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 Boat Design Forums How to calculate how much weight a hull can carry?

#1
11-19-2008, 03:48 AM
 thuffam Junior Member Join Date: Nov 2007 Rep: 10 Posts: 7 Location: NZ
How to calculate how much weight a hull can carry?

Hi there

How do you calculate how much weight a boat can comfortably carry? And is this directly related to displacement?

I would like to build a small outrigger (ideally with a length to beam ratio of 8:1 or more - something easy to paddle) for my kids (to carry one child of up to 40kg). A beam of about 30cm would be suitable for them and I'm guessing the height of the hull would be between 30 and 40cm. So I'd like to know how to calculate what length would cater for this.

Thanks very much
Tim
#2
11-19-2008, 04:57 AM
 Landlubber Senior Member Join Date: Jun 2007 Rep: 1802 Posts: 2,642 Location: Brisbane
Yes it is the displacement of the vessel and the weight of the crew, total displacement. The volume displaced in litres is the weight of the total in kgs (basically).

Work out how heavy everything is, and work out the displacement of the bare boat and you can then see what is known as TPI (tons per inch) for the waterlines required to float at.

If your boat floated now at say 3 inches, and the topsides were 12 inches, you could displace a few inches quite comfortably and still have plenty of freeboard.

I can explain in greater technical detail if required, but i think your question is fairly basic, so I have kept my answer the same. Good luck.
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#3
11-19-2008, 03:06 PM
 Fanie Fanie Join Date: Oct 2007 Rep: 2484 Posts: 4,572 Location: Colonial "South Africa"
Pity you're in NZ though Tim, the US guys's boats can float much more pounds than the Kg's we have to float.
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#4
11-19-2008, 09:04 PM
 PAR Yacht Designer/Builder Join Date: Nov 2003 Rep: 3967 Posts: 17,812 Location: Eustis, FL
Actually, what you need to figure out isn't displacement, but pounds per inch of immersion. All boats will have a specific rate of "sink" as weight is applied. How much "sink" is hull form (shape) and volume specific. After a fair amount of weight is added to any particular hull, it then starts to dramatically lose stability and you're in trouble.

It would be helpful if we knew what design your were working with, as it's likely the information you need, is available from the designer or manufacture. If you're attempting to design something, with as basic a question, I'd suggest you have some studying to perform, before you should risk your children on it. I'm not suggesting you would, but calculating volume is a fundamental task in design work and if this is beyond your grasp currently, then you have a way to go yet. It's not especially hard, but does require some understanding.
#5
11-20-2008, 01:10 PM
 thuffam Junior Member Join Date: Nov 2007 Rep: 10 Posts: 7 Location: NZ

I've been studying yacht design books and online for the past year and have a 'fair' grasp of things - but I'm still a bit cloudy with displacement - as I've read several articles that refer to this as the load carrying capacity which I beleive not strictly true. My understanding is that displacement is how much the hull displaces on its own.

I beleive it's possible to work out all the stats of the desired hull from the spec I've noted eg: expected load to carry, width, height (draft + freeboard) and min length to weight ratio (prismatic coefficient?).

I've looked at many other plans for outriggers, kayaks and canoes so I have an idea about which shape.

I've been using FreeShip which makes it fairly quick to define a shape and to calculate hydrostatics. So really I guess I'd like to know - what's the ratio between displacement and how much the hull can carry?

Thanks again
Tim
#6
11-20-2008, 07:00 PM
 PAR Yacht Designer/Builder Join Date: Nov 2003 Rep: 3967 Posts: 17,812 Location: Eustis, FL
There isn't any "typical" ratio, the actual figure would be application specific.

You could establish a data base for other craft of similar shape, size and configuration and arrange for a very general guide, which could serve as a starting point.

I'm not really sure what you're looking for.

You know basically what shape you want and how much you'd like it to carry. You can calculate the sinkage rate and then you have to ask yourself is this desirable or do you want more or less volume.

Prismatic Coefficient (Cp) is a number that suggests the fullness of the ends of a boat. Determining the proper Cp revolves around the desired or expected speed and it's S/L (speed/length ratio). It's usually a good idea to error on the high side for a Cp, then have too low of one, from a performance point of view.
#7
11-23-2008, 09:05 PM
 Ike Senior Member Join Date: Apr 2006 Rep: 1669 Posts: 1,974 Location: Washington
If I were doing this I would use ABYC's standard for capacity of canoes. You start by determining maximum weight capacity This is essentially a canoe body

a. Canoes - Maximum stated capacity shall be less than or equal to that load which yields seven (7) inches (17.8 Cm)
(17.8 cm) of freeboard as shown on Figure 1;

I can't put the illustration here but what it amounts to is; determine the displacement with seven inches of freeboard left. This can be done by calculating the immersed volume and multiplying by 62.4 lbs/cu ft or simply adding weight until you have 7 inches left. The amount of weight added is the maximum weight capacity. The persons capacity can't exceed the maximum weight capacity. If you have an engine and battery subtract their weight from maximum weight capacity. Then divide the maximum weight capactiy by 165 lbs (74.8 kg)
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#8
11-24-2008, 10:00 AM
 erik818 Senior Member Join Date: Feb 2007 Rep: 310 Posts: 236 Location: Sweden
Tim,
I consider myself a good engineer, but at best an enthusiastic amateur when it comes to boat design and building. To me "displacement" only means amount of water displaced, weight in other words. Displacement obviously varies with the load, so you have to specify if you mean empty hull displacement, maximum load displacement etc. Even if there is some sort of consensus on the meaning of "displacement" without qualifiers it was obviously lost on me.

A maximum beam of 0.3m and a length of 8*0.3 m = 2.4 m will give you a surface area less than 0.72 sqm (rectangle). A more reasonable canoe shape would probably give you 0.5 sqm or so. Assume a total weight for boat + kid of 100 kg and you find that the draft will be 20 cm.

This is a very simplistic way to calculate and assumes a flat bottom, but you get a rough estimate of the draft. I would choose to make the canoe larger e.g. 0.4 * 3 m to reduce draft, but once again I'm only an amateur. Why not look at a similar design for adults and scale it for less load to get an estimate on suitable freeboard for your design? The outriger should take care of stability problems.

Erik
#9
11-27-2008, 12:22 AM
 oldsailor7 Senior Member Join Date: May 2008 Rep: 438 Posts: 2,085 Location: Sydney Australia
The most simple way to determine the max safe load your boat can carry is to measure the volume of the immersed portion of the hull, in cubic feet, at the designed max waterline. Multiply that volume by 63 for fresh water or 64 for seawater and deduct the deadweight of the boat. The result will give you the max safe load you can carry, in Lbs.
#10
11-27-2008, 03:28 AM
 Guest625101138 Previous Member
Quote:
 Originally Posted by thuffam Hi there How do you calculate how much weight a boat can comfortably carry? And is this directly related to displacement? I would like to build a small outrigger (ideally with a length to beam ratio of 8:1 or more - something easy to paddle) for my kids (to carry one child of up to 40kg). A beam of about 30cm would be suitable for them and I'm guessing the height of the hull would be between 30 and 40cm. So I'd like to know how to calculate what length would cater for this. Thanks very much Tim
A nice slender form will move easily. I prefer two outriggers to one because you can set them up to have no load but still prevent capsize. A single outrigger relies on the load you place on it initially and if you want stability to avoid a flip it will carry quite a load so adds a lot of drag.

Remember your kids are growing. 40kg may be light on for a couple of years down the track.

Working with decimetres makes it easy. Lets say 3m long 0.3m wide and 0.1m draft. The displacement will be:
D = 30 x 3 x 1 x 0.6 = 54kg
(The 0.6 is referred to prismatic coefficient and will be around this number for an efficient slender hull based on rectangular cross section.)

This would be the bare minimum as the boat itself will weigh around 15kg. You need reserve buoyancy so sides need to be about 0.2m high.

I would make some allowance for growth. Maybe go to 3.6m long, 0.35m wide and design draft remains at 0.1m. You can do the sum.

Rick W.
#11
12-04-2008, 04:06 AM
 thuffam Junior Member Join Date: Nov 2007 Rep: 10 Posts: 7 Location: NZ
Awesome thanks Rick - exactly what I was after!
Cheers
Tim
#12
02-27-2016, 05:01 PM
 Larry Bartlett Join Date: Feb 2016 Rep: 10 Posts: 3 Location: Fairbanks
Capacity Computation Formula

Hey folks, i read your threads often and this is one i might be able to shed some light on.

There is a technical formula to help calculate watercraft capacity, but you'll need to know some technical figures before you can compute.

Max Load M = (0.75 X V X 1000 - W)

V = Volume (in cubic meters) of the total buoyancy of boat
W = Total weight (in Kilograms) of the boat including all accessories

Problem with this formula is there is what seems to be an arbitrary factor of 0.75, which is thought to represent body weight. In this example, the .75 represents 75 kg (165 lbs).

But, I don't find this formula accurate enough to precisely predict available freeboard of a raft or packraft. I tend to set our design capacities to 50% submerged hulls. Therefore, a boat that has 14" tubes will sink 7" to reveal it's "max capacity." A boat that has 22" tubes will sink to 11" to reach it's "max capacity." The above formula doesn't work unless you substitute the 0.75 for .55 (55 kg = 121.25 lbs).

Example: My new boat (the Kork), is

11' long
44" wide
14" tubes for 107 cm (center tubes sections)
22" tubes for 107 cm (bow and stern sections)
2" drop stitch floor
weight = 39-lbs (17.8 kg)
capacity = 1200 lbs
Total Volume (cubic meters) = 1.026 m3

Technical stuff for formula:

If the textbook formula was correct, this boat would have a 1650-lb capacity. However, the 7" free board rule requires the formula to bend slightly to provide the accurate effects of a safe draft under a maximum load:

0.55 X 1.026 X 1000 - 17.8 = 546.13 kg (1204 lbs)

Anyone can immediately appreciate the subjectivity of mathematical predictions of "capacity" since, in the very end, river character decides your ultimate max capacity, not the boat itself!

Cheers

larry bartlett
#13
02-27-2016, 05:47 PM
 TANSL Senior Member Join Date: Sep 2011 Rep: 300 Posts: 3,840 Location: Spain
Check this formula

M = 1025 * V - W

V = volume of displaced water, in m3
1025 = density of sea water in kg / m3 (1000 for fresh water)
W = weight of the rest of the ship, in kg

Always be careful with the units you use and are the same for all variables.
#14
02-27-2016, 10:11 PM
 Larry Bartlett Join Date: Feb 2016 Rep: 10 Posts: 3 Location: Fairbanks
Isn't that the same formula?
#15
02-28-2016, 04:06 AM
 TANSL Senior Member Join Date: Sep 2011 Rep: 300 Posts: 3,840 Location: Spain
I have not checked. Do it, if you have time. I've only seen that in your formula each variable is measured in units of different systems, which may be not right. I know that my formula is correct.

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