7.2 Maximization of the Range of a Rocket

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1 138 CHAPTER 7. SOME APPLICATIONS The counterintuitive answer that a supersonic aircraft must dive first in order to climb to a iven altitude in minimum time was first discovered by Walter Denham and Art Bryson, Jr. See Ho and Speyer [1990] for historical details. 7. Maximization of the Rane of a Rocket Let us consider the problem of maximizin the rane of a rocket, as illustrated in Fi Here we assume a uniform ravity field (with acceleration ) and inore the effect of dra. We also assume that the thrust acceleration, f(t), is a iven positive function of time and y f θ Fiure 7.7: Maximization of the rane of a rocket. that burnout occurs at t = T. Thus, our problem is to find the control function θ(t) to maximize the rane, R. That is, Maximize: J = R (7.33) subject to: x ẋ = V x ẏ = V y V x = f cos θ V y = f sin θ (7.34a) (7.34b) (7.34c) (7.34d) where f(t) = ṁi spm /m (7.35) is a predetermined function of time and I spm is specific impulse based on mass flow rate, i.e. effective exhaust velocity.

2 7.. MAXIMIZATION OF THE RANGE OF A ROCKET 139 For the I.C.s we assume: x(0) = y(0) = 0, V x (0) = V y (0) = 0 (7.36) In the analysis of this problem we follow the approach of Lawden [1963]. If (x 1, y 1, V x1, V y1 ) is the state at burnout (t=t), then the rane is found as follows. To determine the duration of the coastin arc, t c, we write the familiar equation for the fallin body: d = 1 t c + V y1 t c + y 1 (7.37) where d is the altitude of the body above the round. Settin d = 0 for the time of impact we have 1 t c + V y1 t c + y 1 = 0 (7.38) so that (choosin the larer root) ( ) t c = V y1 + Vy1 + y 1 / (7.39) Therefore, the rane is: R = x 1 + V x1 t c = x 1 + V x1 ( ) V y1 + Vy1 + y 1 (7.40) Now our cost function is in terms of the state variables at burnout which, in turn, can be manipulated by usin the control θ(t). Of course we don t yet know the state at burnout, nor do we know the duration of the coastin arc. Let us apply the necessary conditions as follows. 1. The Hamiltonian is. The Euler-Larane equations are H = λ x V x + λ y V y + λ Vx f cos θ + λ Vy (f sin θ ) (7.41) λ x = H x = 0, λy = H y = 0 (7.4a) so that λ x = c 1, λ y = c (7.4b) and λ Vx = H Vx = λ x, λvy = H Vy = λ y (7.4c)

3 140 CHAPTER 7. SOME APPLICATIONS so where c 1, c, c 3, and c 4 are constants. λ Vx = c 1 t + c 3, λ Vy = c t + c 4 (7.4d) Followin the procedure described after Eq. (4.66) we write the Hamiltonian in Eq. (7.41) as H = λ x V x + λ y V y λ Vy + fµ T σ (7.43) where µ = [ λvx λ Vy ] (7.44) and σ = [ cos θ sin θ ] (7.45) where σ is a unit vector. Because we want to maximize J (the rane) we choose the control to maximize H; so µ T σ = µ and σ = µ µ (7.46) for which cos θ = λ Vx λ Vx + λ Vy (7.47) sin θ = λ Vy λ V x + λ V y (7.48) 3. From the differential form of the transversality condition we have H f dt f λ T f dx f + dφ = 0 (7.49) Followin Lawden [1963], we evaluate the transversality condition at t 1 = T : H 1 dt 1 λ T 1 dx 1 + dφ(x 1 ) = 0 (7.50) We note that since t 1 = T we have dt 1 = 0; we also note that x 1 is free. Recallin that: φ(x 1 ) = x 1 + V x1 t c = x 1 + V x1 ( ) V y1 + Vy1 + y 1 (7.51)

4 7.. MAXIMIZATION OF THE RANGE OF A ROCKET 141 we have: dφ(x 1 ) = dx 1 + dy 1 + dv x1 + dv y1 (7.5) x 1 y 1 V x1 V y1 Thus from transversality we obtain: ( λx1 dx 1 λ y1 dy 1 λ Vx1dV x1 λ Vy1dV y1 ) + { ( ) dx 1 + V x1 V 1/ y1 + y 1 dy1 + 1 ( ) V y1 + Vy1 + y 1 dv x1 + V x1 Settin the coefficients to zero: [ 1 + V y1 ( V y1 + y 1 ) 1/ ] dv y1 } = 0 (7.53) where we use r V y1 + y 1 λ x1 = 1 λ y1 = V x1 r λ Vx1 = V y1 + r λ Vy1 = V x1 r (r + V y1) (7.54a) (7.54b) (7.54c) (7.54d) (7.54e) to simplify the expressions. Thus, for the control law we have: where tan θ = +λ V y +λ Vx = c t + c 4 c 1 t + c 3 (7.55) c 1 = λ x1 = 1 (7.56a) so that c = λ y1 = V x1 r c 3 = λ Vx1 + c 1 T = V y1 + r c 4 = λ Vy1 + c T = V x1 r (r + V y1) + V x1 r T tan θ = c t + c 4 c 1 t + c 3 = (7.56b) + T (7.56c) ( ) V x1 t + V x1 r+vy1 + T r r t + V y1+r + T (7.56d) (7.57)

5 14 CHAPTER 7. SOME APPLICATIONS which reduces to the simple form: tan θ = V x1 (7.58) r Recallin that r is a constant that depends on V y1 and y 1, we see that the maximum rane is achieved by θ = constant which means that the thrust must be kept at a constant anle to the horizontal, even thouh f = f(t) Interation of Equations of Motion when f is Constant Let us consider the special case of constant acceleration. When f is constant, interation of the equations of motion, Eqs. (7.34a) (7.34d), provides: V x1 = ft cos θ V y1 = (f sin θ )T (7.59a) (7.59b) x 1 = 1 ft cos θ y 1 = 1 (f sin θ )T (7.59d) (7.59c) for the state at t = T. Substitutin the solution into the control law, Eq. (7.58), ives: ft cos θ tan θ = (f sin θ ) T + 1 (f sin θ )T ft cos θ = T f (sin θ sin θ) f cos θ = (7.60) sin θ f sin θ Equation (7.60) implies that: sin θ sin θ f sin θ = cos θ Squarin both sides of Eq. (7.61) provides: or = 1 sin θ (7.61) sin θ(sin θ f sin θ) = 1 sin θ + sin 4 θ (7.6) f sin3 θ sin θ + 1 = 0 (7.63) Equation (7.63) is a transcendental equation which we can solve numerically to determine θ. Lawden [1963] summarizes the solutions for θ as follows.

6 7.. MAXIMIZATION OF THE RANGE OF A ROCKET If f >, then there are four roots for 0 < θ < π (one in each quadrant), but only the positive acute solution θ < π/ is meaninful.. If f <, then there is insufficient thrust. The solutions for θ are in the third and fourth quadrants, correspondin to non-physical cases of fallin below round level. 7.. The Optimal Trajectory Once θ has been calculated, we solve for the state at T (x 1, y 1, V x1, V y1 ) from the interated equations of motion, Eqs. (7.59a) (7.59d). Thus, the optimal trajectory is iven (parametrically) as: x = 1 ft cos θ y = 1 (f sin θ )t (7.64) for 0 t T. Equations (7.64) indicate that the rocket travels alon a straiht line durin thrustin, iven by the fliht path anle γ (see Fi. 7.8). y f V γ θ R max x Fiure 7.8: Solution for maximum rocket rane. tan γ = y x = tan θ f sec θ (7.65) Since ẍ = f cos θ ÿ = f sin θ (7.66a) (7.66b)

7 144 CHAPTER 7. SOME APPLICATIONS the line is traversed at a constant acceleration: a = ẍ + ÿ 7..3 Maximum Rane Equation The maximum rane of the rocket is found from Eq. (7.40): = f f sin θ + (7.67) R = x 1 + V x1 (V y1 + r) (7.68) where r is iven in Eq. (7.54e). Usin Eq. (7.58) to eliminate r from Eq. (7.68) we obtain R = x 1 + V ( x1 V y1 + V ) x1 (7.69) tan θ Substitutin the solutions for x 1, V x1 and V y1 from Eqs. (7.59) ives: R = 1 ft cos θ + 1 ( ft cos θ ft sin θ T + cos θ ) tan θ ft = 1 ft cos θ + f ( ) T sin θ + cos θ cos θ sin θ Thus, the maximum rane of the rocket is R max = ft ( f cot θ 1 cos θ ) (7.70) (7.71) where f is the constant acceleration from the thrust, f >, and θ is the constant thrust anle iven by Eq. (7.63). Table 7.1 shows the result for various nondimensional thrust acceleration levels, f/, increasin from 1 to. The infinite value represents an impulsive thrust and the trajectory is the ballistic projectile path studied in elementary mechanics, for which the maximum rane is achieved by the launch anle, θ, of 45 de. In the limit as f, T 0 and the product f T approaches the launch velocity (the impulsive V ). In Table 7.1 the thrust anle θ is reater than the fliht path anle γ for finite duration thrusts in order to counteract ravity actin downward. The nondimensional rane shown is the actual R max divided by the rane of the ballistic path. The fact that this nondimensional rane is less than one for finite duration thrust is due to ravity loss. 7.3 Time Optimal Launchin of a Satellite We now return to one of the fundamental examples of the text: the time-optimal launchin of a satellite into orbit, illustrated in Fi We continue with the same assumptions of Example 4.5, Launch into circular orbit from flat-earth.

8 7.3. TIME OPTIMAL LAUNCHING OF A SATELLITE 145 Table 7.1: Maximum rane for various nondimensional thrust acceleration levels (continuous thrust, uniform ravity) Thrust Thrust Fliht Path Nondimensional Acceleration Anle Anle Rane f/ θ (de) γ (de) R max /f T Specifically, our problem is as follows. Minimize: subject to: J = t f (7.7) ẋ = V x ẏ = V y V x = f cos θ V y = f sin θ (7.73a) (7.73b) (7.73c) (7.73d) where f(t) = ṁi spm /m (7.74) is the thrust acceleration, a predetermined function of time. The initial conditions are t o = x o = y o = V xo = V yo = 0 (7.75) and the terminal boundary conditions are: y f = h V xf = V c V yf = 0 (7.76a) (7.76b) (7.76c)

Vector Valued Functions

SUGGESTED REFERENCE MATERIAL: Vector Valued Functions As you work throuh the problems listed below, you should reference Chapters. &. of the recommended textbook (or the equivalent chapter in your alternative