
Aerospace Design via Quasiconvex Optimization
=============================================

Consider a triangle, or a wedge, located within a hypersonic flow. A
standard aerospace design optimization problem is to design the wedge to
maximize the lift-to-drag ratio (L/D) (or conversely minimize the D/L
ratio), subject to certain geometric constraints. In this example, the
wedge is known to have a constant hypotenuse, and our job is to choose
its width and height.

The drag-to-lift ratio is given by

.. math::


   \frac{\mathrm{D}}{\mathrm{L}} = \frac{\mathrm{c_d}}{\mathrm{c_l}},

where :math:`\mathrm{c_d}` and :math:`\mathrm{c_l}` are drag and lift
coefficients, respectively, that are obtained by integrating the
projection of the pressure coefficient in directions parallel to, and
perpendicular to, the body.

It turns out that the the drag-to-lift ratio is a quasilinear function,
as we'll now show. We will assume the pressure coefficient is given by
the Newtonian sine-squared law for whetted areas of the body,

.. math::


   \mathrm{c_p} = 2(\hat{v}\cdot\hat{n})^2

and elsewhere :math:`\mathrm{c_p} = 0`. Here, :math:`\hat{v}` is the
free stream direction, which for simplicity we will assume is parallel
to the body so that, :math:`\hat{v} = \langle 1, 0 \rangle`, and
:math:`\hat{n}` is the local unit normal. For a wedge defined by width
:math:`\Delta x`, and height :math:`\Delta y`,

.. math::


   \hat{n} = \langle -\Delta y/s,-\Delta x/s \rangle

where :math:`s` is the hypotenuse length. Therefore,

.. math::


   \mathrm{c_p} = 2((1)(-\Delta y/s)+(0)(-\Delta x/s))^2 = \frac{2 \Delta y^2}{s^2}

The lift and drag coefficients are given by

.. math::


   \begin{align*}
   \mathrm{c_d} &= \frac{1}{c}\int_0^s -\mathrm{c_p}\hat{n}_x \mathrm{d}s \\
   \mathrm{c_l} &= \frac{1}{c}\int_0^s -\mathrm{c_p}\hat{n}_y \mathrm{d}s
   \end{align*}

Where :math:`c` is the reference chord length of the body. Given that
:math:`\hat{n}`, and therefore :math:`\mathrm{c_p}` are constant over
the whetted surface of the body,

.. math::


   \begin{align*}
   \mathrm{c_d} &= -\frac{s}{c}\mathrm{c_p}\hat{n}_x = \frac{s}{c}\frac{2 \Delta y^2}{s^2}\frac{\Delta y}{s} \\
   \mathrm{c_l} &= -\frac{s}{c}\mathrm{c_p}\hat{n}_y = \frac{s}{c}\frac{2 \Delta y^2}{s^2}\frac{\Delta x}{s}
   \end{align*}

Assuming :math:`s=1`, so that :math:`\Delta y = \sqrt{1-\Delta x^2}`,
plugging in the above into the equation for :math:`D/L`, we obtain

.. math::


   \frac{\mathrm{D}}{\mathrm{L}} = \frac{\Delta y}{\Delta x} = \frac{\sqrt{1-\Delta x^2}}{\Delta x} = \sqrt{\frac{1}{\Delta x^2}-1}.

This function is representable as a DQCP, quasilinear function. We plot
it below, and then we write it using DQCP.

.. code:: ipython3

    %matplotlib inline
    
    import matplotlib
    import numpy as np
    import matplotlib.pyplot as plt
    import math
    
    x = np.linspace(.25,1,num=201)
    obj = []
    for i in range(len(x)):
        obj.append(math.sqrt(1/x[i]**2-1))
    
    plt.plot(x,obj)




.. parsed-literal::

    [<matplotlib.lines.Line2D at 0x7f571d7bfdd8>]




.. image:: hypersonic_shape_design_files/hypersonic_shape_design_1_1.png


.. code:: ipython3

    import cvxpy as cp
    
    x = cp.Variable(pos=True)
    obj = cp.sqrt(cp.inv_pos(cp.square(x))-1)
    print("This objective function is", obj.curvature)


.. parsed-literal::

    This objective function is QUASILINEAR


Minimizing this objective function subject to constraints representing
payload requirements is a standard aerospace design problem. In this
case we will consider the constraint that the wedge must be able to
contain a rectangle of given length and width internally along its
hypotenuse. This is representable as a convex constraint.

.. code:: ipython3

    a = .05 # USER INPUT: height of rectangle, should be at most b
    b = .65 # USER INPUT: width of rectangle
    constraint = [a*cp.inv_pos(x)-(1-b)*cp.sqrt(1-cp.square(x))<=0]
    print(constraint)


.. parsed-literal::

    [Inequality(Expression(CONVEX, UNKNOWN, ()))]


.. code:: ipython3

    prob = cp.Problem(cp.Minimize(obj), constraint)
    prob.solve(qcp=True, verbose=True)
    print('Final L/D Ratio = ', 1/obj.value)
    print('Final width of wedge = ', x.value)
    print('Final height of wedge = ', math.sqrt(1-x.value**2))


.. parsed-literal::

    
    ********************************************************************************
    Preparing to bisect problem
    
    minimize 0.0
    subject to 0.05 * var30766 + -0.35 * var30793 <= 0.0
               SOC(reshape(var30747 + var30766, (1,)), Vstack(reshape(var30747 + -var30766, (1, 1)), reshape(2.0 * 1.0, (1, 1))))
               SOC(reshape(var30779 + 1.0, (1,)), Vstack(reshape(var30779 + -1.0, (1, 1)), reshape(2.0 * var30747, (1, 1))))
               SOC(reshape(1.0 + -var30779 + 1.0, (1,)), Vstack(reshape(1.0 + -var30779 + -1.0, (1, 1)), reshape(2.0 * var30793, (1, 1))))
               power(power(power(param30811, 2) + --1.0, -1), 1/2) <= var30747
    
    Finding interval for bisection ...
    initial lower bound: 0.000000
    initial upper bound: 1.000000
    
    (iteration 0) lower bound: 0.000000
    (iteration 0) upper bound: 1.000000
    (iteration 0) query point: 0.500000 
    (iteration 0) query was feasible. Solution(status=optimal, opt_val=0.0, primal_vars={30766: array(1.28425055), 30793: array(0.32048066), 30747: 0.9203698369509382, 30779: array(0.86287821)}, dual_vars={30764: 1.184352986830617e-10, 30775: array([ 7.68139086e-12, -9.11799720e-13, -6.85059567e-12]), 30788: array([ 6.73308751e-11,  7.50722737e-12, -6.55220021e-11]), 30802: array([ 4.04979217e-11,  3.43109122e-11, -1.68754271e-11]), 30835: 1.4165742899966837e-10}, attr={'solve_time': 6.0109e-05, 'setup_time': 4.4997e-05, 'num_iters': 7}))
    
    (iteration 5) lower bound: 0.125000
    (iteration 5) upper bound: 0.156250
    (iteration 5) query point: 0.140625 
    (iteration 5) query was infeasible.
    
    (iteration 10) lower bound: 0.145508
    (iteration 10) upper bound: 0.146484
    (iteration 10) query point: 0.145996 
    (iteration 10) query was feasible. Solution(status=optimal, opt_val=0.0, primal_vars={30766: array(1.01067238), 30793: array(0.14440604), 30747: 0.9895144829793, 30779: array(0.97914383)}, dual_vars={30764: 1.2610785752467482e-05, 30775: array([ 6.37367039e-07,  6.73702792e-09, -6.37322961e-07]), 30788: array([ 1.50627898e-05,  1.58286953e-07, -1.50619494e-05]), 30802: array([ 7.77053008e-06,  7.45051237e-06, -2.20683981e-06]), 30835: 2.948014872712083e-05}, attr={'solve_time': 0.000114922, 'setup_time': 3.6457e-05, 'num_iters': 10}))
    
    (iteration 15) lower bound: 0.145874
    (iteration 15) upper bound: 0.145905
    (iteration 15) query point: 0.145889 
    (iteration 15) query was infeasible.
    
    Bisection completed, with lower bound 0.145897 and upper bound 0.1458979
    ********************************************************************************
    
    Final L/D Ratio =  6.854107648695203
    Final width of wedge =  0.9895238539767502
    Final height of wedge =  0.14436946495363565


Once the solution has been found, we can create a plot to verify that
the rectangle is inscribed within the wedge.

.. code:: ipython3

    y = math.sqrt(1-x.value**2)
    lambda1 = a*x.value/y
    lambda2 = a*x.value**2/y+a*y
    lambda3 = a*x.value-y*(a*x.value/y-b)
    
    plt.plot([0,x.value],[0,0],'b.-')
    plt.plot([0,x.value],[0,-y],'b.-')
    plt.plot([x.value,x.value],[0,-y],'b.-')
    
    pt1 = [lambda1*x.value,-lambda1*y]
    pt2 = [(lambda1+b)*x.value,-(lambda1+b)*y]
    pt3 = [(lambda1+b)*x.value+a*y,-(lambda1+b)*y+a*x.value]
    pt4 = [lambda1*x.value+a*y,-lambda1*y+a*x.value]
    
    plt.plot([pt1[0],pt2[0]],[pt1[1],pt2[1]],'r.-')
    plt.plot([pt2[0],pt3[0]],[pt2[1],pt3[1]],'r.-')
    plt.plot([pt3[0],pt4[0]],[pt3[1],pt4[1]],'r.-')
    plt.plot([pt4[0],pt1[0]],[pt4[1],pt1[1]],'r.-')
    
    plt.axis('equal')




.. parsed-literal::

    (-0.04947620645689951,
     1.0390003355948896,
     -0.15158793820131744,
     0.0072184732476817896)




.. image:: hypersonic_shape_design_files/hypersonic_shape_design_7_1.png

