# question on calculating LCB with simpson's rule

Discussion in 'Stability' started by mcm, Oct 17, 2012.

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### mcmSenior Member

To calculate the Longitudinal Center of Buoyancy, I:
Take the summation of all section areas times their distances from the front of the waterline and times their simpson's multiplier, and then divide that by the summation of all section areas times their simpson's multiplier.

The question is: do i do this division by the second summation before or after i multiplied the second summation by the distance between sections divided by 3 ?

Remember, this question is about calculating Buoyancy not Displacement.

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### TANSLSenior Member

Remember, a body immersed in a fluid experiences a thrust equal to the weight of the fluid displaced. So, at the risk of being wrong, do say that thrust (force, buoyancy) is the displacement (volume) multiplied by the density of the fluid. Calculate the displacement is essential to calculate the buoyancy.
As for Simpson's rule, is very interesting that you practice with it, but keep in mind that for some of modern ships, calculations with it, may result in unacceptable differences with reality.

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### jehardimanSenior Member

Do it by columns and only do it after the summations because you will be later adding appendages so you will need both values anyway.

C1 | C2 | C3 | C4 | C5
Station area | SM | C1*C2 | Lever | C3*C4

Volume = Sum(C3)*1/3 station spacing
Moment = Sum(C5)*1/3 station spacing^2 if Lever is in terms of station spacing
= Sum(C5)*1/3 station spacing if Lever is in absolute distance
LCB = Moment/Volume

The reference for Lever can be any convinent point

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### mcmSenior Member

Thanks, Jehardiman

So i need to multiply both summations by 1/3 station spacing before dividing volume summation into moment summation.

But what do you mean by "station spacing ^2" ? Is that squared ?
And what do you mean by "if lever is in terms of station spacing" vs. "if lever is in absolute distance" ?

I'm using the forward end of the water-line as my reference, and multiplying by the station spacing as they add up from that reference point for my levers.

Tansl, thanks - it's just that displacement is already calculated in terms of volume in both summations.

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### TANSLSenior Member

Attached is a very simple example of calculating the volume below deck of a ship, in case it may help.

#### Attached Files:

• ###### Simpson.pdf
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71.8 KB
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### jehardimanSenior Member

yes...^ is used in BASIC (and other languages/programs) as an exponential, it is faster to use than FORTRAN ** or to form the BBC x2.

No, you don't need to multiply both by 1/3 first, you can factorially remove it but as I stated, ...why? ...you need both values anyway.

To find lever you can either multiply by the actual distance (i.e. 5m) or the station number then multiply by the station spacing (i.e. Staion #5 * 1m = 5m), allowing you to pull the actual units out of the calculation, especially if you apply the offsets as a function of station spacing...helpful for geosim model work.

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### mcmSenior Member

jehardiman,
Since i'm using absolute distance to determine the lever then i just multiply moment summation by 1/3 spacing, and NOT 1/3 spacing^2. - Right ?

i was going by larson and eliasson, and i had failed to see that, YES, they did use 1/3 spacing as a factor to produce the longitudinal moment of inertia.

TANSL,
Thanks for the PDF graphs.

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### jehardimanSenior Member

Correct, the Station area is in m^2 so if the Lever is in units (m) you need to multiply by station spacing (m) only to get the volume moment (i.e. m^4). If the Lever is in stations (unitless) then you need to multiply by station spacing^2 (i.e. m^2) to get the volume moment (m^4).

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