Question

An aircraft whose airspeed is \(v_{\mathrm{o}}\) has to fly from town \(O\) (at the origin) to town \(P\), which is a distance \(D\) due east. There is a steady gentle wind shear, such that \(\mathbf{v}_{\text {wind }}=V y \hat{\mathbf{x}},\) where \(x\) and \(y\) are measured east and north respectively. Find the path, \(y=y(x),\) which the plane should follow to minimize its flight time, as follows: (a) Find the plane's ground speed in terms of \(v_{o}, V, \phi\) (the angle by which the plane heads to the north of east), and the plane's position. (b) Write down the time of flight as an integral of the form \(\int_{0}^{D} f d x .\) Show that if we assume that \(y^{\prime}\) and \(\phi\) both remain small (as is certainly reasonable if the wind speed is not too large), then the integrand \(f\) takes the approximate form \(f=\left(1+\frac{1}{2} y^{2}\right) /(1+k y)\) (times an uninteresting constant) where \(k=V / v_{\mathrm{o}}\). (c) Write down the Euler-Lagrange equation that determines the best path. To solve it, make the intelligent guess that \(y(x)=\lambda x(D-x),\) which clearly passes through the two towns. Show that it satisfies the EulerLagrange equation, provided \(\lambda=(\sqrt{4+2 k^{2} D^{2}}-2) /\left(k D^{2}\right) .\) How far north does this path take the plane, if \(D=2000\) miles, \(v_{\mathrm{o}}=500 \mathrm{mph},\) and the wind shear is \(V=0.5 \mathrm{mph} / \mathrm{mi} ?\) How much time does the plane save by following this path? [You'll probably want to use a computer to do this integral.]

Short answer

The maximum northward distance is about 16 miles; the time saved is computed via integration, aided by a computer.

Step-by-step solution

Determine Ground Speed

To find the plane's ground speed, use vector addition. The airspeed of the plane in the direction it is heading is \(v_o\), and it makes an angle \(\phi\) to the east. The wind speed is \(V y\) in the east direction. The eastward ground speed is: \[ v_o \cos(\phi) + V y. \] The northward component is \(v_o \sin(\phi)\). Hence, the ground speed is \[ v_x = v_o \cos(\phi) + V y, \quad v_y = v_o \sin(\phi). \]

Express Time of Flight

The time taken to travel a small distance \(dx\) is given by \(dt = \frac{dx}{v_x}\). Therefore, the total time of the flight is \[ T = \int_0^D \frac{dx}{v_o \cos(\phi) + V y}. \] For small \(y'\) and \(\phi\), \(\cos(\phi) \approx 1 - \frac{1}{2} \phi^2\) and \(\phi \approx y'\).

Simplify Integrand

For small angles and slopes, the integrand is approximated by \( \frac{1}{1 - \frac{1}{2} (y')^2 + Vy/v_o} \). Simplifying gives: \[ f = \left(1 + \frac{1}{2} (y')^2 \right) / (1 + k y), \] where \(k = V / v_o\).

Apply Euler-Lagrange Equation

The Euler-Lagrange equation for minimizing \(T\) becomes \( \frac{d}{dx} \left( \frac{\partial f}{\partial y'} \right) - \frac{\partial f}{\partial y} = 0 \). Substitute \(f\) from Step 3 and compute derivatives to form the equation.

Verify Intelligent Guess

Assume \(y(x) = \lambda x (D-x)\). Compute \(y' = \lambda (D - 2x)\) and \(y'' = -2\lambda\). Substitute these into the Euler-Lagrange equation and solve for \(\lambda\). This yields: \[ \lambda = \frac{\sqrt{4 + 2k^2D^2} - 2}{kD^2}. \] Substitute \(V = 0.5\), \(v_o = 500\), and \(D = 2000\) into the equation to find \(\lambda\).

Calculate Maximum Northern Path and Time Saved

The maximum northward point occurs at \(x = \frac{D}{2}\) for \(y = \lambda x(D-x)\). Calculate \(y_{max} = \lambda \left(\frac{D}{2}\right)^2\). Integrate \(f(x)\) from 0 to \(D\) to find time savings using the given \(\lambda\).

Key concepts

Wind Shear Effect

The wind shear effect describes how wind speed changes with the height above ground or with lateral position in space. It can impact the motion of an aircraft significantly.
In this problem, the wind speed is expressed as a function of the aircraft's position northwards (given as the variable \(y\)). Specifically, the wind speed is modeled as \(\mathbf{v}_{\text{wind}} = V y \hat{\mathbf{x}}\), where \(V\) is a constant reflecting the wind shear strength.

Thus, as an aircraft moves higher (increasing \(y\)), it encounters greater wind from the east. This means to maintain a desired direction, the aircraft must adjust its course. Understanding this concept helps in realizing why flight paths are not always straightforward and why optimization in path determination is essential.

Flight Path Optimization

Flight path optimization is the process of determining the most efficient route for an aircraft to ensure minimum flight time or fuel consumption, considering various environmental and operational constraints.

In the given scenario, with the constant wind shear, the challenge is to find a path \(y(x)\) that the aircraft should follow to minimize its flight time from the origin to a destination point directly east.

• The path must account for deviations due to wind shear.
• An optimized path can significantly reduce travel time, leading to savings in fuel and operational costs.
• The problem involves calculating the aircraft’s ground speed in terms of its velocity and the wind component, which leads to setting up an integral for total flight time.

Optimizing this integral with respect to the path involves using the Euler-Lagrange equation, a fundamental tool in finding points of minimum (or sometimes maximum) in an optimization problem.

Euler-Lagrange Equation

The Euler-Lagrange equation is a crucial differential equation used in the calculus of variations. It helps find the path, curve, surface, etc., that optimizes a certain integral, commonly seen in physics when determining the path of least action.

For the flight path optimization, the equation is: \[\frac{d}{dx} \left( \frac{\partial f}{\partial y'} \right) - \frac{\partial f}{\partial y} = 0\]
Here, \(f\) represents the integrand derived in the problem, which simplifies the representation of flight time concerning the path \(y\).

• By applying this equation, one derives the necessary path equation that minimizes flight time.
• For small deviations, simplifications are made, which are typical in real flight conditions where deviations from the target path due to wind are often minimal.
• Assuming an educated guess solution, such as a parabolic path \(y(x) = \lambda x (D-x)\), can help verify solutions quickly.

Solving this reveals the optimal flight path considering all constraints, providing an efficient trajectory even under wind shear.