MIT OpenCourseWare http://ocw.mit.edu 16.346 Astrodynamics Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� Lecture 31 The Calculus of Variations & Lunar Landing Guidance The Brachistochrone Problem In a vertical xy -plane a smooth curve y = f(x) connects the origin with a point P (x1,y1) in such a way that the time taken by a particle sliding without friction from O to P along the curve propelled by gravity is as short as possible. What is the curve? Assume the positive y -axis is vertically downward. Then the equation of motion is md2s = mg sin γ = mgdy with ds2 = dx2 + dy2 dt2 ds d2s ds dy ds = gdt2 dt ds dt d �ds�2 =2gdy = ds = � 2gydt dt dt ⇒ dt Then � T � x1 � x1 �� x1ds 1 1+ y2 1 0 dt = T = 0 √2gy = √2g 0 √y dx = √2g 0 F (y, y) dx Deriving Euler’s Equation x1 To minimize the integral I = F (x, y, y) dx let y(x, α)= ym(x)+ α�(x) x0 �� ��� � dI x1 ∂F ∂Fd� x1 ∂F d∂FThen dα = x0 ∂y �(x)+ ∂y dx dx = x0 ∂y − dx ∂y �(x) dx Therefore, from the Fundamental Lemma of the Calculus of Variations ∂F d∂F ∂y − dx ∂y =0 is a Necessary Condition which F must satisfy if the integral I is to be a minimum. Special Case of Euler’s Equation ∂y Also � � � � d dx F − ∂F ∂y y = ∂F ∂x + ∂F ∂y − d dx ∂F ∂y y = ∂F ∂x � �� � =0 which will be zero if F is not a function of x. Therefore F − ∂F y = constant Prob. 11–33 which establishes the necessary condition used to solve the Brachistochrone Problem. 16.346 Astrodynamics Lecture 31Solution of the Brachistochrone Problem If T is to be a minimum, then, using Euler’s Special Case of the Necessary Condition, we have � �� y(1 + y2)=2c or dx = x = y dy2c − y Now let y =2c sin2 θ = c(1 − cos 2θ) so that � x =2c (1 − cos 2θ) dθ = c(2θ − sin 2θ) Therefore, the equation of the curve in parametric form is x = c(φ − sin φ) with φ =2θ y = c(1 − cos φ) and represents a cycloid—the path of a point on a circle of radius c as it rolls along the underside of the x axis. Terminal State Vector Control Find the acceleration vector a(t) to minimize � t1 � t1 J = a(t)2 dt = aT(t)a(t) dt t0 t0 subject to dr = vr(t0)= r0 r(t1)= r1dt dvv(t0)= v0 v(t1)= v1= a dt Define the Admissible Functions: r(t, α)= r m(t)+ αδ(t) δ(t0)= δ(t1)= 0 v(t, α)= vm(t)+ αδ(t) where δ(t0)= δ(t1)= 0 a(t, α)= a m(t)+ αδ(t) δ(t0)= δ(t1)= 0 Then � ��t1 t1 t1 J(α)= aT(t)a (t) dt +2α aT(t)δ (t) dt + α2 δ (t)T δ (t) dt m m m t0 t0 t0 16.346 Astrodynamics Lecture 31A Necessary Condition for t1 t1 t1 J(α)= � aT(� Tmt)a m(t) dt +2α am( t)δ (t) dt + α2 � δ (t)T δ (t) dt t0 t0 t0 to be a minimum is that dJ �� t1=0=2 aTm( t)δ (t) dt dα α=0 t0 Use integration by parts ��t� � 1 t1 t1 T daT (t) dδ(t) t1 daT a (t)δ dt = aT (t)δ(t) � � ���−� mdt =0 − m(t) dδ(t)m m dt t0 t0 t0 dt dt t0dt dt daT m(�1 t)�t� �t1 d2a T 2m(t1 T = − ( ) � + t)� d am(t)δ t δ(t) dt =0+ δ(t) dt dt dt2 tdt2 0 t0 t0Hence dJ ����t1 d2aTm( t)dα � =0 = δ (t) dt =0 2 α=0 ⇒t0 dtAgain using the Fundamental Lemma of the Calculus of Variations it follows thd2aTm( t)= 0T = ⇒ a m(t)= c t + cdt2 1 2Therefore, with t go = t1 − t,wehave 4 6 a m(t)= c1t + c2 = [v1 − v(t)] + {r1 − [r(t)+ v1t go]t t2 go go Lunar-Landing Guidance for Apollo Missions To include the effects of gravity a(t)= aT (t)+ g(r) we could use 4 6 aT (t)= [v1 − v(t)] + {r1 − [r(t)+ v1t go]}−g[r(t)]t t2 go go for the thrust acceleration which would be an exact solution if g were constant. 16.346 Astrodynamics Lecture 31