Section 1: Application of Simpson Rules Flashcards
Basic Terminology:
Ordinates
Intermediate Ordinates
Semi-Ordinates
Appendage
Common Interval
Ordinates, the vertical distance between 2 points on the water plane.
Intermediate Ordinates, in-between 2 ordinates near the appendages.
Semi-Ordinates, an ordinate that is divided by the midline into 2 equal parts.
Appendage, the portions of a vessel extending beyond the main hull outline including such items as rudder, propellers, struts, shafts, shaft bossings, sonar domes and bilge keels.
Common Interval, the spaces in-between ordinates that are at equidistant.
What are the 3 variations of simpsons rules
- Rule 1 - Simpson’s 1/3 rule
- Rule 2 - Simpson’s 3/8 rule
- Rules 3 - Simpson’s 5/3 rule
What is simpson rules utilize for?
To estimate the areas under curves or to calculate underwater volumes, which is simpler than using integration.
How would you increase the accuracy of the answer?
- The accuracy of the answers can be increased by reducing the distance between intervals that are measured.
- Increase the number of ordinates (reduce the intervals).
- Rule 1 is more accurate than Rule 2. Therefore, Rule 1 should be used in preference of Rule 2 where possible.
Find the underwater volume of a vessel having water-plane areas at 1 m intervals from the keel upwards to the water line as follows:
1500; 3400; 3560; 4000 and 4500 square metres.
14240 m3
The breadths of a portion of the vessel’s deck at 5 m intervals are:
7.3; 5; 2.8 and 4 m.
Find the area of the deck between the first two ordinates.
30.708 m2.
The half-breadths of the tank top of a double bottom at equally spaced intervals of 5 m are as follows:
9.0; 8.5; 7.8 and 6.8 m, respectively.
Calculate the area of the tank top.
242.625 m2.
The areas of a vessel’s water-planes commencing at the keel plate to a draft of 6m are as below. The measurements have been taken at equidistant intervals.
1500; 3400; 3560; 4000; and 4500 sqm.
Calculate the vessel’s underwater volume and displacement at 6 m draft in SW.
21360 m3.
21894 t.
An area under a curve (bounded by a curve and a straight line) has the following ordinates starting from forward and spaced at equal intervals of 10 m.
5.0; 11.0 and 13.0 m, respectively.
Calculate:
(a) Its area.
(b) The position of the centroid relative to the first ordinate.
206.667 m2.
11.29 m from 1st ordinate.
The ordinates of a vessel’s water-plane are 9; 16; 19; 19; 17; 10 and 1 metre, respectively. They are equally spaced 20 m apart.
Midway between the first two and last two ordinates are additional ordinates of 14 and 7 m, respectively.
Calculate:
(a) The vessels T.P.C. in both F.W. and S.W. at that draft.
(b) The position of the centre of floatation relative to the middle ordinate.
17.867t in FW; 18.313t in SW.
5.448 m from mid ordinate towards the 9m ordinate (ie.54.552 m from the 9 m end).
The area of a vessel’s water planes commencing from the load water plane and spaced at equidistant intervals down to the inner bottom are:
2500; 2000; 1850; 1550; 1250; 900 and 800sqm, respectively.
Below the inner bottom is a meter depth appendage having a volume of 650 cubic metres and centre of buoyancy 0.64m above the keel. The load draft is 7 metres.
Calculate:
(a) The load-displacement in S.W.
(b) The fresh water allowance
(c) The height of the centre of buoyancy above the keel.
Using Rule 1 9993.75 t 97.5 mm 4.292 m.
Using Rule 2 10045 t.
From the following half ordinates of a water-plane 72 m in length, calculate the area of the w/p and the distance of the COF from the zero station and from amidships.
Station: 0 ½ 1 2 3 4 5 5 ½ 6
½ Ord: 0.2 2.2 4.0 4.8 4.8 4.4 3.2 2.0 0
Area 525.6 m2.
35.032 m from station 0.
0.96 m from amidships towards station 0.
Following is an extraction of data from the M.V. Sample hydrostatic tables.
The displacement at draft 5 m is 12876 tonnes. Calculate the displacement and the vertical position of the centre of buoyancy (VCB or KB) at the draft of 10 m if the KB at 5 m draft is 2.69 m.
TPC in the picture is Area.
30223.079 t.
5.520 m.
