A PRACTICAL GUIDE TO DRAFT SURVEYS TRIMMING CALCULATIONS ETC GLENN J SALDANHA (salmonmrn@gmail.com) PLEASE PRINT THIS OUT USING BOTH SIDES OF THE PAPER - HAVE ADJUSTED THE FILE TO ALLOW EASIER READING WHEN PRINTED & FILED THIS WAY INDEX CHAPTER I - THE - SIDED DRAFT CHAPTER II - THE DRAFT SURVEY 11 CHAPTER III - TRIMMING 22 CHAPTER IV - HOG / SAG 51 CHAPTER V - CONTROLLING DRAFTS 59 CHAPTER VI - MAXIMUM DRAFTS 63 CHAPTER I - THE - SIDED DRAFT In the above diagram L.B.P = Length between Perpendiculars L.B.M = Length between Marks C = LCF = Longitudinal Center of Floatation - distance fm midships c = difference between draft at LCF & draft at midships dF = Ford Draft mark dA = Aft Draft Mark FP = Forward Perpendicular AP = Aft Perpendicular WL = water line when even keel W1L1 = water line when trimmed (in this case by stern) T = TRIM (at the perpendiculars) Ta = APPARENT TRIM (at the marks) A = Distance between the ford draft mark & ford perpendicular a = Difference between draft at ford mark & draft at ford perpendicular B = Distance between the aft draft mark & aft perpendicular b = Difference between draft at aft mark & draft at aft perpendicular ^M = Angle of Trim As all are similar triangles with the same ^M : T = Ta = c = a = b = LBP LBM LCF A B Since Ta = a then a = A x Ta LBM A LBM & Tan ^M Ta = b then b = B x Ta LBM B LBM In the formula for correction to be applied to the ford draft A is a constant value on EACH SHIP as is LBM as is B On a Cape-size vessel A = 1.00 metres and LBM = 248.50 meters Since Correction to be applied to the ford read draft (marks) = a = A x Ta LBM Therefore a = x Ta 248.5 And a = 0.0040241 x Apparent trim And since B = 10.5 meters Therefore b = 10.5 x Ta 248.5 And b = 0.0422535 x Apparent Trim When the Draft Marks are on the Perpendiculars then no correction is to be applied If the ford draft mark is aft of the ford perpendicular and the aft draft mark is ford of the aft perpendicular as in the diagram and on most but not all vessels then, “a” will be subtracted from the read ford draft to get the draft ford and “b” will be added to the read aft draft to get the aft draft WHEN THE VESSEL IS TRIMMED BY THE STERN WHEN A VESSEL IS TRIMMED BY THE HEAD, the correction sign will be opposite After reading the drafts and applying the draft correction then one will proceed to calculate the Mean Quarter Mean Draft using the 6-sided formula Many surveyors not worry about the correction to the read drafts when the trim is very small Actually, in theory, there is a correction to be applied to the read midship drafts as well unless the midship draft is computed by measuring the freeboard from the deck line or computing the draft by measuring the distance from the Loadline Mark to the water line - because the loadline mark is exactly at midships and the midship draft marks are actually displaced either just forward or just aft of midships No matter what is taught in the various colleges of the world with respect to hydrostatic draft, bulk carrier draft calculations the world over is based on the six-sided formula Drafts are read on all six sides Fp - Ford port Fs - Ford stbd Mp - Midship port Ms - Midship stbd Ap - Aft port As - Aft stbd After these drafts are read, the trim correction is applied to the ford & aft drafts & to the midship drafts when relevant (on some vessels the midship draft marks are not exactly at midships) - (same explained later) to convert these read drafts to drafts at the perpendiculars We are now left with drafts – (where “c” stands for “corrected”) Fpc, Fsc, Mpc, Msc, Apc, Asc - which are the same as above except that they are corrected to the perpendiculars or midships as the case may be Using the mean of each set we get the Ford, Aft & Midship Drafts (Fpc+ Fsc) = F (Ford Draft) (Mpc + Msc) = M (Midship Draft) 2 (Apc + Asc) = A (Aft Draft) (F + A) = Mean (Mean Draft) Now to get the draft, reqd for calculating displacement, known as Mean Quarter Mean (MQM) draft we use the six - sided formula (6M + F + A) = MQM OR x Midship draft + Ford Draft + Aft Draft = Mean Quarter Mean Draft Now as (Ford Draft + Aft Draft) = Mean Draft OR ( Ford Draft + Aft Draft) = x Mean Draft Therefore the Sided formula is also: (6M + 2Mean) = MQM OR x Midship Draft + x Mean Draft = Mean Quarter Mean Draft The 6-sided formula is central to all calculations on bulk carriers & must be dinned into your head and understood completely In the first place it is used in all actual all draft calculations, but can also be used in precalculation of cargo figures as the following examples will show It is to be remembered that using the sided formula for actual draft calculation is completely accurate, the following examples, which are useful in pre calculation, makes one assumption which is theoretically incorrect, and that is, that we are assuming the vessel is trimming about midships - in actual fact the vessel trims about the Centre of Flotation which is not necessarily midships, but the pre-calculation examples are shown for when the vessel is completing loading, when in case of bulk carriers we almost always finish even keel, without trim, or very close to even keel, which makes any inaccuracy caused by making the assumption that the vessel is trimming about midships, so very small as to be discarded In any event pre-calculations are only that and the actual picture only obtained, during & after loading Before any examples are given, the formula must be understood properly The Mean Quarter Mean Draft which is used for obtaining displacement concept must be understood properly MQM = M + F + A or MQM = M + Mean 8 And Hog / Sag = Difference between Midship Draft and Mean Draft When Midship Draft is greater than Mean Draft, Vessel is Sagging When Midship Draft is less than Mean Draft, Vessel is Hogging Now it must be understood that if M= 17.0 meters and the vessel is sagging by 10 cms, then the MQM is NOT 16.90 meters - to illustrate Example 1: Given that Midship Draft (M) = 17.0 meters, and vessel is sagging 10 cms Vessel is sagging 10 cms, therefore the Midship Draft is 10 cms more than the Mean Draft Therefore Mean Draft = M - 0.10 = 17.0 - 0.10 = 16.90 meters Therefore as MQM = 6M + 2Mean MQM = 6M + 2(M - 0.10) = 6(17.0) + 2(17.0 - 0.10) 8 = 102 + (16.9) = 102 + 33.8 8 Therefore MQM = 16.975 meters AND THIS MUST BE UNDERSTOOD COMPLETELY BEFORE PROCEEDING FURTHER Example 2: Consider a Cape-size bulk carrier, SILC, of about 151,000 dwt Given that max draft is 17.5 meters, by experience you expect that the vessel with that cargo, loaded in all holds will sag 10 cms, and you wish to sail out even keel - you will by the formula get Midship draft = 17.50 meters (v/l sagged & even keel), Mean draft = 17.40 meters and as v/l to finish even keel, F = 17.40m A = 17.40m and therefore, MQM = (6 x 17.50) +17.40 + 17.40 = 17.475 meters Therefore you will get the displacement for 17.475 meters, apply the Dock water correction, reduce the lightship, constant and deductibles and you will get the cargo to be loaded To expand: the density of Dockwater = 1.023 lightship = 18643 K= 400 (FO: 2000 DO: 100 FW: 200 Unpumpable Ballast: 100) Deductibles = 2400 Displacement in SW at 17.475 Dock water 1.023, displ in DW = 169560.15 = 169560.15 x 1.023 1.025 Displacement in DW = 169229.3 Lightship = - 18643 Deadweight = 150586.3 K +deductibles = - 2800 CARGO TO LOAD = 147786.3 Example 3: Max draft = 17.00 meters V/l to finish with 20 cms trim by stern 12 cms of sag To calculate Ford & Aft drafts and MQM to obtain displacement & therefore cargo To get max cargo have the Midship draft equal to the max draft - remember that in the formula the midship draft is multiplied times, the fore & aft draft only once so it is much better to have the midship draft maximum As v/l is sagging 12 cms, (Midship Draft – Sag) = Mean Draft = 17 - 0.12 = 16.88 Now to calculate the fore & aft draft we assume that the v/l trims about the midships and since v/l is to be trimmed 20 cms & is trimming equally ford & aft & therefore ford draft = 16.88 - (0.20 / 2) = 16.78 & aft draft = 16.88 + (0.20 / 2) = 16.98 MQM = x 17.0 + 16.78 + 16.98 = 16.97 meters 8 Example 4: Cargo to load = 130,000 MT FO: 1800 MT DO: 100 MT FW: 250 MT U/P ballast: 100 MT LTSHIP: 18643 K: 400 Density: 1.025 V/l to sail out even keel, sagged by cms, calculate F, A, M Displacement = 130000 + 1800 + 100 + 250 + 100 + 18643 + 400 = 151293 For SILC, displacement of 151293 in 1.025 corresponds to draft of 15.85 meters Therefore MQM is 15.85 meters Let Midship Draft = X meters Since v/l sagged cms, Mean Draft = (X - 0.08) meters From Formula MQM = 6M + 2Mean = MQM = 6X + (X - 0.08) 8 8(MQM) = 6X + 2X - 0.16 (8 x 15.85) = 8X - 0.16 (126.8) + 0.16 = 8X Therefore X = Midship Drafts = 15.87 meters V/l sagged cms, Mean Draft = 15.87 - 0.08 = 15.79 = F = A (v/l finishing even keel) To confirm in reverse MQM = 6M + 2Mean = 6(15.87) + 2(15.79) 8 MQM = 15.85 meters Example 5: Same conditions as in Example except vessel to finish hogged by 10 cms Let Midship draft = X meters As v/l hogged by 10 cms, Mean Draft = (X+0.10) meters MQM = 6X + (X+0.10) => MQM = 6X + 2X + 0.20 8 (MQM) = 8X + 0.20 (15.85) - 0.20 = 8X X = 15.825 meters = M Since v/l hogged by 10 cms & even keel Mean Draft = F = A = 15.825 + 0.10 = 15.925 You can confirm same by formula MQM = 6M + F + A = x 15.825 + 15.925 + 15.925 8 = 15.85 Example 6: MQM = 15.85m Sag: cms Trim: 14 cms by stern Midship draft = X mts RD: 1.025 Mean draft = (X - 0.08) mts (sag of cms) MQM = 6M + 2Mean => MQM = 6X + 2(X-0.08) 8 MQM = 6X + 2X - 0.16 => MQM +0.16 = 8X => 8(15.85) + 0.16 = 8X Therefore X = 15.87 = Midship draft Sag = 8cms Therefore Mean draft = 15.87 - 0.08 = 15.79 mts Assuming vessel trims abt midships & trim = 14 cms F = 15.79 - 0.07 = 15.72 & A = 15.79 + 0.07 = 15.86 Example 7: If the v/l, in Eg.6 was then to go into RD less than 1.025, say 1.010, you would have to first convert the MQM of 15.85 into the MQM of that density At MQM 15.85 mts displacement in SW = 151293 Equivalent displacement in 1.010 = 151293 x 1.025 = 153539.9 1.010 Therefore MQM in 1.010 = 15.968 mts (from SILC) Assuming that the sag remains same, & very likely it will, MQM = 6X + 2(X - 0.08) where X = Midship draft => 8(15.968) = 6X + 2X - 0.16 Therefore X = 15.988 = Midship draft & Mean Draft = 15.988 - 0.08 = 15.908 Again assuming v/l trims around the midship we will have F=15.838 & A=15.978 However there will be a change of trim due change of density & this will have to be applied to get the actual Ford & Aft Drafts in the DW of density 1.010 If the vessel were to hog instead of sag, say hog by 10 cms then as in Example 5, Assume X = Midship draft and Mean Draft = (X + 0.10) - all other calculations same 10 CHAPTER IV - HOG / SAG 51 HOG / SAG What your hog/sag will be is very difficult to predict - however having loaded these ships over the years, I have observed out the following – please note this is completely unscientific and only refers to the named ships On one Panamax I served on - Navios Bulker if you are using the old loadicator - a new loadicator has been fit in the 2nd half of 1998 - but the old one exists - if you can distribute your weights such that as per the OLD loadicator the Bending Moment at Midships in the Harbour Condition shows 8% sag you will end up with no hog or sag at midships For every 2% the loadicator shows on either side of this % you should get either cm of hog or sag - i.e if the loadicator shows 6% sag at midships you should end up with a cm hog at midships If it shows 12% sag at midships you will probably end up with 2cm sag at midships On another Panamax - Lucky Bulker I think you have to distribute the weights such that as per the loadicator the Bending Moment at Midships in the Harbour Condition is about 8-9 % sag to get no hog or sag at midships - I was last on the Lucky in 95and I am not sure if I am absolutely right but even if I am not I would not be far wrong On the sister Cape-size I sailed on you require to distribute the weights such that as per the loadicator the Bending Moment at Midships in the Harbour Condition is between 15-16% sag - to get no hog or sag at midships - then every about 1.6% on either side of the 15-16% translates into cm of hog or sag at midships Remember there is no scientific explanation of the above but we had discovered this by studying and monitoring over a few years and many voyages PLEASE REMEMBER THAT IN ALL CASES WE ARE TALKING ABOUT THE LOADICATOR INFORMATION OF THE BENDING MOMENT IN HARBOUR CONDITION & this is limited to individual vessels & loadicators If you had noticed I have kept mentioning “no hog or sag at MIDSHIPS” - it is important that I am talking only about no hog or sag at midships and it is here only that it is important - the vessel’s drafts can only be read at the fore and aft and at the midships on both sides - if you can distribute the weights such that after reading the drafts you have no hog or sag at the midships then for all means and purposes you have no hog or sag Have loaded the Cape-size more than twice and finished with no hog or sag by calculation - I knew full well that the v/l was sagged over the length of the v/l, but with my cargo distribution I had no hog or sag at the midship draft marks - the strain gauge clearly showed the v/l to be sagged but the drafts did not & since the only way you can calculate hog or sag is by reading the drafts at the draft marks to all means & purposes we had no hog or sag 52 You will not always be able to distribute the cargo such that you will end up with the figures mentioned above & achieve no hog or sag - at midships - when you are loading alternate holds you will almost definitely end up hogged - or when you are loading all holds and are volumetrically full or very close to you don’t have much choice on how to distribute the cargo - but what you can is that within the limitations of the particular circumstance you work, to get as close to the above figures as you can, or the figures that by experience you have determined for your vessel/loadicator We have found how to be able to make the distribution of cargo work for us in that there is no hog or sag at the draft marks & therefore we have loaded the max.cargo & also have actually a little more cargo on board than the Surveyor’s figure shows so that we have no problem at the discharge port However this same situation sometimes can work against you - when you are in the ballast condition on both Panamax & Cape-size you will be hogged over the length of the v/l - but if you are in the Heavy Ballast condition - i.e with the hold in and also all the ballast tanks you will also in all probability be hogged over the length of the v/l, but because the ballast hold is just off the midships you will be sagged at the midships even though you may be hogged over the length of the v/l - which means that the read midship draft is greater than it actually should be As we use the 6-sided formula this results in the Initial Displacement being higher than it actually is, and there is no correction factor that can be applied here, and the v/l is said to be ‘doubly deflected’ which results in the constant being very much higher by calculation than it actually is if you are in heavy ballast on arrival at the load port and you get a very high constant be sure to tell this to the surveyor and try & bring the constant down nearer normal levels Also, temperature effects the hog or sag If you read the drafts at night, and make no changes at all after that and read the drafts again at midday, it is quite likely that the drafts will not be exactly the same as at night, because the hog or sag has changed This is best illustrated by vessels plying in the Great Lakes, where they are working at a max draft and cannot allow this changed hog or sag to increase their draft & so have the deck water running at all times over the main deck in order that the entire shell plating is at the Lake water temperature Where this may affect you is at ports such as Richards Bay, where this has actually happened to some vessels - the max draft you are permitted to sail with is 17.50 meters, when your summer draft permits & you are allowed to have your midship draft to a max of the Summer mark when that is below 17.50 meters It can happen there that you finish loading at night, working to your max drafts permitted & you & the draft surveyor agree on the drafts & these are all within permissible limitshowever you are unable to sail immediately for whatever reason -such as too much swell at the entrance & you only sail at midday - if somebody from the port authority then reads the draft prior sailing you might have exceeded the max draft - only when you explain the day/night effect to them will they permit you to sail - on the other hand if at the time of completion of loading your midship draft has exceeded your summer loadline they will not permit you to sail even though your MQM may be equal to or less than the summer mark - no allowance is made for hog or sag at the completion of loading by the port - they will only permit you the day/night effect allowance in case you cannot sail on completion For this reason, keep something in hand when transiting the Panama Canal 53 A very important thing to remember is that the while the loadicator calculates & shows the ford & aft draft, (from which you can calculate the mean draft) it cannot take into account hog or sag or anything else - it therefore shows mean, midship & MQM as the same Eg if the loadicator shows F: 16.0 meters A: 17.0 meters - it is showing all figures on basis of displacement at the mean draft of (16.0+17.0) / = 16.50 meters If you are hogged or sagged it cannot take that into account & therefore as far as the loadicator is concerned your mean draft is your midship draft & when that happens so is the MQM Now if you assume that the vessel is going to sag say 10 cms you will have to adjust your figures to show the right displacement - remember displacement is got from MQM Say your midship draft is to be 17.50 meters & you expect to sag 10 cms & are at even keel Your first instinct would then be to work your figures such that the loadicator shows ford draft to be 17.40 & aft draft to be 17.40 meters THIS IS WHERE YOU ARE WRONG If you used the sided formula with drafts of F:17.4 M:17.5 A: 17.4 you would get an MQM of (( (17.4 x 2) + (6 x 17.5) )) / = 17.47 What you would then seek to is to adjust your weights to get F: 17.47 A: 17.47 on the loadicator You would of course declare your drafts to be 17.4 / 17.5/ 17.4 If you expect to be trimmed 50 cms with a midship draft of 17.50 meters and 10cms sag, you would work your weights such that the LOADICATOR shows drafts of F: 17.47 - 0.25 = 17.22 & A: 17.47 + 0.25 = 17.72 Of course your declared drafts would be worked on the basis of Midship - sag = Mean & then trimmed about the Mean which would be 17.5 - 0.10 = 17.40 = Mean & 17.4 - 0.25 = 17.15 = ford draft & 17.65 = aft draft 54 LOSS OF CARGO LOADED DUE TO SAG It has been my experience that should you ask Chief Officers on bulk carriers what is the loss of cargo loaded due to sag – the vast majority, well over 90%, would tell you that it is Sag in cms x TPC – so if the sag was cms and the TPC was 65 tonnes the loss would be x 65 = 325 tonnes THIS IS WRONG!!!! Let us examine using the 6-sided draft and the method we followed of calculating MQM in Chapter I Midship Draft – Mean Draft = Sag (should M’ship Draft be less than Mean Draft it would be Hog) And therefore, Midship Draft – Sag = Mean Draft ……… (a) Also, Mean Draft = Ford Draft + Aft Draft => x Mean Draft = Ford Draft + Aft Draft Let us assume Midship Draft to be 14m and sag to be cms (=0.05m) and TPC = 65 Therefore Mean Draft = 14 – 0.05 = 13.95m (from (a) above) MQM = x Midship Draft + Ford Draft + Aft Draft MQM = x Midship Draft + x Mean Draft (from (b) above) MQM = (6 x 14) + (2 x 13.95) And therefore, MQM = 13.9875 Should there have been no sag, then MQM would have been = 14.0000 Difference 0.0125m = 1.25cms And therefore loss due to sag = 1.25 x 65 = 81.25 MT where TPC = 65 MT Let us assume Midship Draft to be 14m & sag to be 10 cms (=0.10m) and TPC = 65 Therefore Mean Draft = 14 – 0.10 = 13.90m MQM = x Midship Draft + x Mean Draft MQM = (6 x 14) + (2 x 13.90) And therefore, MQM = 13.9750 Should there have been no sag, then MQM would have been = 14.0000 Difference 0.0250m = 2.50cms And therefore loss due to sag = 2.50 x 65 = 162.50 MT where TPC = 65MT Now, that you understand the principle, will share a formula for the same LOSS of CARGO due to SAG = SAG in CMS x TPC Check above Where sag = 5cms Where sag = 10 cms x 65 / = 81.25 MT 10 x 65 / = 161.50 MT 55 LOSS OR GAIN OF CARGO LOADED DUE TO HOG As opposed to SAG, HOG may result in a loss or gain of cargo – WHAT IS HOG? Mean Draft – M’ship Draft = HOG (should M’ship Draft be more than Mean Draft it would be Sag) And therefore, Midship Draft + Sag = Mean Draft In some ports the local authorities will state that while you cannot submerge your respective loadline mark at midship you can so forward and aft (which is logical) as you might have some trim In others they will not even permit you that Please see the note on Loss on the following page AA) GAIN Let us assume Midship Draft to be 14m and hog to be cms (=0.05m) and TPC = 65 Therefore Mean Draft = 14 + 0.05 = 14.05m MQM = x Midship Draft + x Mean Draft MQM = (6 x 14) + (2 x 14.05) And therefore, MQM = 14.0125 Should there have been no hog, then MQM would have been = 14.0000 Difference 0.0125m = 1.25cms And therefore GAIN due to hog = 1.25 x 65 = 81.25 MT Let us assume Midship Draft to be 14m & hog to be 10 cms (=0.10m) and TPC = 65 Therefore Mean Draft = 14 + 0.10 = 14.10m MQM = x Midship Draft + x Mean Draft MQM = (6 x 14) + (2 x 14.10) And therefore, MQM = 14.025 Should there have been no hog, then MQM would have been = 14.0000 Difference 0.0250m = 2.50cms And therefore loss due to hog = 2.50 x 65 = 162.50 MT Now, that you understand the principle, will share a formula for the same GAIN of CARGO due to HOG = HOG in CMS x TPC Check above A Where hog = 5cms B Where hog = 10 cms x 65 / = 81.25 MT 10 x 65 / = 161.50 MT 56 BB) LOSS In many ports, they will not let you submerge your forward and aft drafts beyond the respective loadline draft, when your midship draft is at the loadline There will also be occasions when you have a maximum draft at the load port – So let us explore this Say your maximum draft is 14.0m and you end up even keel with a hog of cms As in this case your mean draft will be more than your midship draft, your mean draft will be the maximum which will be 14.0m, and your Midship Draft will be Mean – Hog = M’ship => 14.0 – 0.05 = 13.95 = M’ship MQM = x Midship Draft + x Mean Draft MQM = (6 x 13.95) + (2 x 14.00) And therefore, MQM = 13.9625 Should there have been no hog, then MQM would have been = 14.0000 Difference 0.0375m = 3.75cms And therefore LOSS due to hog = 3.75 x 65 = 243.75 MT Say your maximum draft is 14.0m and you end up even keel with a hog of 10 cms As in this case your mean draft will be more than your midship draft, your mean draft will be the maximum which will be 14.0m, and your Midship Draft will be Mean – Hog = M’ship => 14.0 – 0.10 = 13.90 = M’ship MQM = x Midship Draft + x Mean Draft MQM = (6 x 13.90) + (2 x 14.00) And therefore, MQM = 13.925 Should there have been no hog, then MQM would have been = 14.0000 Difference 0.0750m = 7.50cms And therefore loss due to hog = 7.50 x 65 = 487.50 MT Now, that you understand the principle, will share a formula for the same LOSS of CARGO due to HOG = x HOG in CMS x TPC Check above C Where hog = 5cms x x 65 / = 243.75 MT D Where hog = 10 cms x 10 x 65 / = 487.50 MT So, therefore when you have a limiting draft at the Load port, you have to worry about both Hog and Sag SAG is the lesser of the evils 57 This page intentionally left blank 58 CHAPTER V - CONTROLLING DRAFTS 59 CONTROLLING DRAFTS When you start making your calculations for a particular voyage it is most important to know where to start i.e where your controlling draft is Sometimes this is quite easy Say, you are loading at Tubarao for Rotterdam when it is summer in the Northern Hemisphere Draft available at Tubarao is 22m, at Rotterdam 23m, your summer draft is 17.52 meters - simple you can load in Tubarao up to the summer mark assuming Tubarao is in the Summer Zone - if it is in the Tropical Zone you will be able to load up to Summer Loadline plus the consumption in the Tropical zone till you reached the Summer Zone Say, you are in Tubarao, loading for Rotterdam, when it is winter in the Northern Hemisphere - you can load such that when you enter the Winter Zone off Cabo Finistere you are at the Winter Mark of 17.16 meters Your controlling draft would therefore be 17.16 meters when entering the Winter Zone & you would have to start your calculations from that point It would then work out that your sailing draft from Tubarao would be Winter Draft + consumption from Tubarao to Finistere When calculating I would start at Finistere and then work backwards to Tubarao Say, you are loading in Tubarao, then discharging at a port where the allowed draft is 16.00 meters - simple - your controlling draft would be at the disport - you would base your cargo to arrive even keel at the disport with 16.00 meters draft after calculating what your fuel etc ROB would be at the disport You are loading at Newcastle, Australia, where max draft is15.50m, your summer draft is 17.52 meters, for Rotterdam where max draft is 23.0m Simple your controlling draft is Newcastle & you would start your calculations to sail out of Newcastle even keel, then work for arrival Rotterdam, taking ballast to avoid arriving down by the head You are on a Panamax size where you are required to load as follows Disport is Eregli, Turkey where max draft is 11.5 meters Your summer draft is 13.25 meters Loading at Saldanha Bay, South Africa Disport is Constanza, Romania You have to load for Constanza all you can, that is possible, assuming you are loading maximum for Eregli Max draft in Constanza 17m All ports & the entire route in the summer zone Max draft in Saldanha Bay 20m In this case the initial controlling draft is Eregli You would have to start your calculations based on first arriving at Eregli with an even keel draft of 11.50 meters after calculating what your fuel etc ROB would be at Eregli Then work your fuel consumption backwards to sail out of Saldanha Bay, at the summer loadline - you would then work out the cargo you can load for Constanza, & calculate the trim you would have to sail out of Saldanha Bay with, such that after fuel consumption en route, & discharging in Constanza you would arrive at Eregli with the even keel draft of 11.50 meters - your arrival draft & trim at Constanza in this case is not really a problem 60 You are loading on the Cape-size, summer draft = 17.52m, at Tubarao, where max.draft is 23m, for Pohang where max draft is 17.40metres Consumption of fuel en route causes a rise of 15cm Both ports & route in the summer zone Here even though the draft available in Tubarao is much higher than in Pohang, the departure draft in Tubarao is the controlling draft You are permitted to load up to your summer mark of 17.52 meters & no more - when you that in Tubarao - your consumption en route to Pohang cause a rise of 15 cms - thus your draft arrival Pohang would be 17.52 - 0.15 = 17.37m Of course you would have to adjust your trim departure Tubarao, to arrive, after the fuel consumption, even keel at Pohang Cape-size situation - summer draft = 17.52 meters - all ports & route in summer zone 1st loading port - Sept Isles, Canada - iron ore lumps Max draft at Sept Isles less than summer draft 2nd loading port Saldanha Bay, South Africa - iron ore pellets Max draft 20m 1st discharge port Kimitsu, Japan - discharge the Saldanha Bay cargo Max draft 20m 2nd discharge port Nagoya, Japan - limiting draft, required to discharge the Sept Isles cargo Required to load max cargo for Nagoya - no bunkering en route This was a situation on SILC (actually the cargo was iron ore concentrates - types) Now, in this calculation, you would have to start with Nagoya, after calculating what your arrival Nagoya bunkers etc would be then work the fuel etc consumption backwards to Sept Isles, via Kimitsu & Saldanha Bay, & then see whether with the max Nagoya cargo on board, you were within the limiting draft in Sept Isles -it was - but if it were not then the limiting draft would be at Sept Isles - in this case it was at Nagoya then after calculating your weights at Sept Isles, work your fuel consumption to Saldanha Bay - on basis of that work your figures to sail out of Saldanha Bay at your summer draft, then discharge the Saldanha Bay cargo at Kimitsu, then after consumption of fuel Kimitsu, to arrive at the max draft at Nagoya Please remember there will be probably be density changes which you have to take into account Also vessel must be within allowable stress limits at all times 61 This page intentionally left blank 62 CHAPTER VI - MAXIMUM DRAFTS 63 WHAT ARE YOUR MAXIMUM DRAFTS? Consider this situation on the SILC C/P terms require you to load a cargo of iron ore from Tubarao, Brasil to Pohang, Korea Max draft in Tubarao - 23.0 meters, max.draft in Pohang - 17.40 meters Consumption of fuel from Tubarao to Pohang = 1656 MT Increase in FW = 100MT Therefore, difference in displacement Tubarao to Pohang = 1656 - 100 = 1556 MT less Notwithstanding the available depth of water at Tubarao, the v/l’s summer draft is 17.52 meters Max.draft at Pohang = 17.40 meters Difference = 0.12 meters = 12cms TPC = 106.38 Therefore difference = 12 x 106.38 MT = 1276.56MT But difference in displacement will be 1556 MT Thus, since the vessel cannot load more than the summer draft the controlling draft in this case will be departure Tubarao - when the vessel can load to the summer draft - if you can adjust your trim then you will arrive at Pohang, in this case, less than 17.40 meters By calculation we have found that to arrive at Pohang even keel, after sailing out of Tubarao at summer draft, the drafts by the loadicator, departure Tubarao are as follows F: 17.12m M: 17.52m A: 17.92m With these drafts after our consumption of fuel & increase of fresh water, the arrival drafts at Pohang are computed to be F: 17.38 M: 17.38 A: 17.38 Relative Density at both ends taken to be 1.025 Thus while loading/trimming what would be the maximum drafts you could load to at ford, at midships and at aft? In this case the midship draft is quite simple - it is the summer loadline so come what may you can load up to a maximum of 17.52m draft at midships Your computed drafts tell you that with a ford draft of 17.12 meters with your F.O/FW you will arrive in Pohang with a ford draft of 17.38 Similarly you would arrive with an aft draft of 17.38 after leaving Tubarao with an aft draft of 17.92 Max.draft at Pohang is 17.40 meters & therefore in theory you would be able to load to a max.draft of 17.14 meters ford & 17.94 meters aft - but I would highly recommend that the cms you have in hand you keep in reserve & not load, in this case, to more than 17.12m ford & 17.92 m aft This means that even if your ford draft is 17.00 you cannot load to an aft draft more than, in theory, 17.94, & in practice 17.92 Or for that matter, you cannot load more than to a ford draft of 17.14m, in theory, & 17.12 meters in practice, even if the aft draft is 17.80 meters With your arrival Pohang midship draft, only cms less than the maximum you not have much sinkage, if any available, to fill ballast to change the trim 64 Also if you were hogged, you would not be able to fill more cargo, such that the ford & aft draft exceed 17.12 & 17.92 respectively, even if your midship draft does not reach 17.52m Should you be hogged say cms, your max ford draft would still be 17.12 meters, your max aft draft would still be 17.92 meters Therefore max mean draft would be (17.12 + 17.92) / = 17.52m With a 5cm hog your midship draft would be 17.52 - 0.05 = 17.47m Using the 6-sided formula, your MQM would be 17.48m - this means less cargo, but there is nothing you can about it as otherwise you will be over draft at the ford & aft ends arrival Pohang Please remember that if the density at Tubarao is not 1.025 then you have to make allowance for that Now consider this situation - say you are on a vessel where the summer draft is 17.25 meters You are loading at a port where max draft is 22m The max draft at disport is 17.10 meters, By calculation you have found that to arrive at the disport at an even keel draft of 17.10m you have to sail out of the load port with drafts F: 17.00 A: 17.20 M: 17.40 In the previous example you had a situation where the max midship draft was clear-cut you could only load to your summer draft In this case you find that you are going to finish with sag - say, your drafts at completion are F: 16.95 M: 17.20 A: 17.35 You know that you have not reached the max possible ford or aft drafts, but can you load any more cargo? Loading more would cause your midship draft also to increase, and you know that your present midship draft is 17.20 meters & the summer draft is 17.25 meters The answer is NO - if you so, while you will comply with your summer loadline, you will be over draft at the disport 65 [...]... completion of trimming 17.025 17.727 This is 70 cms by stern Midship draft from above = 17.45m PLEASE REMEMBER MIDSHIP DRAFT IS NOT MEAN DRAFT 31 CASE 5 Assume the read drafts to be Fp = 16.75 Fs = 16.75 Mp = 17.03 Ms = 17.03 Ap = 17.20 As = 17.20 R.D = 1.022 V/l to finish 60 cms by stern Maximum draft at load port 17.60m V/l required to sail out with 60 cms stern trim to arrive disport even keel Drafts by... 2609 K:393 In almost all cases trimming is governed by drafts not quantity of cargo to be loaded where the draft is a limiting factor - max .draft at the load port or max draft at the disport or the draft when v/l crosses into the Winter zone from Summer Zone or when it crosses from Tropical Zone into Summer zone etc However you can also have a maximum quantity of cargo to load - very often the C/P states... they may refuse to sail the vessel, & with the swell you encounter it can be a notoriously difficult place to read the draft Ideally both of you should be able to read the draft exactly the same, but this is not always the case If there is a difference, be prepared to compromise a little on the fore & aft drafts, but not on the midship drafts - remember the 6 sided formula - the midship draft is multiplied... the following method from the foremen in Brazil who are loading at up to 16,000MT per hour and require to be spot on Am going to show various possible situations on the Cape-size vessel SILC at the stop for draft check prior trimming and then trimming with different sets of holds CASE 1 Assume the read drafts at the draft check to be Fp = 16.70 Fs = 16.70 Mp = 16.82 Ms = 16.84 Therefore, F = 16.70... + 0.506 On completion of trimming 17.437 17.437 This is EVEN KEEL Midship draft from above = 17.50m PLEASE REMEMBER MIDSHIP DRAFT IS NOT MEAN DRAFT 29 CASE 4 Assume the read drafts to be Fp = 16.81 Fs = 16.81 Mp = 17.07 Ms = 17.07 Ap = 17.23 As = 17.23 R.D = 1.0215 V/l to finish 70 cms by stern Reqd midship draft by calculation =17.45 Depth available at this port 22.0 meters Trimming with #4 and #7... time of initial draft survey At that time to arrive at a constant satisfactory to you and the surveyor some adjustments are made to the ballast quantity / density / drafts to arrive at this figure - this is in order mainly not to get a negative constant or a constant very different from the constant that is normal for your vessel If you used the same system (without making adjustments to arrive at a... taken this into account when in your pre-calculation you have precalculated a departure load port midship draft of 17.30 meters The max .draft for the load port has been declared to be 17.60 meters By your pre-cal you have found that to arrive at the disport at the required draft, your midship draft at the load port should be 17.30m & you should be trimmed 60 cms by stern to arrive even keel To calculate... 17.443 17.21 Loading 1358 MT in #7 0.01 0.243 On completion of trimming 17.453 17.453 This is EVEN KEEL Midship draft from above = 17.525m PLEASE REMEMBER MIDSHIP DRAFT IS NOT MEAN DRAFT 27 CASE 3 Assume the read drafts to be Fp = 17.25 Fs = 17.25 Mp = 17.19 Ms = 17.19 Ap = 17.00 As = 17.00 R.D = 1.023 V/l to finish even keel Max .draft: - 17.50m Trimming with #3 and #8 Apparent trim is 17.25 - 17.00 = 0.25... trim x 0.042 = 0.23 x 0.042 = 0.01 = corrn to aft draft (to be added) Therefore corrected drafts are Fc = 16.70 Mc = 16.83 Ac = 16.94 Corrected trim = 16.94 - 16.70 = 0.24m 23 Vessel is in water of R.D = 1.025 and is required to load such that its midship draft is 17.25 meters and finish with a stern trim of 42 cms Max allowed draft = 17.50 meters Assume you are to trim with holds 4 & 8 (This was an actual... upright for draft survey but when it is not the following correction must be applied CORRECTION FOR HEEL IN TONNES = 6 x (TP1 - TP2) x (D1 - D2) Where TP1 = TPC for the deepest draft amidships TP2 = TPC for the shallower draft amidships D1 = Deepest draft amidships (in METRES) D2 = Shallower draft amidships (in METRES) This correction is ALWAYS ADDED to the displacement because the effect of heel is to increase ... F + A) = MQM OR x Midship draft + Ford Draft + Aft Draft = Mean Quarter Mean Draft Now as (Ford Draft + Aft Draft) = Mean Draft OR ( Ford Draft + Aft Draft) = x Mean Draft Therefore the Sided... governed by drafts not quantity of cargo to be loaded where the draft is a limiting factor - max .draft at the load port or max draft at the disport or the draft when v/l crosses into the Winter... the draft to be 16.95 with 300 MT to go, all is well - if however the draft is 16.97 then he knows that he is going to finish over draft so he can stop the loading and make adjustments By trimming