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

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:

where coordinates $x, y, z$ are measured in centimeters ($\text{cm}$).

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

1. Compute the exact height $z_0 = f(10, 5)$ of the fuselage at the sensor installation point.
2. Determine the general analytical expressions for the partial derivatives $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$.
3. Evaluate the partial derivatives at $(10, 5)$ 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 $\hat{n}$ to the fuselage surface at $(10, 5)$, and explain why aerodynamicists use $\hat{n}$ to compute the local static pressure force vector $\vec{F}_p = -p \hat{n} \, dA$.

Step-by-Step Solution:

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

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

2. Compute General Partial Derivatives:

   • Differentiating with respect to $x$ (holding $y$ constant):
     $$\frac{\partial z}{\partial x} = \frac{\partial}{\partial x}(20 - 0.05x^2 - 0.1y^2) = -0.10x$$
   • Differentiating with respect to $y$ (holding $x$ constant):
     $$\frac{\partial z}{\partial y} = \frac{\partial}{\partial y}(20 - 0.05x^2 - 0.1y^2) = -0.20y$$

3. Evaluate and Construct Tangent Plane Equation:

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

4. Extract Normal Vector and Aerodynamic Application:

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