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Imagine we have a line-abreast formation of aircraft, with differing airspeeds, turning around a central pivot point, like points on the radius of a phonograph record, with an equal amount time per 360 degrees of turn for each aircraft. What is the formula for the bank angle required of each aircraft, in terms of the variables 1) turn radius and 2) time period per 360 degrees of turn?

Note that essentially the same situation applies when a slow-flying glider-- maybe even a hang glider-- and a faster, heavier glider are thermalling together and trying to stay exactly opposite each other. Again the turn rate must be the same for each, but the turn radius is not, and the required bank angle for each might be different. Clearly, the difference in turn radius is much more significant in this case, than in the case of high-speed aircraft turning in tight formation.

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  • $\begingroup$ I can roll back the edit if it is felt that it invalidates existing answers; that was not my intent and my feeling is that it does not. Note also the original specification of "differing airspeeds" implied that I was not interested only in the case of a tight formation of very-high-speed airspeed aircraft doing a turn of huge radius. $\endgroup$ – quiet flyer Jun 18 at 15:00
  • $\begingroup$ Your change invalidates Michael's answer. $\endgroup$ – bogl Jun 18 at 15:05
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    $\begingroup$ Well, I guess then I or someone will roll it back at some point. $\endgroup$ – quiet flyer Jun 18 at 15:05
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Turning Rate

All planes flying a turn in line-abreast formation are turning at the same rate $\omega$.

A standard turn is commonly defined as $\omega_1 = 3°\frac{1}{\mathrm{s}}$.

Full Circle Time Period

If planes are turning at the same rate, the time period to complete a complete circle will be the same as well.

Time period: $$ T(\omega) = \frac{360°}{\omega}$$

The time period to complete a circle at this standard turn rate is $T_1 = \frac{360°}{3°}\mathrm{s} = 120\,\mathrm{s}$.

Air Speed

Since the outer planes have to travel a longer distance to complete their larger circle in the same perios, the outer planes have to fly at a higher airspeed then the inner planes.

Air speed: $$ v = \omega r$$

Bank angle

Turn rate: $$\omega = \frac{v}{r}$$

Banking angle: $$\tan{\theta} = \frac{v^2}{rg} = \frac{\omega^2 r}{g}$$

Gravitational constant: $$ g = 9.81 \frac{\mathrm{m}}{\mathrm{s^2}}$$

That gives the banking angle as a function of turning radius: $$\theta(r) = \arctan{\frac{\omega^2 r}{g}}$$

See the purple curve in the graph below for the bank angle in a standard rate turn (3°/s): banking angle graph

This curve seems to imply that one can fly a standard turn (or any other given turn rate) at any radius from 0 to infinity. Therefore I added two more curves to indicate the physical limits:

  • air speed (green) must be in the operating range of the plane
  • g load (blue) must not exceed the maximum acceptable for plane and occupants. (The correct scale for the g load is the purple scale on the left divided by 10)

And finally, the same plot but for a double rate turn (6°/s): enter image description here

Practical Relevance

Typical turn radii are in the order of 1000s of meters. Typical distance (wing span) of planes in a tight formation is in the order of a few 10 meters, that is a few percent of the turn radius.

If you compare the banking angles for two planes flying in tight line-abreast formation, the banking angle difference will have a similar relative difference, that is a few percent, or a fraction of 1° in absolute terms.

[I will calculate and insert an example here, when I have the time.]

For the practical purpose of actual formation flying, pilots will give the arcus tangens a break, and fly seemingly identical banking angles.

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  • $\begingroup$ I am still trying to figure out whether our answers are identical- $\endgroup$ – quiet flyer Jun 18 at 13:27
  • $\begingroup$ It looks like you started out talking about turn rate in terms of degrees per time but than you switched to radians per time-- is that correct? That's what was throwing me for a bit. Yes I see that they are the same now. $\endgroup$ – quiet flyer Jun 18 at 13:33
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    $\begingroup$ suggestion for next time: put the legend in a corner of the image with no lines (in this case, either top left or bottom right) took me a while to understand that "g load" in the first picture. $\endgroup$ – Federico Jun 18 at 13:40
  • $\begingroup$ Hmm-- I was thinking I was the one who didn't need to decide, P (period) was time for one circle. w is a rate and so needs angular units as well as time units. I think the answer could use some clarification when you make the switch, but anyway it is a good answer and I see the math is the same as mine. $\endgroup$ – quiet flyer Jun 18 at 13:43
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For all practical purposes, to maintain a line abreast formation the angle of bank and rate of turn will need to be constant, while airspeed will vary with aircraft on the outside of the turn needing to speed up.

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We're looking for long answers that provide some explanation and context. Don't just give a one-line answer; explain why your answer is right, ideally with citations. Answers that don't include explanations may be removed.

  • $\begingroup$ Why the downvotes? How many voting have actually trained and flown in a line abreast formation? I speak from years of real world practical military pilot experience: You have to match the lead aircraft's AOB whether line abreast or in echelon. If you bank differently than your flight lead you will either turn into him, or diverge away. That is an observable, repeatable fact that doesn't rely on any mathematical equation! $\endgroup$ – Michael Hall Jun 18 at 0:09
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    $\begingroup$ I am not a mathematician, I just fly airplanes. I understand equations, but I don't create them in an attempt to demonstrate knowledge of something I have no experience actually doing. $\endgroup$ – Michael Hall Jun 18 at 0:16
  • $\begingroup$ The point is well taken that in actual formation flight at high speed there will be no detectable difference in required bank angle. Consider however the situation of a hang glider and a fast sailplane thermalling together, trying to stay exactly opposite each other-- i.e. trying to fly at the same turn rate-- here the difference in bank angle will be noticeable. The question was meant as a thought experiment encompassing more than just high-speed aircraft flying in formation. $\endgroup$ – quiet flyer Jun 18 at 14:55
  • $\begingroup$ @quietflyer If they fly different radii, they might end up with same bank angle due different speeds. If they fly same radius, they can’t stay opposite each other at different speeds. The example is not obvious to me!? $\endgroup$ – Cpt Reynolds Jun 18 at 15:14
  • $\begingroup$ @CptReynolds -- flying at different speeds, the only way they will stay opposite is if the faster one uses a larger bank angle, but not so much more so that his radius shrinks to the same as the slower one. His radius must be greater than the slower one. If we take the difference in speeds as a given, it's easy to compute the required radius and bank angle from the equations provided so far. If we just start with a known difference in wings-level stall speed, the problem gets more complex but the faster one still ends up flying at a larger radius and bank angle if they are to stay oppsite. $\endgroup$ – quiet flyer Jun 18 at 15:32
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As @quiet flyer has strong experience in gliders, the question seems to focus on "how do we stay in a thermal with limited radius if my flying speed is different than yours?"

As flight is 3 dimensional, there is a vertical option.

But for the sake of discussion (maybe for a photo shoot) they both need to hold the same altitude and remain at opposite points of a circle.

In order to have the same rate of turn, the faster must have the outside track. Since radius is proportional to the SQUARE of speed, the radius of a 60 mph glider will be more than double that of a 40 mph glider if they use the same bank angle. Note the difference in radii means they do not have to stay opposite in the thermal!

According to the radius of turn formula R = V^2/11.26×tanbankangle one can decrease their turn radius by increasing bank angle. The rub is that higher AOA or greater speed is needed at a higher bank angle to generate adequate vertical lift.

Decreasing turn radius by increasing bank angle and AOA will allow one to "catch up" with the other. ROT=1091xtanbankangle/airspeed. Decreasing bank angle and AOA has opposite effect.

But notice radii formula has V^2 and ROT formula does not square speed. The distance travelled in the different size circles (pi x D) is linear with speed for equal rate of turn.

Radius at constant bank angle is dependent on the SQUARE of speed, whereas ROT is linear with speed. What to do?

  1. Outside plane increases bank angle and AOA to increase ROT.
  2. Inside plane decreases bank angle and AOA to decrease ROT.

As we can see what actually happens is highly dependent on the flight envelope of each aircraft. The inside one is safer by lowering bank angle and AOA. But if speed difference is too much, they cannot remain in circular "formation".

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  • $\begingroup$ For keeping a good visual lookout it is desirable that two aircraft thermalling together stay basically opposite each other. Taking the basic case where one has a significantly faster stall speed than the other and both desire to circle just above stall speed (or more precisely just below stall angle-of-attack) to maximize climb rate, in practice the faster one will need more bank angle to stay opposite the other one, and the faster one will also fly at a greater radius. I really didn't mean to throw in such a wrench in the gears by not mentioning the glider situation sooner! $\endgroup$ – quiet flyer Jun 19 at 4:38
  • $\begingroup$ It all seems a variant of the original question to me, but I can see why all might not agree. $\endgroup$ – quiet flyer Jun 19 at 4:40
  • $\begingroup$ In actual practice while thermalling you decrease your bank angle if you wish to fall back relative to the other glider and increase it if you wish to move forward. $\endgroup$ – quiet flyer Jun 19 at 4:41
  • $\begingroup$ "they both need to hold the same altitude and remain at opposite points of a circle." That would put them at equal radii, though, which is at odds with the following paragraph. I think. The whole "they do not have to stay opposite in thermal", besides being much improved without caps, is a bit unclear. $\endgroup$ – AEhere Jun 19 at 10:51
  • $\begingroup$ Please see edits to answer. $\endgroup$ – Robert DiGiovanni Jun 19 at 12:05
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Using the formulae $F_c = m * \frac{v^2}{r}$ and bank = arctan (Fc / (mg)), we can derive the formula bank = arctan (4*pi^2 *r / (P^2 * g)), where Fc is centripetal force, g is the gravitational constant, and P is the period or the time for one complete turn. The aircraft flying at a larger turn radius must use a larger bank angle than the aircraft flying at a smaller turn radius.

Note that mass or weight does not appear as a variable in the final equations.

Read on only if you want to see the math in more detail:

v=d/t = 2*pi*r/P

bank angle= arctan(horizontal force/ vertical force)

=arctan (centripetal force/weight)

=arctan ((mv^2/r)/(mg))

=arctan ((v^2)/(rmg))

=arctan ((2*pi*r/P)^2/(rg))

=arctan (4*pi^2*r/(g(P^2)))

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  • $\begingroup$ See also related question aviation.stackexchange.com/questions/65617/… $\endgroup$ – quiet flyer Jun 17 at 19:55
  • $\begingroup$ Yes, if... What variable defines the speed? $\endgroup$ – Michael Hall Jun 18 at 0:13
  • $\begingroup$ Not sure I understand the last question. It is easy to see the relationship between v (speed), r (radius), and P (period, or time for one circle.) $\endgroup$ – quiet flyer Jun 18 at 0:33
  • $\begingroup$ Lots of downvotes-- should I modify to show more of the math? $\endgroup$ – quiet flyer Jun 18 at 11:40
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    $\begingroup$ @Koyovis it's mathJax (guess it was an autocorrect-induced typo) $\endgroup$ – Federico Jun 18 at 13:37

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