How much energy (eV) is in a mosquito's wing beat?

We're all familiar with the annoying sound of a mosquito near our ear. I estimate the wing beat frequency to be in the 400 to 800Hz range (to accomodate the Bruce Willis and Danny de Vito of mosquitos :-)

I was wondering if we can guesstimate the energy in a single wing beat (say, to within a factor of 10). I have no intuitive understanding about picojoules, which is why I think a value in eV might be more instructive. From this value, can we derive how many molecules of oxygen the mosquito's metabolism must use per second?

If you make estimates for various parameters (like mosquito mass, ...), please also estimate an error.

• "From this value, can we derive how many molecules of oxygen the mosquito's metabolism must use per second?". I don't think so. The mosquito isn't a 100% efficient machine. Heat loss during metabolism would have to be determined biologically. – udiboy1209 Aug 17 '13 at 19:40
• @udiboy Correct, but at least we'd have a lower limit. – Jens Aug 17 '13 at 19:48
• @Jens: This is a very cute order-of-magnitude estimation problem, +1!. But re udiboy's comment, you could characterize the vibrations of the wing by some value of $Q$. If $K$ is the kinetic energy of the wing, then the metabolic output per cycle is roughly $K/Q$. We don't know if $Q$ is less than 1 or greater than 1 (either seems plausible to me). – Ben Crowell Aug 17 '13 at 20:27

The ball-park answer is 1.5 GeV per wing beat. Here’s how we get that:

To start, take the case of a mosquito hovering in a fixed position with no breeze. This keeps the problem relatively simple, and should give us a pretty good ball-park/order-of-magnitude estimate (even though hovering requires somewhat less energy than, for example, flying upward, or horizontally, or dealing with a breeze). This is the same approach outlined in Biophysics Demystified, pp. 307-312, examining the case of a hummingbird hovering.

To hover in the air, the average of lifting force generated by the wings must be equal (and opposite) the weight (gravitational force) on the mosquito. To simplify the problem further, assume that the mosquito generates lift only during the wing’s downstroke, and that no lift (or force) is generated by the upstroke (backstroke) of the wing.

In practice, many insects can orient their wings so as to produce lift on both the downstroke and upstroke. Hummingbirds do this as well (although typically more lift is generated during the downstroke than the upstroke). The total amount of energy expended to maintain the average lift is approximately the same in either case, the main difference being how that energy is distributed between the downstroke and the upstroke, so the assumption of lift being generated only by the downstroke should not alter the answer (certainly not for us to obtain an order-of-magnitude result; I will explain at the end why this greatly simplifies the problem).

I am also going to assume that the time required for a downstroke is the same as that for an upstroke, and that there is zero time between the downstroke and the upstroke. These assumptions should introduce negligible error for our order-of-magnitude estimate.

Given the assumption of lift generated only during the downstroke, the mosquito will rise due to the lifting force generated by the downstroke, and then fall due to gravity during the upstroke. Since we assume the amount of time is the same for up and down, and there is no lifting force during the upstroke, then the downstroke must generate a lifting force equal to twice the weight of the mosquito (so the average lifting force will be equal the weight). This constraint will ensure that the upward acceleration due to lift during the downstroke will exactly match the downward acceleration due to gravity during the upstroke. This also means that the distance the mosquito moves up during the downstroke will be the equal to the distance the mosquito moves down during the upstroke. On average the mosquito will appear to be hovering, but in reality, based on our assumptions, it is actually bobbing up and down over a very small distance, very rapidly.

Now how fast do the wings beat? Let’s assume 500 Hz. (Various sources online report values ranging from 400 to 600 Hz). This means that 500 times a second our hovering mosquito is bobbing up and down. If we know how far our mosquito bobs up and down, and we know the weight of the mosquito, we can calculate how much energy it takes to lift our mosquito this distance for a single wing beat.

How far? Since, based on our assumptions, the downstroke distance and the upstroke distance are the same, we only need calculate one of them. The simplest one to calculate is how far the mosquito falls during the upstroke. This distance is equal to the average velocity times the amount of time:

$$d = v_{avg} * time$$

We assume that the velocity is zero at the moment the mosquito changes direction, and maximum just before the next change in direction. We know acceleration due to gravity is constant (ignoring air resistance for now) which means the increase in velocity is linear, so we can take the halfway point as the average velocity, thus:
$$v_{avg} = 0.5 * g * time$$ (where g is the acceleration due to gravity, -9.8 m/s^2 )

If the wing beat frequency is 500 Hz, the time for one stroke (up or down) is 1/1000 second, thus:

$$d = (0.5 * g * time) * time$$ or $$d = ( -9.8 m/s^{2} ) * (10^{-3} s ) * (10^{-3} s ) = - 4.9 x 10^{-6} meters = -4.9 μ$$

Thus, $$E = m * g * d$$ $$= ( 5 * 10^{-6} kg ) * ( -9.8 m/s^{2} ) * ( -4.9 * 10^{-6} m )$$ $$= 2.4 * 10^{-10} joules$$ $$= 1.5 * 10^{9} eV = 1.5 GeV$$ Now for the sake of transparency, a little more about our assumptions, and our decision to assume no force is applied during the upstroke. To begin with, we are ignoring air resistance, except that we are assuming the downstroke is completely efficient, that is, the energy to push the wings down is entirely converted into an upward force/motion. This is of course not possible, as some of the air molecules will not resist the wing but will be simply pushed away. On the other hand, assuming the upstroke generates no force also assumes a complete lack of air resistance against the wings, and ignores the very small amount of energy required to simply lift the weight of the wing. In the case of hummingbirds, research using “digital imaging particle velocimetry” (DPIV) has shown that hummingbirds generate approximately 25% of their lifting force during the upstroke, and 75% during the downstroke. This cuts the up and down distance approximately in half (see Biophysics Demystified, p.312). If the upstroke generates more force, say 50% of the lift on the upstroke, and 50% on the downstroke, then the up and down movement will be negligible. This means we can no longer use the distance time the force to determine the energy. Instead we would have to determine how much air gets pushed out of the way as the mosquito hovers. This is next to impossible, certainly without empirical data (perhaps using DPIV) to support our assumptions. This is the reason we assume little or no force during the upstroke, that and the fact that the small amount of research done so far indicates that the amount of lift during the upstroke is, at least, significantly smaller than that during the downstroke. By the way, the result above for our mosquito is in close agreement with that calculated here for a 100 mg insect, with a wing beat frequency of approximately 200 Hz, after taking into account that the slower frequency and heavier insect, both contribute to the need for more energy per wing beat.