General wonderment

[rightwinger26]

So, as I'm sitting here working on a 1/200th Bismarck, I find myself wondering how much the model would weigh if I scaled down 50,300 tons......if I was so inclined. I'm sure I could google the weight conversion, but I figured this would start an interesting topic.

Bill

[Bigasshammm]

251.5 tons.

Whahahaha

[rightwinger26]

251.5 tons.

Whahahaha

I think I have JUST enough old fishing sinkers!!!! Lol

[Alan in Yorktown]

I think I have JUST enough old fishing sinkers!!!! Lol

Depleted uranium works better

[dnl42]

Volumetric scaling is needed, so 6.2875 kg or 13.86 lbs

[habu2]

Volumetric scaling is needed, so 6.2875 kg or 13.86 lbs

that's based on metric tons - right?

[dnl42]

Yes, tonnes.

BTW, the frink calculator does a great job at these conversions :)

Edited December 14, 2014 by David N Lombard

[Lancer512]

I guess I figured it out with ships, but how do you calculate the volumetric weight of aircraft?

[dnl42]

It's the same no matter what, the scale-factor cubed.

[Lancer512]

Okay, thanks for the hint. Seems like I hadn't figured it out with ships, either.

[Rex]

If you are reducing the weight of a warship's displacement "to scale",,,,,,that is not the weight if it were placed on a scale to weigh it.

"displacement in tons" is the weight of the water displaced by the vessel when it is placed in the water.

If it has any good qualities of flotation at all, it will float without moving water in the amount to equal the weight of the ship

even if it is a solid chunk of steel,,,,and it sinks, the water it displaces is lighter than the weight of the steel

in the same vein, if you had a lightweight plastic boat, and it was tethered to the bottom, and was underwater, the water it displaces would be heavier than the plastic boat

[Jennings]

There's also the argument that "scale weight" is irrelevant, because the specific mass of the material used to create the real thing can't be scaled. Aluminum is aluminum is aluminum.

[Lancer512]

So does anybody ever weigh ships? As it is stated in displacement, there does not seem to be a need for the actual weight.

However, with aircraft, I guess it can be more preciseley done by scaling down the weight to get the same the wing loading.

[habu2]

So does anybody ever weigh ships? As it is stated in displacement, there does not seem to be a need for the actual weight.

If you know the density of the water and the volume of the displacement you know the weight - same thing.

In engineering we studied 'dimensional analysis' that accounted for the scaling of densities as well as mass etc. Especially important in flow studies e.g. wind tunnels.

[dnl42]

Archimedes' Principle. As habu2 notes, the water displaced by any floating (or submerged) body at equilibrium is exactly its weight. Equilibrium means it's not sinking or rising. If it weighs more than the displaced volume, it sinks; if it weighs less, it rises. This is usually bad for a ship, but hopefully planned for a submarine. :) Edited December 15, 2014 by David N Lombard

[Exhausted]

Ahh yes, science.

[rightwinger26]

I kind of figured this topic would go this way when I started, although its not my kind of science, so I'm just sitting back enjoying how the conversation is going.

[MoFo]

Or more succinctly, the scale weight could vary between 5.2kg and 6.5kg, depending on load and whether displacement was measured in fresh or sea water (no idea what German standards were; that will get you in the ballpark, at least). Which seems to be roughly the weight of the kit when built

For all the math students out there who've had it hammered home to show your work, the formula is fairly simple.

Displacement weights were pulled from Wikipedia. There are 1016kg per long ton, so (long tonnage) x 1016 gives you the weight in kg. Divide this by 200³ (ie. weight ÷ 200 ÷ 200 ÷ 200) as you are dividing the weight by the length, width and depth. This gives you the scale displacement. Then multiply by water density, which is either 1 for fresh water, or (roughly) 1.03 for sea water.

41000 long tons x 1016kg ÷ 200³ x 1 = minimum weight

49500 long tons x 1016kg ÷ 200³ x 1.03 = maximum weight

Swap around water densities to get min weight in sea water or max weight in fresh water. And multiply your answers by 2.2 to get weight in lbs.

[Janissary]

Another way to think about it as follows:

The 'force' that a liquid applies to an object is equal to the 'weight of the displaced liquid'. I think this simplifies things a little bit. If you observe that the object is in static equilibrium, then the weight of the object must be equal to the force exerted by the liquid. This implies that the weight of the object must be the weight of the displaced liquid. This is true if the object is floating (partially submerged) or fully submerged, but not touching the bottom of the reservoir (i.e., sea floor). If the object sinks, hence the sea floor is contributing to static equilibrium, the force that the liquid applies is still the weight of the displaced liquid. However, in this case the object is heavier. The difference is the force exerted by the sea floor.

For a floating ship, if you can measure the weight of the displaced liquid, the density of the liquid becomes irrelevant. For two liquids with different density, if the object floats in both liquids, then the weight of the object is equal to the weight of the displaced liquids. The difference will be that the *volume* that is displaced will be different for the two liquids:

W_object = W_displaced_liquid_1 = Volume_displaced_liquid_1 * density_liquid_1 * grav_acc.

and

W_object = W_displaced_liquid_2 = Volume_displaced_liquid_2 * density_liquid_2 * grav_acc.

Therefore:

W_displaced_liquid_1 = W_displaced_liquid_2

or

Volume_displaced_liquid_1 * density_liquid_1 = Volume_displaced_liquid_2 * density_liquid_2

So the same ship will displace less sea water but more fresh water (volumetrically). But either way, the weights of the displaced waters will be identical, which is the weight of the ship.

Also, the 200³ conversion is correct. One subtlety is that *all material dimensions* have to scaled down by the 200. This means, the thickness of the metal used in the scale model has to be 1/200th of the thickness of the sheet metal used on the real ship. Also, obviously the material used in the scale model has to be the same material used on the real ship.

[Slartibartfast]

If you can calculate volume of wetted area, then you can calculate displacement. No need to build scales to survive an encounter witha million pound ship.

[habu2]

Swap around water densities to get min weight in sea water or max weight in fresh water.

umm, no... :unsure:/>

So the same ship will displace less sea water but more fresh water (volumetrically). But either way, the weights of the displaced waters will be identical, which is the weight of the ship.

umm, yes! :thumbsup:/>

Also, the 200³ conversion is correct.

here's a twist: what about time? (the fourth dimension) Bismark was built (launched) in ~1939, so what would Einstein say the current scaled weight (err, mass) is? :monkeydance:/>

[Slartibartfast]

It's relative velocity to us is zero, so no dilation effects.

[echolmberg]

The other day I loaded 16 tons and what did I get?

[MoFo]

umm, no... :unsure:/>/>

D'oh... yeah. You'd need to know water density to calculate *actual* weight via displacement, not scale weight via actual weight.

My bad.

[rightwinger26]

here's a twist: what about time? (the fourth dimension) Bismark was built (launched) in ~1939, so what would Einstein say the current scaled weight (err, mass) is? :monkeydance:/>/>

Now we're getting closer to my language. BUT, if this conversation turned down this path, eventually we would end up at Battlestar Bismarck, lol