# Variables in calculation of drag coefficient

Okay, so I looked up drag force equation, and I found that the equation involved the drag coefficient. Then I looked up the drag coefficient, and the equation for it involved the drag force. Eventually this loop stopped as I found this section on a Wikipedia page explaining how to find the drag coefficient of a specific shape. It all makes sense to me, but I don't understand some of the variables they defined to me, specifically the vectors (yes, I know about vectors, unit vectors, and dot products, but that's not what I'm asking). It might just be the fact that I haven't taken an AP physics class yet, but the variables don't seem to define the magnitudes of n or t. As well, by the way it's described, it seems î and n have the same direction, which can't be true or the dot product of t*î would be zero. And yes, I know, Wikipedia isn't always trustworthy 100% of the time (nothing is), but I've found that Wikipedia has, for the most part, provided accurate information regarding math and science. If you'd like to get a better reference for this topic, links to this Wikipedia page are down below (look in "Blunt and streamlined body flows"). If you know what these terms are supposed to mean, please explain to me.

https://en.wikipedia.org/wiki/Drag_coefficient

• This is a case of paralysis by analysis. The data that you listed require a lot of MEASUREMENTS before you can calculate a drag coefficient. It would be much easier to lump all of those measurements into one term (the drag coefficient itself), and set up an experimental procedure whereby you measure the drag force on a specific object, and back-calculate the drag coefficient involved. In other words, skip all the integrals, and go directly with the value that nature is giving you. Commented Jul 14, 2017 at 0:16
• David White- that's not what I was asking for. My interest is theoretical, not experimental. I know all the formulas and they're pretty easy to me (definitely much easier than determining the drag coefcicient of every conceivable shape by performing infinity trials.) Commented Jul 14, 2017 at 0:34
• @David White If you could please just elaborate on the values of t, n, and î, that would be most helpful. If you need further information I've left a link to the website in the question. Thank you! Commented Jul 14, 2017 at 2:25
• The equation would be OK if $\mathbf{\hat{i}}$ were the unit vector in the direction of relative motion between the object and the fluid. Commented Jul 14, 2017 at 12:22

Here is the way in which you should interpret the different symbols:

• $$\widehat{\mathbf{i}}$$ is the unit vector in the direction of relative motion (of the object relative to the mass of air).
• $$\mathbf{n}$$ is the unit outer normal to the spherical surface.
• $$\widehat{\mathbf{t}}$$ is the director vector of the tangential part of the traction vector, whose magnitude is $$T_w$$. That is, the tangential part of $$\mathbf{\sigma} \cdot \mathbf{n}$$, with $$\mathbf{\sigma}$$ the Cauchy stress tensor. Note that the tangential component of an arbitrary vector $$\mathbf{v}$$ can be calculated as $$\mathbf{v}-\mathbf{n} \cdot \mathbf{v}$$.

Notice that the first term is the projection along the direction of motion $$\widehat{\mathbf{i}}$$ of integral of the normal component of the traction vector (the pressure) over the sphere (the rest of components are zero (on average) due to symmetry). This is the pressure drag.

Similarly, the second term represents the integral of the remaining components, also projected along the direction of motion. This is the friction (or skin) drag, which is due to viscous friction.

All vectors $\hat{i},\mathbf{t},\mathbf{n}$ are unit vectors.

It is correct that $\mathbf{t}\cdot\hat{i}=0$. The second term on the RHS of the equation is incorrect.

Second term on the RHS is about finding the total force due to shear stress. More correctly they should have begun with the stress tensor $\mathbf{T}$ (having components $T_{ij}$), found the stress vector acting on the surface element with normal $\hat{i}$, viz. $\mathbf{T}(\hat{i})$. Then $\mathbf{t}\cdot\mathbf{T}(\hat{i})$ gives the magnitude of shear stress, which is nothing but the component of the stress vector $\mathbf{T}(\hat{i})$ along a tangential direction $\mathbf{t}$.

• So, this means that n*î=1? Commented Jul 14, 2017 at 15:58
• @MathMachine Yes, although why they have two symbols to denote the same thing I cant say. Why don't you read Fluid Dynamics by Batchelor which is specifically written for applied mathematicians?
– Deep
Commented Jul 15, 2017 at 3:43
• i is the unit vector in the direction of motion, and t is the unit vector in the direction of the shearstress acting on the surface, the wiki page seems to be wrong with the variable definitions. So ti is not always 0, and in is not always 1. Commented May 5, 2018 at 7:26

5 years later!!

I think the expression above (from Wikipedia) is highly confusing due to the A (they say "$$A$$ is the planform area of the body"). BUT the relevant area $$A$$ completely depends on the type of force! My point of view is that total drag force has two terms: pressure (or form) drag and viscous (or friction or skin) drag, and each one has a reference area, say $$A_p$$ and $$A_\mu$$. For the pressure drag, $$A_p$$ is basically the area of the body projected onto a plane normal to flow (in the direction of flow), whilst $$A_\mu$$ (reference area for the friction term) is basically the area of the body projected normally to the flow (for an airfoil, that's its horizontal projection). Of course, very often only one of the terms is significant and the other one is neglected.

Thus, what should be added are the drag forces, not the drag coefficients:

$$F_{\rm d} = F_{{\rm d},p} + F_{{\rm d},\mu}$$,

where:

$$F_{{\rm d},p} = \frac{1}{2} c_{{\rm d},p} \,\rho \, U_0^2 A_p \quad$$ and

$$F_{{\rm d},\mu} = \frac{1}{2} c_{{\rm d},\mu} \,\rho \, U_0^2 A_\mu \quad$$

being $$U_0$$ the velocity of flow (relative to the body) far, at infinity.