Turning Errors

The gravitational feedback mechanism currently implemented in the AHRS code causes the vertical axis of the display (the down direction of the roll/pitch) to align itself with (to slew towards) the measured acceleration, at a fixed rate. At the moment I’ve set the fixed rate to 0.004 radians per second, which works out at about 14 degrees per minute. This is required for two reasons: firstly to overcome drift in the gyros, which look like a slow continuous roll in a random and changing direction, and secondly to overcome integration and other errors that get included over time into the direction cosine matrix. That does mean that during a turn an error is induced, because then the detectable acceleration vector is the sum of gravity (down) and the sideways centripetal acceleration to the outside of the turn.

In this part I want to look at the nature and size of this error.

If you look at the way a mechanical gyroscopic Attitude Indicator (AI) works you’ll see it has a self-erecting mechanism that works via a set of pendulous vanes. If the spin axis of the gyro isn’t aligned with gravity then small vanes at the bottom of the assembly hang to one side and open up air jets. The air jets blow and push so that the gyroscope axis returns to the vertical. This happens at a fixed rate – usually 5 to 8 degrees per minute. That means that it works in the same way as the AHRS code, and suffers from the same accelerational and turning errors, so the analysis here is just as applicable to the mechanical AI. A qualitative description of the errors induced in a turn would go something like this:

• Initially the gyro axis is vertical and the measurable acceleration is purely vertical due to gravity. The aircraft flies straight, let’s say southwards.
• The aircraft banks into a right turn. Imagine it’s a coordinated turn, so the bank angle is fixed by the speed and rate of turn. The gyro axis is still vertical but the new acceleration tilts to the east. The angle of tilt is the same as the aircraft bank angle.
• Because of the self-erecting mechanism the gyro axis starts to tilt eastwards, towards the acceleration vector.
• Meanwhile the aircraft is in a clockwise turn. The acceleration vector always tilts towards the left wing. So during the turn it moves to south-east, then south, then south-west, then west, and so on.
• The gyro axis continues always to move towards the acceleration vector at the same fixed rate, no matter where they both are. So after starting to swing east, it bends south, then south west, then west.
• Overall, the gyro axis travels outwards from the centre, on some kind of spiral path.

Imagine you are running around the outside of a circle. Beginning at the centre of the circle is a dog that chases you as you go. If the dog is very fast it will run at full speed until it catches up with you, then slow down and stay at your heels. If you run faster than the dog can it will wind up running in a circle that’s smaller than yours, while you run ahead of it. So it is with the acceleration vector and the gyro axis: if the acceleration vector moves around wide circle (i.e. large bank angle or rate of turn) then the gyro axis can’t catch up with it. If the bank angle and rate of turn are small then it can. If the gyro axis ever does catch up to the acceleration of course that means the AI will be indicating straight and level, even if the aircraft is really in a bank.

(a) The dog can run as fast as the man, and catches him. (b) The man runs a little faster than the dog (c) The man runs much faster than the dog. The magenta line is in the direction of the instantaneous velocity of the dog and is tangent to the dog's circle.

In the diagram above you can see this graphically. (As depicted, the aircraft would be at the centre of the circle, flying west turning and to the right. The dog and the man give you an overhead view of the gyro axis and the acceleration vector, respectively.)

On the left (a) is the situation where the gyro axis can tilt as fast as the acceleration vector is moving. The gyro axis catches up with the acceleration vector, just as the dog catches up with the man. The attitude indicator will then wrongly show only straight and level.

In (b) is the situation where the acceleration vector is moving only a little faster than the gyro axis is able to. The angle between the blue lies is relatively small and the gyro axis (dog) is displaced from the vertical (centre) in nearly the same direction as the acceleration vector (man). In this case the error would be mostly showing too little bank, and a small erroneous nose-up pitch.

In (c) the man runs very fast around a large circle, corresponding to a large bank angle and a high rate of turn. The gyro axis is displaced from vertical by the same amount as in (b)  because the dog is running at the same speed, but lags behind the acceleration vector by nearly 90 degrees. The AI will show almost the correct bank angle but most of the error will appear as an erroneous nose-up pitch.

For the sake of completeness I should say that although this is the steady-state, We still don’t know anything about the short term behaviour, and how long – or whether at all – the gyro reaches this steady state. If the aircraft starts straight, level and fast, and rolls quickly into a steep turn then the initial error is in bank and the pitch error appears later as the acceleration vector moves ahead of the gyro axis.

To get a more detailed insight we need to look at the situation mathematically. Unfortunately a complication is that the real geometry is spherical and not planar; it’s as though the man and the dog were running around the inside of a great big hemispherical bowl, where their circular paths are meridians of latitude, and the magenta straight line from the dog to the man is actually a great circle.

Let’s have some terminology:

• Θ will be the  the angular deviation of the acceleration vector from the vertical, the same as the bank angle of the aircraft if the turn is coordinated.
• ω is the rate of turn of the aircraft, in unit of radians per second, a standard rate turn of 180 degrees in one minute means ω = π / 60.
• v is the flight speed, in metres per second. 1 knot is 1.94 m/s.
• g, the surface gravity of the earth: 9.81 m/s²
• θ is the angle of deviation of the gyro axis from the true vertical. It corresponds to the radius of the dog’s circle. It is an angle rather than a distance because of the spherical geometry.
• θ_max is the biggest angle of deviation possible for a given rate of turn. It corresponds to the radius of the biggest circle the dog can run, which depends on the maximum speed of the dog and its rate of turn.  θ can be smaller than θ_max but not bigger.
• λ is the slew-rate of the gyro axis. For the AHRS code it’s set at 0.004 radians / second, or 14°/s. For a mechanical gyro-driven AI I understand it’s usually between 5 and 8 degrees per second. In the dog analogy it corresponds to the fastest speed the dog can run.

Then by geometry and a little trigonometry the following results are true:

• Θ = tan^-1(vω / g) ; 0 ≤ Θ < π/2 This is the usual formula for the angle of bank in a turn.
• θ_max (λ, ω) = λ ≤  ω :  sin^-1 (λ / ω);  λ >  ω : π/2  ; and where 0 ≤ θ < π/2
  note θ_max is a function only of λ and ω
• θ = min (Θ , θ_max)
• the cusp between θ = Θ and θ = θ_max occurs when  tan^-1(vω / g) = sin^-1 (λ / ω)
  As far as I know there’s no analytic way to solve that equation, but it can be solved numerically for any set of parameters.

We can look at this in a variety of ways. Let’s keep a fixed λ and examine the relationship between θ, ω and v. In piloting terms it’s interesting to ask how the long-term error in the AI changes with:

1. Airspeed, if flying a fixed rate turn
2. Rate of turn, if flying at at a fixed airspeed

The first is the easiest to answer. If ω and λ are both fixed then θ_max is fixed too. Because of the form of the inverse tan function, Θ is a monotonic increasing function of v, so as the airspeed increases, θ equals the bank angle, until it reaches this fixed value of θ_max – then  θ remains constant as you fly faster.

If I fly a standard rate turn for a long-enough time, with a regular AI – lets say λ is 5 degrees per minute – then the biggest the roll or pitch error it will settle to is sin^-1 ( 5 / 180) = 1.59°. (In fact, if all the angles involved are small then for a standard rate turn the error is limited approximately to the slew rate (in degrees per minute) divided by π; which in this case is 5° / 3.14159 is 1.59°). This is a maximum error; if you’re flying soooo slowly that a standard rate turn involves a bank of less than 1.59°, then that lower angle is the long-term error instead.

For a turn at half standard rate, the the maximum error is actually bigger: sin^-1 ( 5 / 90) = 3.18°. To obtain that error you’d have to have a bank angle of at least that, and therefore your airspeed would have to be greater than 114 knots. Comparing to the three diagrams (a), (b), and (c) above: fly a half-standard rate turn at 114 knots or below with this AI, and eventually it will show straight-and-level. Fly at 41 knots (say you’re in a microlight!) where your bank angle is fractionally more than 3.18°, and eventually that tiny fraction is all that will be left showing on the AI. But if you could fly a half-rate turn at 4700 knots then you’d be in an 80° bank and you’d have most of that 3° degrees of error show as a incorrect nose-up attitude, while the indicated bank would be correct. (You’d also be pulling about 6g and have a turn radius of 50 miles, so it might be time to talk to you national air force about your aircraft technology, they’d like to hear from you.)

Why is there a bigger potential error in a slower rate turn? Because if you reverse your direction more quickly – standard rate compared to a half-standard rate – you have less time in any one direction for the error to build up. Your easterly error that you got while flying south is more quickly cancelled out because you’re sooner flying north again. And in terms of the dog and the man, if the dog has to go around only half a turn in a given time, the radius of its path can be bigger.

Let’s examine the second option. What happens to the AI errors if I peg my airspeed at say 120 knots, and vary my rate of turn? The long term error will equal the bank angle, up to a maximum angle, and then it go down as I tighten the turn. It’s probably easiest to look at this graphically, so here is some data plotted out:

The black line shows θ_max for λ = 5º/min. The coloured lines are the bank angles at speeds from 80 to 200 knots, for a range of turn rates. The long term gyro error for any rate of turn is the lower of the black line and which ever coloured line applies (according to speed).

Angles of bank and theta_max for different speeds, vs. turn rate.
Red=80kts, green=100kts, blue=120kts, yellow=140kts, magenta=200kts

A few observations:

• The biggest errors are for very fast flight combined with slow turn rates. Fly a 0.75-degree per second turn at 200 knots for a sufficient time and eventually your AI will show close to wings-level and therefore be in error by 7 degrees.
• For standard rate turns at all reasonable air speeds, long-term errors are limited by slew rate and not bank angle.
• For standard rate turns, the long-term error is a small proportion of the  true bank angle. The AI remains approximately accurate, and certainly useful, even for very long duration turns. It’s more accurate for standard rate turns at higher speeds.
• For turns of half standard rate, the long term error is a significant proportion of the bank angle. The AI will indicate significantly closer to level flight than it should. Only when the speed gets above 200kts does the AI maintain a reasonable indication of the correct bank angle for a half-rate turn.

As a final example let’s calculate what happens in every pilot’s favourite test maneuver, the steep turn. In this case the aircraft is a continuous turn at 45° of bank, travelling at say 100 knots. That being the case, the rate of turn is about 3.65 times standard rate, just short of 11° turn per second, or 660° / min.   With the same AI with λ = 5º/min, the maximum error is clearly small and slew-rate limited, indeed to a value of about 0.4°. So I don’t think you need to worry about AI errors while practicing steep turns.

This section has been all about the long term maximum error. But we haven’t any real idea how the error to build up. And how much of the error accumulates in a fixed angle of turn – a course reversal or 180° turn for instance. I hope to look at that in the next section.