6
$\begingroup$

Scenario

Lets imagine a 50kts wind from east. Now lets imagine an aircraft flying at 200kts (True Air Speed) east that make a 1.15G turn -90º to north. Maintaining in all moments the given constant speed.

Now, lets imagine the same flight but with 200kts constant ground speed.

enter image description here

In the first case, the speed is 200kts TAS= 150kts GS, and after the turn TAS=GS=200kts In the second case, the speed is 200kts GS = 250kts TAS, and after the turn, TAS=GS=200kts

Any mistake until now?

The question: The question is how that influence the turning circle? In which of the two cases the turning circle is flatten/stretched and in which is circular if any? What is the approximate shape of them?

Further explanation

Considering my very weak understanding, the turning circle G depend on absolute speed (~ Ground speed), and thus, turning with a constant ground speed imply a circular turning. Which is independent of the wind. Of course, the pilot would have to accelerate/decelerate to maintain this speed. With TAS, the real speed is variable, so in our case, I expect to see a turning circle which is horizontally flatten.

But, using a planning system, the result I observe is the reverse: the circle is constant (independent of wind, circular) on TAS and variable (dependent on wind, shape is not drawn) in GS.

Edited: TAS=GS=200kts: I just realized that it is not really true, cause the aircraft need to head slightly east to compensate the wind, so probably the TAS will be ~206 and GS 200. Even if that do not make a big difference for the question.

$\endgroup$
  • $\begingroup$ Hello Adrian, welcome to Aviation.SE! $\endgroup$ – DeltaLima Nov 30 '15 at 8:23
  • 1
    $\begingroup$ At the end of your turn, is your heading or your track angle North? $\endgroup$ – DeltaLima Nov 30 '15 at 12:33
  • $\begingroup$ I would say track, but I guess any would fit as an example. :-) $\endgroup$ – Adrian Maire Nov 30 '15 at 12:54
  • $\begingroup$ You are missing what bank angle you are turning at. Rate of turn is predicated on true airspeed while radius of turn is predicated on groundspeed. If the bank angle is constant for the constant GS exercise, the TAS would make it non-circular. If you vary the bank angle with TAS, the radius will be a perfect circle. $\endgroup$ – wbeard52 Dec 1 '15 at 2:10
  • $\begingroup$ @wbeard52 I though G and Bank-Angle was nearly equivalent. In any case, yes, I am supposing a model in which the turning bank-angle is constant. $\endgroup$ – Adrian Maire Dec 1 '15 at 7:26
2
$\begingroup$

To better understand the physics of a turn, maybe it helps to look at the centripetal force in a different way. This force is actually pulling the aircraft sideways into the new direction of movement. When banking into the turn on an easterly course, the centripetal force will accelerate the aircraft to the north and, towards the end of the turn, will decelerate its easterly speed, such that it will have a northerly speed when the turn is complete. The centripetal force is the means by which ground speed gets converted from a purely easterly speed to a northerly speed.

If you fly with 200 kts in a 50 kts headwind and want to fly a northerly track after finishing the turn, you simply stop turning after having decelerated the aircraft by only 150 kts. You don't fly a quarter circle of 90°, but only one of 75.5°; the angle where the cosine has shrunk to 0.25. After that, your easterly speed component is reduced by 150 kts and your northerly speed component has increased to 193.6 kts. When viewed by another plane flying in the same headwind, you have flown a segment of a circle measuring 75.5°. When seen by an observer on the ground, your ground track is a compressed circle segment.

Starting with 250 kts will make you perform the turn at a higher speed, but with the same technique.

Any mistake until now?

Yes! Your assumption that your ground speed would be 200 kts is wrong. At all times, you will maintain a TAS of 200 kts, and the ground speed after the turn in the first case is only 193.6 kts.

In the second case, you start with a TAS of 250 kts and maintain that over the full turn. Your ground speed after having achieved a northern track is now 242 kts.

The question is how that influence the turning circle? In which of the two cases the turning circle is flatten/stretched and in which is circular if any? What is the approximate shape of them?

That depends on the observer. Your ground track is compressed in easterly direction because you started at an easterly speed component of 150 kts but used the aerodynamic forces of 200 kts for turning. You stopped turning after having yawed the airplane by only 75.5°, so you now fly a correction angle of 14.5° to achieve a purely northern track. The observer will see you accelerate from 150 kts ground speed to 193.6 kts ground speed (and from 200 kts to 242 kts in the second case).

The shape of the circle will only look round for an observer flying in the same headwind.

In a tailwind, you would keep turning until you have completed a circle segment of 104.5°, so you can let the centripetal force act longer in easterly direction for a ground speed change of 250 kts in easterly direction. Since it will reduce the aircraft's northerly speed component over the last 14.5°, the centripetal force will slow the northerly speed component to 193.6 kts over the last 14.5° of the turn after it had reached 200 kts after turning 90°. Now the ground track is stretched in easterly direction.

Below I added a graph with three ground tracks and airplane symbols overlaid to show the correction angle at the specific point of the turn. All three cases perform a 90° turn and fly initially parallel to the wind direction. I used 50m/s TAS and ±20m/s wind speed; the axes are in meters. The blue line is the case without wind.

Ground track when turning in wind

Ground track when turning in wind (own work). A turn from a headwind course takes much less time and azimuth change than one from a tailwind course. Only the ground track without wind is a circle, all other ground tracks are ellipses. In the headwind case the axis parallel to the wind direction is the shorter axis and in the tailwind case it is the longer of the two axes of the ellipse.

If you insist on keeping the ground speed constant, you will need to speed up when turning into the wind rsp. to slow down when turning out of the wind. Your second case with 250 KTAS while heading east and 206 kts when flying a northern track is very unusual. This answer gives you the basic equations. The time to fly a 75.5° turn with 30° bank angle (= 1.15 g) is $$t_{75.5°} = \frac{v\cdot\pi\cdot75.5°}{g\cdot\sqrt{n_z^2-1}\cdot180°}$$

which yields 30.4 s for 250 kts (= 118.6 m/s) and 25.1 s for 206 kts, so you will spend 27.75 s turning when the deceleration is linear. The deceleration along the flight path is 0.455 m/s², so you will need to throttle down during the turn and throttle up when the deceleration phase is over. Good luck keeping the ball centered while turning and changing the p-factor at the same time!

$\endgroup$
  • $\begingroup$ "In the second case, you start with a TAS of 250 kts and maintain that over the full turn. Your ground speed after having achieved a northern track is now 242 kts" That is the case 1 with different speed. Case 2 is: you fly constant 200kts GS, meaning a TAS gradient (non-linear) between 250kts to 206kts $\endgroup$ – Adrian Maire Dec 1 '15 at 7:32
  • $\begingroup$ @AdrianMaire: If you want to fly constant GS in wind, you will need to change airspeed while turning. Where do you take the power to do so? Or the drag to slow down quickly enough? Do you accept to dive or to climb to speed up/decelerate? This is highly unusual; what I describe is what you get taught in flight school and what most pilots do. $\endgroup$ – Peter Kämpf Dec 1 '15 at 8:15
  • $\begingroup$ well, it is a model, so I do not care really about how easy is to fly it. It is more for understanding theoretical behavior for a software. If the Pilot do not want constant GS, he just has to set TAS, or plan GS and fly an approximation. You could also guess a special futurist aircraft that is able to electronically maintain GS by adjusting engine and aero-brakes. $\endgroup$ – Adrian Maire Dec 1 '15 at 8:42
  • $\begingroup$ @AdrianMaire, it is not as much about hard as about only possible up to certain limit. While the wings can generate significant force, and thus acceleration, perpendicular to the relative wind (all aircraft can withstand at least 2.5G, of which 2.29G can be directed sideways in a level flight in suitable bank), you can rarely generate more than about 0.3G forward (thrust-to-weight 1:3) and even less backward (speed brakes are not all that effective)). So in a strong wind, you can only maintain ground speed in a rather loose turn. $\endgroup$ – Jan Hudec Aug 4 '16 at 19:07
4
$\begingroup$

enter image description here

If you maintain a constant bank angle (30 degrees for a 1.15 g turn) and maintain constant true airspeed of 200 Knots, your path will be as shown in the chart by the yellow line.

If you maintain a constant ground speed (150 knots) and a constant acceleration 1.15, then your path will be as described by the grey line. It will be as if there is no wind, however it requires thrust variations and bank angle variations.

$\endgroup$
2
$\begingroup$

The radius of the turning circle depends on the True Air Speed, not the Ground Speed. TAS is the true measure of aircraft performance and is used in every situation when actual performance needs to be measured.

Consider the forces on an aircraft in a coordinated turn with a banking angle of $\phi$

Coordinated turn

Image from code7700.com

Now, the load factor n on the aircraft is simply the ratio of the lift to the weight.

$n = \frac{L}{W} = \frac{1}{cos\phi}$

The centripetal force acting on the aircraft can be given by,

$F = L sin\phi$,

which is equal to,

$F = m . \frac{v^{2}}{R}$,

where $v$ is the TAS and $R$ is the radius of turn.

Solving, we get,

$R = \frac{v^{2}}{g . tan\phi}$,

or in terms of load factor,

$R = \frac{v^{2}}{g . \sqrt{n^{2} - 1}}$

Note that in all these cases, the velocity $v$ is the TAS, not the ground speed. So, for a 1.15g (or n equals any other value) turn, greater the speed, bigger the turn radius, at a constant bank angle.


I'll just list out the various speeds used for qualifying aircraft performance

  • Indicated Air Speed is the the speed of an aircraft as shown on its pitot static airspeed indicator, calibrated to reflect standard atmosphere adiabatic compressible flow at sea level. Simply put, it is what the airspeed indicator shows.

  • Calibrated Airspeed is the indicated airspeed corrected for various errors (like instrument and position errors)

  • Equivalent Air speed is the airspeed at sea level in the ISA at which the dynamic pressure is the same as the dynamic pressure at the true airspeed (TAS) and altitude at which the aircraft is flying.

  • True Air Speed is the speed of the aircraft relative to the atmosphere. This is the speed used for measuring the aircraft performance. The TAS, along with the heading, gives the aircraft velocity relative to the atmosphere.

  • Ground Speed is the vector sum of aircraft velocity and wind velocity (at the flight altitude). This gives the speed of the aircraft relative to the ground.

Air Speeds

Image from stackexchange.com

The ground speed is rarely used in measurement of aircraft performance directly. The performance measurements are made with reference to the atmosphere as the frame of reference. This means that the True air speed, which gives the velocity with respect to the atmosphere is used.

-------------------------------------------------------------------------------For an observer on ground, the turn will be a circle if the TAS is equal to the ground speed i.e. if there is no wind.

If there is a head wind, the turn will be squeezed i.e. the turning circle will be egg shaped, with the squished side in the direction of motion. For tail wind, the squished side will be in the opposite direction.

The case where headwind is equal to TAS is the limiting case- basically, the aircraft will appear to the ground observer to turn along a line.

It will be like some one on shore seeing a swimmer turning in a water body where the water currents vary.

$\endgroup$
  • $\begingroup$ Great explanation, thanks you a lot. I still have a question to understand fully: Why in the centripetal force formula (F=m*v^2/r), you interpret v as TAS and not absolute speed? I mean, for lift the important speed is TAS, but for centripetal force, i would expect Real speed. E.G supposing 100kts head wind and 100kts aircraft TAS speed, the aircraft is currently not moving, so how could he need a radius to turn (meaning infinite turning time)? Instead I would expect 0 bank-angle turning over itself. $\endgroup$ – Adrian Maire Nov 30 '15 at 12:03
  • $\begingroup$ Sorry if I sound pushy, that is not my intention. I just do not want to lets any understanding hole. $\endgroup$ – Adrian Maire Nov 30 '15 at 12:21
  • $\begingroup$ @AdrianMaire I think you've answered the question yourself. Taking ground speed means you've selected the wrong frame of reference. If you consider TAS = headwind, the aircraft is not moving wrt ground, and should not require any thrust. But it does. $\endgroup$ – aeroalias Nov 30 '15 at 13:00
  • $\begingroup$ @AdrianMaire Centrifugal force is an inertial force, meaning that it is caused by the motion of the frame of reference itself and not by any external force. For example, if one stands on the ground and watch children spinning on a merry-go-round, then in the stationary frame of reference their outward acceleration is caused simply by their inertia. In our frame, which is external to the rotating frame, there is no centrifugal force at work. But in the rotating frame of reference of the children, there is a centrifugal force. $\endgroup$ – aeroalias Nov 30 '15 at 13:01
  • 1
    $\begingroup$ False: "TAS is the true measure of aircraft performance and is used in every situation when actual performance needs to be measured." Nearly all limitation and performance data in the POH/AFM is based on indicated airspeed (source PHAK, 8-2) is in indicated airspeed. faa.gov/regulations_policies/handbooks_manuals/aviation/… $\endgroup$ – rbp Nov 30 '15 at 13:29
1
$\begingroup$

Ignoring the wind::

Your first case has a constant TAS, so you should expect a constant radius (i.e., circular) path. (Speed isn't changing, bank angle or 'g' isn't changing, so constant radius)

Your second case requires the plane to reduce its TAS from 250kts to 200kts without changing the bank angle. That will affect the no-wind path to make a "decreasing radius turn". (@Aeroalias's answer shows the relationship between radius and velocity). For a given bank angle, a lower air speed will result in a smaller turn radius. Imagine a 1.15g turn in the SR-71 vs the same 1.15g turn in a Piper Cub. Since you're slowing down during the turn, your no-wind path will look more like the letter "J" than a quarter-circle.

Not ignoring the wind:

Up until now I've talked about the no-wind situation, because the airplane doesn't really know or care about the ground. It's just swimming through a large mass of air. "Wind" is what we call that air mass when is moving relative to the ground.

For the first case, the wind will "squish" the ground track of the latter 1/2 of your turn to make it somewhat "J" shaped.

For the 2nd case, the wind will "squish" the ground track of the latter 1/2 of your J-shaped turn to make it appear to be sharper (less circular).

$\endgroup$
  • $\begingroup$ with a 50 knot westerly, its not going to be constant radius, relative to the ground. it will be declining radius. as the plane turns, it will get pushed further downwind. remember your turns around a point, and the change in bank angle to keep a constant distance? flightsimbooks.com/flightsimhandbook/93-1.jpg $\endgroup$ – rbp Nov 30 '15 at 17:47
  • $\begingroup$ @rbp - Isn't that what I said, in the part starting with "For the first case,..."? $\endgroup$ – Dan Pichelman Nov 30 '15 at 17:56
  • $\begingroup$ your answer is not very clear. one thing that might help would be to edit the question itself so the cases are clearer, and then answer each case $\endgroup$ – rbp Nov 30 '15 at 17:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.