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Old 10-09-2004, 09:32 PM
Marvin Martian Marvin Martian is offline
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mathmatical formula for water line of aluminum 25 footer

Will someone please tell me how to calculate where the water line of my father's 25 foot aluminum fishing boat will be, he is not done building it yet and would like to find that out?!
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Old 10-10-2004, 11:56 AM
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Tad Tad is offline
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Marvin,

Good question, the answer needs some input from you though. Why do you want to know about the waterline, is it a custom design, or a modified design, or are you just curious? Do you have any indication as to where it might be from the designer? Do you need it to decide where to boottop will be, or through hull fittings?

The only way to find where she will float and how she will float (trim) accurately, is to launch the boat. And remember, she will float a little deeper in fresh water than in salt.

If you have a hull lines drawing and know the weight and center of every piece of the boat, you can calculate her floatation quite accurately (but not as close as floating the boat will). But it may not be worth it, depending on your need for accuracy.

So, you want to take a guess at where it will float? First you need to guess what she weighs. The designer’s number may be close, but it also may be meaningless, you changed power, or fuel tank size, or material thickness. You could guess at the major weights, take the weight of aluminum your dad bought, and deduct something for offcuts. Add the engine and fuel (6.1 lb/gal for gas & 7.2 for diesel) and batteries and interior plywood and about 20% for extras. If your dad always fishes with four 200 pound buddies and 12 cases of beer, you might want to add that in too.

So you have a total, let's say 5500 pounds. If the boat was a box and all weights were exactly around the center, it would be easy to find the waterline.

In salt water 5500 pounds is 86 cubic feet. (64 pounds per cubic foot)

25' by 8' is 200 square feet, 86 divided by 200 is .43 feet, or about 5".

If she was a box 25' by 8' and weighing 5500 pounds she would float 5" deep.

But....she's no box, if the boat is vee-bottomed she is a more of a prism, a three dimensional triangle. We could guess that the prism might take up about 70% of our box, which would be a prismatic coefficient of 0.7.

If the waterline is about 23' long and max waterline beam is 8' then we take the box, 23' by 8' = 184 sq ft. multiplied by our .7 = 128.8.

86 divided by 128.8 = 0.667' or about 8"

In fresh water use 62 pounds per cubic foot.

Best of luck. Tad
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