Displacement of Scooters at Depth - Spun off from the A&I Discussion about Nothernone

[W W Meixner]

Folks...

Everyones calculation skills are impressive...even the wrong calculations...fact remains...it is at best...all speculation...

Fact also remains...our dear friend and brother remains lost for reasons unknown...

This photo accompanied my final Cameron Donaldson ''Go Fund Me'' update...

I would like to believe Cameron was competent enough to successfully recover from a DPV malfunction...

Somone once told me that the only thing a Jeep is good for is getting you stuck...deeper...further away from home...

Getting you stuck deeper...further away from home...I can see a scooter accomplishing both of these quite effectively...

I've owned neither a Jeep or DPV...so I'm unable to comment on either...

I saw a DPV once...it took two divers...struggling their guts out...to remove it from the back of an SUV...and carry it to the water...I saw everything I needed or wanted to know about a DPV..

Warren...

 

[Dan]

I would think completely flooded or nose off it would be 20lbs or so negative because of batteries and motor but i dont know my maths is weak


Edit

looks like if this math is correct a 450lb cast iron cannon weighs 180lbs less in sea water

so weight in sea water would be 270 which is a 40 percent reduction in weight

Weight of an object in water

Because the cannon displaces 180 lbs (81.7 kg) of sea water when it sunk to the bottom of the sea. That means the cannon volume was 81.7 liters at the time. So the buoyancy force is 180 lbs. Buoyancy force is up and the force of gravity (the cannon weight) is down. The negative buoyancy then = The weight of the cannon in the sea water = 450 - 180 = 270 lbs.

Weight of an object in water

The submerged weight = air weight minus the buoyant force.

This is a good example of how to estimate negative buoyancy and how much lift force you would need to have on lift bag. Thank you for posting.

 

[Akimbo]

The negative buoyancy then = The weight of the cannon in the sea water = 450 - 180 = 270 lbs.

I think your numbers are a little off. 180 Lbs of negative buoyancy on a 450 Lb metal object would mean the metal has a mass that is only 2.5 x higher than seawater. Sea water is about 64 Lbs/Ft³. Iron is 491 Lbs/Ft³ (7x sea water) and Bronze is about 541 Lbs/Ft³ (8.5x sea water).

Your approach is correct in that you must account for the displacement of materials. For example, a typical block of cured concrete is about 150 Lbs/Ft³ or about 2.3 heavier than seawater. A 150 Lb 1'x1'x1' block of concrete only weighs 86 Lbs in sea water (150-64).

 

[Dan]

I think your numbers are a little off. 180 Lbs of negative buoyancy on a 450 Lb metal object would mean the metal has a mass that is only 2.5 x higher than seawater. Sea water is about 64 Lbs/Ft³. Iron is 491 Lbs/Ft³ (7x sea water) and Bronze is about 541 Lbs/Ft³ (8.5x sea water).

Your approach is correct in that you must account for the displacement of materials. For example, a typical block of cured concrete is about 150 Lbs/Ft³ or about 2.3 heavier than seawater. A 150 Lb 1'x1'x1' block of concrete only weighs 86 Lbs in sea water (150-64).

Good catch. I didn't check how @Jiminy got that 450 lbs. After checking his source, the 450 is iron density in lb/cft, not the actual weight of the cannon, which is 1257 lbs.

Here is the actual example that he got the number from Weight of an object in water

Re: Weight of an object in water
An object in water is lighter than in air, the clearest example is go an jump in the water, The human body loses almost 100% of its weight when immersed.

Weight is determined by the density of an object times the volume of that object.

e.g. lead is so dense it is not affected significantly underwater. Aluminium will lose 38% of its weight underwater.


Scott your stones example depend upon what type of stones as some are denser then others. So you need to know the density of the object. Lets look at your cannon for an example. I'll assume its made of cast iron and its weight per cubic foot would then be 450 pounds (wrought iron would be 485 pounds bronze would be 544) lets not get clever and think that it has lost some of its weight over the years, it has but there is no calculation for that.

So we have the density number for the calculation, now we need the volume of the object. so lets say the cannon is 8 feet long and for the sake of this calculation say it was 8 inches diameter uniform along the cannon then you have the bore say 3inches and is 7 ft long. So lets find the volume.



1/ Get the volume of the cannon in this case keep all dimensions in inches

The Volume of the cannon is two calculations, the whole volume minus the bore volume.

Volume = pye*r2*Length 3.142 * (4” * 4”) = 50.272 * 96” = 4826.11cubic inches


Bore Volume 3.142 * (1.5” * 1.5”) = 7.0695 * 84” = 593.84 cubic inches


4826.11 – 593.84 Volume = 4232.27cubic inches

2/Convert to cubic ft and multiply by the density of iron 450

4232.27 / 1728 = 2.79 cubic ft * 450 = 1256.86 lbs


Next we need to get the weight in sea water

The submerged weight = air weight minus the buoyant force.

To get the submerged weight then you need to take the 2.79 cubic ft and make the following calculation using the density of iron 450

450-64.043 = 385.96 385.96*2.79 = 1076.83 lbs



Hence the cannon weighs 180lbs less in sea water.

 

[Charles2]

Scooters and Buoyancy....If a scooter implodes, it will lose volume, which will make it much more negative than a scooter that has flooded but maintained its volume.

Seems as if this entire discussion was begun because of this erroneous statement by @boulderjohn.

In an effort to correct this statement, several have offered physics lessons, as well as the thoughts of some guy named Archie Medes. All have been bantered around in justification of explanations that sometimes even missed the mark themselves. Throughout all of the ten or so pages that follows the original post, one thing is certain - most of us are better at SCUBA than physics.

 

[Jiminy]

hahaha i cant even use other peoples math correctly

scooters have been stuck in my mind since read the story about the french guys.

 

[fsardone]

Hello everybody,
I did not feel contributing while this was in A&I but now it feels better.
This is one of the classic problems encountered at Uni while I was studying Phisics1.

Calculate how much a slab of ice would sink in water .... and how much more if a polar bear would be on it ...

Fact is for a homogenous solid item not to sink its specific weight (also known as density expressed in kg/m3 for those using engineering units and not strange imperial stuff...) than the fluid in which is immersed. If its specific weight is more it will receive a lift equal to the volume of fluid it displace.

In the above example, if a cube of 1 mt of side (1 cubic meter) is fully immersed in distilled water at 1 atm and 4 degrees it will receive a lift of 1000kgf (I should use Newton and Pascal) (2200 lbf). Wether this cube floats depends if it weight more or less than the 1000kg it receives from archimedean force. In case it floats it will stay a bit above water ... it is a linear 1st degree equation to calculate how high. Than add the polar bear ....

In case of an open tank the water filling it is not displaced .... so the only lift if equal to the volume of the metal x density of water.

Finally the case of the tank filled with gas:
You have to add the weight of the gas
Let’s use an s80 which has a weight of 14.3 kg Aluminium as a density between 2700 and 2800 kg/m3 which means that the volume of aaluminium is 0.0052 cubic meters (5.2 lt) which means than the entire tanks is 11.1 lt (internal volume) plus the 5.2 lt of aluminium. Which gives a positive lift of 16.3 kg-14.3kg which is 2 kg or 4.4 lbf (add valve andf reg and you get a pretty neutral or slightly buoyant stage). This is a tank with 1 bar of air (I ignore the weight of 11 liters of air here).
But I add 200 bar of air I get 200bar 0.0111 m3 x 1kg/m3 about 2 kg of air in the stage. This will make a full stage with reg negatively buoyant when full and neutral or positive at the end. This is why sometime reb divers dive with more helium than needed (we do not use it up) or lower pressure than full in the stages.

Scooter.
My scooter a Suex XJ14 weights 20 kg (44 lb) so it has to have a volume of .... 20 lt. In order to receive a lift of 20 kg. It has an air filled space in the battery compartment, and another one where the motor is installed. The two are sealed. If the seal of the shaft fails only the motor compartment will flood. If the nose oring does the same will happen for the battery compartment. In both cases the scooter will rapidly become unmanageable and needs to be dumped. In case of implosion the scooter would weight pretty much 20 kg because the volume of the metal, plastic and other component would not give any significant lift being denser than water.
If somebody really wants to calculate it, it can be done by submerging the sccoter with the to be flooded compartment open in a graduated container and measuring how much water is displaced. The weight of that water is the lift a flooded scooter would have. My guess is less than 5-6 kg.

My $0.02 in this one.

 

[Akimbo]

Since we are on the subject, a lot of diver's have not made the mental connected between the weight of water and ambient pressure underwater. The US Navy orders their pressure gauges for hyperbaric chambers and pneumo-fathometers calibrated in FSW (Feet of SeaWater). The conversion factor on the purchase order/specification is 0.445 PSI/FSW*. That is based on the average weight of seawater of 64.1 Lbs/Ft³. A cubic foot is 12"x12"x12". That means the area of one side is 144 Square Inches (12"x12"). Therefore, a column of 1"x1"x12" of seawater would weigh 0.445 Lbs (rounded) and exert 0.445 PSI of pressure.

The metric system is a lot easier for diving related calculations (and most others) since:

• 1 Metric ton = 1 cubic meter (or 1,000 Liters) of freshwater at 0° C (2,200 Lbs)
• 1 Liter of freshwater at 0° C = 1 Kilogram (2.2 Lbs)
• 10 Meters of freshwater is very close to 1 Atmosphere of pressure, which is pretty close to 1 Bar, one of the several Metric units of pressure.

Based on the average density that the US Navy uses, the ratio between Freshwater and Seawater is 0.97187305% or a little less than 3%

Fresh water = 62.2970625 Lbs/Ft³ Fresh Water at 0° C (32° F)

* Keep in mind that divers actually care about pressure far more than the linear distance underwater. The reason is that decompression calculations are based on pressure, as well as gas compression. The main value of the linear distance underwater is you can use nautical charts and fathometers that display depth in Fathoms (6'), FSW, or MSW (Meters of Sea Water) for dive planning.

 

[Storker]

The metric system is a lot easier for diving related calculations (and most others) since:

• 1 Metric ton = 1 cubic meter (or 1,000 Liters) of freshwater at 0° C (2,200 Lbs)
• 1 Liter of freshwater at 0° C = 1 Kilogram (2.2 Lbs)
• 10 Meters of freshwater is very close to 1 Atmosphere of pressure, which is pretty close to 1 Bar, one of the several Metric units of pressure.

Based on the average density that the US Navy uses, the ratio between Freshwater and Seawater is 0.97187305% or a little less than 3%

IOW, pretty much negligible for most practical purposes. At least within scuba. Pretty much the same as for 1 bar = 1 ATA. Except for that 3.25% difference between barA and atmA.

 

[fsardone]

The metric system is a lot easier for diving related calculations (and most others) since:

• 1 Metric ton = 1 cubic meter (or 1,000 Liters) of freshwater at 0° C (2,200 Lbs)
• 1 Liter of freshwater at 0° C = 1 Kilogram (2.2 Lbs)
• 10 Meters of freshwater is very close to 1 Atmosphere of pressure, which is pretty close to 1 Bar, one of the several Metric units of pressure.

Based on the average density that the US Navy uses, the ratio between Freshwater and Seawater is 0.97187305% or a little less than 3%

Hello Akimbo,
Fully agree except I believe water temp for density purpose is 4 degrees because at that temp density is maximum (the polar molecules pack more tightly) and also because at 0 degrees you would have ice and not water!

Cheers.