Originally Posted by awblain
The energy to lose in the wheel or cable braking is 0.5 x mass x (airspeed - headwind)-squared. This assumes the headwind is in the same direction as the airspeed vector. If not, you need to replace (airspeed-headwind) with the modulus of the vector sum of the airspeed, and the windspeed measured in the frame of the deck/runway
So the result is 33 Megajoules from my post 27.
I changed it, thinking that energy is not a vector but a number, effectively I disliked my decomposition in horizontal and vertical. But assuming headwind is parallel to airspeed vector looks like that too : airspeed is descending the slope, headwind is horizontal !
I agree with your argument headwind/downwind in my post 32...??
Method seems still unclear with some contradictions..
Am I right with the slope?