Friday, November 23, 2012

More James Bond Physics

More James Bond Physics

One of my Thanksgiving traditions is the annual James Bond Marathon the day after our epic meal. By the time we hit The Spy Who Loved Me, I had seen enough bad physics to make Professor Metzler blush. However, this particular film had an interesting scene in which Bond drives his Lotus Esprit S1 into the sea, where it transforms into a submersible. I was curious what sort of Buoyant force acted on this car.

So here's what we know:
The density of water is 1000 kg/m^3

The dimensions of the Lotus Esprit: 4.191 m x 1.861 m x 1.511 m
Therefore the volume of the car (treating the car as a rectangular prism): 11.785 m^3

Force of gravity: 9.8 m/s^2

Buoyant Force = (density fluid)(Volume)(g)
                        = (1000 kg/m^3)(11.785 m^3)(9.8 m/s^2)
                        = 115493 N

This is a very large force acting upon the car (The value is likely inflated since we treated the volume of the car as a rectangular prism instead of its true shape).

This made me wonder, what kind of drag force acts on the car while it moves underwater. Since the car is moving relatively quickly, let's use the equation for fast-moving objects:

Drag Force = (1/2)(Coefficient of Drag)(density of fluid)(velocity^2)(cross-sectional area)

Now, we don't know the drag coefficient for this car. HOWEVER, I managed to find an equation for the drag coefficient for any object.

Coefficient of Drag = 2(Drag Force)/(density fluid)(velocity)^2(Cross-sectional area)

We assume that the velocity of the car is approx. 10 m/s, and use the width and height for the cross sectional area.
                                = 2(115493 N)/(1000 kg/m^3)(10 m/s)^2(1.861 m x 1.511 m)
                                = .821

Since we now have a value for the drag coefficient, we can solve for the drag force acting on the car:

Drag Force = (1/2)(Coefficient of Drag)(density of fluid)(velocity^2)(Cross-sectional Area)
                   = (1/2)(.821)(1000 kg/m^3)(10m/s)^2(1.861 m x 1.511 m)
                   = 115431 N

Clearly, this is a rather large force acting against the car's motion. However, this value is more than likely inflated; the cross sectional area is supposed to be the area of the front of the car, rather than the maximum area (which I used for the purposes of my calculations).

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