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#1
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| stabilety questions.. Hello. I have big problems whit how to calculate stabilety. Im not wery good on math.. what i want is to calculate the rightening moment RM in newton m Nm. I know the position of center of gravaty G. And i Know Center of boyancy b for different degrees of heel. And the Metacetric hight. I plan devide the "boat" in 1 meter sections. and the add the Nm together.. And the formula in the book say. RM=g*m*arm.. Later on it says someting like: sence gravety is the accelaretion of m/s2.. 9.81. that i understand in a way. but is it an expalation to the rm=g*m*arm , ore do i have to use this accelation ov garaty in any other way..? Here is som exampel numbers.. (metric) The mass in ton on a one 1 m section is 1.2 ton, the gz arm is 0.5m at 15 degrees heel. In this exampel how do i do? 1.2 * 1.2 * 0.5 = 0.72 ton/m , that is abaout 7200 newton meter i guess.. Is this correct. If not can some one please show me how to put this NUMBERS not letters in a formula.. I have tried to find oute abaute this, but all explanations ill find is mere ore less to understand how stabilety works in princip. Tanks in advance.. Daniel |
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#2
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| The formula is correct. Righting moment = lever arm * vessel mass * gravitational acceleration, RM = GZ * m * g. It's common to omit the acceleration, but since you want your RM in Nm you have to use it (otherwise it would be kgm). Your example calculation, however, is wrong. Assuming your numbers for GZ (0.5 m) and m (1.2 t = 1200 kg) are right: RM = 0.5 m * 1200 kg * 9.81 m/sē = 5886 Nm. Oh, and you don't have to divide the boat into sections.
__________________ All the best, Olav GER-1045 ____/)_/)/)___ |
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#3
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| Thank u Olov, It makes sens to me now.. I got totally confused by the expalation of gravaty.. But i know that NM is almost 10 times KgM, I couldent se trhe forest for all the trees i guess! a few more exampels whith numbers in the books would make life easier, at least for me ! I thought that the number i got looked big but read that the stabilety goes up "squered" whit mass, thats not really true then, or am i still getting it wrong.. If u dont have to devide the boat in section then i assume u still have to know the the lengt of the gz arm for the hole boat.. and multply that whith tha total mass and gravety.. But the length of the gz arm and the mass are different in different part of the boat ( if its not a barge ore a big ship whit simular sections) i guess? Its a sailingboat im thinking of.. Thansk again for a wery undersandeble answer.. Cheres |
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