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I am trying to see how much I can reduce the torque on a rudder post by changing the foil shape. I think I need a rudder with a centre-of-effort further ahead than average but I don't know what rudder series to use.

Balancing the rudder (in a spade rudder style) would be much simpler (and more efficient) - however, there is a design requirement for a sharply angled skeg for this particular project. The rudder will be attached/trailing directly behind a skeg.

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As a Naval Architect, this is a question that I am often asked. The centre of effort is very close to the geometric centre of the spade as you know.

Is there a fixed skeg or keel immediately in front of the rudder? That affects the hydrodynamics significantly because the hydrofoil would then have an effective length that is the skeg width (aligned with the fluid flow) plus the rudder chord length. While not precise, this approximation will allow a better calculation of Rn (Reynolds Number) to calculate lift.

The rudder torque is the response to the hydrodynamic pressure on the moving part of the hydrofoil. A perfect spade rudder would be long and thin (think aircraft wing), with its axis approximately 1/3 to 1/2 chord from the leading edge at the centre of lift. Unfortunately with ships the rudder tends to be short and wide and much less efficient, so to achieve the same force, a larger profile area is needed ( F= 1/2 PAV^2) P=fluid density, A = Foil profile area, V=fluid velocity.

Adding a skeg achieves exactly the same result as an aircraft wing with ailerons. The ailerons change the lift characteristic of the foil in two ways - firstly the fluid flow is diverted, changing the relationship of the pressure over the top and bottom surface; secondly the centre of lift moves forward slightly as the aerofoil deflection increases, and therefore increasing effort is needed to operate the aileron until stall is achieved. However the actual effort required with a skeg vs. a free spade rudder is significantly lower since the fixed skeg is part of the foil and carries most of the load.

So the answer to your question depends on the physical geometry required, the fluid flow rate, fluid density, and the aerofoil shape itself.

Some of the low speed Selig or NACA series foils may be suitable - eg NACA 0015 with the movable portion of the foil (rudder) at 50-60% chord from the LE of the skeg. Obviously a leading edge radius on the rudder and a trailing edge radius on the skeg are required. The gap between should be minimised. NACA0015

Useful resources:- http://www.mh-aerotools.de http://airfoiltools.com

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  • $\begingroup$ Thank you very much for your response! The rudder will indeed be attached/trailing directly behind a skeg. $\endgroup$ – avs Nov 15 '16 at 21:37
  • $\begingroup$ Where does the 50-60% chord come from - is that a general rule of thumb? Related - but different question (sorry if that isn't allowed): Is there any resources/rules-of-thumb for the required side force of the rudder+skeg configuration in order for the boat to turn at a certain rate? I have found some formulas that take raw rudder data - but they don't account for a skeg. I have also found equations to calculate the raw force from a rudder + skeg - just trying to tie it all in. $\endgroup$ – avs Nov 15 '16 at 21:49
  • $\begingroup$ Essentially an ideal skeg+rudder is two foils, paired so that the overall profile viewed as a waterline section looks like a single foil. $\endgroup$ – Donald Gibson Nov 20 '16 at 1:38
  • $\begingroup$ A balanced spade is more of a deflector which operates at very high angles of attack with high torque and undesirable drag characteristics at high angles due to the stall characteristics. The skeg/rudder when correctly designed, maintains flow over the foil with relatively small induced drag, even at fairly large angles of attack. 50-60% - is an arbitrary figure. Some designs call for a tall, narrow skeg paired with a larger rudder for example on racing yachts. In which case the rudder post would be around 20% from the LE of the skeg. $\endgroup$ – Donald Gibson Nov 20 '16 at 2:11
  • $\begingroup$ That implies that the rudder is carrying the majority of the lateral load. A more symmetrical arrangement reduces the torque on the rudder post and reduces the drag effects even at high angles of attack. To answer the rest of your question - the vessel pivots about a point called the center of lateral resistance (or effort). On many ships that is somewhere close to midships and on any fin keeled yacht it is around the mid length of the fin. Measure the distance between CLR and the apparent centre of the skeg/rudder to determine the moment arm. The rudder/keel/hull shape all affect the math. $\endgroup$ – Donald Gibson Nov 20 '16 at 2:19

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