Wooden Boat Plans Australia | Bruces Pontoons

Wooden Boat Plans Australia

We were planning to use Bruces pontoon boat that he scored somewhere around the Lake Shasta area as the base for our shantyboat.  In fact, it was Bruces pontoon boat score that gave us the greenlight for this project.  Not so sure about this now.  Every boat has a buoyancy, a certain weight that it can support safely.  Bruce writes:

I dont know how easy it is to calculate this. During high school geometry I had a crush on the Castellucio twin sisters (Karyn and Kathleen) and most of my mental energies were taken up by trying to flirt with the two of them rather than on the substance of the coursework. I havent really revisited the subject of geometry (nor, alas, the Castellucios) since that time.

The pontoons are sort of irregular shaped, they are flat on the top and sort of u-shaped on the bottom. The top to bottom dimension is 17 1/2". They are 18 1/2" wide.  As Ive mentioned before, the platform is 16 long. The pontoons extend another 18" to the rear of the platform. They also extend about 24" to the front of the platform. However, in the front they are tapering down to be much more narrow, expecially in that last foot.

Okay, Im gonna make some assumptions about Bruces pontoon boat.  Doing some crazy math that requires me to get out my geometry book again, I guesstimate the total cross-sectional area of the pontoon is about 2 sq ft.  To get the volume, we multiply that times the length. Well say that is 16 feet + another 1.5 ft at the back, and another maybe 1 foot at the front. For 18.5 feet. So the volume of one pontoon is perhaps 37 cu ft.

We just look at one pontoon, because we want to make sure that the capacity we work with never exceeds the floatation of one pontoon, lest a shift in weight drive it under the water and flip the craft. This situation even has an ominous-sounding name:  Pontoon Effect.  Plus, if you think about two pontoons, you want them both no more than half in the water.

To get the floatation, we multiply the volume times the weight of the same amount of water. Thanks, Archimedes.  One cu ft of fresh water weighs 62.4 lbs. So the flotation of one pontoon is 37 cu ft  * 62.4 lbs/cu ft = 2308 lbs. So the weight of the boat, plus the passengers, plus their gear and stuff, plus some margin of safety 10 to 25% should not exceed 2308 lbs.

Probably too tiny for our heavy shantyboat. However, we can build Bruces pontoon out to be a little hillbilly sun porch to accompany our shantyboat.  Something like this...

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Wood Boat Plans And Kits | Math is Hard Calculating a Waterline

Wood Boat Plans And Kits

Who would have thought boatbuilding would involve so much math?

I was reading Glen L. Witts Boatbuilding With Plywood and realized what should have been obvious to me:  The waterline of a boat is calculated beforehand.  I guess it makes sense that boatbuilders dont guesstimate their designs only to drop their boats into the water and see what happens. On top of that, many of the handling characteristics of a boat are built into the design, including balance.

With our cabin shifted back from the center, this would shift the balance toward the back.  Add to that the weight of the motor and fuel and we have a potential problem.  So far, Ive been assuming symmetrical bow and stern as many barge boats feature, but in order to increase buoyancy in back and shift the center of buoyancy forward, I can reduce the rake in back to get more of the hull in the water there.  This explains some of the barge boat that did feature a smaller stern rake.  How would I go about calculating that?  That is something I will have to think about.

But as an interesting exercise, I can calculate the waterline height as a function of the rake angles, length and width of the boat, and overall loaded weight of the boat.

I had to go back to my algebra and trigonometry reference books to look up how tangent and the quadratic equation worked.  The last equation gives us the waterline height h_w as a function of

w = overall width/beam
l = overall length
h = height from bottom to deck (or to the top of the rake)
?b = angle of bow rake
?s = angle of stern rake
Vw = volume at waterline (= the weight of the displacement of loaded boat)

Simply put, the total volume of water displaced is equal to the sum of the water displaced by the bow, stern, and center.  The volume of each of these can be calculated geometrically as a function of our unknown, the height of the waterline.

We then solve for the unknown and get an equation in a quadratic form (the forth one from the bottom).  So we use the quadradic equation (which Ive always hated) to solve for h_w.

Taking our equation for a spin

Lets say the total weight of the boat plus gear plus people plus 25% safety margin is 7000 lbs.  Then the calculated volume of the boat at the waterline is 193,846 cu in.

Well say the boat is 8 foot (96 inches) wide, the length is 20 foot (240 inches), and the height from the bottom to the deck is 2 feet (24 inches).  The bow rake angle is 45° and the stern rake is a modest 10°. 

So plugging in the numbers, and taking the plus-or-minus of the quadratic formula into account, I get:

hw = 370 inches or -9.28 inches

So either my boat will have a waterline 31 feet above the keel (that is to say, the boat will be underwater), or it will float 9 inches out of the water. No wonder I always dreaded the math part of a real-world problem.

Checking my math... ah I forgot a negative sign!  New solutions:

hw = 9.28 inches or -370 inches

Thats much better.  If we throw out the negative solution, we have a waterline 9 and a quarter inches above the keel.  Cool.

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