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Subsections


11 Hang Glider

Maximize the final horizontal position of a hang glider while in the presence of a thermal updraft.

Formulation

The formulation of this problem follows [9]. The equations of motion for the hang glider are

\begin{displaymath}
x'' = \frac{1}{m}(-L\sin (\eta )-D\cos (\eta )) , \qquad
y'' = \frac{1}{m}(L\cos (\eta )-D\sin (\eta )) - g ,
\end{displaymath} (8)

where $ ( x, y ) $ is the position of the glider, $m$ is the mass of the glider, $g$ is the gravitational constant, and the function $ \eta $ is defined by

\begin{displaymath}
\sin (\eta ) = \frac{w( x,y')}{v(x,x',y')} , \qquad
\cos (\eta ) = \frac{x'}{v(x,x',y')} ,
\end{displaymath}

where

\begin{displaymath}
v(x,x',y') = \sqrt{x'^{2}+w(x,y')^{2}}, \qquad
w(x,y') = y'- u (x) ,
\end{displaymath}


\begin{displaymath}
u (x) = u_{c}(1-r(x)) \exp ( -r(x) ), \qquad
r(x) = \left ( \frac{x}{r_c}-2.5 \right ) ^{2} ,
\end{displaymath}

and constants $ u_{c} = 2.5 $ and $ r_c = 100 $. The updraft function $ u $ is positive in a neighborhood of $ x = 2.5 \, r_c $ but drops to zero exponentially away from $ x = 2.5 \, r_c $. The functions $D$ and $L$ are defined by

\begin{displaymath}
D(x,x',y',c_{L}) = \frac{1}{2} \left ( c_{0} + c_1 c_{L}^{2}...
...
L(x,x',y',c_{L}) = \frac{1}{2}c_{L} \rho S v (x,x',y') ^{2} ,
\end{displaymath}

where $S$ is the wing area, $ \rho $ is the air density, $ c_{L} $ is the aerodynamic lift coefficient, and $ c_{0} + c_1 c_{L}^{2} $ is the drag coefficient. For this glider

\begin{displaymath}
c_{0} = 0.034 , \quad
c_1 = 0.069662 , \qquad
S = 14 , \quad
\rho = 1.13 .
\end{displaymath}

The aerodynamic lift coefficient $ c_L $ must satisfy the bounds

\begin{displaymath}
0 \leq c_{L} (t) \leq c_{\max} ,
\end{displaymath}

and we also impose the natural bounds $ x \ge 0 $ and $ x' \ge 0 $. In this problem $ c_{\max} = 1.4 $, $ m = 100 $, $ g = 9.81 $, and the boundary conditions are

\begin{displaymath}
x(0) = 0 , \qquad y(0) = 1000 , \qquad y(t_{f}) = 900,
\end{displaymath}


\begin{displaymath}
x'(0) = x'(t_{f}) = 13.23 , \qquad
y'(0) = y'(t_{f}) = -1.288 .
\end{displaymath}

Discretization is done with a uniform time step and the trapezoidal rule over \( n_h \) intervals. Data for this problem is shown in Table 11.1.



Table 11.1: Hang glider problem data
Variables \( 5( n_h + 1 ) + 1 \)
Constraints \( 4 n_h \)
Bounds \( 3( n_h + 1 ) \)
Linear equality constraints 0
Linear inequality constraints 0
Nonlinear equality constraints \( 4 n_h \)
Nonlinear inequality constraints 0
Nonzeros in \( \nabla ^{2}f(x) \) 0
Nonzeros in \( c'(x) \) \( 25 n_h \)

Performance

Results for the AMPL implementation are shown in Table 11.2. For starting points we use $ t_f = 1 $ and the functions $ x' = x'(0) $, $ y' = y'(0) $, and

\begin{displaymath}
x(t) = x(0) + x'(0) \left ( \frac{t}{t_f} \right ), \qquad
y...
...left ( y(t_f) - y(0) \right ) \left ( \frac{t}{t_f} \right ) ,
\end{displaymath}

evaluated at the grid points. The initial value for the control is $ c_L(t) = c_{\max} $.

MINOS fails to produce a solution for any of the problem versions we present it, declaring each an infeasible problem (or bad starting guess).



Table 11.2: Performance on hang glider problem
Solver $n_h=50$ $n_h=100$ $n_h=200$ $n_h=400$
LANCELOT $\ddagger$ 211.76 s 693.74 s $\ddagger$
$f$ $\ddagger$ 1.25461e+03 1.24889e+03 $\ddagger$
$c$ violation $\ddagger$ 9.32090e-08 2.86060e-07 $\ddagger$
iterations $\ddagger$ 383 539 $\ddagger$
LOQO $\ddagger$ 2174.8 s 2601.83 s $\ddagger$
$f$ $\ddagger$ 1.25461e+03 1.24880e+03 $\ddagger$
$c$ violation $\ddagger$ 2.1e-11 1.6e-12 $\ddagger$
iterations $\ddagger$ 14257 7206 $\ddagger$
MINOS 28.06 s 95.8 s 206.62 s 732.06 s
$f$ 2.12853e+04$\dagger$ 4.55001e+05$\dagger$ 7.47275e+03$\dagger$ 5.93037e+03$\dagger$
$c$ violation 3.2e+03$\dagger$ 6.0e+06$\dagger$ 6.4e+02$\dagger$ 6.1e+02$\dagger$
iterations 83 124 118 161
SNOPT 11.14 s 44.04 s 240.36 s 1268.67 s
$f$ 1.28239e+03 1.25461e+03 1.24889e+03 1.24797e+03
$c$ violation 1.7e-10 1.8e-10 1.2e-11 5.1e-11
iterations 73 94 125 166
$\dagger$ Errors or warnings. $\ddagger$ Timed out.

Figure 11.1 shows the altitude and control function $ c_L $ as a function of time. The glider starts at an altitude of $ y(0) = 1000 $ and descends until the glider meets the updraft centered at $ x = 250 $. As a result the glider climbs and then descends to the desired final altitude of $ y(t_f) = 900 $ at time $ t_f = 105 $.

Figure 11.1: Altitude and control $ c_L$ for the hang glider problem
\includegraphics[width=2.5in]{ps/glider_y.eps} \includegraphics[width=2.5in]{ps/glider_cL.eps}

Figure 11.2 shows velocities $ x' $ and $ y' $ as a function of time. Note, in particular, the erratic behavior of the velocities while the control is in the bang-region where $ c_L(t) = c_{\max} $.

Figure 11.2: Velocities $ x' $ and $ y' $ for the hang glider problem
\includegraphics[width=2.5in]{ps/glider_vx.eps} \includegraphics[width=2.5in]{ps/glider_vy.eps}


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Liz Dolan
2001-01-02