Aerodynamics exercises
======================

.. include:: /_generated/nav/aerodynamics.rst

Three problems built on the :doc:`aerodynamics gallery </api/gallery/aerodynamics/index>`.
Try each one before opening its solution. The solutions run when the
documentation is built, and the test suite checks every number they print.

1. Stall speed at altitude
--------------------------

A light aircraft weighs 10.7 kN, has a wing area of 16.2 m² and a maximum
lift coefficient of 1.6 with flaps. Find its stall speed at sea level and
at 3000 m in the International Standard Atmosphere. By what factor does it
rise, and why does the indicated airspeed at the stall not change?

*Builds on:* :doc:`/api/gallery/aerodynamics/forces/plot_04_wright_smeaton_stall_speed`
and :doc:`/api/gallery/aerodynamics/atmosphere/plot_01_international_standard_atmosphere`.

.. dropdown:: Solution

   .. plot::
      :context: reset

      import numpy as np

      from engineeringkit.aerodynamics import isa_atmosphere, stall_speed

      W, S, Clmax = 10.7e3, 16.2, 1.6
      air = isa_atmosphere(np.array([0.0, 3000.0]))
      vs = stall_speed(W, air.density, S, Clmax)
      for h, rho, v in zip(air.altitude, air.density, vs):
          print(f"{h:4.0f} m: density {rho:.4f} kg/m^3, stall speed {v:.2f} m/s")
      print(f"ratio {vs[1] / vs[0]:.4f}")
      #    0 m: density 1.2250 kg/m^3, stall speed 25.96 m/s
      # 3000 m: density 0.9091 kg/m^3, stall speed 30.14 m/s
      # ratio 1.1608
      assert np.isclose(vs[1] / vs[0], np.sqrt(air.density[0] / air.density[1]))

   The true stall speed scales as :math:`1/\sqrt{\rho}`, 16% higher at
   3000 m. The airspeed indicator measures dynamic pressure
   :math:`\tfrac12\rho V^2`, and the stall happens at the same dynamic
   pressure, :math:`W/(S\,C_{L,\max})`, at any altitude, so the indicated
   stall speed is the same.

2. The best glide
-----------------

A sailplane has a zero-lift drag coefficient of 0.012 and a wing of
aspect ratio 20 with a span efficiency of 0.9. At what lift coefficient
does it glide farthest, and what is its glide ratio? Check against the
closed form :math:`(L/D)_{\max} = \tfrac12\sqrt{\pi e AR/C_{D0}}`, and show
that at the best glide the induced drag equals the zero-lift drag.

*Builds on:* :doc:`/api/gallery/aerodynamics/wings/plot_03_oswald_drag_polar`.

.. dropdown:: Solution

   .. plot::
      :context: reset

      import matplotlib.pyplot as plt
      import numpy as np

      from engineeringkit.aerodynamics import best_glide_cl, drag_polar, glide_ratio, induced_drag_coefficient

      Cd0, AR, e = 0.012, 20.0, 0.9
      cl_best = best_glide_cl(Cd0, AR, e)
      ld_best = glide_ratio(cl_best, drag_polar(cl_best, Cd0, AR, e))
      print(f"best-glide CL {cl_best:.3f}, glide ratio {ld_best:.1f}")
      # best-glide CL 0.824, glide ratio 34.3
      assert np.isclose(ld_best, 0.5 * np.sqrt(np.pi * e * AR / Cd0))
      assert np.isclose(induced_drag_coefficient(cl_best, AR, e), Cd0)

      cl = np.linspace(0.1, 1.6, 151)
      fig, ax = plt.subplots(figsize=(5, 3.2))
      ax.plot(cl, glide_ratio(cl, drag_polar(cl, Cd0, AR, e)))
      ax.plot(cl_best, ld_best, "o")
      ax.set(xlabel="lift coefficient", ylabel="L/D")
      fig.tight_layout()

   A glide ratio of 34 is typical of a club-class sailplane. Flying
   faster than the best glide (lower :math:`C_L`) loses glide ratio
   slowly; flying slower loses it fast.

3. A normal shock in a supersonic intake
----------------------------------------

Air at Mach 2 meets a normal shock. Find the Mach number and the static
pressure ratio behind it. Then compare with the pressure recovered by an
ideal isentropic diffuser: what fraction of the stagnation pressure does
the shock destroy?

*Builds on:* :doc:`/api/gallery/aerodynamics/compressible/plot_05_rankine_hugoniot_normal_shock`
and :doc:`/api/gallery/aerodynamics/compressible/plot_03_saint_venant_stagnation`.

.. dropdown:: Solution

   .. plot::
      :context: reset

      import numpy as np

      from engineeringkit.aerodynamics import normal_shock_mach2, normal_shock_pressure_ratio, stagnation_pressure

      M1 = 2.0
      M2 = normal_shock_mach2(M1)
      p2_p1 = normal_shock_pressure_ratio(M1)
      p01, p02 = stagnation_pressure(1.0, M1), stagnation_pressure(p2_p1, M2)
      print(f"M2 = {M2:.4f}, p2/p1 = {p2_p1:.3f}")
      print(f"stagnation pressure recovery p02/p01 = {p02 / p01:.4f}")
      # M2 = 0.5774, p2/p1 = 4.500
      # stagnation pressure recovery p02/p01 = 0.7209
      assert np.isclose(M2, np.sqrt(1 / 3)) and np.isclose(p2_p1, 4.5)

   The shock slows the flow to subsonic speed but destroys 28% of the
   stagnation pressure, which is why supersonic intakes slow the flow
   through a series of oblique shocks before a final weak normal one.