Flotation
Example of Basic Flotation Calculations
Assume an Inboard/Outdrive (Sterndrive) runabout
with the following specifications:
| Length Overall |
18' - 6" |
| Beam |
7' - 3" |
| Propulsion |
210 HP Inboard/Outdrive |
| Machinery weights |
1,075 lb. |
| 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 |
2,513 lb. |
From the Applicability section, we determine
that this boat will have to comply with the Basic Flotation
requirement. This means we need only float the boat with equipment
and a certain amount of the passenger load.
In Basic Flotation we consider the entire boat
to be submerged. There is no swamped water line, and consequently
all the components will be affected by the conversion factors
in Table 4.1. We will now run through the calculations to
determine how much flotation will be required. We will use
Table 4.1 for the weight conversion factors of materials.
Step 1: Flotation needed to support the submerged
boat:
Formula:
Fb =([Wh
x Kl] + [Wd x K2]+ .69We
) ÷ B
Let’s identify the components and determine
their weights when submerged:
| |
|
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
= Submerged weight of hull |
169.8
lb. |
|
| |
|
|
|
|
|
|
|
|
| Wd |
= |
Weight of fiberglass
deck |
245 |
x |
0.33 |
= |
80.8 |
|
| |
|
+ |
|
|
|
|
|
|
| |
|
Weight of fir plywood
on deck |
55 |
x |
- 0.81 |
= |
-44.5 |
|
| |
|
|
|
|
|
|
|
|
| |
|
Wd
= Submerged weight of deck |
36.3 lb |
|
| |
|
|
|
|
|
|
|
|
| We |
= |
Weight of factory installed
equipment |
228 |
x |
0.63 |
= |
143.6 |
|
| |
|
|
|
|
|
|
|
|
| |
|
We
= Submerged weight of factory installed equipment, etc.
|
143.6
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:
Step 1: Flotation needed to support the swamped
boat
Formula:
Fb = ( [169.8]
+ [36.3] + [143.6] ) ÷ 60.3
| Fb = 5.8
cubic feet of foam |
Step 2: Flotation needed to support the submerged
propulsion equipment
Formula:
Fp = G ÷
B
G = 75% of 1075 lb. (engine, outdrive and
battery) = 806.2
B = 60.3 lb.
Fp = 806.2 ÷ 60.3
| Fp = 13.4
cubic feet of foam |
Step 3: Flotation needed to support the passengers.
Formula:
Fc = 0.25 (C)
÷ B
Fc = 0.25 X 1400
÷ 60.3
| Fc
= 5.8 cubic feet of foam |
Step 4: Total flotation needed for Basic Flotation
Formula:
F = Fb + Fp + Fc
F = 5.8 + 13.4 + 5.8 = 25
| F = 25 cubic feet of foam |
NOTE:
The Basic Flotation requirements do not address where the
foam is located in the boat.
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