74 7474 7474 4.5. Flight2009-05-04 23:52:14 / rev bb931e4b905e4.5 FlightHow far can birds and planes fly? The theory of flight is difficult and in-volves vortices, Bernoulli’s principle, streamlines, and much else. This sec-tion offers an alternative approach: use conservation estimate the energyrequired to generate lift, then minimize the lift and drag contributions tothe energy to find the minimum-energy way to make a trip.4.5.1 LiftInstead of wading into the swamp of vortices, study what does not change.In this case, the vertical component of the plane’s momentum does notchange while it cruises at constant altitude.Because of momentum conservation, a plane must deflect air downward.If it did not, gravity would pull the plane into the ground. By deflecting airdownwards – which generates lift – the plane gets a compensating, upwardrecoil. Finding the necessary recoil leads to finding the energy required toproduce it.Imagine a journey of distance s. I calculate the energy to produce lift inthree steps:1. How much air is deflected downward?2. How fast must that mass be deflected downward in order to give theplane 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 L2where L is the wingspan. Why this area L2, rather than the cross-sectionalarea, is subtle. The reason is that the wings disturb the flow over a distancecomparable to their span (the longest length). So when the plane travels adistance s, it deflects a mass of airmair∼ ρL2s.The downward speed imparted to that mass must take away enough mo-mentum to compensate for the downward momentum imparted by grav-ity. Traveling a distance s takes time s/v, in which time gravity imparts adownward momentum Mgs/v to the plane. Thereforemairvdown∼Mgsv75 7575 75Chapter 4. Symmetry and conservation 752009-05-04 23:52:14 / rev bb931e4b905esovdown∼Mgsvmair∼MgsρvL2s=MgρvL2.The distance s divides out, which is a good sign: The downward velocityof the air should not depend on an arbitrarily chosen distance!The kinetic energy required to send that much air downwards is mairv2down.That energy factors into (mairvdown)vdown, soElift∼ mairvdown|{z }Mgs/vvdown∼MgsvMgρvL2|{z}vdown=(Mg)2ρv2L2s.Check the dimensions: The numerator is a squared force since Mg is aforce, and the denominator is a force, so the expression is a force times thedistance 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 doublingthe speed of the plane. The fast plane makes the journey in one-half thetime of the original plane. Gravity has only one-half the time to pull theplane down, so the plane needs only one-half the recoil to stay aloft. Sincethe same mass of air is being deflected downward but with half the totalrecoil (momentum), the necessary downward velocity is a factor of 2 lowerfor the fast plane than for the slow plane. This factor of 2 in speed lowersthe energy by a factor of 4, in accordance with the v−2in Elift.4.5.2 Optimization including dragThe energy required to fly includes the energy to generate lift and to fightdrag. I’ll add the lift and drag energies, and choose the speed that mini-mizes the sum.The energy to fight drag is the drag force times the distance. The drag forceis usually written asFdrag∼ ρv2A,where A is the cross-sectional area. The missing dimensionless constant iscd/2:Fdrag=12cdρv2A,76 7676 7676 4.5. Flight2009-05-04 23:52:14 / rev bb931e4b905ewhere cdis the drag coefficient.However, to simplify comparing the energies required for lift and drag, Iinstead write the drag force asFdrag= Cρv2L2,where C is a modified drag coefficient, where the drag is measured relativeto the squared wingspan rather than to the cross-sectional area. For mostflying objects, the squared wingspan is much larger than the cross-sectionalarea, so C is much smaller than cd.With that form for Fdrag, the drag energy isEdrag= Cρv2L2s,and the total energy to fly isE ∼(Mg)2ρv2L2s|{z }Elift+ Cρv2L2s|{z }Edrag.voptimumEdrag∝ v2Elift∝ v−2EtotalEA sketch of the total energy versus velocity showsinteresting features. At low speeds, lift is thedominant consumer because of its v−2depen-dence. At high speeds, drag is the dominantconsumer because of its v2dependence. In be-tween these extremes is an optimum speed voptimum:the speed that minimizes the energy consump-tion for a fixed journey distance s. Going fasteror slower than the optimum speed means con-suming more energy. That extra consumptioncannot always be avoided. A plane is designedso that its cruising speed is its minimum-energy speed. So at takeoff andlanding, when its speed is much less than the minimum-energy speed, aplane requires a lot of power to stay aloft, which is one reason that the en-gines are so loud at takeoff and landing (another reason is probably thatthe engine noise reflects off the ground and back to the plane).The constraint, or assumption, that a plane travels at the minimum-energyspeed simplifies the expression for the total energy. At the minimum-energy speed, the drag and lift energies are equal. So(Mg)2ρv2L2s ∼ Cρv2L2s,77 7777 77Chapter 4. Symmetry and conservation 772009-05-04 23:52:14 / rev bb931e4b905eorMg ∼ C1/2ρv2L2.This constraint simplifies the total energy. Instead of simplifying the sum,simplify just the drag, which neglects only a factor of 2 since drag and liftare roughly equal at the minimum-energy speed. SoE ∼ Edrag∼ Cρv2L2s ∼ C1/2Mgs.This result depends in reasonable ways upon M, g, C, and s. First, liftovercomes gravity, and gravity produces the plane’s weight Mg. So Mgshould show up in the energy, and the energy should, and does, increasewhen Mg increases. Second, a streamlined plane should use less energythan a bluff, blocky plane, so the energy should, and does, increase as themodified drag coefficient C increases. Third, since the flight is at a constantspeed, the energy should be, and is, proportional to the distance traveled s.4.5.3 Explicit computationsTo get an explicit range, estimate the fuel fraction β, the energy density E,and the drag coefficient C. For the fuel fraction I’ll guess β ∼ 0.4. For E,look at the nutrition label on the back of a pack of butter. Butter is almostall fat, and one serving of 11 g provides 100 Cal (those are ‘big calories’). Soits energy density is 9 kcal g−1. In