THE PATH OF A ROTATING SPHERICAL PROJECTILE. 501 



For Fig. 6. 



1. 40,000 200 -00500 '00500 15° -2588 -2588 -9659 -9659 



* * * * * 



26. 16,035 126-6 "00790 -16507 3-523 -0613 4-5617 -9981 25-5497 



* * * # * 



30. 13,940 118-1 -00847 -19809 0-472 -0082 4-6769 -9999 29-5476 



31. 13,472 116-1 -00861 -20670 - 0-360 - -0064 4-6705 -9999 30-5475 



* * * * * 



44. 9,147 95-6 -01046 -33189 -13-854 --2393 3-0442 -9709 43-4147 



* * * * * 



52. 7,850 88-6 -01129 -41952 -24-208 - -4099 -.3650 -9121 50-9412 



I regret that Mr Wood was obliged to give up his calculations before he had worked 

 out more than about a third of the requisite rows of figures for a trajectory differing 

 initially from fig. 5 in the sole particular <£ = 5° instead of 10°. This would have been 

 still more illustrative than fig.5 as a contrast with fig. 6. But a fairly approximate 

 idea of its form is obtained by taking the earlier part of fig.5, regarded as having the 

 dotted line for its base. See a remark in § 22 below, which nearly coincides with this. 



Effect of Wind. 



20. So far, we have supposed that there is no wind. But with wind the conditions 

 are usually very complex, especially as the speed of the wind is generally much greater 

 at a little elevation than close to the ground. Hence I must restrict myself to the case 

 of uniform motion of the air in a horizontal direction. We have in such a case merely 

 to trace, by the processes already illustrated, the path of the ball relatively to the air; 

 and thence we easily obtain the path relatively to the earth. Here, of course, it is 

 absolutely necessary to calculate the time of passing through each part of the trajectory 

 relative to the air. If the wind be in the plane of projection, and its speed U, the 

 relative speed with which the ball starts has horizontal and vertical components 

 Fcosa— U, and Fsin a, respectively. Thus, relatively to the moving air, the angle of 



elevation is given by 



, Vsina 

 tana = 



Vcosa— U 

 and the speed is 



V'= JV*-2UVcoaa+UK 



The relative trajectory, traced from these data, must now have each of its points 

 displaced forwards by the distance, Ut, through which the air has advanced during the 

 time, t, required to reach that point in the relative path. Of course, for a head-wind, 

 U is negative ; and the points of the relative trajectory must be displaced backwards. 



Figs. 7, 8, 9 illustrate in a completely satisfactory manner, though with somewhat 

 exaggerated speeds and coefficient of resistance, the results of this process. Mr Wood 

 had calculated for me the path in still air, with a = 288 (or, rather, 282), F=300, 

 <p = 6°, k = l/B. Since the time of reaching each point in this path had been incident- 

 ally calculated, it had only to be multiplied by 25, and subtracted from the corre- 



