Ill K INTERNAL WORK OF THE WIND. 



33 



Since K> g in the numerical example, the ordinate y is obtained from formula 

 (4) ; to find the ordinate at the point C we substitute 22.1607 for V and for b in 

 this formula and get : y c = —22.0908 m. 



From the same formula, by making V= 8.319 and b = ~ we get for the 

 ordinate of the point B' : y B , = — 2.74013 in., A being taken as the origin for 

 these values. 



3. From B' to B the aeroplane is carried by its weight, which also retards its 

 ascent; for this stretch we have therefore: -Z- = — g and heuce V = V B ,— gt and 

 8 = Y B ,t-±gt*. 



At the point B, V — 0, and therefore gt = V B ,. Substituting the value for t 

 from this equation in the equation for S we find that : 



1 Vo' 1 (8,319)' ] 



B'B =S-- 



3.52794 in. 



2 g 2 9.8088 



Correlating all the results thus far obtained, we have for the different vertical 

 distances traversed : 



Descent Ascent 



From A to A' 5 097465 m. 

 From A' to C 22.0908 m. 



From C to B' 19.35067 m. 

 From B to B 3.52794 m. 



Total 27.188265 m. Total 32.87861m. 



Thus the finishing point B is about 4.3 meters below the starting point A. 



C ' c 



Direction of the Wind 

 Fig. 5. Fig. 6. Fig. 7. 



(We have not yet calculated the abscissa of the point B as it is not of great 

 importance here. The time required by the aeroplane, however, in going from the 

 point A to the point B is found by the method of approximation previously men- 

 tioned to be about five seconds.) 



If we suppose now that an aeroplane falls into a calm atmosphere under the 

 same conditions as in the preceding case, it will describe the same trajectory and 

 we will have as before: 



V A =0 V A -=10 V c =22.1067 



Vertical distance from A to A : 5.097465 m. 

 Vertical distance from A to C : 22 0908 m. 



