Question 10.7: A spacecraft of mass m and with the dimensions shown in Figu...
A spacecraft of mass m and with the dimensions shown in Figure 10.15 is spinning without precession at the rate \omega_{0} about the z axis of the principal body frame. At the instant shown in part (a) of the figure, the spacecraft initiates a coning maneuver to swing its spin axis through 90°, so that at the end of the maneuver the vehicle is oriented as illustrated in Figure 10.15b . Calculate the total delta-H required, and compare it with that required for the same reorientation without coning. Motion is to be controlled exclusively by the pairs of attitude thrusters shown, all of which have identical thrust T.
According to Figure 9.9c, the moments of inertia about the principal body axes are
A=B=\frac{1}{12} m\left[w^{2}+\left\lgroup \frac{w}{3} \right\rgroup ^{2}\right]=\frac{5}{54} m w^{2} \quad C=\frac{1}{12} m\left(w^{2}+w^{2}\right)=\frac{1}{6} m w^{2}The initial angular momentum H _{G_{1}} points in the spin direction, along the positive z axis of the body frame,
H _{G_{1}}=C \omega_{z} \hat{ k }=\frac{1}{6} m w^{2} \omega_{0} \hat{ k }We can presume that in the initial orientation, the body frame happens to coincide instantaneously with inertial frame XYZ . The coning motion is initiated by briefly firing the pair of thrusters RCS-1 and RCS-2, aligned with the body z axis and lying in the yz plane. The impulsive torque will cause a change \Delta H _{G_{1}} in angular momentum directed normal to the plane of the thrusters, in the positive body x direction. The resultant angular momentum vector must lie at 45° to the x and z axes, bisecting the angle between the initial and final angular momenta. Thus,
\left\|\Delta H _{G_{1}}\right\|=\left\| H _{G_{1}}\right\| \tan 45^{\circ}=\frac{1}{6} m w^{2} \omega_{0}After the coning is underway, the body axes of course move away from the XYZ frame. Since the spacecraft is oblate (C > A), the precession of the spin axis will be opposite to the spin direction, as indicated in Figure 10.15. When the spin axis, after 180° of precession, lines up with the X axis, the thrusters must fire again for the same duration as before so as to produce the angular momentum change \Delta H _{G_{2}}, equal in magnitude but perpendicular to \Delta H _{G_{1}}, so that
H _{G_{1}}+\Delta H _{G_{1}}+\Delta H _{G_{2}}= H _{G_{2}}where
H _{G_{2}}=\left\| H _{G_{1}}\right\| \hat{ I }=\frac{1}{6} m w^{2} \omega_{0} \hat{ k }For this to work, the plane of thrusters RCS-1 and RCS-2 — the yz plane — must be parallel to the XY plane when they fire, as illustrated in Figure 10.15b . Since the thrusters can fire fore or aft, it does not matter which of them ends up on top or bottom. The vehicle must therefore spin through an integral number n of half rotations while it precesses to the desired orientation. That is, the total spin angle ψ between the initial and final configurations is
\psi=n \pi=\omega_{s} t (a)
where \omega_{s} is the spin rate and t is the time for the proper final configuration to be achieved. In the meantime, the precession angle must be φ must be π or 3π or 5π, or, in general,
\phi=(2 m-1) \pi=\omega_{p} t (b)
where m is an integer and t is, of course, the same as that in (a). Eliminating t from both (a) and (b) yields
n \pi=(2 m-1) \pi \frac{\omega_{s}}{\omega_{p}}Substituting Equation 10.23, with θ = π/2, gives
\omega_{p}=\frac{C}{A-C} \frac{\omega_{s}}{\cos \theta} (10.23)
n=(1-2 m) \frac{4}{9} \frac{1}{\sqrt{2}} (c)
Obviously , this equation cannot be valid if both m and n are integers. However, by tabulating n as a function of m we find that when m = 18, n = – 10.999. The minus sign simply reminds us that spin and precession are in opposite directions. Thus, the eighteenth time that the spin axis lines up with the X axis the thrusters may be fired to almost perfectly align the angular momentum vector with the body z axis. The slight misalignment due to the fact that |n| is not precisely 11 would probably occur in reality anyway. Passive or active nutation damping can drive this deviation to zero.
Since \left\| H _{G_{1}}\right\|=\left\| H _{G_{2}}\right\|, we conclude that
\Delta H_{ total }=2\left\lgroup \frac{1}{6} m w^{2} \omega_{0} \right\rgroup =\frac{2}{3} m w^{2} \omega_{0} (d)
An obvious alternative to the coning maneuver is to use thrusters RCS-3 and 4 to despin the craft completely, thrusters RCS-5 and 6 to initiate roll around the y axis and stop it after 90 , and then RCS-3 and 4 to respin the spacecraft to \omega_{0} around the z axis. The combined delta-H for the first and last steps equals that of (d). Additional fuel expenditure is required to start and stop the roll around the y axis. Hence, the coning maneuver is more fuel efficient.
