Before the actual test of the “lift” could be made, it was necessary to know the exact distance of the vertical center of gravity of the model and the extra weights from the knife-edge ‹B›. This was determined by the following method: A known weight was suspended from the arm ‹AB› at some arbitrarily selected distance from the point ‹B›. This weight caused the perpendicular arms ‹AB› and ‹DE› to rotate through an angle, θ, which was measured on the scale ‹KL›. Knowing, then, the weight on the arm ‹AB›, its point of application, the weight of the aerodrome suspended on the arm ‹DE›, and the angle of rotation, it is easy, by a simple application of trigonometric functions, to determine the distance of the center of gravity of the model from the point ‹B›. In a test of Aerodrome No. 6 made on September 23, 1898, the weight suspended from ‹AB› was 10,000 grammes, its point of application 50 cm., the model was weighted to 20,450 grammes, and the angle of rotation, θ, was 7° 2′. Letting y equal the distance of the ‹CG› from ‹B›, we may equate the balanced forces thus: 10,000×50 cos 7° 2′ = 20,450׋y› sin 7° 2′ 10,000×50 cot 7° 2′ = 20,450‹y› ‹y› = 198.2 cm. Having determined this distance, the weight on ‹AB› was removed and the aerodrome was allowed to regain its former position. The distance of the center of thrust from ‹B› was then measured. The engine was next started and the number of revolutions of the propellers counted by a tachometer. The thrust of the propellers, acting perpendicularly to the arm ‹BD›, produced rotation around the point ‹B›, the angle of which was measured as above. In the power test of No. 6, the following data were obtained: ‹W› = weight of aerodrome = 20,450 grammes. θ = angle of lift = 19° 30′. Distance of ‹CG› from center of rotation = 198.2 cm. Distance of center of thrust from center of rotation = 186.3 cm. As the propeller thrust and the weight of the model are forces acting in opposite directions at known distances from a center of rotation, letting ‹L› equal the “dead lift,” we may express the equation thus: ‹W› sin θ×198.2 = ‹L›×186.3, ‹L› = (198.2/186.3)×sin 19° 30′×20,450, ‹L› = 7,263 grammes “dead lift.” The flying weight of Aerodrome No. 6 was 12,064 grammes, and the per cent of this weight lifted was, therefore, 7,263/12,064 = 60.3. This was much more than was necessary for flight, but in order to insure successful flights and avoid delay, the rule was made in 1895 that no aerodrome was to be launched until it had previously demonstrated its ability to generate enough power to maintain for at least two minutes a lift of 50 per cent of the total flying weight. At the same time other important data were obtained, such as the steam-pressure, the time required to raise sufficient steam, the total time of the run, and the general working of the boilers and engines.