Quick, Easy 3-Man Canoe Plans

What does a square have to do with it?

The second girth value is easy enough to derive. For sides of length one unit each, the girth of my section will be 3 units. The area will be the average beam multiplied by the draft.

Bottom width is one unit. Topsides are at 30 degrees, so waterline beam will be 1 + (2 x sin30), which is 2. That means average beam is 1.5.

Depth will be 1 x cos30, or 0.866.

Multiply that by the average beam and you get a cross section area of 1.299.

To make it comparable to your basic 2:1 rectangle, you need to scale it up so that area is 2 square units. That means the scale factor is (2/1.299)^0.5 or 1.2408.

Multiply that by the girth of the intial section (3) and you get 3.722.

 

A two to one rectangle is the same whether sideways or vertical, no?

I meant what I said.

Just more stupid ideas, forget it.

What would you suggest?

 

"ST: Building time is not part of the timed race. We have four hours so whether it can be built in two hours or three and a half it doesn't matter."

Whether it can be built in 4 hours with hand tools is what matters.

 

No. It will have 25% more wetted surface for the same displacement, and one eighth the stability*.

*Very basic stability calc, which will also be affected by other factors. Factors which, in practice, will probably make things worse.

Build the longest, skinniest thing you can get around the corners. Throw the three blokes in it. Paddle like hell.

 

As above.

 

Yes I realise that you'll have floats to stabilise it, but there's no point making the hulls too deep. All that does is slow you down by increasing the wetted surface.

ETA: My assumption was that you wouldn't get a 40 footer around the corners in a reasonable time, so no. You probably wont be able to get the optimum length for low drag. It will be too hard to turn through 120 degrees every hundred yards. Rough guess is that the optimum length is likely to be up around 25 feet. You need to choose the longest boat you think you can turn around the course. After that, the rest of it will fall into place.

 

Square cross-section or 30 degree shear?

 

Well obviously I'd be backing the 30 degree section.

 

For a seven percent reduction in wetted area, right?

 

Yup. That's the idea.

 

And that outweighs the advantage of a narrower hull.

The effort to achieve 20/1 l/b ratio.

 

If you have two hulls of the same length, displacement and prismatic, and one of them has 7% less wetted surface, it will be faster.

ETA: Of course, depending on who you're competing against, it may not matter. You know the race. It's your call. I'm just giving you the lowest drag option.

 

Here, load this up in Delftship and take a look. 20 feet long. 8 inch draft. 4 inch freeboard. Has your 640 lb displacement. No stability. Minimum wetted surface. Prismatic is probably about right. Rockered all the way to the ends so you have some hope of turning it. If Michlet isnt telling lies, should be able to do about 7 knots with three gorillas paddling it.

Not optimised to hell and back. This is just a quick stab at something that should be about right. Would probably try something close to this if it was me, but it's not me so do wotcha like.

 

Other than JavaProp, I can't get any boat design software to run.

It frustrates the hell out of me but something I've learned to live with (or is it "without").

 

I think you are way over thinking this contest, a fast build and brute force it around a few marks will win it. Go do your test build and get all of you out in it and practice, practice, practice. that will do you way more good than eight percent less wetted area. Also, if each of you can loose 10 to 15 lbs by than, it will also give you less wetted area, and less load to drag around in the water.