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Empirical Formula to Calculate the Center of Gravity
Leopard
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Hi, I need empirical formulas. The videos you sent are explaining what is COG and so on. Let's say I have an offshore supply vessel of length 70M x 12M x 4M . I would like to know what could be the LCG & VCG. (approximation)
There is, to my knowledge, no such empirical formula.
If you have the ship afloat, and you have its body lines plan, you can calculate the center of buoyancy, which will give you the longitudinal position of the CoG. As for VCG, the CoG of the lightship ship is usually close to the deck, below. If you adopt that position you will be in a conservative option. From there, calculate the CoG of the remaining weights.
If you don't have that data, I don't see how you could calculate LCG and VCG. Sorry I can not be of more help.
If you have the ship afloat, and you have its body lines plan, you can calculate the center of buoyancy, which will give you the longitudinal position of the CoG. As for VCG, the CoG of the lightship ship is usually close to the deck, below. If you adopt that position you will be in a conservative option. From there, calculate the CoG of the remaining weights.
If you don't have that data, I don't see how you could calculate LCG and VCG. Sorry I can not be of more help.
gonzo
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The process is tedious but simple. You need to find the weight of every element and part of the ship. Then you break it down into simple shapes and find the CG of each of them.
center of gravity calculation for multiple objects - Google Search https://www.google.com/search?q=center+of+gravity+calculation+for+multiple+objects&rlz=1C1GCEA_enUS987US987&ei=hWDmY5C5Epmv0PEPhs2PmAU&oq=center+of+gravity+formula+multipl&gs_lcp=Cgxnd3Mtd2l6LXNlcnAQARgAMgYIABAWEB4yBQgAEIYDMgUIABCGAzIFCAAQhgM6CggAEEcQ1gQQsAM6BwgAELADEEM6BQgAEIAEOgQIABBDOgkIABAWEB4Q8QRKBAhBGABKBAhGGABQjAZY6BJgliNoAXABeACAAUuIAeEDkgEBOJgBAKABAcgBCsABAQ&sclient=gws-wiz-serp#kpvalbx=_omDmY8bvFuO80PEPmJS9uAw_29
The other system is to measure the submerged area of the hull and divide it into small volumes. Then use the method above.
center of gravity calculation for multiple objects - Google Search https://www.google.com/search?q=center+of+gravity+calculation+for+multiple+objects&rlz=1C1GCEA_enUS987US987&ei=hWDmY5C5Epmv0PEPhs2PmAU&oq=center+of+gravity+formula+multipl&gs_lcp=Cgxnd3Mtd2l6LXNlcnAQARgAMgYIABAWEB4yBQgAEIYDMgUIABCGAzIFCAAQhgM6CggAEEcQ1gQQsAM6BwgAELADEEM6BQgAEIAEOgQIABBDOgkIABAWEB4Q8QRKBAhBGABKBAhGGABQjAZY6BJgliNoAXABeACAAUuIAeEDkgEBOJgBAKABAcgBCsABAQ&sclient=gws-wiz-serp#kpvalbx=_omDmY8bvFuO80PEPmJS9uAw_29
The other system is to measure the submerged area of the hull and divide it into small volumes. Then use the method above.
jehardiman
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As TANSL states it is impossible to determine the actual CoG without an experiment. A complete weight report will get as close as the accuracy of the reported weights. The Curves of Form will tell you where it should be, not exactly where it is.
If all you have is the particulars then the only way to estimate where the CoG should be is years, and years of experience with the weight reports and curves of form for that hull type. As I have said before, Naval Architecture can be considered one of the few remaining guild systems and everybody has to do their own weights.
If all you have is the particulars then the only way to estimate where the CoG should be is years, and years of experience with the weight reports and curves of form for that hull type. As I have said before, Naval Architecture can be considered one of the few remaining guild systems and everybody has to do their own weights.
mitchgrunes
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If you balance anything on a small point, that point will lie vertically underneath the center of gravity. Which means it lies on the resulting (vertical) line. You could mark the bottom contact point, and the top place where where that line crosses the object.
Change the orientation of the object, find the new balance point, and that will still be true. That creates a different line, because the contact point is different, and "vertical" is now in a different direction relative to points on the object.
The intersection of those two lines will be the COG.
You could add additional orientations, and get many lines, so you can more accurately estimate things.
I am, BTW, assuming that the object is rigid. I.E., that balancing it on the point doesn't change it's shape.
However, barring special cases like whitewater boats, which have to be made very sturdy for specialized reasons, and some other very small boats, balancing a ship on a small enough point to make useful estimations is likely to damage the ship. Maybe you could do it with a rubber raft, but it wouldn't be rigid.
There is, BTW, another complication, which makes my method approximate, especially for very tall ships. The weight of any element of the object depends (slightly) on the strength "gravity" at that point. Which to a first approximation depends on the distance from the center of the earth, and on the latitude (because "weight" includes the centrifugal force from the spin of the earth). (This is also the reason why the COG is not the same as the center of mass. I think engineers who design skyscrapers make a significant distinction between COG and COM, because the force of gravity varies. I don't know if any ships are tall enough for this effect to be significant.) And if you want to get really picky, you have to include tidal forces - i.e., at various tidal stages, your COG will be slightly different - but not enough to notice. My best guess is an approximation at any given ship orientation, altitude, and position on the earth will be good enough for all practical purposes at other ship orientations, altitudes and positions.
A more significant difference, for sailing craft, is that the shape and weight distribution of the ship depends on where the boom is currently located.
There is another way to estimate the COG, but it takes a lot of work. Create a scale model of the ship, using the same materials, and with the thicknesses of everything scaled to match. Ignoring the probably insignificant changes in the force of gravity I mentioned above, It's COG point will be the same as the corresponding COG point in the full scale ship. Since it is smaller, it might be strong enough to use my method.
But if you are good at computers it might be easier to create a 3D model of the ship in a computer. You create a fine mesh 3 dimensional grid of boxes that encompasses the ship. For each grid box, multiply the weight of the parts of the ship in that grid box by the x-position of that grid box. Sum those values together, and divide by the total weight. That will be the x-position of the COG. You can do the same for the y and z axis.
Technically what you are doing is an integral. I.E., the COG is the (3 dimensional) integral of (the density of the ship times the 3 dimensional vector positions), divided by the total weight. But if you understand that, you probably didn't need to ask this question.
Change the orientation of the object, find the new balance point, and that will still be true. That creates a different line, because the contact point is different, and "vertical" is now in a different direction relative to points on the object.
The intersection of those two lines will be the COG.
You could add additional orientations, and get many lines, so you can more accurately estimate things.
I am, BTW, assuming that the object is rigid. I.E., that balancing it on the point doesn't change it's shape.
However, barring special cases like whitewater boats, which have to be made very sturdy for specialized reasons, and some other very small boats, balancing a ship on a small enough point to make useful estimations is likely to damage the ship. Maybe you could do it with a rubber raft, but it wouldn't be rigid.
There is, BTW, another complication, which makes my method approximate, especially for very tall ships. The weight of any element of the object depends (slightly) on the strength "gravity" at that point. Which to a first approximation depends on the distance from the center of the earth, and on the latitude (because "weight" includes the centrifugal force from the spin of the earth). (This is also the reason why the COG is not the same as the center of mass. I think engineers who design skyscrapers make a significant distinction between COG and COM, because the force of gravity varies. I don't know if any ships are tall enough for this effect to be significant.) And if you want to get really picky, you have to include tidal forces - i.e., at various tidal stages, your COG will be slightly different - but not enough to notice. My best guess is an approximation at any given ship orientation, altitude, and position on the earth will be good enough for all practical purposes at other ship orientations, altitudes and positions.
A more significant difference, for sailing craft, is that the shape and weight distribution of the ship depends on where the boom is currently located.
There is another way to estimate the COG, but it takes a lot of work. Create a scale model of the ship, using the same materials, and with the thicknesses of everything scaled to match. Ignoring the probably insignificant changes in the force of gravity I mentioned above, It's COG point will be the same as the corresponding COG point in the full scale ship. Since it is smaller, it might be strong enough to use my method.
But if you are good at computers it might be easier to create a 3D model of the ship in a computer. You create a fine mesh 3 dimensional grid of boxes that encompasses the ship. For each grid box, multiply the weight of the parts of the ship in that grid box by the x-position of that grid box. Sum those values together, and divide by the total weight. That will be the x-position of the COG. You can do the same for the y and z axis.
Technically what you are doing is an integral. I.E., the COG is the (3 dimensional) integral of (the density of the ship times the 3 dimensional vector positions), divided by the total weight. But if you understand that, you probably didn't need to ask this question.
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mitchgrunes
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I should mention that I know nothing about ship design. That was based on simple physics definitions.
And therefore I made an error that I assume no real ship designer would make: when I said the balance point would be vertically under the COG, I assumed that there was no wind acting on the object (i.e., the ship). So you could only use my methods on a windless day, or indoors. For a sail boat, I also ignored the way the wind distorts the shape of the sail, because that distorts the weight distribution of the sail. Likewise for the way in which it bends the mast. And I ignored the fact that the fuel, crew, passengers and cargo would shift around when you re-orient the ship. Etc.
Ugh. The real world is so much more difficult than theory.
And therefore I made an error that I assume no real ship designer would make: when I said the balance point would be vertically under the COG, I assumed that there was no wind acting on the object (i.e., the ship). So you could only use my methods on a windless day, or indoors. For a sail boat, I also ignored the way the wind distorts the shape of the sail, because that distorts the weight distribution of the sail. Likewise for the way in which it bends the mast. And I ignored the fact that the fuel, crew, passengers and cargo would shift around when you re-orient the ship. Etc.
Ugh. The real world is so much more difficult than theory.
gonzo
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I think that calculating the submerged volume, if the ship is out of the water, or from the construction plans is the easier method. The OP is asking for a rough estimate, which can be accomplished by calculating the volumes by counting squares.
jehardiman
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Sigh...
In days of yore, when we still had to turn the crank and slip the slide; where spreadsheets needed to be drafted; and a weight deck was exactly that; there were many arcane ways of graphically and mechanically finding the LCB, TCB, and VCB. However, the CG's where much more elusive creatures, due to their being non-homogeneous unlike water and spread, like vermin, in the very fabric of the vessel. There was never any graphical method available by the quill to determine the LCG, TCG, or VCG without summoning, through various mystic arithmetical arts, the individual motes of data distributed throughout the various construction and outfitting parchments and blue toned diazos. Some days we scribed the data on to pre-printed paper, the columns spread across the sheets like graves waiting to be filled with dead numbers. Other days we sat at an infernal machine that gruesomely punched holes in a thin tablet. These tablets were then collected, stacked in a precise near-sorcerous order and carefully carried in supplication to the wizard in his air-conditioned lair to be fed to the whirling maw of the reader, only to appear hours or days later as an unending scroll of numerical columns with three values, entombed in asterix's at the end. Though we have now chained the genii to the silicon, the penitent must still kowtow to the keyboard and still tease, in monastic fashion, quaint gems such as density from the enigmatic vendor sales brochure. To find the enlightenment of the CG narvinna, the disciple must toil and search still using his mathematical skills.
Really, you still have to do your own weights.
In days of yore, when we still had to turn the crank and slip the slide; where spreadsheets needed to be drafted; and a weight deck was exactly that; there were many arcane ways of graphically and mechanically finding the LCB, TCB, and VCB. However, the CG's where much more elusive creatures, due to their being non-homogeneous unlike water and spread, like vermin, in the very fabric of the vessel. There was never any graphical method available by the quill to determine the LCG, TCG, or VCG without summoning, through various mystic arithmetical arts, the individual motes of data distributed throughout the various construction and outfitting parchments and blue toned diazos. Some days we scribed the data on to pre-printed paper, the columns spread across the sheets like graves waiting to be filled with dead numbers. Other days we sat at an infernal machine that gruesomely punched holes in a thin tablet. These tablets were then collected, stacked in a precise near-sorcerous order and carefully carried in supplication to the wizard in his air-conditioned lair to be fed to the whirling maw of the reader, only to appear hours or days later as an unending scroll of numerical columns with three values, entombed in asterix's at the end. Though we have now chained the genii to the silicon, the penitent must still kowtow to the keyboard and still tease, in monastic fashion, quaint gems such as density from the enigmatic vendor sales brochure. To find the enlightenment of the CG narvinna, the disciple must toil and search still using his mathematical skills.
Really, you still have to do your own weights.
Don't forget to weigh the paint.
rwatson
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I should mention that I know nothing about ship design. That was based on simple physics definitions.
And therefore I made an error that I assume no real ship designer would make:.....c.
This recent video brought up a whole NEW set of considerations for me too.
Might be of interest.
philSweet
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... With apologies to Eugene Field.
Wynken, Blynken, and Nod one night
Sailed off in a wooden shoe,—
Sailed on a river of tallied weights
Got into a bit of a stew.
"Where are you going, and what do you wish?"
The old moon asked the three.
"We have come to determine for once and for certain
a Center of Gravity."
With Numbers in cumbersome columns
Weights ranged in wearisome rows
With Volumes and Densities in such immensities
embattled, confront we our Foe.
Said Wynken,
Blynken,
And Nod.
Wynken, Blynken, and Nod one night
Sailed off in a wooden shoe,—
Sailed on a river of tallied weights
Got into a bit of a stew.
"Where are you going, and what do you wish?"
The old moon asked the three.
"We have come to determine for once and for certain
a Center of Gravity."
With Numbers in cumbersome columns
Weights ranged in wearisome rows
With Volumes and Densities in such immensities
embattled, confront we our Foe.
Said Wynken,
Blynken,
And Nod.
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