Originally Posted by
Dutch M
Let me change the challenge a little:
- Land your glider with a airspeed of 30.25 m/s and a headwind of 30 m/s (just enough to move a little bit related to the ground).
- The presumed TAS based Ekin would be 0.5xmx915 =458 m J
- The moment the glider does touch down, everything becomes ground speed based and the ground speed based Ekin would be 0.5xmx0.25x0.25 = 0.03 m J.
Where did the Ekin of (458 - 0.03) m J go ?????
It got nowhere. The only energy that got converted into heat was the 1/2*m*(0,25m/s)^2 during braking on the ground.
With regard to the air the remaining 1/2*m*(30m/s)^2 is still in the glider. Being fixed on the ground it is effectively being towed through the air at 30m/s by the ground with regard to the air. See my example of the microlight.
If the wind breezes up by 1m/s to 31 m/s it will lift off again. Or push it forward at 1m/s and it will take off.
If the air is still you will have to push at 31m/s. Big difference for you, no difference for the glider for its ability to lift off and fly. The difference for the glider is how it will move in relation to the 'other' reference system, i.e. the ground.
That's why aircraft prefer to take off and land against the wind. Mother Earth gives a certain amount of kinetic energy to fly to the Aircraft for free. For the same Aircraft you need less energy (thrust*distance) to leave the ground. And upon landing you have to dissipate less energy in your brakes. Still in between you were equally able to fly as if you took off and landed with the wind.
The trick is that you need to understand that for an Aircraft once it has left the ground, the movement of the ground below is totally irrelevant for the ability to stay in the air. And it stays irrelevant until it contacts the ground again. In between only its relative movement through the air is relevant.