Simple question relating to canoe displacement

[cumbrian_matt]

Firstly i'd like to introduce myself, im a 23yr old guy living in Cumbria, England. I am an electronics engineer, keen model maker and sailor. I am about to make my first attempt a building a boat though.

Due to a tight budget I am looking to build a simple open canoe style craft for my first attempt. If this works well then it will give me confidence to invest in a more challenging project.

I have found a design, called a Dory, but I am not sure about the weight of person this boat is designed to hold. I have seen other designs very similar which are designed really for children.

Is there a simple formula that can be used to calculate the displacement of a simple hull?? I have tried to find one on google without much success, but then i'm not 100% sure what to look for.

Attached is a copy of the boat plan......if it's relevant the hull material would be 1/4" or 3/8" Plywood

Any help much appreciated,

Matt

[redcoopers]

Yeah, there is a very simple formula: just integrate.

Archimedes principle tells us that the displacement of the hull is proportional to the volume of water it displaces. So, all we do is find the volume of water it occupies, and multiply by density:

displacement = volume * density.

If you're sailing in fresh water, then density = 62.4 lbs/ft^3. If salt water, then 64 lbs/ft^3.

Just taking a quick look at your boat, if I needed fast calculations, I would just move all of my proportions into simple rectangles and triangles to find the volumes. Let's just assume that the waterline of the boat is 4" from the keel.

Now, I would make the forward half of the boat into a triangle:
Area_forward = 22" * 5' * 1/2 = 3300 in^2
If I assume that the hull is "wall-sided", meaning, the sides are pretty much vertical, then we can just take Area_forward * depth_of_hull = 3300 * 4 = 13200 in^3.

Let's make the aft section into a trapezoid so that:
Area_aft = (22" + 6")/2 * 5' = 4200 in^2
Again, assuming that the boat is wall-sided, the aft volume is 4200in^2 * 4in = 16800 in^3.

Adding the forward and aft sections, we find that the approximate volume is 30000 in^3 = 17.361 ft^3.

Therefore, again assuming the boat will float in 4" of water, your very rough approximation of displacement is 1100 lbs.

-Jon

[cumbrian_matt]

thanks for the help, although I think your estimations sound a little high because if the base was 8ft long a 2ft wide rectangular with a 1ft sidewall then the volume would only be 16 ft^3 wouldnt it, and it's a lot smaller than that.

The maths makes sense though, i'll try to make a good estimation of the base area and go through it from there. Sounds like it will be suitable for a single adult to use though.

Drawing a rectangle around the base which is 8ft x 22" (1.8ft). and then estimating the area to be around 3/4 of the rectangle it comes out around 11ft^2. Then if i'm looking to have the waterline at 4", 11ft x 0.33ft gives a volume of 3.63ft^3. Multiply that by the water density gives a displacement of 226lbs. 226lbs sounds more realistic to me, but have I made any errors in my maths to reach that value?

Cheers

Matt

[Kevin Lester]

If I calculate roughly, using metric. length 2.4 x breadth 0.4 (average guess) x depth 0.1 (roughly 4 inches) it gives me 0.096 cubic metres. Since one cubic metre weighs one ton it would displace roughly 96 kg (very approximate) to convert to pounds kg x 2.2.
I get around 210 pounds.

The 1100 pounds is a little high for four inch displacement.

[duluthboats]

If the bottom is flat I get about 60 lbs displacement per inch. ( I didn’t account for flair) Most dories have some rocker so I will guess Kevin has it pretty close for 4".

Now you must add the weight of the boat, the gear, and passengers to detremin where the water line will be.

Gary

[cumbrian_matt]

cheers guys, I fully understand the maths now, so thats a great help in choosing what design I decide to build. I haven't considered the rocker in my calculations but then neither have I considered sidewall flair so it should hopefully be close to balancing the values.

Cheers Kevin for the reassurance as I got 226lbs, and as Gary mentioned I am aware that the weight of the boat has to be taken away from this too.

If I cant get a 10ft piece of ply and have to make a joint along the base, then I will scale the design up a little or even go for a 2 man canoe.

Thanks for all your help,

Matt