OAR LENGTH FORMULAE! (preceded by rant of reasonable length)

Boatthink:

So it recently occurred to me that the least surface area for the most buoyancy would entail a part of a sphere under the water line. Anyone who crunches a few displacement and area numbers can see that this is the case. If we think 2 dimensionally, then a cylindrical hull section gives the same result. If we vary from this model that maxmizes displacement while minimizing frictionable surface area below the waterline, it should achieve tracking, wave handling, stability, and/or speed. There are really no other reasons to increase hull area at any scale because the displacement to hull surface area ratio is the main fraction that one would want to maximize. Why then, would anyone make a square hull? Ease of manufacture is no excuse for making a substandard craft. That said, a spherical hull is not optimal, once other desirable hull functions are considered. We want the thing to track, and this is more important in a paddled craft because the center of effort is so far off the centerline of drag.

Width to length ratio holds a close second to wetted surface area when it comes to “drag factors in paddled boats”. I say this because at speeds below 7kts, and total weights around 180-700lbs. hull lift from planing is a negligible factor, unless one is paddling on mercury, which would be metal as fuck.

Anyway, rowing is not what it aughttabe. Slippage on double oars should be rather high at low speeds while accelerating, and should tend toward an asymptote at hull speed under sustainable force. Those huge blades and associated health probs are bullshit in my humble opinion. These problems are totally avoidable in pleasurecraft design, and you wouldn’t even lose that much overall speed, compared to the same effort expenditure in a coxless rowing scull. Sliding seats and oarlocks that are cantilevered out beyond the whale are both kinda cool, but they are both wastes of money and weight in the type of utilitarian, lightweight, pleasurecraft that I aim to produce.

With a 32″ oarlock spread and 6’11” oars that overlap the centerline for a crossarm rowing style, impressive flatwater speeds over 4.7 kts are easily within reach of the average pleasurecraft paddler. The only thing that is needed is thoughtful hull design. I wonder who is gonna do that…

Assuming you already have a boat or most of a design, how would one know how long to make the oars? There are several formulas for this online. None of them take into account the height of the oarlocks above the waterline, the amount of inboard space available to move the handle around in, or the height of the rower. The oarlength formulas I’ve found consider the beam and guess the rest. These methods are not useless, but when applied to many types of rowboats, they are not better than guessing or eyeballing, and certainly not better than sitting in a cardboard box with a broom and a tape measure. Don’t laugh, I’ve totally done that. If your oars are too long, you may not be able to move them efficiently inboard or outboard. The mechanical advantage may be such that you can only do long, slow strokes (sic) and have trouble recovering and catching each time. If they are too short, the rowing ergonomics will be even worse. You may have to stand up to reach the water and dump water in your boat on each recovery. This is extra crappy in cold weather. The WORLD NEEDS A SIMPLE FORMULA THAT GIVES ROWBOAT HEIGHTS AND LENGTHS FOR ANY BOAT THOUGH, WITH A MINIMAL AMOUNT OF EYEBALLING/GUESSWORK…So I worked out some simple algebra, maybe not as simple as 3x 1/2 the beam + 6″, but manageable for anyone who can apply Pythagorean theory for right triangles. I’m kinda happy that I came up with this because I dig this kind of math and problem solving:

Image

ITZNU’S OARLENGTH FORMULAE!

An oar on a gunnel with an imaginary line and a bunch of variables for line lengths

This shape is made of two directly proportional right triangles and it rotates around axis “c” in 4 dimensions. I’m calling every bit of line “c” inboard of the oarlock line “b“, and everything outside the oarlock is “line a“. Radii “d” and “p” change constantly, within predictable parameters, to describe ellipses, circles, and straight lines. They are presented here as circles to represent an average. This is why 2 dimensional and 3 dimensional rowing models can never be accurate. The radius of the oarpull circle is line “p” the radius of the handle pull circle is line “d“. The bases of the right triangles together make line “c” which is parallel to the waterline and intersects the oarlock center. The two hypotenuses h1 and h2, together, represent the oar itself, from the handle’s center of effort (point γ) to the center of the oarlock (point α), and then out to the wetted blade’s center of area (point β).

Variables of varying certainty

So there’s variable’s like a, that can only be found by their algebraic relationship to the other variables, then there’s variables like d, that are based on measurements that you can fudge, but not decide on.

line b roughly equals half the beam between oarlocks.

line d must be < distance from rowers shoulders to the tops of their legs while rowing. For most people in most seats, this is around 12″

if this number is too big, you’ll be really cramped while rowing, or be stuck pulling the oars at a weird angle. If it is too small, and p is correspondingly tiny, you won’t be able to raise the blades high enough to get them out of waves and whitecaps will screw you up. Maybe it ain’t so bad in a racing scull that is only used on glassy water but a good boat should be versatile and be able to paddle through surf and storm.

lines d and p are directly proportional to each other and are always perpendicular to the waterline.

ALGEBRA and the Useful Bits

b= 1/2 beam at oarlocks

a+b=c

d/b=p/a

√(d^2+b^2)=h1

√(p^2+a^2)=h2

h1+h2= oarlength

The first step

would be to take this directly proportional equation

d/b=p/a

and start plugging in known things

like 1/2 beam for b

and an approximation of sitting rower’s handlestroke for d

an approximation of the boat’s total draft from a bit under the waterline to the oarlock gives p

Make sure all your units are the same. The equation now only has one variable and may look something like this:

12/17=10/a

finding a is basic algebra. 17/12≈1.41666

a≈14.1667

The second step

is to plug these values into the pythagorean formula to find h1 and h2

√(144+289)=h1

√(20^2+14.1667^2)=h2…

√(400+200.69444)≈24.5091

h1≈20.8087″, or about 1.75′

h2≈24.5091″, or about 2′

The third step

is to make all the obvious connections:

a+b=c

c≈ 31.166666″ here, which is not that useful, come to think of it

h1+h2= oarlength!

in the example case, there is 3.75 ‘ of diagonal distance from the centerline of the boat in the height of the paddler’s comfortable stroke position, over the gunnel, and straight to the waterline on glassy water. This means that a 4’10″‘ oar would work comfy and fast, with around 10″ of blade in the water without reaching… maybe 5’ would be better. h1 should remain around 2’2″, adding only 5″ max. for a full handle in crossarm rowing stye. If you want to sacrifice a wee bit of leverage to make sure your oars can’t bump each other, be more strict with your h1 measurement and go for 1’9″. h2 measurement can have as much as 2′ added too it without really hurting anything but weight balance and narrow river ability. Comfy oars should have similar weights on each side of the oarlock, long oars can be hard to manage in a river, and good lumber is expensive. So there really are disadvantages to having oars too long. The formula tells us that the oars must be between 4’10” and 5’4″ for optimal rowing.

2b=spread,

this is the beam at the oarlock, probably close to max beam overall. It’s 34″ here. That seems a bit wide for a tiny boat but remember that this is from the oarlocks, not the gunnels; also, the beam at the waterline is the important one when calculating hull drag and this can still be under 20″ with a light paddler. Rowboats should have a bit of extra width because you need the room between oarlocks for leverage.

h1/h2≈handle force/blade force (this is for calculating leverage, or mechanical advantage)

14″/46″ pulled through 12″ with 5 lbs. of force in 2.5 seconds is about 1.52 lbs. pulling through 39.43″ in 2.5 seconds on each side of the boat! That’s like 120 lbs. of force per short, easy pull. These leverage equations show why adding length to oars is often desirable, why gunnels and oarlocks need stout construction, and why good oars are faster than a kayak paddle with the same effort and same hull.

and there you have it. I hope that lots of people use this for free and think “that Itznu guy is so cool, my oars work great”.

Advertisements

Say stuff

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: