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Aerospace Systems

, Volume 2, Issue 2, pp 83–95 | Cite as

Increasing relative angular velocity for air combat in zero gravity

  • Bill DengEmail author
  • Timothy Collier
Original Paper
  • 232 Downloads

Abstract

Angular velocity plays a critical role in determining the outcome of a close-range aerial engagement between two identical fighter aircraft pitching at full deflection. In a zero gravity environment, a pursuer may exploit its ability to roll to increase its relative angular velocity against a pitching opponent. In this paper, we present a repeatable maneuver for an unmanned fighter aircraft that increases its relative angular velocity. Additionally, we provide maneuvers for aligning an aircraft’s trajectory with a desired target trajectory.

Keywords

Flight control Air combat Trajectory planning Unmanned systems 

List of symbols

t

Time (s)

p

The roll rate of our aircraft (rad/s)

q

The pitch rate of our aircraft (rad/s)

y

The amount of pitch rate we give up (instantaneously) at the start of our advantage maneuver (rad/s)

m

Mass of our aircraft (kg)

v

Speed of our aircraft (m/s)

\(v_\mathrm{f}\)

Speed of air hitting the wing in the normal direction, caused by rolling (m/s)

r

Radius of our turn circle (m)

\(F_\mathrm{c}\)

Centripetal force (lift) acting on our aircraft (\(\hbox {kg}\,\hbox {ms}^{-2}\))

I

Moment of inertia of our aircraft about the roll axis (\(\hbox {kg}\,\hbox {m}^2\))

\(\tau \)

Torque generated from wings (N m/rad)

W

Wingspan of our aircraft (m)

\(W_\mathrm{d}\)

Width of a wing. Used to approximate drag (m)

\(C_\mathrm{d}\)

Drag coefficient of our aircraft about the roll axis (dimensionless)

\(C_{\tau }\)

Drag coefficient of torque drag from rolling the aircraft (dimensionless)

\(\rho \)

Air density (\(\hbox {kg}\,\hbox {m}^{-3}\))

1 Introduction

As the capabilities of autonomous flight control systems become more advanced, studies of optimal strategies and policies that unmanned aircraft can implement in air combat become increasingly relevant. In particular, an aerial engagement between two unmanned fighter aircraft may involve extended periods of pitching at full deflection, since considerations for g-force tolerance are limited to only those for the physical integrity of the aircraft.

We are interested in the scenario where we have two unmanned fighter aircraft engaged in a turn fight. If the two aircraft are identical, then the aircraft may continue turning indefinitely (disregarding fuel). This would occur if both aircraft are pitching at the maximum rate, without using roll or yaw. Conventionally, the attacker may perform a high Yo-Yo or a low Yo-Yo maneuver to break this stalemate situation. These energy maneuvers are known to be effective [1], as they exploit the trade-offs between altitude and airspeed. In this paper, we investigate the possibility of exploiting the geometry of the flight paths instead, provided that we disregard gravity. We show that this is indeed possible. Future work will include the development of methods needed to augment the maneuvers presented in this paper, to accommodate for gravity.

In our investigation, we will assume that the enemy aircraft maintains a defensive policy of continuing to pitch at the maximum rate. This is the same assumption made when an attacker performs a high/low Yo-Yo maneuver. We note that under this assumption, it is possible for the attacker to simply fly away from the defender, and fly back again to achieve a firing solution. It is reasonable to assume that a defender will be able to detect such an extreme increase in distance between the two aircraft, and capitalize on it to either escape, or reset the dogfight to a neutral scenario. Therefore, we will aim to develop a maneuver that remains close to the defender. In particular, our maneuver will maintain the maximum pitch rate. We will also control the roll rate, so that the lateral separation of the turn circles, as well as the angle off tail (AOT), is reduced. This increases the likelihood of being able to remain in pursuit of the defender, in the eventuality that they break away from the turn circle (see Definition 2.1). Thus, our maneuver will only affect the roll rate of the attacker. Additionally, our maneuver will be repeatable, in the sense that the attacker will return to the same turn circle, so that in effect their relative angular velocity will have been increased. In other words, our maneuver will have essentially allowed the attacker to have turned faster than the defender; i.e., the repeatability property of the maneuver will allow the attacker to repeat the maneuver until a firing solution has been achieved.

We note that the ability to roll is of critical importance to our maneuver; in a two-dimensional scenario, we would not be able to achieve the lateral separation required without a reduction in pitch rate. At this point, it might be conjectured that there exists a certain threshold for the roll rate; i.e., the attacker might be required to roll at a rate above a certain threshold in order to compensate for the fact that it is now pitching at an oblique angle to the defender’s turn circle. We show that in fact our maneuver works for any positive roll rate in our kinematic model. However, when the maximum lift constraint is considered, we show that there is indeed a threshold for which the roll rate must exceed in order for the relative angular velocity to be increased.

Theoretical models of pursuit and evasion such as those presented in [2, 3, 4] have not provided an effective solution to the problem of three-dimensional aerial combat, as their large computational time complexities make real-time decision-making unfeasible. In [5], a reduced-order mathematical model is used to determine the outcome of an aerial engagement in two-dimensional space, where the aircraft have asymmetric performance characteristics, and are assumed to turn optimally. The work [6] also uses a two-dimensional model where optimal play is assumed, with the aim of classifying roles of two aircraft given initial starting conditions. In contrast, we do not aim to develop a general solution, but rather a possible maneuver that could be performed in three-dimensional space, assuming that the defender does not turn optimally. Furthermore, the roles of the aircraft will have already been determined in our scenario. Additionally, in our scenario, the aircraft will have identical performance characteristics, as we are interested in whether flight geometry alone is sufficient for the attacker to gain advantage.

Approaches using artificial intelligence and numerical optimization have yielded promising results for the problem of three-dimensional aerial combat. The work [7] uses rule-based systems and a decision tree to create an agent that can recognize various situations and choose suitable maneuvers. The agent in [8] uses fuzzy logic and genetic algorithms to control an aircraft in a simulation. The work [9] models the problem as a two-player (two fighters), zero-sum, sequential-interaction game. At each time step, there are 14 possible maneuvers that a fighter may choose from, and the minimax decision rule is applied to determine which maneuver is executed. The maneuvers are derived from a guidance law, which generates lead, pure, and lag pursuit together with climb and dive maneuvers. The heuristic function takes into account the angles between the velocity vector and the line of sight, distance between the aircraft, and the optimal distance for a firing solution. In contrast to these approaches, in our paper, we are interested in the development of the maneuvers themselves, rather than the higher-level strategy in an aerial engagement. We do not use any explicit heuristic functions; we instead plan out the entire trajectory of the maneuver based on initial conditions. Our approach will also avoid artificial intelligence techniques such as neural networks, genetic algorithms, and decision trees, as we desire to elucidate the geometric mechanisms that give rise to our maneuvers.

The work [10] applied approximate dynamic programming to make real-time decisions in air combat. This involved the use of a heuristic scoring function. The approach was tested against several different maneuvering policies. These policies determined if the defender would attempt pure pursuit or maintain maximum pitch rate. The methods used were shown to be effective in a two-dimensional setting, and the results were obtained under experimental conditions using remote-controlled aircraft. However, as level flight was assumed, this method does not utilize the ability to roll to perform more effective maneuvers. Furthermore, the performance characteristics of the attacker were superior to those of the defender; the attacker could turn at a higher rate than the defender. In contrast, we investigate the scenario where the attacker and defender have the same turn rate.

The work [11] approaches the aerial combat problem by developing a guidance law. This guidance law uses virtual pursuit points to guide the attacker along lead, pure, and lag pursuit curves. The guidance law analyses variables such as velocity, turn performance, energy-maneuverability, aspect angle, and distance to the defender to determine where to place the virtual pursuit point. This is a point in three-dimensional space near the defender which the attacker follows using pure pursuit guidance. Depending on the placement of this virtual pursuit point, the resulting trajectories could result in pure, lead, or lag pursuit curves, or basic fighter maneuvers (BFMs). This method was shown to be effective in simulating BFMs such as high Yo-Yos and low Yo-Yos. In contrast, we develop maneuvers which follow an explicit path based on initial conditions, rather than a guidance law which continuously adjusts the trajectory based on observed changes in variables. Future work will include incorporating our maneuvers into more general guidance laws.

Dubins paths were also considered for this problem. In a two-dimensional setting, for a vehicle with a constant forward speed and a fixed turn rate, the minimum-distance path between two configurations is called a Dubins path. The three-dimensional case is more complicated. The work [12] describes a three-dimensional Dubins airplane model, and presents a method to obtain a target orientation. This was expanded upon in the work [13]. A kinematic model with trigonometric equations was used to describe the Dubins paths for the airplane. This extended the two-dimensional Dubins vehicle by adding an altitude component to the coordinate vector. Using this model, the Dubins paths were determined by finding the intersection of two two-dimensional manifolds. The resulting Dubins paths consisted of helical paths and straight line segments. This method was shown to be effective in a simulation. However, these methods were not applied to an aerial combat scenario. In our case, Dubins paths are not applicable. In particular, in our scenario the target orientation is constantly changing, as the defender will be a moving target. Furthermore, a Dubins path does not take into account our desire to remain close to the turn circle.

We will approach the problem heuristically. The main benefit of our approach is that it exposes the underlying geometric intuition, which allows for the possibility of developing more generalized maneuvers. Our approach also has a longer planning horizon than time-discretized optimization methods, as it is able to operate based on more information from the maneuvering policy.

In this paper, we will use a kinematic model to describe the combat scenario. We assume that the trajectories of the turning aircraft form circles, which allows us to efficiently represent them in terms of trigonometric functions. This also gives us symmetric differential equations that have solutions with elegant properties which will be used to derive our maneuver. Our kinematic model is derived from the analysis of the trajectory of an aircraft under the forces of lift, drag, and thrust, without gravity. We then utilize the symmetric properties of our model to construct a maneuver that an aircraft can perform to gain advantage. This maneuver is composed of spiral segments that result from the application of a continuous roll whilst pitching. Additionally, we also consider how the maximum lift constraint affects the roll rate in our maneuver. We then further develop our theory to include methods for manipulating arbitrary initial trajectories into a scenario where our maneuver can be performed. Finally, we extend our results to account for situations in which the aircraft are yawing.

For simplicity, we set the radius of the turn circle to 1 unit in our kinematic model, where a unit is an arbitrary distance. As a result, our figures will have dimensionless axes, since the results will scale accordingly for any given speed and turn radius.

2 Kinematic model and assumptions

In this section, we introduce our model and assumptions. We shall disregard gravity and other physical variables, and approach the problem from a kinematic standpoint. Although gravity is an important practical consideration, this paper is primarily interested in exploring the theoretical possibilities that arise purely from the geometry of flight paths. In our model, we have two identical fighter aircraft. They are each represented by a position and orientation in three-dimensional Euclidean space. Their forward speeds are equal and will always remain constant. The initial positions and orientations of the aircraft are such that one has a distinct advantage over the other; it is pursuing the other and wishes to increase its advantage. In our discussion, we assume the perspective of the pursuer. This will typically be referred to as our aircraft. Consequently, the other aircraft will be referred to as the enemy aircraft.

In addition to their forward speeds being constant, both aircraft will always be pitching upwards at a constant rate. In our model, distance is measured in units and time is measured in seconds. For simplicity, we set the speed to one unit per second, and the pitch rate q to one radian per second. The variable t will be used to represent time.

Given our initial conditions, it is clear that the trajectories of the aircraft will each be a circle. We will be using the idea of turn circles extensively in the following sections. Turn circles have been used in other works such as [14, 15, 16] in order to calculate various maneuvers in unmanned systems.

Definition 2.1

A turn circle is the locus of an aircraft pitching upwards at a constant rate and moving with constant forward speed. A parameterized turn circle is a function \(C:\mathbb {R} \rightarrow \mathbb {R}^3\) such that C(t) is the position of the aircraft at time t. The polarity of an aircraft is the direction it is traveling with respect to its turn circle; i.e., either clockwise or anticlockwise.

We can now specify our initial conditions as follows: the turn circles of our aircraft and the enemy aircraft are initially identical, with the position of our aircraft being slightly behind the enemy aircraft.

Without loss of generality, we can set the parameterized turn circles of the enemy aircraft and our aircraft, respectively, to \(\left( \cos {(t+\gamma )},\sin {(t+\gamma )},0\right) \) and \(\left( \cos {(t)},\sin {(t)},0\right) \), for some small \(\gamma > 0\).

We seek a maneuver that brings our aircraft closer to the enemy aircraft. We will impose the additional restriction that upon completion of the maneuver, our aircraft must have the same turn circle as it did previously, with the same polarity. Additionally, we shall only permit adjustments of roll rate in the execution of the maneuver. We will also limit the maximum roll rate to M where \(M>0\).

Observe that since we are required to return to our original turn circle, we can measure the advantage we have gained at the end of the maneuver in terms of the parameterized turns circles of our aircraft and the enemy aircraft.

The vectors that describe the orientation of our aircraft are shown in Fig. 1. We will be using the Frenet–Serret frame. The unit vector \(\mathbf {T}(t)\) is the velocity at time t, \(\mathbf {N}(t)\) is the normal unit vector, and \(\mathbf {B}(t)\) is the binormal unit vector. We also introduce the vector \(\mathbf {k} = p \mathbf {T} + q \mathbf {B}\), which will play an important role in the following sections, as well as the vector \(\mathbf {N} = \mathbf {B}\times \mathbf {T}\), which will be used when we extend our results to include yaw.
Fig. 1

Vectors describing the orientation of our aircraft. \(\mathbf {T}\) is the forward velocity, \(\mathbf {B}\) is the direction outwards from the aircraft to the right, parallel to the pitch axis of the aircraft. Note that \(\mathbf {N} = \mathbf {B}\times \mathbf {T}\) and \(\mathbf {k} = p \mathbf {T} + q \mathbf {B}\)

Definition 2.2

A roll is a rotation of \(\mathbf {B}\) about \(\mathbf {T}\). An advantage maneuver (or simply maneuver) is a function \(f:[0,T] \rightarrow [-M,M]\), where \(T>0\) such that if our aircraft rolls at f(t) radians per second at time t, then the turn circle of our aircraft at \(t = T\) is identical to the turn circle of our aircraft at \(t = 0\).

Definition 2.3

If the parameterized turn circle of our aircraft before applying an advantage maneuver is \(\left( \cos {(t)},\sin {(t)},0\right) \), and afterwards is \(\left( \cos {(t+x)},\sin {(t+x)},0\right) \) for some \(0 \le x < 2\pi \), then we refer to the value x as the advantage or advantage gained from the maneuver.

3 Instantaneous roll example

In this section, we will consider how instantaneous rolls affect the movement of our aircraft to illustrate the geometric mechanism behind the advantage maneuver introduced in the following section.

Let our aircraft’s parameterized turn circle be
$$\begin{aligned} C_0(t) = \left( \cos (t), \sin (t), 0 \right) \end{aligned}$$
(1)
and let the components of \(C_0\) be denoted by \(C_{0,i}\), i.e.
$$\begin{aligned} C_{0,1} = \cos (t), C_{0,2} = \sin (t),C_{0,3} = 0. \end{aligned}$$
(2)

Definition 3.1

An instantaneous roll \(R_{t_1,\mu }\) is a transformation that rotates a parameterized turn circle around its tangent vector at a particular time \(t_1\) by \(\mu \) radians. In particular, \(\mu > 0\) corresponds to an aircraft rolling right.

Let \(W(\mathbf {T},\mu )\) be the Rodrigues rotation matrix around a unit vector \(\mathbf {T}\) by an angle of \(\mu \). Then
$$\begin{aligned} \begin{aligned} C_1&= R_{t_1,\mu }(C_0) \\&= W\left( C_0'(t_1) ,\mu \right) (C_0(t)-C_0(t_1))+C_0(t_1) \end{aligned} \end{aligned}$$
(3)
is the parameterized turn circle of our aircraft after applying an instantaneous roll of \(\mu \) radians at \(t_1\).

Definition 3.2

Let \(C_0\) be a parameterized turn circle. An instantaneous advantage maneuver (or simply instant maneuver) is a sequence of instantaneous rolls \(R_{t_0,{\mu }_0},R_{t_1,{\mu }_1},\ldots ,R_{t_n,{\mu }_n}\) which combined give an advantage gained of \(x \ge 0.\) In particular, if \(C_{n+1} = R_{t_n,{\mu }_n}(R_{t_{n-1},{\mu }_{n-1}}(\ldots R_{t_0,{\mu }_0}(C_0)\ldots )\), then \(C_{n+1} (t) = C_0(t+x)\). We extend the definition of advantage gained to include instant maneuvers.

We now give an example of an instantaneous advantage maneuver \(R_{t_0,{\mu }_0},R_{t_1,{\mu }_1},\ldots ,R_{t_5,{\mu }_5}\).

Let
$$\begin{aligned} \begin{aligned} C_1&= R_{t_0,{\mu }_0}(C_0) = R_{0,\frac{\pi }{8}}(C_0),\\ C_2&= R_{t_1,{\mu }_1}(C_1) = R_{\frac{\pi }{2},\frac{\pi }{8}}(C_1). \end{aligned} \end{aligned}$$
(4)
Solving \(C_{2,3} '(t_{\beta }) = 0\), we obtain two solutions for \(t_{\beta }\). We will choose
$$\begin{aligned} t_{\beta } =\pi +\arctan \left( \cos \frac{\pi }{8}\right) \end{aligned}$$
(5)
for reasons that will soon become clear. Let
$$\begin{aligned} \begin{aligned} \delta&= \arctan \frac{7-4\sqrt{2}}{2\sqrt{52+14\sqrt{2}}},\\ {\mu }_{\beta }&= \arctan \sqrt{23-16 \sqrt{2}},\\ C_3&=R_{t_2,{\mu }_2}(C_2) = R_{t_{\beta },{\mu }_{\beta }}(C_2),\\ C_4&=R_{t_3,{\mu }_3}(C_3) = R_{3\pi -x,-\frac{\pi }{8}}(C_3),\\ C_5&=R_{t_4,{\mu }_4}(C_4)= R_{3\pi -x+\frac{\pi }{2},-\frac{\pi }{8}}(C_4),\\ C_6&=R_{t_5,{\mu }_5}(C_5) = R_{3\pi -x+t_{\beta },-{\mu }_{\beta }}(C_5). \end{aligned} \end{aligned}$$
(6)
This instant maneuver describes an aircraft initially rolling right by an angle of \(\frac{\pi }{8}\), then \(\frac{\pi }{8}\) again. The next roll is chosen such that it occurs either at the maximum (with respect to the z-axis; i.e., \(C_{2,3}\), the third component of \(C_2\)) or minimum point of \(C_2\). In this example, we have chosen the maximum point \(C_2(t_{\beta })\). This then allowed us to choose \({\mu }_{\beta }\) so that \(C_{3,3}\) is constant; i.e., our aircraft is now moving horizontally in the image of \({C_3}\), as it did in \(C_0\). Hence by symmetry, we can choose appropriate values for \(t_3,t_4,t_5\) and apply the instantaneous rolls \(R_{t_3,-\frac{\pi }{8}},R_{t_4,-\frac{\pi }{8}},R_{t_5,-{\mu }_{\beta }}\) to bring our aircraft back to its original turn circle; i.e. \(C_6(t) = C_0(t+x)\).
In this case, our advantage gained is given by
$$\begin{aligned} x = 2\delta . \end{aligned}$$
(7)
Thus, through a sequence of instantaneous rolls, our aircraft has effectively traveled with higher angular velocity than it otherwise would have if it remained in its original turn circle \(C_0\) (Fig. 2).
Fig. 2

The curve connecting \(C_0(0)\) (red point) and \(C_2(t_{\beta })\) (blue point) is the trajectory of our aircraft from \(t=0\) to \(t=t_{\beta }\) during the instant maneuver. The circles \(C_0\) and \(C_3\) are shown in blue and purple, respectively. The circular arcs of \(C_1\) and \(C_2\) are shown in green and red, respectively (color figure online)

3.1 Explanation of advantage gained

We now give a geometric explanation for the origin of this advantage gained. Let the midpoint of \(C_3\) be given by
$$\begin{aligned} Q(C_3) = \frac{C_3(0)+C_3(\pi )}{2}. \end{aligned}$$
(8)
Now let
$$\begin{aligned} (v_1,v_2,0) = C_3(0)-Q(C_3). \end{aligned}$$
(9)
The value of \(\delta \) was computed by taking
$$\begin{aligned} \delta = \arctan \left( \frac{v_2}{v_1}\right) . \end{aligned}$$
(10)
This is because \(C_3(t)-Q(C_3) = C_0(t+\delta )\); i.e., the time parameter of \(C_3\) has shifted by \(\delta \). This implies that our aircraft has gained advantage. We now explain intuitively the cause of this shift. This shift is caused by the second roll \(R_{\frac{\pi }{2},\frac{\pi }{8}}\). At time \(t = \frac{\pi }{2}\), our aircraft is in the circle \(C_1\). Observe that the circle \(C_1\) is oblique to \(C_0\). This implies that the velocity vector \(C_1'(\frac{\pi }{2})\) does not lie in the horizontal plane. Hence a rotation of the circle \(C_1\) about the vector \(C_1'(\frac{\pi }{2})\) will also rotate \(C_1\) by some amount about the vertical axis (0, 0, 1). After adjusting for positional offsets, we find that this amount is \(\delta .\)
Next, our aircraft reaches the highest point on \(C_2\) at \(t_{\beta }\). At this point, \(C_{2,3} '(t_{\beta }) = 0\); i.e., the velocity vector lies in the horizontal plane. Thus, after applying a rotation of \({\mu }_{\beta }\) to \(C_2\) we obtain a horizontal circle \(C_3\), where
$$\begin{aligned} C_3(t)= C_0(t+\delta )+Q(C_3). \end{aligned}$$
(11)
(A turn circle is horizontal if the plane containing it is parallel to the plane spanned by \(\{\left( 1,0,0 \right) , \left( 0,1,0\right) \}\).)
By symmetry, \(R_{t_4,{\mu }_4}\) also shifts t by \(\delta \). Then, the end result is that the advantage gained is
$$\begin{aligned} x = 2\delta . \end{aligned}$$
(12)

4 Continuous roll maneuver

In this section, we will derive an advantage maneuver comprised of continuous rolls. This derivation follows from the natural inclination to integrate the discrete sequences of instantaneous rolls discussed above. In order to do this, we will first derive equations that give the precise position and orientation of our aircraft over time if we apply a constant roll rate p to our aircraft.

Let q be the pitch rate of our aircraft. Suppose we now also apply a constant roll rate to our aircraft. Then, let p be the roll rate.

If we consider an infinitesimal change in time dt, then the changes in \(\mathbf {T}\) and \(\mathbf {B}\) are given by
$$\begin{aligned} \begin{aligned} {\text {d}}\mathbf {T}&= q \mathbf {B} \times \mathbf {T} {\text {d}}t \implies \frac{{\text {d}}\mathbf {T}}{{\text {d}}t} = q \mathbf {B} \times \mathbf {T},\\ {\text {d}}\mathbf {B}&= p \mathbf {T} \times \mathbf {B} {\text {d}}t \implies \frac{{\text {d}}\mathbf {B}}{{\text {d}}t} = p \mathbf {T} \times \mathbf {B}. \end{aligned} \end{aligned}$$
(13)
Observe that
$$\begin{aligned} \frac{{\text {d}}\mathbf {T}}{{\text {d}}t} = -\frac{q}{p} \frac{{\text {d}}\mathbf {B}}{{\text {d}}t} \text { and } \frac{{\text {d}}\mathbf {B}}{{\text {d}}t} = -\frac{p}{q} \frac{{\text {d}}\mathbf {T}}{{\text {d}}t}. \end{aligned}$$
(14)
Integrating (see Sect. 8.1.1) both sides with respect to t and solving for the constant of integration, we obtain
$$\begin{aligned} \frac{{\text {d}}\mathbf {T}}{{\text {d}}t}= & {} \mathbf {k}\times \mathbf {T}, \end{aligned}$$
(15)
$$\begin{aligned} \frac{{\text {d}}\mathbf {B}}{{\text {d}}t}= & {} \mathbf {k}\times \mathbf {B}, \end{aligned}$$
(16)
where
$$\begin{aligned} \mathbf {k} = p \mathbf {T}(0) + q \mathbf {B}(0). \end{aligned}$$
We will refer to \(\mathbf {k}\) as the orientation vector. We solve equation (15) to obtain
$$\begin{aligned} \mathbf {T}(t)= & {} \frac{\mathbf {k}\cdot \mathbf {T}(0)}{|\mathbf {k}|^2}\mathbf {k}+\left( \mathbf {T}(0)-\mathbf {k}\frac{\mathbf {k}\cdot \mathbf {T}(0)}{|\mathbf {k}|^2}\right) \cos \left( |\mathbf {k}|\,t\right) \nonumber \\&+\frac{\mathbf {k}}{|\mathbf {k}|}\times \mathbf {T}(0)\sin \left( |\mathbf {k}|\,t\right) . \end{aligned}$$
(17)
Note that \(\mathbf {T}\) moves at constant speed in a circle around \(\mathbf {k}\) with period \(\frac{2\pi }{\left| \mathbf {k}\right| }\). Integrating \(\mathbf {T}\) over time yields a spiral in the direction of \(\mathbf {k}\) (or \(-\mathbf {k}\) if \(p < 0\)) with period \(\frac{2\pi }{\left| \mathbf {k}\right| }\). This implies that our aircraft resets to the same orientation it had when it started rolling every \(\frac{2\pi }{\left| \mathbf {k}\right| }\) seconds. This means that at the points of reset, our aircraft is in a parallel turn circle, and furthermore that it is in the same position on the turn circle as it was when we first started rolling. (Two turn circles are said to be parallel if they are identical under translation.) Hence, by symmetry, if we stop rolling at one of these points of reset and continue along the parallel turn circle for exactly half its period, then start rolling at the same rate we initially did but in the opposite direction, we will return to our original turn circle, opposite the point where we first left it. Thus, our advantage maneuver is given by
$$\begin{aligned} f_p (t) = {\left\{ \begin{array}{ll} p &{} \text {if } 0 \le t< \frac{2\pi }{\left| \mathbf {k}\right| } \\ 0 &{} \text {if } \frac{2\pi }{\left| \mathbf {k}\right| } \le t< \frac{2\pi }{\left| \mathbf {k}\right| } + \frac{\pi }{q} \\ -p &{} \text {if } \frac{2\pi }{\left| \mathbf {k}\right| } + \frac{\pi }{q}\le t\le \frac{4\pi }{\left| \mathbf {k}\right| } + \frac{\pi }{q} \\ \end{array}\right. }. \end{aligned}$$
(18)
We will often refer to the three intervals of f and their images by their geometric forms. For example, we may call the interval \([0,\frac{2\pi }{\left| \mathbf {k}\right| }]\) the initial spiral or initial spiral period.
Geometrically, our maneuver is a spiral, followed by a semicircle, followed by a spiral. Thus, we have traveled 2.5 revolutions with respect to our original turn circle. However, the enemy aircraft has only traveled \(\frac{4\pi q}{\left| \mathbf {k}\right| } + \pi \) radians around the turn circle. Subtracting this amount from \(5\pi \) and simplifying, we find that the advantage gained is
$$\begin{aligned} 4\pi \left( 1-\frac{q}{\sqrt{q^2+p^2}}\right) . \end{aligned}$$
(19)

4.1 Example of an advantage maneuver

Let
$$\begin{aligned} C(t) = \left( \cos (t-\pi /2),\sin (t-\pi /2)+1,0\right) \end{aligned}$$
(20)
be our aircraft’s parameterized turn circle and \(f_{p_0}\) be our advantage maneuver where \(p_0 = 0.15\).
Then, the advantage gained is
$$\begin{aligned} 4\pi \left( 1-\frac{1}{\sqrt{1^2+0.15^2}}\right) \approx 0.14 \end{aligned}$$
(21)
and the trajectory of our aircraft during the maneuver is shown in Fig. 3.
Fig. 3

The curve connecting C(0) (red point) and \(C(5\pi )\) (blue point) is the trajectory of our aircraft from \(t=0\) to \(t=\frac{4\pi }{\left| \mathbf {k}\right| } + \frac{\pi }{q}\) during the maneuver. The initial spiral is shown in green, the final spiral is shown in red. The semicircle where \(f_{p_0}(t)=0\) is shown in purple. The purple point is \(C\left( \frac{4\pi }{\left| \mathbf {k}\right| } + \frac{\pi }{q}\right) \); i.e. the position our aircraft would have been at if it stayed in C. \(\mathbf {B}(0)\) is the orange vector, \(\mathbf {T}(0)\) is the blue vector. The dashed circle is our original turn circle C (color figure online)

4.2 Displacement of the advantage maneuver

We can measure the displacement \(\mathbf {h}\) of the advantage maneuver by the distance between the original turn circle and the turn circle of our aircraft when \(f(t) = 0\).

In the \(\{\mathbf {T},\mathbf {B}\}\) basis, \(\mathbf {k} = \left( p, q \right) .\) The rate of displacement of our aircraft’s turn circle is given by the projection of \(\mathbf {T}\) onto \(\mathbf {k}\), i.e.
$$\begin{aligned} \frac{\mathbf {T} \cdot \mathbf {k}}{\mathbf {k}\cdot \mathbf {k}} \mathbf {k}. \end{aligned}$$
(22)
Since our maneuver begins with a spiral of exactly one period, our displacement is given by multiplying the rate by the period \(\frac{2\pi }{\left| \mathbf {k}\right| }\). This simplifies to the following expression:
$$\begin{aligned} \mathbf {h} = \left( h_1(p), h_2(p) \right) = \Biggl ( \frac{2\pi p^2}{(p^2+q^2)^\frac{3}{2}} ,\frac{2\pi pq}{(p^2+q^2)^\frac{3}{2}} \Biggr ), \end{aligned}$$
(23)
which has magnitude \( \frac{2\pi p}{p^2+q^2}\).

5 Consideration of the maximum lift constraint

A potential concern about this maneuver is that it violates the maximum lift constraint that dictates the minimum turn radius in horizontal flight. The practical solution to this involves reducing the pitch rate of our aircraft by a small amount, which will enable our aircraft to roll, and thus perform the maneuver. In this section, we show that the maneuver will still result in advantage being gained, provided that our aircraft reaches a certain roll rate.

Given that our aircraft is turning at its maximum pitch rate q in zero gravity, if we perform an advantage maneuver with roll rate p, we gain advantage equal to
$$\begin{aligned} 4\pi \biggl ( 1-\frac{q}{\sqrt{p^2+q^2}}\biggr ). \end{aligned}$$
(24)
If we give up y in pitch rate during the spiral parts of our maneuver (i.e., the parts where we are rolling), then the advantage gained is
$$\begin{aligned} 4\pi \biggl ( 1-\frac{q}{\sqrt{p^2+(q-y)^2}}\biggr ). \end{aligned}$$
(25)
This means that we require
$$\begin{aligned} p > \sqrt{2qy-y^2} \end{aligned}$$
(26)
in order for any advantage to be gained.

5.1 Calculation of p(t)

Suppose now that p is not instantaneously applied, but rather calculated from the force we gain by reducing our pitch rate by y. Firstly, we work out this force. We have
$$\begin{aligned} F_\mathrm{c} = mvq \end{aligned}$$
(27)
from the centripetal force formula. If we reduce q by y, we see that the force is reduced by
$$\begin{aligned} mvy. \end{aligned}$$
(28)
We assume that our forward speed is kept constant. We then assume that the force mvy is evenly distributed across both wings, through aileron deflection in opposite directions. That is, on the port wing, we apply an upwards (relative to the aircraft) force of \(\frac{mvy}{2}\), and on the starboard wing we apply a downwards force of \(\frac{mvy}{2}.\) This is assuming that the aircraft is able to convert the force with 100% efficiency, and that the force is evenly distributed across both wings. In reality, since the wing shape is not uniform, the distribution of force will also not be uniform, and some force will also be lost. For one wing, the torque generated is given by
$$\begin{aligned}&\lim _{n \rightarrow \infty } \sum _{i=1}^n\biggl ( \frac{1}{n}\cdot \frac{mvy}{2}\biggr )\biggl ( \frac{i}{n}\cdot \frac{W}{2}\biggr ) \end{aligned}$$
(29)
$$\begin{aligned}&=\frac{Wmvy}{8}, \end{aligned}$$
(30)
using the formula for torque. Then, the total torque is given by
$$\begin{aligned} \tau = \frac{Wmvy}{4}. \end{aligned}$$
(31)
So if we start the maneuver at \(t=0\), then \(p(0) = 0\), and
$$\begin{aligned} p'(t) = \frac{\tau }{I}, \end{aligned}$$
(32)
using the formula for angular acceleration, given torque and moment of inertia. However, we now need to account for the angular drag experienced by the rolling aircraft. This drag slows down our roll rate; i.e., it reduces p. We will approximate angular drag by integrating the linear drag equation over all points. At a point on the wing that is x meters away from the roll axis, the drag experienced by that point is given by
$$\begin{aligned} \begin{aligned} \frac{1}{2} C_\mathrm{d} v_\mathrm{f}^2 \rho = \frac{1}{2} C_\mathrm{d} (xp(t))^2 \rho = \frac{1}{2} C_\mathrm{d} \rho (p(t))^2 x^2. \end{aligned} \end{aligned}$$
(33)
So, the torque at this point is given by
$$\begin{aligned} = \frac{1}{2} C_\mathrm{d} \rho (p(t))^2 x^3. \end{aligned}$$
(34)
Then, the torque drag on one wing is given by
$$\begin{aligned} \frac{1}{2} W_\mathrm{d} C_\mathrm{d} \rho (p(t))^2 \int _0^{W/2}x^3 dx = \frac{1}{2} W_\mathrm{d} C_\mathrm{d} \rho (p(t))^2 \frac{W}{64}. \end{aligned}$$
(35)
So, the total torque drag is
$$\begin{aligned} W_\mathrm{d} C_\mathrm{d} \rho \frac{W}{64}(p(t))^2. \end{aligned}$$
(36)
We let
$$\begin{aligned} C_{\tau } = W_\mathrm{d} C_\mathrm{d} \rho \frac{W}{64}, \end{aligned}$$
(37)
so, the reduction in angular acceleration is given by
$$\begin{aligned} \frac{C_{\tau }}{I}(p(t))^2. \end{aligned}$$
(38)
So, our final expression for \(p'\) is
$$\begin{aligned} p'(t) = \frac{\tau }{I}-\frac{C_{\tau }}{I}(p(t))^2. \end{aligned}$$
(39)
This has solution
$$\begin{aligned} p(t) = \sqrt{\frac{\tau }{C_{\tau }}} \tanh {\biggl (\frac{\sqrt{C_{\tau }\tau }}{I}t\biggr )}. \end{aligned}$$
(40)
We now compute p(t) for the F-16C with the following values Table 1.
Table 1

Values for the F-16C

 

Value

Notes

q

0.2478

 

y

0.002478

1% of maximum pitch rate

m

11793

 

v

295

 

I

12710

Approximate; scaled up by increasing mass density

W

9.8

 

\(W_\mathrm{d}\)

4

Rough approximation

\(C_\mathrm{d}\)

1.28

\(C_\mathrm{d}\) of flat plane in 3D space

\(C_{\tau }\)

903.92

 

\(\rho \)

1.225

 
We see that p(t) rapidly tends towards
$$\begin{aligned} \sqrt{\frac{\tau }{C_{\tau }}} \approx 4.83. \end{aligned}$$
The real-life maximum roll rate of the F-16C is approximately 4.18. So, our model gives a reasonable approximation. More importantly, p(t) quickly (\(\approx \) 0.03s) exceeds
$$\begin{aligned} \sqrt{2qy-y^2} \approx 0.035, \end{aligned}$$
(41)
so that we do in fact gain advantage. (When calculating advantage, we will ignore the discrepancies in the short time it takes for p(t) to accelerate past \(\sqrt{2qy-y^2}\).) However, our model yields unrealistic values for high values of y, since in reality the conversion rate of lift to roll is not 100%, and is less efficient as the pitch rate is reduced further, due to the performance limits of ailerons.

6 Connecting two parallel turn circles

We have assumed thus far that our aircraft and the enemy aircraft start in the same turn circle. We now consider the case where both aircraft start in parallel turn circles spaced arbitrarily far apart. Trajectory generation methods are given in [11, 17]; however, the symmetry of our model allows for a geometric method to be obtained. Furthermore, our method maintains full pitch deflection, which is desirable as it does not decrease our relative angular velocity. The formula for \(\mathbf {h}\) allows us to derive a method of translating our turn circle towards any given parallel turn circle. Without loss of generality, we can assume that the turn circles are horizontal (i.e., the third component of C(t) is constant for all t), and that the target turn circle is centered at the origin. Additionally, assume that both aircraft have the same polarity.

Consider the vertical plane P centered at the origin and intersecting the center of our aircraft’s turn circle. If there is more than one such P, choose one arbitrarily. Working in coordinates with respect to P, label the center of our aircraft’s turn circle \(\mathbf {g}_O\) and the center of the target turn circle \(\mathbf {g}_T\). Our goal is to shift \(\mathbf {g}_O\) to \(\mathbf {g}_T\). Let H be the set of the four functions
$$\begin{aligned} \begin{aligned} \mathbf {h}_1(p)&= \left( h_1(p),h_2(p) \right) , 0 \le p \le M\\ \mathbf {h}_2(p)&= \left( -h_1(p),h_2(p) \right) , 0 \le p \le M\\ \mathbf {h}_3(p)&= \left( h_1(-p),-h_2(-p) \right) , -M \le p \le 0\\ \mathbf {h}_4(p)&= \left( -h_1(-p),-h_2(-p) \right) , -M \le p \le 0. \end{aligned} \end{aligned}$$
(42)
Let \(\mathbf {k}(p,t) = q\mathbf {B}(t)+p\mathbf {T}(t)\) so that \(\left| \mathbf {k}(p)\right| = \sqrt{p^2+q^2}\). Let \(S_O\) be the set of the four functions \(\{ \mathbf {g}_O + \mathbf {h}_i(p) \}, 1 \le i \le 4.\) Let \(S_T\) be defined similarly. Then, the union of the images of all the functions in \(S_O\) are all the admissible centers within one spiral period for our aircraft in P; i.e., it is possible for our aircraft to translate the center of its turn circle to any point in the union by choosing an appropriate p and rolling for \(\frac{2\pi }{\left| \mathbf {k}(p)\right| }\) seconds, whilst ensuring the turn circle it ends up in is parallel to the original starting turn circle. Then, the procedure is as follows:
  1. 1.
    If possible, choose \(\mathbf {s}_O \in S_O\) and \(\mathbf {s}_T \in S_T\) such that \(\text {Im } \mathbf {s}_O \cap \text {Im } \mathbf {s}_T \ne \emptyset .\)
    1. (a)

      If an intersection exists, arbitrarily choose one and label it \(\mathbf {x}\) and proceed to Step 2.

       
    2. (b)

      Otherwise choose \(\mathbf {s}_O\) and \(p_0\) such that \(\left| \mathbf {s}_O(p_0)-\mathbf {g}_T\right| \) is minimized. Apply the roll rate \(p_0\) to our aircraft for \(\frac{2\pi }{\left| \mathbf {k}(p_0)\right| }\) seconds, beginning when \(\mathbf {T}(t)\) intersects P. Note that \(\mathbf {T}(t)\) intersects P at two points in time; choose the point \(t_0\) such that the angle between \(\mathbf {s}_O(p_0)-\mathbf {g}_O\) and \(\mathbf {T}(t_0)\) is smaller. After applying the roll rate, \(\mathbf {T}\) and \(\mathbf {B}\) now refer to our aircraft’s orientation in its new turn circle. Redefine \(S_O\) and \(S_T\) for our aircraft’s new position, then go back to Step 1.

       
     
  2. 2.

    Let \(p_1, p_2\) be such that \(\mathbf {s}_O (p_1) = \mathbf {s}_T (p_2) = \mathbf {x}.\) Choose \(t_1\) such that \(\mathbf {T}(t_1)\) intersects P and the angle between \(\mathbf {s}_O(p_1)-\mathbf {g}_O\) and \(\mathbf {T}(t_1)\) is minimized. Apply the roll rate \(p_1\) to our aircraft for \(\frac{2\pi }{\left| \mathbf {k}(p_1)\right| }\) seconds starting at \(t_1\). If \(\mathbf {x} = g_T\), we are done. Otherwise proceed to Step 3.

     
  3. 3.

    Since an intersection \(\mathbf {x}\) exists, by symmetry of the set H there exists \(\mathbf {s} \in \{\mathbf {s}_O(p_1) + \mathbf {h}_i(p)\}\) such that \(\mathbf {s}(-p_2) = \mathbf {g}_T\). Choose \(t_2\) such that \(\mathbf {T}(t_2)\) intersects P and the angle between \(\mathbf {s}(-p_2)-\mathbf {s}_O (p_1)\) and \(\mathbf {T}(t_2)\) is minimized. Apply the roll rate \(-p_2\) to our aircraft for \(\frac{2\pi }{\left| \mathbf {k}(p_2)\right| }\) seconds starting at \(t_2\).

     
Our aircraft’s turn circle now coincides with the target turn circle.

6.1 Example of connecting two parallel turn circles

Let \(M = 2\), and
$$\begin{aligned} \begin{aligned} C(t)&= \left( \cos (t) +4 , \sin (t) + 1, 4 \right) ,\\ T(t)&= \left( \cos (t), \sin (t) + 1, 0 \right) , \end{aligned} \end{aligned}$$
(43)
be our aircraft’s parameterized turn circle and the target parameterized turn circle, respectively.
Then,
$$\begin{aligned} \begin{aligned} t_1&= \frac{\pi }{2},\\ \mathbf {s}_T (p)&= \mathbf {h}_1 (p),\\ \mathbf {s}_O (p)&= \left( 4,4 \right) + \mathbf {h}_4 (p),\\ \mathbf {s} (p)&= \mathbf {s}_O(p_1) + \mathbf {h}_4 (p),\\ \mathbf {x}&\approx \left( 2.41, 1.59 \right) ,\\ p_1&\approx -0.66,\\ p_2&\approx 1.52,\\ t_2&= t_1 + \frac{2\pi }{\left| \mathbf {k}(p_1)\right| }. \end{aligned} \end{aligned}$$
(44)
The procedure is then applying \(p_1\) for \(\frac{2\pi }{\left| \mathbf {k}(p_1)\right| }\) seconds, followed by immediately applying \(-p_2\) for \(\frac{2\pi }{\left| \mathbf {k}(p_2)\right| }\) seconds. The intersection \(\mathbf {x}\) is shown in Fig. 4. The trajectory of our aircraft’s position throughout the procedure is shown in Fig. 5.
Fig. 4

A subset of admissible centers for our turn circle starting from turn circles centered at the three points via the advantage maneuver. This is the image of the curves of \(\mathbf {s}_T(p)\) (blue) and \(\mathbf {s}_O(p)\) (yellow) for \(0\le p \le M\). \(\mathbf {g}_O\) is the red point, \(\mathbf {g}_T\) is the blue point. \(\mathbf {x}\) is the purple point (color figure online)

Fig. 5

The corresponding maneuver to move our aircraft between the points shown in Fig. 4. Position of our aircraft over time. Initial point (red), final point (blue). Applying \(p_1\) (green curve), then applying \(-p_2\) (red curve). The circles T(t) and C(t) are shown in blue. \(\mathbf {B}(t_1)\) is the orange vector, \(\mathbf {T}(t_1)\) is the blue vector. \(\mathbf {g}_O, \mathbf {g}_T,\) and \(\mathbf {x}\) are shown as they would appear when projected in the direction of the Y-axis (color figure online)

7 Rotating a non-parallel turn circle

In Sect. 6, we established a procedure for translating our aircraft’s turn circle into a parallel turn circle. If we consider the case where the target turn circle is not parallel, then we can reduce the problem of making the turn circles coincide to the parallel case if we are able to rotate our aircraft’s turn circle to be parallel. In this section, we describe a procedure to perform such a rotation.

Let our aircraft’s parameterized turn circle be C(t). We desire to be parallel to the turn circle T(t). Observe that we only need to rotate our aircraft’s \(\mathbf {B}\) vector until
$$\begin{aligned} \mathbf {B} = T(0)\times T\left( \frac{\pi }{2}\right) , \end{aligned}$$
(45)
where \(T(0)\times T\left( \frac{\pi }{2}\right) \) is the corresponding vector \(\mathbf {B}_T\) of T(t). Choose \(t_0\) such that \(\mathbf {B}_T\) lies in the plane P spanned by \(\{\mathbf {B}(t_0),\mathbf {T}(t_0)\}\) and the angle between \(\mathbf {T}(t_0)\) and \(\mathbf {B}_T\) is minimized. Let \(\alpha \) be the angle between \(\mathbf {B}(t_0)\) and \(\mathbf {B}_T\). Observe that if we roll at \(p > 0\)radians per second, then we are rotating \(\mathbf {B}(t_0)\) around \(\mathbf {k}(p,t_0)\) so that \(\mathbf {B}(t_1) = \mathbf {B}\left( t_0+\frac{\pi }{\left| \mathbf {k}(p)\right| }\right) \in P\). Then, the angle between \(\mathbf {B}(t_1)\) and \(\mathbf {B}_T\) is given by
$$\begin{aligned} \alpha -2\arctan \frac{p}{q}. \end{aligned}$$
(46)
Then, the procedure is as follows:
  1. 1.
    Choose the smallest \(n\in \mathbb {Z}^+\) giving \(p_0 \in [0,M]\) such that
    $$\begin{aligned} 2\arctan \frac{p_0}{q} = \frac{\alpha }{n}. \end{aligned}$$
    (47)
     
  2. 2.

    Apply \(p_0\) for \(\frac{\pi }{\left| \mathbf {k}(p_0)\right| }\) seconds.

     
  3. 3.
    Repeat the following steps \(n-1\) times.
    1. (a)

      Continue on the current turn circle for \(\frac{\pi }{q}\) seconds.

       
    2. (b)

      Apply \(p_0\) for \(\frac{\pi }{\left| \mathbf {k}(p_0)\right| }\) seconds.

       
     
Our \(\mathbf {B}\) vector is now equal to \(\mathbf {B}_T\). Geometrically, the trajectory of our aircraft during this procedure alternates between half-spirals and semicircles, repeatedly rotating \(\mathbf {B}\) by \(\frac{\alpha }{n}\).

7.1 Example of rotating a non-parallel turn circle

Let
$$\begin{aligned} \begin{aligned} M&= 1.5,\\ t_0&= 0,\\ \mathbf {T}(t_0)&= \left( 1,0,0 \right) ,\\ \mathbf {B}(t_0)&= \left( 0,0,1 \right) ,\\ \mathbf {B}_T&= \left( 0.8,0,-0.6 \right) . \end{aligned} \end{aligned}$$
(48)
Then,
$$\begin{aligned} \begin{aligned} \alpha&= 2\arctan 2,\\ n&= 2,\\ p_0&= \tan \left( \frac{\arctan 2}{2}\right) . \end{aligned} \end{aligned}$$
(49)
Then, we apply \(p_0\) for \(\frac{\pi }{\left| \mathbf {k}(p_0)\right| }\) seconds, continue on the resulting turn circle for \(\frac{\pi }{q}\) seconds, and apply \(p_0\) for \(\frac{\pi }{\left| \mathbf {k}(p_0)\right| }\) seconds again. The trajectory of our aircraft’s position throughout the procedure is shown in Fig. 7, and the trajectory of our aircraft’s \(\mathbf {B}\) vector is shown in Fig. 6.
Fig. 6

The direction of the \(\mathbf {B}\) vector (orange) of our aircraft over time under the rotation procedure. Initial direction (vector on the red point), final direction (vector on the blue point). Intermediate points when applying \(p_0\) (green curve), then applying \(p_0\) again (red curve) (color figure online)

Fig. 7

The curve connecting the red point and the blue point is the position of our aircraft over time under the rotation procedure. Initial point (red), final point (blue). Applying \(p_0\) (green curve), then continuing on the circle (blue curve), then applying \(p_0\) again (red curve) (color figure online)

8 Generalizing to include yaw

The maneuvers described in this paper are still applicable if both aircraft are yawing at a constant rate, in addition to pitching at a constant rate. We find that if our aircraft is yawing at a constant rate, the resulting trajectory is still a circle. We can then extend the definition of a turn circle to include “yawing at a constant rate.” Yawing is preferable as it reduces the period and radius of the turn circle. Although an aircraft cannot typically achieve a significant rate of yaw, we nonetheless show that the concepts expounded upon previously extend elegantly to include yaw. Furthermore, it will follow from our results that even a small amount of yaw will be beneficial to gaining advantage.

Let r be the yaw rate, and \(\mathbf {N}\) be the unit vector \(\mathbf {B}\times \mathbf {T}\).

8.1 Derivations of formulas

We now give derivations that include yaw for the formulas presented in this paper.

8.1.1 Integrating to obtain \(\mathbf {k} \times \mathbf {T}\)

The following differential equations describe the motion of our aircraft:
$$\begin{aligned} \frac{{\text {d}}\mathbf {T}}{{\text {d}}t}= & {} q \mathbf {B} \times \mathbf {T} + r \mathbf {N} \times \mathbf {T}, \end{aligned}$$
(50)
$$\begin{aligned} \frac{{\text {d}}\mathbf {B}}{{\text {d}}t}= & {} p \mathbf {T} \times \mathbf {B} + r \mathbf {N} \times \mathbf {B},\nonumber \\ \frac{{\text {d}}\mathbf {N}}{{\text {d}}t}= & {} q \mathbf {B} \times \mathbf {N} + p \mathbf {T} \times \mathbf {N}. \end{aligned}$$
(51)
Observe that due to the anticommutativity of the cross product, we have
$$\begin{aligned} p\frac{{\text {d}}\mathbf {T}}{dt} + q\frac{d\mathbf {B}}{{\text {d}}t} + r\frac{{\text {d}}\mathbf {N}}{dt} = 0. \end{aligned}$$
(52)
Integrating both sides with respect to t, we obtain
$$\begin{aligned}&p \mathbf {T}(t) +q \mathbf {B}(t) + r \mathbf {N}(t) = \mathbf {k}, \end{aligned}$$
(53)
$$\begin{aligned}&\mathbf {k} = p \mathbf {T}(0) +q \mathbf {B}(0) + r \mathbf {N}(0). \end{aligned}$$
(54)
Rearrange (5354) to obtain
$$\begin{aligned} \mathbf {B} = -\frac{p}{q}\mathbf {T}-\frac{ r}{q}\mathbf {N}+\frac{\mathbf {k}}{q} \end{aligned}$$
(55)
and substitute into (5051) to obtain
$$\begin{aligned} \begin{aligned} \frac{{\text {d}}\mathbf {T}}{{\text {d}}t}&= q \left( -\frac{p}{q}\mathbf {T}-\frac{ r}{q}\mathbf {N}+\frac{\mathbf {k}}{q}\right) \times \mathbf {T} + r \mathbf {N} \times \mathbf {T}\\&= (-p\mathbf {T}- r\mathbf {N}+\mathbf {k}) \times \mathbf {T} + r \mathbf {N} \times \mathbf {T}\\&= \mathbf {k}\times \mathbf {T} \end{aligned} \end{aligned}$$
(56)
as desired.
Hence, by symmetry, we also have
$$\begin{aligned} \frac{{\text {d}}\mathbf {B}}{{\text {d}}t} = \mathbf {k}\times \mathbf {B} \text { and } \frac{{\text {d}}\mathbf {N}}{{\text {d}}t} = \mathbf {k}\times \mathbf {N}. \end{aligned}$$
(57)
Thus we can still apply the solution given in (17), which will still yield a circle.

8.1.2 Yawing reduces the size of the turn circle

If \(p = 0\) and \(r \ne 0\), then
$$\begin{aligned} \begin{aligned} \mathbf {T}(0) \cdot \mathbf {k}&= \mathbf {T}(0) \cdot (p \mathbf {T}(0) +q \mathbf {B}(0) + r \mathbf {N}(0))\\&= 0 \end{aligned} \end{aligned}$$
(58)
since \(\mathbf {T},\mathbf {B},\mathbf {N}\) are always orthogonal. Hence, \(\mathbf {T}(0)\) is orthogonal to \(\mathbf {k}\). But we know from (17) that \(\mathbf {T}\) orbits \(\mathbf {k}\); i.e., the angle between \(\mathbf {T}\) and \(\mathbf {k}\) is constant. Then the projection of \(\mathbf {T}\) onto \(\mathbf {k}\) is always 0 and thus, the locus of our aircraft is still a circle, even with constant yaw. Without yaw, the period of our turn circle is given by \(\frac{2\pi }{q}\). However, with yaw the period is given by
$$\begin{aligned} \frac{2\pi }{\left| \mathbf {k}\right| }= \frac{2\pi }{\sqrt{q^2 + r^2}} < \frac{2\pi }{q} \end{aligned}$$
(59)
so, our turn circle is tighter than it would be if we did not yaw.

8.2 Non-reversal of yaw

Although we are constantly yawing in one direction, our advantage maneuver function retains the same structure. Furthermore, we do not need to define an additional function to control yaw; i.e. yawing remains constant throughout the maneuver.

This is shown by considering \(\mathbf {T},\mathbf {B},\) and \(\mathbf {N}\) over the various stages in the maneuver. After the first spiral period, the three vectors are unchanged. However, after traveling around the semicircle, we have the new vectors \(\mathbf {T}_1,\mathbf {B}_1,\) and \(\mathbf {N}_1\). During the semicircle, the normal vector to the turn circle is given by
$$\begin{aligned} \mathbf {k}_1 = q \mathbf {B}_1 + r \mathbf {N}_1. \end{aligned}$$
(60)
This is necessarily the same as the normal vector \(\mathbf {k}_0\) at \(t=0\); if we let \(\mathbf {B}_0= \mathbf {B}(0),\mathbf {T}_0= \mathbf {T}(0)\) and \(\mathbf {N}_0= \mathbf {N}(0)\), then
$$\begin{aligned} \mathbf {k}_0 = q \mathbf {B}_0 + r \mathbf {N}_0 = \mathbf {k}_1 = q \mathbf {B}_1 + r \mathbf {N}_1. \end{aligned}$$
(61)
Since our aircraft traveled in a semicircle, we have
$$\begin{aligned} \mathbf {T}_1 = -\mathbf {T}_0. \end{aligned}$$
(62)
Without loss of generality, assume \(p > 0\). If \(\mathbf {k} = p \mathbf {T}_0 + \mathbf {k}_0\) is the direction of the initial spiral, then we wish to show that the direction of the spiral on the third interval of the maneuver is \(-\mathbf {k}\). Let the orientation vector of the latter spiral be \(\mathbf {k}_L\). Let the roll rate of this spiral be \(p_L = -p\). Then
$$\begin{aligned} \begin{aligned} \mathbf {k}_L&= p_L \mathbf {T}_1+\mathbf {k}_1\\&=p\mathbf {T}_0+\mathbf {k}_0\\&=\mathbf {k}. \end{aligned} \end{aligned}$$
(63)
Since \(p_L < 0\), the direction of the spiral is \(-\mathbf {k}_L = -\mathbf {k}\) as desired.

8.3 Advantage gained

Let \(|\mathbf {k}_0| = \sqrt{q^2+r^2}.\) With the addition of yaw, our advantage maneuver is given by
$$\begin{aligned} f_p (t) ={\left\{ \begin{array}{ll} p &{} \text {if } 0 \le t< \frac{2\pi }{\left| \mathbf {k}\right| }\\ 0 &{} \text {if } \frac{2\pi }{\left| \mathbf {k}\right| } \le t< \frac{2\pi }{\left| \mathbf {k}\right| } + \frac{\pi }{|\mathbf {k}_0|}\\ -p &{} \text {if } \frac{2\pi }{\left| \mathbf {k}\right| } + \frac{\pi }{|\mathbf {k}_0|}\le t\le \frac{4\pi }{\left| \mathbf {k}\right| } + \frac{\pi }{|\mathbf {k}_0|}\\ \end{array}\right. }. \end{aligned}$$
(64)
Let A(p) be the advantage gained if we choose p for our advantage maneuver function. As our maneuver function has the same structure as before, we still calculate advantage gained by subtracting the amount the enemy aircraft has traveled during our maneuver from 2.5 revolutions. Then, we have (Fig. 8)
$$\begin{aligned} \begin{aligned} A(p)&= 4\pi \left( 1-\frac{|\mathbf {k}_0|}{|\mathbf {k}|}\right) \\&=4\pi \left( 1-\sqrt{\frac{q^2+r^2}{q^2+r^2+p^2}}\right) . \end{aligned} \end{aligned}$$
(65)
Fig. 8

Advantage gained with respect to the roll rate p. The red curve corresponds to advantage gained without yaw, i.e. \(A(p) = 4\pi \left( 1-\frac{1}{\sqrt{1+p^2}}\right) \), and the blue curve is with yaw; \(A(p) = 4\pi \left( 1-\sqrt{\frac{1+0.5^2}{1+0.5^2+p^2}}\right) \) (color figure online)

8.4 Displacement

To calculate displacement, we will work in the \(\{\mathbf {T},(q\mathbf {B}+ r\mathbf {N})/\sqrt{q^2+ r^2}\}\) basis. Then \(\mathbf {k} = \left( p, \sqrt{q^2+ r^2} \right) .\) We then calculate \(\mathbf {h}\) in the same manner as before and obtain
$$\begin{aligned} \mathbf {h} = \Biggl (\frac{2\pi p^2}{(p^2+q^2+ r^2)^\frac{3}{2}} ,\frac{2\pi p\sqrt{q^2+ r^2}}{(p^2+q^2+ r^2)^\frac{3}{2}} \Biggr ), \end{aligned}$$
(66)
which has magnitude \( \frac{2\pi p}{p^2+q^2+ r^2}\).

8.5 Connecting two parallel turn circles

The procedure is analogous. Assume without loss of generality that the turn circles are horizontal; i.e., the vertical axis of P is the direction of \(q\mathbf {B}+ r\mathbf {N}\).

8.6 Rotation of a non-parallel turn circle

The procedure is analogous. We rotate \(q\mathbf {B}(0)+ r\mathbf {N}(0)\) around \(\mathbf {k}\) instead. Replace \(2\arctan \frac{p}{q}\) with \(2\arctan \frac{p}{\sqrt{q^2+ r^2}}\).

9 Simulation results

The continuous roll maneuver was tested in a six-degree-of-freedom flight simulator. The aircraft used for the simulation is the FA-22 Raptor in X-Plane 10. Gravity was set to zero. The aircraft was programmed to roll at 10% deflection for the spiral parts of the maneuver. The duration of each spiral part was approximately 20.35 s, and the semicircular arc had a duration of 13.5 s. The resulting advantage gained was approximately 0.43 radians. Discrepancies were observed between the flight path predicted by the kinematic model and the flight path in the simulation. This is because the actual flight dynamics are significantly more complicated than our simplified kinematic model. In particular, the effects of adverse yaw were noticeable, and a 5% deflection of the rudder was used to mitigate this effect. Changes in air density with respect to altitude were unaccounted for. Additionally, it was observed that at higher roll rates, there was a significant loss in airspeed; so, the value of 10% was chosen for aileron deflection.

10 Consideration of gravity

Several issues arise when gravity is introduced to our model. Firstly, the initial horizontal turn circles now represent banked turns, whereas previously the lift vector was parallel to the ground. When the aircraft starts to roll, gravity will cause a reduction in airspeed. This will change the performance characteristics of the flight control surfaces. This will also affect the wings’ capacity to generate lift, and hence its turn performance. Since the direction of gravity remains constant, but the direction of the lift vector is rotating, the trajectory is no longer a spiral. Due to these complexities, the inclusion of gravity has been left for future research.

11 Conclusion

This notion of turn circles and thinking geometrically proves effective for the case of aircraft pitching at a constant rate, and that it is by moving out of the circle and obtaining an oblique angle relative to the other aircraft’s circle that it is possible to increase relative angular velocity. This paper formulates an advantage maneuver utilizing this idea. This maneuver can be repeatedly applied to gain advantage in the given combat scenario. Additionally, this paper provides a procedure that transforms any given initial position and orientation of a fighter aircraft to that in which an advantage maneuver can be applied. We have also shown that the results extend similarly to the case where the aircraft are yawing at a constant rate. Future work includes the consideration of additional variables, such as airspeed reduction and adverse yaw, as well as adapting the maneuver for a gravitational field.

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Copyright information

© Shanghai Jiao Tong University 2019

Authors and Affiliations

  1. 1.University of New South WalesSydneyAustralia
  2. 2.University of SydneySydneyAustralia

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