User:Eas4200c.f08.aeris.guan/929lecture

NACA Airfoil and Torsion of Uniform, Non-circular Bars
In homework 3, we are examining the NACA 4 digit airfoil series by using Matlab to do the analysis. Below are some pictures of some examples of the NACA airfoil. The one to the left describes fluid flow around the airfoil cross section (source: rit.edu) and the one to the right is a simple and crude cross sectional representation of the NACA airfoil with a (y,z) coordinate system where the y axis passes through the trailing edge. The y-axis is discretized in "ns" segments. The red circle highlights the infinitesimally small segment that will be analyzed. 

Here is the picture of the enlarged segment. As a crucial goal while analyzing the NACA airfoil is to determine the "average" area, we must devise a method to do this. It is possible to use a summation of infinitesimally small trapezoids, however, this method is not very elegant. In turn, we will use a method that utilizes triangles to determine the quadrature of the airfoil. The triangle will be a cross product a vector from an observation point (can be positioned anywhere) to a point on the airfoil and a vector of infinitesimally small length along the airfoil. These two vectors enclose a triangle of infinitesimally small area dA. This is shown in the picture below.  Torsion of uniform, non-circular bars are also examined in homework 3. This is the case in which warping will occur as opposed to the case of uniform, circular bars. Warping is the axial displacement along the x-axis of the bar (along the length of the bar) of a point on the deformed (rotated) cross section. The below pictures describe the bar and the bar's cross section when torsion is applied. Point P is a point on the (y,z) plane which moves to P' as the torque is applied. Beta describes the initial angular displacement measured from the y-axis and alpha is the angular displacement due to the torsion. Alpha is also measured from the y-axis.  Below is the magnified version of point P and its movement and displacement on the (y,z) plane of the cross section. $$u_y$$ and $$u_z$$ are the components of the displacement vector $$\vec {PP'}$$.  Why is the vector PP' perpendicular to OP? Isn't it more intuitive that one would draw an arc with radius |OP| and plot point P' at an angle alpha away from vector OP? The perpendicular nature of PP' to OP is because the angle alpha is very small. The proof is below. Let there be two vectors OP and OP' of equal length (|OP|=|OP'|=R). The displacement differences in the points P and P' are denoted by $$u_z$$and $$u_y$$ where z denotes vertical displacement and y denotes horizontal displacement. 

$$\displaystyle Rsin\alpha \cong R\alpha (\alpha small)$$ $$\displaystyle R(1-cos\alpha) \cong 0 (first order)$$ Thus, it is easy to see that since the horizontal displacement of point P' from point P is zero, point P' directly above point P in this case. That makes vectors OP and PP' perpendicular.