Negative bouyancy/sinking a car

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The discussion focuses on calculating the weight needed to keep a 1997 BMW 328i submerged in a 5-foot deep pool after removing its engine, transmission, hood, and trunk, leaving a body weight of 2,000 pounds. The interior cabin volume is approximately 1047.2 cubic feet, which translates to about 29.65 cubic meters when converted. Participants emphasize the importance of accounting for the car's buoyancy, the weight of added ballast like sand or steel, and the need to remove air pockets for accurate calculations. The buoyant force of the water is calculated to be around 8,736 pounds, indicating this amount is necessary to achieve negative buoyancy. The conversation highlights the complexities of mixing measurement units and the need for precise calculations in practical applications.
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Hello all and thanks in advance for your contributions.

I am calculating how much weight it will take to keep a car on the floor of a 5' deep pool. The engine and transmission have been removed as well as the hood and trunk. The remaining body weighs 2,000 lb. The cab will be sealed air tight. I will have valves in the firewall so I can take on water in order to sink it and will then pump the water out.
The vehicle is a 1997 BMW 328i
Interior cabin space is 7' long x 5' wide x 4' tall equaling 1047.2 cubic feet
overall vehicle dimensions are 175" long x 67" wide x 55" tall
The water is 5' deep so the roof of the vehicle will be about level with the waters surface.
I intend to either place sand bags or use plate steel as ballasts to weight it in the trunk area and engine compartment if feasible.
I will be filling and emptying the cabin.
A hole will be placed in the rooftop for access.

Thank you
 
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Welcome to PF!

We try to help people (of all ages) learn to do the problems themselves. The math of this is relatively simple, it is just a matter of organizing the buoyancies and weights. Why don't you give it a shot and tell us what you get or where you get hung up and we'll nudge you in the right direction.
 
Lets do it!
density of fluid is 1000 kilograms/meter3
volume is 317.7 cubic meters
force of gravity is 9.18 Newtons/kilogram

The equation is Fb=VxDxFg
so 317.7 x 1000 x 9.18 = 2,916,486 pounds to keep the car under water?
 
Water weight displaced x cubic feet of air in cabin space
62.4 pounds per square foot of fresh water
7 x 5 x 4 = 140 square feet x 62.4 = 8,736 pounds of boyancy
Therefor requiring a minimum of 8,736 pounds to submerge this vehicle given we remove all other air pockets such as trunk fenders ect.
Tee ta
 
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A couple of things. First, you are mixing English and SI units. But not to fear: a pound is 2.2 kg, so you can replace the 9.8 with 2.2.

Also, since it appears that you actually intend to test this (!), I think you'll need to be more precise: you should calculate/include the buoyancy of the car itself. Hint: You have the weight. Assuming it is almost all steel, what is the volume of steel? You'll need to do the same thing with the sand. You basically have three components:

Car
Sand
Air

Each has a buoyancy (volume) and a weight. The only one who's weight it is probably reasonable to ignore is the air's.
 
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I don't know where to start on calculating the volume of the steel Russ. The surface area x thickness? I would have to make estimates for multiple layers and such.
The volume of the sand will be determined by how much weight is necessary to achieve negative buoyancy for the project and yes, I actually intend to test this. That is why I came here for guidance.
 
No, the steel isn't difficult, you just look at it from a different direction: you have the weight and you can look up the (weight) density. v=w/d

By volume of sand, I do mean actual volume for finding its own buoyancy: If you calculate that you need 1,000 pounds of force to make the car sink and you use 1,000 lb of sand, you will likely be disappointed to find that the car is nowhere close to sinking. The math does start to get a little tricky with that, since you will need to combine some of the equations you are now figuring out separately, but it still isn't too bad. We'll get there...
 
EB3 Effects said:
Interior cabin space is 7' long x 5' wide x 4' tall equaling 1047.2 cubic feet

EB3 Effects said:
volume is 317.7 cubic meters

To convert from cubic feet to cubic meters you multiply by what conversion factor?
 
  • #10
russ_watters said:
A couple of things. First, you are mixing English and SI units. But not to fear: a pound is 2.2 kg, so you can replace the 9.8 with 2.2.

Also, since it appears that you actually intend to test this (!), I think you'll need to be more precise: you should calculate/include the buoyancy of the car itself. Hint: You have the weight. Assuming it is almost all steel, what is the volume of steel? You'll need to do the same thing with the sand. You basically have three components:

Car
Sand
Air

Each has a buoyancy (volume) and a weight. The only one who's weight it is probably reasonable to ignore is the air's.
A kilo is around about 2.2 pounds.
Watch out if you buy bananas:smile:
 
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  • #11
EB3 Effects said:
Lets do it!
density of fluid is 1000 kilograms/meter3
volume is 317.7 cubic meters
force of gravity is 9.18 Newtons/kilogram

The equation is Fb=VxDxFg
so 317.7 x 1000 x 9.18 = 2,916,486 pounds to keep the car under water?
You've taken the volume of the cabin, 1047 cubic feet, and divided by 3.28 feet / meter, and come out with the wrong answer for the volume in cubic meters.

1 cubic meter = 3.283 ft.3 = 35.31 cubic feet

Therefore, the volume of the cabin is 1047 / 35.31 = 29.65 cubic meters, approximately.

Now you have something a little less super tankerish to work with.
 
  • #12
Buckleymanor said:
A kilo is around about 2.2 pounds.
Watch out if you buy bananas:smile:
Yeah, I mathed it right, but wrote it wrong.
 
  • #13
Craig Byrom said:
Water weight displaced x cubic feet of air in cabin space
62.4 pounds per square foot of fresh water
7 x 5 x 4 = 140 square feet x 62.4 = 8,736 pounds of boyancy
Therefor requiring a minimum of 8,736 pounds to submerge this vehicle given we remove all other air pockets such as trunk fenders ect.
Tee ta
This is making sense to me. I don't understand the mathmatic symbols in some of the prior posts, so it becomes difficult to learn the formula on my way to a solution.
Sorry boys, but I'm a 52 year old left brained visual thinker.
I appreciate your contributions.
Everett
 
  • #14
Craig Byrom said:
Water weight displaced x cubic feet of air in cabin space
62.4 pounds per square foot of fresh water
7 x 5 x 4 = 140 square feet x 62.4 = 8,736 pounds of boyancy
Therefor requiring a minimum of 8,736 pounds to submerge this vehicle given we remove all other air pockets such as trunk fenders ect.
Tee ta
It's 62.4 pounds of water per cubic foot.

A space measuring 7 feet by 5 feet by 4 feet has a volume of 140 cubic feet
 
  • #15
Copy that. The end result is the same though, yes?
 
  • #16
The volume of the cabin cannot be more than a few cubic meters.
7x5x4=140 cubic feet. (and not thousand).

30 cubic meters would be close to the space enclosed by a bedroom in a house.
 

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