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Formula for 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

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    $\begingroup$ 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. $\endgroup$ – David White Jul 14 '17 at 0:16
  • $\begingroup$ 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.) $\endgroup$ – Math Machine Jul 14 '17 at 0:34
  • $\begingroup$ @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! $\endgroup$ – Math Machine Jul 14 '17 at 2:25
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    $\begingroup$ 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. $\endgroup$ – Chet Miller Jul 14 '17 at 12:22
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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 $\mathbf{t}$.

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  • $\begingroup$ So, this means that n*î=1? $\endgroup$ – Math Machine Jul 14 '17 at 15:58
  • $\begingroup$ @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? $\endgroup$ – Deep Jul 15 '17 at 3:43
  • $\begingroup$ 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. $\endgroup$ – Orbit May 5 '18 at 7:26

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