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

*Hugh* - Well I suppose if we’re trying to figure out how fast you need to go[img float=right]/forum/copies/RTEmagicC_800px-WorldOfOutlawsLateModel.jpg.jpg[/img] in a car and somebody grabbed a hold at the back, and you’re try to lift them up off the ground, it’s all about aerodynamic drag.

The aerodynamic drag, there's a formula for it which is the half *ρ V*^{2} multiplied by frontal area, multiplied by drag coefficient.

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

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

A is the frontal area of the person. Now, let’s make a rough guess, it’s 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. It’s 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. It’s impossible to know what the drag coefficient would be, but let’s say 0.5 – 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, it’s such a complicated airflow behind the car and there are questions about when does a drag become lift? Once the person’s 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.