The wash-down effect of the wings on the tail is proportional to the chord and not to the span, and for this reason an increase in span does not always necessitate an increase in the length of the body. An adequate damping effect requires a large surface at the end of a long lever arm.
Balancing the Aeroplane. Figs. 1 to 6 show the principles involved in the balancing of the aeroplane. In Fig. 1 a number of weights 1'-2'-3' and 5M are supported on a beam, the load being balanced on the fulcrum point M. The load 2' being directly over the fulcrum, has no influence on the balance, but load 1’ at the left tends to turn the mass in a left-hand direction, while 3' and 5M tend to give it a right-hand rotation. This turning tendency depends upon the weights of the bodies and their distance from the fulcrum. The turning tendency or "Moment" is measured by the product of the weight and the distance from the fulcrum. If weight 1' should be 10 pounds, and its distance A' from the fulcrum should be 20 inches, then it would cause a left-hand moment of 10 x 20 = 200 inch pounds. If the system is to be in balance, then the left-hand moment of 1’ should be equal to the sum of the moments of 3’ and 5M. Thus: 1’ x A = (3' x B) + (5M x C').
Figs. 1-6. Methods of Balancing an Aeroplane About Center of Lift.
The application of this principle as applied to a monoplane is shown by Fig. 4, in which X-X is the center of pressure or lift. The center of lift corresponds to the fulcrum in Fig. 1, and the weights of the aeroplane masses and their distance from the center of lift are shown by the same letter as in Fig. 1. The engine 1' is at the right of the C. P. by the distance A, while the fuel tank 2 is placed on the C. P. in the same way that the weight 2' in Fig. 1 is placed directly over the fulcrum. By placing the tank in this position, the balance is not affected by the emptying of the fuel since it exerts no moment. The chassis G acting through the distance E is in the same direction as the engine load. The body 5 with its center of gravity at M acts through the distance C, while the weight of the pilot 3 exerts a right-hand moment with the lever arm length B. If the moments of all these weights are not in equilibrium, an additional force must be exerted by the tail V.
Fig. 2 shows an additional weight 4' that corresponds to the weight of the passenger 4 in Fig. 5. This tends to increase the right turning moment unless the fulcrum is moved toward the new load. In Fig. 2 the fulcrum M remains at the same point as in Fig. 1, hence the system requires a new force P’ acting up at the end of the beam. If the load was in equilibrium before the addition of 4', then the force P’ must be such that P’ x T’ = 4" x D’. In the equivalent Fig. 5, the center of gravity has moved from its former position at S to the new position at R, the extent of the motion being indicated by U. To hold this in equilibrium, an upward force P must be exerted by the elevator at Y, the lever arm being equal to (T + U).
Fig. 6 shows the single-seater, but under a new condition, the center of pressure having moved back from X-X to Z. To hold the aeroplane in equilibrium, a downward force must be provided by the tail V which will cause a right-hand moment equal to the product of the entire weight and the distance U. For every shift in the center of pressure, there must be a corresponding moment provided by the elevator surface. The condition is shown by the simple loaded beam of Fig. 3. In this case the fulcrum has been moved from M to N, a distance equal to the center of pressure movement in Fig. 6. This requires a downward force P' to maintain equilibrium.
Center of Pressure Calculation. Fig. 7 is a diagram showing the method of calculating the center of gravity. The reference line R is shown below the elevators and is drawn parallel to the center of pressure line W-W, the latter line being assumed to pass through the center of gravity. The line R may be located at any convenient point, as at the propeller flange or elsewhere, but for clearness in illustration it is located to the rear of the aeroplane. The weight of each item is multiplied by the distance of its center of gravity from the line R, these products are added, and the sum is then divided by the total weight of the machine. The result of this division gives the distance of the center of gravity from the line R. Thus, if the center of gravity of the body (11) is located at (10), then the product of the body weight multiplied by the distance B will give the moment of the body about the line R. The weight of the motor (2) multiplied by the distance F gives the moment of the motor about R, and so on through the list of items.
Center Of Gravity Table