: I am considering building one of Putz's Walrus's and have gone through the
: archives, but may have missed the answers to my questions.

: First question: What are the beam widths of each Walrus (W1 and W2)?

the beam widths are going to be dependent on how high of a sidewall you have. as i'll explain further along, with your weight you will probably sink lower in the water (which will improve your stability a bit) than the design calls for, so to have enough of the boat above water ( this is called "freeboard") so that you don't swamp, you'll want to make the side height a bit greater than Putz calls for. I don't have the book in front of me, but I think he has sides that are about 7 inches wide. You might want to change this to 8 or 9 inches, depending on how long you decide to make the boat. Once you've picked the sidewall height you can draw out the middle form on graph paper and measure the beam.

: Since I am a heavier paddler (285 lbs.), I believe
: I need a beam of about 30" for either sized boat. If neither of the
: Walrus's have close to a 30" beam (I would consider 28" to be
: close, 26" to not be as close as I would like) has anyone increased
: the beam?

Changing the beam on the Walrus might be a bit dicey. Putz apparently had a naval architect redraw the Skeene version of the 17 foot walrus, tweeking it to make the thing stable. After that, the 18.5 foot version seems to be just a simple enlargement of all the measurements by a fixed percentage. It got longer, as well as wider and higher. If you are going to increase the beam you might want to work with the numbers for the 18.5 foot boat and jusut decrease the spacing between the forms to reach the length you chose.

If you want some other beam than that, then I'd suggest you work with the numbers for the 17 footer and enlarge or reduce those. I think you'ld get a better shot at a successful design working that way.

As for the "numbers" for these two boats. Putz expresses them in his book in what I consideer to be a rather awkward fashion. I've recalculated the directions he gives and posted the results at:
http://www.geocities.com/eureka/business/7882/walrus18.htm for the 18.5 foot boat and at http://www.geocities.com/eureka/business/7882/walrus17.htm for the 17 foot boat.

What I have are simple coordinates for each form which you can easily plot directly onto the plywood you are cutting your frames from. the angles and shapes of the forms are identical with Putz's , but I eliminated about 20 inches of unnecessary height from the base of each form, so that you can save a sheet of plywood, and thus reduce the cost.

What I did not include in the list of coordinates was the location of the chine that defines the sidewall height. Measure along the side of the frame and mark it where you wish.

: I may be opening up a proverbial can-o'-worms here, but is there a rule of
: thumb for "Greenland" style kayaks, like Putz's Walrus, Clark
: Craft's BK10, etc. for length vs. beam width for load capacity. I ask
: because I compared CC's BK10 to Folbot's Yukon kayak and they are rated
: for the same load (300#). Either CC is overly optimistic, Folbot is overly
: cautious or something in the middle because the BK10 is 11' long with a
: 28" beam vs. the Yukon being 13' with a 30" beam. Or perhaps
: they really are that different despite them being very similar looking to
: the casual observer.

Okay, there is no can of worms here, just more calculus and engineering than most people want to play with. ( Would that be a pit of snakes rather than a can of worms?) If you want a perfect answer, you can pop the dimensions of any of these boats into a boat design program and let the computer take care of figuring out the stats.

You've suggested a relationship between displacement and the length and width of the boats. For a perfect answer that is not enough. you would need to add to this the amount of curature of the bottom of the boat, and the amount, and location of curvature of the sides of the boat. Then you would have to use this data to calculate the volume of the boat which was underwater -- and you'ld need to supply the distance underwater that the boat could safely go. (If a boat that is 12 inches high suddenly goes 12.25 inches nder water, it fills with water and is swamped, unless the cockpit coaming is staying above the water, so there is a certain limit here.)

I'm lousy at doing calculus in my head to solve these questions, so I use a quick and dirty approximation which relies on high school geometry. Here is how I guestimate displacement:

First, I assume that the boat in question is flat bottomed and shaped like a long diamond. The length of the boat is the length of the diamond, and the beam is the width of the diamond. Second I assume that the designed displacement is based on a 4 inch waterline, that is, with the designed weight loaded in, 4 inches of the boat will be submerged. Note, this is not the maximum load. As you add more weight to the boat it will sink deeper into the water. If the boat has high sides it can carry more weight, as it can sink deeper without water coming in over the sides. Maximum safe weight is when you have very little freeboard. Maximum displacement is the quantity of water it takes to fill the boat. You don't want to load most boats that much, but some designs ARE made with very small displacements so that the paddler can sink under the water and use the aerodynamic shape (that should probably be hydrodynamic shape) of the boat while submerged. Finally, i assume that water weighs aobut 60 pounds per cubic foot. If you do the calculations in the metric system you can calculte the displaement in liters and assume that each liter weighs 1 kilogram. In reality there are differences between fresh and saltwater.

Now for the math: Think of the diamond shape of the boat as being made of two triangles, connected base to base. The area of one triangle is one half of the base of that triangle, times the height. With an 18 foot boat, each triangle would be 9 feet tall, and the base of each triangle would be the beam of the boat. For convenience let's say that beam is 24 inches (an even 2 feet). Half of that base (or beam) would be 1 foot. Multiplying that quantity times the length of 9 feet gives me an area of 9 square feet for one of the triangles. Since I have two triangles making up the diamond shape, I have a area of the "footprint" of this severely shaped boat, which works out to 18 square feet. If I am working with a designed waterline of 4 inches then I am looking at a depth of 1/3 of a foot. to get the volume of this craft I multipy 18 times 1/3 and end up with 6 cubic feet as the underwater volume. With each cubic foot of water weighing 60 pounds, such a boat would displace 360 pounds of water. subtract the actual weight of the boat (lets say 50 pounds) and you can carry a 310 pound load.

Here is another way you cna use that information. if the boat displaces 360 pounds at a depth of 4 inches, then it is displacing 90 pounds for every inch which is submerged. So, should you add another 180 pounds to this boat, ( giving a load of 540 pounds) it would submerge another 2 inches.

OK, so those are just rough approximations. There are no boats shaped like the model I suggested. Real boats have bottoms which are V shaped or round. Think of slowly lowering such a boat into the water. When the bottomw was submerged jsut 1 inch the wiater would not cover half full width of the bottom. lowering the boat further, say to 2 inches, would get the water almost out to the beam. Between the hree inch depth and the 4 inch depth the water under the boat is almost the same shape, with it being the designed shape at 4 inches.

If we assume that a REAL boat has properties like this, then we can simply average the displacement, from 0 inches to 4 inches and come closer to the actual displacement. In this case, we have 0 displacement at 0 inches of depth, and 360 pounds of displacement at 4 inches. The average would be 180 pounds. Of course the shape of a kayak or canoe is not like a triangle, but is curved out, adding to the displacement, so you could guess that such a kayak would have a displacement of anywhere from 180 to 360 pounds at a depth of 4 inches. Depending on how you view the lines of the boat ( slim or full, rounded bottome or flatter bottom) you can pick a number closer to one extreme or the other.

What does not change very much, though, is the fact that as these 18 foot boats sink deeper into the water they will displace close to 90 pounds for every inch. One the rounded bottom of the hull is submerged you only have to deal with the sides -- and for many boats they are rather straight. Here is another rule of thumb: with a 24 inch beam, an 18 foot boat will displace 90 pounds for each additional inch it is submerged. That is 5 pounds for every foot of length. So, a 16 foot boat which is pushed down to a depth of 5 inches, or an inch beyond its designed waterline, should displace about 80 pounds more, and a 14 foot boat should displace 70 pounds more than the designed displacement.

If the beam is wider then the displacement also increases.

We worked with a 2 foot beam for the first example. Lets look at something closer to what you are thinking of. Most canoes are 34 to 36 inches wide, and i'll consider that to be a maximum. You would be well under that width with a 30 or 28 inch wide kayak. 30 inces is 2 1/2 feet, and 28 is 2 1/3rd feet.

A diamond shaped boat (almost a kayak) which is 2.5 feet wide and 14 feet long would have the area of 17.5 square feet -- which is fairly close to the 18 square feet of the 18 foot by 2 foot shape we started with. the displacement at 1/3rd of a foot (4 inches) would be a bit over 5.8 cubic feet, or about 348 pounds. Again, depending on the shape of the bottom of the hull this could range from about 170 to 340 pounds in real life. And each additional 85 pounds of load (about a quarter of that 348 pound displacement of the diamond) would submerge the boat another inch.

With the narrower width of 2.33 feet (28 inches) and 14 feet long your diamond shape would be 16.33 square feet, and the displacement at 4 inches would be about 5.4 cubic feet, or 324 pounds. On real boats you'ld see a variation of design displacements for such boats from 160 to 300 pounds, and they woudl sink another inch for ech 80 pounds of additional load.

Oh, a 14 foot boat which was 24 inches wide would have a footprint of 14 square feet, a displacement at 4 inches of 4.66 cubic feet, or about 280 pounds, giving a "real boat" range of somewhere between 140 to 280 pounds, and each additional 70 pounds would push the boat down an extra inch.

Now, lets take all of this information and these approximations and plan your boat. If you make the boat to the smallest dimensions iv' calculated above-- that would be a 14 foot boat with a 24 inch beam, then at the "light" extreme it would have a design displacement of 140 pounds. If you sat in such a boat, (your weight at 285, the boat at 40 pounds, and 25 pounds of gear, then you would probably ride three inches lower in the water than the boat was designed for. The bottom of your boat would be 7 inches under water. Your "fix" for this condition would be to make the sidewalls of YOUR boat 3 inches higher than the original design. That would give you back the designed freeboard. since you are riding further underwater, your ceneter of gravity is much lower, so your stability should increase.

With the 28 inch beam on a 14 foot boat the extreme "light end" of the displacement is 170 pounds, so with the same 350 pound load (285 of you plus 40 for the boat plus 25 for gear) you would sink down 2 additional inches, giving a total depth of 6 inches. to compensate somewhat for your weight you would want to build this boat with a sidewall height which is 2 inches greater than the original design. That is an inch less than the previous example.

Now that you see how I approximate things, I'll let you calculate any other boat design yourself.

:If it matters, I am planning on using it mostly in flat water, inland rivers
: and streams, although possibly where one river near me is close to going
: to a large bay and is approx. 1 mile across.

You may find that some designs which have sufficient freeboard for rough water will not need much modification for use by a heavier-than-designed-for paddler when paddled in calm waters. Remember that in those cases where the deck is well sealed to the hull you won't get water into the boat until the water comes up to the height of the cockpit coaming -- which can be several inches above the sidewall.

I do not suggest that you build an undersized boat, though. Technically you would be overloaded, and that is very unsafe. But, on the other hand, the original paddlers of these boats obviously ignored such matters. They were able to bring back heavy game on their fishing and hunting trips. Even so, we don't have any data though on how many went out and did not return because of an overload. They may have done this as a matter of survival. If you just want a nice quiet paddling experience, build the boat big enough to support you safely.

Good luck on your project. Hope this helps.

PGJ