Cd constant... or speed dependant?
 

Hi guys,

I didn't do a full theoretical investigation on this subject, so please forgive me for this lazy topic / question.

I've read in at least one thread here, that Cd is said to be quasi-constant at legal road speeds.
However, this doesn't align with my gut-feeling. If I'd compare a relatively short teardrop shape with a very long one, I imagine they can perform equally well at low speeds.
Maybe at 30 mph or so, even the short teardrop could get fully attached, laminar flow. But at 60 mph, I guess you'd need a longer teardrop to achieve a fully attached flow...?

At 30 mph, no full-size car is operating in laminar flow; the Reynolds number at any point on the body except the very front is too high for that.

When flow trips from laminar to turbulent, drag suddenly drops and then continues to rise on a velocity-dependent curve.

In this turbulent-flow regime, Cd is proportional to the inverse square of velocity--but this variation is so small that it can be considered constant for road-vehicle speeds.

Hi Vman, my 30 mph and 60 mph were just some random figures, just for some scale in my comparison.
I was comparing ideal teardrops / raindrops. Those do get laminar flow, don't they?

As far as I can project my comparison to full-size cars: I would imagine that two different shapes could have similar flow attachment at a low speed (thus similar Cd), but vastly different attachment at high speeds...?

It all depends on Reynolds number (Re), which is the ratio of inertial force to viscous force of the fluid. At sub-critical Re, flow is laminar; above sub-critical Re, flow is turbulent. Shape doesn't come into that; it's purely a function of fluid viscosity, fluid velocity, and test length.

You're probably thinking of attached versus unattached flow, a common mistake. Yes, an ideal teardrop should have attached flow at the speeds we talk about here.

that's a nice graph, it's rather tangible for me.

It doesn't compare the influence of speed however.

I guess skin friction in passenger cars is a few factors smaller than the drag caused by the wake?

Going back to your original question--

I think you're also confusing Cd with drag force, when these are not the same; Cd is a factor in drag force and so is the square of velocity.

A short teardrop and long teardrop will probably see similar drag force at a low speed, but not because Cd has changed or because the flow characteristics have changed. Assuming your "low speed" metric has Re large enough that the flow field is turbulent, there will be little difference between that and "high speed." But because of the v^2 factor, the velocity makes more difference than Cd anyway.

For example: Consider a short teardrop of cross-sectional area 1.0 m^2 and Cd 0.12, and another of the same area but Cd 0.06. At 60 mph (26.8 m/s) the teardrops will see 52.8 N and 26.3 N respectively. At 30 mph (13.4 m/s), the teardrops will see 13.2 N and 6.6 N respectively, a much smaller difference--but it isn't because their Cd has changed, it's because the velocity has.

Is that what you're asking?

Paraphrasing aerohead here: Reynolds number determines the low speed end of aerodynamic drag, depending on barometric pressure and temperature, at ~25mph for vehicle sized objects. Scale models need to be in water instead of air to compensate for scale.

At the high end, transonic shockwaves cap the range of speeds when the formulae apply.

Cd is constant, up to ~250MPH, all else being equal, as I understand it.

The problem as I understand comes to play when the size of the object is scaled down to fit in a wind tunnel. I still can't wrap my head around it but apparently you have to increase the wind speed on a scaled model to get an accurate Cd measurement. To me it would seem if you had say a 1/10th scale model, you would run it at 1/10h the speed or 6 mph vs 60 mph but it's not the case. If speeds had so little effect on Cd then it wouldn't matter and models would be just as good as the real thing without messing with speeds.