{"id":9354,"date":"2013-09-20T06:00:14","date_gmt":"2013-09-19T20:00:14","guid":{"rendered":"http:\/\/www.aspistrategist.ru\/?p=9354"},"modified":"2013-10-01T09:02:51","modified_gmt":"2013-09-30T23:02:51","slug":"geek-of-the-week-frederick-lanchester-and-why-quantity-is-a-quality","status":"publish","type":"post","link":"https:\/\/www.aspistrategist.ru\/geek-of-the-week-frederick-lanchester-and-why-quantity-is-a-quality\/","title":{"rendered":"Geek of the week: Frederick Lanchester and why quantity has a quality all of its own"},"content":{"rendered":"

\"FE2b_aircraft_1916\"<\/a>In 1916, English mathematician (and poet, singer, pioneer aerodynamicist and designer of combustion engines) Frederick Lanchester<\/a> turned his mind to the subject of aerial warfare. In particular, he realised that the nature of war in the air\u2014a novelty at the time\u2014was fundamentally different to that of the slaughter underway on the ground below.<\/p>\n

When two massed armies clash on the ground along a wide front, to a good approximation the losses on each side depend on how large both armies are. Crudely put, the size of an army dictates how much fire it can throw the other way, but also how many targets it presents to the other guys. In (almost) mathematical terms, the losses to such \u2018unaimed fire\u2019 go like this:<\/p>\n

Army A losses = constant x (size of Army A) x (size of Army B)\u00a0\u00a0\u00a0\u00a0 (1)<\/p><\/blockquote>\n

where the constant depends on the effectiveness of Army B\u2019s weapons against Army A. There\u2019s a similar expression for B\u2019s losses. (For calculus loving Strategist<\/i> readers, there\u2019s an appendix below. Normal people should just read on.)<\/p>\n

Lanchester\u2019s great insight was to realise that while an artillery shell lobbed towards another army always has the chance of hitting something or someone, air warfare wasn\u2019t like that\u2014aeroplanes could only be shooting at one target at a time. That meant that your chance of being shot down depended only on the number of other guys there were to shoot at you. In the case of such \u2018aimed fire\u2019 the loss equation looks like this:<\/p>\n

Air Force A losses = constant x (size of Air Force B)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 (2)<\/p><\/blockquote>\n

It turns out that the difference between equations (1) and (2) is significant. If two massed armies are armed with equally effective weapons, they\u2019ll grind each other down in equal measure, and the only way to win is to have a force bigger than the other. Even then, both sides will suffer the same losses in the process. If Army A is twice the size of Army B, it will lose half its strength by the time it has overrun B. (At least in principle\u2014in practice battles often stop before sides suffer casualties of those magnitudes.) These observations go a long way towards explaining the appalling battlefield losses of WWI.<\/p>\n

But the air combat equation gives a qualitatively different result. It\u2019s the square<\/i> of the numbers that matters now, making extra numbers especially worthwhile. Doubling the force size quadruples its combat weight.<\/p>\n

For example, if A-force can bring twice the number of equally effective aircraft to a fight, then it can completely defeat B-force while only losing a few of its own number (13%). Being able to concentrate forces<\/a> provides a huge advantage in aimed fire situations. Perhaps more striking is what happens when B-Force has aircraft twice as effective at shooting down A-force planes than the other way around. Even in that case, bringing twice the number to the fight allows A-Force to win, and to lose \u2018only\u2019 30% of its number in destroying B-force. The table below gives some numerical examples.<\/p>\n\n\n\n\n\n\n\n
<\/td>\n\n

Force A<\/b><\/p>\n<\/td>\n

\n

Force B<\/b><\/p>\n<\/td>\n<\/tr>\n

<\/td>\n\n

Beginning<\/b><\/p>\n<\/td>\n

\n

End<\/b><\/p>\n<\/td>\n

\n

Beginning<\/b><\/p>\n<\/td>\n

\n

End<\/b><\/p>\n<\/td>\n<\/tr>\n

Unaimed mass land combat<\/b><\/td>\n\n

200<\/p>\n<\/td>\n

\n

100<\/p>\n<\/td>\n

\n

100<\/p>\n<\/td>\n

\n

0<\/p>\n<\/td>\n<\/tr>\n

Aimed air combat (equal)<\/b><\/td>\n\n

20<\/p>\n<\/td>\n

\n

17<\/p>\n<\/td>\n

\n

10<\/p>\n<\/td>\n

\n

0<\/p>\n<\/td>\n<\/tr>\n

Aimed air combat
\n(B twice as good)<\/b><\/td>\n
\n

20<\/p>\n<\/td>\n

\n

14<\/p>\n<\/td>\n

\n

10<\/p>\n<\/td>\n

\n

0<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

These are simple mathematical models, and real warfare is more complicated than the assumptions that underpin them; tactics, posture (defensive or offensive) mobility, and other factors matter. And the models work best when the sides are roughly equal in terms of weapon effectiveness.<\/p>\n

But they\u2019re not without application either. For example, when the RAND Corporation modelled an air war over the Taiwan Strait<\/a> a few years ago, they produced a set of numbers that allow us to test Lanchester\u2019s models. From the RAND data, the American forces had a clear aircraft-on-aircraft superiority over Chinese forces, ranging from a whopping 27:1 for the F-22 Raptor to better than two and a half to one for the F\/A-18 Super Hornets. When the number of sorties by each is taken into account, the US and Taiwanese aircraft collectively had an average effectiveness of almost five times the Chinese ones.<\/p>\n\n\n\n\n\n\n\n\n
<\/td>\nAdvantage<\/b><\/td>\n<\/tr>\n
F-22<\/td>\n\n

27 : 1<\/p>\n<\/td>\n<\/tr>\n

F-15<\/td>\n\n

4.6 : 1<\/p>\n<\/td>\n<\/tr>\n

F\/A-18<\/td>\n\n

2.6 : 1<\/p>\n<\/td>\n<\/tr>\n

Taiwanese<\/td>\n\n

2.3 : 1<\/p>\n<\/td>\n<\/tr>\n

Average<\/b><\/td>\n\n

4.7 : 1<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

So the US wins, right? Well, no\u2014despite the numbers above, the RAND shows that China wins in a great many of the simulations. The reason, as Lanchester would have known, lies in the number of sorties generated. RAND puts it this way:<\/p>\n

The effects of China\u2019s missile barrage are dramatically apparent: on the first day, the PLAAF generates about 3.7 times as many sorties as do the United States and Taiwan combined. \u00a0\u2026 China\u2019s ability to suppress or close\u2026 bases could give the PLA Air Force an almost overwhelming numerical advantage that\u2014coupled with the rough qualitative parity that now exists between the two sides\u2014could allow China to attain air superiority over Taiwan and the strait.<\/p><\/blockquote>\n

The US response to this calculation could be to renew production of the F-22, which I imagine would make the F-22 cheer squad very happy. But it would be wrong\u2014buying more expensive \u2018silver bullet\u2019 platforms actually misses the point of this sort of analysis. The rational Chinese response depends on the relative cost of improving the quality of its aircraft to reduce the 27:1 advantage and further increasing numbers. They don\u2019t have to achieve parity\u2014if they get close enough, numbers will do the rest, as they do in RAND’s analysis. Success in any future air campaign will be about numbers, persistence and technical quality\u2014probably in that order.<\/p>\n

Geeky appendix<\/b><\/p>\n

For the un-aimed fire case the coupled differential equations are (graphic, click to enlarge):\"lanchester<\/a>The quadratic form of those solutions (‘Lanchester square law’) is responsible for the dramatic difference in the effect of numbers. You can find more examples, including some explicit solutions of Lanchester’s equations, here<\/a>. A more detailed discussion of the circumstances in which Lanchester’s models apply to real battles can be found here <\/a>(PDF, calculus heavy).<\/p>\n

Andrew Davies is a senior analyst for defence capability at ASPI and executive editor of\u00a0<\/em>The Strategist. Image courtesy of the Imperial War Museum<\/a>.<\/em><\/p>\n","protected":false},"excerpt":{"rendered":"

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