I recently attended the DSTL Symposium on Historical Analysis (of defence related topics). One of the talks was "Dogger Bank, Weighing the fog of war", based on [1] and summarised in this article in Significance [2] and the slides from the DSTL presentation (which are not very informative). Unfortunately [1] is behind a paywall and so unless you know how to obtain copies of such papers without paying you are a bit stuck (I did request a copy form one of the authors on their Research Gate page but have had no response). There is also a longer version of the talk given to the Naval War College on YouTube. There is also a related paper [3] on Jutland which also looks interesting, but which I have not yet had time to procure or study, but expect I will get around to it in the near future.
As the paper deals with matters - Bayesian Statistics, Lanchester Like Models of Combat, Monte-Carlo Simulation, Operational Analysis, Naval History, ... - in which I have both a professional and personal interest I have acquired a copy of the original paper. The paper is quite interesting but I find it difficult to follow when describing the combat model (worrying since I am some sort of expert on this sort of stuff). Before I go any further I should say that I like the general idea of this type of analysis, and that I am inclined to agree with the conclusions in [1].
The basic model of combat is idealised to an exchange of fire one turret at a time (not sure what target selection, or turret scheduling rules are applied, presumably those that reproduce what happened in the actual battle - Tiger we are looking at you here. I suspect a next event type simulation with the next turret/gun ready to fire being the critical timing events is employed). But the ships are characterised by gun data: "Rate of fire", "Number of turrets", "Accuracy", "Effectiveness", and the ship itself by "Resilience", "Flash" and "Disablement".
"Accuracy" appears to be the hit probability (assumed independent shot-to-shot) and reflects the actual hit proportions observed at Dogger Bank and Jutland. The "Accuracy" is 3% for the Battle Cruiser Fleet, and 5% for the 1st Scouting Group which seems about right (I would have preferred two digits of precision here) and is consistent with my own models of naval gun fire at a gross level.
As the paper deals with matters - Bayesian Statistics, Lanchester Like Models of Combat, Monte-Carlo Simulation, Operational Analysis, Naval History, ... - in which I have both a professional and personal interest I have acquired a copy of the original paper. The paper is quite interesting but I find it difficult to follow when describing the combat model (worrying since I am some sort of expert on this sort of stuff). Before I go any further I should say that I like the general idea of this type of analysis, and that I am inclined to agree with the conclusions in [1].
The basic model of combat is idealised to an exchange of fire one turret at a time (not sure what target selection, or turret scheduling rules are applied, presumably those that reproduce what happened in the actual battle - Tiger we are looking at you here. I suspect a next event type simulation with the next turret/gun ready to fire being the critical timing events is employed). But the ships are characterised by gun data: "Rate of fire", "Number of turrets", "Accuracy", "Effectiveness", and the ship itself by "Resilience", "Flash" and "Disablement".
"Accuracy" appears to be the hit probability (assumed independent shot-to-shot) and reflects the actual hit proportions observed at Dogger Bank and Jutland. The "Accuracy" is 3% for the Battle Cruiser Fleet, and 5% for the 1st Scouting Group which seems about right (I would have preferred two digits of precision here) and is consistent with my own models of naval gun fire at a gross level.
Now we have the following enigmatic statement (Italic text added by current author for clarification.):
"The ship to fire then has certain probability of hitting its target "Accuracy"(Table 1); if a hit is achieved, the probability of damage is then computed. This probability is computed by multiplying two factors, which quantify respectively the effectiveness of the shell of the attacker "Effectiveness" and that of the armor of the defender (1 D resilience) "Resilience" to arrive at a probability. Damage here means probability of affecting the Lanchestrian unit, the turret.
This I must assume contains a typo, the probability of damage must go up with "Effectiveness" and down with "Resilience". I assume what is meant is something like (there seems to be something wrong with our bloody LaTeX today):
\[ P(\rm{damage}|\rm{hit})= \frac{\rm{effectivness\ of\ attacking\ gun}}{\rm{reslilience\ of\ defending\ ship}}\]
or more generally, that there exists some increasing function mapping \([-\infty,\infty]\) to \([0,1]\) such that:
\[ P(\rm{damage}|\rm{hit})=f\left( \frac{\rm{effectivness\ of\ attacking\ gun}}{\rm{reslilience\ of\ defending\ ship}}\right)\]
However there is a problem here; According to MacKay et al [1] Campbell [4] tells us that at Jutland the probability of a hit causing the loss of a turret is ~12% independent of nationality or of the individual ship, and the product form rather than that above is set up to approximately give this. In addition there must be all sorts of caveats on Campbell's assessment, as I'm sure that at least he does not wish it to apply to medium calibre hits on battleships. Note there is an implicit assumption that a damage causing (penetrating?) large calibre hit (on a turret) will knock the turret out, not an unreasonable assumption. (The authors have just confirmed 2017-09-28, by email that there is indeed a typo they used \( (1-\rm{resilience}) \))
OK.. lets look at how this would look at the battle of the Falkland Islands. Invincible took about 12 hits from 21cm projectiles according to Massie[6]. These are the same mass as those from Blucher but at 87% of the muzzle velocity which reduces the effectiveness from Blucher's 0.27 to ~0.20*, so the probability of a hit knocking out a turret on Invincible goes from ~10.3% to ~7.7%. Then the number of turrets knocked out by 12 hits has a binomial distribution B(12,0.077), and the probability that no turrets were knocked out is ~38%. Not overwhelmingly in agreement with the actual result, but also not sufficiently in disagreement to raise any eyebrows.
Anyway this post is becoming over long and I will end it here with the statement that I do not believe the Blucher's 21cm guns had anything like the relatively effectiveness compared to 11", 12" or 13.5" guns proposed in [1]. Which of course calls into question how the effectiveness used are calculated.
I expect I will have more to say about this paper in subsequent posts.
*How the effectiveness is obtained is not clear, the reference [5] does not give a value (unless I am reading wrongly) but just formulae for calculating T/D, penetration normalised to projectile diameter. Now it might be that the article has changed since Mackay et al retrieved it, but probably not. It appears that the "Effectiveness" is close to being proportional to \( \frac{W V^2}{D^3}\), where \(W\) is the projectile mass, \(V\) the muzzle velocity and \( D \) the projectile diameter. Thus for geometrically similar projectiles the MoE is proportional to \(V^2\). I have used this relation in scaling the effectiveness of the East Asiatic Squadron's 21cm guns from that given in [1] for the Blucher's. I can see an argument for using a MoE \(\propto \frac{W V^2}{D^2}\), that is proportional to the kinetic energy per unit area of the projectile cross section, but using this this would make no difference to the argument here.
\[ P(\rm{damage}|\rm{hit})= \frac{\rm{effectivness\ of\ attacking\ gun}}{\rm{reslilience\ of\ defending\ ship}}\]
or more generally, that there exists some increasing function mapping \([-\infty,\infty]\) to \([0,1]\) such that:
\[ P(\rm{damage}|\rm{hit})=f\left( \frac{\rm{effectivness\ of\ attacking\ gun}}{\rm{reslilience\ of\ defending\ ship}}\right)\]
However there is a problem here; According to MacKay et al [1] Campbell [4] tells us that at Jutland the probability of a hit causing the loss of a turret is ~12% independent of nationality or of the individual ship, and the product form rather than that above is set up to approximately give this. In addition there must be all sorts of caveats on Campbell's assessment, as I'm sure that at least he does not wish it to apply to medium calibre hits on battleships. Note there is an implicit assumption that a damage causing (penetrating?) large calibre hit (on a turret) will knock the turret out, not an unreasonable assumption. (The authors have just confirmed 2017-09-28, by email that there is indeed a typo they used \( (1-\rm{resilience}) \))
OK.. lets look at how this would look at the battle of the Falkland Islands. Invincible took about 12 hits from 21cm projectiles according to Massie[6]. These are the same mass as those from Blucher but at 87% of the muzzle velocity which reduces the effectiveness from Blucher's 0.27 to ~0.20*, so the probability of a hit knocking out a turret on Invincible goes from ~10.3% to ~7.7%. Then the number of turrets knocked out by 12 hits has a binomial distribution B(12,0.077), and the probability that no turrets were knocked out is ~38%. Not overwhelmingly in agreement with the actual result, but also not sufficiently in disagreement to raise any eyebrows.
Anyway this post is becoming over long and I will end it here with the statement that I do not believe the Blucher's 21cm guns had anything like the relatively effectiveness compared to 11", 12" or 13.5" guns proposed in [1]. Which of course calls into question how the effectiveness used are calculated.
I expect I will have more to say about this paper in subsequent posts.
*How the effectiveness is obtained is not clear, the reference [5] does not give a value (unless I am reading wrongly) but just formulae for calculating T/D, penetration normalised to projectile diameter. Now it might be that the article has changed since Mackay et al retrieved it, but probably not. It appears that the "Effectiveness" is close to being proportional to \( \frac{W V^2}{D^3}\), where \(W\) is the projectile mass, \(V\) the muzzle velocity and \( D \) the projectile diameter. Thus for geometrically similar projectiles the MoE is proportional to \(V^2\). I have used this relation in scaling the effectiveness of the East Asiatic Squadron's 21cm guns from that given in [1] for the Blucher's. I can see an argument for using a MoE \(\propto \frac{W V^2}{D^2}\), that is proportional to the kinetic energy per unit area of the projectile cross section, but using this this would make no difference to the argument here.
References:
1. MacKay N, Price C, Wood J, Weighing the fog of war:Illustrating the power of Bayesian methods for historical analysis through the Battle of the Dogger Bank, Historical Methods 2016 49 (2) pp80-91
2. MacKay N, Price C, Wood J, Dogger Bank:Weighing the fog of war, Significance 2017 14 (3) pp14-19
3. MacKay N, Price C, Wood J, Weight of Shell Must Tell: A Lanchestrian Reappraisal of the Battle of Jutland, History 2016 101 (347) pp536-563
4. Campbell, J., Jutland: An analysis of the fighting. Conway Maritime Press, 1986.
5. Okun, N. Major Historical Naval Armor Penetration Formulae.
http://www.navweaps.com/index_nathan/Hstfrmla. 2001
6. Massie R. K., Castles Of Steel: Britain, Germany and the Winning of The Great War at Sea , Jonathan Cape, 2004
3. MacKay N, Price C, Wood J, Weight of Shell Must Tell: A Lanchestrian Reappraisal of the Battle of Jutland, History 2016 101 (347) pp536-563
4. Campbell, J., Jutland: An analysis of the fighting. Conway Maritime Press, 1986.
5. Okun, N. Major Historical Naval Armor Penetration Formulae.
http://www.navweaps.com/index_nathan/Hstfrmla. 2001
6. Massie R. K., Castles Of Steel: Britain, Germany and the Winning of The Great War at Sea , Jonathan Cape, 2004
