# Free Surface Moment of an Anti-Roll Tank

Discussion in 'Stability' started by RVK, Sep 14, 2012.

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### RVKNaval Architect

FSM can be calculated in two methods

1) loss of GM = (Moment of Inertia of the tank plan * Density of the fluid)/Deadweight

2) loss in "righting arm" = shift in the C.G of the tank (parallel to the waterline) * Mass of the fluid = d*w/g = moment of transference.
(As given in PNA)

While evaluating the stability of an AHTS,

Which method should we consider to calculate the loss of GM in that loading condition for this anti-roll tank?
(PNA recommends Method (2), but the for my tank in consideration FSM is much higher in method (1)-which seems to be the more worst condition.)

Also how do we correspond this "loss in righting arm" in method 2 to the loss in GM?

Does limiting angle (from KG limiting curves) come into picture while calculating the loss in GM?

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### jehardimanSenior Member

Actually, 1 and 2 are the same for small angles were the tank bottom is not exposed. In practice the two methods are BOTH used, but for different things. Remember that GM and righting moment are two different things. The calculation using free surface (method 1) is only used to reduce the zero heel GM (i.e. initial small angle stability). Method 2 is used to calculate the reduction to the righting arm curves (i.e. large angle ultimate stability). See PNA (1988 ed) Chapter 2, Section 5, Para 5.10.

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### RVKNaval Architect

Thank You Sir,

So as per section 5.10 , we calculate the Free surface effect for the anti-roll tank and as per method 2. Then we add this value to the effect of all other tanks and divide with displacement.

But I am referring to a calculation which has been done exactly as mentioned above but before addition the FSM( moment of transference) has been divided by sin φ (limiting angle arrived from limiting curves)......I am confused here......?

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