Original BMatthews post in black

<span style="color: rgb(255,0,0)">Flythesky Comments in red.

</span><span style="color: #339966">Bmatthews comments in green
__________________________________________________ __
</span>Yep, the figure you gave are nonsense because you're mixing air, wind and ground speeds up in a salad of confusion. None of your numbers add up to anything that makes sense.

You said-"There is a 20kt wind from left to right and I start from the left at 20kts "

That implies that you are approching the square loop with a tailwind of 20 kts plus a forward airspeed of 20 knots because you're flying downwind for a total ground speed of 40 kts.

You then said- "(I) pull straight up directly iin front of me. I maintain a straight line and slow to 20 kts at the top where I turn into the wind"

Well, since you were already doing 20 kts AIRSPEED when you were flying downwind to enter into the square loop you can't very well slow down in the vetical line to 20 kts. <span style="color: rgb(255,0,0)">I was going 40 kts ground speed, did I somehow lose all of this speed when I pulled up?</span> <span style="color: rgb(255,0,0)">The air (wind) is staionary in respect to the plane now.<span style="color: #339966">Like we've been saying you can't use GROUNDspeed for any of this. T</span><span style="color: #339966">he model only sees and works with AIRspeed. Inthe case of your scenario itcarriesits20 knots of airspeed around the corner less something for drag in the turn. The momentum from the tailwind along with the windhitting the top of the model pushes it downwind at20 knts while the model is climbing. That part of the 40 knots stays horizontal and continues toblowthe model downwind at a constant 20 knots. The actual ground speed of the model can be shown from the vector addition of the air and wind speeds for any point around the loop. You picked an extreme scenario with the wind and flying speed being the same. No one with half a shred of concern for their model would normally flyunder such nasty conditions so you very likely havenotseen how such a wind can affect the vertical lineof a model in flight. But in the case of this scenario if you turn the model so the fuselage isverticaland fly with a 20 kt upward speed in a 20 knot wind the model'sflight path will actuallybe a 45 degree angle upwards and inclinedin the downwind direction. It is no different from theskew of a ground track when flying horizontal in the wind. If you flew your 20 kt model cross wind then it would appear to be crabbing throughthe air at a 45degree yaw angle from the ground when you pass overhead. The crab angle in the square loop just happens tobe upward in this scenario.To actuallytravel in avertical trackas seen from the ground you would have to pitch the modelpast verticalto a 135 degree angle so it crabbed upwind in a true vertical line as seen from the ground but the fuselage of the model would be at a 45degree angle pointed backinto the wind. I took more time to dwell on this first point because it's the root of all the misunderstanding in the rest of the scenario. </span></span> This is the start of the confusion. But forget that for the moment. So the airplane flew along at 20kt airspeed and pulled vertical and maintained a 20kt airspeed up the line (perhaps thanks to adding power?). <span style="color: rgb(255,0,0)">Throttle management is to understood but not defined.<span style="color: #339966">Fair enough.</span></span>That's fine. As it did so the wind hitting the top of the model along with it's momentum from the 40kt ground speed kept it drifting belly first downwind at 20 kts all while the model was heading up at 20kts. <span style="color: rgb(255,0,0)">I do not understand why you would think that this is happening. By definition a vertical line has no ground speed.<span style="color: #339966">We need to consider who's frame of reference isseeing the vertical line. There's no "definition" about it. Forthe model flying along inthe moving river of air at it's happy 20 kts it turns through a 90 degree angle and proceeds upward just fine from ITSframe of reference. Butit is doing this in a moving body of air. You on the ground wouldsee the model crabbing upward and downwind on a45 degree upward track as described above. Pattern flyers in competition deal with this issue all the time in winds. They have to use rudder and elevator to cant the model into the wind so that theupward track LOOKS verticalto the judges.But they deal with up to maybe 15mph winds while flying a model that is moving along atmaybe more like 60 to 70 mph. So the amount they have to cant thefuselage over into the wind to produce a vertical track isn't anything like what we are looking at in your scenario.</span></span>

Now you turn the next corner to inverted to fly back into the wind where you said- "...at the top where I turn into the wind, lose 5 kts and continue at 15 kts plus the added wind speed of 20kt for an air speed of 35kts."

So why is this corner costing you 5 kts where the other one didn't? <span style="color: rgb(255,0,0)">The speed loss was not specified but you can see by my previous statement that the plane is slowing down. <span style="color: #339966">Then to be consistent it should be described as slowing down in all the corners since that is the more realistic scenario. </span></span>Let's skip any drag related speed losses in the corners for now even though they certainly occur. So the model flips inverted from the 20kt upline to flying at 20 kts of AIRSPEED into the 20kt headwind that is still coming from the left as per your original statement where both the model AND the wind are coming from the left at 20kts each. <span style="color: rgb(255,0,0)">The plane is traveling up at 20kts when it turned inverted and lost 5kts. This speed is in relation to the ground now<span style="color: #339966">- no itisn't, the model does notknow there is any ground down there other than the effect of gravity on it. You can only consider airspeed since that is all that the model can feel Andif the model entered the cornerwith 20 kts it'll come out of the turn with 20 kts or 15 if we are allowing for somespeed loss in the corners-</span>and is flying into a 20kt wind. The speed of the air over the plane is now 35kts.<span style="color: #339966">- Again, no it isn't. The modelwas climbing at 20 kts,but was also being blowndownwind at 20 knotsat the same timeand the resulting upward track ends up being a45 degree inclined path.When it turns from vertical to inverted and ifit loses 5 kts fromturning the corner it isnow flying at 15 kts of airspeed. And since this is occuring in a 20 kt wind it would actually appear to the pilot on the ground to be going backwards at 5 kts.But if the model isn'tstalling it'll continue tofly up there in the air just fine.Thewind cannot add actual flying speed to the model and raise it to the 35kts you suggest. That just won't happen. Given that we agree that throttle control will be used to hold the model at 20 kts and that we lose 5 kts in the turns the model's AIRspeed will never drop lower than 15 kts or rise to higher than 20 kts. It can't since the wind speed cannot suddenly add to the model's speed at any time.. </span></span>So you see where our and your confusion is coming from?<span style="color: rgb(255,0,0)"> I am not confused by my scenario. </span>So anyway, our model has just rotated to inverted and heading back the way it entered and facing the 20 kt wind from that direction. So it sits in the air upside down at zero ground speed apparently defying gravity but is actually very happy because it is still plowing through the passing air at its happy speed of 20 knts.

<span style="color: rgb(255,0,0)">You are saying that the upward speed has somehow disappered? Underlying this discussion you seem to be saying that the momentum of a plane continues to move a plane in it's original direction after turning in a new direction. If you dive to the ground and level out do you continue to descend toward the ground at the original diving speed?</span> <span style="color: #339966">The model's FLYINGspeed stays with it as it goes around the corners. But we're dealing with the model doing this in a fast moving river of air with us standing on the stationary bank. You chose to use a square loop as your example.But let's try a slightly different scenario. A radio control glider is winched up and moves upwind from the release point. It's a blustery day with a wind of 15 mph and the model is an old floater that flies best at 15 mph. The pilot feels like he found a thermal so he turns and then sets the trims to hold the model in asteadycirclethat stays in the thermal. The model drifts back overhead. What will the flight path look like from the ground? It won't be a circle. It'll be look like a series of half circle C shapes joined at the points. The points being where the model is pointed directly into the wind and sitting up there with zero ground speed. As it turns off the wind it'll crab to the side and downwind.

Now let's do a square turn overhead with this same glider to bring the horizontal example into line with your looping example. The model comes off the winch and sits there pointed into the wind because it's flying at 15 mph airspeed into a 15 mph wind. The pilot turns the model so the fuselage is pointed 90 degrees to the right. But due to the wind the ground track seen by the pilot from below for this sideways straight leg will be coming back towards him and to the right at a 45 degree line. So far so good? It's simple ground track calculation for wind. So then he turns another 90 degrees to fly dead downwind at the same 15mph of AIRSPEED. But from the ground we see what appears to be 30 kts due to the wind passing by and carrying the model with it. Note that the model doesn't care a bit about this. It is still only seeing 15 mph of air passing by. It is us on the ground that are lead astray due to the model flying in the body of air that is also passing by. Now he again turns 90 to the right from downwind so he's actually headed to the left. Again the model appears to be flying crabbed at a 45 degree yaw angle due to our perspective . And finally we turn back into the wind where the model again comes to a halt as seen from the ground as it is again flying in a 15 mph headwind. The final shape of our "square" as seen from the ground ends up being a trapezoid with two angled sides, a closed short side and an open long side since the model was not able to get any headway back upwind.

Simlarly your 20 kt square loop ASSEENFROMTHEGROUNDwhen started from a path that included a 20 kt tailwind but where the pilot only rotates the model using 90 degree turns will be a fast low line to a 45 degree downwind climb to a stationary inverted hang at the top then a 45 degree down and downwind line to the recovery that continues downwind. The final shape would be a path that looks like ____/\____ The inverted part taking place at the peak of the /\ part. Because of the wind if you want to make the line truly vertical your first rotation would have to be 135 degrees so the model climbs pointed into the wind at a 45 degree angle. At the top you rotate 45 degrees to inverted and just sit therebecause of the wind and flying speed being the same but pointedagainst each other.Andthen to come back down you would only rotate byanother45 so it crabs down on a vertical track as seen from the ground but the fuselage is at a 45 pointed into the wind. And finally to return to level you rotate it through another 135 degrees. All of this would make your square loop look like this ______|______ because there's no actual ground speed while at the top of the loop.

</span>So from here you pull the last corner for the downline where you said- "I proceed to the down ward turn and dive straight down while chopping the throttle to a speed of 20kts. At the bottom of the leg I pull level, lose 0kts because gravity helped in keeping the speed up in the turn and proceed at 20kts minus the wind speed of 20kts coming from behind for a net air speed of 0kts and hope I can accelerate to a flying air speed before I hit the ground."

Nope, not the case at all. The model does come down at 20 kts like you say but it also is now being pushed by the drift so by the time you get to the bottom line where you pull back to level the model is being pushed canopy first "upwards" at 20 kts. <span style="color: rgb(255,0,0)">Again, Ican't understand why you think this. You are saying that a plane can't fly in a vertical line in a wind. Ithink it can because Ido it when Ifly.<span style="color: #339966">In calmer winds where the model's airspeed is much faster than the wind theangle of cant required to produce a track that appears vertical from the ground isn't much. Same as the drift isn't much and it can be missed by a sport flyer. But get someone in your club that has flown or judged pattern competition events and ask them to watch your verticals for signs oftrack pushing due tothe wind. They'll have the experience to seeit or to see that you've got your model canted intothe wind to counter the effect. Or try flying your model on a day where you'd ratherjust stay on the ground due to the high wind conditions. Go off to the side rather than directly up or downwindso you'll see the wind's effect.Then try toturn the model through a 90 degreecorner and go verticalas if you're doing a reallyaccurate wingoverand see the effect. The downwind drift will become very apparent. Try it again and see how far over you have to yaw or pitch the model to actually achieve a ground percieved truly vertical track. I think the angle will shock you.. Even onwindy but tolerabledays if you do this off to the sideof the wind where the effect is morenoticable to you instead of directly upwind or downwindyou'llsee how thewind pushes the actual flight trackl over at an angle while the model is pointed directly up.</span></span>Also it maintains it's speed through the corners because we're ignoring the drag losses in the turns so it pulls out with both 20 kts of airspeed PLUS the 20kts of wind speed from the tail and is immediately flying at 40 kts with out any sign of sag towards the ground.