Gravity Turn Concept Curvilinear Coordinate System Gravity Turn Manoeuvre concept Solutions for Constant Pitch Rate
Inclined Motion Concept In reality, vertical motion is used only for a very small part of the overall ascent mission and for the most part, ascent trajectory is inclined & curvilinear in nature. This is mainly because one of the terminal constraint is that the inclination of the velocity vector with respect to the local horizon is required to be close to zero. Considering Earth s curvature, the rocket needs to undergo large flight path angle changes (~110 0 ) during the ascent mission. This requirement calls for a different methodology of trajectory design & solution.
Effect of Inclination A curvilinear flight path requires motion in a plane and therefore, needs models for a planar motion. Also, thrust is used mainly for velocity increments and is always along the flight path, so that a normal force is needed to produce the curvilinear path. Consider a rocket having inclination with the vertical. L Y θ u s, v s F c X T mg u n, v n
Curvilinear Motion Model Consider the following schematic of a planar motion. s ds V = lim = V = sɺ uˆs t 0 t dt Acceleration: d d a= V = (Vuˆs ) dt dt duˆs ɺ ɺ a = Vus + V = Vus + V θɺun dt a = a s + an ( )
Planar Motion Equations The equations of planar motion are as follows. dv mas = m = mg ɺ 0I sp mg cosθ dt dθ man = mv = mg sin θ dt In this case, the resulting trajectory is called gravity turn trajectory, as g alone is responsible for (dθ/dt). The above non-linear time-varying differential equations contains three unknowns i.e. V, θ and m(t) (the design input), for which no general solutions exist.
Special Analytical Solutions Special analytical solutions to the gravity turn equations are possible which, while restricting the overall degree of freedom, provide immense practical utility. In addition, lot of insight can be obtained by analyzing the equations themselves. In a kinematic sense, the gravity turn equations can be rewritten as, mg ɺ 0Isp g sin ( t) Vɺ ɶ θ = gɶ cos θ ( t); ɺ θ = m( t) V ( t) Also, it is possible to obtain V and θ, if m(t) is specified, or vice versa. This is the basis for Pitch Program in launch vehicle mission design.
Case 1: Constant Pitch Rate In this case, rocket is commanded to track a specified pitch rate i.e. (dθ/dt), which is achieved through an independent pitch rate tracking control system. This results in the second equation providing the velocity solution, which is then used in the first equation to obtain the required burn profile {m(t)} or pitch program. ɺ gɶ sinθ θ = q0 θ ( t) = q0t + θ0; V ( t) = ; Vɺ = gɶ cosθ q dm 2gɶ cosθdt m 2gɶ = = θ m g I m q g I 0 ln (sin sin 0) 0 sp 0 0 sp 0 θ
Case 1: Constant Pitch Rate As q 0 is constant at all times including the initial time, we can write, q gɶ sinθ = 0 or θ = 0, V =0 not admissible. 0 0 0 0 V0 This means that gravity turn manoeuvre can be started only from a non-zero pitch down angle, after it acquires a minimum forward speed. This requirement is usually met by giving a pitch kick to the vehicle at appropriate time to initiate manoeuvre and usually happens after acquiring some altitude.
Case 1: Constant Pitch Rate The altitude profile can be obtained by resolving the velocity V in vertical direction as follows. dh dh V cosθ g sinθ cosθ = V cosθ = = ɶ dt d θ q q 2 2 0 0 2 0 0 dh g sin 2θ g h( θ ) (cos 2θ cos 2 θ ) h dθ = ɶ 2q = ɶ 4q + 0 0 Can θ(t) be more than 90 0? What would such a condition represent? What is the impact on the burn rate and total propellant mass?
Case 1: Constant Pitch Rate Another trajectory parameter of interest is the final flight path angle, which can be evaluated as, θ b 1 0 0 0 sin g q I sp m ln sinθ = + 0 2gɶ mb Burnout time & horizontal distance are as follows. θb ( θb θ0) dx 1 tb = ; = V sin x x0 = g 2 ( 1 cos 2 ) d q0 dt 2q ɶ 0 θ θ θ θ ( sin 2θ sin 2θ ) gɶ b 0 x( θ ) = 2 ( θb θ0 ) + x( θ0) 2q0 2 0
Constant Pitch Rate Example First stage of the Chinese Long March rocket has the following lift off parameters. m 0 = 79.4 Tons, m p = 60 Tons, I sp = 241 s, g 0 = 9.81m/s 2, Payload mass = 9.4 Tons, β 0 = 600 kg/s (until t i ), t i = 10s, (1) Determine the trajectory parameters at end of 10s. V i = 0.0876 km/s, h i = 0.426 km, (2) Determine terminal parameters in case the rocket executes the gravity turn for a further 90s. θ i = 5 o. V t = 0.827 km/s, h t = 35.0 km, m t = 39.3 Tons, θ t = 55.3 o, q 0 = 0.559 o /s, t b = 90 s, x t = 25.6 km.
Constant Pitch Rate Example (3) Also, determine if all the propellant can be burnt to reach 90 o? If yes, give final burnout parameters. If no, give reasons as well as the final burnout mass. No. m t = 36.5 Tons. (4) What should beθ i if all fuel is to be burnt? (θ b = 90 o ) q 0 = 0.32 o /s,θ i = 2.87 o, t b = 272.3 s.
Summary Gravity turn trajectories take much longer time, but result in lower velocities in denser atmosphere and also reduce the energy loss due to gravity. Constant pitch rate solution is simple to obtain in closed form, though requiring initial conditions consistent with the amount of propellant to be burnt.