Dear mr Mall,

The mass of a floating body (ship + bunkers + stores + consumables + cargo)

equals the mass of the displaced volume. (Law of Archimedes)

By measuring the draft, either from draft marks directly or through the distance between the water level and marks, and using the ships frame drawing (carene), converted into the draft table, it is possible to determine the displaced volume of water.

Multiplying the displaced volume by the water density gives the total mass of the ship.

(ship + bunkers + stores + consumables + cargo)

The total weight of the ship (ship + bunkers + stores + consumables + cargo) is measured before loading + all readings of tanks, bunkers etc.

After unloading, this procedure is executed again.

The difference of those two measurements is the cargo + the changes in bunkers, consumables etc.

Errors can occur in:

-difference between draft table (based on drawings) and as built

-draft reading due to f.i. waves

Draft marks are normally 10 cm -> accuracy +/- 1 cm

-accuracy in water density estimate ( saltiness and temperature)

-accuracy in determining bunkers, stores etc.

Due to the shape of the carene (underwater body), whereby forward- and aft ship are differently shaped, a trim correction on the draft measurement has to be done.

Nowadays, the pressure at keel level is measured at 6 locations and those readings are processed in a computer.

This method eliminates the necessity for a water density measurement.

This automatic system is still calibrated to a manual draft survey.

An eventual error in this calibration becomes a system error.

The bunkers, stores etc still have to be monitored during loading and unloading.

Official draft measurements should be executed by authorized pesons, agreed by both parties involved.

In case a measuring installation on shore is available, the draft survey can be checked against this output.

In case the sea going ship is divided over a number of barges, each barge can be measured

separately and the sum of the barges should equal the ship’s reading.

( It normally does not, I experienced differences of 1%)

take care

teus ■

Hi.

Checking the draft survey report you should be careful with any figure in it. All data in this report we can divide onto three types: incoming, working, and outgoing information. The most important is incoming data. There are: draft readings, ballast soundings, density of sea (river) water, density of ballast water, and quantity of other liquids on board. The rest information you can recalculate by yourself having hydrostatic and sounding data from that ship.

Important outgoing information is: ship’s displacement, weight of deductible liquids, and constant. The difference between these three figures taken from initial and final surveys gives you the weight of laden or discharged cargo.

The accuracy of draft survey normally is 0,5%. But it’s very rough figure. From my experience, if draft survey was properly done, the difference between draft and shore-scale figures for handy-size (50000 MT DWT) vessel is about 50 MT in still water.

The effect of density variation can be easily calculated. Almost all ships’ hydrostatic tables prepared for so-called “sea density” witch is 1.025 kg/L. So density change only for 1g/L for ship with 60000 MT displacement will give correction: 60000*(1-1.024/1.025) = 58.5 MT. If surveyor use certified hydrometer for determining the density of water, usually it doesn’t cause any mistakes.

How to ensure the accuracy of draft survey having only draft survey report in your hand? I think it’s most important for you but most difficult to answer question. There are hundred ways to make mistake in draft survey whether accidentally or intentionally. They can be made while draft reading during the big waves, ballast sounding when tanks overflowed, during density measuring, calculations etc. The first signal for surveyor that he probably mistaken it is a constant, if it differs from previous draft survey reports. To recap all let’s say that you should pay attention if:

1) Determined constant differs from normal constant for this vessel.

2) Initial and final constants are different enough.

3) Unusual sea or ballast water density.

4) Big difference between declared and determined weight.

But all of this is only reason to ask the questions and doesn’t mean for 100% that any mistake has a place.

Best regards,

Vladimir. ■

5% inaccuracy?! – no, definitely no. It’s absolutely inaccurate draft survey. I can be agreed with Teus about 1% if sea is in worry condition, but 2500 MT mistake for 50000 MT of total weight it’s too much, pls believe.

About the weighfeeder, balance, weighing scale or whatever you call it. Any weight measuring device can be of cyclic (interrupted) or continues weighing type. In first case it is more complicated, and more expensive but very accurate device. Our company and many others who works with the grains use such kind of weighing scales and we always use shore figure for bill of lading. Though, we do draft survey if receiver or ship owner has requested it. So, as I told in first message, the usual difference between measured weight and draft survey result for 50000 DWT vessel is less than 50 MT and abt 15-30 mt for the vessels with less than 10000 DWT, but it's for sheltered port. If you will load/discharge ship directly from/into the trucks or wagons using the truck’s or railroad’s weighing scales, this difference will be greater. It is not a standard but more likes to an “informal international” agreement: if the difference between shore weight and draft survey result less than 0.5% - nobody blames, in other case master of ship can refuse shore figure or issue letter of protest with corresponding explanations.

I know some terminals in our area that are working with continue type of weighing devices. In such case measuring unit under the belt of conveyor weighs the mass of cargo with belt in run. I don’t know about others, but all that I know are very inaccurate (5-10% inaccuracy) and used only for internal mass estimation. They are using draft survey results for official documents too. ■

Hello, mall.

The main steps of draft survey as follows:

1. Reading the draft

2. Correction to perpendiculars

3. Meandraft calculation for hull deformation

4. Calculation of corresponding displacement

5. Displacement correction for the effects of trim and list (First, second trim correction, list correction, last if required)

6. Displacement correction for water density

7. Measurement of liquids on board. ie ballast sounding (measuring)

8. Soundings trim correction

9. Liquids volume calculation

10. Measuring the density of sounded/measured liquids

11. Calculation of weight of liquids in ship’s tanks

11. Determination of constant: Const=Displacement-Light Ship-Weight of Liquids

12. Weight of loaded/discharged cargo calculation as the difference between final and initial draft surveys.

It’s difficult to say abt trim, because I don’t understand clear what trim correction you mean: correction for perpendiculars or displacement correction?

The water density correction it’s very easy. When you have displacement corrected for trim (and list if you need), the density correction will as follows: Displacement corrected for density = (Displacement * Density of Surrounding Water) / Density Shown in Hydrostatic Tables. For example: Displacement = 10500 MT. Surrounding water density = 1.002 (river). Hydrostatic tables prepared for 1.025. Displacement corrected for density = 10500*1.002/1.025=10264 MT.

I can’t describe detailed all process because it’s a whole book. But you can try to found some literature. I can advice UN Standard “Code of Uniform Standards and Procedures for the Performance of Draft Surveys of Coal Cargoes”. Or maybe it would be better to find the GOOD prepared draft survey report from some vessel with all formulas and explanations. I can share it with you if you can’t find it.

But if you are talking about some certain report, you can send it (or part of it) to me and probably I’ll able to give you a most comprehensive explanation.

Best regards,

Vladimir. ■

If it will help you, below you can find the formulas for trim corrections. They are for metric system.

First trim correction:

FTC = ( TRIM * LCF * TPC * 100 ) / LBP

Correction has sign ”+” if: Trim by head and LCF forward of midship or Trim by stern and LCF aft of midship. In other case “-“.

Second trim correction:

STC = ( TRIM * TRIM * 50 * dm/dz ) / LBP,

dm/dz = ( MCT(at draft+0.5m) – MCT(at draft-0.5m),

the sign is always positive.



Where:

TRIM – difference between drafts at fwd and aft perpendiculars

LCF – length to the center of flotation from midship

TPC – tones per centimeter

LBP – Length between perpendiculars

MCT – moment to change trim

You should take TRIM from draft readings after perpendicular’s correction, LBP from ship’s particulars, TPC, LCF and MCT from hydrostatic tables of the ship. ■

tq

I have a paper written by a Naval Architect consultancy on the accuracy of draft survey, which is public domain, which I have attached. We have permission to distribute the document since it is beleived that users should have no illusions about the accuracy of ships draft survey.

It was commissioned by CETOA, the the Coal Export Terminal Operators Association. In the case of coal, it said for old ships expect about 1% systematic error +/- 1% random, for the newer automated ships, about 0.5% +/- 0.5%.

A good belt weigher can do 0.25% pretty easily, draft survey is pretty rough really. Good belt weighers dont just happen however, many peoples/companies claim they can do them, few deliver.

Attachments

06200070 (PDF)

■

Dear Hash.

You self have partly given an answer to your question about procedure of perpendicular correction. In your step 2 you wrote “Apply the necessary correction for Trim & List (FWD & AFT)” and than in sections 3-6 you have explained how to do it, but not exactly correct. This procedure is not so difficult if you clearly understand what ship’s perpendiculars are and what draft marks are.

For example we have next ship: LBP = 150m, AFT Draft = 6m, FWD Draft = 2m, AFT draft mark is 1m to forward from aft perpendicular, and FWD draft mark is 0.5m to aft from forward perpendicular.

First we calculate: Trim = Aft draft – Fwd draft = 6-2=4m; Distance between draft mark, as you call it LBM = 150-1-0.5 = 148.5m (all marks are inward form their perpendiculars).

Correction for aft perpendicular = Trim / LBM * Distance between draft mark and perpendicular = 4/148.5*1=0.026m.

Correction for fwd perpendicular = 4/148.5*0.5 = 0.013m.

The sign of correction can be whether positive or negative at any perpendicular. If ship trimmed to aft and draft mark situated to aft from perpendicular or if ship trimmed to fwd and draft mark situated to fwd from perpendicular the sign of correction will be negative “-“. In two other cases it will be positive “+”.

In our case ship trimmed to aft. Aft mark is on fwd from perpendicular, so sign is “+”. Aft daft corrected to perpendicular = 6+0.026 = 6.026m. Fwd mark is on aft from its perpendicular, so sign is “-”. Fwd daft corrected to perpendicular = 2-0.013 = 1.987m.

Well, hope I have poured some light on your question.

Best regards,

Vladimir. ■

Hi all, can some one guide me thru how the formula for the second trim correction is derived? ■

pls kindly help how to make hog/sag corrections? ■

Hi, Benny.

Hog or sag correction (or correction for hull deformation) in draft survey is known as Mean of Means Correction.

When you have Fwd, Mid, and Aft Mean Drafts corrected to their perpendiculars, you have to use (6*Mid+Fwd+Aft)/8 or (3*Mid+(Fwd+Aft)/2)/4 or other formula to find the MMC value.

Mean of Means Corrected is the mean draft of the ship corrected for hogging or sagging. ■

a little info for hog/sag ...

its possible to have a large hog/sag, just depends on the size of vessel and/or if its in a partly loaded/discharged condition or when hatchcovers are open or closed !!!

!!!! ALWAYS DO THE FINAL DRAUGHTSURVEY IN SAME CONDITION AS YOU DID ON THE INITIAL SURVEY - open or closed hatchcovers - and if possible in same ballast and bunker condition (forget the last part if you're on a seavessel) !!!!

for coasters its normal fom 0 to 15 cm and with the old flat bottom russian ships (simular as the sormovskys, volgabalts, ladogas .... even upto 40 cm and above)

what i try to explain here is that hog/sag a good but not a real accurate personal checkpoint is while calculating the drs. ■

hi all,

can someone explain if there's any formula or way to calculate ship's constant??? thnks ■

Originally Posted by DraftMaster

Dear method,

According to our research if vessel has hog/sag more than 7 cm an extra error approximately 2% from displacement appears.

To solve this problem we design advanced software for russian vessels.

Dear DraftMaster,

i cant wait to get my hands on such software, BUT ..... call me oldshool or not ..... i still like and still perfer to calculate by hand, theres nothing more accurate and workintens as calculate by hand.... ■

Originally Posted by DraftMaster

Dear itakor78,

Ship's constant is difference between actual weight of empty ship now and weight of empty ship indicated in ship documents (lightship).

There is only one way to calculate constant - to do draft survey.

Please remember that actually ship's constant is variable value. For example constant will change if ship catch some silt into ballast tanks. For this reason while draft survey all measurings and calculations are effected twice - before and after loading.

yes,

make note that some C/O put there (aprox) ships constant including the deductions for calculation, DONT accept or calculate that - its wrong !!!! , the ships constant as D.M. explained only can be calculated when a total drs has been carried out. ■

Dear Sirs,

I'm looking for information, how to optimize methods of draft survey, or survey of liquid cargoes.

For example how to make taking draft or ballast measurement more acurated...

I'll be appreciate for any information (it's for my final work at Academy) ■

Originally Posted by DraftMaster

Therefore hand calculations are impossible.

wanna bet ?

just joking here m8, but still i perfer to calculate by hand.

btw i looked up to youre website but its all in rusian, can i translate it somewhere inside the site or is it possible to post a link to a rapidshare or megaupload packed file for me pls ?

sorry for the late reply on this one but work is killing me ..... softly

kind regards from antwerpen ■

Originally Posted by DraftMaster

For information in english please visit http://www.shipsurvey.ru/en/draftmaster/

many thanks for this !!! n1 ■

Vessel constant can be calculated when vessel is unloaded by using the following method:

Vessel displacement - deductables (ballast, freshwater & fuel) - light ship = constant. ■

Dear allanis77DM, Draftmaster,

Indeed, draft readings are influenced by strong currents.

However, this effect is also related to the water depth in combination with the ship’s draft (keel clearance) and is called squat.

A draft survey consists of 2 readings, 1 loaded and 1 unloaded.

When unloading in shallow water:

The loaded draft reading will result in a higher cargo mass, due to the fact that the ship is deeper in the water, because the current sucks the ship down.

The unloaded draft reading, however, is almost correct, because the keel clearance is then much more.

This would result in a higher unloaded cargo mass and allanis77DM is 200 tons short.

When loading in shallow water, the effect is opposite and when the draft survey at the unloading site is then correct, a calculated shortage will occur.

A 5000 dwt ship has a length of approx. 100m and a width of approx 15m, resulting in a waterline area of 1500 m2

With a loaded waterline coefficient of 0.8, this gives an area of 1200 m2

The resulting TPC is then 1200/100 = 12 tons/cm immersion.

Assuming a current under the keel of 3 knots (3 * 1852 / 3600 = 1.54 m/sec), the following pressure decrease is estimated:

dp = * 1000 * 1.54^2 = 1187 Pa (N/m2)

The assumed bottom area coefficient is approx 0.6, resulting in a bottom area of approx 900 m2.

The pressure at the keel reduces then with 1187 * 900 = 1068300 n # 106.8 tons, compensated by an increased draft.

If the current under the ship is 6 knots (2*3 knots), then the squat causes 1.41*106.8= 150 tons cargo error.

Resume:

The mentioned discrepancy of 200 tons at the unloading site, due to shallow water and current (if that is the cause), can only be caused by a shallow water (resulting in a very small keel clearance) and high current situation at the loading site.

The keel clearance must be very small to generate enough squat.

The clearance between the ship and the quay wall is also a key parameter that can influence the current underneath the ship.

Have a nice day

Teus ■

Dear Draftmaster,

At the moment of loading, the master just notices a certain draft.

If the master is qualified and experienced, he should be aware of the effect of shallow waters and high currents.

However, when he sails from the loading port, he must also notice that the ship’s draft is suddenly less.

Arriving at the loading port, the draft survey is repeated again and when there is no shallow water and high current situation, the out-loaded cargo mass is correct, but not corresponding with the draft survey at the loading site.

Comparing the 2 draft surveys will show where it went wrong.

A squat at very small keel clearance and high velocities of 15 cm ( ft) does not seem impossible to me and actually, that is what we discussed here.

(I remember an old article that squat measurements on large bulk carriers entering the port of Rotterdam showed a squat up to 1 foot)

See also:

http://chl.erdc.usace.army.mil/libra...hetn-ix-19.pdf

and

http://www.pilotmag.co.uk/2008/02/01/squat-part-1/

The squat effect exists and I tried to quantify that effect.

Whether it is the cause of Allanis’s problem, we do not know.

Interesting subject.

Take care

Teus ■

Dear Teus and Draftmaster

Your comments on the ship squad and underkeel current influence are great

Beside, some point need to be considered - as most of inaccuracy are due to ship condition, i.e:

- Blocking on the sounding pipe: likely a sounding bob parted and left in the sounding pipe that can cause to an error = bob length in cm * TPC of tanks - may be large if tank is DB and large tanks

- Parted to the bob touch plate of sounding pipe, caused to sounding bob passed over touch point and got into tank bottom: big sounding error

- Air void in the TST, specially in case of high trim - while TST overflow

- Sometime, crew delibrately block sounding pipes to cheat shore rep.

In our country, most of inaccuracy to draft survey are due to defective of vessels.

With regards

Nguyen Pico ■

if have a case with 1.3º list having a difference of 0.73 m between midships drafts.

I've an old formula 6(tpc1-tpc2) x (d1-d2) being d drafts amidships and tpc the tpc at drafts on each side, any comment ■

Hi

I have a question

What kind of problem can the hydrostatic particulars (water displacement table) have when the vessel is f.e. half loaded? Can we be sure that calibration that is made on every vessel is precise on every meter?

Thanks in advance. ■

Hi Mall,

I am daily involved in shipments of various bulk cargoes Worldwide and we do on all our ships a full draft survey at loading port and also at discharging port by fully independant surveyors. Sofar we always find that abt 20 pct of the cases the ships are discharging less than loading and stated in the Bs/l and for about 80 pct of the cases we found that the ships actually discharged slightly more than loaded onboard. Please always arrange for proper surveys so you can sleep at night ! ■

Dear Members,

Am seeking advise from members.....

Our company manufacture granular urea in bulk. While conducting draft survey always the discrepency arise in ship constant. Allmost all time the ship's constant calculated by Draft surveyor is much more than what constant declared by Ship.

Kindly give me advise the following doubts.

1. Importance of constant in draft survey?

2. If the declared constant by ship is more as per calculated constant then what will be the impact? Are we loosing cargo?

regds

R/M ■

Hi,

i am trying to understan this below formula

(3 x deflection (cm) x tpc)/4

load master it shows this formula about bending correction. the ship is capesize 170.000 deadweight and if 10 cm hog or sag hog/sag corection about 899 tons

always in draft survey i taken account this corection or not i am not sure

as i know bending correction (hog /sag corection) before the caculation of mmm

draft mid +(1/3 x hog)

i confused can anybody cleared about this matter? ■

Dear captain.tr,

First some theory.

When a ship is not evenly loaded according the difference in buoyancy between amidships and the ends, then the ship (considered as a beam) bends (hog or sag).

Assume we have more weight at the ends and the amidships draft stays the same after hogging.

Then the buoyancy has increased with the extra immersion at the ends, although the real weight did not increase.

Therefore the ship comes amidships out of the water and the draft reading amidships should be increased for weight calculations to get to the right cargo content.

As already indicated, the effect is depending on the buoyancy at the ends of the ship in comparison to the buoyancy amidships, considered at the actual waterline.

Therefore, the correction is ships form (waterline shape) related and is different for each ship and draft.

General formulas are for generally the same type of ships.

The amidships draft correction (regardless the used formula) should give the same result.

This formula seems to be based on the assumption that the hogging/sagging curve is a parabola, where the average height of a cut is 2/3 of the maximum height. (1-2/3=1/3)

The delta(draft) = 1/3*hog

The delta(weight) = delta(draft) * tpc

The delta(weight) = 1/3 * hog * tpc

Can also be written as:

3/4 x deflection (cm) * tpc/4

The delta(weight) = 3/4*hog * tpc/4

These 2 delta(weight)’s should be equal.

Thus:

1/3 * hog * tpc = 3/4*hog * tpc/4

1/3 * hog = 3/4*hog * 1/4

16/3 * hog = 3*hog

5.333 is unequal to 3

This inequality proves that the formulas are not the same.

The conclusion is that each ship must have its own hog/sag correction table and general formulas for a certain type of ship are approximations.

Have a nice day

Teus ■