## Wind drag on car flag

## Wind drag on car flag

(OP)

Hi,

I need to calculate the moment, which develops by the wind on the bottom of flag bar (mounted on car), during driving in 120 km/hr speed.

Flag size: 4 cm x 25 cm, 2 mm thick. Flag bar length is 50 cm. Flag installed on the top of the bar. The flag made of rigid Nylon.

Please your advice.

Thanks,

Aharon

I need to calculate the moment, which develops by the wind on the bottom of flag bar (mounted on car), during driving in 120 km/hr speed.

Flag size: 4 cm x 25 cm, 2 mm thick. Flag bar length is 50 cm. Flag installed on the top of the bar. The flag made of rigid Nylon.

Please your advice.

Thanks,

Aharon

#### Recommended for you

## RE: Wind drag on car flag

This sounds like something a professor would dream up. Here's how I would approach this problem:

Let's assign some values:

rho = density of air at STP = 1.2 kg/m3

mu = viscosity of air at STP = 1.8e-5 Ns/m2

V = velocity of vehicle = 33.33 m/s (120 km/hr)

L = length of flag = 0.25 m

b = width of flag = 0.04 m

t = thickness of flag = 0.002 m

First calculate the Reynolds number, ReL = rho*V*L/mu = 5.6e5. This Reynolds number is close to the transition from laminar flow to turbulent flow for a flat plate. Since this problem is much easier tackled assuming laminar flow, that's what I would do.

Now, there are two sources of drag. The primary source is stagnation of the flow at the leading edge of the blunt flag. For a thin plate, the coefficient of drag, C, is 2.0. Drag is calculated by D = C*0.5*rho*V2*A, where A is the frontal area (t*b). I get a drag force of 0.11 N.

The other source of drag is due to the boundary layer on either side of the plate. Assuming laminar flow and a Blasius-type boundary layer, the coefficient of drag can be estimated by C = 1.328/Re0.5 (square root of Reynolds number) = 1.8e-3. Drag is now calculated by D = C*0.5*rho*V2*A, where A is the flag area (L*b). I get a drag force of 0.012 N on one side, making the drag force due to both boundary layers 0.024 N.

This gives a total drag force of 0.13 N. I would assume this is acting half way up the flag, or 0.48 m from the base of the flag pole, which gives a moment of 0.062 Nm.

Remember, there are a couple of important assumptions here. The first is an assumption of laminar flow, which is borderline at this Re. The other is that the flow that stagnates at the leading edge of the flag does not disturb the formation of a Blasius-type boundary layer along the flag's length. Probably not the case, but a relatively accurate assumption (I think) none-the-less.

Hope this helps,

Haf

## RE: Wind drag on car flag

## RE: Wind drag on car flag

## RE: Wind drag on car flag

## RE: Wind drag on car flag

My above assumption of laminar flow is borderline, as stated above. With a smooth surface the transiton Reynolds number is about 3e6. A regular, non-smooth surface has a transition around 5e5.

## RE: Wind drag on car flag

## RE: Wind drag on car flag

National Association of Architectural Metal Manufacturers' Guide Specifications for the Design of Metal Flagpoles, NAAMM Standard FP-1-90[/]## RE: Wind drag on car flag

Thanks.

To your opinion, can I use a common antenna (of car) as a base for this flag? Take in account the vibration it will make during driving in high speed. Antenna diameter is around 6 mm.

Aharon

## RE: Wind drag on car flag

Well, off hand I would say no. However, this isn't based on any calculations. I've seen a lot of people driving around lately with flags that are darn near ripping their antennae off. I'd put it in one of your windows. You could also conduct a simple experiment by attaching the flag and driving on a road where you can gradually increase speed. If your antanna looks like it's about to give up the ghost before you get to top speed, you'll know if it works...more of an experimental approach.

If you're intent on calculating this, here's one thing you can do. Treat the antenna as a cantilever beam with a load acting on the end. For a cantilever beam loaded under this condition, deflection is calculated by the equation Px^2(3L-x)/(6EI), where

P = load

x = location along beam (max deflection is when x = L)

L = length of beam

E = modulus of elasticity

I = moment of inertia

The moment of inertia depends on the cross-section of the beam. If the beam has a solid circular cross-section, I = pi*D^4/64. If the beam has a thin circular ring cross-section, I = pi*D^3*t/8, where D is the diameter of the beam and t is the thickness of the wall (pi is the greek letter = 3.14159...).

You're flag sounds like it has a very large aspect ratio (4 cm high and 25 cm long). Are you sure that's right? Sounds more like a banner than a flag.

By the way, if an answer to one of your posts (or any other post, for that matter) is helpful in your opinion, it's nice to mark it as helpful (just click the link to the left under the post). It's hard to bring this up without sounding greedy, but this is a good way to let the author (and other readers) know that the post helped you out. I think it also makes the author more likely to help out in the future if he knows his posts are useful.

Good luck; I hope your antenna survives. Generally, they're not that expensive to replace anyway.

Haf

## RE: Wind drag on car flag

For details, see:

Flow-induced Vibrationby Robert D. Blevins;Format: Hardcover, 2nd ed., 477pp.

ISBN: 0894648233

Publisher: Krieger Publishing Company

Pub. Date: May 1994

Edition Desc: 2nd ed

On a practical note, the 1999 Chevy S-10 pickup I drive has fairly large (~+/- 5 mm guestimate) amplitude (2nd mode of vibration) of the radio antenna at 48 MPH. This is despite the antenna having helical ribs which are used to counteract vortex shedding. Its interesting to watch, but I've never heard of a radio antenna such as this failing from fatigue.

## RE: Wind drag on car flag

Thanks again.

Once I’ll have some results from my trial, I’ll let you know.

Aharon