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.
Added: That's 47MJ with no headwind, and thus a deckspeed of 143kt. If the ship steams at 27kt, then the deckspeed is V=116 kts, and then T~31MJ.
The ratio is (116/143)-squared ~ 0.66.
If there's a small descent angle theta, the groundspeed is less than the total speed, and so you could perhaps justify dividing V by a factor of cos(theta).
However, for 3 degrees, cos(theta) is only about 1-0.5x(1/20-squared) or ~0.9986, so it has the same effect as adding 30kg to the landing weight, or increasing the landing speed by 0.1kt at 143kt.
If you treat the wind and airspeed as vectors, then any three-dimensional relationship between them is accounted for automatically.
Last edited by awblain; 4th Jun 2013 at 02:05.