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Supersonic Nose Cone Aerodynamic Pressure (Aerospace & Mechanical Engineering) Solution

Formulas & Definitions

An aerospace aerodynamicist evaluates the supersonic airflow across the nose cone of a high-speed atmospheric flight test vehicle. In wind tunnel validation studies, the upper curved aerodynamic profile of the fuselage nose cone is modeled by the paraboloid of revolution:

z=20−0.05x2−0.1y2z = 20 - 0.05x^2 - 0.1y^2
where coordinates x,y,zx, y, z are measured in centimeters (cm\text{cm}).

During supersonic Mach 2 flight, a pitot-static pressure sensor probe is installed at the surface location (x0,y0)=(10,5) cm(x_0, y_0) = (10, 5)\text{ cm}.

  1. Compute the exact height z0=f(10,5)z_0 = f(10, 5) of the fuselage at the sensor installation point.
  2. Determine the general analytical expressions for the partial derivatives ∂z∂x\frac{\partial z}{\partial x} and ∂z∂y\frac{\partial z}{\partial y}.
  3. Evaluate the partial derivatives at (10,5)(10, 5) and state the equation of the tangent plane to the fuselage surface at this location in standard form.
  4. Extract the outward-pointing unit normal vector n^\hat{n} to the fuselage surface at (10,5)(10, 5), and explain why aerodynamicists use n^\hat{n} to compute the local static pressure force vector F⃗p=−pn^ dA\vec{F}_p = -p \hat{n} \, dA.

Step-by-Step Solution:

  1. Compute Fuselage Height at (10,5)(10, 5):

    z0=f(10,5)=20−0.05(10)2−0.1(5)2z_0 = f(10, 5) = 20 - 0.05(10)^2 - 0.1(5)^2
    z0=20−0.05(100)−0.1(25)=20−5.0−2.5=12.5 cmz_0 = 20 - 0.05(100) - 0.1(25) = 20 - 5.0 - 2.5 = 12.5\text{ cm}

  2. Compute General Partial Derivatives:

    • Differentiating with respect to xx (holding yy constant):
      ∂z∂x=∂∂x(20−0.05x2−0.1y2)=−0.10x\frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(20 - 0.05x^2 - 0.1y^2) = -0.10x
    • Differentiating with respect to yy (holding xx constant):
      ∂z∂y=∂∂y(20−0.05x2−0.1y2)=−0.20y\frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(20 - 0.05x^2 - 0.1y^2) = -0.20y

  3. Evaluate and Construct Tangent Plane Equation:

    • At (x0,y0)=(10,5)(x_0, y_0) = (10, 5):
      fx(10,5)=−0.10(10)=−1.0f_x(10, 5) = -0.10(10) = -1.0
      fy(10,5)=−0.20(5)=−1.0f_y(10, 5) = -0.20(5) = -1.0
    • State the Tangent Plane Formula (defined in Tangent Planes):
      z=z0+fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0)z = z_0 + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)
    • Substitute values:
      z=12.5−1.0(x−10)−1.0(y−5)z = 12.5 - 1.0(x - 10) - 1.0(y - 5)
      z=12.5−x+10−y+5z = 12.5 - x + 10 - y + 5
      z=−x−y+27.5z = -x - y + 27.5
    • In standard linear form:
      x+y+z=27.5x + y + z = 27.5

  4. Extract Normal Vector and Aerodynamic Application:

    • From Tangent Planes, the upward outward normal vector is:
      n⃗up=⟨−fx,−fy,1⟩=⟨−(−1.0),−(−1.0),1⟩=⟨1.0,1.0,1.0⟩=i^+j^+k^\vec{n}_{\text{up}} = \langle -f_x, -f_y, 1 \rangle = \langle -(-1.0), -(-1.0), 1 \rangle = \langle 1.0, 1.0, 1.0 \rangle = \hat{i} + \hat{j} + \hat{k}
    • Compute its magnitude:
      ∥n⃗∥=12+12+12=3\|\vec{n}\| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}
    • The outward unit normal vector is:
      n^=⟨13,13,13⟩\hat{n} = \left\langle \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right\rangle
    • Aerodynamic Application: Air pressure acts purely compressive and strictly normal to any solid immersed boundary. Aerodynamicists integrate −pn^ dA-p \hat{n} \, dA over the surface to decompose the resultant aerodynamic load into total vehicle lift (vertical +z+z component) and wave drag (streamwise −x-x component).