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Post by dorn on Nov 3, 2019 1:37:09 GMT -6
IMHO, 'crossing the T' is nothing but a simple exercise in mathematics: The side that's doing the crossing will be able to bring more guns to bear, and thus inflict more pain on the enemy - and since wars (and thus battles) are won by the side that inflicts the most pain on the enemy - the side that's doing the crossing will automatically win. The reduced chance of hitting is still offset by the additional number of guns able to fire on the enemy. Hell, let's make a mathematical model - Lanchester style: Side A is 'crossing the T' on side B. They both have 8 ships with guns in 4x2 configuration - however, Side A will, naturally, have all it's guns available - but Side B will only have 1/2 of it's guns initially available - so, basically a 2x2 configuration. Let's take into account the reduced chance of hitting by giving Side A 10 guns to make a hit - while Side B needs only 8 guns to make a hit. Each hit will take out a turret (with 2 guns in it) - so, basically, Side A will lose a ship in 4 hits, and Side B a ship with 2 hits. Yes, I'm aware that Side B's ships will be a lot less damaged than Side A's, but their turrets will be out of action - and thus they will (basically) be sitting ducks - to be taken out later. With those rules in place - let's fight! Round 1: Side A has 8x4x2 = 64 guns available = 6.4 hits - let's say 6, with the .4 carried over. Side B has 8x2x2 = 32 guns available = 4 hits total. So, basically, Side B has just lost 3 ships (5 ships remaining) - and Side A only 1 (7 remanining). Round 2: Side A has 7x4x2 = 56 guns available = 5.6 +0.4 (leftover from last round) = 6 hits Side B has 5x2x2 = 20 guns available = 2.5 hits - so, 2 hits, with 0.5 carried over. So, Side B again loses 3 ships (2 remaining) - while Side A loses 0.5 ships (let's say damaged) Round 3: Side A has 6.5x4x2 = 52 guns available = 5.2 hits & more than enough to wipe out the enemy. Side B has 2x2x2 = 8 guns available = 1 hit And the Side B has been wiped out - while Side A has barely had a quarter of it's strenght taken out. I'd take a victory like that any day of the week. There are another points. The side which enemy crossed the T usually cannot fire from all ships as they are too far or have obstructed firing solution. This add additional advantage to crossing T. But large column of ships is needed for that. 2 or 3 ships can easily and quickly adapt by changing course after the first sighting.
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Post by tortugapower on Nov 3, 2019 18:20:33 GMT -6
IMHO, 'crossing the T' is nothing but a simple exercise in mathematics: The side that's doing the crossing will be able to bring more guns to bear, and thus inflict more pain on the enemy - and since wars (and thus battles) are won by the side that inflicts the most pain on the enemy - the side that's doing the crossing will automatically win. The reduced chance of hitting is still offset by the additional number of guns able to fire on the enemy. Hell, let's make a mathematical model - Lanchester style: Side A is 'crossing the T' on side B. They both have 8 ships with guns in 4x2 configuration - however, Side A will, naturally, have all it's guns available - but Side B will only have 1/2 of it's guns initially available - so, basically a 2x2 configuration. Let's take into account the reduced chance of hitting by giving Side A 10 guns to make a hit - while Side B needs only 8 guns to make a hit.I don't disagree with your result based on your math, but your math is made up (as in, I don't think it has a real-world basis). I underlined the line of importance.
The math is trivial: if you are firing half your guns, then you need to hit twice as often per gun. The difficult question is: what are the percentages in real life?
That's what my theory-crafting is trying to do: come up with some numbers that will allow us to see how effective crossing fire is versus crossed fire.
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Post by srndacful on Nov 3, 2019 23:18:41 GMT -6
dorn: you are absolutely right - in any kind of battle, the 'crossing the T' advantage would not last more than a round - still, that round might just prove decisive, since it still gives Side A the chance to strike a significant blow that will start a 'death spiral' for Side B. OTOH, it's now irrelevant since: tortugapower: quite right - I was pulling my math out of my a$$, contributing nothing to the discussion, and I hereby apologise for derailing the thread. That said: have you taken a look at the dispersal patterns of land artillery? I've done a quick Google search and it came up with www.poeland.com/tanks/artillery/dispersion.html. I don't know just how much it'd be helpful but ... Best of luck and cheers!
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Post by director on Nov 4, 2019 0:41:24 GMT -6
There is also the issue of shells fired from multiple ships creating splashes that confuse the gunnery direction of firing ships.
A related argument was heatedly fought-over in the 1700s up to the Napoleonic Wars, concerning whether troops should be manuevered tactically in line or column.
One side argued that the line allowed the greatest number of men to fire upon an advancing enemy, which only the fire from a similar line could counter. The difficulty here is that it is difficult to advance in a line abreast; this seems to translate into naval tactics also.
The other side argued that a unit advancing in a column could not only move more quickly than a line but also equal or exceed its firepower.
---------------------------------------------------------
II II II II II II II
You will have to imagine the column of 'I' characters centered on the line - board won't let me do it. Now imagine the left 'I' firing slightly to his left and the right 'I' doing the same to the right so that everyone in the approaching column engages the line.
These proponents were undeterred by parade-ground exercises showing it was not possible to advance while keeping up a steady fire slightly to one side or the other - the formation entirely collapses while soldiers attempt to do it. The argument was finally resolved by the Napoleonic Wars, which saw the French (and then everyone) manuever in column (as was traditional) but do so right up to the engagement zone or even into it (which was not), forming quickly into line for firepower or remaining in column for assault.
In a naval sense, there simply is no sane, competent admiral who will accept the enemy crossing his T if he can do anything to prevent it (with the manuever of the Austrian fleet at Lissa being a desperate, 'nothing else to try' expedient that tests the rule - Tegetthof would have fought a conventional engagement if he'd had the materiel to do so).
So: for my part, it doesn't matter much whether it is best to cross the T or have your T crossed; the human animal rebels against steaming head-long into the enemy's guns and will turn aside to return effective fire unless otherwise prevented.
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Post by griffin01 on Nov 5, 2019 8:18:48 GMT -6
IMHO, 'crossing the T' is nothing but a simple exercise in mathematics: The side that's doing the crossing will be able to bring more guns to bear, and thus inflict more pain on the enemy - and since wars (and thus battles) are won by the side that inflicts the most pain on the enemy - the side that's doing the crossing will automatically win. The reduced chance of hitting is still offset by the additional number of guns able to fire on the enemy. Hell, let's make a mathematical model - Lanchester style: Is it, though? Do we have any substantial empirical evidence (because the attrition coefficients for Lanchester-style models are usually derived from empirical analysis, AFAIK) that double the amount of guns would suffice to make up for the difficulty aiming? Moreover, how do you determine the amount of hits to be allocated? The amount of hits needed to sink/disable a ship? Your approach is mathematically sound, I suppose, but makes a set of assumptions that might not necessarily be in place when considering a practical situation. If you don't mind me nitpicking, I'd also dare say that winning battles does not necessarily correlate with winning wars - as certain Pyrrhus once said, "One more such victory and I will be undone".
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Post by srndacful on Nov 5, 2019 23:37:21 GMT -6
IMHO, 'crossing the T' is nothing but a simple exercise in mathematics: The side that's doing the crossing will be able to bring more guns to bear, and thus inflict more pain on the enemy - and since wars (and thus battles) are won by the side that inflicts the most pain on the enemy - the side that's doing the crossing will automatically win. The reduced chance of hitting is still offset by the additional number of guns able to fire on the enemy. Hell, let's make a mathematical model - Lanchester style: Is it, though? Do we have any substantial empirical evidence (because the attrition coefficients for Lanchester-style models are usually derived from empirical analysis, AFAIK) that double the amount of guns would suffice to make up for the difficulty aiming? Moreover, how do you determine the amount of hits to be allocated? The amount of hits needed to sink/disable a ship? Your approach is mathematically sound, I suppose, but makes a set of assumptions that might not necessarily be in place when considering a practical situation. If you don't mind me nitpicking, I'd also dare say that winning battles does not necessarily correlate with winning wars - as certain Pyrrhus once said, "One more such victory and I will be undone". As a matter of fact - and looking at the page I found and posted in the answer to Tortuga's post above - I now think that I might have seriously miscalculated and that the odds of hitting are even more in favour of the side doing the 'crossing the T'. Based on a gun's dispersion pattern, it certainly seems to be much easier to hit a ship when it's facing towards you, than when it's facing sideways - even if we take the height (and the chance of hitting the superstructure even if it misses the hull) into account. As for being a nitpicker - being a kind of a nitpicker myself, I certainly don't mind. With that said, though, I'll invite you to take a better look at that part of my post above, and note that I said 'wars are won by the side that inflicts more pain on the enemy' - which is the most important part here because, as I'm sure you'll agree, winning wars is the most important part of the whole story -with battles being only the means to achieve this end. So, with that in mind, would you kindly tell me which side suffered more pain: Pyrrhus for losing 10.000 out of his 100.000 potentially recruitable men, or Rome for losing 20.000 of it's 400.000 potentially recruitable men? (note: numbers mostly derived from my a$$ - but with a serious attempt to make it as close to reality as my mk.1 eyeball will let me)
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Post by dorn on Nov 6, 2019 1:19:51 GMT -6
Is it, though? Do we have any substantial empirical evidence (because the attrition coefficients for Lanchester-style models are usually derived from empirical analysis, AFAIK) that double the amount of guns would suffice to make up for the difficulty aiming? Moreover, how do you determine the amount of hits to be allocated? The amount of hits needed to sink/disable a ship? Your approach is mathematically sound, I suppose, but makes a set of assumptions that might not necessarily be in place when considering a practical situation. If you don't mind me nitpicking, I'd also dare say that winning battles does not necessarily correlate with winning wars - as certain Pyrrhus once said, "One more such victory and I will be undone". As a matter of fact - and looking at the page I found and posted in the answer to Tortuga's post above - I now think that I might have seriously miscalculated and that the odds of hitting are even more in favour of the side doing the 'crossing the T'. Based on a gun's dispersion pattern, it certainly seems to be much easier to hit a ship when it's facing towards you, than when it's facing sideways - even if we take the height (and the chance of hitting the superstructure even if it misses the hull) into account. As for being a nitpicker - being a kind of a nitpicker myself, I certainly don't mind. With that said, though, I'll invite you to take a better look at that part of my post above, and note that I said 'wars are won by the side that inflicts more pain on the enemy' - which is the most important part here because, as I'm sure you'll agree, winning wars is the most important part of the whole story -with battles being only the means to achieve this end. So, with that in mind, would you kindly tell me which side suffered more pain: Pyrrhus for losing 10.000 out of his 100.000 potentially recruitable men, or Rome for losing 20.000 of it's 400.000 potentially recruitable men? (note: numbers mostly derived from my a$$ - but with a serious attempt to make it as close to reality as my mk.1 eyeball will let me) Guns dispersion patterns are not the most important thing. As William noted the evaluation of distance is big issue especially as ship move towards you. If error of such ranging is so high dispersion patterns of several hundreds of yards does not matter too much. Try some arcade game and usually you have issue more with range and you exactly know target position and speed, in reality you just guess both.
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Post by stevethecat on Nov 6, 2019 3:03:20 GMT -6
In regards to accuracy in firing at a head on target it seems a bit of a trade off.
You are aiming at a smaller target, but you have removed the horizontal travel from your aiming calculations, you only need to adjust for vertical for distance.
So it should be much easier to bracket the target but harder to get hits. And those hits might not be as effective at sinking the ship, although hitting the turrets and bridge from front on might remove the target as a threat long before it sinks. Ala Bismarck after Rodney ended its existence as a warship long before the ship sank by destroying the turrets.
The in battle advantage of that is that for two equal ships the broadside ship will have more shells bracketing but more will bounce and splash. While the front on ship will have less shells firing and more difficulty bracketing the target.
But that's 1 v 1, consider 2 battle lines of QE class, 5 in each.
The broadside line could be firing 40 shells at one target, a smaller target or not that poor ship is grandly screwed. In return the other line can fire 4 shells. Not entirely fair! (Especially considering that many ships could not fire over their bows.)
The classic advantage to crossing the T was that raking fire down a ships length was utterly devastating, while that aspect has become obsolete it has been replaced with maximising your firepower while minimising the enemies. A few more shells bounce? Seems a small price to pay for an overwhelming firepower advantage.
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Post by tortugapower on Nov 6, 2019 21:40:17 GMT -6
Theory Post #2A: Vertical vs. Horizontal Error, Revisited
We have a very trusting group, which is great. Unfortunately, I betrayed that trust by not vetting my own plots.
My previous post (now edited) had a significant error, with respect to horizontal versus vertical dispersion as a function of gun elevation. It showed that horizontal dispersion was greater than vertical at around 10 degrees. Below is the correct plot. I'm using equal variation (+/- 0.1 degrees) for both horizontal and vertical here. With equal errors for the both turret directions (elevation and rotation), crossover is ~33 degrees = ~33 km = ~20 miles -- much farther than most engagements. As mentioned before, there is every reason to believe that vertical error should vary more than horizontal. I have also managed to generate splashdown patterns. These are really fun to look at! Let's take a look at the dispersion pattern for three elevations, and I'll remind us what range that would be targeting. Elevation of 5 degrees = 10 km / 6 milesI'll have to add 10 degrees and 35 degrees in the next post. There is a limit of only two attachments per post!
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Post by tortugapower on Nov 6, 2019 21:47:57 GMT -6
Theory Post #2B: Vertical vs. Horizontal Error, Revisited (II)(Continuing from previous, splashdown at various elevations/ranges) Elevation of 10 degrees = 18 km / 11 miles Elevation of 35 degrees = 39 km / 24 mileNote: the axes are not perfectly to scale, but they are close. So you are (nearly) seeing splashdown dispersion as a helicopter pilot above the target location would. So, this shows that, even for equal errors in elevation and rotation (which is probably the worst-case scenario for horizontal/rotational error), splashdown variation is a lot greater along the direction of shell travel until you get to very, very long ranges.
There are two really big considerations missing from this picture of error so far. 1. Shell trajectories (i.e. splashdown does not tell the full story) 2. Human errors (i.e. in ranging). My goal will be to address #1 in the next theory post. Sorry again for the previous incorrect plot.
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Post by vidboi on Nov 8, 2019 6:08:41 GMT -6
Great thread and interesting stuff, I'll be looking forward to the next post. One thing I believe you are missing though is the error in shell velocity (due to errors in shell and propellant weights) which should increase the dispersion along the length of travel. Also, for reference data, the USN carried out a study in 1944 evaluating the accuracy of the Iowa-class against targets of the same size in a similar manner to what you're doing here. A quick summary of the results can be found here: www.navweaps.com/Weapons/WNUS_16-50_mk7.php#Accuracy_During_World_War_II
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Post by tortugapower on Nov 11, 2019 17:06:15 GMT -6
Great thread and interesting stuff, I'll be looking forward to the next post. One thing I believe you are missing though is the error in shell velocity (due to errors in shell and propellant weights) which should increase the dispersion along the length of travel. Also, for reference data, the USN carried out a study in 1944 evaluating the accuracy of the Iowa-class against targets of the same size in a similar manner to what you're doing here. A quick summary of the results can be found here: www.navweaps.com/Weapons/WNUS_16-50_mk7.php#Accuracy_During_World_War_IIThank you! The navweaps articles pair very well and is great background for this. (I've been directed there previously, and it's where I'm getting a lot of my data.) "The data used to calculate these probabilities came from target practice shoots carried out by the battleships." So long as they are using real-sized target profiles, my data should match it. The weird thing to me is: why does their ratio (of broadside vs. end-on) first get better, then get worse? That doesn't make sense from a shell dispersion analysis standpoint, unless I'm missing something.
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Post by vidboi on Nov 12, 2019 15:11:56 GMT -6
"The data used to calculate these probabilities came from target practice shoots carried out by the battleships." So long as they are using real-sized target profiles, my data should match it. The weird thing to me is: why does their ratio (of broadside vs. end-on) first get better, then get worse? That doesn't make sense from a shell dispersion analysis standpoint, unless I'm missing something. I believe it's due to the how the shot distribution and target profiles change with range. At short ranges, the dispersion is small compared to the size of the target, so the ratio of chance to hit is lower as it's easy to hit a head on target. As ranges and dispersion increase the target profile becomes more important, causing the peak in the ratio. As the dispersion becomes large with respect to the ship, increases in dispersion with range decrease the chance the hit for both head-on and broadside targets uniformly, so the effect on ratio becomes smaller. At this point the difference in the target hit area (i.e. the area on the surface where a shell landing would pass through the target) takes over, and this ratio becomes smaller as the angle of fall increases, explaining the decrease in the hit chance ratio at longer ranges.
Annoyingly I can't find the original source anywhere
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Post by atparmentier on Nov 14, 2019 13:43:05 GMT -6
Great thread and interesting stuff, I'll be looking forward to the next post. One thing I believe you are missing though is the error in shell velocity (due to errors in shell and propellant weights) which should increase the dispersion along the length of travel. Also, for reference data, the USN carried out a study in 1944 evaluating the accuracy of the Iowa-class against targets of the same size in a similar manner to what you're doing here. A quick summary of the results can be found here: www.navweaps.com/Weapons/WNUS_16-50_mk7.php#Accuracy_During_World_War_IIAre those percentages for one gun or a broadside? Because if it's for one gun it would mean on the lower end that 9 guns to a vertical target have about 0.30 more hits per salvo than 6 gun to a broadside target.
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Post by evil4zerggin on Nov 16, 2019 3:30:54 GMT -6
If our main concern is how aspect affects the hit probability, then there is a simpler way to analyze this while also taking into account danger space etc.: compute the angle of fall and then consider the ship's silhouette as viewed from that angle. For example, you can take a 3D model of a ship and zoom out a fair distance in perspective projection, or use orthographic projection. If the dispersion is much larger than the ship you could assume that it is roughly uniform over the area of the ship and therefore the chance of a hit is proportional to the size of the silhouette; otherwise, you could compute the dispersion pattern normal to the direction of shell travel and overlay it. This allows us to illustrate the following: "The data used to calculate these probabilities came from target practice shoots carried out by the battleships." So long as they are using real-sized target profiles, my data should match it. The weird thing to me is: why does their ratio (of broadside vs. end-on) first get better, then get worse? That doesn't make sense from a shell dispersion analysis standpoint, unless I'm missing something. I believe it's due to the how the shot distribution and target profiles change with range. At short ranges, the dispersion is small compared to the size of the target, so the ratio of chance to hit is lower as it's easy to hit a head on target. As ranges and dispersion increase the target profile becomes more important, causing the peak in the ratio. As the dispersion becomes large with respect to the ship, increases in dispersion with range decrease the chance the hit for both head-on and broadside targets uniformly, so the effect on ratio becomes smaller. At this point the difference in the target hit area (i.e. the area on the surface where a shell landing would pass through the target) takes over, and this ratio becomes smaller as the angle of fall increases, explaining the decrease in the hit chance ratio at longer ranges.
Annoyingly I can't find the original source anywhere At close range the dispersion is not much wider than the ship, so the broadside is only a bit more likely to be hit than the front. At medium range the dispersion increases, covering more of the broadside, but the front of the ship was already covered so this relatively benefits the shot at the broadside. At long range the shell sees more of the deck, which, unlike the vertical surfaces, is the same area for both front and broadside. Model: gamemodels3d.com/games/worldofwarships/vehicles/pasb008
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