weighting a scale model

Discussion in 'Boat Design' started by jmargush, Mar 15, 2013.

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jmargushNew Member

I am working on a 5th scale model of a project I hope to build full scale.

It is a 21' wooden boat.

My question is if I want to weight the model and float it to see how it floats is the real weight divisible by 5 or is the relationship for scale weight different that the dimensional relationship?

ex.
The current design looks like the wood should weigh around 1100, fuel 200, motor 500, misc. 100 and passengers 900 for a total of 2800.

so is the scale weight - fuel 200, +motor 500,+ misc. 100
and passengers 900=1700/5 or something differnt that divisible by 5?

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Mr EfficiencySenior Member

To sit on the same waterline, it will need to be 1/125th of the full size weight, including the tiny passengers. It is five times longer, five times wider, and five times the depth. So the model will need to be around 22.4 pounds gross, if the waterline is consistent with the full-size object.

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Mr E laid it out correctly. I will explain in a bit more detail.. In case you build models at one fifth or other scales you can determine scale weight by using the scale factor raised to the third power. In the case of the one fifth scale model simply take the third power of 5 which is 5 x 5 x 5 =125. At a scale of one eigth your figure would be 8^3 = 512 That is the useful figure for that scale.

In the fifth scale case,say the finished boat with everything aboard will weigh 2500 pounds. divide 2500 by 125 (the cube of the scale) to get 20 pounds of model weight.. If you had used eigth scale then 2500/512 = 4.88 pounds. When you think about this arithmetic, it is easy to see that there is much advantage in using a large scale model, that is one with a smaller fractional number. You have done so.

This works in the other direction too. Say that your model weighs 22.5 pounds ...simply multiply 22.5 times 125 to get a full sized boat weight of 2812 pounds. This works equally well with the metric system of measurement.

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RobjlSenior Member

Lcg ?

"My question is if I want to weight the model and float it to see how it floats"
It's not quite that simple, you also need to know where the longitudinal centre of gravity of your vessel is so you can replicate it in your model, otherwise you may build a full size boat that is up or down in the bow, ie it doesn't float on its waterline.
The longitudinal centre of gravity in your model is unlikely to be in the same place as the full size vessel.
Have you considered this?
It can be calculated

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jmargushNew Member

By longitudinal CL do you mean the CL that divides the boat in half side to side or the cl perpendicular to that one?

How do you calculate that?

Is this the same thing as center of gravity? or is that different since the boat is being floated and not just dependent on gravity?

don't you have a small amount of control by where you place the occupants and fuel tank?

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DMacPhersonSenior Member

Not to put too fine a point on it, but remember that it is the immersed volume that is a cubic of the scale ratio. Weight is a cubic relationship only if the density of the water is the same between model and full scale. Otherwise, if the full scale boat is to be in salt water and you are floating your model in fresh water, you will need to reduce your model weight by approximately 3%. Probably not significant for this work, but it can be important in other circumstances.

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TANSLSenior Member

jmargush,
All engaged in designing "things" that move, have to make an estimate of the weight of the "thing" and the position of its center of gravity. And this is done without creating a scale model and throw it into the water. Precisely the experience of the designer, his rigor in the calculations, get the forecast is correct and therefore the operation of the "thing" is successful.
On boats like yours these calculations, though laborious, are so simple to make them profitable and avoid building an expensive model.
Consider this option if this is more convenient

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RobjlSenior Member

Lcg

LCG is the longitudinal centre of gravity.
It is usually on the centreline of the vessel.
On most vessels, depending on your required speed, it will be a bit aft of the halfway point of your waterline length. Eg if your boat has a 30' waterline length the centre of gravity will be about 16' along that waterline. Ignore overhangs fore and aft in calculating this.
In your design stage you need to calculate where the LCG will be. It is only basic maths. You do need the know the weights of every item that goes into the boat and where it will be.
Can you use Xcell ?
The method of making a model and dragging it through the water is useful only if the LCG is in the right place.
Even then I would still not build a boat just on the basis of a model.
I suggest you use a yacht design program.
Some more details of what you are proposing would help.

Good luck.

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gonzoSenior Member

A computer design program or a model are only useful if you have the knowledge to interpret them. Models have been used successfully for centuries. They are really cheap to build. They take some time and skill though. To have the weight and CG proportional on a model takes a lot of care, but it is routinely done for tank testing.

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rwatsonSenior Member

To make it simple in an excel spreadsheet, the formulae =(K30/(5*5*5))

where K30 is the full size weight, will giove you the 1:5 model weight
eg engine weight at cell K30 of 130 kilos, will give you 1.04 kilos for the model

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TANSLSenior Member

that means "...will give you..."

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