Question

consider an infinitely thin airfoil defined by: y = 0.218*sin(πx) 0.06*sin(2πx) where the leading and trailing edges are at x = 0 m and x = 1 m, respectively. Find the components of the unit vector oriented outward normal to the upper surface, at x = 0.136 m. Give the answer to three decimal places. X component Y component

Answer 1

To find the normal vector components at a specific point on the airfoil, we differentiate the airfoil equation to find the slope of the tangent, calculate the negative reciprocal to get the normal slope, and use trigonometric relationships to find the normalized horizontal and vertical components.

Step-by-step explanation:

To find the components of the unit vector normal to the upper surface of the airfoil described by the equation y = 0.218sin(πx) - 0.06sin(2πx), we first need to determine the slope of the tangent to the airfoil at x = 0.136 m. Calculating the derivative dy/dx will give us the slope at any point x along the airfoil. At the specific point x = 0.136 m, we find the derivative, which will be the negative reciprocal of the slope of the normal vector at that point since the normal vector is perpendicular to the tangent.

After computing the slope of the normal vector, we can represent the normal vector in terms of its horizontal (X) and vertical (Y) components using trigonometric relationships. Since we need a unit vector, we will divide these components by the magnitude of the vector to normalize it. Therefore, the final step is to ensure that the vector components have a unit length by dividing each component by the vector's magnitude.