Bats, Bugs, and the Moon

When hobby astronomers talk bugs, it is usually to commiserate about being stung and bitten at the telescope or to share advice on how to avoid this. But bugs affect not only skin sensation, they can contribute to the visual experience, albeit indirectly, as a lure for bigger fish. 
When we observe the night sky through a telescope, we peek through swarms of bugs without even noticing because, fireflies aside, bugs are too dark, too small or too much out of focus to register. Their mammalian predators, being many times larger, are a different matter. Bats, with some luck, are eminently visible. You can see them at the right time of year by observing the moon, as I happened to be doing on 8 September, 2022, from my backyard in Lexington, Kentucky. 
At around 10 PM, every thirty seconds or so, a bat silhouette would flutter across the moon’s face. Some silhouettes were large and out of focus, most were small and sharper. The smallest had a wingspan of probably no more than 1/25 of the diameter of the moon. These bats had to be far away and high up in the sky. Most of them were flying reasonably straight paths, which would suggest migration. If this were the case, I would expect them to fly in more or less the same direction, but they didn't: they kept crisscrossing the moon’s face, and at apparently quite different altitudes. Since bats aren’t known for soaring high in search of a mate, I reckon there must have been bugs up there to make the climb worth their while. I did see one or two bats fly erratic paths much like the ones they fly closer to the ground in pursuit of prey.
How high is high? It’s easy to calculate the altitude above ground level (AGL) using this formula (derived in the appendix):
AGL = wingspan × sin (moon's altitude angle) ÷ [2⋅tan (silhouette-to-moon ratio × 0.25°)]
The moon's altitude angle can be obtained from a stargazing app; on this occasion, it was about 24°. The ratio of bat silhouette to moon diameter I estimated to be 1/25; the 0.25° in the formula comes from the moon's angular size of 0.5°. The wildcard is the bat’s wingspan, which varies from species to species. The calculated AGL is directly proportional to it. Let’s assume that the species I saw was the big brown bat, which is said to be the most common one in my neighborhood and has an average wingspan of 13". Cranking these numbers through the formula gives an AGL of 1,262 ft (385 m).
I had never heard of bats flying that high and began to doubt my estimates. More solid photographic evidence was called for. I had little success trying to cobble together a scope attachment for my phone. But I luckily remembered that I had some adapters in the closet that I had once used to attach a real camera to the telescope for some improvised bird photography. The next night’s skies were clear, and the bats were out again, in lesser force than the night before, but enough for a decent sample. Below is the video to prove it. The moon jumps around whenever I push my non-motorized scope to follow it. I wish the bats were in better focus, but the moon was my only focusing target, and without a good idea of the bats' distance I didn't know how far out I should rack the focuser. They look sharper when seen directly through the telescope than they appear in the video because the eye can accommodate a little while the camera cannot. 
I located the frames in which the bats were most stretched out and took wingspan measurements at 4x magnification. This puts most of the silhouettes in the clip at between 1/30 and 1/60 of the moon’s diameter. (My earlier estimate of 1/25 wasn’t far off.) The video timestamps yield accurate numbers for the moon’s changing altitude during the hour I was filming (actual recording time was maybe 20 minutes; there was a lot of fiddling involved as I did this for the first time). For an assumed wingspan of 13", these numbers imply an AGL range of about 1,500 ft to 3,500 ft (457 m to 1067 m). You’ll notice some shadows in the video that are much closer. Some that I didn’t include were closer still. The high end of the range comes from the fact that, by the end of the hour, the moon had risen from 24° to 28.5°.
And then there is the highflier. Fifty-one seconds into the clip, a tiny silhouette enters at one o’clock, heads diagonally across the moon and, eighteen seconds later, exits at seven o’clock.  You may have to replay the clip in full-screen mode to see it. Below is a crop showing a speck in the center heading South across the mare serenitatis. The speck is at best 1/120 of the moon’s diameter (I tried to err on the side of size overestimation so as to not overstate the AGL). The moon’s altitude at the time was 24.5°. Again assuming a 13" wingspan, the corresponding AGL is an astounding 6,178 ft (1,883 m). 
Now what if this wasn’t a 13" big brown bat? For any other wingspan v (in inches), the AGL is obtained by multiplying 1,883 m by the factor v/13". How big or small are alternative candidates? I contacted Gary McCracken, who is one of the foremost experts on high-flying bats. The species he proposed are the hoary bat with a wingspan of 12" to 15", the 13" red bat, the 11.5" silver-haired bat, and the 12" Brazilian free-tailed bat. If it was any one of these, it couldn't have flown much below 6,000 ft because it wouldn't have been much smaller (and in fact possibly bigger and thus higher) than a big brown bat. 
Decades ago, bats in Texas were reported to fly at such altitudes, but then everything is bigger in Texas. (
The question now is, What kinds of bugs are up there, and what are they doing there besides nourishing the bats – if indeed bugs are the reason for the bats' ascent?
To fix ideas, let
w = the bat’s actual wingspan
d = the bat’s distance along the line of sight
𝛂 = the visual angle subtended by the bat’s outstretched wings (assumed to be orthogonal to the line of sight)
𝛃 = the moon’s altitude angle at the time of observation
a = the bat’s altitude above ground (presupposing that the ground below the bat is at the same level as the ground under the observer; if it isn’t, add in a correction term)
We also need
0.5° = the visual angle subtended by the moon
r = the ratio of the bat’s wingspan to the moon’s diameter.
Then we have
(1)     𝛂 = r⋅0.5°                                by elementary geometry
(2)     tan (𝛂/2) = w/2d                     from the first diagram
(3)     sin 𝛃 = a/d                              from the second diagram
(4)     d = w/[2⋅tan (r⋅0.25°)]           by using (1) to substitute for 𝛂 in (2) and rearranging terms
and finally
(5)     a = w × sin 𝛃/[2⋅tan (r⋅0.25°)]        by using (4) to substitute for d in (3) and rearranging
When you use this equation, keep in mind that the altitude comes out in the units of the bat’s wingspan and wants to be converted into something more useful.
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