Minimum Radius Turn back.

[rookiepilot]

My question would apply to an EFATO situation, (with some altitude) or especially attempting an emergency minimum radius turn back within a narrow valley / box canyon.

As we all know stalling speed increases dramatically past a certain point with bank angle in a level turn —- limiting the radius of the turn.

What about a descending turn? How much difference would it make in reducing the banked stall speed through unloading the wing, in that theoretical canyon turn back? Any studies been done?

I suppose the optimum for a minimum radius turn would be a wingover maneuver.

[‘Bob’]

Immelmann if you have excess energy.

Split-S if you have excess altitude.

Hammerheads and wingovers are just asking for trouble.

All of these are asking for trouble.

What were we talking about?

Oh… right… A far more likely scenario is both pilots getting food poisoning and you’re the only pilot on board who can fly the plane.

[photofly]

What about a descending turn? How much difference would it make in reducing the banked stall speed through unloading the wing, in that theoretical canyon turn back?

The wing is only unloaded in the brief instant that you accelerate downwards from level flight into a descent. In a steady descent, there is no unloading.

I suppose the optimum for a minimum radius turn would be a wingover maneuver.

If you have a lot of excess airspeed there are lots of things you can do. Excess airspeed is one manifestation of excess energy, excess altitude would be another, and you can switch between them. An EFATO with 300 knots on the clock will enable you to zoom climb about 2500' while you slow to 100 knots.

I don't think Red Bull air racers have to worry about the turnback in the event of an engine failure during a race, as their high airspeed can be quickly exchanged for altitude from which to conduct a power-off approach.

However in an EFATO in a piston single you don't have a lot of excess energy; there's very little airspeed to give up in an attempt to wingover or anything similar.

[digits_]

What about a descending turn? How much difference would it make in reducing the banked stall speed through unloading the wing, in that theoretical canyon turn back? Any studies been done?

What study do you need? At 0g you can't stall. So unloading the wing is critical if you want to make very steep turns.

I suppose the optimum for a minimum radius turn would be a wingover maneuver.

Nose down while you roll 180 degrees through a split S is probably the thightest trun radius you can get if you have a lot of altitude. Not really what one would recommend in real life.

Realistically I'd unload the wing, nose down, with 60 degrees of bank in a descending turn and then ease out of the dive. That would probably end up looking close to a wing over style maneuver. You'll need a lot of altitude though. The steeper you want to roll or pull, the more altitude required.

[photofly]

What study do you need? At 0g you can't stall. So unloading the wing is critical if you want to make very steep turns.

This is true. However at 0g, the aircraft will of course descend under the full influence of gravity. From 500agl it would take slight less than six seconds to impact the ground at zero g, with a rate of decent at impact of nearly 11,000 feet per minute, which somewhat negates the purpose of a minimum radius turn.

If you decide to pull (push?) a zero-g turn for all of 2 seconds, you'll lose 64 feet in the descent, but your rate of descent after those two seconds will be 4000 feet per minute. You will then have to pull significantly more than 1g to reduce your rate of descent to something like a 600fpm glide for the rest of the approach, and you will lose a lot of energy (height and/or airspeed) recovering from that dive that you wouldn't have had to lose if you'd made a smooth descent instead of trying aerobatics at 500agl.

3 seconds of zero-g turn begun from level flight and you've lost 160', and your rate of descent is now nearly 6000' fpm. Assuming your EFATO was at 500', With a zero radius turn you have used one third of your height, you still haven't made any progress towards the place you wanted to land (which was the purpose of turning around) and you are now 340 agl with your rate of descent six times too large to register on a typical VSI. Good luck.

(Here's Bob Hoover doing something like what is being suggested here, from about the 09:30 mark:
https://www.youtube.com/watch?v=tnl3JZwnUZA)

[cargocowboy]

You may be interested to read "Contact Flying" by Jim Dulin. He discusses this maneuver at length, he calls it the energy management turn. Not necessarily for EFATO; the technique can be applied in various scenarios. There are other useful techniques in the book for those of us flying low powered aircraft.

[rookiepilot]

I think I’m more interested in, with a different technique, would an otherwise fatal turn up a blind box canyon…..could a tighter minimum radius be successfully accomplished……by giving up 1-2000 feet of altitude. So a powered scenario…..as opposed to the EFATO turnback.

[digits_]

Full power, close to 90 degrees bank, pull and descend, roll out on your reciprocal heading. That should get you close I would say. You could probably do that within 2000 ft altitude loss without excessive forces in something like a c17w. Still not legal though.

[PilotDAR]

Oh… right… A far more likely scenario is both pilots getting food poisoning and you’re the only pilot on board who can fly the plane.

Hey, don't joke, that happened on one of my first right seat flights in the Cheyenne, Timmins to Thunder Bay at night. My captain was better the next morning...

Okay, continue the topic at hand...

[photofly]

I think I’m more interested in, with a different technique, would an otherwise fatal turn up a blind box canyon…..could a tighter minimum radius be successfully accomplished……by giving up 1-2000 feet of altitude. So a powered scenario…..as opposed to the EFATO turnback.

How about a hammerhead?

I don’t do a lot of mountain flying as it’s flat round here, but a wingover costs you no altitude at all. That’s what I’d do. A minimum radius turn needs the lowest possible airspeed, and the quickest way to slow down is to climb, rapidly. Put the two together and you end up with the first half of a wingover, or even a chandelle where you keep the altitude and exit in slow flight… that’s what it was invented for.

But doesn’t the canyon turn issue involve being stuck under a ceiling? I thought it was axiomatic that you can’t climb.

[youhavecontrol]

Unless your surrounded by steep cliffs, most descending turns would cause the valley to narrow, I'd imagine... making a descending turn no move advantageous.

The emergency canyon turn, when done correctly, can be done in a very very small radius (assuming typical small plane), which to me makes this hypothetical scenario only relevant if one has already painted themselves into such a corner that their judgement in this situation is already toast... at which point I should hope an improvised maneuver is avoided.

[rookiepilot]

Unless your surrounded by steep cliffs, most descending turns would cause the valley to narrow, I'd imagine... making a descending turn no move advantageous.

The emergency canyon turn, when done correctly, can be done in a very very small radius (assuming typical small plane), which to me makes this hypothetical scenario only relevant if one has already painted themselves into such a corner that their judgement in this situation is already toast... at which point I should hope an improvised maneuver is avoided.

Yep, yep, yep. I get it. I’m not asking any of that. It’s not a post inviting comments on the stupidity of flying up a blind canyon.

I’m just curious if it changes the math, and if it doesn’t, why not, as a descending turn unloads the wing, vs maintaining altitude.

[photofly]

A descending turn doesn’t unload the wing. The load in a steady descent is exactly the same as in level flight, and as in a climb.

A falling turn unloads the wing, but as I pointed out before, you pay it back when you stop accelerating downwards (into a steady descent), and then pay it back again when you want to level off.

To unload the wing for any significant time and to any significant extent means building up a very rapid rate of descent.

[the-minister31]

He probably means to not load the wings as much as if you were to maintain altitude (about 1g) so you don't raise the stall speed as much

Of course you still have to give up a lot of altitude (probably 400-500 ft) and you are gonna pay for when you have to stop the dive, but not as much.

As a side note, raising the nose to convert airspeed to altitude, wingover on the stall buzzer and easing out of the dive gave me good results when trying stuff out in the planes I have flown. You have to give up a bit of energy, but not as bad as an unloaded turn or a hammerhead :p. But from experience, nothing is more efficient then a "mini-maxi" turn. Full power, bank and load the aircraft until you get an intermittent stall buzzer.

[JasonE]

How ironic, I was thinking about this thread while doing wingovers during some basic aerobatics training yesterday.

It would take some serious balls to do a properly executed 180 wingover back to the field at 500ft. I need to go back out and practice that now! (At higher altitude....)

[fleetcanuck]

If you would like to see some well researched information on the EFATO turn back please have a look at recent Charlie Precort (sp?) columns in Sport Aviation. There is a reason it is generally referred to as the 'impossible turn'.

[photofly]

It would take some serious balls to do a properly executed 180 wingover back to the field at 500ft. I need to go back out and practice that now! (At higher altitude....)

the wingover was suggested as a "canyon turn" when you have power and airspeed, and presumably some headroom to climb, but need to reverse course in a short distance (which seems like a reasonable use for it). If you're heading "back to the field" at 500agl, you don't have any excess airspeed or power to play with and it's probably not the manoeuvre to try!

[JasonE]

It all depends on how much energy you have available I guess (speed and altitude). I never depart at best angle (unless actually required) and when your wings are unloaded there is no stall speed.

It will be fun to try at altitude (and likely fail) though. I was also practicing spin recovery with minimum altitude loss yesterday.

[‘Bob’]

When your wings are unloaded that means you are in a ballistic arc.

Unless you want to continue that arc all the way to the ground, you need to load the wings up again. If you don’t have enough airspeed to load them up without stalling, your recovery point will be several hundred feet underground.

The other thing is that unloaded wings aren’t doing anything. You need to have them loaded to turn the aircraft unless you are doing a hammerhead or something. Lots of people who do wingovers set off the stall horn at some point… hardly an unloaded wing.

That’s why airspeed is life.

[photofly]

It all depends on how much energy you have available I guess (speed and altitude). I never depart at best angle (unless actually required) and when your wings are unloaded there is no stall speed.

Experiment! commencing at Vy (or your habitual climb speed after takeoff) with a hard floor at some sensible altitude, then close the throttle, count to two and a half and then see if a wingover can be achieved.

The other thing is that unloaded wings aren’t doing anything. You need to have them loaded to turn the aircraft unless you are doing a hammerhead or something.

That's an interesting point. In the thought experiment where you instantaneously unload the wing then rapidly yaw the plane with rudder to face back the way you came, the aircraft is now facing the right direction but still travelling in the original direction, now backwards

Going back to your point about doing a "regular" steep turn but unloading the wing in the turn, to avoid the stall speed rising:

Let's say you want to pull only 1.0g all the way around a turn, so your stall speed is whatever it is in level flight, but you want to make that turn in the same radius and rate of turn as if it were a 45°bank at 1.4g (a familiar "steep turn") - what would that look like? In other words, you want to make a steep turn in a glide, but modify it by unloading the wing back to 1.0g. Well, your regular 1.4g steep turn has 1.0g of horizontal acceleration. If that's the entirety of your g budget then you can only turn at that same rate by rolling to 90° of bank (knife edge) and your vertical acceleration will be whatever it is after 13 seconds of free fall. Which is 25,000 fpm downwards.

[rookiepilot]

He probably means to not load the wings as much as if you were to maintain altitude (about 1g) so you don't raise the stall speed as much

Correct. I meant relative to a steep turn while maintaining altitude.

[photofly]

He probably means to not load the wings as much as if you were to maintain altitude (about 1g) so you don't raise the stall speed as much

Correct. I meant relative to a steep turn while maintaining altitude.

Adding this from the edit to my previous answer ...

I don't think you win.

Let's say you want to pull only 1.0g all the way around a turn, so your stall speed is whatever it is in level flight, but you want to make that turn in the same radius and rate of turn as if it were a 45°bank at 1.4g (a familiar "steep turn") - what would that look like?

In other words, you want to make a steep turn in a glide, but modify it by unloading the wing back to 1.0g. Well, your regular 1.4g steep turn has 1.0g of horizontal acceleration. If that's the entirety of your g budget then you can only turn at that same rate by rolling to 90° of bank (knife edge) and "pulling" 1g, now all horizontal: your vertical acceleration will be whatever it is after 13 seconds of free fall. Which is 25,000 fpm downwards.

Your vertical acceleration (downwards) will be at (1 - sqrt(g_budget^2 - g_horizontal^2)) x gravity.

If your g budget is say 1.2g, and you want to use 1.0g for the turn, You still get a lot of vertical acceleration that you don't want: 0.75g's worth. That gives you a rate of descent of 19,000 fpm after 13 seconds.

Putting it another way, enough g to stop you falling doesn't leaves you almost nothing with which to make the turn, leading to a very very slow rate of turn - unless you load up the wing.

[JasonE]

I won't be able to do it for a few weeks, but I have access to an aerobatic plane with a G meter. I'll get out the Gopro and document my attempts.

[rookiepilot]

He probably means to not load the wings as much as if you were to maintain altitude (about 1g) so you don't raise the stall speed as much

Correct. I meant relative to a steep turn while maintaining altitude.

Adding this from the edit to my previous answer ...

I don't think you win.

Let's say you want to pull only 1.0g all the way around a turn, so your stall speed is whatever it is in level flight, but you want to make that turn in the same radius and rate of turn as if it were a 45°bank at 1.4g (a familiar "steep turn") - what would that look like?

In other words, you want to make a steep turn in a glide, but modify it by unloading the wing back to 1.0g. Well, your regular 1.4g steep turn has 1.0g of horizontal acceleration. If that's the entirety of your g budget then you can only turn at that same rate by rolling to 90° of bank (knife edge) and "pulling" 1g, now all horizontal: your vertical acceleration will be whatever it is after 13 seconds of free fall. Which is 25,000 fpm downwards.

Your vertical acceleration (downwards) will be at (1 - sqrt(g_budget^2 - g_horizontal^2)) x gravity.

If your g budget is say 1.2g, and you want to use 1.0g for the turn, You still get a lot of vertical acceleration that you don't want: 0.75g's worth. That gives you a rate of descent of 19,000 fpm after 13 seconds.

Putting it another way, enough g to stop you falling doesn't leaves you almost nothing with which to make the turn, leading to a very very slow rate of turn - unless you load up the wing.

Interesting. Thanks.

What about doing your numbers with a 60 degree bank?

I’ve flown in valleys out west a few times, (with lots of room) I’m also just curious —- if I ever had to — what the best turnback technique would be.

[AirFrame]

Your vertical acceleration (downwards) will be at (1 - sqrt(g_budget^2 - g_horizontal^2)) x gravity.

If your g budget is say 1.2g, and you want to use 1.0g for the turn, You still get a lot of vertical acceleration that you don't want: 0.75g's worth. That gives you a rate of descent of 19,000 fpm after 13 seconds.

This is only true if your descent is unrestricted, which it isn't. I'd grant you that if you were pointed straight nose down in a slippery airplane like an RV or a Glasair, you may accelerate faster, but still not unrestricted.

Given the situation, where you have a component of motion forward to start with, and have an airframe that's extremely draggy in the vertical direction, your vertical speed isn't going to increase anywhere near that much.