Originally Posted by awblain #35
Re: 27
For an M-tonne aircraft landing at a groundspeed of V kts (~V/2 m/s), the kinetic energy
T ~ 125.M.[V-squared](in joules).
If V=143 kts and M=18, I reckon that T~47MJ.
47 or 33? (Your posts #35 #33) That is the question.
Originally Posted by awblain #33
Do it all in the frame of the deck/runway, since that's where you need to end up with zero kinetic energy, unless you want a trip to the hospital.
We don’t reckon absolute kinetic energy ( we are still flying with the hospital and solar system toward Vega at 20 km/sec and kinetic energy is not zero
) , kinetic energy changing in an other form of energy. So we have to evaluate ke at time 1 and time 2, (in any reference frame, and do the difference between ke1 and ke2.
With the Ship doing its 27 kts own wind, the Ship the arresting gear system and the traped F-4 have a GPS ref speed of 27 kts.
But the plane is stopped on the deck when the airspeed indicator shows still 27 kts should the wind be wind or Ship's speed
With a speed of zero the plane would be in the rounddown.Could one of our Aircraft Carrier Pilot confirm that thought (or deny, I am ready to accept that I should be wrong).
When the Ship does its own wind we have a balistic problem to land added to the classical relative wind problem, aswell as a parachutist landing on a truck for a movie or for fun.
And when the Ship is doing its own wind