1. What is the reason that a streamlined body has a drag coefficient that is lesser than the drag coefficient of a streamlined half-body, when the latter has a completely flat bottom, while the former has bulges on either side of the centerline?

  2. Why do long cylinders have a lesser drag coefficient than short cylinders? Length means more surface for the flow to move against, but it seems that a shorter object with an otherwise same shape, has more drag to it.


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    $\begingroup$ You very much should retract your acceptance of the posted answer. It is quite wrong, but by accepting it you have discouraged others from writing better answers. As a general rule, unless its very obvious an answer is correct, it's best to wait several hours, perhaps even a day, before marking an answer as accepted. Patience vastly improves the odds of receiving better answers. $\endgroup$ – David Hammen Jul 24 '18 at 15:15

The first thing you need to do when comparing drag coefficients is to check the reference areas. Even though it is not clear from the picture you link to, it is pretty obvious that the frontal area was used. This means for your first question that the half body creates roughly as much drag force as the full body (which has twice the frontal area).

Next, you need to know the Reynolds number, surface roughness and the level of turbulence to have a better understanding of how the drag coefficient had been measured. Again, all those details are missing. Now consider that the half-body is mounted on a flat surface. Most likely, the flat plate alone was tested first and then the half-body added, and the drag of the whole plate plus half-body measured. I suspect that the surface area covered by the half-body was subtracted when the additional drag of the plate was eliminated from the measurement, but I have no way of knowing. In all, giving a single number without more details on how the result was obtained is unsound and potentially misleading.

Your second question is easier to answer. Essentially, a longer body has a lower drag coefficient because it creates a smaller frontal area of separated flow. When flowing around the forward edge of the cylinder, the flow will separate and follow a curved path that is dictated by the pressure difference between outer flow and the separated region right after the forward edge. The long cylinder will prevent air from behind the cylinder to flow forward and fill up the separated region, so the pressure difference is stronger, leading to a higher curvature of the flow past the edge and eventual reattachment before the end of the long cylinder is reached. On the other hand, the separation field past the edge of the short cylinder joins with the separated flow behind the cylinder and the flow looks like that around a flat disk (see this answer for an extensive list of possible drag coefficients - and yes, Mr. Hoerner gives the valid Reynolds number range and an extensive list of citations for his single-figure numbers).


Drag is not friction between the surface of the body and the fluid. The assumption is usually that the fluid is not moving relative to the surface at the surface itself. This is the basis of boundary layer theory. D'alembert's paradox shows that laminar flow does not produce drag. The upshot is that the drag is due to power loss into turbulence - ultimately into heating the fluid. Viscous losses are friction losses, but not friction of the fluid rubbing on the surface. Hence, larger surface area does not in and of itself lead to larger drag.

The streamlined body is designed to produce the low turbulence by balancing the fluid streamlines coming of the back of the body so that they have the same velocity on detaching and do not tend to collide with each other. The half streamlined body has the fluid coming around the top with some downward velocity, while the fluid on the bottom, in principle, is going straight through. The two streams could be expected to merge and create turbulence. It often helps here to recall that the fluid is actually atoms, and think about what they are doing.

Navier Stokes theory is based on the idea of the existence of a velocity field. But, in turbulence, there is no velocity field - as the particles are going every which way and crashing into each other. These collisions lead to creation of entropy, and heating, and at another level, the emission of photons adding to the heat bath.

  • $\begingroup$ In turbulence, there can be a velocity distribution, as in the kinetic theory of gas. But, there is no velocity field in the sense of a one velocity to rule them all at each time and place. $\endgroup$ – Ponder Stibbons Jul 23 '18 at 23:25
  • $\begingroup$ There's a lot wrong with this answer. In the 1st 3 lines alone: 1-Friction between the surface of the body and the fluid is indeed responsible for at least part of the drag; 2-Laminar flow does create drag; 3-D'alembert's paradox is about inviscid flow, not laminar flow. Need I continue? $\endgroup$ – D. Halsey Jul 23 '18 at 23:50
  • $\begingroup$ I know D'Alembert's paradox, but, yes, I did not say my piece well; however, I think you are rather harsh and dismissive. At the molecular level, the force on the body is due to collisions with atoms. When the atoms near the surface of the body collide with the body, they bounce off and are no longer Laminar at that level. So, the drag force due to laminar flow is due-to/associated-with at least microscopic turbulence that is a thermodynamic increase in heat. But, this can be ram pressure, and does not have to be skin friction. Ultimately, it's just atoms and the void. Is that better? $\endgroup$ – Ponder Stibbons Jul 24 '18 at 5:24
  • $\begingroup$ No. The physics required to address the OP's question involve continuum mechanics, not "atoms and the void." Turbulence is a macroscopic phenomenon. Discussing microscopic turbulence at molecular levels of laminar flows ignores the far more relevant fact that all turbulent boundary layers have laminar sublayers close to the wall. I don't mean to appear harsh, but your answer and comment are both seriously off the mark. $\endgroup$ – D. Halsey Jul 24 '18 at 15:50

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