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MIT 6 055J - The Art of Approximation in Science and Engineering

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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.40 4040 402008-01-14 22:31:34 / rev 55add9943bf16.055 / Art of approximation 40 the contact force from the track. In choosing the shape of the track, you affect the contact force on the roller coaster, and thereby its acceleration, velocity, and position. There are an infinity of possible tracks, and we do not want to analyze each one to find the forces and acceleration. An invariant, energy, simplifies the analysis. No matter what tricks the track does, the kinetic plus potential energy 12mv + mgh 2is constant. The roller coaster starts with v = 0 and height hstart; it can never rise above that height without violating the constancy of the energy. The invariant – the conserved quantity – solves the problem in one step, avoiding an endless analysis of an infinity of possible paths. The moral of this section is: When there is change, look for what does not change. 6.2 Flight How far can birds and planes fly? The theory of flight is difficult and involves vortices, Bernoulli’s principle, streamlines, and much else. This section offers an alternative ap-proach: use conservation estimate the energy required to generate lift, then minimize the lift and drag contributions to the energy to find the minimum-energy way to make a trip. 6.2.1 Lift Instead of wading into the swamp of vortices, study what does not change. In this case, the vertical component of the plane’s momentum does not change while it cruises at constant altitude. Because of momentum conservation, a plane must deflect air downward. If it did not, grav-ity would pull the plane into the ground. By deflecting air downwards – which generates lift – the plane gets a compensating, upward recoil. Finding the necessary recoil leads to finding the energy required to produce it. Imagine a journey of distance s. I calculate the energy to produce lift in three steps: 1. How much air is deflected downward? 2. How fast must that mass be deflected downward in order to give the plane the needed recoil? 3. How much kinetic energy is imparted to that air? The plane is moving forward at speed v, and it deflects air over an area L2 where L is the wingspan. Why this area L2, rather than the cross-sectional area, is subtle. The reason is that the wings disturb the flow over a distance comparable to their span (the longest length). So when the plane travels a distance s, it deflects a mass of air mair ∼ ρL2s.41 4141 412008-01-14 22:31:34 / rev 55add9943bf1|{z} 41 6 Box models and conservation The downward speed imparted to that mass must take away enough momentum to com-pensate for the downward momentum imparted by gravity. Traveling a distance s takes time s/v, in which time gravity imparts a downward momentum Mgs/v to the plane. There-fore mairvdown ∼ Mgs v so vdown ∼ Mgs vmair ∼ Mgs ρvL2s = Mg ρvL2. The distance s divides out, which is a good sign: The downward velocity of the air should not depend on an arbitrarily chosen distance! The kinetic energy required to send that much air downwards is mairv2 That energy down. factors into (mairvdown)vdown, so Mgs Mg (Mg)2 Elift ∼ mairvdown vdown ∼ = s. | {z } v ρvL2 ρv2L2 Mgs/v vdown Check the dimensions: The numerator is a squared force since Mg is a force, and the de-nominator is a force, so the expression is a force times the distance s. So the result is an energy. Interestingly, the energy to produce lift decreases with increasing speed. Here is a scaling argument to make that result plausible. Imagine doubling the speed of the plane. The fast plane makes the journey in one-half the time of the original plane. Gravity has only one-half the time to pull the plane down, so the plane needs only one-half the recoil to stay aloft. Since the same mass of air is being deflected downward but with half the total recoil (momentum), the necessary downward velocity is a factor of 2 lower for the fast plane than for the slow plane. This factor of 2 in speed lowers the energy by a factor of 4, in accordance with the v−2 in Elift. 6.2.2 Optimization including drag The energy required to fly includes the energy to generate lift and to fight drag. I’ll add the lift and drag energies, and choose the speed that minimizes the sum. The energy to fight drag is the drag force times the distance. The drag force is usually written as Fdrag ∼ ρv2A, where A is the cross-sectional area. The missing dimensionless constant is cd/2: 1Fdrag = 2cdρv2A,42 4242 422008-01-14 22:31:34 / rev 55add9943bf1| {z } 6.055 / Art of approximation 42 where cd is the drag coefficient. However, to simplify comparing the energies required for lift and drag, I instead write the drag force as Fdrag = Cρv2L2 , where C is a modified drag coefficient, where the drag is measured relative to the squared wingspan rather than to the cross-sectional area. For most flying objects, the squared wingspan is much larger than the cross-sectional area, so C is much smaller than cd. With that form for Fdrag, the drag energy is Edrag = Cρv2L2s, and the total energy to fly is (Mg)2 s + Cρv2L2s .E ∼ ρv2L2| {z } EdragElift A sketch of the total energy versus velocity shows interesting voptimumEdrag∝ v2Elift∝ v−2EtotalEfeatures. At low speeds, lift is the dominant consumer be-cause of its v−2 dependence. At high speeds, drag is the dom-inant consumer because of its v2 dependence. In between these extremes is an optimum speed voptimum: the speed that minimizes the energy consumption for a fixed journey dis-tance s. Going faster or slower than the optimum speed means consuming more energy. That extra consumption can-not always be avoided. A plane is designed so that its cruis-ing speed is its minimum-energy speed. So at takeoff and landing, when its speed is much less than the minimum-energy speed, a plane requires a lot of power to stay aloft, which is one reason that the engines are so loud at takeoff and landing (another reason is probably that the engine noise reflects off the ground and back to the plane). The constraint, or assumption, that a plane travels at the minimum-energy speed simplifies the expression for the total energy. At the minimum-energy speed, the drag and lift


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