I did something very simple which I found convincing to me and others ~ I said "Hey, how many degrees does that contrail cover over there in the part where the pro-missile crowd claims it all started".
Now you see I have to start with that question because to me it looks like the contrail goes over the horizon so I don't really see where it started, but I can start with the oldest, earliest visible part of the contrail.
If that part of the contrail is "x degrees" wide, then how many feet is that". This is a trivial question to answer with nothing more than basic trigonometry (an ancient process used by the Egyptians to redraw property lines after the Spring Flood of the Nile).
We know that the horizon for the camera is "Y miles" at a given altitude. If we assume the copter is on the ground at the average elevation of Los Angeles (less the Santa Monica mountains) then the horizon is about 35 miles away. If the copter is at 1500 feet, the horizon is at about 42 miles, and if the copter is at 2500 feet, the horizon is at about 62 miles. The higher the copter the further the horizon.
The claim is that there's been a missile launch at some point between the copter and the horizon.
Now there are missiliers here who have argued that maybe the missile was launched BEYOND the horizon which is why we don't see the normal surplus of smoke at the point of departure ~ instead we see some clouds at some appreciable altitude over the ocean ~ and that's not what I'm talking about. I'm talking about the "base" of the contrail column seen from where it is clearest, with an assumption any possible launch were made this side of the horizon (where the Earth looks like it falls off our screen).
Although tornadic winds might be sufficient to scatter the contrail in a few seconds, if this is a missile and there's no source of very high winds, we are looking at a missile contrail just a few seconds old ~ and it's wide because the missile is big, not because the wind is blowing hard.
Obviously I can't know what part of the horizon is covered by that contrail if I have no idea what part of the horizon is covered by my camera. Is this a wideview camera, or telephoto, or what.
Here's where we get into "ranges" and that'll give us an idea of how big the missile had to be for its engines to produce a contrail the width of the one we see.
(NOTE: There's an easier way to do this that's used all the time by the Army to lay in mortar or artillery fire, but that would require us to know where the specific settings on the telephoto unit on that camera were set so I won't go that way).
What we are going to have to do is find out what the minimum width is, as well as a maximum width, and by knowing only an estimated distance of 35, 42 or 62 miles from the helicopter to the "missile".
We know, by definition, that the circumference of the horizon is divided into 360 degrees. The camera, accordingly, is bringing in some part of that 360 degrees ~ somewhere from 1 to 360.
Given our knowledge of cameras, we can be pretty sure this one is taking in no more than 180 degrees ~ but when we look to the sides of the picture framed for the video it's most likely no greater than 120 degrees, and for argument's sake I'll even take 60 degrees ~ or, if you want, 30 degrees for a tight telephoto picture.
We know that Circumference divided by Diameter always equals 3.14159265. C/D = Pi
In all cases if we know two factors we can find the third. We don't presently know C, but we know D, and we know Pi. D is going to be twice 35 miles, twice 42 miles, or twice 62 miles with the camera at the center of the circle.
That means that the size of the circle is going to be the double the distance of the camera from the "missile" (proxied by "horizon") times Pi.
My original computation suggested that the width of the exhaust (at the base of the rocket) was 3220 feet with an aperture of 120 degrees and a distance of 35 miles. A bit of refinement reveals that if the camera has a telephoto lens yielding effective aperture width of 30 degrees, then at a distance of 35 miles the missile exhaust is about 4800 feet (or almost a mile wide).
If you increase the aperture width to 180 degrees (half the sky) then at a distance of 35 miles if the exhaust looks like it's as much as 1% of the horizon, it's even more enormous ~ GINORMOUS even.
if the rocket has an exhaust nearly a mile wide I think we have something else ~ like an airplane coming at us.
BTW, all the missile assumptions have it that the widest part of the contrail is nearest the observer. The airplane assumptions have it that the widest part of the contrail is furthest the observer and has simply expanded by being blown by the wind over a period of time, not the few seconds expected with a missile launch.
You can't get a mile wide contrail out of a rocket unless it's really big!
What can more can I say? You see a problem that you've calculated on paper; I see perfect normality within the context of real observed comparisions. I hate to keep asking this because it sounds so snotty, but honestly ... how many missile launches have you seen live? Perhaps you've already answered me on that and I've forgotten the answer; if so, please excuse the repeat query. But really -- how many launches have you observed live and from what range of distances? MORE to the point, how often have you used binocs to observe both airliners AND missile launches within the scale/distance showin in both the still shots and the video? See, all the math and figuring and time-lapse photos in the world become houses of cards when compared to the reality of that. There is zero liklihood of someone equipped with binoculars confusing a missile launch with an airliner contrail, and therefore zero liklihood of a professional camerman with zoom lenses making that mistake.
Good stuff!