Wells has a very blunt approach to melees - if troops come into contact, they wipe each other out. Thus two equal-sized units will just eliminate each other, and if a unit of m men is involved in a melee with a smaller unit of strength n, the only survivors of this awfulness will be (m - n) men from the first unit.
Obviously it works, and it is quick, but it does seem a bit crude. I also had a quick squint in Peter Young's Charge! - the melee procedure there is refined by the introduction of probability (dice) and limiting the number of rounds in each melee, but otherwise the line of descent is clear to see.
I dug out the Theorems of Frederick W Lanchester, just to check. If you are familiar with Lanchester then read no further, but FWL was an English engineer and mathematician, who died in 1946 (same year as Wells), and he is most famous as an automotive inventor. I wish to mention, in passing, that my Uncle Harold had a green Lanchester 10 saloon when I was a boy, and a big solid thing it was, too - the Lanchester marque was swallowed by the British Daimler company during the 1950s. More relevantly, in 1916 – 3 years after the publication of Little Wars - Dr Lanchester produced a mathematical analysis of warfare, and the two best known elements of this are his Linear Law (an abstraction of ancient-style warfare) and his n-Square Law. If you look these up on the internet, you'll find so much diverse explanation that it is hard to believe that it all relates to the same ideas. In the interests of providing yet more redundant information on a subject which has already been hammered to death, I'll attempt to provide yet another lightweight view on Lanchester's Laws!
The Linear Law
Imagine two groups of warriors, armed only with (say) a club. One of them has m men, the other has n (a smaller number). Men can only fight one-against-one, so anyone who has no-one to fight presumably stands and watches, cheering (or placing bets?). Anyway, in this remarkably organised and chivalrous form of melee, they match off into n fighting pairs. Assume that each man has an even chance of winning his fight. On a given word or command, there is an almighty thwack! and the casualties are removed. On average, we would expect each side to lose n/2 men. So the (m - n/2) survivors of the first group will now fight the n/2 survivors of the second group. Since n/2 is the smaller group, there will be n/2 fights, of which each side will lose n/4.
And so on. If you are keen on the theory of finite differences, you can solve this as the sum of a series. If not, you can put in some real numbers and do it on a spreadsheet.
If the first force is 1000 men and the second force is 500, and each man has an equal chance of winning each fight, then - on average - we find that we can expect the first force to wipe out the second, while themselves losing 500 men. If the forces were of equal strength, they would eliminate each other.
Step forward, Mr Wells [applause] - in this rather stilted form of combat your rule is exactly correct. It does rather gloss over what the unemployed warriors would be doing during each thwack, for example, and it also ignores the possibility that someone might decide this was a bad idea before reaching the point of actual annihilation. Otherwise, nice job.
The n-Square Law
Let's now consider a more modern form of warfare, in which all the troops on one side are able to kill any of the troops on the other side - perhaps they are all armed with long-range automatic weapons.
If, again, there are m one side and n on the other, and if each man in the first side has a killing rate (the number of enemy troops he can kill in 1 unit of time) equal to Km (this allows for his own effectiveness, the defensive capability of his opponents, and any other relevant contextual factors), and the other side has a killing rate of Kn, then the actual rates of loss will be proportional to the numbers of men
Now, the armies would be considered equally matched if they are wiping each other out (proportionally) at the same rate - i.e. would both be reduced to half strength at the same time, for example.
In this special case, we have m Km = n Kn at any instant; if we substitute in (1) & (2) and integrate, we get
This is Lanchester’s n-Square Law – in words, the total fighting strengths of two forces are equal when their fighting effectiveness (killing rate), multiplied by the square of the numerical strength, is equal.Thus a force which has half the numerical strength of the opposition would have to have four times its killing rate to be equally matched.
Let's look, again, at our 1000-man force attacking a 500-man one, with equal values of Kn & Km. Running it on a spreadsheet demonstrates that we should expect the 500 man force to be eliminated for a cost of 130-something casualties to the other side. This is very different from the HG Wells situation. It also demonstrates the importance of concentrating your armies, thus:
if two identical 1000-man forces engage, then Lanchester’s n-Square Law has them eliminate each other. However, if one side divides (or is divided) into two 500-man forces, then the 1000-strong enemy force will eliminate the first 500 men and still have 870 or so troops left, which is more than enough to eliminate the second 500 men. Napoleon was right, even though he never met Dr Lanchester.
And that is quite enough of that.
umh, my copy of Little Wars must be a bit different, It has the pictures and diagrams but the melee rule is also slightly different as it a force of 2:1 captures its enemy if it is unsupported. So if 500 toy soldiers, ok that's unlikely, say 5 toy soldiers attack 10. The 5 would be captured at no loss to the 10 apart from 1 guard to escort the prisoners, but if they left 2 men in support and attacked with 3 then each side would lose 3 (or 3,000 which is closer). On the next turn the remaining 7 could of course move up and snap up the remaining 2 at no cost if they stood around and no one else intervened. If the numbers are not 2:1, they kill each other off 1 for 1 until one side is 2:1 and then the small group surrenders. To change the equation slightly if 8 say faced 10, they would kill each other off 1 for 1 until 1 side outnumbers the other by 2:1 at which point the small side surrenders. So 8 vs 10 would mean the small side has 6 dead and 2 captured vs 6 dead from the other side, unless there were supports zzzzzzzzz. I'm not about to feed 800 vs 1,000 into Lancaster's theory to see what the result ought to be.
ReplyDeleteOne day I must actually play a game of Little Wars, with perhaps a slight modification to the actual cannon fire.
-Ross
Section III of the version I have sets out what are described as "the rules of the perfect battlegame as we play it in an ordinary room" - it is all a bit confusing, since the book, as you know, describes the initial form of the game and then discusses the evolution and refinement of the rules.
ReplyDeleteUnder "Hand to Hand Fighting and Capturing", the melee system is given as I described - the only additional rule is if the forces are unequal and the smaller force is "isolated" (definition given in the rules, based on proximity and numbers of friendly troops), in which case some of the losses will be prisoners.
Example Wells gives is 11 men fighting 9. "Isolation" variant gives (11 - 9) = 2 prisoners, and the balance of the 9 losses will be casualties, so each side will lose 7 dead, leaving the larger side with 4 men guarding 2 prisoners, and the smaller side with nothing.
I had a squint through for the 2:1 rule you mention, but didn't find it - I'm sure you're right - my eyesight doesn't do joined-up reading any more! What seems pretty clear is that Wells didn't want fiddly melees wasting time that could be better used crawling round the floor firing toy cannons.
The idea of fighting battles on the floor has always appealed, though my knees would not appreciate the idea. But think of the space - wow. For a while, I thought of doing sieges on the floor. On the other hand, think of the collateral damage every time someone walked through. And then there's dogs and stuff. I was looking at some photos of a Little Wars type action in a vast and rather barren church hall - looked very like these domino-fall world record attempts - a couple of appropriately nerdy guys painstakingly setting the troops in neat rows while their friends (including, dare I mention it, some young ladies - obviously of Wells' "more intelligent sort") conducted an independent social gathering in a remote corner.
I'd certainly like to see the game played. Are there any LittleWars type blogs out there?
Cheers - thanks for thoughts, as ever
Tony
Ross - as I pressed the Publish button I realised that the process where the men are killed off until a 2:1 advantage arises is described in the book - it's arithmetically exactly the same as the "isolation" variant I described.
ReplyDeleteIf m, n are the sizes if the units (m > n), and if k, p are the numbers of killed and prisoners, then, doing it the 2:1 way, we want to remove k dead, one by one, until the 2:1 superiority holds
then m - k = 2 (n - k)
k = 2n - m
and p is the balance of the n losers
p = n - (2n - m) = m - n
Doing it the way I described, you know straight away that p is the difference between the two forces
p = m - n
and the balance of the n casualties (for each side) must be
k = n - (m - n) = 2n - m
which is the same as before. Apologies if this is completely obvious by inspection, but it would have bugged me for the rest of the day if I hadn't proved it!
Tony
Actually I think he describes it both ways at different times. The sad thing is that I didn't get up and check the book, just quoted out of my head.
ReplyDeleteHelps explains why there is so little room for useful info.
-Ross