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Name: Anonmyous
Status: Other
Age: 40s
Location: N/A
Country: N/A
Date: October 2002

We are a volunteer fire departmen dive team.We recover stolen and dumped automobiles for the police department using lift bags. Is their a way to compute the cubic foot displacement of an automobile? This would assist us to determine the amount of cubic feet of air required to lift the car.

This is difficult to determine very exactly because autos all have different shapes. But here is how I would approach the problem: 1. You can find out the weight of a car, truck, etc. by year/made/model either from your state Department of Motor Vehicles or from the manufacturer. There may even be a book tabulating this data. I don't know. 2. As a starting point assume the vehicle is an ellipsoid of revolution. The volume, V(formula), is: V(formula) = 4/3(pi)BA^2 where A = the length of the vehicle, and B = the width.

For this calculation you can approximate (pi) = 3.1416 by "3" so the volume formula becomes:

V(formula) = 4BA^2. For each make/model/year that you recover you can develop a data base of the actual displacement vs. the displacement calculated from the formula V(formula) = 4BA^2 so add a "correction factor = K" to the formula. Then: V(actual) = 4KBA^2 The ratio V(actual)/V(formula) = K.

The hope is that V(formula) takes into account most of the volume, so that "K's" will cluster around the same value. Keep a record of your recovery history (or look back into the files and see if actual displacements were recorded) to get a value for the factor "K".

Vince Calder

I am not quite sure I understand the question. I think you are asking how much the weight of the car is decreased when it is underwater due to the buoyancy of the water displaced by the car. I can think of several ways to determine the displacement of an automobile, but none that would be very useful to you. One would be to lower the car into a big pot and see how much the water level rises. Then the buoyant force would be that change in level (in feet) times the area of the pot (in square feet) times the density of water (62.3 lb/cubic foot.

Another way would be to hoist the car with a crane equipped with a scale and see how much the weight recorded by the scale decreases as the car enters the water.

Maybe the best way is to estimate the volume of water displaced by the car (in cubic feet) and multiply by the density of water. Note that if a bubble of air is trapped anywhere in the car (most notably, perhaps, by the roof of the passenger compartment), the effective weight of the car will be decreased by the volume of that bubble times the density of water. Maybe the car manufacturer can give you a reasonable estimate of the volume of water likely to be displaced by an underwater car.

Wish I could be more helpful.

Best, Dick Plano

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