# Rudder angle and rake calculation

Discussion in 'Boat Design' started by rsford, Mar 23, 2009.

1. Joined: Mar 2009
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### rsfordJunior Member

Hello,

I'm doing a tiller to wheel conversion on a Fenwick Williams design 24' gaff yawl (34' LOA) wood double-ender in order to create more room in the small cockpit. This is a full keel boat with an outboard rudder hung off the sternpost, with 26 degrees of rake in the axis of the rudder (the top of the rudder axis is aft of the bottom to the tune of 26 degrees from vertical). I'm designing an inboard hydraulic system to be mounted to the very heavy framing members under the aft deck, with the hydraulic ram going through the hull and attaching to a 6" tiller arm to be fabricated and bolted across the full fore and aft width of the rudder at that point (hull opening 12" above the waterline). A small custom-made bellows will attach to the outboard part of the ram and then to the hull at the pentration point, thereby preventing water from splashing in through the ram's hole.

My question is simply this: how do I calculate how much rudder angle I need, measured perpendicular to the rudder axis, to give an effective rudder angle of 36 degrees measured horizontally. Undoubtedly, as the rudder axis rake angle increases, the effective rudder angle decreases, but I think this must be a non-linear relationship. Does anyone know of a simple formula for calculating this?

I could guestimate, but my boat is small and any excessive rudder angle will translate into lost inches of cockpit space due to accommodation for an excessively long hydraulic ram. Thus my desire to calculate this correctly.

Thanks for any help or suggestions.

Bob Ford

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### daiquiriEngineering and Design

From a purely geometrical point of view:
If "I" is the rudder angle measured around the inclined axis and "V" is the rudder angle measured around the vertical axis, the relationship is
V = I cos(26°) i.e. V = 0.899 I.
In your case V = 36° so I = 36°/0.899 = 40°.

But the hydrodynamic point of view is very different. The rudder "seen" by the flow is not the same in two cases. If you take the same rudder and incline it's axis by 26° with respect to the vertical, you have that a velocity perpendicular to the rudder (the one which produces lift) has diminished by a factor of (approximately!) cos(26°)=0.899. Or, conversely, the "apparent" chord parallel to the flow is augmented respect to the chord of a vertical rudder, by the factor of cos(26°). Also, consider the fact that aspect ratio of common non-racing rudders is generally low, hence tip effects and cross-flows are important and will radically change the hydrodynamic characteristics of the rudder.

So it is true that with the formula above you have given you rudder a geometrical equivalent of a vertical-axis 36°, but it is not true that by doing so you have restored the hydrodynamic forces which would act on an identical but vertical rudder.

3. ### Submarine TomPrevious Member

Bob,

I would take a sliding T-bevel and measure your horizontal, true rudder angle for each 5 degrees of tiller and make a graph. Then you could interpolate from the graph your true rudder angle for any given tiller input.

This seems too simple so I may be missing something here.

Are you on Lake Union?

Tom

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### rsfordJunior Member

Thanks for the replies. I feel a little dumb that I couldn't come up with the formula myself since I was always really good at math (but it's been a LONG time since I took geometry). Anyway, thanks for the formula. The hydrodynamics is definitely over my head, so would you recommend increasing the true rudder angle a few degrees to compensate for the dynamic factors you mentioned, Daiquiri?

Tom, I'm at Elliott Bay Marina now, but I might move to Lake Union later in the season. The boat currently has just electric propulsion with rather limited range, so Lake Union might be a better spot until I can afford the time and money for installation of a new diesel to make it a hybrid.

Thanks again for your help. I just asked Google my question and was led to this site.

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### daiquiriEngineering and Design

I think you are safe to increase rudder angle by at least 5° (without knowing further details about your rudder). But not for the reasons stated above, which are hard to quantify.
The actual value of rudder max angle depends on your boat's turning speed. That's because the effective angle of attack is diminished by a transverse speed component due to rotation, which can be approximately calculated as
Vt = Omega x Distance (from rudder to boat's C.G.)
Omega is the turning speed:
Omega = (deg/second) / 57.3.
If Beta is the rudder deflection, the geometric angle of attack (AoA) will then be:
AoA = Beta - arctan(Vt/Vboat)
All the angles are to be expressed in radians (which is degrees/57.3).
So, faster the turn, smaller the angle of attack seen by rudder - and that allows you to increase the rudder max. angle by a correspondent quantity without incurring the stall. The only precaution will be to deflect the rudder gradually, from max. 35° at the beginning of the maneuver up to max available angle once the boat has attained a steady rotaation. Fast-turning boats can have (and require) a rudder max angle up to 45°-50° or even more (up to 60°, depending on rudder form, section and aspect ratio).

P.S.
An example (I'll have to convert each number to SI units in order to obtain correct results):
Say that you have a boat of some 40 ft (12.2 m) sailing at 8 kts (4.12 m/s). Assume a distance from the rudder stock to the C.G. of, say, 15 ft (4.6 m). The rudder's geometry is such that it stalls at AoA=35°.
At the moment you start pushing the tiller your maximum rudder angle shoudn't exceed 35° or otherwise it will stall. That's because the rudder "sees" the water arriving at angle 0° respect to boat's centreline (keel line).
Once the boat has attained a turn rate of, say 15 degrees/sec the rotation speed will be:
Omega = 15°/57.3 = 0.26 rad/sec.
This will create a transverse velocity component at the rudder, with a value of:
Vt = 0.26 * 4.6m = 1.20 m/s.
The water no more arrives to the rudder form the centerline direction, but is rotated sideways by an angle of arctan(Vt/Vboat). Therefore, the effective AoA, seen by the rudder will now be:
AoA = 35° - arctan(1.20/4.12) = 35° - 16.2° = 18.8°.
The side force given by the rudder will therefore be smaller (by nearly 50% !) than at the beginning of the turn.
It means that constant rudder angle of 35° is not the optimum if you want to increase the turn rate.
In order to restore the initial side force (and rudder effectivness), you'll have to increase the rudder angle untill the effective AoA nears the maximum value of 35° again.
In other words, you'll have to increase the tiller angle by 16.2°. Therefore, the maximum rudder angle for this boat should not be 35°, but (approximately) 35°+16° = 51°.

Hope this example will make things easier to understand.

Last edited: Mar 25, 2009
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### rsfordJunior Member

Ok. Thanks a lot, daiquiri. Sounds like I certainly need at least 40 degrees and should try for 45 if I can fit the longer hydraulic cylinder under the aft deck. Actually even 40 degrees would be an improvement over the tiller; since it had to have a big loop in it to go around the mizzen mast, its travel was limited to about 34 degrees, and indeed turning has been a bit sluggish. So hopefully the hydraulic steering will improve performance as well as allow reclaiming some more room in the cockpit.

Thanks again.

Bob Ford

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put 2 rudder stops at 36 degrees, & measure the ram's travel, are you on a lake, because 12" freeboard is not enough for the sea, & you have to think of that ram's hole as your freeboard, is it being done under survey, I doubt it, no surveyor can pass that seal unless it's pressure tested, its your weak point, when she buries her stern, that water wont be splashing in, it will come in under pressure, & in bad weather you dont want to have to think about that rubber seal, good luck

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### rsfordJunior Member

Thanks for your comments, Peter. I was planning on a rather substantial custom-made bellows to be sealed aft around the ram with hose clamp type fitting and at the forward end up against the hull with flat flange sort of configuration to be sealed watertight with fasteners and sealant. However, this is not anything I've ever seen. I read in Practical Sailor about a successful conversion to wheel steering of a double-ender with sternpost hung rudder using transmission steering equipment with a similar draglink to the rudder going through the hull and being sealed with an improvised rubber seal, something over which I thought the bellows idea represented an improvement.

Still, your concerns are welcome and a good reason to pause and reconsider other options, including transmission steering with the gearbox mounted just below the aft deck and its shaft going up through the aft deck through a heavy bearing, with the attached lever arm articulating with a draglink going though a slot cut in the 5" bulwark below the caprail. This would preserve the sheer line and eliminate the water problem since any water coming in the slot would just drain off the deck in the usual fashion. The down side would be the cost of the transmission steering gear, which looks like it would be about double that of hydraulic equipment.

9. Joined: Mar 2009
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I 'm not saying theres anything wrong with your low ram & seal, it may last 20 years, but as you know the enemies are those twin gangsters, frost & sun, they'll try to break down that rubber , I would advise you to get it passed by a surveyor for insurance, you could mock up a plywood yoke on the rudder head with cramps an old idea but it allows you versatility in cheaper design, just to see if it suits your eye which is all important, & consider cable or chain in tubes to wheel steering, an lod idea but it allows you cheaper versatility in design, i cant say whats best because i dont know your boat

10. Joined: Mar 2009
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### rsfordJunior Member

Thanks Peter. Lot's to think about. I think this year I'll just enjoy the boat with its tiller and worry about converting to a wheel next winter.

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