Science Questions

How fast do I have to go to be lifted off the ground?

Sun, 20th Feb 2011

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Question

Jane Brittain-Long asked:

Hi, I have been trying to find out an answer to this question ever since I did Physics O-level in 1985!

If you were an average man, say 6 feet tall and weighing about 12 stone, if you took hold of a spoiler on the back of a car, how fast would the car have to go to ensure you were actually lifted off your feet and flying behind it? I know this is very† hypothetical but I want to know!!! There must be a mass/speed equation going on but I am not a scientist and my two boys want to know, too!

regards,

 

Jane

 

Answer

We posed this question to Dr Hugh Hunt from the University of Cambridge...

Hugh - Well I suppose if weíre trying to figure out how fast you need to gorace car in a car and somebody grabbed a hold at the back, and youíre try to lift them up off the ground, itís all about aerodynamic drag. 

The aerodynamic drag, there's a formula for it which is the half ρ V2 multiplied by frontal area, multiplied by drag coefficient. 

Now ρ is the density of air and that's 1.2 kg/cubic metre Ė that's easy. 

V is the speed of the car in meters per second. 

A is the frontal area of the person.  Now, letís make a rough guess, itís a 10-stone person, so they're reasonably slim, a couple of meters high, say on average, 20 centimetres wide, wider in the middle, thinner at the legs. Itís as good guess as any, so that gives them a frontal area of 0.4 square meters.

The drag coefficient, well, we need to make another guess here because the air behind the car is very turbulent.  Itís impossible to know what the drag coefficient would be, but letís say 0.5 Ė that's a reasonable figure, I think.

So if you do these sums, so the drag force is roughly equal to the weight of the person, then you end up needing to drive at about 160 miles an hour.

Well, is that reasonable?  Look, itís such a complicated airflow behind the car and there are questions about when does a drag become lift?  Once the personís out at an angle of 45 degrees then you might start thinking, we need to calculate the lift on the person rather than the drag on a person.  We could have lots of arguments over this over a few beers if you want, but at least that's my start up calculation.

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Terminal velocity for an adult is about 125mph in the spread-eagled position with belly (down) into the air flow  ...

http://www.youtube.com/watch?v=gnXEDAYCsFs

But head down, (head first into the air flow), the terminal velocity is 160 - 200mph, according to wikipedia

RD, Sun, 13th Feb 2011

You'd be 90% horizontal if the car kept accelerating at about 10g (98.1 m/s/s), although that sort of acceleration is a bit unlikely, even with a spoiler  Geezer, Mon, 14th Feb 2011

Don't forget to hold the spoiler upside down. Bored chemist, Mon, 14th Feb 2011

I don't think it is an acceleration problem as much as a lift and drag problem.

For example, if you tied a ribbon onto your outside mirror.  At rest, it will fall towards the ground.  Once you start moving, the ribbon moves towards a horizontal.

Acceleration would be most useful if you took the guy to a circular track  

The lift should be easy enough to calculate if you ignore the car.

There also might be some amount of ground-effect that would provide extra lift, except that the car would have a vortex behind it that would tend to reduce a person's effective buoyancy.  CliffordK, Thu, 17th Feb 2011



Au contraire. 

If it wasn't for acceleration, there would be no need for lift and drag. Geezer, Thu, 17th Feb 2011

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