CuriousRambler
Contributor
All else being equal, an aluminum plate and a carbon plate (and a tungsten plate, and a stainless plate, and a uranium plate) will displace identical amounts of water. Displacement is purely based on volume of the object being submerged - mass and/or density have no impact on it. (Edited to add: this is assuming you're fully submerging a solid object. Once you have empty voids encapsulated in material, or if you're only partially submerged, the calculations start to change) So, assuming the plates are of identical design (thickness, profile, etc) displacement is the same. If displacement is identical, the buoyant force is technically identical (counter-intuitive, I know...hear me out) so you have to account for the mass of the plate. If the plate's mass is greater than the force of the buoyancy, it sinks (i.e. it "adds weight" to your kit), if the plate's mass is less than the force of buoyancy, it floats. To be neutrally buoyant, your total mass must equal your total buoyant force.
Buoyancy can be calculated by Force of Buoyancy= Submerged Volume x Fluid Density x Force of Gravity
Let's assume submerged volume = 1m^3 for all plates, since we're assuming they're identical design - this may vary in the real world, and obviously our plates are much much smaller than a cubic meter..just simplifying math for comparison's sake.
Density of sea water is roughly 1,025kg/m^3 - again, this number doesn't actually matter since you're diving different plates in the same water, we could use any number here.
Gravity is 9.81m/s^2, which is equal to 9.81newtons/kg, so:
Force of buoyancy for a given plate is going to be identical, at 1m^3*1025kg/m^3*9.81N/kg ==> 10,055.25 Newtons (again, we've got a BIG plate here)
Now we look at the density of the material, internet search shows carbon fiber ~1.55 to 1.70 g/cm^3, let's call that 1.6, or 1,600 kg/m^3. Aluminum is about 2700 kg/m^3. Back to our easy math, we've assumed 1m^3 plates:
CF "plate" is 1600kg (mass) * 9.81N/kg (gravity) = 15,696N (weight)
Al "plate" is 2700kg (mass) * 9.81N/kg (gravity) = 26,487N (weight)
Compare to buoyant force of 10,055.25N:
10,055.25 - 15,696 = 5,640.75N negative for CF
10,055.25 - 26,487 = 16,431.75N negative for Al
So the short answer is "carbon fiber is roughly 66% less negatively buoyant than aluminum." So, if you've found you need 1lb of lead to sink your aluminum plate, you'll probably need a bit less than another pound of lead for a carbon fiber plate.
All this is assuming my internet-provided density for CF is remotely correct; I'm not a materials guy and don't know how far off that number may be in reality. Also reasonably likely I botched my math somewhere..
Buoyancy can be calculated by Force of Buoyancy= Submerged Volume x Fluid Density x Force of Gravity
Let's assume submerged volume = 1m^3 for all plates, since we're assuming they're identical design - this may vary in the real world, and obviously our plates are much much smaller than a cubic meter..just simplifying math for comparison's sake.
Density of sea water is roughly 1,025kg/m^3 - again, this number doesn't actually matter since you're diving different plates in the same water, we could use any number here.
Gravity is 9.81m/s^2, which is equal to 9.81newtons/kg, so:
Force of buoyancy for a given plate is going to be identical, at 1m^3*1025kg/m^3*9.81N/kg ==> 10,055.25 Newtons (again, we've got a BIG plate here)
Now we look at the density of the material, internet search shows carbon fiber ~1.55 to 1.70 g/cm^3, let's call that 1.6, or 1,600 kg/m^3. Aluminum is about 2700 kg/m^3. Back to our easy math, we've assumed 1m^3 plates:
CF "plate" is 1600kg (mass) * 9.81N/kg (gravity) = 15,696N (weight)
Al "plate" is 2700kg (mass) * 9.81N/kg (gravity) = 26,487N (weight)
Compare to buoyant force of 10,055.25N:
10,055.25 - 15,696 = 5,640.75N negative for CF
10,055.25 - 26,487 = 16,431.75N negative for Al
So the short answer is "carbon fiber is roughly 66% less negatively buoyant than aluminum." So, if you've found you need 1lb of lead to sink your aluminum plate, you'll probably need a bit less than another pound of lead for a carbon fiber plate.
All this is assuming my internet-provided density for CF is remotely correct; I'm not a materials guy and don't know how far off that number may be in reality. Also reasonably likely I botched my math somewhere..