Energy-State Approximation and Minimum-Time Climb

This section develops the mathematical foundations of the energy-state approximation for aircraft performance optimization. The central application is the minimum-time climb of a supersonic aircraft. The method became historically important because it transformed a difficult high-dimensional trajectory optimization problem into a lower-dimensional problem with a strong physical interpretation.

The key idea is to replace speed and altitude by a total specific energy variable and then use the rate of change of this energy as the primary performance quantity.

The chapter develops:

the longitudinal aircraft equations of motion;
the total specific energy equation;
the reduced-order energy-state approximation;
the minimum-time climb problem;
the Hamiltonian and optimality conditions;
the interpretation of maximizing excess power;
the non-intuitive dive–climb trajectory;
numerical solution strategies;
limitations of the approximation.

Historical Context¶

The energy-state approximation was used in classic aerospace optimization studies during the 1960s. At that time, computational resources were extremely limited, so simplified models were essential.

The central engineering problem was:

How should a high-performance aircraft vary its speed, altitude, and flight path angle to reach a specified terminal energy state in minimum time?

A direct climb is not necessarily optimal. In many supersonic aircraft, aerodynamic drag rises sharply near Mach 1. An optimal trajectory may therefore climb, dive to accelerate through the transonic region, and then climb again.

Longitudinal Aircraft Model¶

Consider planar longitudinal motion. Let

$V$ be the airspeed;
$\gamma$ be the flight-path angle;
$h$ be altitude;
$x$ be horizontal range;
$m$ be mass;
$T$ be thrust;
$D$ be drag;
$L$ be lift;
$\alpha$ be angle of attack;
$g$ be gravitational acceleration.

A standard point-mass model is

\begin{aligned} m\dot V &= T\cos\alpha-D-mg\sin\gamma,\\ mV\dot\gamma &= T\sin\alpha+L-mg\cos\gamma,\\ \dot h &= V\sin\gamma,\\ \dot x &= V\cos\gamma,\\ \dot m &= -\dot m_f. \end{aligned}

(1)

The first two equations describe kinetics. The last two describe kinematics.

Aerodynamic Model¶

Dynamic pressure is

q=\frac{1}{2}\rho(h)V^2.

(2)

Lift and drag are

\begin{aligned} L &= qSC_L(M,\alpha),\\ D &= qSC_D(M,\alpha), \end{aligned}

(3)

where $S$ is reference area and $M$ is Mach number.

A common aerodynamic approximation is

C_D=C_{D0}(M)+K(M)C_L^2.

(4)

Then

D=qS\left[C_{D0}(M)+K(M)C_L^2\right].

(5)

The thrust model may be represented as

T=T(V,h,\eta),

(6)

where $\eta$ is throttle.

Total Specific Energy¶

Define total specific mechanical energy

E = \frac{V^2}{2}+gh.

(7)

This quantity combines kinetic and potential energy per unit mass.

Differentiate:

\dot E = V\dot V+g\dot h.

(8)

Using

\dot h=V\sin\gamma

(9)

and

\dot V = \frac{T\cos\alpha-D}{m} -g\sin\gamma,

(10)

we obtain

\begin{aligned} \dot E &= V\left[ \frac{T\cos\alpha-D}{m} -g\sin\gamma \right] + gV\sin\gamma. \end{aligned}

(11)

The gravity terms cancel:

\boxed{ \dot E = \frac{V}{m} \left( T\cos\alpha-D \right). }

(12)

For small angle of attack,

\cos\alpha\approx1,

(13)

\boxed{ \dot E \approx \frac{V}{m}(T-D). }

(14)

The quantity

P_s=\dot E

(15)

is called the specific excess power.

Physical Interpretation of Specific Excess Power¶

Specific excess power measures the rate at which the aircraft can gain total energy.

If $T>D$ , then $\dot E>0$ and the aircraft gains energy.
If $T=D$ , then $\dot E=0$ and total energy remains constant.
If $T<D$ , then $\dot E<0$ and the aircraft loses energy.

The aircraft may exchange kinetic and potential energy without changing total energy. For example, during a constant-energy dive,

\frac{V^2}{2}+gh=\text{constant}.

(16)

Differentiating gives

V\dot V+g\dot h=0.

(17)

Thus a decrease in altitude causes an increase in speed.

Energy-State Approximation¶

The energy-state approximation is based on the following assumptions:

Motion is nearly horizontal.
Flight-path angle changes slowly.
Normal acceleration is small:

V\dot\gamma\approx0.

(18)

Thrust normal to the flight path is negligible:

T\sin\alpha\approx0.

(19)

The aircraft is approximately in normal-force equilibrium:

L\approx mg.

(20)

Speed and altitude can be parameterized through total energy.

Under these assumptions, the flight-path angle dynamics are neglected.

The reduced equations become

\begin{aligned} \dot E &= \frac{V}{m}(T-D),\\ \dot h &= V\sin\gamma,\\ \dot x &= V\cos\gamma. \end{aligned}

(21)

For a pure performance optimization in the $(V,h)$ plane, the energy equation may be used as the dominant dynamic equation.

Energy Curves in the Speed–Altitude Plane¶

For constant energy,

E=\frac{V^2}{2}+gh.

(22)

Solving for altitude,

\boxed{ h = \frac{E}{g} - \frac{V^2}{2g}. }

(23)

Thus constant-energy curves are parabolic in the $(V,h)$ plane.

The slope is

\frac{\,\mathrm{d} h}{\,\mathrm{d} V} = -\frac{V}{g}.

(24)

Moving to higher speed along a constant-energy curve requires a decrease in altitude.

Minimum-Time Climb Problem¶

Suppose the aircraft starts at

V(0)=V_0, \qquad h(0)=h_0,

(25)

and must reach

V(t_f)=V_f, \qquad h(t_f)=h_f.

(26)

The final energy is

E_f = \frac{V_f^2}{2}+gh_f.

(27)

The objective is

\boxed{ \min t_f = \min \int_0^{t_f}1\,\mathrm{d} t. }

(28)

In the reduced energy-state problem,

\dot E = P_s(E,V) = \frac{V}{m} \left[ T(E,V)-D(E,V) \right].

(29)

The control is effectively the speed $V$ selected at each energy level, subject to feasible flight-envelope constraints:

V_{\min}(E)\le V\le V_{\max}(E).

(30)

Time as an Integral over Energy¶

Since

\dot E=\frac{\,\mathrm{d} E}{\,\mathrm{d} t},

(31)

we have

\,\mathrm{d} t = \frac{\,\mathrm{d} E}{\dot E}.

(32)

Therefore,

t_f = \int_{E_0}^{E_f} \frac{1}{\dot E(E,V)} \,\mathrm{d} E.

(33)

Thus,

\boxed{ t_f = \int_{E_0}^{E_f} \frac{m} {V[T(E,V)-D(E,V)]} \,\mathrm{d} E. }

(34)

At each energy level, minimizing the integrand requires maximizing $\dot E$ .

Therefore,

\boxed{ V^*(E) = \arg\max_V \dot E(E,V). }

(35)

Equivalently,

\boxed{ V^*(E) = \arg\max_V \frac{V}{m} [T(E,V)-D(E,V)]. }

(36)

This is the mathematical basis for the statement that minimum climb time is obtained by maximizing specific excess power at each energy level.

Hamiltonian Derivation¶

Consider the reduced one-state problem

\dot E=P_s(E,V),

(37)

with objective

J=\int_0^{t_f}1\,\mathrm{d} t.

(38)

The Hamiltonian is

\mathcal{H} = 1+\lambda_E P_s(E,V).

(39)

For an interior optimum,

\frac{\partial\mathcal{H}}{\partial V}=0.

(40)

Therefore,

\lambda_E \frac{\partial P_s}{\partial V} = 0.

(41)

For a nontrivial extremal,

\lambda_E\neq0,

(42)

\boxed{ \frac{\partial P_s}{\partial V}=0. }

(43)

This confirms that the optimal speed at each energy level is a stationary point of specific excess power.

To ensure it is a maximum,

\boxed{ \frac{\partial^2P_s}{\partial V^2}<0. }

(44)

Costate Equation¶

The costate equation is

\dot\lambda_E = -\frac{\partial\mathcal{H}}{\partial E}.

(45)

Hence,

\boxed{ \dot\lambda_E = -\lambda_E \frac{\partial P_s}{\partial E}. }

(46)

This linear differential equation has solution

\lambda_E(t) = \lambda_E(0) \exp\left[ -\int_0^t \frac{\partial P_s}{\partial E} \,\mathrm{d}\tau \right].

(47)

Therefore, if $\lambda_E(0)\neq0$ , the costate cannot cross zero.

Free-Final-Time Condition¶

If $t_f$ is free and there is no terminal cost, then

\mathcal{H}(t_f)=0.

(48)

Thus,

1+\lambda_E(t_f)P_s(E_f,V_f)=0.

(49)

Therefore,

\boxed{ \lambda_E(t_f) = -\frac{1}{P_s(E_f,V_f)}. }

(50)

For a feasible climb with positive excess power,

P_s>0,

(51)

\lambda_E(t_f)<0.

(52)

Since $\lambda_E$ does not change sign,

\lambda_E(t)<0

(53)

along the trajectory.

Then minimizing

\mathcal{H}=1+\lambda_EP_s

(54)

requires maximizing $P_s$ .

Optimal Speed Schedule¶

The optimal energy-state speed schedule is determined by

\boxed{ \frac{\partial}{\partial V} \left[ V(T-D) \right] = 0. }

(55)

If mass is approximately constant,

\begin{aligned} 0 &= T-D + V\left( \frac{\partial T}{\partial V} - \frac{\partial D}{\partial V} \right). \end{aligned}

(56)

Hence,

\boxed{ T-D + V(T_V-D_V) = 0. }

(57)

This equation determines the speed giving maximum specific excess power at a prescribed energy.

Why the Optimal Trajectory May Dive¶

Near Mach 1, wave drag may increase sharply. A direct climb through this region can be slow because:

thrust margin is low;
drag is high;
excess power is small;
climbing consumes energy as potential energy.

A dive allows the aircraft to convert potential energy into kinetic energy:

\Delta\left(\frac{V^2}{2}\right) = -g\Delta h.

(58)

If $\Delta h<0$ , then

\Delta(V^2/2)>0.

(59)

Thus the aircraft can accelerate through the high-drag transonic region more quickly by sacrificing altitude.

After passing through the transonic region, the aircraft may regain altitude where excess power is more favorable.

Approximate Constant-Energy Dive¶

During an approximately constant-energy dive,

\dot E\approx0.

(60)

Therefore,

T\approx D.

(61)

Also,

V\dot V+g\dot h\approx0.

(62)

Hence,

\boxed{ \dot h \approx -\frac{V}{g}\dot V. }

(63)

Integrating,

\boxed{ h_2-h_1 = -\frac{V_2^2-V_1^2}{2g}. }

(64)

This relation quantifies the altitude sacrificed to gain speed.

Full Model versus Energy-State Approximation¶

The full longitudinal model includes:

\begin{aligned} \dot V &= \frac{T\cos\alpha-D}{m} -g\sin\gamma,\\ \dot\gamma &= \frac{T\sin\alpha+L}{mV} -\frac{g\cos\gamma}{V},\\ \dot h &= V\sin\gamma. \end{aligned}

(65)

The energy-state approximation reduces the problem to

\dot E = \frac{V}{m}(T-D).

(66)

The approximate solution is useful as:

an initial guess for a full indirect method;
an initial guess for direct collocation;
a means of explaining the qualitative trajectory;
a reduced-order onboard guidance law;
a method for generating performance envelopes.

Initial Guess Construction¶

A practical initial guess may be built in stages:

accelerate on the runway;
climb toward a transonic energy level;
dive approximately along a constant-energy contour;
perform a shallow energy-building climb;
complete a steep terminal climb.

The approximate path is then refined using a full optimal control solver.

Direct Collocation Formulation¶

Let the state be

\boldsymbol{x}= \begin{bmatrix} V & \gamma & h & x & m \end{bmatrix}^{\mathsf{T}},

(67)

and the control be

\boldsymbol{u} = \begin{bmatrix} \alpha & \eta \end{bmatrix}^{\mathsf{T}}.

(68)

The minimum-time problem is

\min_{\boldsymbol{x}(\cdot),\boldsymbol{u}(\cdot),t_f} t_f

(69)

subject to the full dynamics, endpoint constraints, and path constraints.

Using normalized time

\tau\in[0,1], \qquad t=t_0+\tau(t_f-t_0),

(70)

the dynamics become

\frac{\,\mathrm{d}\boldsymbol{x}}{\,\mathrm{d}\tau} = (t_f-t_0)\boldsymbol{f}(\boldsymbol{x},\boldsymbol{u}).

(71)

A collocation method enforces defects

\boldsymbol{\Delta}_k=0

(72)

at all mesh intervals.

Indirect Formulation¶

For the full model, define

\mathcal{H} = 1 + \lambda_V\dot V + \lambda_\gamma\dot\gamma + \lambda_h\dot h + \lambda_x\dot x + \lambda_m\dot m.

(73)

The costates satisfy

\dot{\boldsymbol{\lambda}} = -\frac{\partial\mathcal{H}}{\partial\boldsymbol{x}}.

(74)

The interior controls satisfy

\begin{aligned} \frac{\partial\mathcal{H}}{\partial\alpha} &=0,\\ \frac{\partial\mathcal{H}}{\partial\eta} &=0. \end{aligned}

(75)

The resulting two-point boundary-value problem is highly nonlinear. The energy-state solution is valuable because it supplies a physically meaningful initial guess.

Flight Envelope Constraints¶

The optimization must respect limits such as:

\begin{aligned} \alpha_{\min} &\le\alpha\le\alpha_{\max},\\ 0 &\le\eta\le1,\\ q &\le q_{\max},\\ n &\le n_{\max},\\ M &\le M_{\max},\\ h &\ge0. \end{aligned}

(76)

Dynamic pressure is

q=\frac12\rho V^2.

(77)

Load factor is approximately

n=\frac{L}{mg}.

(78)

These constraints may force the solution away from the unconstrained maximum-excess-power curve.

Numerical Procedure¶

A robust workflow is:

construct aerodynamic and propulsion lookup tables;
interpolate $T(V,h)$ and $D(V,h,\alpha)$ ;
compute energy contours;
compute the maximum-excess-power curve;
build an energy-state initial trajectory;
transcribe the full model using direct collocation;
solve the nonlinear program;
refine the mesh;
verify endpoint and path constraints;
compare with the reduced-order solution.

MATLAB Skeleton¶

function esa_demo()

    g = 9.81;
    m = 12000;

    Egrid = linspace(0.5e5,3.0e6,250);
    Vgrid = linspace(80,700,500);

    Vstar = nan(size(Egrid));
    Psmax = nan(size(Egrid));

    for i = 1:length(Egrid)

        E = Egrid(i);
        h = (E - 0.5*Vgrid.^2)/g;

        feasible = h >= 0;

        T = thrust_model(Vgrid,h);
        D = drag_model(Vgrid,h);

        Ps = Vgrid.*(T-D)/m;
        Ps(~feasible) = -Inf;

        [Psmax(i),idx] = max(Ps);
        Vstar(i) = Vgrid(idx);
    end

    figure;
    plot(Egrid,Vstar,'LineWidth',1.5);
    xlabel('Specific energy E');
    ylabel('Optimal speed V^*(E)');
    grid on;

    figure;
    plot(Egrid,Psmax,'LineWidth',1.5);
    xlabel('Specific energy E');
    ylabel('Maximum specific excess power');
    grid on;
end

function T = thrust_model(V,h)
    T0 = 90000;
    T = T0 .* exp(-h/18000) .* max(0.4,1-2e-4*V);
end

function D = drag_model(V,h)
    rho = 1.225 .* exp(-h/8500);
    S = 50;
    CD0 = 0.02 + 0.04*exp(-((V-340)/45).^2);
    D = 0.5 .* rho .* V.^2 .* S .* CD0;
end

Interpretation of the Famous Dive–Climb Trajectory¶

The non-intuitive trajectory can be understood as follows:

At low speed, the aircraft accelerates and climbs.
Near transonic speed, drag increases rapidly.
A direct climb allocates energy to altitude when speed is urgently needed.
A dive converts potential energy into kinetic energy.
The aircraft crosses the transonic region faster.
At higher supersonic speed, the excess-power characteristics improve.
The aircraft then climbs to the final altitude and speed.

The optimal trajectory therefore minimizes time through energy management rather than altitude monotonicity.

Approximation Error¶

The energy-state approximation introduces error because:

$\gamma$ dynamics are neglected;
lift equilibrium may not hold exactly;
thrust may not align with velocity;
mass may change;
path constraints may become active;
engine and aerodynamic models may be nonsmooth;
rapid maneuvers violate quasi-steady assumptions.

The approximation should therefore be validated against the full model.

Verification Tests¶

A credible solution should satisfy:

terminal speed and altitude constraints;
all path constraints;
monotonic mesh convergence;
agreement between numerical and finite-difference derivatives;
Hamiltonian consistency for indirect solutions;
agreement between approximate and refined trajectory structure;
physical feasibility across the entire flight envelope.

Common Errors¶

confusing total energy with kinetic energy;
omitting the cancellation of gravity terms in $\dot E$ ;
treating speed as an unconstrained algebraic variable;
maximizing thrust rather than excess power;
ignoring drag variation near Mach 1;
assuming altitude must increase monotonically;
using incomplete lookup tables without checking extrapolation;
interpreting the reduced model as exact;
failing to verify full-model dynamics;
using an arbitrary initial guess for an indirect method.

Summary¶

Total specific energy is

E=\frac{V^2}{2}+gh.

(79)

Its rate of change is

\dot E=\frac{V}{m}(T-D).

(80)

Minimum climb time can be written as

t_f=\int_{E_0}^{E_f}\frac{1}{\dot E}\,\mathrm{d} E.

(81)

Therefore, the energy-state approximation chooses

V^*(E)=\arg\max_V\dot E(E,V).

(82)

Constant-energy curves describe exchange between speed and altitude.
A dive may be optimal because it accelerates the aircraft through a high-drag region.
The reduced solution provides an excellent initial guess for the full optimal control problem.
The method remains one of the clearest examples of physics-informed trajectory optimization.