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 
aud solving it with respect to V, we have ; 
Vapit! a/ prt 3 Axton EAL i abt SUP ee 
py taj (Cy eee (19) 
and supposing the same things in (18), it will become 
mKAV § 
2g (P+ Pe 
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, v, = 0, and the expressions for V and the force expended in an unit 
Of time, viz. (19) and (20) will become respectively ‘ 
PPV —uP+ a (V+ pu)? } eveeeeseeceeerseeee ee (20) 
\E gtr 208 
pP—q *wKA 
>) Cees eeeeeteoeecssecseseeeesse(al) 
wPtq, /Ptr 2Ws 
P-Pt’ eee ORS 
ee cccerccccccceccccccesseseers (2a) 
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 mcrease 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%, and is alsoinversely asthe ./density of theair. Let us take, as an example, the data given by M. 
Chabrier with respect to the Swallow. W = 0.01526 kil. w= 1.25kil. 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 andq < 1, and by giving different values to g, we 
i = obtain the corresponding values of p. 
Let g=}, then when (22) is a minimum. 
p = 3.073 
V= 8,18 = 26.8304 feet. 
(22) = W 10.36 = 33.784 feet. 
2. Let g = Jeececccvccccccccscccccc ep = 1.866. 
V = 6,9] = 22.6648 feet. 
. (22)==:W 8m.39 = 27.5192 feet. 
De Oo LRt Joc cc ees cccccccvecccccvep = 1.6 
V = 6,81 = 20 3368 feet. 
(22) = W 7.95 = 26.0660 feet. 
We may observe that the several values assigned to g 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 
.'é rts 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 wing 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°= sy then the length of this are will be found 
o by the following proportion 1 : 3 8s : 2% = 6.2832 inches; the velocity divided by this 
length is = 43.85, being the number of arcs which would be described by the descending wing in 
_ one second, but the ascending wing takes 1.866 seconds to describe the same number of arcs, conse- 
oe «sgt 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, 
el is 14.9, or nearly 15 flappings per second; but actual observation proves this to be 
ly three times the real number; a similar discrepancy is observed in the case of the Rook, 
; 
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A 
