Originally Posted by
Strongresolve
Again I must say, that the exact translation of the word that the controler used to alert of runway condition is Damp, he said "pista humeda."
Thanks, I have put your remark in a comments in the translation. Was this already your second remark - didn't see your first!
Originally Posted by
DingerX
in the 4 seconds it's on camera, the aircraft travels 5 plane lengths. Length of an A320 is 37 meters and change, so back-of-the-envelope gives 170 kph, or approximately 100 kt.
That approach is seriously, seriously flawed!
Let's make this clear in an assumed picture: the camera looks in a right angle to the runway, is fixed and the lens system makes sure, that all angles in the pictures are depicted correct from corner to corner. The camera looks from 30 degrees right to 30 degrees left, the camera is say 150 meters off the runway.
As the airplane comes into the view at the right hand corner of the picture, it is therefore at a distance to the camera of 150/cos(30 degrees)=173 meters, and it has to travel a distance of 150*sin(30 degrees)=75 meters to reach the center of the picture. Because of the larger distance the airplane appears smaller than it would be in the center due to the perspective, and because of the angle the distance of 37 meters in reality appears on the picture as 37*sin(30 degrees)=18.5 meters. The same applies for the left hand corner of the picture. The total distance on the runway centerline from right to left corner would be 150 meters or 8.1 times the airplane length measured at the corner or roughly 4 times the airplane length measured in the center of the picture.
Now, let's put the camera at an angle to the runway, say 30 degrees to the left. The center line of the runway at the middle of the picture would now be 173 meters away (the camera still 150 meters off the runway in the same position, just rotated). The airplane appears in the picture at its closest position to the camera, 150 meters distance and departs the picture at an angle of 60 degrees (as we have another right angle in the triangle, this time not in the center but at the right hand corner of the picture, we can use Pythagoras without any doubt), the airplane travels therefore 86.5 (173*sin(30)) meters until it hits the center of the picture. The distance of the runway center line at the left hand corner of the picture would appear at 300 meters (150/cos(60)), the total distance travelled from right to left hand corner of the picture would therefore be 260 meters (compare 86.5 meters from right corner to center point!). You can not apply any linear measurement in this view at all.
One can not simply say therefore, the airplane could be put 5 times into the picture from appearance to disappearance, 4 seconds, the airplane is 37 meters long and thus we see at a speed of 5*37/4 meters/second or 166 kph. The perspective of the picture and the angles of even a still camera (not to mention a panning one!) introduce gross errors to that approach.
Now, we aren't even able to determine the angles between the camera and the runway centerline with any degree of certainty in this scenario, but we would need to do so for every single frame of the video to be able to use those frames for the measurements. So the panning introduces a lot of additional difficulty into that determination.
All of this with an optimal camera, which a surveillance camera usually isn't (that aims to view a broader angle than would appear on the pic). You'll note, that the exact distance to the airplane is needed and the exact angles to be able to compute the actual distances covered. And we need to know the exact time stamps of the pictures to be able to really compute the time - that video doesn't provide times at all. We don't know, whether frames have been inserted or removed for a "smooth video", what time intervals were between the various pictures of the camera (maybe it was indeed 1 picture per second as it may look like).
There is a reason, why photogrammetry is a science!
Servus, Simon