Stability

Next to last kick at the stability cat

Well it turns out that it does work for small round bottom boats and boats without two parallel sides. I wouldn't say that it is one hundred per cent accurate but for the boats we're dealing with it's close enough. Here's how we go about it .

This is the hull we dealt with last time,

So I took that hull shape and rounded off the corners, gives a sort of canoe shape,

We'll keep the centre of gravity the same but the waterline needs adjusting to account for the change in underwater shape. Once that is done then we're going to create a trapezoid, first by cutting off the round parts, those volumes are so small, and equal in area, 50 sq inches, that it makes little difference to the calculation at this juncture.

Now we're going to join the upper ends of those lines that cut off the round part to make a trapezoid and a triangle and then find the centre of those bodies as we did previously.

Once we have both centres then we need to do some calculation, first we'll join both centres with a line. Our centre of buoyancy is somewhere on that line. But where.

In my last post I sent you to Jim Michalak's page here, http://www.jimsboats.com/1dec15.htm where he sets out how to calculate the center of effort of sails. We can do the same thing here. So lets name the triangle area small or AS and the trapezoid AL. We'll call the line joining the two centres LS-LL. Now we measure the area of AS and AL and the length of LS-LL

AS is 131.45 sqin and AL is 270.75 sqin, LS-LL is 12”. Divide AS into AL and add 1, that come out to 3.06. Divide LS-LL by 3.06, which is 3.9. The center of buoyancy is 3.9 inches from LL.

I can hear the mathematicians out there saying that's not right! But here's an example using a lever and fulcrum.

How much weight do we have to hang off the end of the nine foot arm to balance the lever assuming the lever weighs nothing? (which is what we're actually doing when we calculate the relative distance of the centres of our areas)

10x3/9 = 3.33 lbs

So now lets turn that around and say where do we put the fulcrum to balance a 10 lb weight and a 3.33 lb weight on a 12 foot lever?

((10/3.33)+1) divided into 12' = 3' from the heavy end.

Getting back to our vessel, at 5 degrees of heel the righting arm is 2.5” and GM is 2' 4.8”

How does that compare with our square bottom vessel? At the same degree of heel the square boat had a righting arm of 1” and a GM of a foot however if you overlay the two at the waterline,the centres of buoyancy are almost identical.

Next time a quick look at a greater degree of heel and a rounder bottom vessel and one without parallel sides.