henra,
I should clarify that my post #546 was only indicating
an initial symbolic notation of the Kinetic energy as a function of speed, with the speeds, which are easy reference points to the BEA text, which mentions the two IAS values 275 and 211. That was a correct notation, and a correct equation. Further calculations based on that equation are correct, as long as as the expansion of the functions include the correct speeds.
Originally Posted by
henra Post #580
Originally Posted by
airtren Post #546
dEk = Ek (275knots) - Ek(215knots) = dEp (2500ft)
Not wanting to nitpick, just for the sake of correctness, it has to be noted that the 275kts and 215kts were IAS.
For any kind of energy calculation you have to take TAS.
But what attracts my attention more in your post, is the TAS values.
I am not sure what considerations you've made in calculating, or getting the TAS. Can you elaborate?
One of the AF 447 BEA reports indicates a TAS of 461 knots at FL350, while yours is 495 knots.
Even further, and more importantly, you're indicating a calculated "delta h" based on the two TAS of 495 and 390, of 4200 ft, and you're reasoning that the larger "height" value (of 4200ft versus 2500ft, a difference of 1700ft) is
explicable by the increased drag during the ascent/climb.
Originally Posted by
henra Post #580
In this case that would be approx. 495kts and 390kts, repectively.
Edit:
This corresponds to ~4200ft but is still very well explicable by the increased drag of the maneuvers. So I don't contradict the statement that the kinetic energy figures are plausible.
You expend further on that in your reply to JD_EE.
JD_EE is pointing out an element that I interpret as being in line with my further explanation bellow. He is also up to something, further in the second 1/2 of his post,
but without elaborating....
Originally Posted by
henra Post #612
Originally Posted by
jd_ee, Post #601
henra, grity mentioned the aircraft's thrust. That's a stable system. So the real answer is probably between the two calculations you give. I'd suspect it is closer to the square of the delta velocity. (In fact I did a long ways back.)
In fact if you want to be exact, grity is right. The equilibrium of drag and thrust is only valid for one speed. So when the AC slows down in the climb, the 1g drag decreases. Therefore the thrust of the engines will add some energy to the equation.
However, this Delta of energy is negligable compared to the Deltas in energy state we are talking here. The effects of increased drag by maneuvering and changing bank angles will by far exceed the Excess thrust due to the speed decrease.
It's right that the Thrust and Drag are valid for one speed.
But here is further:
Based on the BEA text, we know 275 knots (FL350), and 211 knots (FL375), were speeds that resulted from Real Time measurements (and/or calculations based on Real Time measurements).
Thus they implicitly include the Real Time AF 447 Thrust and Drag, in Real Time conditions, at FL350, FL375, and the climb/ascent in between. And so should be the TAS.
Also based on the BEA text, we know the delta height of 2500ft, which is also a Real Time measured element, and thus it includes the Thrust and Drag that were present in getting to that height.
Pointing out, or explaining that a calculated approx 4200 ft height's discrepancy of 1700ft is due to Drag, or Thrust, means including Drag or Thrust a second time in the equation. That does not seem to be correct.
So, I think there is a need for a different explanation....
....
The equation bellow, which is a next step from the one I've posted, is correct.
This is to say that the earlier posted warning that
v1^2 - v2^2 is not equal to (v1-v2)^2
was absolutely correct.
Originally Posted by henra Post #612
Between the two calculations I compared there is no doubt about which one is the physically correct one:
it is 1/2(m) v1^2 (Kinetic Energy before the climb) minus 1/2 (m) v2^2 (Kinetic Energy after the climb).