Question:

How is the angle of equilibrium determined by the RA command when
/FSM is used?  For example, when HMMT is present and the
equilibrium angle is not zero, how does the increased VCG due to
the FSM adjustment affect the equilibrium angle?

Answer:

The FSM adjustment is made by moving the CG upward perpendicular
to the waterplane.  Therefore it does not destroy equilibrium.

The only way to have RA /FSM start at equilibrium is to base the CG
adjustment for FSM on the fully SOLVEd (equilibrium) condition
(including tank CG shifts).  This is done as follows:

  (set condition)
  SOLVE
  STATUS WEIGHT /FSM
  RA /FSM

If this is not what you want; i.e. if you want to use FSM rather
than CG shifts in determining RA "equilibrium", you can fix the
heel at, say, zero and do a partial solve:

  (set condition)
  HEEL = 0
  SOLVE TRIM
  STATUS WEIGHT /FSM
  RA /FSM

In this case the RA curve will start at zero heel, not equilibrium;
but you can reference EQU in your LIMIT commands.

Note that the STATUS WEIGHT /FSM command shows how the CG
modification used by RA /FSM is based on the free surface moment.


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