MOTION. 427 



motion ; but according to our experiments the whilst the areas of their wings are respectively 

 weight of the Pigeon is 4347.344 grains; of 0.6198, 1.11, and 0.054 feet : hence, we see 

 the Rook 4170.25; and of the Canary 229; that in these instances, as probably in other 



and solving it with respect to V, we have 



V = p 1+1 P+l\* V * >>+1 "8+2 



p-q 'V \p-q/ P- 



and supposing the same things in (18), it will become 



Here we may substitute for V its value found in equation (19). If the bird merely supports itself 

 in the air without rising or falling, t/ = o, and the expressions for V and the force expended in an unit 

 of time, viz. (19) and (20) will become respectively 



p q 9 K A 



P* P 9*fp 9 w K A 



We may suppose q to be known from the figure of the bird and the power which it has of expanding 

 the wing in a depression, and closing it during an elevation, so as to increase and diminish the surface 

 and resistance alternately; and in order to find the least quantity of force necessary to sustain the 

 bird, we must give to p such a value as will make the expression (22) a minimum. 



All other things being the same, we see that the quantity of force expended is proportional to the 

 ,/W 3 , and is also inversely as the ^/densitvof the air. Let us take, as an example, the data given by M. 

 Chabrier with respect to the Swallow. W = 0.01526 kil. <rr= 1.25 kil. A = 0.0086 met. car. g=. 

 9.80422 met., now from experiment it is found that a plane surface very nearly equal to the surface of a 

 swallow's wing will produce a resistance in air equal to the weight of a column of fluid, whose 

 base is the plane and altitude between one and a half, and twice the height due to the velocity with 

 which the plane is moving ; if, therefore, we suppose the resistance of the wing to be a little more on 

 account of the concavity of its surface, and consider the altitude just twice the height due to the 

 velocity, K will = 2. It is also evident that p > 1 and q < 1, and by giving different values to q, we 

 may obtain the corresponding values of p. 



1. Let 9 = 2, then when (22) is a minimum. 



p =3.073 



V ss 8 m.l 8 = 26.8304 feet. 

 (22) = W 10m.36 = 33.784 feet. 



2. Let q = J p = 1.866. 



V = 6n.91 = 22.6648 feet. 

 (22)=W 8n.39 = 27.5192 feet. 



3. Let q =. I p = 1.6 



V = 6"i.81 =203368 feet. 

 (22) = W 7<n.95 = 26.0660 feet. 



We may observe that the several values assigned to 9 do not produce any great variations in the 

 values of V and (22), we may, therefore, with tolerable accuracy conclude that a bird which just 

 supports itself in the air, expends as much force every second as would be sufficient to raise its own 

 weight a height of 8 metres or 27.5192 feet. The mean velocity of the descending wing is about 

 7 metres or 22.96 feet per second, and the velocity of the ascending w ing is about one-half of it. 

 Let us now take the distance of the centre of the swallow's wing, from its axis of motion, to be three 



inches, and that it oscillates in an arc of 120= then the length of this arc will be found 



o 



2 T 



by the follow ing proportion 1 : 3 :: : 2 TT = 6.2832 inches; the velocity divided by this 



3 



length is = 43.85, being the number of arcs which would be described by the descending w ing in 

 one second, but the ascending wing takes 1.866 seconds to describe the same number of arcs, conse- 

 quently this number, divided by the sum of the times taken by each wing, namely, 2.866, gives 15.3, 

 the number of flappings, each consisting of two arcs, per second. 



Such is the result by Chabrier's formula. If we apply the same formula to the case of the Pigeon, 

 the result is 14.9, or nearly 15 flappings per second; but actual observation proves this to be 

 nearly three timas the real number; a similar discrepancy is observed in the case of the Rook, 



