230 



Suppose now the body of the bird to be inclined at an angle Q to the 

 hoi'izon, and moving through the air with a velocity of v. feet a second, it 

 would rise (omitting the force of gravity for the present) by the angle at which 

 it was flying V. tan Q feet a second, and the resistance of the air to its wings 

 would give it a further upward movement of v sin Q cos Q feet a second, so 

 that the total rise of the bird would be v (tan Q + sin Q cos Q) feet per 

 second ; but we have already seen that the terminal velocity of the bird is 

 20 feet a second, so that in order to find the velocity at which the albatros 

 must fly at an angle of Q to the horizon in order to make the upward move- 

 ment just sufficient to counteract the force of gravity, or in other words to 

 maintain a horizontal line of flight, we have 



v (tan Q + sin Q cos Q) = 20 



20 



tan Q + sin Q cos Q 



20 cos Q 

 sin Q (1 + cos 2 Q) 



If we take Q to be 5°, we shall find that v = 1 1 6 ; and if we take it to be 

 10°, we get v = 58. It appears, therefore, that if an albatros starts with a 

 velocity of 1 1 6 feet a second, while sailing at an angle of 5° to. the horizon, he 

 could maintain a constant height above the sea level until his velocity was; 

 reduced to 58 feet a second, by gradually increasing the angle at which he 

 was flying to 10°. 



I will now compare the actual known resistance, of the air to a round shot, 

 to what ought to be the resistance to an albatros to allow it to sail for half an 

 hour without using its wings, and only reducing its velocity from 116 feet to 

 58 feet per second. 



The formula for calculating the resistance to round shot as given by 

 Parcelet, is 



R- 0-0006 A v 2 



where A is the resisting area in square feet. 



If now we take the area offered to the air by the front surface of the bird 

 to be 0'66 square feet, and its mean velocity at 87 feet per second, we have by 

 the round shot formula 



R = 0-0006x0-66x87 2 



= 3 lbs. nearly, which is evidently too great. 



I Avill now estimate roughly the real resistance the albatros ought to have 

 met with in order to enable it to sail for half an hour. 



Taking the average velocity at 87, it is evident that in half an hour 

 it would traverse 156,600 feet. Now at starting it would have accumulated 



W 2 17x1162 oroo 



— y z = — - — — — = 3599 units of work, 



2g 2 x 32 



at finishing it would still have unexpended 

 17 x 58 2 



2x32 



= 900 units of work. 



So that substracting one from the other, 2699 units of work have been con- 

 sumed in going 156,600 feet, and the resistance overcome would be 0*017 lb. 

 per foot, or only JL_ of that calculated by the round shot formula. This, how- 



