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MIT 6 055J - Proportional reasoning

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MIT OpenCourseWare http://ocw.mit.edu 6.055J / 2.038J The Art of Approximation in Science and EngineeringSpring 2008For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.35 3535 352008-01-14 22:31:34 / rev 55add9943bf15 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 ρAh 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 ∼ l2 and m ∝ l3. What about the jump height h? The simplest analysis predicts that all animals have the same jump height, so h ∝ l0. Therefore the numerator ρAh is ∝ l1, the denominator m is ∝ l3, and Edrag l2 = l−1 .Erequired ∝ l3 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 Edrag ρAh ρl2h Erequired ∼ m ∼ ρanimall3 . It simplifies to Edrag ρ h .Erequired ∼ ρanimal lAs 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 ρanimal ∼ 103 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 Edrag 1 kg m−3 1 m Erequired ∼ 103 kg m−3 10−3m ∼ 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 Cycling36 3636 362008-01-14 22:31:34 / rev 55add9943bf136 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. What is the world-record cycling speed? Before looking it up, predict it using armchair proportional reasoning. The first task is to define the kind of world record. Let’s say that the cycling is on a level ground using a regular bicycle, although faster speeds are possible using special bicycles or going downhill. To estimate the speed, make a model of where the energy goes. It goes into rolling resis-tance, into friction in the chain and gears, and into drag. At low speeds, the rolling resis-tance and chain friction are probably important. But the importance of drag rises rapidly with speed, so at high-enough speeds, drag is the dominant consumer of energy. For simplicity, assume that drag is the only consumer of energy. The maximum speed happens when the power supplied by the rider equals the power consumed by drag. The problem therefore divides into two estimates: the power consumed by drag and the power that an athlete can supply. The drag power Pdrag is related to the drag force: Pdrag = Fdragv ∼ ρv3A. It indeed rises rapidly with velocity, supporting the initial assumption that drag is the im-portant effect at world-record speeds. Setting Pdrag = Pathlete gives !1/3Pathlete vmax ∼ ρA To estimate how much power an athlete can supply, I ran up one flight of stairs leading from the MIT Infinite Corridor. The Infinite Corridor, being an old building, has spacious high ceilings, so the vertical climb is perhaps h ∼ 4 m (a typical house is 3 m per storey). Leaping up the stairs as fast as I could, I needed t ∼ 5 s for the climb. My mass is 60 kg, so my power output was Pauthor ∼ potential energy supplied time to deliver it = mgh 60 kg × 10 m s−2 × 4 m ∼ 500 W. t ∼ 5 s Pathlete should be higher than this peak power since most authors are not Olympic athletes. Fortunately I’d like to predict the endurance record. An Olympic athlete’s long-term power might well be comparable to my peak power. So I use Pathlete = 500 W. The remaining item is the cyclist’s cross-sectional area A. Divide the area into width and height. The width is a body width, perhaps 0.4 m. A racing cyclist crouches, so the height is maybe 1 m rather than a full 2 m. So A∼ 0.4 m2. Here is the tree that represents this analysis:37 3737 372008-01-14 22:31:34 / rev 55add9943bf15 Proportional reasoning 37 vmaxPathlete∼ Estairs/tstairsEstairs2400 Jm60 kgg10 m s−2h4 mtstairs5 sPdrag∼ ρv3Aρ1 kg m−3vA0.4 m2Now combine the estimates to find the maximum speed. Putting in numbers gives !1/3 !1/3Pathlete 500 W vmax ∼ ρA ∼ 1 kg m−3 × 0.4 m2 . The cube root might suggest using a calculator. However, massaging the numbers simpli-fies the arithmetic enough to do it mentally. If only the power were 400 W or, instead, if the area were 0.5 m! Therefore, in the words of Captain Jean-Luc Picard, ‘make it so’. The cube root becomes easy: !1/3400 W vmax ∼∼ 1 kg m−3 × 0.4 m2 ∼ (1000 m3 s−3)1/3 = 10 m s−1 . So the world record should be, if this analysis has any correct physics in it, around 10 m s−1 or 22 mph. The world one-hour record – where the contestant cycles as far as possible in one hour – is 49.7 km or 30.9 mi. The estimate based on drag is reasonable! 5.4.4 Swimming The last section’s analysis of cycling helps predict the world-record speed for swimming. The last section showed that !1/3Pathlete vmax ∼ ρA . To evaluate the maximum speed for swimming, one could put in a new ρ and A directly into that formula. However, that method replicates the work of multiplying, dividing, and cube-rooting the various values. Instead it is instructive to scale the numerical result for cycling by looking at how the maxi-mum speed depends on the


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