Catamaran math problem

[geoffkryten]

I have a math question and no math skills so a numerical answer is what I’m hoping for.
The question: how many pounds of buoyancy would it take at the head of the mast to keep the mast afloat in a capsize? Assume no wind, the cat was rolled by the freakiest wave in the history of the world. The mast float is a self inflating airbag.
Length 25 feet
Beam 14.9 feet
Mast length 27 feet
Weight 1700 lbs.

 

[bajansailor]

Welcome to the Forum Geoff.
What are your sketching / drawing skills like?
Try drawing a simple midship section of your cat to scale on a piece of paper.
How deep are the hulls? I am guessing maybe about 4 feet?
And how wide? If this is also about 4 feet, draw a square box as the section for each hull.
Then draw the mast height to scale on the midship section.
Along with a rough idea of a float at the top of the mast.
Now, rotate your piece of paper 90 degrees, to simulate the cat capsized to 90 degrees.
However she is actually going to go over a bit more than 90 degrees, as the mast float is currently still more than 7' above the water surface.
Rotate your piece of paper a bit more, until the mast float is now effectively kissing the water surface.
You can consider the weight of the cat to be acting vertically downwards through it's Centre of Gravity (COG). And because the cat is symmetrical, the centre of gravity should be on the centreline of the vessel.
Lets now assume (very roughly) that this centre of gravity is (say) at the mast base on the bridge deck. You would have to do a detailed weight estimate to get a more accurate estimate, but this will do for now.
Do you know the principles involved in taking moments?
It is a bit like two people on a see-saw - if I am 100 lbs and my weight is 5' from the pivot of the see-saw, then my moment about the pivot is 5' x 100 lbs = 500 foot pounds.
If you weigh 200 lbs, then if you sit 2.5' (2' 6") from the pivot, then we will be balanced level, as your moment will be 2.5 x 200 = 500 foot pounds.
The weight of the cat (1,700 lbs) will be acting vertically downwards through the COG (which we assume is at the mast base) trying to capsize the cat some more, once it heels more than 90 degrees.
Meanwhile, the hull in the water has buoyancy, and this is a force acting upwards through the Centre of Buoyancy, stopping the hull from sinking further.
Lets assume for simplicity for now that the COB is in the middle of the hull cross section.
We now take moments again, same as what we did on the see-saw.
The capsizing moment is the weight of the complete hull (1700 lbs, acting downwards through the COG) multiplied by the horizontal distance of the COG from the COB of the hull in the water.
We can calculate the capsizing moment (you can measure the lever arm from your scale drawing).
This has to be resisted by the buoyancy bag at the masthead.
And the moment of this resistance is the buoyancy of the bag (in lbs) multiplied by the distance of the centre of the bag from the COB of the immersed hull.
We know this distance (by again measuring on your rough scale drawing), and so can then calculate how much buoyancy is required in the bag for the resulting see-saw to be in equilibrium.

 

[geoffkryten]

Thank you for that wonderful answer. Hulls are about six feet tall at the center and about four feet wide. I’m assuming no water in the submerged hull so it sits high, but since the boat is about 100 degrees it wants to turtle unless it can be prevented. Again I’m grateful for your answer and I will try to wrap my head around it but if anyone has an answer in the form of a number, I’d appreciate that too. Is turtle the right phrase? I mean 180 capsize.

 

[bajansailor]

"Is turtle the right phrase? I mean 180 capsize."

It will take a lot more effort to right the cat from 180 degrees, compared to righting her from say 100 degrees.
So you need to ensure that the float at the top of the mast has enough buoynacy to ensure that she does not go past 100 degrees, or the angle when it hits the water surface.
But note that these theoretical calculations are simplified for flat calm water, and no wind - the odds are that the wind blowing on the underside of the bridgedeck when she is over at say 100 degrees will create a further capsizing moment.
And then you have wave action to consider as well.
But how about try doing a calculation yourself, and see what you come up with?
And then post your workings on here?