Flotation
Example of Level Flotation Calculations
Assume an outboard engine-powered runabout with
the following specifications:
| Length Overall |
18’- 6” |
| Beam |
7’- 3” |
| Propulsion |
140 HP Outboard engine |
| Engine weight |
457 lb. (including controls and battery) |
| Fuel |
Portable fuel tank |
| Maximum weight capacity |
1,400 lb. |
| Maximum persons capacity |
1,100 lb. or 8 persons |
| Dry Hull weight 800 lb. |
(fiberglass 650 lb. + plywood 150 lb.) |
| Dry deck weight |
300 lb. (fiberglass 245 lb. + plywood
55 lb.) |
| Deck hardware |
228 lb. (Mostly aluminum) |
| Hull hardware |
110 lb. (Aluminum 80 lb. + stainless
steel 30 lb.) |
| Total weight |
1,895 lb. |
From the Applicability section, we determine
that this boat must comply with the Level Flotation requirements.
It is an outboard powered, more than 2 HP, mono-hull boat
under 20 feet in length, and the requirements include a Level
Flotation system and some tests to determine its compliance.
In Level Flotation we must establish a swamped
waterline, or the position in which the boat will float after
preconditioning and swamped for 18 hours (See regulation).
We will assume in this example that this swamped line is at
the hull sheer or deck-to-hull joint; therefore, the hull
will be considered swamped and its component’s weight
converted to submerged weight, while the deck will be considered
as dry weight since it will be out of the water.
We will now run through the calculation steps
to determine the amount of flotation material necessary, and
where to install it.
Step 1: Flotation material needed to support
the swamped boat
Formula:
Fb = [(Wh
x K) + Wd] ÷ B
Let’s identify the boat’s components:
| |
|
Part
Description |
Weight |
|
Factor
(K)
Table 4.1 |
|
Submerged
Weight |
| |
|
|
|
|
|
|
|
| Wh |
= |
Weight of fiberglass
hull |
650 |
x |
0.33 |
= |
214.5 |
| |
|
+ |
|
|
|
|
|
| |
|
Weight of hull fir plywood |
150 |
x |
-0.81 |
=
|
-121.5 |
| |
|
+ |
|
|
|
|
|
| |
|
Weight of hull aluminum
hardware |
80 |
x |
0.63 |
= |
50.4 |
| |
|
+ |
|
|
|
|
|
| |
|
Weight of hull steel
hardware |
30 |
x |
0.88 |
=
|
26.4 |
| |
|
|
|
|
|
|
|
|
|
Wh
= Swamped weight of hull |
169.8
lb. |
| |
|
|
|
|
|
|
|
| Wd |
= |
Weight
of fiberglass deck |
245 |
| |
|
+ |
|
|
|
|
|
|
|
Weight
of fir plywood on deck |
55 |
| |
|
+ |
|
|
|
|
|
|
|
Weight
of factory installed equipment (dry) |
228
|
| |
|
|
|
|
|
|
|
|
|
Wd
= Weight of dry deck |
528
lb. |
| |
|
|
|
|
|
|
|
B will be calculated as follows (assuming use
of Polyurethane foam of 2.0 lb. density):
B = 62.4 – 2.0 = 60.4; then allow for
5% moisture absorption (2.0 X 0.05 = 0.1)
Substituting in the formula above:
Step 1: Flotation needed to support the submerged
boat
Fb = (169.8 +
528) ÷ 60.3 = 11.57
| Fb =11.6
cubic feet of foam |
This foam should be installed under the floors,
symmetrically distributed about the boat’sbalance point.
Step 2: Determine the flotation material needed
to support the swamped propulsion equipment (Fp)
Formula:
Fp = S ÷
B
Where:
S = The swamped weight of the maximum horsepower
capacity engine for which the boat is rated on the capacity
label, plus the submerged weight of the battery.
B = The buoyancy of the flotation material in pounds per
cubic foot.
The boat has been rated for a 140 HP outboard
engine. When the boat is swamped, this engine will be partially
submerged to the power head or the cowling - approximately.
Table 4 (in Appendix A) gives us the weights needed here.
Look at the Table. The 140 HP engine falls in the 80.1-145.0
range and, while the dry weight of the engine may be more,
the number to use is the swamped (Column 2) weight. To this
we must add the weight of the battery, and since it is going
to be submerged, we use the submerged weight.
Therefore, S = 352 + 25 = 377 lb.
And, B = 60.3 as calculated before
Substituting in the formula above:
| Fp
= 377 ÷ 60.3 = 6.3 cubic feet of foam |
This is the portion of the total foam that must
be carefully located inside the volume formed by the portion
of the boat forward of the top of the transom, where the engine
is mounted.
NOTE:
The outboard engine weights in 33 CFR Subpart H (see Appendix
A Table 4 of this Guideline), are outdated and much lighter
than modern 4 stroke outboard engines. The reader is encouraged
to refer to ABYC Standard S-30, Outboard Engines and Related
Equipment Weights, for the latest outboard engine weight table.
It is strongly recommended here that manufacturers use this
new table of weights and be conservative on this important
point.
Step 3: Determine the flotation material needed
to support the persons capacity (Fc)
Formula:
| Fc = .5 (first
550 lb. of persons capacity) + .125 (remaining persons
capacity) + .25 (Clev - persons
capacity) |
B |
Where:
Clev = C (maximum
weight capacity) minus weight in column 6, Table 4 for the
maximum horsepower rating of the boat. (Appendix
A)
B = Buoyancy of flotation material in pounds per cubic foot.
Be careful here. Let’s substitute in the
formula one step at a time to avoid confusion. Let’s
add the weight components first and then divide by B.
Look at the specifications. The boat has been
rated for:
- Persons capacity in pounds = 1,100 lb.
- Maximum weight capacity = 1,400 lb.
| Fc |
= |
.5 of the first 550 lb. of persons capacity
= .50 X 550 = 275 lb. |
| |
|
+ |
| |
|
.125 of the remaining persons capacity = .125 ( 1,100
– 550) = 68.75 lb. |
| |
|
+ |
| |
|
.25 of Clev minus the persons
capacity (column 6 Table 4 – Appendix A) = .25 [(1,400
– 550)– 1,100] |
| |
|
|
| |
|
= - 62.5 (Since this is less than
0, use 0) |
| |
|
|
| |
|
Fc =
(275 lb. + 68.75 lb. + 0) ÷ 60.3 = 5.70 cubic feet
of foam |
| |
|
|
This foam should be distributed along the hull
sides and under the deck gunnels in the passenger carrying
areas.
Step 4: Determine the total flotation material
needed for Level Flotation
Formula:
F = Fb + Fp
+ Fc
| F = 11.6 + 6.3 + 5.7 = 23.6
cubic ft. of foam flotation |
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