Steering Effort on Sailboat w/ Hydraulic Rudder Control

[Marinovich91]

Greetings to everyone,

Recently I bought sailboat with rudder which is actuated by hydraulic oil by cylnder and pump at steering wheel.

I have been trying to calculate steering effort(feedback) on my hydraulic operated sailboat if I change my current cylnder,but nusucessfully.

Problem is that with my current specs I don't need to use any effort to stear the boat and there are too many turns(6) on 70cm diameter steering wheel.

With sailboat it is necesary to have effort when sailing, and to achieve this I need to downsize the cylnder which is connected directly to rudder over tiller arm 18.But downsizeing cylnder I am increasing effort, but I don't want to be in situation to use both my hands to steer.

As well on my steering wheel is connected my autopilot which can't handle more than 5Nm of force, so before buying anything I decided to make calculation,to know by how much and how effort is changed.
In order to calculate approximation (+- 200 grams) found couple of formulas and putted them down in excell sheet down below, for each rudder angle of attack(different drag force), necesary tiller arm and different speed(since speed drops automaticly when you steer trough water).

As you can see firstly I was calculateing force of drag for each angle of attack, and than transvered forces to hydraulic cylnder, and than moment on pump shaft.

Problem comes when I try to calculate drive power of pump in wats and put it Newton-meters per second...all goes wrong.At this point I am not sure what I am doing wrong.

For example I get,

I need to use 0,06kg/s on my steering wheel for 3 seconds to achive feedback of ruder at the end of 3 seconds of 0,54kg... but 0.06+0.06+0.06=0.18kg

If anyone can help me I would be grateful.

In attachment you can find excel sheet of all calculations, which includes 3 cylnders with different angles of attack, but it is same formula for all rows.

Below are related images of system.
System overview

Rudder transformation to calculate how much rudder is balanced and effective drag area

Drag coeficient trend

Current cylnder
P.S.
It may be important to say that rudder is hydrofoil shaped
Can anyone confirm I used correct formulas for calculation of steering effort
P=(p*Q)/(600*n1)= (watt)

 

[Lnewqban]

Welcome, Marinovich91
How did you calculate the “power to give on the wheel”?

 

[Marinovich91]

Welcome, Marinovich91
How did you calculate the “power to give on the wheel”?

By useing this formula P=(p*Q)/(600*n1)

Or,

Drive power = pressure * volume flow /600/total efficiency
Result is kW * 1000 = W = Nm/s(?)

(Nm/s)/10= kg*m/s (this is on 1m radius)
For 35cm radius (70diameter) = (kg*m/s)/0.35m

Pressure = 3,95bar

For Q i used this formula;

Volume flow = pump capacity*rpm*volumetric eficiency/1000

Pump capacity is 28cc
I took that rpm is 1
Volumetric eff is 0,95
Total eff. is 0.82

Firstly I was calculateing volume flow for 3 revolutions and later dividing result by 2,95s(how fast I can turn wheel per second) to get volumetric flow in seconds.

 

[jack action]

If I understood you correctly, you want to find the force you will apply at the wheel or maybe simply the torque torque.

As you can see firstly I was calculateing force of drag for each angle of attack, and than transvered forces to hydraulic cylnder, and than moment on pump shaft.

This is how I would start.

Problem comes when I try to calculate drive power of pump [...]

I'm not sure why you need to know the power for this problem.

If you know the force $F_p$ acting on the piston (based on the rudder force), then the system pressure $p$ will be $p =\frac{F_p}{A_p}$, where $A_p$ is the area of the piston.

Then the torque $T$ of the motor is $T = \frac{pD}{2\pi}$, where $D$ is the displacement of the motor (per revolution).

To get the force applied to the steering wheel $F_s$, then $F_s = \frac{T}{r}$, where $r$ is the radius of the wheel (not the diameter).

Units could be (without adding conversion factors):

Variable​	SI​	US​
$F_p$​	N​	lb​
$A_p$​	cm²​	in²​
$D$​	cc/rev​	in³/rev​
$r$​	cm​	in​

Let me know if I did not understand the problem correctly, because I don't see the need for power.

Also try to use the correct terms to make sure we are talking about the same things. You use the terms force, moment, power all with weird units.

Force => N or lb, etc.
Moment, Torque => N.m or lb.in, etc. (force X distance)
Power => W or hp, N.m/s or lb.ft/s, etc. (force X velocity, torque X angular velocity)

 

[Baluncore]

Problem is that with my current specs I don't need to use any effort to stear the boat and there are too many turns(6) on 70cm diameter steering wheel.

I would have thought that 6 turns would be about right for accurate sailing. For a balanced rig, the rudder should not stall in the water, so you will only need to move the rudder by about ±20°.

During normal conditions, with a well balanced rudder, it should take little force on the wheel. You may need the assistance of the current hydraulic ratio to handle the boat at speed in a storm.

The hydraulic ratio would be optimised for different conditions. Are you sailing in confined waters or across an ocean? I believe you should sail the boat under different real conditions before changing the ratio designed and built.

The easiest solution would be to abandon the torque or force approach, and change your computation to the rate of wheel turn. If it takes 6 turns of the wheel now and you want to do it in 3 turns, then the ratio of 3 / 6 = 0.5 suggests you need a hydraulic actuator with half the cross sectional area. Remember to subtract the rod areas from the cylinder areas when computing the change. The forces change in proportion to the rate.

How does the autopilot connect to the system?
How is the oil level maintained?

 

[Marinovich91]

Thanks for your reply.

Also try to use the correct terms to make sure we are talking about the same things. You use the terms force, moment, power all with weird units.

I was converting all to kilograms in order to know how much force I am going to feel on wheel. Becouse I am used to Kg, and know how much it is.

Let me know if I did not understand the problem correctly, because I don't see the need for power.

If let say on 10° of ruder I get 10Nm of feedback force. Work to achieve this feedback force grately depends on how fast you will turn with wheel, Since boat will remain in higher velocity if turned faster. Am I correct?

It also depends how big cylnder you are useing since:

Seastar needs 3 turns from middle to full left or right, which I can achieve in 3 seconds, but with vetus I can make it in 1 second(1,2rev) , thus same work will be made in less time.

During normal conditions, with a well balanced rudder, it should take little force on the wheel. You may need the assistance of the current hydraulic ratio to handle the boat at speed in a storm.

That is why I am trying to make all calculations, since now I don't feel any force at wheel and I should. It is 10m boat for local seas.(Not oceans)

Normaly on sailing you always need to give little counter force trough water, it is constant balance(battle) of speed and direction of sailing, and by feeling that force I would know how much resistance I am making on rudder. Like all sailboats do.

Actually this is first time local people see hydraulyc steering on sailboat. Usually is system of wire ropes an pulleys.Now is closed system, and oil level is constant( if there arent leaks)
There is filling hole on the top of pump.

Autopilot is connected to wheel over "friction" belt which is actuated by planetary gear actuated small electromotor.

True this inner wheel

 

[Baluncore]

With sailboat it is necesary to have effort when sailing, ...

I do not believe that effort is necessary. Maybe your previous experience was with poorly designed sail rigs with unbalanced rudders.

You should be able to steer the boat by balancing the rig, while adjusting the set of the sails. You then do not need to apply any significant force to the steering, which will minimise drag through the water. You really need to practice balancing the rig before changing the hydraulic ratio of the rudder mechanism.

If you have an inboard engine driving a propeller, then a balanced rudder is essential. It is therefor important to study the steering ratio while the engine is operating. The foil profile of the rudder generates hydrodynamic 'lift', which applies the side force on the rudder post that turns the boat. That only requires very minor changes to the steering.

 

[Marinovich91]

Now there isn't any effort needed to steer the boat at any speed or wave, I can't feel where is the rudder since have 6 turns from full left to full right...and when wave catches me I am steering on memory to maintain direction I want.

I want to feel something but not too much, as well to reduce turns becouse now is the battle to steer the boat especially on waves.

This is old modification done by somebody, it is not made in factory, I just wat to improve it.

Rudder is well balanced I think, only 1/3 is effective area.

 

[jack action]

If let say on 10° of ruder I get 10Nm of feedback force. Work to achieve this feedback force grately depends on how fast you will turn with wheel, Since boat will remain in higher velocity if turned faster. Am I correct?

If you have a feedback torque of 10 N.m, the feedback torque is 10 N.m. If you maintain that feedback torque for a longer distance (more turns), you will require more work. But I fail to see the relevance for your problem.

Seastar needs 3 turns from middle to full left or right, which I can achieve in 3 seconds, but with vetus I can make it in 1 second(1,2rev) , thus same work will be made in less time.

Yes, same work in less time, which means more power required. But you are looking for the feedback torque and you already know that the vetus torque will be 3 / 1.2 = 2.5 times greater than the one from Seastar; Why look any further then?

 

[Marinovich91]

Yes, same work in less time, which means more power required. But you are looking for the feedback torque and you already know that the vetus torque will be 3 / 1.2 = 2.5 times greater than the one from Seastar; Why look any further then?

That I actually what I am asking...

It is linear curve, not a exponential?

When I contacted sestar tehincian he told me that effort increases drastricly. That is why I started doing calculations in the first place. I tought it is an exponential increase, so I wanted to know by how much.

 

[Tom.G]

You might want to read this article before making any changes:
https://www.cruisingworld.com/how/rudder-loads-modern-cruiser/

(above found with:
https://www.google.com/search?&q=boat+rudder+steering+force+vs+angle)

 

[sandy stone]

As a longtime sailor, I would like to comment here. Most sailors absolutely want some feedback at the helm, which helps them to feel the boat's way upwind. There is also the factor of having a kind of failsafe action if the sailor looses control of the helm - the boat, left to its own devices, will hopefully head up into the wind and slow to a stop as the wind is spilled from the sails. One other aspect is that when sailing upwind it is generally thought that up to about a 5 degree rudder is beneficial to the boat's progress, since it causes a net force on the boat into the wind, without excessive drag. Boat designers go to some effort to produce this feedback, known as weather helm, and most sailors will try to trim their sails to produce a slight rudder angle, not completely nullify it.

The OP mentions three complete turns of the wheel to put the rudder hard over, I think on a 70 cm diameter wheel. That sounds wildly excessive, as I can put the rudder hard over on my boat with 1/2 turn, on a ~1 m wheel. I don't think he mentioned the size of his boat; mine is 37 feet (a little over 11 m). He also says he has a balanced rudder, which should not require a lot of steering effort, or mechanical advantage in the steering mechanism.

Finally, I have to say that hydraulic steering is generally considered the worst possible arrangement on a sailboat, precisely because of the inherent lack of feedback, or feel. Also there is usually some internal leakage in the hydraulic pump at the wheel hub, or the cylinder at the rudder, causing drift in the wheel position required to maintain the rudder angle. For this reason hydraulic steering is rarely used, except in cases where there is not a feasible way to mount a mechanical steering gear.

I think the optimum solution for the OP would be to replace the hydraulic steering altogether, if mechanically and financially possible. Failing that, I would certainly investigate ways to reduce the mechanical advantage in the system. If the steering arm connection on the rudder post is beefy enough, maybe move the cylinder attachment point closer to the rudder shaft.

 

[jack action]

That I actually what I am asking...

It is linear curve, not a exponential?

When I contacted sestar tehincian he told me that effort increases drastricly. That is why I started doing calculations in the first place. I tought it is an exponential increase, so I wanted to know by how much.

Changing the piston or motor size will affect the effort linearly, just like changing the wheel radius.

I suspect the technician was referring to the angle of the rudder, where the drag force will not increase linearly with the angle of attack. Therefore, at 2 revolutions of the wheel you should expect more than twice the force that you experience at 1 revolution of the wheel.

 

[Marinovich91]

The OP mentions three complete turns of the wheel to put the rudder hard over, I think on a 70 cm diameter wheel. That sounds wildly excessive, as I can put the rudder hard over on my boat with 1/2 turn, on a ~1 m wheel. I don't think he mentioned the size of his boat; mine is 37 feet (a little over 11 m). He also says he has a balanced rudder, which should not require a lot of steering effort, or mechanical advantage in the steering mechanism.

Finally, I have to say that hydraulic steering is generally considered the worst possible arrangement on a sailboat...

That is true, but modifying system for mechanical would be too complex vs changeng only clylnder...

I know it won't be perfect but for now I am looking just for improvement.

By your opinion and experience, what you think would be the best feedback, let say at full speed from 0° to 10°, 0° to 20° and 0° to 30° in kilograms messured at wheel, in an emergency steering situation, like just trying to avoid tree in water 15m away.

Just need some approximation, to know what to aim for... As you can see from exell sheet with smallest cylnder I get 0.41kg at 5°...with 10°,0.5kg...
Is it too much?
Sailboat I own is 31 foot or 9.80m...

Cylnder from which material should I buy - brass or aluminium?

Jack thanks

 

[sandy stone]

This is just a wild guess, completely unsupported by calculations. On a 31 foot boat traveling 6 knots, abruptly putting the rudder over to 30 degrees with a 70 cm wheel - with mechanical steering and typical steering ratio I would expect a force at the wheel rim of maybe 15 to 20 kg (I suppose technically about 150 - 200 N). Obviously YMMV, especially in regard to the steering ratio.

Have you considered a sleeved-cable type steering system? Not ideal, but possibly an improvement.

 

[Tom.G]

Cylnder from which material should I buy - brass or aluminium?

There are very few Aluminium alloys that withstand a marine enviroment, I suggest Stainless Steel or Brass. Look at the rest of the fittings on the boat.