Turn radius of slipping/skidding turns

[Revtach]

Can anyone help me understand if slipping/skidding turns affect the turn radius?

For a given bank angle, will a slipping turn result in a larger turn radius than in a coordinated turn?

For the same bank angle, will a skidding turn result in a smaller turn radius that in a coordinated turn?

It seems to me that the lower and higher rates of turn in a slips/skids would cause the radius to be larger/smaller compared to coordinated turns.

Conversely, I have been told that a pilots attempt to tighten a turn with rudder input, for example during base to final in an attempt to not overshoot, is in vain since the dangerous skidding turn they enter into doesn't help prevent overshooting.

Hoping to get clarification. Thanks.

 

[nauga]

Greatly simplified: Slipping in a bank aids the vertical force needed to maintain level flight (since the fuse is providing some), so the wing provides less and as a result centripetal acceleration will be lower and the radius larger. Skidding in a bank works against the vertical force needed to maintain level flight (since the fuse is essentially producing a downward force), so the wing must provide more and as a result centripetal acceleration will be higher and the radius smaller.

Nauga,
who checks his math, but not always before posting

 

[Dan Thomas]

Skidding in a bank works against the vertical force needed to maintain level flight (since the fuse is essentially producing a downward force), so the wing must provide more and as a result centripetal acceleration will be higher and the radius smaller.

Anytime I deliberately skidded a turn (at altitude) it seemed to take a lot longer to make the turn than if it was just coordinated. The length of time implies a larger radius. Is there some authoritative source for all this?

Skidding the base-to-final risks a stall/spin at low altitude, which is normally not survivable.

 

[MauleSkinner]

Conversely, I have been told that a pilots attempt to tighten a turn with rudder input, for example during base to final in an attempt to not overshoot, is in vain since the dangerous skidding turn they enter into doesn't help prevent overshooting.

The reason the skidding turn is bad isn’t because it doesn’t help prevent overshooting, it’s because if you die in a stall/spin crash turning final nobody cares which side of centerline you’re on.

 

[luvflyin]

Thrust also is in the equation. How much I dunno. In a skid the thrust is pointing to the inside of the circle. In a slip it is pointing to the outside of the circle.

 

[dtuuri]

I would say the answer to your first two questions is found in the very definition of the terms:

Slipping turn. An uncoordinated turn in which the aircraft is banked too much for the rate of turn (and vice versa for skids).​

I.e., for a "given bank angle" a slip has a slower rate of turn than a skid. Why? Because there's less deflective force available, since the relative wind is coming from inside the turn. During a skid, the relative wind is from the outside, deflecting the flight path more. So, a slip has a greater radius by my reasoning (airspeed the same).

As to adding rudder to tighten the turn to final, well duh, it sure will. A spin is about as tight a turn as you can ever have.

 

[nauga]

Is there some authoritative source for all this?

Newton? A slip or skid (just 'sideslip' in engineering terms) generates sideforce (sideways lift) in the direction of the sideslip, i.e. nose left of the left of the velocity vector generates sideforce to the left. In a left bank, i.e. left turn, that's a skid, and sideforce is directed inside the turn and inclined down. The horizontal component of the sideforce combined with the additional horizontal component of the increased lift (required to overcome the downward component of sideforce) will decrease the radius of turn if *true* (not *indicated*) airspeed is held constant. Reverse the sign on all the sideforce components for a slip and see that the radius increases. For constant airspeed, decreasing radius means increased turn rate.

A 'definitive source' will usually have a free-body diagram representing the above. I tend to distrust sources that state it without proof...just like I've done above .

Nauga,
who is intuitively obvious to even the most casual observer

 

[EdFred]

Isn't a spin more or less a really really tight skidding turn? What's the radius on that compared to a coordinated turn?

 

[Revtach]

Thank you all. This confirms what I believed to be true.

I did dig deeper into the question earlier today and found another individual asserting that a skidding turn would not help you prevent overshooting on final: Skidded Turn Aerodynamics - APS Emergency Maneuver Training - YouTube

He mentions it at the 6:45 mark.

Would you infer from that assertion that a skid therefore does not decrease turn radius? That is the conclusion I come to and perhaps that is the mistake that causes my confusion in the first place.

 

[Salty]

Hate to disagree with some of the experts, but I fail to see how a skidding turn can fail to decrease the turn radius, all other variables including the dread stall spin ignored. Or maybe the experts agree and I’m just too stupid to understand them. Lol

 

[GeorgeC]

Isn't a spin more or less a really really tight skidding turn? What's the radius on that compared to a coordinated turn?

It's not flying, it's falling with style.

 

[RussR]

This is the kind of thing where the discussion gets complicated, but if you just jump in the plane it's so obvious.

Fly a 30 degree banked left turn, normal coordination.

Now fly with 30 degrees of left bank, but with full right rudder (slipping turn). It will be immediately obvious that it's going to take you a long time to get around that turn.

Now do the same but with full left rudder (skidding turn). The turn rate will be much faster.

Since bank angle and speed are held constant (or close enough), a higher turn rate means a smaller radius, and vice versa.

 

[dtuuri]

Would you infer from that assertion that a skid therefore does not decrease turn radius? That is the conclusion I come to and perhaps that is the mistake that causes my confusion in the first place.

Full disclosure: I didn't watch the video.

Rate of turn and TAS set the circumference of the turn, right? (Rate X time = distance.) In a skid the rate of turn increases, so if the TAS is unchanged the circumference and radius both will decrease. Ergo, skids will decrease radius. But, DON"T DO IT!

 

[geezer]

Simple flight test:
First climb to at least 5,000 feet above ground level, trim to level flight, cruise airspeed.
In straight and level flight, smoothly depress the right rudder as you bank to the left.
Do not pull back on the yoke to maintain altitude, as that would alter the rate of turn.
The bank should be the correct amount to keep the plane traveling straight ahead.
When you reach full right rudder, and are still flying straight ahead, you are in a slip with an infinitely large turning radius, thus proving that slips slow the turn.
Just for fun, then continue banking further to the left, until you have a standard rate turn.
The bank angle will now be in the order of 60 degrees.
The rate of climb will be in the region of minus 1,000 FPS or more.
Hold the bank, and adjust the rudder input for a coordinated turn, bank unchanged.
The turn coordinator will be pegged. Reproving that a coordinated turn turns faster than a slipping turn.
About here, you will begin to realize why the altitude was needed for this fun experiment.

 

[Dana]

Isn't a spin more or less a really really tight skidding turn? What's the radius on that compared to a coordinated turn?

No, a spin is what happens immediately after that really really tight skidding turn.