Originally Posted by
DingerX
I was merely giving the methodology for "how" those numbers were arrived at, not making any statement about their veracity; indeed I did note the problems with moving cameras and angular speed.
Alright, taken note of, and we are in agreement here.
I do not find your initial arguments conclusive, namely that if it were going at 100 kts at that point, it would not have overrun.
Not quite ;-)
If the airplane had slowed down to 100 knots already, from an approximate 148 knots or so, it would have exercised quite some braking already. If that braking action continued for some time, it would not have overrun. In fact, even on an ungrooved wet runway a deceleration of 1.5 meters/second/second (3 knots/s) is well normal, and I used that deceleration rate.
First, you are factually wrong. The tape is time-stamped to the thousandth of a second. Second, the camera's perspective is good enough that, even with the panning, you can overlay two images without noticeable problems from distortion, which I've done. Third, you are intuitively correct: at the point it comes on the screen, it is going notably faster than 100 kts.
Yep, absolutely - we are in agreement of the speed being higher than 100 knots.
And you are right, that there are timestamps overlaid by the copyright notice of the TV station that obviously broadcast it. I hadn't seen those until your remarks.
Can you recheck your time stamps, especially as you write 18.928 and then compute 0.593 seconds to 20.421, please?
The timestamps are hard to decipher, but the airplane is not visible at ...19.421, is fully visible at ...20.028, is last visible at ...22.031 and is entirely out of the picture at ... 22.431, so that makes it maximum 3 seconds visible.
Now add the angle between runway and camera to that scenario ...
If I however use 130 knots as a speed at that point 650 meters past the runway threshold with 1000 meters to go, a deceleration rate of 1.5 m/s/s would be insufficient to stop, stopping distance would be 1500 meters (departing the runway end at a speed of 38 meters/second=74 knots). A deceleration rate of 2.2 meters/second/second (~4.4 knots/s) would be necessary to stop at the runway end (which may not be available on a wet runway).
Assuming conservatively that's forty meters, that would come out 243 kph or 131 kts. I'm sure greater precision (and perhaps an even higher ground speed) can be had.
Photogrammetry can certainly clear that up. That was my main point.
So: 1) camera motion can be compensated for
No doubt about that.
2) The image gives us precise time measurements, and 3) you'd still need to calculate angular effects, but back-of-the-envelope shows it's at least 130 kts.
Agreed.
Servus, Simon