Impossible Turn Practice

[BillTIZ]

We teach "the impossible turn" in gliders at 200ft. Granted we have on avg a 23/1 Ld ratio. Not the 7/1 of a Cessna with a windmilling prop.

The rules are all the same, get the nose down, get the turn started and keep it coordinated. Sink rate will increase in the turn due to decreased vertical lift vector.

Best glide plus 5 is a good number, what is your stall speed at 45 degrees of bank, use that plus 5 and keep it coordinated. Practice at altitude and see how much you lose in a 225 degree turn and then a 45 degree turn back to line up.

I had the power roll back to barely idle 2yrs ago. I had turned crosswind and just reached TPA when it happened, started a 180 towards the airport, call tower that I am returning, no power, evaluate my position and decided a base to final land opposite direction was best. Tower cleared me to land any runway. One departing behind me kept it low and cleared below my base easily as I transitioned across to the parallel runway. A non event.

 

[flyingcheesehead]

I also intend to practice this. If the "impossible turn " is Possible is practiced, it can become one of the tools available in an emergency. It is definitely something that needs practice. Dave

Practice at altitude first...

IIRC, AOPA did a study on this a few years ago - I think they were simulating a 172 - And even after significant practice, it really was impossible below about 700 AGL. Those who had not had a lot of practice on the maneuver very recently required significantly more altitude.

Of course, it depends on the airplane. But, it would seem that most airplanes and pilots can't do it at the 500 AGL number I've seen a couple times on this thread.

Also, remember that if you started at the beginning of the runway, if you had a reasonably significant headwind (15+ knots?) you'll overshoot if you do execute the turn successfully, and you'll touch down at a much higher groundspeed due to the tailwind - That's bad, it means you'll have a lot more energy to dissipate (more airframe damage, and likely more occupant injuries as well).

 

[mantakos]

... Also, remember that if you started at the beginning of the runway, if you had a reasonably significant headwind (15+ knots?) you'll overshoot if you do execute the turn successfully...

Headwind is generally helpful, as the hard part is making it back to the airport grounds. I don't think "overshooting" is a major concern, if I understand your meaning, it would take one heck of a headwind to overshoot the runway entirely, given the assistance of flaps and slipping to lose energy.

But one thing I'd add to the discussion of "possible or not" is that this is typically described as "did the wheels touch down on the runway", but runways are typically surrounded by a significant amount of space cleared of obstructions, a clearing that might not be easily found in the surrounding area, and "win" is defined more by "everybody survives" than as "my wheels didn't get dirty", so I don't know that reaching the runway is the sole determinant of success.
-harry

 

[TripleZ]

Headwind is generally helpful, as the hard part is making it back to the airport grounds. I don't think "overshooting" is a major concern, if I understand your meaning, it would take one heck of a headwind to overshoot the runway entirely, given the assistance of flaps and slipping to lose energy.

Except at some point in the turn, that head wind component transitions into a tailwind component and airspeed will drop 2x. To gain more airspeed, you then must sacrifice even more altitude to speed up.

I think this again, is why the impossible turn is impossible. The pilot would be attempting to make a rapid turn, increased loading, near stall, then loosing airspeed due to a change in wind component, with little to no altitude to sacrifice for speed (natural reaction is to pull up when you see the ground coming at you). Its asking for a stall.

I would expect a no-wind situation to be the best. Even at that, a very difficult maneuver to pull off.

 

[mantakos]

Except at some point in the turn, that head wind component transitions into a tailwind component...

Yes, thus effectively extending your glide, helping to carry you back to the airport.

... and airspeed will drop 2x...

Google the "downwind turn myth".
-harry

 

[tonycondon]

i've done a lot of "impossible" turns in gliders. usually the problem is not making it back to the runway, usually the problem is getting down and stopped by the end of the runway on the downwind landing (especially if you started the takeoff at the beginning of the runway).

This is with marginal climb rates compared to a typical airplane climb rate and about the same glide angle on the way back. I don't like "practicing" them in anything more than about 10 knots wind. One result of all this practice i've done is that in almost any situation where i have clear landing areas straight ahead i basically make that my emergency plan up to the point where i have altitude to do an abbreviated pattern and land into the wind.

 

[dmccormack]

Except at some point in the turn, that head wind component transitions into a tailwind component and airspeed will drop 2x. To gain more airspeed, you then must sacrifice even more altitude to speed up.

I think this again, is why the impossible turn is impossible. The pilot would be attempting to make a rapid turn, increased loading, near stall, then loosing airspeed due to a change in wind component, with little to no altitude to sacrifice for speed (natural reaction is to pull up when you see the ground coming at you). Its asking for a stall.

I would expect a no-wind situation to be the best. Even at that, a very difficult maneuver to pull off.

A headwind is preferable, as you are closer to the runway environment for the same altitude gain in no wind (vector it out and see for yourself).

I'm not sure what you mean by "airspeed will drop 2x" ..?

In a descending turn you decrease G loading.

 

[Everskyward]

Headwind is generally helpful, as the hard part is making it back to the airport grounds. I don't think "overshooting" is a major concern, if I understand your meaning, it would take one heck of a headwind to overshoot the runway entirely, given the assistance of flaps and slipping to lose energy.

I don't see how headwind would be that helpful, maybe in getting back to the airport grounds but not in landing. If you have a 10 knot headwind taking off you would have a 10 knot tailwind landing. That's a 20 knot difference in touchdown speed. Although I usually don't read through all the "impossible turn" threads I notice that I haven't heard anyone mention the fact that some other airplane may have taken the runway behind you.

 

[mantakos]

I don't see how headwind would be that helpful, maybe in getting back to the airport grounds but not in landing...

Getting back to the airport is the way in which it is helpful. It is a detriment to your landing distance.

In terms of survivability we're mostly interested in not running into something solid at flying speeds.

... Although I usually don't read through all the "impossible turn" threads I notice that I haven't heard anyone mention the fact that some other airplane may have taken the runway behind you.

That's certainly a concern, but this process will take 30 - 45 seconds or so, and you have some opportunity to make a call on the radio to give a heads up to following traffic.
-harry

 

[gprellwitz]

That's certainly a concern, but this process will take 30 - 45 seconds or so, and you have some opportunity to make a call on the radio to give a heads up to following traffic.
-harry

Aviate, Navigate, Communicate. I'd expect you would likely have your hands more than full just doing the first, so it's unlikely that the third will be an option.

I was once coming in to the field during primary training and there was a plane inbound NORDO doing a precautionary landing because his engine wasn't running quite right on takeoff. He wasn't on the radio until he was down. It was luck that I had already decided that I didn't like the look of my approach and had called a go-around.

 

[Palmpilot]

Also, remember that if you started at the beginning of the runway, if you had a reasonably significant headwind (15+ knots?) you'll overshoot if you do execute the turn successfully, and you'll touch down at a much higher groundspeed due to the tailwind - That's bad, it means you'll have a lot more energy to dissipate (more airframe damage, and likely more occupant injuries as well).

The touchdown point can be adjusted with flaps and/or slipping if necessary. The length of the runway needs to figure in there somewhere too.

As for the danger of a too-fast landing, I think one needs to compare that to the survivability of the obstacle environment off the departure end of the runway before making a decision. Is the latter an area dense with tall buildings, or flat open fields? My instructor said that rolling off the end of a runway is almost certainly better than crashing into a building or ditching.

 

[Palmpilot]

Except at some point in the turn, that head wind component transitions into a tailwind component and airspeed will drop 2x. To gain more airspeed, you then must sacrifice even more altitude to speed up.

Airplanes fly in accordance with their motion relative to the airmass they are in, and the motion of that airmass relative to the ground has no effect on airspeed whatsover. It ONLY affects groundspeed.

I have flown a lot of 180 degree turns, and I have NEVER seen a decrease in airspeed from it.

Try this experiment: Go flying on a day with winds aloft that are strong enough to make a difference. At cruise altitude, make a 360 degree turn while maintaining altitude and a constant bank angle, and see if you see any correlation between your airspeed and the direction of the wind.

 

[TripleZ]

Yes, thus effectively extending your glide, helping to carry you back to the airport.

Google the "downwind turn myth".
-harry

A headwind is preferable, as you are closer to the runway environment for the same altitude gain in no wind (vector it out and see for yourself).

I'm not sure what you mean by "airspeed will drop 2x" ..?

In a descending turn you decrease G loading.

"In a descending turn, g load is decreased." Agreed, as long as you match bank g loading for decent -g loading.

Googled "downwind turn myth", and it appears to be inconclusive.


If we take an example, a 180 degree turn, no power, NO wind, ground speed / airspeed of 65knots. Let's say for a 45 degree bank, the turn requires 300ft of altitude to complete the turn with an ending airspeed /groundspeed of 65knots.

Take the same turn with a 10knot headwind. Groundspeed is 55knots, airspeed 65kts. With the SAME ALTITUDE LOSS / Constant bank angle, you will end the turn with a ground speed of 55kts, and airspeed of 45kts. There is no additional energy put into the airframe by the wind, to increase ground speed. 50% of the turn is spent loosing speed, 50% of the turn gains speed, for a net neutral.

For a plane to maintain upwind airspeed to downwind airspeed, it must accelerate from 55kts to 75kts. That energy has to come from somewhere. Wind, is nothing but drag on an airframe, it does not "push" the plane. Without an engine, the only energy available to the pilot is the potential energy of altitude. You sacrifice altitude for speed.



Right / wrong??

 

[Palmpilot]

I don't think "overshooting" is a major concern, if I understand your meaning, it would take one heck of a headwind to overshoot the runway entirely, given the assistance of flaps and slipping to lose energy.

My instructor calculated that with a 30 knot headwind/tailwind component, using full flaps after the turn would cause a 172 or 182 to touch down about at the point of liftoff.

 

[Palmpilot]

"In a descending turn, g load is decreased." Agreed, as long as you match bank g loading for decent -g loading.

Googled "downwind turn myth", and it appears to be inconclusive.


If we take an example, a 180 degree turn, no power, NO wind, ground speed / airspeed of 65knots. Let's say for a 45 degree bank, the turn requires 300ft of altitude to complete the turn with an ending airspeed /groundspeed of 65knots.

Take the same turn with a 10knot headwind. Groundspeed is 55knots, airspeed 65kts. With the SAME ALTITUDE LOSS / Constant bank angle, you will end the turn with a ground speed of 55kts, and airspeed of 45kts. There is no additional energy put into the airframe by the wind, to increase ground speed. 50% of the turn is spent loosing speed, 50% of the turn gains speed, for a net neutral.

For a plane to maintain upwind airspeed to downwind airspeed, it must accelerate from 55kts to 75kts. That energy has to come from somewhere. Wind, is nothing but drag on an airframe, it does not "push" the plane. Without an engine, the only energy available to the pilot is the potential energy of altitude. You sacrifice altitude for speed.



Right / wrong??

Wrong.

Under the Theory of Special Relativity, which has been well-confirmed experimentally, the laws of physics are the same for all uniformly-moving reference frames. The analysis is a lot simpler if you use the airmass as your reference frame instead of the earth. It is then easy to see that the relative motion between the airmass and the earth is irrelevant. Only the motion of the airplane relative to the airmass has any effect on the airplane. It is also clear that in that reference frame, the kinetic energy of the plane remains the same regardless of direction, as long as a constant airspeed is maintained.

You can use the earth as your reference frame if you must, but it complicates the analysis, thereby introducing more opportunity for error. I would have to give it more thought to do the analysis that way, but I have faith that we are not going to disprove the Theory of Special Relativity after all these years.

 

[Capt. Geoffrey Thorpe]

"In a descending turn, g load is decreased." Agreed, as long as you match bank g loading for decent -g loading.

Googled "downwind turn myth", and it appears to be inconclusive.


If we take an example, a 180 degree turn, no power, NO wind, ground speed / airspeed of 65knots. Let's say for a 45 degree bank, the turn requires 300ft of altitude to complete the turn with an ending airspeed /groundspeed of 65knots.

Take the same turn with a 10knot headwind. Groundspeed is 55knots, airspeed 65kts. With the SAME ALTITUDE LOSS / Constant bank angle, you will end the turn with a ground speed of 55kts, and airspeed of 45kts. There is no additional energy put into the airframe by the wind, to increase ground speed. 50% of the turn is spent loosing speed, 50% of the turn gains speed, for a net neutral.

For a plane to maintain upwind airspeed to downwind airspeed, it must accelerate from 55kts to 75kts. That energy has to come from somewhere. Wind, is nothing but drag on an airframe, it does not "push" the plane. Without an engine, the only energy available to the pilot is the potential energy of altitude. You sacrifice altitude for speed.



Right / wrong??

wrong.

In the no wind case you are traveling at 65 knots in one direction (lets say north) at the begining and at the end you are traveling at 65 knots in the other direction (south) for a net north to south acceleration of 130 knots.

In the second case you go from 55 to 75 knots (WRT the ground) for the same 130 knot acceleration.

Velocity is a vector, not a scalar. It has direction.

So you really go form 65 north to 0 north to -65 north and at the same time from 0 east to 65 east to 0 east.

 

[flyingcheesehead]

If we take an example, a 180 degree turn, no power, NO wind, ground speed / airspeed of 65knots. Let's say for a 45 degree bank, the turn requires 300ft of altitude to complete the turn with an ending airspeed /groundspeed of 65knots.

Take the same turn with a 10knot headwind. Groundspeed is 55knots, airspeed 65kts. With the SAME ALTITUDE LOSS / Constant bank angle, you will end the turn with a ground speed of 55kts, and airspeed of 45kts. There is no additional energy put into the airframe by the wind, to increase ground speed. 50% of the turn is spent loosing speed, 50% of the turn gains speed, for a net neutral.

For a plane to maintain upwind airspeed to downwind airspeed, it must accelerate from 55kts to 75kts. That energy has to come from somewhere. Wind, is nothing but drag on an airframe, it does not "push" the plane. Without an engine, the only energy available to the pilot is the potential energy of altitude. You sacrifice altitude for speed.

Right / wrong??

Wrong.

Your airspeed in the turn will remain roughly the same. Your groundspeed will increase from 55 knots to 75 knots.

When you did turns around a point in your training, did you have to adjust power to maintain airspeed? No. Did you have to adjust your bank to correct for groundspeed? Yes.

 

[flyingcheesehead]

In a descending turn you decrease G loading.

Well, not exactly.

In a downward-accelerating turn, you decrease G-loading. At a constant rate of descent, the G-loading will be the same as in a level turn.

 

[denverpilot]

When you did turns around a point in your training, did you have to adjust power to maintain airspeed? No. Did you have to adjust your bank to correct for groundspeed? Yes.

When I did them too close and exceeded 45 degrees of bank on the downwind side, the drag I had induced meant ... yes, I needed to add power.

That's just kidding of course, I didn't do that...

But the steeper the bank the more drag and less vertical component of lift... so yes, you might need to put in power on the downwind side if you're doing turns about a point in a 100 knot wind.

But... that's 'cause you have the airplane standing on a wingtip and have little upward lift being generated and you're hauling back on the yoke to get around the turn and trying to pull the wings off, vertically!

GRIN!!!

For "normal values of turns-about-a-point"... no power change required.

 

[mantakos]

... For a plane to maintain upwind airspeed to downwind airspeed, it must accelerate from 55kts to 75kts...

Physics offers us our choice of non-accelerating frames of reference. If we choose the frame of reference that travels along with the wind, the picture is much simpler.

But if we choose the frame of reference anchored to the ground, then you are correct that we must accelerate from 55kts to 75kts, that we have more kinetic energy when going downwind, and that this energy must come from somewhere.

Wind, is nothing but drag on an airframe, it does not "push" the plane.

Tell that to Dorothy's house; wind picked it up and pushed it all the way to Oz.

The idea that the plane doesn't "feel" wind, that there is just the force of drag, and that drag is the same whether you're going upwind or downwind is perfectly true ... if we choose a frame of reference that moves along with the wind. If, instead, we choose a frame of reference anchored to the ground, as you have chosen, then the plane certainly does "feel" the force of the wind. It would take greater power to travel upwind at 100kts (relative to the ground) than it would take to travel downwind at 100kts (relative to the ground). Since our engine is putting out constant power, we travel slower upwind than downwind (relative to the ground).

If we're determined to fly a constant 100kts (groundspeed), we'd find that we'd need to put out a lot more engine power upwind than downwind, and we might be tempted to declare that the force of drag is much greater upwind. The reason this sounds weird is because it's weird for us to use a frame of reference anchored to the ground.

Without an engine, the only energy available to the pilot is the potential energy of altitude.

If I stand in a field on a very windy day, holding a balloon in my hand, and I release the balloon, it will accelerate quickly and fly away. It started out with no kinetic energy, but quickly picked up some. Where did that energy come from? It came from the wind.
-harry

 

[Capt. Geoffrey Thorpe]

Physics offers us our choice of non-accelerating frames of reference. If we choose the frame of reference that travels along with the wind, the picture is much simpler.

But if we choose the frame of reference anchored to the ground, then you are correct that we must accelerate from 55kts to 75kts, that we have more kinetic energy when going downwind, and that this energy must come from somewhere

You accelerate from 55 to negative 75. The "extra" kinetic energy (WRT ground) comes from the wind. The change in velocity (and change in kinetic energy) is the same in either case.

 

[Palmpilot]

I think it's pretty clear by this point that when you use the air mass as your reference frame, the turn causes no change in the kinetic energy of the airplane regardless of the wind, whereas if you use the earth as your reference frame in the presence of wind, there is a change in the kinetic energy.

So why the difference between the two analyses?

Some possibilities:

1. The airplane converts more potential energy into kinetic energy in the turn than it does when gliding straight ahead, but that's the same regardless of which reference frame you use, so that can't be the answer.

2. The drag force does work* on the airplane, where work is the force times the distance traveled in the direction of the force. The distance traveled during the turn is different depending on which reference frame you use, so this is a possibility.

3. The horizontal component of lift during the turn also does work on the airplane, and again, this is the force times the distance traveled in the direction of the force, and again, the distance traveled during the turn is different depending on which reference frame you use, so this is also a possibility.

So it looks to me like either 2 or 3, or a combination of them, are candidates for how to explain the difference in kinetic energy between the two reference frames. Writing equations for those is more work than I have time to do, however, so I guess I'll just have to hang my hat on the Principle of Relativity (which actually predates Einstein by nearly 300 years).

http://en.wikipedia.org/wiki/Principle_of_relativity

* Note that I'm using "work" in the sense it is used in physics:

http://en.wikipedia.org/wiki/Work_(physics)

 

[TripleZ]

I think it's pretty clear by this point that when you use the air mass as your reference frame, the turn causes no change in the kinetic energy of the airplane regardless of the wind, whereas if you use the earth as your reference frame in the presence of wind, there is a change in the kinetic energy.

So why the difference between the two analyses?

Some possibilities:

1. The airplane converts more potential energy into kinetic energy in the turn than it does when gliding straight ahead, but that's the same regardless of which reference frame you use, so that can't be the answer.
...............................

Im working on the equations...... I DO want to understand.

Using the fluid as the "Zero" point, a 180 degree constant rate turn is a perfect semi-circle. Using the ground as "zero", a constant rate turn looks like a hook for constant airspeed. And would look like a perfect semi-circle if you instead had constant ground speed.

For the problem using a turn in wind, you HAVE to use ground as reference, since the planes energy is relative to the ground. In gusty conditions, if the wind suddenly goes to 0, your airspeed drops. You don't immediately move with the fluid, due to inertia, so the zero reference can't be the fluid.

If I stand in a field on a very windy day, holding a balloon in my hand, and I release the balloon, it will accelerate quickly and fly away. It started out with no kinetic energy, but quickly picked up some. Where did that energy come from? It came from the wind.

But the wind relative to the plane is ALWAYS a headwind, airspeed. You can have more or less airspeed (more or less drag). But the wind will never generate positive work.


With a turning motor, propeller efficiencies, etc. I can see how in low wind conditions a turn around a point, you don't need to change throttle position. (Doesn't mean engine RPM, thrust, or altitude isn't changing slightly)

But for the glider, Im still not seeing how a turn downwind can be completed without additional altitude loss over calm conditions. It might only be 50 or 100ft, but you have to get that energy from somewhere.


To be continued..... gotta drive home.

 

[dmccormack]

Well, not exactly.

In a downward-accelerating turn, you decrease G-loading. At a constant rate of descent, the G-loading will be the same as in a level turn.

And what are you doing during the "impossible turn" practice?

I immediately push forward and turn -- simultaneously.

Guess what happens to the G-load?

 

[RyanShort1]

And what are you doing during the "impossible turn" practice?

I immediately push forward and turn -- simultaneously.

Guess what happens to the G-load?

Exactly...

ANY time your engine quits in a climb you NEED to push over initially to avoid slowing down too quickly, then you can establish best glide, or whatever airspeed you intend to hold on the way down. Same is true going straight ahead or turning. Ya gotta maintain airspeed. That is the #1 thing that looked to have gone wrong in the video of the biplane crashing posted earlier in the thread.

Ryan

 

[denverpilot]

What's missing from the discussion is inertia. The speed changes relative to the Earth, take time in the aircraft inside the air mass. Most of the time that'd be pretty "instantaneous" but in a huge relative wind change... (think microbursts and low-level wind-shear).

 

[TripleZ]

What's missing from the discussion is inertia. The speed changes relative to the Earth, take time in the aircraft inside the air mass. Most of the time that'd be pretty "instantaneous" but in a huge relative wind change... (think microbursts and low-level wind-shear).

Bingo!!

Ok.... I crunched through the math. Taking a Cessna 172, 65knot airspeed, 2300lb gross. Making a standard rate, 180 degree turn, assuming constant airspeed, guessing at drag coefficient and frontal area (drops out between the two scenarios anyway,)........................................

With no wind, the drop is from 500ft to 86ft.
WITH a 15 knot headwind, the drop is from 500ft to 51ft.

The ONLY difference is caused by the inertial effect to accelerate the airframe from 50knots to 80kts.

So turning upwind to downwind has a whopping 35ft effect on altitude if you're un-powered and maintaining airspeed.

 

[Capt. Geoffrey Thorpe]

I think it's pretty clear by this point that when you use the air mass as your reference frame, the turn causes no change in the kinetic energy of the airplane regardless of the wind,

Wrong. Velocity has direction. If you started out with X kinetic energy, you end up with -X after a 180 turn.

 

[Palmpilot]

For the problem using a turn in wind, you HAVE to use ground as reference, since the planes energy is relative to the ground.

Sorry, no. Kinetic energy (which is the kind you're talking about when you're deriving it from the airplane's speed) is relative to whatever inertial reference frame you use to do the analysis. The Principle of Relativity applies to airplanes. I would stake my life on it.

(Note: in this discussion, an "inertial" reference frame is one that is not accelerating or decelerating.)

In gusty conditions, if the wind suddenly goes to 0, your airspeed drops. You don't immediately move with the fluid, due to inertia, so the zero reference can't be the fluid.

In gusty conditions the air mass is no longer an inertial reference frame, but in that case, you can choose an inertial reference frame whose motion is the average of the air's motion, and an analysis using that reference frame will be valid.

Furthermore, an analysis based on a non-gusty day will be valid for the purposes of this discussion. All introducing wind gusts into the discussion accomplishes is to make the analysis unnecessarily complicated. It does NOT prove that the earth is the only valid reference frame.

In order for your analysis to be correct, the Principle of Relativity would have to be wrong. That means you are going up against physicists all the way from Galileo to Einstein.

But the wind relative to the plane is ALWAYS a headwind, airspeed. You can have more or less airspeed (more or less drag). But the wind will never generate positive work.

Then where does the kinetic energy of a sailboat come from?

That having been said, I suspect that the energy required to reverse the plane's direction comes out of the potential energy that is lost when the airplane descends, not from the wind.

 

[Palmpilot]

Wrong. Velocity has direction. If you started out with X kinetic energy, you end up with -X after a 180 turn.

Velocity has direction, but energy does not. Velocity is a vector, but energy is a scalar quantity, not a vector. There is no negative energy.

Look at the equation for kinetic energy:

e = mv^2

If the velocity is negative, the energy is still positive, because the square of a negative number is a positive number.

That having been said, it does take energy to reverse the direction of travel. We know this because the force required to change the plane's direction of travel does work on the airframe according to the distance traveled in the direction of the force. Since the only available source of energy is gravitational potential energy, the plane descends faster in the turn than when it is gliding straight ahead.

 

[mantakos]

... energy is a scalar quantity, not a vector...

A scalar quantity measured relative to your frame of reference. In one frame of reference an object might be standing still, in another frame of reference the same object might be in motion. So what's its kinetic energy? Depends on what frame of reference you choose.
-harry

 

[brcase]

What is missing from the kintec question is the assumption that the wind doesn't add any energy to the airframe. With a reference the wind/airmass it does not. With reference to the ground into the wind it is robbing energy from the airframe slowing it down from its maximum kinetic energy it could obtain with no wind. The moment you start your turn the wind starts accelerating the airframe to a higher ground speed. By the time you have turned 180 degrees you your ground speed will have doubled.

Think of an airplane doing 50kts in a 50kts wind, with ref to the ground it has Zero Kinetic energy. Once you turn it will quickly accerlerate to a 100kt Ground Speed. But the airspeed will be any different than if there was no wind.

I said it earlier but the big problem with the down wind landing is the illusion of speed and you will have a very strong tendancy, (Like the leans during instrument flying) especially if you coming it low to the runway, to pull back and subconciously slow down, this is how you will lose airspeed in the down windturn. If you don't realize you are doing it we will be reading about you in another NTSB Stall/Spin Report.

Brian
CFIIG/ASEL

 

[Palmpilot]

Think of an airplane doing 50kts in a 50kts wind, with ref to the ground it has Zero Kinetic energy. Once you turn it will quickly accerlerate to a 100kt Ground Speed. But the airspeed will be any different than if there was no wind.

I think you mean "the airspeed will not be any different than if there was no wind."

I once hovered a Cutlass at 11,000 feet over Puget Sound. I decided to pick an altitude with less of a headwind!

 

[mantakos]

... With no wind, the drop is from 500ft to 86ft.
WITH a 15 knot headwind, the drop is from 500ft to 51ft...

Are you suggesting that the plane will lose more altitude doing an "upwind to downwind" turn as compared to a "no wind" turn? This is impossible.
-harry

 

[skidoo]

Are you suggesting that the plane will lose more altitude doing an "upwind to downwind" turn as compared to a "no wind" turn? This is impossible.
-harry

Maybe he is figuring that with the wind he would be further down the runway which may have an elevation 35 ft higher. :wink2:

 

[TripleZ]

Are you suggesting that the plane will lose more altitude doing an "upwind to downwind" turn as compared to a "no wind" turn? This is impossible.
-harry

Maybe he is figuring that with the wind he would be further down the runway which may have an elevation 35 ft higher. :wink2:

No, the math says an upwind to downwind turn will require a larger drop in altitude in order to accelerate the airframe from 50kts to 80kts (15kt headwind).

The 35ft drop is the required loss in potential energy mass*gravity*altitude to account for the change in kinetic energy of the plane 1/2*mass*Velocity^2.

Velocity of the airframe changes from 50kts (upwind) to 80kts (downwind). That energy has to come from somewhere.


If you don't account for the kinetic energy of the plane, flying a headwind vs. no wind has NO effect on the turn. Once you do...... you realize that a slight drop in altitude is required.




The fact that the difference is so slight...... is why when you fly by the seat of your pants, you never notice the difference. And for a point turn, the altitude lost for the upwind to downwind turn, is regained for the downwind to upwind change.

This again assumes CONSTANT airspeed. As airspeed varies up and down, the effect of Drag has a much much higher effect on altitude.

 

[dmccormack]

No, the math says an upwind to downwind turn will require a larger drop in altitude in order to accelerate the airframe from 50kts to 80kts (15kt headwind).

The 35ft drop is the required loss in potential energy mass*gravity*altitude to account for the change in kinetic energy of the plane 1/2*mass*Velocity^2.

Velocity of the airframe changes from 50kts (upwind) to 80kts (downwind). That energy has to come from somewhere.


If you don't account for the kinetic energy of the plane, flying a headwind vs. no wind has NO effect on the turn. Once you do...... you realize that a slight drop in altitude is required.




The fact that the difference is so slight...... is why when you fly by the seat of your pants, you never notice the difference. And for a point turn, the altitude lost for the upwind to downwind turn, is regained for the downwind to upwind change.

This again assumes CONSTANT airspeed. As airspeed varies up and down, the effect of Drag has a much much higher effect on altitude.

I'm thinking Bob Hoover never figured out the math...

 

[mantakos]

No, the math says an upwind to downwind turn will require a larger drop in altitude in order to accelerate the airframe from 50kts to 80kts (15kt headwind)....

Ok, so corner the math and ask it the following questions.

Bob Hoover, while pouring a glass of iced tea, flies a 180 at constant airspeed and constant rate of descent and constant bank while in a 10kt (or whatever) wind.

Now, observer A uses a frame of reference that moves along with the wind. He sees the plane do a nice descending semi-circle. He sees the plane descending at the same rate as Bob does, he sees the maneuver taking "t" seconds, he multiplies the two to determine the amount of altitude lost, that all matches up with how much lower the plane appears to him to be, all of which matches what Bob sees on his altimeter.

Bob lands, has another glass of iced tea, visits the men's room and waits for the wind to die down. Now, in zero wind, Bob does the same 180. He flies it identically, at the same airspeed, the same rate of descent, the same bank, etc. Observer A is now using a ground-based frame of reference. He watches Bob perform the maneuver a second time, he sees the exact same descending semi-circle, the same rate of descent, he sees that the maneuver takes exactly the same amount of time as it did the first time, he multiplies the two together to calculate how much altitude was lost and comes up with the same answer he calculated after the first maneuver. He confirms that this calculation matches how much lower the plane appears to him in his frame of reference.

You math says that Bob's plane lost more altitude on the first maneuver than the second. But the observer noted that both maneuvers took the same amount of time and were flown at the same rate of descent and thus lost the same altitude. Your math says this is impossible. So how does math explain this?

Math would seem to have to insist that it is not possible for the two maneuvers to be flown at both the same rate of descent and take the same time. But to the plane, the aerodynamics of both maneuvers are identical, so why would one take longer than the other, or result in a higher rate of descent?
-harry

 

[TripleZ]

Ok, so corner the math and ask it the following questions.

Bob Hoover, while pouring a glass of iced tea, flies a 180 at constant airspeed and constant rate of descent and constant bank while in a 10kt (or whatever) wind.

Now, observer A uses a frame of reference that moves along with the wind. He sees the plane do a nice descending semi-circle. He sees the plane descending at the same rate as Bob does, he sees the maneuver taking "t" seconds, he multiplies the two to determine the amount of altitude lost, that all matches up with how much lower the plane appears to him to be, all of which matches what Bob sees on his altimeter.

Bob lands, has another glass of iced tea, visits the men's room and waits for the wind to die down. Now, in zero wind, Bob does the same 180. He flies it identically, at the same airspeed, the same rate of descent, the same bank, etc. Observer A is now using a ground-based frame of reference. He watches Bob perform the maneuver a second time, he sees the exact same descending semi-circle, the same rate of descent, he sees that the maneuver takes exactly the same amount of time as it did the first time, he multiplies the two together to calculate how much altitude was lost and comes up with the same answer he calculated after the first maneuver. He confirms that this calculation matches how much lower the plane appears to him in his frame of reference.

You math says that Bob's plane lost more altitude on the first maneuver than the second. But the observer noted that both maneuvers took the same amount of time and were flown at the same rate of descent and thus lost the same altitude. Your math says this is impossible. So how does math explain this?

Math would seem to have to insist that it is not possible for the two maneuvers to be flown at both the same rate of descent and take the same time. But to the plane, the aerodynamics of both maneuvers are identical, so why would one take longer than the other, or result in a higher rate of descent?
-harry

Because my math is not a fictional character. It would be nice if someone observed this. But in reality, even in a strong headwind the difference in altitude is minimal to the entire change. So the casual observer, seeing a 7% difference in altitude (WITH NO MEASURING EQUIPMENT), won't ever "see" the difference.

On top of this, how often do pilots have the opportunity to do two perfectly identical turns with no power. Both with wind and without wind?

If you use the fluid as the point of reference, you are ignoring inertia of the plane. Remember, in all of these examples we have been using pure straight line wind speeds. In reality, the wind gusts and changes speed.

In order to make up the airspeed from a drop in wind speed....... you drop the nose (loose altitude) or add power. You add energy to accelerate the mass. It can come from either altitude or the engine.

The plane does not instantaneously change speed with the wind. In frames of the turn, you accelerate the mass from 50kts to 80kts. In terms of a wind gust, you may only need to accelerate 15kts (small drop in altitude).

 

[mantakos]

... If you use the fluid as the point of reference, you are ignoring inertia of the plane...

Please explain what this means in mathematical terms. Are you saying that one choice of frame of reference is more correct than another for the purposes of this problem?

Can you address the questions I raised in my previous posting? You're saying the two maneuvers result in different altitude loss. Do they take a different amount of time to complete?
-harry