180° turn; distance calculations

GRG55 said:
I like the OP's thought process!

That's the thing about OSH. You can't count on a "standard pattern". On 27, for example, you might get sent out over the lake for spacing for slower airplanes ahead on final, or you might get asked to turn inside them (in which instance it won't be a square pattern for any of the faster airplanes). You might have to land on the threshold, or long on another colored dot. And they might change the aiming point when you are on short final. I've dealt with all of these arriving at OSH in the Aztec.

When I was there in 2015 someone piloting a piston Malibu overshot the turn to final and tried to save the landing. Stalled it cross-controlled, fortunately low enough to pancake it roughly wings level onto the runway, starting a fire as the engine separated. Bad scene. Just one example.
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I read the NTSB report on the Malibu. Sounds like he may have been in a ‘skid,’ not cross controlled. If cross controlled, in a slip, the left wing would have stalled first. Maybe it did but not soon enough to lift the right wing before the wing tip impacting the ground.
 
I thought that 180 degree turn was 1% of your groundspeed, SR turn? That means, 100 kts GS is 1nm for the 180.

I’m sorta used to doing ‘whatever it takes’ that I sometimes forget the rules of thumb.
 
luvflyin said:
Ok. What am I missing here. Isn't a skid to much rudder or to little aileron, but both in the same direction? Not opposite?
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Lateral stability would cause the bank angle to increase when the excess rudder is (mis)applied, thus opposite aileron is (mis)applied.
 
dmspilot said:
Lateral stability would cause the bank angle to increase when the excess rudder is (mis)applied, thus opposite aileron is (mis)applied.
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I'm still not getting it. Take the classic overshoot final scenario. I over shoot, lets make it left traffic. So I give it left rudder to get the plane back to center line. But I don't add enough left aileron along with the rudder to keep it coordinated, ball centered. I'm now in a skid. Call it to much left rudder or not enough left aileron, but I have left of both. This is 'cross controlled'??
 
luvflyin said:
I'm still not getting it. Take the classic overshoot final scenario. I over shoot, lets make it left traffic. So I give it left rudder to get the plane back to center line. But I don't add enough left aileron along with the rudder to keep it coordinated, ball centered. I'm now in a skid. Call it to much left rudder or not enough left aileron, but I have left of both. This is 'cross controlled'??
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Left rudder and right aileron.
 
luvflyin said:
I'm still not getting it. Take the classic overshoot final scenario. I over shoot, lets make it left traffic. So I give it left rudder to get the plane back to center line. But I don't add enough left aileron along with the rudder to keep it coordinated, ball centered. I'm now in a skid. Call it to much left rudder or not enough left aileron, but I have left of both. This is 'cross controlled'??
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When a pilot sees an impending overshoot to final they invariably skid around the corner out of fear of steepening the bank and raising stall speed. The skid causes an overbank which they stop with opposite aileron. This ironically sets them up for an aggravated stall where the lowered wing stalls first, in my opinion caused by fuselage induced turbulence due to the skid. Others would say the lowered aileron raises the angle of attack, but I've neutralized the ailerons before the break and it doesn't change the stall behavior one iota.
 
Let'sgoflying! said:
I am thinking 25° bank, but perhaps this is airplane specific and tied to turn rate?
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It's funny you mentioned 25 degrees. While the Rate of Turn calculation is v^2 / 11.26 tan ( angle of bank), a funny thing happens at 25 degrees... The entire equation can be rounded and simplified to:

( NM/min ) ^ 2 / 9 = radius of turn in NM

So you can then memorize it as:

( NM/min ) ^ 2 / 9 = radius of turn in NM
or
(NM/min) ^2 * .111 = radius of turn in NM
or you can just simplify and get a "good enough" measurement by dividing by 10 instead of 9.
(NM/min) ^ 2 / 10 = radius of turn in NM
or
(NM/min) ^ 2 * .1 = radius of turn in NM


So at 600 kts and 25 degrees bank, you would have
(10NM/Min) ^ 2 / 9 = 100 / 9 = 11.11NM
(10NM/Min) ^ 2 * .111 = 100 * .111 = 11.11NM
(10NM/Min) ^ 2 / 10 = 100 / 10 = 10NM
(10NM/Min) ^ 2 * .1 = 100 * .1 = 10NM

At 60 kts and 25 degrees bank you would have:
(1NM/Min) ^ 2 / 9 = 100 / 9 = 1.11NM
(1NM/Min) ^ 2 * .111 = 100 * .111 = 1.11NM
(1NM/Min) ^ 2 / 10 = 100 / 10 = 1NM
(1NM/Min) ^ 2 * .1 = 100 * .1 = 1NM

25 degrees is actually a pretty good bank angle to do the calculation on given that in an instrument environment ICAO indicates you should execute turns:
A) At a bank angle of 25 degrees or
B) A rate of turn equivalent to 3 degrees per second (standard rate turn)
whichever requires the lesser degree of bank.


So your maximum bank angle is 25 degrees if you're fast and if you're slow enough to establish a standard rate turn (or half standard rate in aircraft flying 250+kts and equipped with a 4-min turn coordinator), you can use the other equations for circumference of a circle to calculate quickly. (Airspeed * Time = arc distance flown = circumference = 2 * Pi * radius * degrees in turn / 360 ; we'll consider 3 "close enough" for Pi which gives us Airspeed * Time = radius * degrees in turn / 60).

If you're over 25 degrees or less than standard rate then you're stuck with the V^2 / 11.26 tan(angle of bank) = radius of turn equation which is just annoyingly cumbersome and doesn't have an easy way to simplify.
 
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@dtuuri & @dmspilot , I think I’m getting the picture now. You don’t get out of the skid, but by putting in a little opposite control pressure to ‘lessen’ the skid, so to speak, you are cross controlled. You haven’t flipped it all the way over to the other side into a slip yet, but you have little up aileron on the right wing as you try to reduce the bank angle or just stop the rate of roll to the left while still holding left rudder. Is that about right?
 
luvflyin said:
@dtuuri & @dmspilot , I think I’m getting the picture now. You don’t get out of the skid, but by putting in a little opposite control pressure to ‘lessen’ the skid, so to speak, you are cross controlled. You haven’t flipped it all the way over to the other side into a slip yet, but you have little up aileron on the right wing as you try to reduce the bank angle or just stop the rate of roll to the left while still holding left rudder. Is that about right?
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Even without the over-banking tendency and instead just letting it start to roll toward the earth side rudder and left turning tendencies, if you increase angle of attack at that point to above the AoA for a stall, one wing (the down toward earth wing) will stall and it’ll spin in the direction of the down (toward earth) rudder (skid).

Stepping on the ground (rudder toward earth and low wing) is “pro-spin” rudder input, stepping on the sky (rudder away from the ground is generally anti-spin and Skyhawks and other docile airplanes don’t like to spin out of slips. The nose will fall through and keep enough airspeed over both wings.

Also note that prop driven aircraft in the US typically spin better to the left than right. All of the left turning tendencies are essentially pro-spin to the left as mentioned above. Add even more bottom rudder and they’ll go right into it.

Going right is harder and some trainer airplanes need some coaxing by entering a little faster and pulling to above stall AoA a little quicker (almost an accelerated stall but not quite) while adding earth side rudder to the right and opposite aileron to get it to barely roll over kinda on its back and into the rotation.

Spinning Skyhawks isn’t as common as it once was. They don’t really want to go to the right as much as to the left, and if you remember to “step on the sky” if cross-controlled at all, they won’t want to start rotation, the nose will tend to fall first as it mushes around the corner in a slip.

Skid (step on the ground) and get the AoA up fairly quickly and they’ll go either way. But they’ll go left easier.

Now think about the normal left hand pattern. Left turns. Pilot overshoots final and is low. They pull back instinctively without adding enough power and at the same time step on the ground trying to hurry the nose around with yaw... perfect setup for the only spin a Skyhawk likes to do easily.

Make that a right hand pattern and they push the power up sooner, and the airplane ends up slipping around the corner or being close to coordinated instead. :)
 
luvflyin said:
@dtuuri & @dmspilot , I think I’m getting the picture now. You don’t get out of the skid, but by putting in a little opposite control pressure to ‘lessen’ the skid, so to speak, you are cross controlled. You haven’t flipped it all the way over to the other side into a slip yet, but you have little up aileron on the right wing as you try to reduce the bank angle or just stop the rate of roll to the left while still holding left rudder. Is that about right?
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I find pictures to be worth 1000 words, to coin a phrase. See page 126 here. While that plane is shown in level flight it could as easily be drawn in a turn either way by tilting the horizon one way or the other. I.e., it could be either skidding or slipping, depending on your point of view. Assume the drawing is made just as the plane hangs on the precipice of a stall and assume the pilot has recognized that fact and made the ailerons neutral (to eliminate a confounding factor). What happens next is the trailing wing stalls. It stalls before the leading wing despite having an apparent lower angle of attack due to the dihedral of the wings. Why does this happen? I don't exactly know for sure. It does, though, and I'd bet my life on it. I think (and Bill Kershner did too) it's caused by turbulence across the fuselage disturbing the laminar flow over the wing. Maybe it's a local change in the relative wind direction against the trailing wing as it bends under the fuselage and swoops up toward the bottom of the trailing wing (that doesn't explain why the same thing happens in a low wing). Who knows? Maybe Nauga knows? For Leighton Collins' treatise on all this, see here, page 337.

EDIT: I see my post here didn't answer your question. Yes, it's about right. :)
 
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I love math as much as the next guy, but holy crap talk about over complicating things.

I bet if someone asked some of you for a cookie recipe, you'd go all the way back to describe how the planting needs to be done for the wheat to make the flour along with raising the chickens for the eggs, etc...
 
I saved a bookmark to the derivation for future reference!
 
EdFred said:
I love math as much as the next guy, but holy crap talk about over complicating things.

I bet if someone asked some of you for a cookie recipe, you'd go all the way back to describe how the planting needs to be done for the wheat to make the flour along with raising the chickens for the eggs, etc...
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First we’d need to know if the chicken or the egg came first
 
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