding 8 large freighters, I draw out the escort and sink them (since my long-range gunnery is quite superior) then close in to torpedo the merchies.
In the thread-leading instance, I had already expended all my torpedoes and almost all my 4" shells, and hoped a couple thousand 40mm rounds could finish the job. Unfortunately, even if they do cause damage to ships, I don't think they can really get below the waterline

[TorpX]

I think you're looking at it bass ackwards. I'm miles away from being a math whiz, but the square root is the number which multiplied by itself is the original number. Example bursting charge 16 pounds, square root of 16 is 4, so the "proportional effect" of a 16 pound charge would be 4 pounds, in other words weaker, not stronger. Dunno if that's true or not, I gave up trying to figure out algebra and women in 1968.



I think he got the math right; it's just the 'rule' itself is questionable.

If you have a model, with a 'medium' shell with 1 lb. of TNT, you could give it a base damage of 1. A large shell with 4 lbs. of TNT, would give damage of 2, and so on. A bomb with 100 lbs. of TNT, would give damage of 10. This sounds alright, until you ask if 10x medium shell hits are really as effective as one large bomb hit. Imo, this idea fails when examined closely.

This sort of concept may make sense, if you are applying it to a damage model, where ships are defined in very simple terms, having a total number of hit points, without any specific damage zones, critical hit areas, etc. The idea being, additional damage to areas already hit/destroyed becomes superfluous, so there is a diminishing return on larger types of ordnance, and scattered hits are assumed to damage more sections.

However, in a more developed model (as in SH4), you already have limits as to how much damage a zone will take, and a floodable zone can only flood to a certain level, and no more. So, too much fire directed to a small part of a ship is likely to be wasted.

[Sniper297]

No, I meant YOU were looking at the math wrong.

"And, if it were really true, why would navies even bother with torpedoes and bombs; just use small or medium size guns, and shower the enemy with high volume fire, to do the job"

If it is true that means LESS damage from deck guns and small caliber cannon, not more, so they wouldn't bother with deck guns / Bofors / Oerlikon at all.

"Being pragmatic you can lay waste to the superstructure reducing it to twisted iron but if there are no holes below the WL the ship will not sink."

Depends on what else is in the superstructure and if you can start a fire. I was in bootcamp in 1977, and we were taught that fire is a bigger danger than flooding on a warship. Mainly because of this disaster 10 years before;

http://en.wikipedia.org/wiki/1967_USS_Forrestal_fire

we spent about 3 of the 9 weeks training in firefighting and only 2 days in flooding control.

To me this looked promising;

http://www.subsim.com/radioroom/showthread.php?t=180920

But in my tests it didn't work very well.

We saw this training film;

https://www.youtube.com/watch?v=U6NnfRT_OZA

Bear in mind the ZUNI rocket didn't travel far enough to arm so the warhead didn't explode, but it did make a metal to metal spark which ignited the jet fuel. In addition to the film we also learned firsthand how quickly a fire can spread and get out of control. In WWII they hadn't learned this lesson yet, so only a small percentage of a warship's crew had any training at all in firefighting, and the merchant crews had none.

Again it depends on the cargo, but it should be possible to start a fire with 40mm which would quickly spread and destroy the target, flooding is irrelevant because if it blows up it's gonna sink.

[Crannogman]

So to compare various submarine shells, we would convert their bursting charge into a blast coefficient- per the NavWeaps site, that coefficient is the square root of the bursting charge's mass.

900g 40mm shell charge = 68g, BC = 8.35

5900g 3" shell charge = 340g, BC = 18.44

14970g 4" shell charge = 630g, BC = 25.10

24460g 5" shell charge = 3305g, BC = 57.49

47600g 6" shell charge = 3218g, BC = 56.73

This suggests that the bursting charge from a 5" or 6" gun is equivalent to 7 40mm shells, which seems absurd on the face of it. No doubt the kinetic energy of the entire projectile is also important, and is directly proportional to the shell's mass. So it's not merely a question of whether the square root rule is valid, but what portion of projectile damage is due to kinetic energy and what portion is due to blast effects.

If nobody has any better sources, I might try calling the Naval Surface Warfare Center

[Hambone307]

What I'm gathering is just stick to using my torpedos. All this math is making my head hurt... :doh:

:har:

[Sniper297]

Tossing all the math out, the ideal would be to set shells.zon and the ammo supply so you can sink one or two small to medium merchant ships per patrol with gunfire alone, leaving torpedoes as your primary weapon.

[TorpX]

No, I meant YOU were looking at the math wrong.


If you disagree with my ideas, fine, but I KNOW I got the math right.

Again, if small shells are so omnipotent, why not equip cruisers and destroyers with them. They are lighter than 4 or 5 inch guns, and so is the ammunition. It is hard to find examples of significant ships that were sunk (or blown up) with small shells. It is not hard to find examples of ships that were destroyed or sunk with torpedoes and bombs. The reason the small 20mm and 40mm guns were there was for AAA. The reason for the 4 in. guns (at least originally) was to take out small torpedo boats when they tried to attack the big ships.



This suggests that the bursting charge from a 5" or 6" gun is equivalent to 7 40mm shells, which seems absurd on the face of it. No doubt the kinetic energy of the entire projectile is also important, and is directly proportional to the shell's mass. So it's not merely a question of whether the square root rule is valid, but what portion of projectile damage is due to kinetic energy and what portion is due to blast effects.

Certainly, the KE must come into play, but if your 4 in. AP shot produces a 4.2 (?) in. hole, doubling it's KE will not change much. Fragmentation effects are important when it comes to personnel casualties, but are not going to sink ships. Fragments will not have nearly the penetration of an intact shell/shot, though they are deadly to the crews.

[Crannogman]

Doubling the kinetic energy allows the projectile fragments to penetrate farther into/through the target. So maybe you'll get two holes though the hull and maybe on through the boiler to boot. Even the bursting shells had a solid cap that could travel into deeper into the ship.

I guess the major problem with the 40mm is getting hits below the waterline. Still, they might be useful for counterbattery fire.

I use a fire damage mod as well, can't say that it does much

[Sniper297]

Transmitting and receiving on different frequencies or something here.

"I KNOW I got the math right.

Again, if small shells are so omnipotent"

The math they're talking about is REDUCED effectiveness, not INCREASED, rather than the square root making small shells omnipotent, the square root would make them less potent. If the effect was the square of the weight rather than the square root, that would make them more effective.

Again I don't know where the formula came from, but if it's true it would support your opinion that small caliber shells aren't very potent rather than contradicting your opinion. If the math called for the square of the charge then 16 ounces multiplied by 16 would be the equivalent of 256 ounces, more powerful. That's not what it says, it says the square ROOT - the square root of 16 is 4, so a 16 ounce shell is the equivalent of 4 ounces, less powerful. That's assuming the formula is correct, which I have no clue.

That's what's confusing me, you're saying the formula must be wrong because it would make the shells MORE powerful than real life, but the square ROOT would make them LESS powerful?

[Crannogman]

You can't take it based on a single shell - again, the square root is for comparing damage between different bursting charge weights. So, if you double the blasting charge, you only get 1.4x the damage; in order to double the damage, you have to quadruple the weight of the blasting charge. Hence you may perceive diminished returns from increased size of gun and shell.

Even should this square root rule hold true, a larger gun or shell typically delivers more kinetic energy and at longer ranges. Thus, there is still an advantage to carrying larger guns, especially for penetrating armored targets.

[Sniper297]

There is that factor. One of the considerations in the planning of the Pearl Harbor attack was that dive bombing was more accurate, but high level horizontal bombing gave more velocity to the bomb, meaning better armor penetration with the same bomb. Eventually they went with a mix of both high level and dive bombing, plus the torpedo bombers.

[TorpX]

Transmitting and receiving on different frequencies or something here.

"I KNOW I got the math right.

Again, if small shells are so omnipotent"

The math they're talking about is REDUCED effectiveness, not INCREASED, rather than the square root making small shells omnipotent, the square root would make them less potent. If the effect was the square of the weight rather than the square root, that would make them more effective.

Again I don't know where the formula came from, but if it's true it would support your opinion that small caliber shells aren't very potent rather than contradicting your opinion. If the math called for the square of the charge then 16 ounces multiplied by 16 would be the equivalent of 256 ounces, more powerful. That's not what it says, it says the square ROOT - the square root of 16 is 4, so a 16 ounce shell is the equivalent of 4 ounces, less powerful. That's assuming the formula is correct, which I have no clue.

That's what's confusing me, you're saying the formula must be wrong because it would make the shells MORE powerful than real life, but the square ROOT would make them LESS powerful?

Taking the square root diminishes the difference between types of ordnance. It is not the individual numbers, but the ratios that matter.



If the 'effectiveness' is proportional to the bursting charge itself:

notional small AA shell ______ 0.25 lb. charge _____ 0.25
notional medium shell _______ 1.0 lb. charge ______ 1.0
notional bomb _____________ 100 lb. charge ______ 100.

So, we have a ratio of .25 to 1.0 to 100, (or 1:4:400). That is a bomb is as effective as 100 medium shells or 400 small shells.


If the 'effectiveness' is proportional to the square root of the bursting charge:

notional small AA shell ______ 0.25 lb. charge _____ 0.5
notional medium shell _______ 1.0 lb. charge _____ 1.0
notional bomb _____________ 100 lb. charge _____ 10.0

This gives a ratio of 0.5 to 1.0 to 10.0, (or 1:2:10). Then a bomb is as effective as 10 medium shells or 20 small shells. The ratio of these is, at the extreme, only 20 to 1, where before, it was 400 to 1. If you only need 20 small shells to equal a large bomb, it is more powerful, than if you need 400.

My criticism of this 'square root idea', is that you can get to the point of destroying Yamato-sized ships with glorified heavy machineguns, because small shells can be thrown out in great volumes.

[Sniper297]

Doesn't work that way, instead of decimals convert to ounces.

.25 pound = 4 ounces. Square root of 4 ounces = 2 ounces.
1 pound = 16 ounces. Square root of 16 ounces = 4 ounces.
Square root of 100 pounds is 10 pounds, which is 160 ounces. So the increase over a 1 pound charge would be 40 times the bang, not 10 times the bang. It would be 80 times the bang of a .25 pound (4 ounce) charge.

Again I have no idea if the formula itself is valid, what I'm saying is IF the formula is valid it would make smaller charges less effective rather than more effective per weight.

To me it seems more logical that a 100 pound bomb would have 100 times the bang of a 1 pound bomb, but 40 times the bang wouldn't make the increase meaningless.

https://scontent-lga.xx.fbcdn.net/hphotos-xpa1/v/t1.0-9/11257058_1255041181191999_9129141592636843616_n.jp g?oh=723367144cc76a8777779d1787c9a146&oe=55CFB253

[TorpX]

Doesn't work that way, instead of decimals convert to ounces.

Oh, but it does work that way.

.25 pound = 4 ounces. Square root of 4 ounces = 2 ounces.
1 pound = 16 ounces. Square root of 16 ounces = 4 ounces.
Square root of 100 pounds is 10 pounds, which is 160 ounces. So the increase over a 1 pound charge would be 40 times the bang, not 10 times the bang. It would be 80 times the bang of a .25 pound (4 ounce) charge.
The statement in red isn't valid.

In that line, you took the sq. root first. You didn't do that in the first two lines.

You must first convert the 100 lbs. to ounces, then take the square root (same as you did in the upper 2 lines). Thus, 100 lbs. = 1600 oz. The square root of 1600 is 40. The ratio of 40 oz. to 2 oz. is 20 to 1. This is the same answer you get if you use pounds, as I did.

[Sniper297]

Well, that's why math never made any sense to me, if any formula is valid it should be the same regardless of whether you use grams, ounces, pounds, or tons. Square root doesn't.

2000 pound bomb square root is 44.7, ridiculously low effect which makes me question where the square root theory came from in the first place.

2000 pounds is one ton - square root of 1 is 1, so a 1 ton bomb has the equivalent power of a 1 ton bomb, which is a lot more than 44.7 pounds. A 1000 pound bomb is 1/2 ton, the square root of 0.5 is 0.7, so a half ton bomb has the explosive power of a 3/4 ton bomb? The whole thing is too screwy for words, if you can't convert back and forth and still get the same values the formula just won't work.

[TorpX]

Well, you can use any unit, but you have to use the same unit across the board. That is, you must convert to the selected unit first, then take the root. Otherwise, you end up with a hash.

Basically, this is because when we take the sq. root of a 100 lb. charge, we are creating an arbitrary unit. Instead of pounds, we have sq. root pounds. Just as you can't add linear feet, sq. feet, and cubic feet, you can't mix sq. root lbs., ordinary pounds, and sq. root oz., etc. Defining a function in terms of an arbitrary unit, in itself, doesn't invalidate the idea, but one must be careful to be consistent in the units used.

Where did they get the idea? I suspect they are using a convention from land artillery used in wargames. If you are throwing shells at infantry companies, or troops formations, a bigger bomb doesn't necessarily get you a much better return. The troops are spread out, so you might be better off using a scattering of small mortar shells, instead of one large shell. In this context, the idea makes sense. But, for buildings or ships, I think it is misplaced.

[Sniper297]

That would make sense, the whole reason why rockeyes were invented in the first place. Instead of a single 500 pound mark 82 against a spread out formation you drop a rockeye cluster bomb which spreads 247 1.3 pound bomblets over something like 2500 square yards. Navy used those against small fast boats (kind of a shotgun where you only need one pellet to hit), but they wouldn't be very effective against warships.

"Basically, this is because when we take the sq. root of a 100 lb. charge, we are creating an arbitrary unit. Instead of pounds, we have sq. root pounds. Just as you can't add linear feet, sq. feet, and cubic feet, you can't mix sq. root lbs., ordinary pounds, and sq. root oz., etc. Defining a function in terms of an arbitrary unit, in itself, doesn't invalidate the idea, but one must be careful to be consistent in the units used."

But that's the problem with the whole theory, 2000 pounds is 1 ton is 32,000 ounces, if you divide by 2 or multiply by itself it doesn't matter where you start or where you end, the result will be the same value whenever you convert it. Doesn't work that way with a square root, and if you have to create an arbitrary unit and know what that unit is - and it doesn't work exactly the same whatever unit is used - then the formula is flawed.