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Lanchester Models and the Battle of Britain

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Lanchester Models and the Battle of Britain Lanchester models and the Battle of Britain Ian Johnson and Niall MacKay Naval Research Logistics 58 (2011) 210-222, memorial volume for Rick Rosenthal Frederick Lanchester (1868-1946) Automotive and aeronautical engineer who built the first British motor car Frederick Lanchester (1868-1946) Automotive and aeronautical engineer who built the first British motor car Invented the carburettor, disc brakes, the accelerator pedal; theory of lift and drag. Lanchester’s equations (1914) The aimed-fire model: G(t) Green units fight R(t) Red units. dG = rR dt − Green’s instantaneous loss-rate is proportional to Red numbers dR = gG dt − and vice versa. Lanchester’s equations (1914) The aimed-fire model: G(t) Green units fight R(t) Red units. dG = rR dt − Green’s instantaneous loss-rate is proportional to Red numbers dR = gG dt − and vice versa. Divide: dR gG = or rR dR = gG dG dG rR and integrate: 1 2 1 2 rR = gG + constant 2 2 throughout the battle. Implications of Lanchester’s equations 2 2 rR gG − tells us how to combine numbers (R, G) and effectiveness (r, g). Implications of Lanchester’s equations 2 2 rR gG − tells us how to combine numbers (R, G) and effectiveness (r, g). Numbers win: suppose we begin with twice as many Reds as Greens, R0 = 2G 0, but that Greens are three times more effective, g = 3r . Then 2 2 2 2 2 rR gG = r(2G 0) 3rG 0 = rG 0 > 0, − − and Red wins: the battle ends with G = 0, R = G 0. Implications of Lanchester’s equations 2 2 rR gG − tells us how to combine numbers (R, G) and effectiveness (r, g). Numbers win: suppose we begin with twice as many Reds as Greens, R0 = 2G 0, but that Greens are three times more effective, g = 3r . Then 2 2 2 2 2 rR gG = r(2G 0) 3rG 0 = rG 0 > 0, − − and Red wins: the battle ends with G = 0, R = G 0. Concentration is good: If Red divides its forces, and Green fights each half in turn, Green wins the first battle, with p2/3 80% of G 0 remaining, ' Green wins the second battle, with p1/3 60% of G 0 remaining. ' Some variants of Lanchester’s equations Ancient warfare, along a fixed, narrow battle-line with N(t) fighting on each side: dG dR = rN = gN dt − dt − Modern warfare, but with hidden targets (the unaimed-fire model): dG dR = rRG = gGR dt − dt − Either way, dR/dG is now fixed, and the constant quantity is rR gG , − which is much more intuitive: fighting strength is just numbers effectiveness. × Asymmetric warfare Green attacks, Red defends: dG dR R = rR, = gG dt − dt − R0 and 1 2 rRR0 gG − 2 is conserved, so that – Green benefits more from numbers and concentration, but – needs g to be twice as great, or G 0 to be √2 times as great, as in a symmetric aimed-fire battle. Asymmetric warfare Green attacks, Red defends: dG dR R = rR, = gG dt − dt − R0 and 1 2 rRR0 gG − 2 is conserved, so that – Green benefits more from numbers and concentration, but – needs g to be twice as great, or G 0 to be √2 times as great, as in a symmetric aimed-fire battle. Red has a defender’s advantage A generalized Lanchester model fits loss-rates to powers of own and enemy numbers: dG dR = rRr 1 G g 2 = gG g 1 Rr 2 dt − dt − A generalized Lanchester model fits loss-rates to powers of own and enemy numbers: dG dR = rRr 1 G g 2 = gG g 1 Rr 2 dt − dt − Divide and re-arrange: − − gG g 1 g 2 dG = rRr 1 r 2 dR Integrate: the conserved quantity is r g Rρ G γ . ρ − γ where ρ =1+ r 1 r 2 and γ =1+ g 1 g 2 − − A generalized Lanchester model fits loss-rates to powers of own and enemy numbers: dG dR = rRr 1 G g 2 = gG g 1 Rr 2 dt − dt − Divide and re-arrange: − − gG g 1 g 2 dG = rRr 1 r 2 dR Integrate: the conserved quantity is r g Rρ G γ . ρ − γ where ρ =1+ r 1 r 2 and γ =1+ g 1 g 2, the exponents, − − capture the conditions of battle: – Red should concentrate its force if ρ > 1, divide if ρ < 1. – if ρ < γ then Red has a defender’s advantage, by a factor γ/ρ Do Lanchester equations describe real warfare? (either sets of battles or time-series within battles) Do Lanchester equations describe real warfare? (either sets of battles or time-series within battles) Fits to land- and all-arms battles (US Civil War, Iwo Jima, Kursk, Ardennes, Korean War) are poor. Do Lanchester equations describe real warfare? (either sets of battles or time-series within battles) Fits to land- and all-arms battles (US Civil War, Iwo Jima, Kursk, Ardennes, Korean War) are poor. Of course: Lanchester’s equations are temporally and spatially homogeneous, with no command-and-control or variation in terrain or tactics. Do Lanchester equations describe real warfare? (either sets of battles or time-series within battles) Fits to land- and all-arms battles (US Civil War, Iwo Jima, Kursk, Ardennes, Korean War) are poor. Of course: Lanchester’s equations are temporally and spatially homogeneous, with no command-and-control or variation in terrain or tactics. F W Lanchester Aircraft in Warfare: the dawn of the fourth arm (London: Constable & Co., 1916) How about a (purely) aerial battle? The Battle of Britain A battle of attrition and intended annihilation, in which one day’s fighting was much like another, the single-seat fighters on each side were well-matched, and all units were seeking engagement. The Battle of Britain A battle of attrition and intended annihilation, in which one day’s fighting was much like another, the single-seat fighters on each side were well-matched, and all units were seeking engagement. Take daily loss-rates for RAF (δR) and Luftwaffe (δG ) aircraft and fit to RAF (R) and Luftwaffe (G) daily sortie numbers: Find the parameters r, r 1, r 2, g, g 1, g 2 for which the data best fit δR = gG g 1 Rr 2 , δG = rRr 1 G g 2 by linear regression onto log δR = log g+g 1 log G+r 2 log R , log δG = log r+r 1 log R+g2 log G Are the days independent? Losses / Sorties 0.01 0.02 0.03 0.04 0.05 0.06 0 10 20 30 40 50 Day Losses per sortie (δG + δR)/(G + R) vs day RAF losses dR dR 5 10 15 20 25 30 35 5 10 15 20 25 30 35 500 1000 1500 2000 200 400 600 800 1000 1200 G R δR vs G δR vs R RAF losses log(dR) log(dR) 1.0 1.5 2.0 2.5 3.0 3.5 1.0 1.5 2.0 2.5 3.0 3.5 4.5 5.0 5.5 6.0 6.5 7.0 7.5 5.5 6.0 6.5 7.0 log(G) log(R) log δR vs log G log δR vs log R RAF losses dR 1 12±0 17 0 18±0 25 = gG . R . dt − 1 2 2 = gG . (ΣR = 0.66) − RAF losses dR 1 12±0 17 0 18±0 25 = gG . R . dt − 1 2 2 = gG . (ΣR = 0.66) − Hooray for Lanchester! Luftwaffe losses log(dG) log(dG) 1.5 2.0 2.5 3.0 3.5 4.0 1.5 2.0 2.5 3.0 3.5 4.0 5.5 6.0 6.5 7.0 4.5 5.0 5.5 6.0 6.5 7.0 7.5 log(R) log(G) log δG vs log R log δG vs log G Luftwaffe losses dG 0 00±0 25 0 86±0 18 = rR . G . dt − 0 9 2 = gG . (ΣR = 0.49) − Luftwaffe losses dG 0 00±0 25 0 86±0 18 = rR . G . dt − 0 9 2 = gG . (ΣR = 0.49) − Not so good. In fact, fitting to R alone, dG 0 87±0 22 R . dt ∝ explains only ΣR2 = 0.24. Subtleties I G and R are highly correlated (0.74): log R 5.5 6.0 6.5 7.0 4 5 6 7 log G log R vs log G and so the overall powers in the loss-rates, g 1 + r 2 and r 1 + g 2, are better-determined than their constituents: variation is less significant along the lines of constant g 1 + r 2 and r 1 + g 2 than orthogonal to them. Subtleties II When g 1 + r 2 = 1 or r 1 + g 2 = 1, autonomous battles (‘raids’) 6 6 should not be aggregated into daily data. If they are, the effect is to push the overall powers g 1 + r 2 and r 1 + g 2 away from their true values and towards one, and to reduce the quality of the fit. Subtleties II Example: y = x2 y 0 10 20 30 40 50 1 2 3 4 5 6 7 x has log y = 2 log x, of course. Subtleties II Example: y = x2 and sums of these: e.g. not only (3, 9) but also (1 + 2, 1 + 4) = (3, 5) and (1+1+1, 1 + 1 + 1) = (3, 3). y 0 10 20 30 40 50 1 2 3 4 5 6 7 x and the best fit is now log y = 1.5 log x, with ΣR2 = 0.6.
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