29 Aircraft Equations of Motion

Introduction

Unlike a two-dimensional terrestrial vehicle, an aircraft can move along an almost infinite number of possible three-dimensional paths. An aircraft may undergo accelerated motions along its pitch and/or roll and/or yaw axes, as shown in the figure below. In practice, however, any aircraft’s flight path will be limited to values within its aerodynamic performance and structural stress envelopes. In this regard, not all aircraft are created equally, nor will they have unlimited flight capabilities. For example, the number of feasible flight paths possible with an airliner will not be the same as those for a fighter jet, nor would they be expected to be based on their design purpose.

 

An aircraft can pitch, roll, yaw, and move about all three axes to follow an almost unlimited number of possible curvilinear flight paths.

To help analyze aircraft motion and performance, the general equations of motion for an aircraft in flight must be established. These equations help expose fundamental performance characteristics during steady flight and some special cases of maneuvering flight, including turns and pull-ups. These results also help appreciate the factors that can, and inevitably will, limit the aircraft’s flight capabilities, either aerodynamically, structurally, or both. Structural limits are usually defined in terms of a maneuvering flight envelope, which maps out the combinations of limiting airspeeds and maximum load factors within which the aircraft can safely fly.

Learning Objectives

  • Know about the terminology and conventions used to describe the motion of an airplane, including its pitch, roll, and yaw.
  • Set up the equations for the forces on an airplane following a general flight path and when undergoing simple maneuvers.
  • Understand the meaning and significance of the “load factor” on an airplane and calculate the load factors on an aircraft in steady turns and pull-up maneuvers.
  • Appreciate the significance of a maneuvering envelope in terms of limiting airspeeds and allowable load factors.

Assumptions

To analyze an aircraft in flight, the equations that describe its motion must first be set down in terms of lift, weight, drag, and propulsive force (i.e., thrust). The overall approach is relatively easy but requires the careful application of the principles of statics and dynamics. The objective is to describe the airplane’s movement through the atmosphere using equations that physically describe its curvilinear motion, thereby allowing its performance and other capabilities to be evaluated.

In this initial analysis, the aircraft can be replaced by a point mass at the center of gravity (c.g.) following a curvilinear flight path, as indicated in the figure below. Remember that when an object moves along a curved flight path, the motion is called curvilinear compared to the case where it moves in a straight line path, which is rectilinear.

 

An airplane following a curvilinear flight path with pitch angle \gamma and bank angle \phi. Resolving the forces to the center of gravity is convenient for analysis, assuming moment equilibrium.

The aerodynamics of the wings, empennage, airframe, etc., are not considered in individual detail but can be represented as total lift, drag, and pitching moments on the aircraft. Experience shows that the lift and drag of an entire airplane can be analyzed with a high confidence level using a composite aerodynamic drag polar (i.e., the relationship between lift coefficient and drag coefficient) if this can be suitably obtained or even assumed.

As discussed earlier, the most common and representative drag polar for an airplane at subsonic flight speeds up to the point of wing stall is

(1)   \begin{equation*} C_D = C_{D_{0}} + \frac{{C_L}^2}{\pi A\!R e} \end{equation*}

The first term in the preceding equation is the non-lifting profile/parasitic drag component, and the second term is the induced drag, with A\!R being the aspect ratio of the wing and e being its Oswald’s efficiency factor (e < 1). Polars are available for various airplanes or can be estimated based on historical data in cases where the polar may not be known.

It is often convenient to have an analytic relationship between the C_L and C_D coefficients if general equations for thrust and or power for flight are to be determined. However, another method may be used in some cases, such as a table look-up process, where the coefficients may be specified as discrete values as functions of the angle of attack and Mach number.

At first, the propulsion system details need not be considered, but it must be recognized that not all propulsion systems will have the same characteristics and limitations. Nevertheless, the propulsive system must eventually be considered for all forms of flight analysis, at the very least in terms of thrust produced and/or power available, as well as the specific fuel consumption, i.e., the engine’s fuel efficiency in producing thrust or power.

In summary, it is possible to proceed to analyze the motion of an aircraft by stating some basic assumptions that will allow the development of the equations of motion and expose the primary influencing parameters and their dependencies:

  • The distributed weight of the aircraft can be replaced by a center of gravity (c.g.) location and assume that the entire weight acts at the c.g., i.e., a point mass assumption.
  • As previously discussed, the aerodynamic representation uses integrated quantities such as lift coefficient C_L and drag coefficient C_D as a drag polar.
  • The complexity of the propulsive device is recognized; however, it is considered only in terms of its thrust production (or power supplied) and its specific fuel consumption, although a dependency on the throttle setting may also be specified.
  • The aircraft is in pitching, rolling, and yawing moment equilibrium, i.e., operating in balanced or “trimmed” flight such that all of the moments about the c.g. are zero.

Primary Forces on an Airplane

The four net (resultant) forces involved in the flight are the lift L, weight W, drag D, and thrust T, as shown in the figure below.

  • The lift is given the symbol L and, by definition, acts perpendicular to this flight path, i.e., in the direction of the free-stream or flight path velocity, V_{\infty}.
  • The drag is given the symbol D and acts in a direction parallel to the flight path velocity.
  • The weight W is concentrated at the center of gravity (c.g.) and acts toward the center of the Earth.

 

In the level flight trim condition, the forces and moments on the airplane will be in perfect balance.

Straight & Level Flight

Most airplanes will spend much of their flight time in straight-and-level, unaccelerated flight conditions. In this case, the forces on the airplane are in balance, i.e., a static equilibrium.

 

Trim condition for straight and level flight where lift equals weight and thrust equals drag.

For horizontal equilibrium, then

(2)   \begin{equation*} T \cos \epsilon - D = 0 \end{equation*}

and for vertical equilibrium

(3)   \begin{equation*} L + T \sin \epsilon - W = 0 \end{equation*}

Notice that the angle \epsilon denotes the line of action of the propulsive thrust force, which may differ from the flight direction for various reasons. When \epsilon = 0, which is generally small anyway and so a reasonable assumption to make, then

(4)   \begin{equation*} L = W \quad \mbox{and} \quad D = T \end{equation*}

Worked Example #1 – Steady level flight

Consider the aircraft in the figure shown below. The aircraft’s weight can be assumed to act at its center of gravity. What must be the balance of forces and moments on the aircraft in straight and level, unaccelerated flight at a constant airspeed? If L_T = a \, L_W where a is a constant, what is the value of L_W? Find also the relationship between l_2 and l_1.

For vertical force equilibrium, then

    \[ L_W- L_T = W \]

We are given that L_T = a \, L_W so

    \[ L_W- a \ L_W = W \]

or

    \[ L_W ( 1- a ) = W \]

so

    \[ L_W = \frac{W}{(1 - a)} \]

For pitching moment equilibrium, then

    \[ L_W \ l_1 = L_T \ l_2 = a \ L_W \ l_2 \]

so

    \[ \frac{l_1}{l_2} = a \]

Climbing Flight

The figure below shows an airplane climbing relative to the surface of the Earth (assumed here as the horizontal reference) where \gamma can be viewed as the climb or flight path angle. Notice that the angles in the diagram are exaggerated for clarity compared to a typical climb, but the airplane could be maneuvering. For descending flight, the value of \gamma would, of course, be negative.

 

An airplane in a steady climb along a rectilinear flight path is typical of a standard operational climb.

Consider first the particular airplane motion in a pure vertical plane with no turning. In the direction parallel to the direction of flight, then

(5)   \begin{equation*} F_{\parallel} = T \cos \epsilon - D - W \sin \gamma \end{equation*}

Therefore, the acceleration parallel to the curvilinear flight path will be

(6)   \begin{equation*} a_{\parallel} = \frac{d V_{\infty}}{dt} \end{equation*}

and so

(7)   \begin{equation*} F_{\parallel} = \left(\frac{W}{g}\right) a_{\parallel} \end{equation*}

where the “mass” of the airplane M equals W/g. Therefore,

(8)   \begin{equation*} \left( \frac{W}{g} \right) \frac{d V_{\infty}}{dt} = T \cos \epsilon - D - W \sin \gamma \end{equation*}

In the direction perpendicular to the flight path, the forces are

(9)   \begin{equation*} F_{\perp} = L \cos \phi + T \sin \epsilon \cos \phi - W \cos \gamma \end{equation*}

and the acceleration perpendicular to the flight path is

(10)   \begin{equation*} a_{\perp} = \frac{V_{\infty}^2}{R_v} \end{equation*}

where R_v is the instantaneous radius of curvature of the flight path in the vertical plane. Therefore, because F_{\perp} = (W/g)a_{\perp} then

(11)   \begin{equation*} \left(\frac{W}{g}\right) \frac{V_{\infty}^2}{R_v} = L \cos \phi + T \sin \epsilon \cos \phi - W \cos \gamma \end{equation*}

These are general results and apply to any flight path in a vertical plane.

Circular Flight Path

Consider now the forces on an airplane following a completely circular path of radius R_v = constant in a vertical plane flying at a constant airspeed, as shown in the figure below. Notice that when continuing this pull-up, the airplane would perform a complete loop in a vertical plane, which is a particular case. In proceeding, it is reasonable to assume that \epsilon = 0.

 

The balance of forces on an airplane in a pull-up flight maneuver in a vertical plane.

Vertical equilibrium, in this case, requires that

(12)   \begin{equation*} L - W = \left( \frac{W}{g} \right) \frac{V_{\infty}^2}{R_v} \end{equation*}

Therefore, the lift required is

(13)   \begin{equation*} L = \left( \frac{W}{g} \right) \frac{V_{\infty}^2}{R_v} + W = \left( 1 + \frac{V_{\infty}^2}{g R_v} \right) W = n W \end{equation*}

i.e., the lift on the airplane must be greater than its weight, where the load factor n is

(14)   \begin{equation*} n = \left( 1 + \frac{V_{\infty}^2}{g R_v} \right) \end{equation*}

The excess lift is related to the load factor n such that L = nW, i.e., the number of effective g loadings. So, it can be seen that for a given radius of the flight path, the load factor increases with the square of the airspeed. For a given airspeed, the load factor is inversely proportional to the radius, i.e., a faster and/or tighter flight path will produce a higher load factor.

The radius of curvature R_v of the flight path, in this case, will be

(15)   \begin{equation*} R_v = \frac{V_{\infty}^2}{g (n - 1)} \end{equation*}

For a given load factor, the radius of the flight path increases quickly with the square of the airspeed. This result explains the development of tactics needed for combat maneuvers used by military aircraft, where tighter maneuvers must be performed at lower airspeeds. The corresponding angular rate in the maneuver is

(16)   \begin{equation*} \omega = \frac{d\gamma}{dt} = \frac{g (n - 1)}{V_{\infty}} \end{equation*}

which again confirms that tighter vertical maneuvers are best flown at lower airspeeds.

Perfect Circular Loop Maneuver

As previously discussed, lift on the wing must be sufficient to overcome the airplane’s weight and produce the centripetal acceleration to execute a circular flight path in a vertical plane. The lift required is

(17)   \begin{equation*} L = \left( \frac{W}{g} \right) \frac{V_{\infty}^2}{R_v} + W \cos \psi = \left( \cos \psi + \frac{V_{\infty}^2}{g R} \right) W = n W \end{equation*}

where \psi = 0^{\circ} at the bottom of the loop and \psi = 180^{\circ} at the top. The corresponding load factor will be

(18)   \begin{equation*} n = \left( \cos\psi + \frac{V_{\infty}^2}{g R_v} \right) \end{equation*}

At the bottom of the loop, the lift must be greater than the weight to overcome both the weight and create the centrifugal force so

(19)   \begin{equation*} n = \left(1 + \frac{V_{\infty}^2}{g R_v} \right) \end{equation*}

At the top of the loop, the weight helps in the direction of the centrifugal force, so the load factor is

(20)   \begin{equation*} n = \left(-1 + \frac{V_{\infty}^2}{g R_v} \right) \end{equation*}

and so this is less than at the bottom of the loop.

Worked Example #2 – Loop-the-loop!

Consider a flight maneuver where an aerobatic airplane is inverted at the top of a perfectly circular loop performed at a constant airspeed. Show how to determine the load factor on the airplane as it flies around the loop. Will the “g” forces on the pilot be higher or lower at the top of the loop than when the airplane is at the bottom?

Let R_v be the radius of the loop maneuver. During the loop, the lift on the wing must be sufficient to overcome the airplane’s weight and produce the centripetal acceleration to execute a circular flight path in a vertical plane. The lift required is

    \[ L = \left( \frac{W}{g} \right) \frac{V_{\infty}^2}{R_v} + Mg \cos \psi = \left( \cos \phi + \frac{V_{\infty}^2}{g R_v} \right) W = n W \]

where the mass of the airplane is M (= W/g) and so W \cos \psi is the vertical component of the weight. Notice that \psi = 0 at the bottom of the loop and \psi = 180 degrees at the top of the loop. The corresponding load factor will be

    \[ n = \left( \cos\psi + \frac{V_{\infty}^2}{g R_v} \right) \]

Remember that we can relate the excess lift on the airplane to the load factor n such that L = nW, i.e., the number of effective g values. At the bottom of the loop, the lift must be greater than the weight to overcome both the weight and create the centrifugal force so

    \[ n = \left(1 + \frac{V_{\infty}^2}{g R_v} \right) \]

At the top of the loop, the weight helps in the direction of the centrifugal force, so the load factor is

    \[ n = \left(-1 + \frac{V_{\infty}^2}{g R_v} \right) \]

so it will be less than at the bottom of the loop. In the particular case where

    \[ \frac{V_{\infty}^2}{g R_v} = 1 \]

then the pilot will feel “weightless,” i.e., n = 0. We also notice that the load factor is a function of airspeed V_{\infty} and the radius of the loop R_v. This means that if the pilot flies a loop at a higher airspeed or tightens the loop to get a smaller radius R_v, the airplane and the pilot will experience higher load factors.

Turning Flight

Consider the forces on an airplane in a pure horizontal turn with a bank angle \phi and at a constant airspeed V_{\infty}, as shown in the figure below. To perform a turn, the airplane must be banked at an angle \phi such that a component of the wing lift creates the necessary inward force to balance the outward centrifugal force, i.e., the effects of the centripetal acceleration. In proceeding, it is again possible to assume that \epsilon = 0.

 

The balance of forces on an airplane in a horizontal (no climb or descent) banked turn.

Vertical equilibrium requires that

(21)   \begin{equation*} L \cos \phi = W \end{equation*}

and horizontal equilibrium requires

(22)   \begin{equation*} L \sin \phi = \left(\frac{W}{g} \right) \frac{V_{\infty}^2}{R_h} \end{equation*}

where R_h, in this case, is the radius of curvature of the turn.

It is apparent then that to perform a turn, the lift on the wing of the airplane must be greater than its weight, i.e., L > W, to create the necessary aerodynamic force not only to balance the weight of the airplane but also to produce the inward radial force to create the needed centripetal acceleration to execute a turn.

Solving for the lift required gives

(23)   \begin{equation*} L = \frac{W}{\cos \phi} \end{equation*}

In this case, the load factor is

(24)   \begin{equation*} n = \frac{W }{W \cos \phi} = \frac{1}{\cos \phi} = \sec \phi \end{equation*}

This result shows that the load factor must increase with the inverse of the cosine of the bank angle. For example, a 60^{\circ} banked turn will correspond to a load factor of two.

The corresponding radius of curvature of the flight path can be solved using

(25)   \begin{equation*} R_h = \frac{V_{\infty}^2}{g \sqrt{n^2 -1}} \end{equation*}

and the rate of turn (angular velocity) in the turn is given by

(26)   \begin{equation*} \omega = \frac{d\gamma}{dt} = \frac{g \sqrt{n^2 -1}}{V_{\infty}} \end{equation*}

Worked Example #3 – Steady banked turn maneuver

Consider a fighter aircraft of mass M = 20,000 kg in a steady banked turn with angle \phi = 65 degrees at a constant altitude.

  1. Find the total lift needed on the wing and the corresponding load factor n.
  2. Determine an expression for the centripetal force needed to execute the turn in terms of airspeed and radius of turn.
  3. If the airspeed is 250 knots, what will be the radius of the turn?
  4. If the maximum allowable load factor on the aircraft is n = 7, what will be the corresponding maximum permissible bank angle and new turn radius at this airspeed?

1. The total lift needed on the wing will be

    \[ L = \frac{W}{\cos \phi} = \frac{Mg}{\cos \phi} = \frac{20,000 \times 9.81}{\cos 65^{\circ}} = 464.25~\mbox{kN} \]

and the corresponding load factor is

    \[ n = \frac{1}{\cos 65^{\circ}} = 2.37 \]

2. Concerning the figure above, the equation for the inward centripetal force CF to produce this level turn is

    \[ CF= L \sin \phi = \frac{M V_{\infty}^2}{R_h} \]

where R_h is the radius of the turn.

3. An airspeed of 250 kts is equal to 128.6 m/s, so rearranging the previous equation gives the radius of the turn as

    \[ R_h = \frac{M V_{\infty}^2}{\sin \phi \ L} = \frac{20,000 \times 128.6^2}{\sin 65^{\circ} \times 464,249} = 786.2~\mbox{m} \]

4. We are given that n_{\rm max} = 7, so

    \[ \phi_{\rm max} = \sin^{-1} \left( \frac{1}{n_{\rm max}} \right) = 81.8^{\circ} \]

and the maximum lift on the wing will be

    \[ L_{\rm max} = \frac{Mg}{\cos \phi_{\rm max}} = \frac{20,000 \times 9.81}{\cos 81.8^{\circ}} = 1.376\times 10^6~\mbox{N} \]

The radius of the turn in this condition is

    \[ R_h = \frac{M V_{\infty}^2}{\sin \phi_{\rm max} \ L} = \frac{20,000 \times 128.6^2}{\sin 81.8^{\circ} \times 1.376\times 10^6} = 242.9~\mbox{m} \]

Summary of the Equations of Flight

In summary, the following general equations apply to the motion of an airplane:

(27)   \begin{eqnarray*} \mbox{$\parallel$ to flight path:} \quad && \hspace*{-7mm} \left(\frac{W}{g}\right) \frac{d V_{\infty}}{dt} = T \cos \epsilon - D - W \sin \gamma \\[12pt] \mbox{$\perp$ to flight path:} \quad && \hspace*{-7mm} \left(\frac{W}{g}\right) \frac{V_{\infty}^2}{R_v} = L \cos \phi + T \sin \epsilon \cos \phi - W \cos \gamma \\[12pt] \mbox{Horizontal plane:} \quad && \hspace*{-7mm} \left(\frac{W}{g}\right) \frac{(V_{\infty} \cos \gamma)^2}{R_h} = L \sin \phi + T \sin \epsilon \sin \phi \end{eqnarray*}

where R_v = the radius of curvature of the flight path in a vertical plane, and R_h = the radius of curvature of the flight path in a horizontal plane.

In many cases, the line of action of the thrust vector relative to the flight path is small, so it is reasonable to assume that \epsilon = 0 in the forgoing equations, i.e.,

(28)   \begin{eqnarray*} \mbox{$\parallel$ to flight path:} \quad && \hspace*{-7mm} \left(\frac{W}{g}\right) \frac{d V_{\infty}}{dt} = T - D - W \sin \gamma \\[10pt] \mbox{$\perp$ to flight path:} \quad && \hspace*{-7mm} \left(\frac{W}{g}\right) \frac{V_{\infty}^2}{R_v} = L \cos \phi - W \cos \gamma \\[10pt] \mbox{Horizontal plane:} \quad && \hspace*{-7mm} \left(\frac{W}{g}\right) \frac{(V_{\infty} \cos \theta)^2}{R_h} = L \sin \phi \end{eqnarray*}

It should be remembered that in accelerated flight, the lift will not equal the weight of the airplane because of the need for the wing to create whatever lift value is needed to produce the accelerations to follow the required flight path. The resulting lift force may be greater or less than the airplane’s weight, so the load factor can be positive or negative during flight.

Limiting Airspeeds & Load Factors

An airplane (or the pilot and crew) cannot withstand infinite load factors, and the airplane will be either aerodynamically or structurally limited, or both. An airspeed/load factor diagram (a so-called V–n diagram) is one form of operating envelope for an airplane. This diagram maps out the conditions for flight without the airplane stalling or exceeding its structural strength limits.

The figure below shows a representative V–n diagram for an airplane as a function of its airspeed (flight Mach number may also be used). The green area is the normal flight envelope, with the orange and red zones denoting structural overload conditions.

 

Representative V–n diagram in terms of load factor versus airspeed.

Notice that the “stall limit” traces out one corner of the operating envelope, which is the load factor that can be attained in normal (upright flight) before the wing stalls, denoted by the region between points A and B. Point B corresponds to a level flight stall, and Point B is an accelerated stall with a limiting load factor. The other stall limit is the corresponding maximum attainable load factor before the wing stalls when the airplane is inverted, denoted by Point C; obviously, not all airplanes will be capable of inverted flight.

There is an airspeed called the “corner” airspeed, where the airplane will operate at the edge of the stall and pull the maximum load factor. This condition is identified as point A and is called the maximum maneuvering airspeed, called V_A. In very turbulent or gusty atmospheric conditions, it is essential that the airplane not be structurally overstressed and must be flown at or below V_A to prevent atmospheric gusts from reaching a threshold where they may structurally overload the airframe. The maximum of “never exceed” airspeed V_{\rm NE} is the airspeed where the maximum aerodynamic pressures are produced on the airframe.

The maximum structural attainable load factor an airplane is designed to withstand depends on the particular airplane and precisely what it is intended to do. The minimum achievable positive load factor for most airplanes (the so-called limit load) is usually 3.8, although the FARs contain more specific requirements for different types of civil airplanes. Aerobatic and military fighter airplanes are designed to tolerate much higher load factors, often between minus 10 and plus 12. To the limit load factors, 50% is added for structural design purposes (i.e., a margin of safety), which becomes known as the ultimate load.

Summary & Closure

In the analysis of airplane performance, it has been shown how the general equations of motion for an airplane in flight can be readily derived by following the basic principles of statics and dynamics. These equations have helped expose some fundamental results for steady-level and maneuvering flight. In addition, they have set a rational basis for determining variations in the airplane’s flight performance and potential limitations. Much of the analysis of civil airplanes will be for steady-level flight, small angles of displacement, and mild maneuvers. However, for military aircraft such as fighters, their flight maneuvers may be more aggressive and include various types of aerobatics with large displacements and rates. In such cases, the load factors produced on the aircraft may be significant, and the aircraft may fly close to its aerodynamic and/or structural limits.

5-Question Self-Assessment Quickquiz

For Further Thought or Discussion

  • Consider an acrobatic airplane in a roll maneuver with a constant angular velocity. What factors will affect the maximum possible roll rate?
  • Consider a pull-down maneuver where an aircraft is inverted at the top of a loop. Show how to obtain the load factor.
  • It is claimed that a small general aviation Cessna airplane can “out-maneuver” an F-16 fighter airplane. What does this mean, and is there any truth in this claim?
  • The ability to perform a banked turn will be limited by wing stall. Explain.
  • What factors may limit an aircraft’s maximum and minimum attainable load factor? Hint: Not all of these factors may have an engineering basis.
  • How might the equations of motion differ for different flight regimes, e.g., subsonic, supersonic, or hypersonic?
  • What are the challenges in modeling and solving the aircraft equations of motion in real-time scenarios, such as during flight simulations or control systems design?
  • How do the aircraft equations of motion relate to the concepts of stability and control? How are these concepts incorporated into the design and operation of aircraft?

Other Useful Online Resources

To learn more about flight maneuvers and aircraft limitations, take a look at some of these online resources:

  • Read the Code of Federal Regulations on the flight maneuvering envelope §25.333.
  • Video on flight maneuvers from ERAU.
  • Load factors on an airplane are explained in this video.
  • A video presentation giving a simplified explanation of the V-n diagram.

License

Icon for the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License

Introduction to Aerospace Flight Vehicles Copyright © 2022, 2023 by J. Gordon Leishman is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License, except where otherwise noted.

Digital Object Identifier (DOI)

https://doi.org/https://doi.org/10.15394/eaglepub.2022.1066.n27