In designing aeroplanes for flying machines, we should not lose sight of the fact that area alone is not sufficient. Our planes must have a certain length of entering edge--that is, the length of the front edge must bear a certain relation to the load lifted. An aeroplane 10 feet square will not lift half as much for the energy consumed as one 2 feet wide and 50 feet long; therefore, we must have our planes as long as possible from port to starboard. At all speeds of 40 miles per hour or less, there should be at least 1 foot of entering edge for every 4 lbs. carried. However, at higher speeds, the length may be reduced as the square of the speed increases. An aeroplane 1 foot square will not lift one-tenth as much as one that is 1 foot wide and 10 feet long. This is because the air slips off at the ends, but this can be prevented by a thin flange, or _à la_ Hargrave’s kites. An aeroplane 2 feet wide and 100 feet long placed at an angle of 1 in 10, and driven edgewise through the air at a velocity of 40 miles per hour, will lift 2·5 lbs. per square foot. But as we find a plane 100 feet in length too long to deal with, we may cut it into two or more pieces and place them one above the other--superposed. This enables us to reduce the width of our machine without reducing its lifting effect; we still have 100 feet of entering edge, we still have 200 feet of lifting surface, and we know that each foot will lift 2·5 lbs. at the speed we propose to travel. 200 × 2·5 = 500; therefore our total lifting effect is 500 lbs., and the screw thrust required to push our aeroplane through the air is one-tenth of this, because the angle above the horizontal is 1 in 10. We, therefore, divide what Prof. Langley has so aptly called the “lift” by 10; 500/10 = 50. It will be understood that the vertical component is the lift, and the horizontal component the drift, the expression “drift” also being a term first applied by Prof. Langley. Our proposed speed is 40 miles per hour, or 3,520 feet in a minute of time. If we multiply the drift in pounds by the number of feet travelled in a minute of time, and divide the product thus obtained by 33,000, we ascertain the H.P. required-- 50 × 3,520 ---------- = 5·33. 33,000