Question 8.2: Consider a four-phase, 8/6 VRM. If the stator phases are exc......

Figure 8.7 shows in schematic form an 8/6 VRM. The details of the pole shapes are not of importance for this example and thus the rotor and stator poles are shown simply as arrows indicating their locations. The figure shows the rotor aligned with the stator phase-1 poles. This position corresponds to that which would occur if there were no load on the rotor and the stator phase-1 windings were excited, since it corresponds to a position of maximum phase-1 flux linkage.

Consider next that the excitation on phase 1 is removed and phase 2 is excited. At this point, the stator flux wave has rotated 45° in the clockwise direction. Similarly, as the excitation on phase 2 is removed and phase 3 is excited, the stator flux wave will move an additional 45° clockwise. Thus the angular velocity ω_s of the stator flux wave can be calculated quite simply as π/4 rad (45°) divided by T_0/4 ~sec, or ω_s = π/T_0 ~rad/sec.
Note, however, that this is not the angular velocity of the rotor itself. As the phase-1 excitation is removed and phase 2 is excited, the rotor will move in such a fashion as to maximize the phase-2 flux linkages. In this case, Fig. 8.7 shows that the rotor will move 15° counterclockwise since the nearest rotor poles to phase 2 are actually 15° ahead of the phase-2 poles. Thus the angular velocity of the rotor can be calculated as -π/12 rad (15°, with the minus sign indicating counterclockwise rotation) divided by T_0/4 ~sec, or ω_m = -π/(3T_0) ~rad/sec.
In this case, the rotor travels at one-third the angular velocity of the stator excitation and in the opposite direction!