32

P. Mukkavilli et al.

The power required for maintaining hovering to lift the insect body weight against

gravitational pull can be calculated and average force



Favg

needed for two wings

during downstroke is also two times that of an insect weight (i.e. Favg = 2W).

Generally, the pressure applied by the wing will be uniformly distributed over the

total wing area and hence, the force generated by the individual wing will be acting on

a single point (i.e. at the midsection of the wing). Total work done during downstroke

(wings traverse to a vertical distance d) can be given as

Work = Favg × d = 2Wd.

(3.11)

Work also means a force should move an object in the direction of the force. The

component of the force in the direction of the movement does any work.

As an example by considering an insect weighing 100 mg (0.1 gm) with wings

of 1 cm long, vertical distance d as 0.57 cm, we get the work done by two wings

as equal to Work = 2 × 0.1 × 980 × 0.57 = 112 erg. Now we can calculate the

energy required by an insect with mass 0.1 gm to raise by a height of 0.1 mm during

downstroke which is given by E = mgh = 0.1 × 980 × 10⁻² = 0.98 erg.

The results obtained here show that the energy spent for a downstroke is almost

negligible and the maximum share of the energy developed due to downstroke is

converted into kinetic energy.

The calculation of power required by the wing per second can be calculated as an

output power,

P = 112 erg × 110



s⁻¹

= 1.23 × 10^{4 erg}

s

= 1.23 × 10⁻³ Watts [1w = 10⁷ ergs/s].

Considering the rotary motion of the insect wing, the kinetic energy of the flying

insect during each stroke is given by

K E = ¹

2 ^Iw2

max^,

(3.12)

where

I

is the mass moment of Inertia of the wing.

wmax

is the maximum angular velocity during each wing stroke.

Here moment of inertia of the wing will be given by

I = ^ml2

3 ^,

(3.13)

where

l

is the length of the wing;

m

is the mass of the wing (≈10^–3 g)