{"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":"
<\/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