Dave Gerr - Boat Mechanical Systems Handbook-How to Design, Install, and Recognize Proper Systems in Boats

the free-surface effect. Separate and dis-

tinct from the free-surface effect is the

effect of the total weight (mass) of the fluid

in the tank and the vertical location of its

center of gravity. High tanks will reduce

stability when full, while low tanks will in-

crease stability when full. This is the effect

of the mass of the fluid and notfree-surface

effect, which is what we’re dealing with

here.

TABLE 6-2.SPECIFIC GRAVITY OF

COMMON LIQUIDS

(At 60°F, 15.5°C)

Liquid Specific Gravity

Diesel 0. 852

Gasoline 0. 727

Fresh Water 1. 000

Salt Water 1. 028

Lube Oil 0. 921

Example:Say you have a boat with a cal-

culated GM 1° of 3.58 ft. at a displacement of

2,156 cu. ft. (61.6 tons, 137,984 lb.) with half-

full tanks. GZ 1°, for this load condition, is

then 0.062 ft., and RM 1° is 0.062 ft.× 137,984

lb. =8,555 ft. lb. The boat has four tanks—

2 for diesel and 2 for fresh water. (The diesel

and water tanks, in this case, are identical

port and starboard.) You would find the effec-

tive GM by using the following formula (you

can find the formula for other shapes in stan-

dard engineering texts).

Formula 6-5a. Moment of Inertia of a

Rectangular Plane Area (English)

I =bh^3 ÷12

Where (for evaluating tank free surface)

b =the length of the tank fore-and-aft, ft.

h =the width of the tank athwartships, ft.

In our case, the diesel tanks are 9 feet

long by 6.8 feet wide athwartships. Thus the

moment of inertia of each tank is

I =9 ft.× (6.8 ft.)^3 ÷ 12 =235.8 ft.^4

Then

Multiply by 2 for both diesel tanks =0.18 ft.

GMred (diesel)

The water tanks are 2.5 feet long and

6.5 feet wide athwartships. Thus the moment

of inertia for each tank is

I =2.5 ft.× (6.5 ft.)^3 ÷ 12 =57.21 ft.^4

Then

Multiply by 2 for both water tanks =0.052 ft.

GMred (water)

Now

and

This is the loss in stability due to free-

surface effect and can be translated into

reduced GZ or RM using the preceding rela-

tionship between GZ and GM.

In our case, stability after free-surface

effect is

GZ 1°=3.35 ft. GM× sin 1°=0.058 ft.

RM 1°=0.058 ft. GZ 1°× 137,984 lb.

Disp. =8,067 ft.lb.

This is a reduction in righting moment of

6% due to free-surface effect.

Or

In metrics, this works out as follows:

Say you have a boat with a calculated

GM 1° of 1.091 m at a displacement of 61.05

m^3 (62.6 Mtons, 62,588 kg) with half-full

tanks. GZ 1° for this load condition is then

0.0190 m, and RM 1° is 0.0190 m× 62,588 kg

=1,195 kgm. The boat has four tanks—2 for

diesel and 2 for fresh water. (The diesel and

water tanks, in this case, are identical port

and starboard.) You would find the effective

GM by using the following formula (you can

find the formula for other shapes in standard

engineering texts).

GM 3.58 ft. GM 0.23 ft.

GM

eff s

redT 3.35 f

=−

= tt.

GM 0.18 ft. GM 0.052 ft.

GM

redT =+red (diesel)

rred (water)=0.23 ft.

GM

57.21 ft.

2,156 ft.

1.000

1.02

red

4

= 3

⎛

⎝⎜

⎞

⎠⎟ 88

0.026 ft.

(for each water tank)

⎛

⎝⎜

⎞

⎠⎟

=

PART TWO:FUEL SYSTEMS

GM

235.8 ft.

2,156 ft.

0.852

1.02

red

4

= 3

⎛

⎝⎜

⎞

⎠⎟ 88

0.09 ft.(for each diesel tank)

⎛

⎝⎜

⎞

⎠⎟

=

Formula 6-5a.