132 Captain F. W. Hutton on the Mechanical Principles 



which velocity it would maintain until it fell into the sea. This 

 is called its " terminal velocity." 



Let AB represent the axis of the body of the bird flying in 



Fig. 3. 



the direction B A and at an angle AE H with the horizon, Let 

 C D represent the wings of the bird making an angle C E H with 

 the horizon. Take the line H E to represent the velocity at 

 which the bird is flying, or the number of feet it passes through 

 the air in one second. From H draw the perpendicular H A ; 

 this line will represent the distance which the bird will rise (omit- 

 ting for the present the force of gravity) by means of the angle 

 at which he is flying to the horizon. But the force of the wind 

 H E acting upon the inclined wings C D will be resolved into 

 two forces, one of which, H K, will be parallel to the wings and 

 so have no effect on them, while the other, KE, will be at 

 right angles to them. This force will be again resolved into two 

 others at right angles to one another — one, K L, opposing the 

 forward movement of the bird, and the other, L E, causing it to 

 rise; so that the total amount that the bird will rise per second 

 will be LE -f HA feet. But we have previously seen that it will 

 fall by the action of gravity 30 feet a second ; so that in order 

 that it may fly horizontally, without either rising or falling, 

 LE + HA must equal 30 ; and we want to find what must be 

 the length of HE, or, in other words, the velocity of the bird to 

 do this. 



Now KE = HEsinCEH, because CEH is equal to EHK, 

 and LE is equal to KE cos CEH, because KEL is also equal to 

 CEH. Therefore 



LE = HE sin CEH cos CEH, and AH equals HE tan AEH ; 



.-. HE tan AEH + HE sin CEH cos CEH =30, 



HE (tan AEH + sin CEH cos CEH) =30 ; 



. TT p_ 30 



* ' tan AEH + sin CEH . cos CEH 



If, now, we take AEH = and CEH = 15°, we shall find that 

 HE equals 115. If we take AEH = 7°and CEH = 22°, we find 



