Calculating Drag in 3D with Different Areas & Coefficients

[TimeHorse]

Okay, thanks Hikaru and K² for your help on the accelerated drag problem. Now I have another drag related problem I wonder if you you help me with.

Say a car is put in a wind tunnel and is measured to have the following attributes:

Frontal Area: A_x
Side Area (Right): A_y
Frontal Coefficient of Drag: Cd_x
Side Coefficient of Drag (Right): Cd_y

Where the Frontal and Side forces are measured independently.

So say a wind is applied to this car with an overall speed of 20 m/s such that 16 m/s is applied in the X direction and 12 is applied in the Y. Can I simply compute:

$$\vec{F_{drag}} = \left(\frac{1}{2} \rho A_x Cd_x 256 \frac{m^2}{s^2}, \frac{1}{2} \rho A_y Cd_y 144 \frac{m^2}{s^2}, 0 \right)$$

Or do I have to recompute A and Cd in the wind tunnel for wind incident at approximately 53° to figure out the forces involved in this situation?

Thanks!

 

[hikaru1221]

Hi TimeHorse,

I am also curious about this, and have been waiting for an answer from a specialist. But it seems that the thread has been sleeping for quite long, so I'll try to wake it up though I know nothing about this topic

Let's look at a simpler case: a sphere of radius R. Consider the drag force:
1 - when the wind blows at speed v directly to the sphere:
$$F_1=\frac{1}{2}\pi R^2\rho Cd_1v^2$$
2- when the wind blows at speed $$v\sqrt{2}$$ (this looks like the combination of 2 flows of speed v; see the figure):
$$F_2=\pi R^2\rho Cd_2v^2$$

If $$F_2=F_1\sqrt{2}$$ (i.e. your "vector sum" rule is valid) then $$Cd_2=Cd_1/\sqrt{2}$$ (*). As you know, the drag coefficient is a function of the Reynolds number $$Re=\rho Lv/\eta$$, and this may lead to (*). However, the function remains unknown. According to Wikipedia, at certain range of the values of speed, dimension, viscosity, etc, Cd may be treated as a constant depending only on the shape, which violates with (*). Have a look at this: http://en.wikipedia.org/wiki/Drag_coefficient

Though the speed you gave us is not too large, it's not too small for us to apply Stokes' law, so I assume the quadratic drag equation holds validity in your case. Therefore I think the "vector sum" rule is not valid.

However as I said, I know no further than you do, so don't rely on my opinion

 

[TimeHorse]

Thanks Hikaru!

Let's look at a simpler case: a sphere of radius R. Consider the drag force:
1 - when the wind blows at speed v directly to the sphere:
$$F_1=\frac{1}{2}\pi R^2\rho Cd_1v^2$$
2- when the wind blows at speed $$v\sqrt{2}$$ (this looks like the combination of 2 flows of speed v; see the figure):
$$F_2=\pi R^2\rho Cd_2v^2$$

Interesting. But, working in another direction, let's assume that for a given velocity, C_d is constant for a sphere since that is a fully symmetrical shape (which as we can see is empirically calculated to have a drag co-efficient of 0.47). It's interesting to note that the drag co-efficient quoted is only accurate to 2 significant digits, as far as we can tell. One could therefore assume that there is a, say, 2%-4% relative error built into the wikipedia values and those quoted for cars, even if the value for a Sphere can be calculated using Stokes' Law. This may mean that the quoted C_d is valid for a variety of speeds as long as the correct errors are propagated.

But anyway, if a wind blows at a velocity of v at a 45° along a Cartesian plane bisecting the sphere at its equator, then the force in that direction is indeed given by:

$$F_{45^\circ}=\frac{1}{2}\pi R^2 \rho C_d \nu^2$$

Now I believe, but since I it's been so long since I studied for a BSc in Physics, I could be quite wrong, that it is the force vector that can be safely split along the Cartesian plane. Thus, along the x and y axes:

$$F_{x} = \frac{\sqrt{2}}{2}F_{45^\circ} = \frac{\sqrt{2}}{4}\pi R^2 \rho C_d \nu^2$$
$$F_{y} = \frac{\sqrt{2}}{2}F_{45^\circ} = \frac{\sqrt{2}}{4}\pi R^2 \rho C_d \nu^2$$

The question then becomes, how do you calculate the wind speed in the x and y direction. If we naïvely assume the Cartesian vectors for velocity are also given by the $$\frac{\sqrt{2}}{2} \nu$$, we would end up with:

$$F_{x} = \frac{1}{2}\pi R^2 \rho C_d \nu_x^2 = \frac{1}{4}\pi R^2 \rho C_d \nu^2 \neq \frac{\sqrt{2}}{4}\pi R^2 \rho C_d \nu^2$$
$$F_{y} = \frac{1}{2}\pi R^2 \rho C_d \nu_y^2 = \frac{1}{4}\pi R^2 \rho C_d \nu^2 \neq \frac{\sqrt{2}}{4}\pi R^2 \rho C_d \nu^2$$

Therefore, we can see that the assumption that an orthogonally split velocity can be applied to the drag formula is itself flawed, since clearly C_d is not going to vary by $$\sqrt{2}$$ because the velocity is $$\frac{\sqrt{2}}{2}$$ slower than before.

We could get around this if we assume that when applying the velocity to the formula that we apply the formula in reverse to calculate the orthogonal velocity (this time I'll make it more generic, using some incident angle $$\theta$$ in the Cartesian plane):

$$F_{x} = F cos(\theta) = \frac{1}{2}\pi R^2 \rho C_d \nu^2 cos(\theta) = \frac{1}{2}\pi R^2 \rho C_d \nu_{x}^2$$
$$F_{y} = F sin(\theta) = \frac{1}{2}\pi R^2 \rho C_d \nu^2 sin(\theta) = \frac{1}{2}\pi R^2 \rho C_d \nu_{y}^2$$

Thus:

$$\nu^2 cos(\theta) = \nu_{x}^2$$
$$\nu^2 sin(\theta) = \nu_{y}^2$$

If you apply this formula to calculate the apparent wind velocity, the forces sum properly. I think this is the correct assumption since we know the C_d should be fairly constant for a regular sphere.

Now, how you sum the forces of wind incident in 2 different directions is a whole other can of worms that I will investigate further and get back to you.

Though the speed you gave us is not too large, it's not too small for us to apply Stokes' law, so I assume the quadratic drag equation holds validity in your case. Therefore I think the "vector sum" rule is not valid.

Yes.

 

[hikaru1221]

So the reason we mainly focus on is that Cd = const. And it seems that if the "vector sum" rule doesn't apply to a sphere, it is inapplicable to other complex cases like your car.

Interesting. But, working in another direction, let's assume that for a given velocity, C_d is constant for a sphere since that is a fully symmetrical shape (which as we can see is empirically calculated to have a drag co-efficient of 0.47). It's interesting to note that the drag co-efficient quoted is only accurate to 2 significant digits, as far as we can tell. One could therefore assume that there is a, say, 2%-4% relative error built into the wikipedia values and those quoted for cars, even if the value for a Sphere can be calculated using Stokes' Law. This may mean that the quoted C_d is valid for a variety of speeds as long as the correct errors are propagated.

Stokes' Law? The law is valid only for small speeds, so I don't think it can be used to calculate Cd in a wide range.

The only fluid-mechanics-focused book I have also provides experimental analysis on drag on a sphere. They measured Cd versus Re of a sphere, where $$Re=\frac{2R\rho v}{\eta}$$ and what they got is a graph (see figure attached).

Their analysis points out that the variation can be divided into 4 main ranges, corresponding to 4 regions in the graph:

_ Region (I): $$Cd=\frac{24}{Re}$$ which is valid for Re<1.

_ Region (II): $$Cd=A(1+\frac{a}{\sqrt{Re}})^2$$ where a = 9.06 and Aa² = 24.

_ Region (III): For 10³ < Re < (5 x 10⁵), Cd is almost unchanged and is about 0.43-0.47 (just as Wiki says!)

_Region (IV): There is a sudden change at the critical value of Re, and after that, Cd remains unchanged.

Since we mainly focus on the range where Cd=const, I'll consider region (III) only. In this region, 10³ < Re < (5 x 10⁵). At room temperature, air viscosity is about $$\eta = 1.8 \times 10^{-5} Pas$$ and density of air is about $$\rho = 1.2kg/m^3$$. Consider a super big ball of R=1m, which might be compared to dimensions of a car. For 10³ < Re < (5 x 10⁵), we have v is in the range from 0.0075 m/s to 3.75 m/s. I'm quite surprised at this result. However, this convinces me that if we go out of this range, to a greater speed like 20m/s, we still get Cd=const (as in region (IV)).

Now I believe, but since I it's been so long since I studied for a BSc in Physics, I could be quite wrong, that it is the force vector that can be safely split along the Cartesian plane. Thus, along the x and y axes:

$$F_{x} = \frac{\sqrt{2}}{2}F_{45^\circ} = \frac{\sqrt{2}}{4}\pi R^2 \rho C_d \nu^2$$
$$F_{y} = \frac{\sqrt{2}}{2}F_{45^\circ} = \frac{\sqrt{2}}{4}\pi R^2 \rho C_d \nu^2$$

You're still right at there

 

[TimeHorse]

You're still right at there

Thanks! :)

So I agree that we're probably in the Region III range for a car although clearly when velocities are really small there are no doubt other rather complex drag equations since C_d is not constant in the Region I and Region II regimes, but as you point out this 0.0167 mph is barely perceptible with respect to a car's motion, I think we can safely rule those out. Of course I too am surprised at the variability in Region IV where the car spends most of it's time with respect to if it was modeled as a Sphere. Of course, car's are much more complicated than Spheres and probably have more than just 4 regions, but I think you're right in getting things nailed down in the simple case and working upward.

So if we take the generic result for F_drag in the x and y directions, it looks like we get the following for a wind incident at some angle $$\psi$$ with velocity $$\left|\nu\right|$$ with x and y components $$\nu_x$$ and $$\nu_y$$:

$$F_x = \frac{1}{2} \rho A C_d cos\left(\psi\right) \nu^2 = \frac{1}{2} \rho A C_d \frac{\nu_x}{\left|\nu\right|} \nu^2 = \frac{1}{2} \rho A C_d \nu_x \left|\nu\right|$$
$$F_y = \frac{1}{2} \rho A C_d sin\left(\psi\right) \nu^2 = \frac{1}{2} \rho A C_d \frac{\nu_y}{\left|\nu\right|} \nu^2 = \frac{1}{2} \rho A C_d \nu_y \left|\nu\right|$$

AFAICT. So, if that's true then how does one compute P = Fv? Which v does one use to compute power? It seems to me if P = P_x + P_y, meaning the total power required to overcome drag is given by the sum of the power applied in the X direction and the power applied in the Y direction. Now, in order to get $$F_x \nu_{?} + F_y \nu_{??} = F v$$, we need to choose some $$\nu_{?}$$ and $$\nu_{??}$$ such that this be true. Fortunately, we know that $$\nu_x^2 + \nu_y^2 = \nu^2$$ as we are dealing with flat space here. So clearly we get:

$$P_x = F_x \nu_x$$
$$P_y = F_y \nu_y$$

Which does seem to make sense since I'm projecting Force onto the Cartesian axes so when I take the velocity component I am only taking the portion of it also projected onto that axis to form the vector dot product. Ah. Bliss! It's so nice when math and physics agree with common sense! :)

Of course, this still doesn't answer whether C_d can be considered in the Cartesian way I initially described. All this gets me is to revise my original result:

$$\vec{F_{drag}} = \left(\frac{1}{2} \rho A_x Cd_x 320 \frac{m^2}{s^2}, \frac{1}{2} \rho A_y Cd_y 240 \frac{m^2}{s^2}, 0 \right)$$

 

[hikaru1221]

For the power things, you're right. Math agrees with your physics sense

Of course, this still doesn't answer whether C_d can be considered in the Cartesian way I initially described. All this gets me is to revise my original result:

$$\vec{F_{drag}} = \left(\frac{1}{2} \rho A_x Cd_x 320 \frac{m^2}{s^2}, \frac{1}{2} \rho A_y Cd_y 240 \frac{m^2}{s^2}, 0 \right)$$

I think now we have no $$Cd_x$$ , $$Cd_y$$ , $$A_x$$ , $$A_y$$ here. All we know is $$\vec{F}=\frac{1}{2}Cd.A.\rho .v^2\frac{\vec{v}}{|v|}$$, then we have its x and y components:
$$F_x=\frac{1}{2}Cd.A.\rho .v^2\frac{v_x}{|v|}$$
$$F_y=\frac{1}{2}Cd.A.\rho .v^2\frac{v_y}{|v|}$$
Those 2 equations were actually deduced by you in post #5

 

[K^2]

Basically, because V² drag is due to turbulence and turbulence is non-linear, in general, you need to find a separate drag coefficient for every possible direction of relative wind.

Of course, that's a major pain, and if you are interested in modeling a vehicle, you probably want to avoid that.

Usually, you can approximate drag using a tensor coefficient of drag. In that case, the general formula looks like this:

$$\vec{F_d} = - \vec{v} \cdot (C_d \vec{v}) \hat{v}$$

The C_d will be an NxN matrix where N is the number of dimensions of the problem. Naturally, it may have cross-terms in your chosen coordinate system. To construct that tensor experimentally, first find directions of relative wind under which you get maximum and minimum values for your drag. These two directions should be 2 of your coordinates. The 3rd will be perpendicular to these 2. In this coordinate system, C_d should be diagonal, and you can get the 3 diagonal terms by measuring drag with relative wind along these 3 directions.

For a car, I would imagine that maximum drag would be against the bottom/top, and minimum from the front, which greatly simplifies the problem. The forward/back, up/down, and left/right are your 3 coordinates in which C_d is diagonal, and you just need to make measurements with wind along these directions.

Keep in mind that you'll really find that drag from front and from back are going to be different. Furthermore, the car's body will generate lift. So for a complete description, you'll have to account for that as well.