1/72 Seawolf Ballast tank size -are my calculations correct?

[Slats]

I have been reading David's excellent accompanying text with the new 3.5" SD kit.
David recommends a ballast tank of around 5" in length on Joel's 1/72 Permit.

I was wondering if its correct mathematically to simply up-scale the ballast tank volume to the 1/72 Seawolf relative to Permit?

I.e.
Using 1:1 submerged displacement of Seawolf vs Permit, the ratio is around 209%.
That is Seawolf submerged displacement is 209% that of a Permit class.

Using this ratio could I compute the volume of the ballast tank for the Permit using the 5" length (V=TTr^2) and multiple that volume by 209% and then work backwards to get the length of the Seawolf ballast tank?

Clearly an assumption being made by my reasoning here is that Seawolf and Permit have identical reserve buoyancy proportions - which is likely not to be the case, but is it close enough for the purposes here of estimating ballast tank size?

I.e. Permit ballast tank volume using 3.5" SD in cubic cm=
3.5" = 8.89cm
r= 8.89cm / 2 = 4.445cm
TTr^2 = PIE x (4.445 x 4.445) =62.07cm
Length (L) = 5" = 12.7cm
Volume (V) =TTr2 x L
V= 62.07 x 12.7 = 788 ml

Seawolf submerged displacement 9,138t
Permit submerged displacement 4,369t
ratio 209%

Estimated Seawolf ballast tank volume 209% larger than Permit.
209% x 788ml = 1646 ml
1646 ml / TTr2 = Length (L)
L= 1646 / (62.07) = 26.53 cm = 10.44"

another easier calc based on the same principle is simply multiply the Permit tank's length 5" x 2.09 = 10.45"

All that stated - is this valid? What other methods should / need be applied?

Please advise.
Thanks
John

[oztruck]

Good question. Is there a formula to work by? or is it trial and error. I am getting ready to work on my Lindberg "P"*. class Jap sub and need to know how big the ballast tank should be on it.

Cheers Chris.

#pretend

[Subculture]

Very difficult to calculate a teardrop shaped hull like the Permit class.

I have this link bookmarked in my favourites; very useful for working out complicated volumes, a real time saver- http://grapevine.abe.msstate.edu/~ft...vol/index.html

You could try the 'barrel' formula at that site, then work out what percentage of the barrel is above the waterline. Then you need to know the laminate thickness, and take into account a bit of variation in that department, as you're building it as a wet hull.

Allow a 10-15% fudge factor in your calculations, and that might not be enough.

Ultimately, any calculation can give you a rough estimate, but fine tuning will need to be empirical based.

Andy

[He Who Shall Not Be Named]

Ballast tank water weight is equal to the weight of the displaced water pushed aside by the above waterline structures.

Here's a way to figure ballast water weight without having to work out topside structure volume (what submarine is a collection of simple geometric shapes, anyway!? ...).

Let's start with a simple example:

If the items above waterline have a specific gravity (density, relative to fresh water) of 1, i.e. those structures have the same density as fresh water, then there is a direct coloration between the weight of the above waterline items and the weight of the water the ballast tank has to hold to pull those structures down till their buoyant force is counteracted by the weight of water taken on.

Simple.

But, as a practical matter, the upper portion of the hull, above the waterline, is all to often attached to inseparable portions that extend below the waterline. And the materials from which those structures are made is likely to posses different densities. What to do?

When you weight the assembly you'll have to figure the ratio of above waterline structure to below waterline structure and subtract lower structure weight from total weight to get the weight of the above waterline structures. Math time. Welcome to planet Earth!

Some more reality check here:

GRP and polystyrene materials have a specific gravities a bit greater than 1, they are a bit denser. Anywhere from 1.04 to 1.8. So, what do you do when the structures above water line are denser or less dense than fresh water?

You identify each type material (polyurethane resin, white metal, brass metal, GRP, polystyrene, Acrylic, etc.) work out the weight of each group, multiply the weight times the specific gravity of the material. That gives you the weight of the fresh water that group displaces. Add 'em all together and you have the weight of fresh water all the above waterline elements displace, and that number represents the weight of water the ballast tank has to take on in order for the boat to achieve near neutral buoyancy in submerged trim.

Good practice to take that number and plug in an additional ten-percent -- It's easier to fix a boats trim with a ballast tank holding too much ballast water on board, than too little. But, don't overdo it: excessive ballast water puts an unnecessary burden on your means of getting the water out and the additional volume takes up valuable internal real-estate!

Hell, John, I don't know if prototype displacement figures can be directly compared. Your assumption sounds right to me. You hate Geometry too, don't you!

David,

[Slats]

Thanks Andy and David. Always a case of suck it and see, but how did you arrive at the figure of 5" BT length for Joel's Permit - your math method above (for weight of materials above the water line) - or experimentation?

Yes I hate geometry too. It shows doesn't it!

Thanks

J