
Hi Pete,
>> I would think the altitude of the plane would influence how this looked as well as focal length used.
The altitude would matter, in the context of distance to the camera. I mentioned that in my discussion of the application of the distance formula. Double the distance and the computed speed doubles.
The point I tried to raise is that the object would have to be very close (a few hundred yards or less) in order for the computed speed to be somewhere around the speed of typical aircraft.
An example... let's assume a jet descending, flaps down, at 150mph, which I think is a reasonable number for a minimum speed.
150mph = 220 feet per second (fps)
Now let's estimate a maximum possible speed, and that would be the speed of sound, 1126 fps. Aviation is not allowed to break the sound barrier in populated areas. Commercial aircraft can do about mach 0.85 or about 950 fps. I doubt a military hot dog pilot would risk going much more than mach 0.90 to 0.95 around LA, so mach 0.85 is within 10% of the max for any conventional aircraft.
At 220 fps, the camera to subject distance to cover 1/4 degree in 1/200s = 252 feet.
At 950 fps, the camera to subject distance to cover 1/4 degree = 1088 feet.
The numbers suggest that if that were a plane, it passed within 252  1088 feet from the shooter. If that plane passed within 1100 feet of Harry he would have noticed it . In a big way .
That is why I suggested that if it were a plane, it would have been a mile or two distant, minimum. But then the computed speeds are way too high.
Focal length does not matter here because the distance computation only requires angular diameter of the "subject" (the streak) and distance. If the focal length doubled, the moon is twice as big, the streak is twice as long, and the streak is still half the diameter of the moon (roughly), so it is still roughly 0.25 degrees long.
Similarly, suppose you wanted to photograph a 6 foot tall man in Silhouette against the moon, such that he exactly spanned the diameter of the moon.
At what distance would you shoot that scene?
The distance formula gives you a precise number:
d=206265*s/a
d = subject distance s = size (6') a = 1800 arcseconds (average)
Answer: 688 feet
(this is a useful formula to tuck away. A person(s) in front of the moon is a classic image)
The focal length does not matter. Short focal lengths will deliver a tiny moon and a tiny man. Long focal lengths will deliver a big moon and tall man but in all cases, at a 688 foot working distance from the man, the image results in a man the height of the moon. Same principle.
I'm not sure I am even disagreeing with you, except I do not believe the streak was made by an aircraft sans contrail. While I accept it could be a contrail, it begs the question of the opaqueness of contrails.
One other reason it is not a solid object...
Let's assume a solid object 10% as long as the black trail. If so, then each moon pixel behind that streak was blocked for only 10% of the exposure time. If the object length is 20% of the streak then the blockage only lasted 20% of the exposure time.
The result would be a "ghost image" similar to a person walking through a long exposure scene (we all did that one when we got our first SLR, right?). The image would record 8090% moon and only 1020% (in density) of our mystery object.
The fact that the streak is very dark, black or nearly so, precludes the possibility of an object of small size relative to the length of the trail.
The more I think about it, only a contrail makes sense. And the streak looks like a rapidly dissipating contrail. _________________________________ Neil
my Nikonians gallery. 