Many people here have probably heard of the controversy regarding Elon Musk's claim that "stealth aircraft could be easily detected using IR cameras and ML" (sorry, this is the last mention of that person in this post). Recently I came across a video claiming that aircraft can, indeed, be detected from distances as large as 30 km using an array of smartphone-grade cameras and a pair of simple algorithms (motion detection + triangulation). I've seen a video by an astronomer (UPD.: the video title was "Lowlight Cameras Can't Beat Stealth") explaining that, because of atmospheric turbulence and Rayleigh limit (UPD.: He didn't use the Rayleigh limit), and also some considerations regarding information theory which I don't remember (UPD.: Basically throwing the Shannon limit at the viewer), such detection is not, in fact, feasible, without providing any calculations (UPD.: Providing some computations regarding resolution per pixel). So, upon seeing the linked video, I myself tried to estimate the maximal distance from which an aircraft can be detected using a camera similar to that of a smartphone (calculations are provided below). However, my optics course was quite a while ago, I didn't particularly ace it, and it's not my specialty, so I was only able to perform a very rough estimate based on the Rayleigh limit.
First, as far as I could understand from my little research on the topic, the actual limiting factor is usually not the Rayleigh limit, but the atmospheric turbulence, and I wasn't able to understand how to calculate the maximal distance at which an object of a fixed size can be resolved using a given optical system. Second, there's also the factor of atmospheric extinction of light: I am not quite sure how to include it in calculations. Third, since we aim to detect movement, is resolution even that relevant? Yes, we have a dizzy twinkling mess of pixels instead of an aircraft image, but we don't really need an aircraft image in this context, do we? All we need is to tell that something is moving out there with a given speed.
I understand that many people are allergic to this whole story, but I believe the topic itself is quite interesting and deserves a proper discussion, regardless of the personal qualities of people involved. Now, to the calculations:
The Rayleigh criterion states that two points separated by an angle of $\theta$ can be resolved by an optical system including an aperture of diameter $D$, if $$\theta \ge \approx 1.22 \frac{\lambda}{D}$$ where $\lambda$ is the wavelength of the light detected by the system. The angular size of an object having a linear size $d$ is $$\theta = 2 \arctan\left(\frac{d}{2R}\right)$$ where $R$ is the distance from which the object is observed. The size of an aircraft is much smaller than the distance from which we observe it, so $d << R$ and we can approximate $$\theta \approx \frac{d}{R}$$ so, given a fixed minimal angular resolution $\theta$ and a fixed $d$, $$R\le \frac{d}{\theta} \le \frac{Dd}{1.22\lambda}$$
I was able to find information that the aperture of a typical smartphone camera is approximately 0.8 to 3.6 mm in diameter, depending on the usage mode, so that is our $D$. The length of an F-35, according to Wikipedia, is about 15 m - this is our $d$. The visible light wavelengths are from about $\lambda_{\text{IR}}=380 \text{ nm}$ to about $\lambda_{\text{UV}}=750 \text{ nm}$, we'll take 500 nm for an estimate. Then $$R_{500 \text{ nm}}\le \frac{3.6 \text{ mm} \cdot 15 \text{m}}{1.22 \cdot 500 \text{ nm}}= \frac{3.6 \cdot 1.5}{1.22 \cdot 5} \cdot 10^{-3+1+7}\text{m} \approx 88.5 \text{ km}$$ which is obviously quite a lot. So Rayleigh criterion doesn't appear to rule the idea out even for a smartphone camera.
Now, from the Wikipedia article I understood that a point light source gets distorted by the atmospheric turbulence into a "disc" with angular size $$\theta \approx \frac{\lambda}{r_{0}}$$ where $r_{0}$ is the "field parameter": $$r_{0}=\left[ 0.423 k^2 \int_{\text{Path}} C_{n}^2(z')dz' \right]^{-3/5}$$ where $\text{Path}$ is the path along which the light travels, $k$ is the wave number $\frac{2\pi}{\lambda}$, and $C_{n}(z)$ is atmospheric turbulence strength, which depends mostly on altitude, so the formula can be rewritten as
$$r_{0}=(\cos \zeta)^{3/5}\left[ 0.423 k^2 \int_{\text{Vertical path}}C_{n}^2(z)dz \right]^{-3/5}$$ There is no, as far as I understand, a universally accepted mathematical model describing $C_{n}(z)$, but Wikipedia links to Hufnagel-Valley model as a good approximation. According to this model, $$C_{n}^2(h) = 0.00594\left( \frac{v}{27} \right)^2 (h/10^{5})^{10}e^{-h/1000}+2.7 \cdot 10^{-16}e^{-h/1500}+C_{0}e^{-h/100}$$
Here $v$ is RMS of "upper atmospheric" (whatever that means) wind speed, and $C_{0}$ is the value of $C_{n}^2$ at the ground level (in whichever way it is measured), $h$ is height above the Earth's surface. It bothers me somewhat (like tsunami warning bothers a staff member of a nuclear power plant) that we see a dimensional quantity being exponentiated, and I've just lost hope of tracking the dimensions in the overall formula, so I calculate everything in SI and hope that is what the guys who wrote the formula intended.
In another article, I was able to find (Figure 2.1 at page 25) a typical value of $C_{n}^{2}(0)$, $1.70 \cdot 10^{-13} \text{m}^{2/3}$, and a typical value of $v=21 \text{m/s}$. Suppose our F-35 flies at the altitude of 5 km. Then $$r_{0}=\left(\cos\arctan \frac{5 \text{ km}}{L}\right)^{3/5}\cdot C_{5000, \lambda}^{-3/5}$$ where $$C_{5000, \lambda}=0.423\frac{4\pi^2}{\lambda^2}\int_{0}^{5000}C_n^{2}(h)dh$$
According to Worlfram Alpha, the definite integral on the right is $1.74 \cdot 10^{-11}$ of whatever units. Taking again $\lambda = 500 \text{ nm}$, and converting nanometers to meters, we have $$C_{5000,500\text{ nm}}=0.423 \frac{39.48}{25}\cdot 1.74 \cdot 10^{14-11} \approx 1162$$ (I suppose this quantity is dimensionless, because we exponentiate it again. That would mean the previous quantity has the dimension of $1 / \text{m}^2$, but I don't understand how that corresponds to the dimension of the constant $[C_{0}^2]=\text{m}^{2/3}$). Finally, $$C_{5000,500\text{ nm}}^{-3/5}=0.0144$$ Suppose our F-35 is 50 km away from the observation point. Then $$r_{0}=\left( \cos \arctan \frac{1}{10} \right)^{3/5}\cdot 0.0144 = 0.997 \cdot 0.0144 \approx 0.0144$$ and the angular size of the disk into which each point of our F-35 is distorted is $$\theta = \frac{5}{1.4} \cdot 10^{-7+2}=3.57 \cdot 10^{-5} \text{ rad}$$ which is... 7 arcseconds? I guess, that's not too bad? At this point I struggle to determine how this would affect the ability to detect movement (let alone the validity of the calculations above). So I invite anyone who understands optical systems better than me to explain how does one actually calculate the limit resolution in this situation. And what are those information theory considerations?
P.S. How much more focused do you need this question to be, if you have a perfectly fine question about Legolas being able or not being able to see the Rohirrims? If anything, this question is much more specific, since we are talking about specific real-world technology in a very similar, but much more specific context. I feel that the actual reason people voted to close is the general allergy towards everything related to Elon Musk, but Musk's personal qualities aren't the question's fault. I can remove the mention of his name and of the controversy, since they aren't relevant to the question apart from being the motivation to ask it.
