Math Check: Hull Armor Volumes

[aramis]

Am abstracting out. Wondering if someone could check the math.

Volume of hull = 14* Td.
SA of cubic hull (bigger than a sphere, which is why I used it - safe assumption will be over) ~= 6* V^(2/3)/100
V(1cm)=SA/100 kl
Direct to V(1cm) kl= 0.06 * (14 * TD)^(2/3)
Kl= 0.06 * 14^(2/3) * Td^(2/3)
Kl~=0.06* (~5.81) * Td^(2/3)
Kl~=.34*Td(2/3)
Td= .34/14 * 1/14^(2/3) * Td^(2/3)
Td~= 0.024 * .172 * Td^(2/3)
Td~= 0.0041 * Td^(2/3)

Did I reduce it right? (This should, if I did it right, directly produce a rough estimate of the tonnage of a 1cm shell on a roughly cubic volume. It's 215am local...and I can't keep track.)

 

[Theo D Lite]

Aramis,

If I've read this right you are trying to estimate the volume of hull in a cubic hull of xdT.

Here's my take:

Volume of Ship (dTons) = x
Volume of Cube (cu m) = L^3 (where L = side length in metres)
Volume of ship (cu m) = 14x
Therefore L^3 = 14x, L = (14x)^(1/3) = 2.41(x^(1/3))

Surface Area of Cube (sq m) = 6L^2

If Hull Thickness = 0.01m then,
Hull Volume (cu m) = 0.01 * 6L^2

In truth there is a very minor over estimate built in here as the corners are being counted for volume on all adjoining faces which is not the case in reality. The error is however very minor.

Given Hull Volume (cu m) = 0.06L^2
and L = 2.41(x^(1/3))

Then Hull Volume (cu m) = 0.06*(2.41(x^(1/3)))^2 = 0.348(x^(2/3))

Thus Hull Displacement (dTons) = 0.0249*(x^(2/3))

I've had to edit this three times so far checking for errors!

Edited at 21:00 local for error in wrongly squaring 0.06 when substituting L in Hull Volume

All the best

 

[Fritz_Brown]

Or, you can use inner and outer dimensions:

Lo = length of side (outside) in meters
Li = length of side (inside) in meters
(where Li = Lo - (Hull Thickness * 2) )

Hull V = Lo^3 - Li^3

Lot shorter and fewer possibilities for mistakes combining formulas. And, it works for any shape for which you have a volume formula.
Edit AND, it accounts for the corners! end edit

 

[aramis]

Guys, I need to go from tons , not TO tons. I;'m NOT looking for alternate routes. I'm loking for did I do the math right.

 

[Fritz_Brown]

Define your variables, please - I don't have a clue what your equations are working toward.

 

[Anthony]

Originally posted by Aramis:
Am abstracting out. Wondering if someone could check the math.

Running through the math, and trying to make sense of what you're saying:

Volume of hull = 14* Td.
Correct.
SA of cubic hull (bigger than a sphere, which is why I used it - safe assumption will be over) ~= 6* V^(2/3)/100
SA is 6 * V^2/3. You seem to be incorporating several steps right here.
V(1cm)=SA/100 kl
Correct.
V(1cm) kl= 0.06 * (14 * TD)^(2/3)
Kl= 0.06 * 14^(2/3) * Td^(2/3)
Kl~=0.06* (~5.81) * Td^(2/3)
Kl~=.34*Td(2/3)
All correct.
Td= .34/14 * 1/14^(2/3) * Td^(2/3)
Incorrect. It should be Td = .34/14 * Td^2/3
Td~= 0.024 * .172 * Td^(2/3)
Td~= 0.0041 * Td^(2/3)
Eliminating the incorrect 1/14^2/3, we get 0.024 * Td^2/3.

 

[Theo D Lite]

Originally posted by Aramis:
Guys, I need to go from tons , not TO tons. I;'m NOT looking for alternate routes. I'm loking for did I do the math right.

Sorry, have I misunderstood?

I looked at your last equation which had a term dT on both sides and interpretted this to mean that you were looking to arrive at a formula that would allow you to estimate the displacement of the hull from the total displacement of the ship.

I posted my workings because when I looked at it I didn't get to the same result as you. Of course if I was working towards a different goal then I would have a different answer to you.

Assuming we are both working towards the same goal the fact that we have arrived at different results suggests that one of us has made an error. Others can no doubt check both sets of workings and shed some light on the differences.

In the meantime if I have got hold of the wrong end of the stick then I apologise. Perhaps with a clearer brief we can start over?

Regards

<EDIT>
Since making this post I have spotted a third error in my original calculations (yes looks very red faced now!). I think my result now agrees with Anthonys above. Sorry for the confusion.
</EDIT>

 

[aramis]

Ok, thanks Anthony, Theo.
I came to the same catch; kept alternating between 0.025 and 0.0043.

When I redid it in a different sequence, I came up with .025 (rounded). WHich is close enough for my purposes.

Anthony: that was exactly the kind of answer I was looking for.