From a scientific point of view it doesn't matter what reference system you use, you should always be able to satisfy the laws of physics and get the same answer.
The simplest and most useful reference system in aviation is that parcel of air that an aircraft flies in. You could also use a point on the ground or even infinity as the origin of your reference system to analyse aircraft aerodynamics, but the maths is more complicated.
Take for example the police radar trap. The radar measures the velocity at which you are approaching the gun ie gives you a direct measurement of your speed. However, if you could measure your velocity relative to the centre of the earth and compare it with the velocity of the policeman relative to the same point you will get the same result - a £60 fine and three points!
But Overpitched has decided to use a point on the earth for his (can I assume that you are male?) reference point. Nothing wrong with that, except that you will get different answers because you are looking at momentums and energies associated with the aircraft and ground interacting - something we all try to avoid. I think what is puzzling Overpitched is that there is a change in the magnitude of aircraft momentum between up and downwind without a force to generate that change. That got me puzzled too, but I think I have the answer:
Lets assume an aircraft speed of 100kts, a wind speed of 30 knots and an aircraft weight of 1000kg. The reference point is an observer on the ground.
If the aircraft is flying directly into the wind, the ground speed is 70kts, giving momentum RELATIVE TO the Earth of 70,000kt.kg. In the same way down wind the momentum is 130,000kt.kg. But why the difference - how has the momentum changed by 60,000kt.kg by turning down wind without any extra forces acting on the system? We have to remember that we are using the earth as a reference system so when we apply aerodynamic theory to the aircraft we have to do it relative to the ground.
If we maintain a constant total rotor thrust vector by maintaining power and disk attitude, the only other horizontal force acting on a helicopter is air resistance, which is depends on velocity of the air flowing over and around the airframe. When using the air as a reference system, TAS is used to calculate skin friction and form drag. But we are using the earth as a reference system so we have got to account for the motion of the aircraft reative to the air and the air relative to the ground.
Into wind, the skin friction and form drag (relative to the earth) are calculated from the speed of the aircraft through the air PLUS the speed of the air over the ground (wind). In this case 100 + 30 = 130 kts.
Down wind, the skin friction and form drag (relative to the earth) are calculated from the speed of the aircraft through the air MINUS the speed of the air over the ground (wind). In this case 100 - 30 = 70 kts.
Hence the skin friction RELATIVE TO THE EARTH changes and that accounts for the change in the magnitude of momentum when using the ground as a reference system.
It is interesting in the above example that regardless of the reference system you use, the change in momentum for a 180 degree turn is the same - 200,000kt.kg. That's because in the turn the momentum is changed by the centripetal force acting towards the centre of the turn. This centripetal force is generated by the horizontal component of total rotor thrust and is therefore independent of wind or ground.
It makes sense to me, but I am still going to duck for incoming!
