perfectly normal everyday things that are misinterpreted (e.g. GOFAST is a goose)
Again, this seems pretty unlikely - given that the radar readings from the Nimitz verified what the FLIR camera was seeing. Geese don't create a large enough radar signature to register.
There's equal evidence for alien craft as for there is for these two theories - which is to say none at all.
You're right--I'm being hasty since this could have been a seagull. You can just make out its precious little wings flapping before the tracking system got a solid lock on it. Based on the system readouts and some high school math, the object is ~1 meter in size and is flying at 2.5km altitude. Very impressive technology tracking a very unimpressive target. But given the fact this was released as one of the smoking guns a while ago, it casts doubt that there was any competence (or even any attempt at competence) in the military's "analysis" of these sorts of videos. The interesting question is why is this sort of pixelated-bukkake is being released in the first place.
Can you explain the high school math you've used to draw your conclusions?
https://youtu.be/mfhAC2YiYHs?t=683
Thank you! I was wondering how the distance part of the triangle was being calculated.
Let's check the math:
F18
|\
|*\
|
|
|
---------------- object
|
|
|
|
|
80007620 m (narrator made a mistake in his ft to m conversion - 25000 ft is 7620 m, not 8000)
We know that the FLIR is looking down 35 degrees in one of the video clips so the angle of * is roughly 55 degrees.
We can calculate the field of view accurately! The narrator said that we've got +/- 1.5 degrees as a field of view, and I'm trusting him on this as I don't have my FLIR tech spec any more. So what exactly is the field of view then?
Calculate min FOV
sin(55 + 1.5) = 7620/inFront
inFront = 9477 m
Calculate max FOV
sin(55 - 1.5) = 7620/inFront
inFront = 9592 m
FOV = Max - Min = 114 m
But we also know that the angle of the FLIR is changing during the video. It goes to -22 degrees as it's tracking the object, while the plane altitude stays the same.
Calculate min FOV
sin(68 + 1.5) = 7620/inFront
inFront = 8193.5 m
Calculate max FOV
sin(68 - 1.5) = 7620/inFront
inFront = 8309 m
FOV = Max - Min = 116 m
So the field of view is half what the narrator is claiming.
cos* = adj/hyp
cos55 = 7620/distance
distance = 13,275 m
cos* = adj/hyp
cos68 = 7620/distance
distance = 17,318 m
So ground is between 13,275 m and 17318 m away from the plane over the video.
OK, carrying on. Mach .62 is 210 m/s, so in 1.5 seconds the plane moves 315m - that part checks out.
At this point though, I don't follow his math for calculating the height of the object which the narrator doesn't explain. Can you help me out?
So first thing is I don't think the calculation done was intended to be be exact (hence rounding to 8,000m) but was more of a back-of-the-envelope deal. I was a physics major so if the answer is within half an order of magnitude, it is considered "good enough" (also: cows are spheres). My reconstruction of some of the video commentary that is not fully explicit in the first image below.
First part of first image is what is used to get field of view. I'm maybe not following all of your FOV math (math in website text boxes doesn't always convey) but I just did the case where the camera is looking straight down with a 1.5 degree field of view. I'm not clear on the terms of art around field of view but took that to be the
total angular size of the image, so 1.5 degrees is 0.75 degrees in either direction from straight down. Simple trig results in 200m.
Second part of first image shows the use of parallax and similar triangles to determine height of the object if we known height of the plane. I get a slightly higher answer using this approach than in the video (almost 3km vs 2.5km).
The third diagram shows how using parallax does not change the result if the camera is inclined relative to the vertical. the triangles are still similar and all of the angular dependence cancels out. I think this is what was meant in the video. However, I had thought perhaps this meant that the dependence of effective field of view as a function of camera angle also cancelled out in the overall calculation, which I now think is not true...
It turns out the FLIR system is giving a range readout in miles all the time, so there is a very simple way to determine object altitude. I show the construction of this in the second image. The result is ~4.5km, which is 50% higher than using the approach in the video. This is such a straight-forward way of doing it that it must be closer to being right than using features in the background sweeping across in 1.5 seconds.
I also think I know why the two approaches result in different altitudes. The first one assumes the field of view when looking at an angle over the horizontal plane of the ocean is close to that when looking straight down. This is wrong, since the field of view in absolute terms grows as you look further out towards the horizon. Because of this and the complex 3D motion of the jet relative to the spot being examined (the feature sweeps from the corner in a diagonal direction), I think the supposition the ocean is 200m across in the view is incorrect and is on the low side. Adjusting for the actual field of view should bring this result into closer correspondence with the second simpler method mentioned.
What is clear in either case from a semi-order of magnitude estimate is that the object is at a decent altitude and is fairly small. Also, since we know the velocity of the jet, the angles to the object, and the distance to the object, it should be possible to calculate its 3D motion in space for the duration of the FLIR system lock & footage available. More back-of-the-envelope calculations suggest the object is not moving very fast but is the speed given in terms of air-speed or ground speed? In any case, it is
maybe a high flying bird, an errant weather balloon, or something as exotic as a drone. If we can't decide between these likeliest options conclusively, I can seen why it might be formally labeled a UAP.
