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We need a good mental picture. So, let’s break the sub down into two components (Principle of Superposition), the portion above water line and the portion always wet. You can think of these two as separate submarines that drive very close together.
First] H [/u]), it would sink up to middle bar. (---H---) The buoyant force of the perfect foam is equal to the weight of the water in the top part of the "H". I hope you see a comparison to tank size and Dry/Hull weight.
Yup....like I said earlier, the ballast tank needs to be able to displace at least the same weight in water as the weight of everything at/above the waterline.
Imagine you have a large model submarine, perfectly ballasted, with an ideally sized ballast tank and with the correct surface waterline and so on.
You then tie an inflated balloon to the top of your conning tower. The change to the weight of your large model is insignificant. If you like, fill the balloon with hydrogen and then add tiny weights to completely cancel out any change in the weight of your model.
Then try and dive...
Do you doubt that you will now have to swallow a whole lot more water to get below the surface? Your ballast tank would need to grow in size by an amount equal to the volume of the balloon, because this is the amount of water it displaces when it submerges.
Here's a simple way of looking at things, which works for me (simple minded).
Any material heavier than water will sink to the bottom, no matter how big, thick, or thin it is.
Just about any material we use to build a submarine, except most woods, are heavier than water. The only reason the plastic foam provides bouyancy, is because it's filled full of air.
The only we can prevent the boat sinking to the bottom, is by creating a structure, that's mostly filled with air i.e. a ships hull, or the cylinder we use inside a submarine.
Stop, you're both right! http://www.subcommittee.com/forum/icon_biggrin.gif
Stop, you're both right!
Archimedes' said
a body immersed in a fluid is buoyed up by a force equal to the weight of the displaced fluid.
If you read between the lines here, you'll realize that both volume and weight are factors of concern depending on wether you wish to sink or float, and that you cannot globally decouple one from the other.
Gotland, let us consider another box of the same dimension (occupying 2l ov volume) but weighing 1.5kg. If weight is truely not a factor in this principle than box 2 should have the same waterline as in your example. In actuality, it will sink until only the top 1/4 of the box was protruding from the surface. Why? Because a box weighing 1.5kg will sink in the water until it has displaced a volume of water equivalent to 1.5kg, which is 1.5l. (1.5l of a 2l box will be submerged, thus only 1/4 of the box will be above the surface.) Now go back box 1 (having a 1kg weight in the box) and place a 0.5kg weight on top of the box- I assure you that it too will sink to 3/4 submerged. For this example, consider the 0.5kg weight to be your superstructure. Since the superstructure is not in the water, it's volume is irrelevant. When surface running only the weight (not the volume) of the superstructure is a factor, and this weight (not volume) must be countered by the hull which displaces an equivalent weight (by volume) of water. However, since only targets run at the surface all day long...
Lets go back to box 1 (having a 2l volume) and look what happens when filled with 1l of water. It weighs 2kg in total, so it's neutrally buoyant and will hover all day long. If box 2 were also to take on 1l in ballast its total weight would climb to 2.5kg. Since it only displaces 2kg of water Box 2 is now negatively buoyant and will sink to the bottom. If we now introduce Box 3, which also weighs 2.5kg but has a volume of 2.5l (coincidentally displacing 2.5kg of water) we'd find that like box 1 it too is neutrally buoyant, despite having a larger volume than box 1. When submerged, the total volume of the sub, both above and below the waterline, must be made to weigh the same as the water it displaces.
A light weight superstructure will require less buoyant force to support when surface running and will also help to not unnecessarily raise the CG of the ship. However, an excessively light weight superstructure (for the volume of water it displaces) will require more weight (or ballast) submerge below the surface. Keep it light and occupy as little volume as possible.
You are right, that box with 1,5 kg on bord is 3/4 in the water. But now we need only 0,5 Liter of water to dive.
By the way I tried to explain that the weight of the things above the waterline is not important for the size of the ballast system. Nothing else!
But however, the important thing is the volume of all things above the waterline. Believe me, the weight is only important for the center of gravity of the boat. (I am calculating my subs since 25 years this way, it works perfect)
>>> And the weight of that displaced fluid is determined by the volume (i.e. displacement) of the above-surface superstructure, not its weight.
This is only true if the above-surface superstructure is hollow (and sealed) for some reason. This is not the typical scenario for r/c subs and introduces an aspect to this discussion that doesn't normally apply....unless you are building a dry-hull boat, in which case the principle trimming exercise will be to add lots of weight to counter the inherent buoyancy of the boat. Talk about heavy subs!
...and since the original question was posed in terms of building a 1/25th scale Type VII, I think it is a reasonable assumption that this will be a wet-hull boat!
>>> By the way I tried to explain that the weight of the things above the waterline is not important for the size of the ballast system. Nothing else!
Ummm......nope.
Kevin's got it basically correct.
Taking this discussion a bit deeper.....
For wet-hull subs, the weight of the above surface superstructure is what needs to be taken into account.
Volume comes in when calculating how large the ballast tank needs to be. When the boat is submerged, the part that was the above surface superstructure will displace an amount of water equal to its weight. This volume of water will normally not be enough to counter that weight, so it needs to be made up with the ballast tank.
So the volume of the tank needs to be the delta between the volume of water already displaced (by what's above the waterline) and the additional volume of water needed to counter the total weight of above surface superstructure. The weight represented by both the (volume of) water in the ballast tank plus the (volume of) water displaced by the submergence of the above surface superstructure needs to equal the weight of everything above the waterline.
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