URGENT QUESTION!....HELP!

If you have a multimillion dollar deal riding on this, you should HIRE an engineer to work on it for you.

Homework........Plan ahead.

As far as I can see it is 2xF times torque arm length.

Reasoned as follows ..

1.The two opposing forces will be equal and opposite...thus the centre buoy will remain effectively 'fixed'.

2.If the Centre buoy is fixed then the torque from one outer buoy is Fx arm length.

3. As we stated in (1) the forces are equal and opposite ...thus the torque from the second arm is also Fx arm length.
Therefore total torque is 2xFx arm length.

But hey whadda I know I'm a cat?
The whole system presupposes the lengths are tuned to the prevailing wavelength....

Is that the scheme of an waves actuated electric generator?

I am not sure to totally understand your question but I imagine that there following possibilities:

a wave comes from let say right.

the right buoy will go up on the wave slope and the torque arm will turn around its pivot and transmit part of the lifting force to the central and left buoys. The left one tries to go deeper and generate a lifting force pushing the right one deeper till the sum of the two actions is zero. The arm is loaded in bending by this force only a&nd is the above mentioned force x arm length.

the wave goes further to the left and will reach the central buoy depending on wave length the right buoy can be either still on the ascending slope or already over the edge. If it is still on the slope the lifting of the central buoy will add itself to the effect from the right one so that the bending will decrease. If the right buoy goes over the ridge then it goes down and tries with the central buoy as support to lift the left one the central buoy is pushed in water and again a position is reached with sum of moments = zero. Loading as above depends only on the differences.

If the wave length allows it there is a position when the 2 external buoys are on the ridge and the central between 2 adjacent waves. The weight of the central buoy will load via the arm the external ones till an equilibrium occurs. So that the maximal bending moment can be the buoyancy of a lateral buoy x arm length and no more.

It does sound more like is a 2 x F x Torque Arm Distance to me but then again the torque is relative to the wave peaks variant since they are not all the same always technicaly. Some are going to be taller or shorter on their peak definetly. This will be also established by an average waves peaks due experience and observation during time frame. 8 hours or 24 hours for example. Also if you would like more power out of such set then will probably going to need couples more Buoyances and if more power of motion need out of it then by inject few pounds of air pressure on them would probably make them to float much stronger than what it is actually doing. More air pressure on them will make float much stronger I believe.

Easy Does It,

MC

Hello Guest,

I can tell you the correct answer, once your deposit of US$1,000,000 has been verified by my Associate in the Bank of Nigeria at Lagos.

As you have a multimillion dollar deal riding on this, the fee will not be much out of your profits.

But there is a problem: You omitted your email address in your desperate first Post.

Please reply urgently, with

Kind Regards....

Yes for sure duh..! That do sound goody. Waooo.. Hee Haa Hee Hoouuh..! 'DEAL or NO DEAL' I'm telling you I will get into the town librarie looking for this perfect answer for those 1000000.00 bucks but then again ahhhhhhh...! I'm not so sure, I will'nt find it out in there. I will keep on looking at the Internet insted. There are a cool "Mathematics" website just for numbers and formulas crunchers, unbeliavable, how much numbers and formulas stuff you seing there. It's like a rocket science detective in a mission. I don't Know my friend. 'Deal or NO DEAL' ?

Respectfuly Have a good deal,

MC