Aeroplane Construction and Operation

by John B. Rathbun

Fig. 2 is a form of wing in which the spar F' also forms the entering edge, thus eliminating one part of the wing. One objection to this construction is that the front spar must necessarily be shallower than the spar shown in Fig. 1. The rear spar is in the usual location at S', the two spars being connected through the usual end bow A'. The trailing edge T’ may be either a thin strip, or it may be a thin cable as indicated. This wing is similar to the wing used on the early Wright machines, and is still used by Farman, Voisin and other European manufacturers of biplanes. Usually the trailing ends of the ribs overhang the rear spar for quite a distance, in this type of wing, giving a flexible trailing edge. The front and rear interplane struts (m) and (n) are shown, the former connecting with the front spar at a point near the entering edge.

Fig. 3. Sub-Rib Construction, the Sub-Ribs (r) Are Placed at the Entering Edge.

Fig. 3 shows the usual construction except that short "Sub-ribs" marked (r) are placed between the main ribs R at the entering edge. These short ribs increase the support and accuracy of the curve at the entering edge, or else allow wider spacing of the main ribs R. The fabric must be well supported at this point, not only to maintain the best efficiency of the aerofoil, but to relieve the stress on the fabric, as it is here (Top surface) that the greatest suction pressure comes. Should there be a rip or tear near the entering edge, in the lower surface, the upper fabric will be subjected to both the pressure underneath and the vacuum above. This adds fully 25 per cent to the load on the upper facing.

The main spars may be of wood or steel tubing, although the former material is generally used. They are of a variety of forms, the "I" beam section, solid rectangular, hollow box, or a combination of plate and I sections, the total object being to obtain the greatest strength with the least possible weight. When made up of several pieces of wood they are known as "Built up" spars.

Fig. 4. Effects of C.P. Movement on Spar Loading

The load on the spars varies with the total weight carried, and also with the movement of the center of pressure due to changes in the angle of incidence. When the center of pressure moves to any extent, the loads on the two spars may vary between wide limits, and in extreme cases, either spar may carry the full load. This is shown clearly by Fig. 4, a section taken through the wing. The front spar F and the rear spar S are spaced by the distance L, the respective spar loads being indicated by Y and Z. As before explained, the center of pressure moves forward at large angles (CP), while at small angles it moves back say to position (CP’). Should it move back as far as CP-2, the load will come directly under the rear spar and this member will therefore carry the entire load. When at the forward position CP, the greater part of the load will come on the front spar, and only a small portion will now come on S. In the same way, when at a small angle of incidence, the center of pressure will be at CP’, a distance (K) from the rear spar. The greater part of the load will now be on S. The action is the same as if the entire weight W or lift, were concentrated at the center of pressure.

When intermediate between the two spars, the center of pressure causes a bending moment in the rib R, and is at a maximum when the CP is midway between the two spars. It will be seen that the C. P. movement has an important effect outside of the question of stability, and this travel must be taken into careful consideration when the strength of the spars is calculated. To find the load on the rear spar, for example, with the center of pressure at CP, multiply the lift W by the distance P, and divide by the spar spacing L. This will give the load Z. With the C. P. in the same position, the load on the front spar will be the difference between the total lift W and the load on the rear spar, or Y = W-Z. With the load at CP’, the load on the front spar will be: Y = WxK/L, and the load on S will be Z = W - Y.

For example, we will assume that the lift W = 1000 pounds, and the distance P = 12 inches. The spar spacing L = 30 inches and the center pressure is at CP. The load Z on the rear spar, will be: Z = WxP/L = 1000 x 12/30 = 400 pounds. The load on the front spar can be found from the formula, Y = W – Z = 1000–400 = 600 pounds.