Let ‹af› represent the resultant of the vertical components of the pressure on the wings; the horizontal component will lie in the line ‹ae›. [Illustration: FIG. 4. Diagram showing relation under certain conditions of thrust, C. P. and C. G.] Let the center of gravity be in the line ‹bd›, and the resultant thrust of the propellers be represented by ‹cd›. Let ‹W› = weight of aerodrome. Let ‹T› = thrust of propellers. Then if we neglect the horizontal hull resistance, which is small in comparison with the weight, equilibrium obtains when ‹W›×‹ab› = ‹T›×‹bd›. Second case. The diagram (Fig. 5) represents the same system of forces as Fig. 4, but in this case the point of support is directly over the center of gravity ‹g›, when the axis of the aerodrome is horizontal. Let ‹W› = weight of aerodrome. Let ‹T› = thrust of propellers. Let ‹R› = distance of ‹CG›_2 below ‹CP›_2 = ‹ag›. Let ‹S› = distance of thrust-line below ‹CP›_2 = ‹ad›. If now the aerodrome under the action of the propellers be supposed to turn about the ‹CP›_2 (or, ‹a›) through an angle α, so that ‹g› takes the position ‹g′›, we [p049] obtain by the decomposition of the force of gravity an element ‹g′k› = ‹W› sin α which acts in a direction parallel to the thrust-line. If we again neglect the horizontal hull resistance, equilibrium will be obtained when ‹kg′›×‹ag′› = ‹T›×‹ad′› or ‹WR› sin α = ‹TS› ∴ α = sin^{−1}×(TS/WR) [Illustration: Fig. 5. Diagram showing relation under certain conditions of thrust, C. P. and C. G.] The practical application of these rules is greatly limited by the uncertainty that attaches to the actual position of the center of pressure, and this fact and also the numerical values involved may be illustrated by examples. CONDITION OF AERODROME NO. 6, NOVEMBER 28, 1896 The weight was 12.5 kilos. On November 28, the steam pressure was less than 100 pounds, and the thrust may be taken at 4.5 kilos. The distance ‹bd› was 25 cm. Hence 12.5׋ab› = 4.5x25 cm. ‹ab› = 9 cm. This appears to give the position of ‹CP›_1, but ‹CP›_1 is a resultant of the pressure on both wings, and its position is determined by the empirical rule just cited. We [p050] cannot tell in fact, then, with exactness how to adjust the wings so that ‹CG›_1−‹CP›_1 may be 9 cm., and equilibrium was in fact obtained in flight when (the empirically determined) ‹CG›_1−‹CP›_1 = 3 cm. Again, let it be supposed that ‹CP›_1 was really over ‹CG›_1 . . . . The distance of the center of gravity below the center of pressure is 43 cm. = ‹R›. Then α = sin^{−1} {(4.5×25)/(12.5×43)} = 12° nearly.