Examples of how the deadweight is decided at the test

### Example 1:

The gear and the catch amount to 300 kg at present.

With a density of 2 kg/litres, the deadweights will amount to a volume of 150 litres.

Steel has a density of 8 kg/litre. Consequently, the 300 kg deadweights can be replaced by 180 kg iron (22.5 litre) + 150-22.5 = 127.5 litre = 127.5 kg water.

Thus, the contribution from iron and water will give the necessary volume of 150 litres and the weight will amount to 180 kg + 127.50 = 307.50 kg.

## Example 2:

A vessel is approved for seven persons and a maximum in total of 630 kg deadweight.

The seven persons of 75 kg, corresponding to 525 kg, will when seated on the thwarts be partly submerged when the vessel is filled with water. Consequently, these seven persons can instead be replaced by 7 X 40 kg weight on the thwarts.

If the outboard motor and the tank weigh 50 kg, iron is placed instead of this weight. Then, the residual weight for gear and catch will amount to 630 – 525 – 50 = 55 kg. These 55 kg can be replaced by 0.6 x 55 = 33 kg iron placed at the bottom.

The 630 kg for which the vessel is approved are thus converted into steel with the following weight:

7 persons x 40 kg, 280 kg placed on thwarts. Outboard motor, 1 x 50 kg, 50 kg located astern. Gear/catch 0.6 x 55 kg, 33 kg located at the bottom. A total of 363 kg iron/steel.

### Example 3:

The point of departure is the same vessel as the one used in example 2, but during fishing activities the vessel will, in practice, often carry a maximum of 2 persons corresponding to 150 kg, whereby the vessel’s proportion for fishing gear and catch can be increased to 630 – 150 – 50 = 430 kg. In order to reflect this condition, the deadweights during the test are com- pounded thus:

2 persons x 40 kg, 80 kg located on the thwarts. Outboard motor, 1 x 50 kg, 50 kg located astern. Gear/catch 0.6 x 430 kg, 258 kg located at the bottom. A total of 388 kg iron/steel