I was reading Glen L. Witt's

*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 don't 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, I've 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.

_{w}as a function of

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

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 I've always hated) to solve for h

_{w}.

#### Taking our equation for a spin

Let's 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.We'll 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:

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.h_{w}= 370 inches or -9.28 inches

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

That's much better. If we throw out the negative solution, we have a waterline 9 and a quarter inches above the keel. Cool.h_{w}= 9.28 inches or -370 inches