How wide should it be?...a method proposed

[Stephen Ditmore]

Disclaimer: The following method mixes valid naval architecture with assumptions made in order to simplify the problem. At this point, it is conjecture, intended as an aid in preliminary design. It is not a substitute for preparing righting arm curves at various loading conditions and comparing them against IMO or 46 CFR Subchapter S criteria once a fully developed hydrostatic table and detailed weight study are available.

When possible, results should be compared with existing vessels of similar size & type known to perform satisfactorily.

Applicability: The following postulates a target GM of 0.70 meters (2.30 feet). In my opinion this would likely be applicable to:
Monohull powerboats, esp. those operating at displacement and semi-displacement speeds
Boats with preliminary freeboard at midship at least 30% of preliminary beam
Boats with a preliminary profile (windage) area not exceeding preliminary waterplane area * (preliminary beam / 2)^0.5 (meters)
or
preliminary waterplane area * (preliminary beam / 6.56)^0.5 (feet)
O.K., got my preliminary design -- but is the beam enough? Will it be stable?

How 'bout we take the inboard profile (showing machinery and the like), make a quick list of weights and centers, and do a moments calculation to estimate the overall center of gravity. What might be misleading is that, in profile view, I'm looking at the deck(s) on edge. So let's draw half the planform of each deck, centered at its average height, superimposed on the profile.

We take our best shot at where the center of gravity is
We take our best shot at where the center of buoyancy is (d/3 below waterline is usually a good guess)
We measure the vertical distance between them: BG

How wide should my boat be at the waterline? I suggest:

B(wl) = [12d*(BG+0.7)]^0.5 (meters)
or
B(wl) = [12d*(BG+2.3)]^0.5 (feet)

Where:
B(wl) = waterline beam
d = fairbody draft, midship
BG = vertical distance from center of buoyancy to center of gravity. If center of gravity is below center of buoyancy, this value is negative.

How did I come up with that? I'll tell you tomorrow. Meantime, if you have a second, do me a favor: check it against some boats you're familiar with, & let me know if it seems right.

Thanks.

[SailDesign]

Stephen,
If I assume right, you're going to be horribly smug about this tomorrow, aren't you? :)
I strongly suspect the formula of calculating something on the order of 1=1, thinly disguised as hydrostatic magic. I will endeavour to crack it before you reveal the "secret"
Steve

[Stephen Ditmore]

Thanks for the response, Steve.

First hint: I used the method of approximating I(t) = LB^3*Cwp^2 / 12
where:
I(t) = transverse moment of inertia
Cwp = waterplane coefficient
L = waterline length
B = waterline beam

I make one other important (and reasonable) assumption that allows me to cancel. The rest is just substitution.

[Stephen Ditmore]

I start with the approximation
I(t) = LB^3*Cwp^2 / 12
and with
BM = I(t) / Displacement(as a volume)

Substituting, we get
BM = (LB^3*Cwp^2 / 12) / (Cb*LBd)
BM = (LB^3*Cwp^2) / (12*Cb*LBd)

Now here's that assumption: I assume Cwp^2 and Cb are equal. I think you'll find these two values are, in fact, usually very close.

Canceling, we get
BM = B^2 / 12d

BM = BG + GM
target GM = 0.7 meters
BM = BG + 0.7
B^2 / 12d = BG + 0.7
B^2 = 12d*(BG+0.7)
B = [12d*(BG+0.7)]^0.5

Definitions:
BM = vertical distance from center of buoyancy to metacenter
GM = vertical distance from center of gravity to metacenter
BG = vertical distance from center of buoyancy to center of gravity
I(t) = transverse moment of inertia
Cwp = waterplane coefficient
Cb = block coefficient
L = waterline length
B = waterline beam
d = fairbody draft, midship

Sorry to spoil the guessing game, but I have a couple minutes now, and it was never my intent to be mysterious. On the contrary, it's my hope to make stability less so.

My assumption, Cp = Cwp^2, may not hold for some shapes, such as flat-bottomed sharpies with fine ends in plan, and with very rectangular underwater sections throughout. In that case, all the boaty shape is in one plane, so Cp = Cwp (not Cwp^2). It's interesting that Steve Baker (SailDesign) happens to have designed some sharpie style powerboats. Steve's boats do not appear to be entirely flat-bottomed. Nonetheless, here (at the end) is the long form of the equation, containing the coefficients:

BM = (LB^3*Cwp^2) / (12*Cb*LBd)
BM = Cwp^2*B^2 / Cb*12d
BM = BG + 0.7
Cwp^2*B^2 / Cb*12d = BG + 0.7
Cwp^2*B^2 = Cb*12d*(BG+0.7)
B^2 = Cb*12d*(BG+0.7) / Cwp^2
B^2 = (Cb/Cwp^2)*12d*(BG+0.7)
B = [(Cb/Cwp^2)*12d*(BG+0.7)]^0.5

O.K. -- now (until someone finds a mistake) I can go eat lunch & bask in my smugness! How're you enjoying it, Steve, as the longest running design award winner in the distinguished history of BoatDesign.net?

[SailDesign]

Good thinking there, Stephen. The whole Cb = Cwp^2 thing is interesting, and is where I got stumped.
How're you enjoying it, Steve, as the longest running design award winner in the distinguished history of BoatDesign.net?
I had just assumed that Jeff had been lazy recently, and not picked a successor yet. Jeff? You there? :)

[Stephen Ditmore]

Now if someone wanted to be REALLY clever they'd show the relationship between Cb/Cwp^2 and KB (which arose earlier as...

....we take our best shot at where the center of buoyancy is (d/3 below waterline is usually a good guess)

...) in light of various possible curves of the waterline plane areas (from DWL down), and then would rewrite the equation for beam using KG instead of BG (or alternatively, using DWL as the reference).

Too complicated? Another possibility: show how to take a vertical offset from DWL some stipulated distance out from centerline, and adjust the constant (12) to yield a better approximation.

[Guillermo]

Interesting, Stephen.

Using approximations:
BM = B^2 / 12d
KB = d/3

We may get a system to approximate GM and so be able to estimate KG.

GM is estimated getting,
first:
GZ(20) = (BWL / T)2/14.84 (BWL in feet and T in seconds)
GZ(30) = (BWL / T)^2/11.00
Where
T = 6.28*( I /(82.43*LWL*(.82*beam)^3))^.5
I = (disp^1.744 )/35.5

and then:
G(10) = a*(10^2)+b*(10)+c
getting a, b and c from an interpolation formula based in GZ(20) and GZ(30)
(quite long to write, but I can try to post it later, when checked thoroughly)

Having this, we approximate
GMo = GZ(10)/sin(10)

Now, being BG= BM-GM
finding KG is inmediate as KB+BG

Seems to work reasonably well as per a quick comprobation with some heavy boats' data. Have to do more testing, but that will be not today...