What is the lifting power of styrofoam when submerged, or put underwater?

Question

I need to build a floating platform, so I need to use styrofoam, and I need to know the lifting power per cubic inch, foot, of yard. Thanks!

Answer 1

Water weighs about 62.4 pounds per cubic foot. Therefore the lifting power will be the difference in weight between a cubic foot of water and the amount of styrofoam necessary to displace a cubic foot of water. If the styrofoam is of a type that does not absorb water (perhaps in spaces between grains?) a cubic foot of it will displace a cubic foot of water. Note that the styrofoam must be completely submerged to develop maximum lifting power. Any styrofoam above water in your design will not give lift but may be necessary to take up any load added to the platform. Good luck.

Answer 2

You need several measurements before you begin, but I will get you on the right track.
Density of water 1 g/cm^3

Density = Mass / Volume

When a object floats
B - w(o) = 0
B - w(o) = [D(f) - D(o)] * V(o) * gravity

Example
Say you have 1 kg of an object and it has a volume of 1 cubic meter
D(o) =(1 kg / 1 m^3) * (1 m^3 / 1000000 cm^3) * (1000 g / 1 kg)
D(o) = 0.001 g / cm^3

B - w(o) = [D(f) - D(o)] * V(o) * gravity
B - 1000 g = [1 g/cm^3 - 0.001 g / cm^3] * 1000000 cm^3 * 9.8 m / s^2

B = 9790200 g * m / s^2 = 9790.2 N
1 N = 0.22481 lb
B = 2200.93 pounds of force

To find out what you need for your styrofoam is to use the weight you are trying to float, look up/ calculate the density of the styrofoam being used and work backwards. You can also use some basic tables to switch between SI units and American Standard.
Formula to figure it all out is:
F - (m(s) * 9.8 m / s^2)= (1 g/cm^3 - D(s)) * V(s) * 9.8 m/s^2
where
F = Force needed (weight to be floated)
m(s) = mass of styrofoam
D(s) = density of styrofoam
V(s) = volume of styrofoam

When you have the density, can solve for either mass or volume using m = V * D

Answer 3

a 1911 40 5 vehicle can fire underwater. another weapons will blow up. any gun can fire in area as there is no choose for air to burn the powder. I watched a try (on American Rifleman i think) the place they fired a 1911 .40 5 handgun underwater (in a pool). Now that's particularly distinctive with the aid of fact the bullet began under water and did no longer wreck the air/water airplane, however the bullets travelled 13-15 ft. Even the testers have been shocked. They have been easily shifting with adequate rigidity to do harm at 7 ft. this is with a rather great and slow projectile. I even have additionally watched law enforcement officials at a party fire a 40 5 1911 underwater