Hi All,
quote:
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In nil wind if I was flying thru the sky(space) heading south at 100 knots in a 206 at 1400 kg my momentum would be M X V = 140,000 kt/kg
Add a 50 kt tailwind I'm now flying thru the sky at 150 knots in a 206 at 1400 kg my momentum is 210,000kt/kg
The practical application from the pilots point of view is that in both cases he is doing 100kias. From a physics point of view the momentum of the aircraft would increase turning downwind & decrease when turning upwind and that can only be caused in this case by accelleration. Which means unballanced forces.
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As CSC rightly pointed out it always depends on the reference frame, e.g it depends on where you set your zero energy datum. Overpitched, you set yours fixed to the earth and thereby you are right in your calculations and with increased tailwind your energy will increase - with respect to your datum - the earth.
To illustrate, if you where to hit a mountain (which is bound to your reference point - the earth) you are quite right in saying that with a 50kt tailwind your impact will be a lot harder than without because your energy with regards to the reference point was increased.
In all energy systems we can chose the datum as we wish as long as we stick to this particular datum all the way. This is very important and I think the reason why you are a bit confused.
Lets look at speed vectors for a moment and to make it easier only at the speed vector along the aircrafts fligthpath at this moment in time.
I'll take your above example. You are travelling due south (although this has no relevance) at 100kts with nil wind. You are executing a rate one turn for 180 degrees to end up due North. Your speed vector is therefore changing from 100kts facing south to 100 kts facing north.
This means that in the turn, to overcome the forward momentum of the aircraft which obviously wants to keep moving in that direction (see Newton), the turning force (which is the changed Thrust vector at the rotor hub) has to decelerate the aircraft from 100 kts to zero speed (which would be at 90 degrees e.g either east or west depending on which way you go) and back to 100kts facing the other way. Remember I am only looking at the speed vector along the North-South flightpath axis and not at the aircraft airspeed! The deceleration to zero and back to 100kts equals an overall speed and therefore energy change of 200kts.
Now let's look at the 50kts tailwind scenario. Same principle applies. You start a rate one turn and this time (with reference point earth) the speed is 150kts. At 90 degrees (east or west) your speed is now not zero but 50kts going south due to the wind (still only looking at speed along North South axis) so in the first 90 degrees the speed has changed from 150kts to 50kts (100kts difference). The next 90 degrees of the turn changes the speed vector from 50kts south to 50 kts north at the end of the turn facing due north. This again equals a speed change of 100kts. So overall the speed and also energy change has been 200kts.
This shows that for the aircraft the same amount of energy is required to turn 180 degrees regardles of the wind. No extra power necessary at all. The only difference looking from our reference point (the earth) will be the shape and endpoint of the turn as we obviously will end up a lot further downwind in the second scenario.
I have recently tried this going around from a rig here in the north sea with winds of 65 kts and 75kts airspeed without any loss of height. If what you are suggesting would be true, the speed would have changed from 75kts to 10kts going downwind - below TL and the power requirement for that would have been very much higher.
Hope this helps, a drawing would probably clear things up a lot quicker. Maybe someone could do a nice graph?
Woolf