B) Wind Surface moments

W e need to know what our wind surface moment is. We need to know it for 3 area surfaces:

• 0˚ to 30˚
• 0˚ to 40˚
• 40˚ to 30˚

We will make use of the following formulas:

For 0˚ to 30˚ the formula: (3/8)*(10/57.3)*(3*GZ10° + 3*GZ20° + 1*GZ30°)

For 0˚ to 40˚ the formula: (3/8)*(10/57.3)*(4*GZ10° + 2*GZ20° + 4*GZ30° + 1*GZ40°)

For 40˚ to 30˚ the formula: (0˚ to 40˚) – (0˚ to 30˚)

Step 5: Rolling period

W e also want to know our rolling period. For this we make use of the following formula:

» T = (2*C*B) / √ GM final with

» C = 0.373 + 0.023*(B/d) – 0.043*(L/100)

B Sirrah = 19.4mtr d Sirrah = 9.45mtr L Sirrah = 125.5mtr

Onboard we have a table where you can see what the rolling period is for the calculated GM. The values in this table are calculated with the following formula:

» T = √ ((bc*B)^2 / GM final) where bc (block coefficient = 0.665)

Step 6: Checking the results against the IMO-rules

A fter calculating everything we need to check if the stability is OK according the IMO-rules.

1. Is GM bigger than 0.15mtr? (Navigia rule bigger than 0.35mtr, also depending on captain, mostly > 0.55mtr)

2. Is our Mean draught less than the summer (7.51mtr FW | 7.34 SW) or winter (7.36mtr FW | 7.19 SW) allowable mean draught?

3a. Are the Wind Surface moments more than the IMO allowable values?

0˚ to 30˚ > 0.055mrad

0˚ to 40˚ > 0.090mrad

40˚ to 30˚ > 0.030mrad

3b. Is the GZ value at 30˚ bigger than 0.20mtr?

3c. Is the maximum GZ value bigger than 25˚?

4. Is our list acceptable? (Depending on captain. Normally list must be 0˚)

5. Do we have a positive deadweight reserve?
Deadweight reserve = maximum displacement – displacement.

A complete workout of a stability example

I n this example we’ll show the complete workout of an example were we use:

• 2 tanks: WB1 tank @ full @ 1.015 t/mі and HFO4 tank @ 75mt
• 2 containers: 1*20 feet in position:090202 @ 20mt and 1*40 feet in position 320384 @ 25mt
• Break bulk
• Light ship

As we have 6 different values we have to recalculate them to get only 1 (one) value.

First an explanation how to get the information we need. We start with the WB1 tank.

WB1 tank

We said that WB1 was full and has a density of 1.015 t/mі. If we look in Appendix A, we see that WB1 is full and has a volume of 173.08mt. We must apply the weight correction because of our density difference.

Doing so:

Correct weight = sounded volume * actual density

Correct weight = 173.08 * 1.015

Correct weight = 175.68mt

Remember that we read the LCG, VCG, TCG and IT for the volume and not for the weight.

Info: according to the tables, a full ballast tank has a FSM value. In reality a full tank does not has a FSM, because there is no slack space in the tank.

HFO 4 tank

HFO 4 has, according to the engineers sounding, 75mt inside. When we look at Appendix B, we search in the weight column for 75mt. As you can see 75mt lies between 74.19mt and 75.28mt. This means that we must make an interpolation calculation to get the right values.

Дата добавления: 2015-11-14