# Stability formulas - who'll check my algebra?

Discussion in 'Boat Design' started by Stephen Ditmore, Dec 15, 2001.

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### Stephen DitmoreSenior Member

This is an attempt to consolidate the several equations naval architects use to calculate initial stability. Is there someone who would be willing to check this formula against known or computer generated data, check my algebra, and report back?

Thanks.

The result of my effort is the formula:

In salt water RM1 = [B*Awp^2 / (670*L)] + (GB*Disp / 57.3)
In fresh water RM1 = [B*Awp^2 / (688*L)] + (GB*Disp / 57.3)
The resulting RM1 will be in meter-tonnes.

note: since Awp will be higher in fresh water, I suspect it won’t make much difference whether the boat is in salt or fresh in most cases.

Glossary of variables:
RM1 = righting moment at 1 deg of heel
L = waterline length in meters (feet*0.3048)
B = waterline beam in meters
Awp = waterplane area in meters^2
Disp = weight of vessel, ½ load condition, in metric tonnes (pounds/2204.6)
GB = vertical distance from the center of gravity to the center of buoyancy, in meters, ½ load condition. (NOTE: IF THE CENTER OF GRAVITY IS ABOVE THE CENTER OF BUOYANCY, THIS VALUE IS NEGATIVE)

My algebra appears below, as do formulae for RM1 in ft-pounds.

Dv = displacement as volume, in meters^3
K = [Awp / (L*B)]^2
I = K*LB^3 / 12
BM = I / Dv
GM = BM + GB
RM1 = GM*Disp / 57.3

Therefore:
I = [(Awp / (L*B)]^2 * LB^3 / 12 = B*Awp^2 / (12*L)
BM = (B*Awp^2 / 12*L) / Dv = B*Awp^2 / (12*L*Dv)

Salt water, metric
BM = B*Awp^2 / (12*L*0.975*Disp)
BM = B*Awp^2 / (11.7*L*Disp)

GM = B*Awp^2 / (11.7*L*Disp) + GB

RM1 = [B*Awp^2 / (11.7*L*Disp) + GB]*Disp / 57.3
RM1 = B*Awp^2 / (11.7*L*57.3) + GB*Disp / 57.3
RM1 = B*Awp^2 / (670*L) + GB*Disp / 57.3

Fresh water, metric
BM = B*Awp^2 / (12*L*Disp)
BM = B*Awp^2 / (12*L*Disp)

GM = B*Awp^2 / (12*L*Disp) + GB

RM1 = [B*Awp^2 / (12*L*Disp) + GB]*Disp / 57.3
RM1 = B*Awp^2 / (12*L*57.3) + GB*Disp / 57.3
RM1 = B*Awp^2 / (688*L) + GB*Disp / 57.3

NOTE!
When using the following, enter values in feet, sq. feet, cu. feet, or pounds.

Salt water, ft-lbs
BM = B*Awp^2 / [12*L*(Disp / 64)]
BM = 64*B*Awp^2 / (12*L*Disp)

GM = 64*B*Awp^2 / (12*L*Disp) + GB

RM1 = [64*B*Awp^2 / (12*L*Disp) + GB]*Disp / 57.3
RM1 = 64*B*Awp^2 / (12*L*57.3) + GB*Disp / 57.3
RM1 = B*Awp^2 / (10.74*L) + GB*Disp / 57.3

Fresh water, ft-lbs
BM = B*Awp^2 / [12*L*(Disp / 62.4)]
BM = 62.4*B*Awp^2 / (12*L*Disp)

GM = 62.4*B*Awp^2 / (12*L*Disp) + GB

RM1 = [62.4*B*Awp^2 / (12*L*Disp) + GB]*Disp / 57.3
RM1 = 62.4*B*Awp^2 / (12*L*57.3) + GB*Disp / 57.3
RM1 = B*Awp^2 / (11.02*L) + GB*Disp / 57.3

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### Steve HollisterJunior Member

Hi Steve,

Since these are empirical (not exact) equations that try to estimate the value of initial stability using a limited set of input values, it's hard to evaluate their correctness unless you can give the sources you used for the equations. One has to evaluate the assumptions made in these sources to see if they all are appropriate for the type of craft you wish to apply them to. Another way to check, as you suggest, is to compare the results from these equations for many test cases using a program that calculates the exact value using the 3D hull shape. This is not as reliable as re-evaluating the original assumptions made in the equations, but it can work.

... Also, is this a typo? At the top of the message, you have

In salt water RM1 = [B*Awp^2 / (670*L)] + (GB*Disp / 57.3)

... But near the bottom, you have

RM1 = B*Awp^2 / (10.74*L) + GB*Disp / 57.3

... I tested a simple round bilge boat in ProSurf and plugged the accurate results in your first equation above and got numbers that are in the same ballpark. My program gives a RM1 value equal to 5.04 m-MT (Calculated at a true heel angle of 1 degree) and your first formula above calculated RM1 = 4.74 m-MT.

... I'm sorry I didn't try to trace through your equations, since I would rather go back to the original sources to try to follow their logic.

... The other thing that makes it more difficult to trace is that you use units-specific values (like the density of water) and combine the constants in the equations. This might make it look like a simpler equation, but it makes it difficult or impossible to do a units check on the equation and to compare the equation with other sources. For programs, my philosophy is to keep all equations as simple and clear as possible so that someone else can look at the equations and know what each variable and constant means and the units used for each. These extra constants and longer equations might seem less efficient, but most computers are more than fast enough and compilers can do a lot to optimize equations behind the scenes.

...I hope this helps.

Steve Hollister

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### Stephen DitmoreSenior Member

Alas, my formula for K involved some wishful thinking on my part. My formulas will give a pretty good approximation, but checking has revealed that the result may be off by a few percent in either direction, as the value for "I" is not exact. Perhaps someone more familiar with the derivation of formulas for "I" can improve upon what I've done.

Thanks, Steve H. Regarding your question about the formula at the end, I should have made it clearer that everything following where it says "NOTE!" about 2/3 of the way down is an addendum for those who want to use feet and pounds to calculate a moment in foot-pounds (instead of meter-tonnes).

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### Steve HollisterJunior Member

That was partly my fault for not paying attention to the details.

K = [Awp / (L*B)]^2
I = K*LB^3 / 12

... Yes, there is a problem trying to estimate the transverse moment of inertia of the waterplane if all that you have is the waterplane area. LB^3/12 is the transverse inertia of a waterplane box that is of length Lwl and beam Bwl. This estimation above says that the real inertia is a squared percentage of this maximum value based on the ratio of Awp/(Lwl*Bwl). This is reasonable assuming that the ends of the waterplane are pointed. Actually, this formula is exact if you compare a triangular (diamond shaped) waterplane that has exactly 1/2 the waterplane area of L*B. If the waterplane area is 1/2 and you square this value, you get .25, which is the exact ratio of inertia values between the rectangle and the diamond shape. Of course, for a real boat, the ratio of real waterplane area and the L*B rectangle will never get as low as 0.5. My guess is that this inertia estimate is on the low side for normal boats.

The true transverse moment of inertia of the waterplane is equal to integrating the true waterplane shape. At each station, for each waterline half-breadth along the hull, you need to multiply the half-breadth times the center of the half-breadth squared. For example, if the waterline half-beam at station 8 is 4 feet (which would make the center of the half breadth equal to 2 feet), then the local value that you have to integrate is 4 * 2^2. After integrating along the length of the waterline, the value is doubled to account for both sides of the waterplane. Don't quote me exactly here because I'm lazy and doing this off the top of my head. What this means is that waterplane area further away from the centerline has a magnified effect on the transverse inertia and therefore, the stability of the vessel. It also means that there is no way to simplify calculation of I easily.

... Without any research, my guess is that you really cannot improve on your estimation of I if all that you have to work with is Lwl, Bwl, and Awp.

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### Steve HollisterJunior Member

I did a test in ProSurf for a typical round bilge hull to test the error in the moment of inertia estimation. For the simple test hull, the estimated inertia using the area ratio was 12.5% too low compared to the exact calculation of inertia. Since BM is approximately equal to I/Vol and BM is the major component of RM, the total error in the formula will be about the same. This also means that your formula is on the consertative side - it will underestimate the righting moment of the boat. I think the error is due to the fact that most waterplane shapes are not triangular or diamond-shaped. They tend to have fine ends and push the area out towards the maximum beam, which would increase the moment of inertia.

I really should stop this fun stuff and get back to work!

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### Stephen DitmoreSenior Member

Revised problem statement

Cy Hamlin's book gives the moment of inertia of a diamond as .022LB^3 - is this correct? Based on that, I've modeled the error for simple geometries as follows:

Problem Statement:
Calculate the moment of inertia (I) of a symmetrical shape about its longitudinal axis given its length (L), its breadth (B) and its area (A). This relates to the calculation of boat stability.

I've put the problem in the form I = [(B*A^2) / (4*L)] * n

section area coeff. = A/(LB) n
For a rectangle: 1 1/3
For a circle: pi/4 1/pi
For a "diamond" shape 1/2 approx. 1/2.84

Is there an equation that would relate these two columns of numbers, and thus give me values for n ?

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### Stephen DitmoreSenior Member

The auto-formatting messed up my columns. Here's the information again.

I've put the problem in the form I = [(B*A^2) / (4*L)] * n

section area coeff. = A/(LB) -------------- n
For a rectangle:
1 ------------------------------------------- 1/3
For a circle:
pi/4 ----------------------------------------- 1/pi
For a "diamond" shape
1/2 -------------------------------- approx. 1/2.84

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### Steve HollisterJunior Member

From my statics book, I get the following

rectangle: I = 1/12*L*B^3 (1/12 = 0.083333)

For a diamond: I = 1/48*L*B^3 (1/48 = 0.02083333)
(it looks like Cy's value of 0.022 is slightly wrong)

For a Circle (L = B): I = 1/4*pi*L*B^3

So, the inertia of the diamond shape is 1/4 the inertia of the rectangle shape, but the area of the diamond is only 1/2 of the rectangle area. This means that the following formula for I works exactly for the case of the diamond.

Waterplane coefficient Cwp = Awp / (Lwl*Bwl)

I = Cwp^2 * 1/12*L*B^3

This is the formula you show in the original post.

The problem with the area (Awp) is that you do not know how it is distributed along the length of the boat. You're equation above assumes that the relationship is a squared relationship, which is exact for the diamond shape and the rectangle shape, but I think underestimates the inertia for typical boat shapes in between.

Using Cy's table of waterplane shapes (on page 70 in my book), you have a list of waterplane coefficients(Ka) versus the value he shows as Kt. To match your formula, Kt should equal Ka^2 / 12. Note that he combines the 1/12 rectangle constant into the value.

1. Rectangle Ka(Cwp) = 1.0 Ka^2/12 = 0.08333 Kt = 0.08333 exact

2. Powerboat Ka(Cwp) = 0.83 Ka^2/12 = 0.0574 Kt = 0.060 low

3. Powerboat Ka(Cwp) = 0.78 Ka^2/12 = 0.507 Kt = 0.050 high

4. Sailboat Ka(Cwp) = 0.67 Ka^2/12 = 0.0374 Kt = .040 low

5. Lemon? Ka(Cwp) = 0.61 Ka^2/12 = 0.0310 Kt = 0.0350 low

6. Diamond Ka(Cwp) = 0.50 Ka^2/12 = 0.020833 Kt = 0.020833 exact

Note again that the value of Kt in Cy's book is slightly off.

Assuming that the values of Kt in his book are exact, then the Ka^2/12 estimate can be both low or high, depending on the shape of the hull's waterplane area. For boats with two "pointy" ends, the estimate is on the low side. If it is a transom stern vessel, then it appears that it is closer to the estimate, but it could be low or high, although I still think that for most boats the estimate will be on the low side.

I don't think you can tweak this estimate any more to reduce the error unless you add a factor for different types of boats. If the hull has two pointy ends, then use one correction factor to increase the estimate of 'I'. For boats with one pointy end, use another factor or perhaps, don't correct.

A lot depends on what you want to do with this equation. This is only for the initial stability for very small angles of heel, since Bm = It/Vol is only an approximation. I have never liked using Bm or Gm, even for ships. It tries to reduce the stability characteristics of a boat to one number. This relates to the initial slope of the righting arm curve, but for small boats, this does not tell you much about the overall shape of the righting arm curve. The formula also assumes that you know Awp,VCG, and VCB, which aren't easy to obtain either.

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### Stephen DitmoreSenior Member

Thanks again. To clarify my purpose, it's to eliminate as many coefficients as possible by substitution, do as much cancelling as possible, and end up with a formula made up purely of boat dimensions. Even if it's only approximate, I think a formula that is as valid as possible and relates directly to the physical boat will contribute to understanding. If you're correct about the diamond then my original formula should be exactly correct for both it and the rectangle. The instance that intrigues me is the circle.

After having looked at this problem a bit now, I like the form
I = [(B*A^2) / (4*L)] * n
better than the conventional formulations. My "n" varies much less than Cy's Kt

section area coeff. = A/(LB) -------------- n
For a circle:
pi/4 ----------------------------------------- 1/pi

This causes me to wonder if replacing the 12 in the denominator with 4*pi wouldn't improve its accuracy for boat hulls.

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### Stephen DitmoreSenior Member

(clarifying note: The "12" in my last sentence refers to the conventional expression for rectangles: LB^3/12, which I took as my basis at the beginning of this process)

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### Stephen DitmoreSenior Member

Oops, but that'd reduce my I, so it'd be going the wrong way according to what you're saying, wouldn't it?

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### Steve HollisterJunior Member

I like your original formula the best. Most will recognize that LB^3/12 is the inertia of the rectangle and will assume that K is some factor that reduces that maximum value for a particular waterplane shape.

K = [Awp / (L*B)]^2
I = K*LB^3 / 12

This K formulation is pretty good, but the question is can you do better for a general collection of vessels?

To recap:

Rectangle: I = LB^3/12
Diamond: I = LB^3/48

The 'K' formulation above is exact for these two extremes of waterplane areas, but it tends to underestimate the inertia for "normal" waterplane areas.

How about a circle ... actually, you need to use an ellipse, since L is usually different from B.

Ellipse: I = PI*LB^3 / 64
Area = PI*L*B/4

The problem is that for a given L and B, I is a fixed value - not dependent on the waterplane area. If you fixed L, but varied B to achieve the desired waterplane area, you would get

B = 4*Awp/(PI*L) (Even though this may not agree with the actual Bwl)

Then, for the ellipse, I = PI*L*[4*Awp/(PI*L)]^3 / 64

This actually is worse, since the Bwl of an ellipse that has the same Awp is usually smaller than the actual one. This has a great effect on the inertia value. For an example round bilge boat:

True I = 304 ft^4
Using K above, I = 286 ft^4
Using the ellipse formulation above, I = 245 ft^4
This error is due to the ellipse having a Bwl = 6.3 ft, as opposed to the actual Bwl of 7.03 ft.

... oops, I got to go - more later.

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### Stephen DitmoreSenior Member

The problem is that for a given L and B, I is a fixed value - not dependent on the waterplane area.

************************************************

Alright, now you're confusing me. For a given shape (we were discussing elipses) waterplane area is also (clearly) proportional to L and B, no? Doesn't that make "I" and Area interdependent, i.e. correlated?

************************************************

If you fixed L, but varied B to achieve the desired waterplane area, you would get

B = 4*Awp/(PI*L) (Even though this may not agree with the actual Bwl)

************************************************

What's the difference between B and the actual Bwl?

************************************************

Then, for the ellipse, I = PI*L*[4*Awp/(PI*L)]^3 / 64

************************************************

I have other work I have to get done now, too, so I'm going to have to look at this later, over a cup of coffee. Thanks, though, I'll try to wade through this when I have a chance, and I appreciate your having taken the time to wade through my mumbo-jumbo.

Cheers,
SD

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### Steve HollisterJunior Member

I had to quit quickly last night before I could go over what I wrote.

The problem is that for a given L and B, I is a fixed value - not dependent on the waterplane area.

************************************************

Alright, now you're confusing me. For a given shape (we were discussing elipses) waterplane area is also (clearly) proportional to L and B, no? Doesn't that make "I" and Area interdependent, i.e. correlated?

... I am being confusing here, I guess. I was looking at using a standard ellipse shape to estimate the moment of inertia of a hull, given a waterplane area (Awp). This means that

Awp = pi/4 * L * B

So, if Awp in an input value and you fix L, then B can be calculated as

B = 4*Awp / (pi*L)

As I calculated, this is a bad estimate, since the B value that is calculated is smaller than the real B value and that really affects the value of I. You could try calculating a new value for L instead, but I don't think that this addresses the real problem.

I guess what I am doing is trying to think (write) out loud and it is perhaps not clear what I am doing. I am trying (as you are) to come up with a good estimate for the transverse moment of inertia using just L, B, and Awp. Your basic formula is pretty good and I understand that you are trying to come up with an additional correction factor (n) that will get it closer.

I think it would be better to come up with a simple type of generic shape that you can use to estimate the value of I. This shape has to try and represent all types of waterplane area shapes. This cannot be done with any great accuracy, but the question is whether you can get a better correction factor than

K = (Awp/L*B)^2

which is exact for the extreme diamond and rectangle shapes.

If you start with a rectangle of size L X B, then draw a simple, geometric waterplane shape inside. This would consis of three parts: a triangle forward to represent the shape of the waterplane near the bow, a parallel middle body rectangle in the middle of the boat, and a triangle aft to represent the shape of the waterplane at the stern. The exact shape is not critical for calculating I, only the transverse distribution of the area.

You could then vary the length of the parallel middle body rectangle to match the input Awp area. Then, since the assumed waterplane shape consists of two triangles and a rectangle, you can calculate the total moment of inertia analytically. This shape would match the diamond shape as Awp goes to 0.5 and would match the rectangle shape as Awp goes to 1. I would wager that in-between 0.5 and 1.0, this method would better represent real-life waterplanes.

I would be easier to draw the shape I propose, but I don't think it matters, since it really didn't help.

If M is the length of the parallel middle body rectangle, then

M = (2*Awp - L*B)/B
and
I = 2[1/3*M*(B/2)^3 + 2*[1/12*(L-M/2)*(B/2)^3]]

This is the moment of inertia of a waterplane shape that consists of a triangle up near the bow, a parallel rectangular mid-body, and a triangle near the stern that has the same length as the forward triangle. M is calculated to get the sum of the areas of the triangles and rectangle equal to Awp. It seems that this would best represent the transverse distribution of the waterplane area for most boats.

However, for my round bilge test case in ProSurf, I get the following:

True, accurate I = 304 ft^4
Using your original K correction, I = 286 ft^4
Using this new fangled method, I = 322 ft^4

Interestingly, the value has the same error amount, but the estimate is too high. I'm surprised that this approach is not more accurate. It might be on average, however, if you compare it to many test cases. In general, I think that you might get some formula for I that minimizes the error for most cases, but it will never eliminate or reduce the maximum potential error.

Also, you have to keep in mind that any righting moment equation that depends on moment of inertia and metacentric height is only an estimate for initial stability. One might also reasonably think that initial stability would be better approximated by the righting moment at 5 or 10 degrees of heel. A lot depends on what information you have and how accurate you want to be. To be accurate, you ned to incline the actual hull or you need the 3D shape of the boat and an accurate VCG.

For designers that use computer programs for design purposes, this is moot, since these programs calculate the exact righting moment values - no estimation is required, except for, perhaps, the VCG. To estimate the initial stability of a vessel in which you do not have the full 3D hull shape, it is always a problem, since you probably don't know the VCG (KG) or Awp or VCB either!

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### Stephen DitmoreSenior Member

I'm indebted to you for this interlocular, Steve, but let me be arguementative on a couple of points.

I agree it's great to have software, but the results are only instantaneous if you've already drawn the boat. One can be either at the front (preliminary) end of the process, just working a few numbers on something that's been sketched on a napkin, or at the back end - looking at a boat in a boatyard and contemplating modifications without benefit of the original drawings.

I agree that assuming all waterplanes are ellipses is not a good model. What I'm looking for is a formula that's more generalizable than that. Using my formulation, the same "n" value that works for circles, 1/pi, also works for ellipses. Here are two examples:

Ellipse, L=4, B=2
Area = pi*LB/4 = 2pi = 6.283
I = pi*LB^3/64 = 1.571
I = (BA^2/4L)*1/pi = BA^2/(4*pi*L) = 1.571

Ellipse, L=8, B=1
Area = pi*LB/4 = 2pi = 6.283
I = pi*LB^3/64 = 8pi/64 = pi/8 = 0.3927
I = (BA^2/4L)*1/pi = BA^2/(4*pi*L) = 0.3927

So I think you're being too quick to throw my formula away. To summarize, the formula:

I = [(B*A^2) / (4*L)] * n
where n = 1/3 for straight line shapes
and n = 1/pi for rounded shapes

works for squares, diamonds, circles and ellipses, exactly. And yes, it can also be written:

I = Cwp^2*LB^3 / 12 (for straight line shapes)
I = Cwp^2*LB^3 / 4pi (for rounded shapes)

The simple case it seems to have trouble with is the combination of a square, say 2x2, with a triangle for a bow, or a triangle on each end, for a total L=4, in which case "n" seems to increase to 1/2.7 (which is perhaps more in line with what you're seeing in your computer runs).

This is where I reach my point of being a bit mystified. I'd like to have a handle on the factors that cause this, and believe that I could come to such an understanding if my calculus was better, alas.

SD

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