I have a question about the drag coefficient in the drag equation.

Let's say I have a rectangular prism oriented such that, looking down on it, the long side is parallel to the y-axis. Moving forward in a fluid at high velocity (Re > 100000), based on what I've read, the front surface area would experience drag with a coefficient of ~1.28. The drag coefficient on the side is equivalent because they are both rectangles, however, it would experience no drag because there is no relative velocity in that direction. Now assume the rectangle rotates, but it is still moving in the direction of the same face, as if it were a boat for example. See the image below:

enter image description here

I'm thinking of it as a vector, so in the first image it moves vertically some velocity, and horizontally at no velocity. In the second, it moves both vertically and horizontally. The magnitude of the velocities should each be the same; let's say each is moving at the object's terminal velocity.

The drag coefficient MUST change (if expressed as a vector like I'm saying) because if it didn't, the object would never be able to reach it's terminal velocity moving at an angle like in the second picture. Obviously the area increases, so the if the drag coefficient was constant, the drag would increase, so the velocity would decrease. So is there a known drag coefficient for a flat plate at an angle of attack $\theta$? Any help in understanding what's happening here would be greatly appreciated!

  • 1
    $\begingroup$ A sketch would help. $\endgroup$ Commented Jul 4, 2013 at 17:38
  • $\begingroup$ Added. Also clarified what I'm asking. $\endgroup$
    – williamg
    Commented Jul 4, 2013 at 23:47

1 Answer 1


Based on your sketch, the block is always moving along it's long axis, in other words, the velocity is always along the direction of your red vector. This means in the picture on the left, there is only vertical velocity while in the picture on the right, there is both vertical and horizontal velocity. This is what you have described, I am only summarizing to make sure the rest of my answer makes sense and is consistent with this understanding.

Now, you claim the drag coefficient of the case on the left is known yet the drag coefficient for the case on the right is unknown. To clarify a point, the drag coefficient is a scalar, not a vector as you claimed it to be. And in this case, the drag coefficient is actually the same in both cases if the block is moving along the vectors shown. Why is this the case?

Consider the area vector of each face to be the area of the face times the surface normal, ie.:

$$\vec{A} = A\cdot\hat{n}$$

In your example on the left, the area normal is in the $\hat{j}$ or Y-normal direction and so is the velocity. This gives you the drag coefficient you cite.

In your example on the right, you now have a component of the area vector in the $\hat{i}$ direction and a component in the $\hat{j}$ direction. But your velocity also has components in both those directions.

You are correct to say that the area increases in the $\hat{j}$ direction -- you now have both a portion of the short side and a portion of the long side exposed to the Y-direction. However, the block is still moving normal to the short side along the red vector, which means the area vector for the long sides are perpendicular to the red vector and that area does not contribute to the so called "frontal area" of the block.

  • $\begingroup$ But in order for me to handle it that way, wouldn't my coordinate system have to rotate with the block? If my coordinate system is fixed then, for example, when the block is moving as described in my second picture, wouldn't the areas be as I depicted? Or is that my error? $\endgroup$
    – williamg
    Commented Jul 6, 2013 at 17:28
  • $\begingroup$ If your coordinate system doesn't rotate with the block, then your area vector and velocity vector will both have components in $\hat{i}$ and $\hat{j}$. But the dot product of the velocity vector with the long-side area vector will still be 0 if it's moving the way you have drawn it. So the area vector in the direction of travel is again the same as the left picture. $\endgroup$
    – tpg2114
    Commented Jul 6, 2013 at 18:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.