mar05 (1)

# mar05 (1) - MIT OpenCourseWare http/ocw.mit.edu 6.055J...

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Unformatted text preview: MIT OpenCourseWare http://ocw.mit.edu 6.055J / 2.038J The Art of Approximation in Science and Engineering Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms . 5 Proportional reasoning 35 Since A is the cross-sectional area of the animal, Ah is the volume of air that it sweeps out in the jump, and ρ Ah is the mass of air swept out in the jump. So the relative importance of drag has a physical interpretation as a ratio of the mass of air displaced to the mass of the animal. To find how this ratio depends on animal size, rewrite it in terms of the animal’s side length l . In terms of side length, A ∼ l 2 and m ∝ l 3 . What about the jump height h ? The simplest analysis predicts that all animals have the same jump height, so h ∝ l . Therefore the numerator ρ Ah is ∝ l 1 , the denominator m is ∝ l 3 , and E drag l 2 = l − 1 . E required ∝ l 3 So, small animals have a large ratio, meaning that drag affects the jumps of small animals more than it affects the jumps of large animals. The missing constant of proportionality means that we cannot say at what size an animal becomes ‘small’ for the purposes of drag. So the calculation so far cannot tell us whether ﬂeas are included among the small animals. The jump data, however, substitutes for the missing constant of proportionality. The ratio is E drag ρ Ah ρ l 2 h E required ∼ m ∼ ρ animal l 3 . It simplifies to E drag ρ h . E required ∼ ρ animal l As a quick check, verify that the dimensions match. The left side is a ratio of energies, so it is dimensionless. The right side is the product of two dimensionless ratios, so it is also dimensionless. The dimensions match. Now put in numbers. A density of air is ρ ∼ 1 kg m − 3 . The density of an animal is roughly the density of water, so ρ animal ∼ 10 3 kg m − 3 . The typical jump height – which is where the data substitutes for the constant of proportionality – is 60 cm or roughly 1 m. A ﬂea’s length is about 1 mm or l ∼ 10 − 3 m. So E drag 1 kg m − 3 1 m E required ∼ 10 3 kg m − 3 10 − 3 m ∼ 1 . The ratio being unity means that if a ﬂea would jump to 60 cm, overcoming drag would require roughly as much as energy as would the jump itself in vacuum. Drag provides a plausible explanation for why ﬂeas do not jump as high as the typical height to which larger animals jump. 5.4.3 Cycling 36 6.055 / Art of approximation This section discusses cycling as an example of how drag affects the performance of people as well as ﬂeas. Those results will be used in the analysis of swimming, the example of the next section....
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mar05 (1) - MIT OpenCourseWare http/ocw.mit.edu 6.055J...

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