# Rig Weight Reduction vs Righting Moment

Discussion in 'Stability' started by Gashmore, Dec 7, 2009.

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### GashmoreJunior Member

Assume a sailboat with displacement of 25,000 lb and RM30 of 80,000Ft Lb. If you remove 90 pounds from the rig 25' above the center of gravity what other factors to you need to know to calculate the effect on RM30? I keep hearing it would increase by 25*90 pounds but I believe it must be a bit more complex than that.

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### jehardimanSenior Member

It is more complicated than that, there is a Sine of 30 degrees term in there also amoung other things. But for the 99.64% answer it would increase by (25*sin (30))*90 = 1125 Ft-Lbs for a "normal boat shaped" hullform.

Last edited: Dec 8, 2009
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### GashmoreJunior Member

Thanks. That will get me close enough. I knew the angle had to be taken into account but I was thinking the total displacement had to figure in some how.

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### jehardimanSenior Member

It does, you just don't need it for the change in RM. The righting moment is equal to GZ*displacement; which is identical to the overturning moment - buoyant moment. By removing the weight and its associated moment, you reduced the overturning moment by some small amount, which therefor increases the righting moment. Draw yourself an FBD with the moment axis at the CG to satisify yourself.

Now in removing the weight, the vessels floats new lines as it has a new displacement, and therefor will have a new GZ, but 90 lbs out of 25,000 is 0.36% so unless you have some weird hullform, just call it 1,000 FT-lbs because it is contrived calculation anyway (i.e. the real seaway will have much more effect on instant RM, which is why righting energy is the required criteria).

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### GashmoreJunior Member

Like I said, that's close enough. With all the stuff the first mate plans to haul aboard the numbers are just theoretical anyway.

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