On Mon, 15 Dec 2003 14:19:01 -0500, Glenn Ashmore
wrote:



Brian Whatcott wrote:

If you are visualizing what I am visualizing, then two single acting
rams can substiture for one double acting ram. /snip/

Actually, there is a difference and two pushing cylinders make it worse.
Think force vectors for a second. Lets assume we have a 72º hard over
to hard over rudder angle and with the rudder amidships the ram
intersects the centerline of the rudder at 45º. /snip/

I am evidently not visualizing what you are visualizing, Glenn.
Why would a ram be set at 45 deg to the long axis of the hull?

If I visualize an aft rudder, with a forward facing tiller. I can fit
either one double acting ram acting from side to side.
Or two single acting rams.
I could place these two rams transversely (like the double acting
ram) but I *could* consider an arrangement that places each ram
connecting to the tiller at 36 degrees aft of the transverse
direction.
Then, at either hard over position of the tiller, the ram is acting at
90 degrees to the tiller, the most advatageous position.
But with moderate hydrodynamic balancing of the rudder surface to
moderate the steering forces is this such a problem?

Brian W

I am not visualizing what you are either. In your original post you
were talking about a transom mounted rudder and hydraulic steering. To
me that usually means a power boat with the hydraulics mounted on the
transom. Maybe you should explain exactly what you have and what you
are trying to do. If this is a sailboat with a tiller why are you going
through the hassle of hydraulic steering?

Brian Whatcott wrote:

On Mon, 15 Dec 2003 14:19:01 -0500, Glenn Ashmore
wrote:



Brian Whatcott wrote:

If you are visualizing what I am visualizing, then two single acting

rams can substiture for one double acting ram. /snip/

Actually, there is a difference and two pushing cylinders make it worse.
Think force vectors for a second. Lets assume we have a 72º hard over
to hard over rudder angle and with the rudder amidships the ram
intersects the centerline of the rudder at 45º. /snip/


I am evidently not visualizing what you are visualizing, Glenn.
Why would a ram be set at 45 deg to the long axis of the hull?

If I visualize an aft rudder, with a forward facing tiller. I can fit
either one double acting ram acting from side to side.
Or two single acting rams.
I could place these two rams transversely (like the double acting
ram) but I *could* consider an arrangement that places each ram
connecting to the tiller at 36 degrees aft of the transverse
direction.
Then, at either hard over position of the tiller, the ram is acting at
90 degrees to the tiller, the most advatageous position.
But with moderate hydrodynamic balancing of the rudder surface to
moderate the steering forces is this such a problem?

Brian W

--
Glenn Ashmore

I'm building a 45' cutter in strip/composite. Watch my progress (or lack
there of) at: http://www.rutuonline.com
Shameless Commercial Division: http://www.spade-anchor-us.com

"Glenn Ashmore" wrote in message
news:vItDb.5348$JD6.3954@lakeread04...
I am not visualizing what you are either. In your original post you
were talking about a transom mounted rudder and hydraulic steering. To
me that usually means a power boat with the hydraulics mounted on the
transom. Maybe you should explain exactly what you have and what you
are trying to do. If this is a sailboat with a tiller why are you going
through the hassle of hydraulic steering?


I am talking about a sailboat., 18,000lbs, 33ft , double ended, with a barn
door for an external rudder. (Notice I didnt use the word transom, I
thought I was simplfying things by describing a transom hung rudder, but
obviously not)

So what I was envisaging was two rams mounted either side of the rudder,
mounted one end on the hull, running aft, to a fixing bracket on either side
of the rudder.

The reason I am contemplating this hydraulic solution, is that the wheel
steering solution using cables that I currently have is rather in-elegant
:)

I do have an emergency tiller that fits on the top of the rudder stock, but
for long journeys, or heavy weather the tiller is way too much work.

regards
garry

garry crothers wrote:


I am talking about a sailboat., 18,000lbs, 33ft , double ended, with a barn
door for an external rudder. (Notice I didnt use the word transom, I
thought I was simplfying things by describing a transom hung rudder, but
obviously not)

So what I was envisaging was two rams mounted either side of the rudder,
mounted one end on the hull, running aft, to a fixing bracket on either side
of the rudder.

That is not a very efficient arrangement. Have you seen such a system
in use? In a parallel setup the force arm is the length of the bracket
and the load arm is the distance from the pivot point to the center of
pressure on the rudder. For example, if the bracket is 6" long and the
center of pressure is 18" aft of the pintle the mechanical advantage is
cut to 1:3. Every Kg of turning force will require 3 Kg of ram force.
If the ram extends 3" the rudder moves 9". That could make steering
extremely sensitive. You would have to make the brackets rather long and
set the cylinders away from the rudder to get a reasonable turning force.

The reason I am contemplating this hydraulic solution, is that the wheel
steering solution using cables that I currently have is rather in-elegant
:)

I don't know that all that hydraulic equipment will be much more graceful.
--
Glenn Ashmore

I'm building a 45' cutter in strip/composite. Watch my progress (or lack
there of) at: http://www.rutuonline.com
Shameless Commercial Division: http://www.spade-anchor-us.com

"Glenn Ashmore" wrote in message
news:HwDDb.5375$JD6.5160@lakeread04...


garry crothers wrote:


I am talking about a sailboat., 18,000lbs, 33ft , double ended, with a
barn
door for an external rudder. (Notice I didnt use the word transom, I
thought I was simplfying things by describing a transom hung rudder, but
obviously not)

So what I was envisaging was two rams mounted either side of the rudder,
mounted one end on the hull, running aft, to a fixing bracket on either
side
of the rudder.

That is not a very efficient arrangement. Have you seen such a system
in use? In a parallel setup the force arm is the length of the bracket
and the load arm is the distance from the pivot point to the center of
pressure on the rudder. For example, if the bracket is 6" long and the
center of pressure is 18" aft of the pintle the mechanical advantage is
cut to 1:3. Every Kg of turning force will require 3 Kg of ram force.
If the ram extends 3" the rudder moves 9". That could make steering
extremely sensitive. You would have to make the brackets rather long and
set the cylinders away from the rudder to get a reasonable turning force.

Perhaps I am missing something in my calculation, taken from Vetus calalogue
I had contemplated using approx 20cm brackets set back from the pintels by
same 20cm distance.
(using Vetus MTC72 ram with a stroke of 225mmm I would need 196mm lever to
give me 2 x 35 degree rotation of rudder)

Area of rudder 0.69 m2
Max Speed 16 Kmh
Force on rudder = 23.3 X 0.69 x (16 x 16)
approx 4000N

Torque = Force on Rudder x Lever.
4000 x 0.2
800Nm


This figure is just outside of the rated capacity of the MTC72, thats why I
was asking if I used two Rams , one on each side of the rudder, can I divide
the torque required by 2.





The reason I am contemplating this hydraulic solution, is that the wheel
steering solution using cables that I currently have is rather
in-elegant
:)

I don't know that all that hydraulic equipment will be much more graceful


It cetainly can be any worse than what I have already.



garry

I was unwise to attempt to help on the basis of too little
information. I hope that Glenn can visualize this
arrangement better.
But better yet would be a diagram to put in front of a person
familar with hydraulic steering layouts, I'd think.

Brian W

On Tue, 16 Dec 2003 14:31:19 -0000, "garry crothers"
wrote:


Perhaps I am missing something in my calculation, taken from Vetus calalogue
I had contemplated using approx 20cm brackets set back from the pintels by
same 20cm distance.
(using Vetus MTC72 ram with a stroke of 225mmm I would need 196mm lever to
give me 2 x 35 degree rotation of rudder)

Area of rudder 0.69 m2
Max Speed 16 Kmh
Force on rudder = 23.3 X 0.69 x (16 x 16)
approx 4000N

Torque = Force on Rudder x Lever.
4000 x 0.2
800Nm


This figure is just outside of the rated capacity of the MTC72, thats why I
was asking if I used two Rams , one on each side of the rudder, can I divide
the torque required by 2.
///
garry

At this point I am kind of lost myself. Steering rams are rated for
operating more or less perpindicular to a tiller arm off the pivot
point. Alternately they are mounted off the transom at an angle as with
an outdrive.

What I think Garry is talking about is mounting the rams off the stern
parallel to the 7.5 sq.ft. rudder to brackets extending about 8" off
either side of the rudder. To me that seems to be a very odd
configuration that can't be calculated using the normal vendor supplied
formulas.

Brian Whatcott wrote:
I was unwise to attempt to help on the basis of too little
information. I hope that Glenn can visualize this
arrangement better.
But better yet would be a diagram to put in front of a person
familar with hydraulic steering layouts, I'd think.

Brian W

On Tue, 16 Dec 2003 14:31:19 -0000, "garry crothers"
wrote:



Perhaps I am missing something in my calculation, taken from Vetus calalogue
I had contemplated using approx 20cm brackets set back from the pintels by
same 20cm distance.
(using Vetus MTC72 ram with a stroke of 225mmm I would need 196mm lever to
give me 2 x 35 degree rotation of rudder)

Area of rudder 0.69 m2
Max Speed 16 Kmh
Force on rudder = 23.3 X 0.69 x (16 x 16)
approx 4000N

Torque = Force on Rudder x Lever.
4000 x 0.2
800Nm


This figure is just outside of the rated capacity of the MTC72, thats why I
was asking if I used two Rams , one on each side of the rudder, can I divide
the torque required by 2.

///

garry




--
Glenn Ashmore

I'm building a 45' cutter in strip/composite. Watch my progress (or lack
there of) at: http://www.rutuonline.com
Shameless Commercial Division: http://www.spade-anchor-us.com

garry crothers wrote:


Perhaps I am missing something in my calculation, taken from Vetus calalogue
I had contemplated using approx 20cm brackets set back from the pintels by
same 20cm distance.
(using Vetus MTC72 ram with a stroke of 225mmm I would need 196mm lever to
give me 2 x 35 degree rotation of rudder)

Area of rudder 0.69 m2
Max Speed 16 Kmh
Force on rudder = 23.3 X 0.69 x (16 x 16)
approx 4000N

Torque = Force on Rudder x Lever.
4000 x 0.2
800Nm


This figure is just outside of the rated capacity of the MTC72, thats why I
was asking if I used two Rams , one on each side of the rudder, can I divide
the torque required by 2.


Took me a little while to figure this out. Being over 50 I have trouble
working with all these newfangled French measurements. I work in feet
knots and pounds. To me KmH is that other dial on the speedometer and a
Newton is a cookie with figs in the middle. :-)

Other than the physical arangement of the rams I see a problem with your
figures.

To get the torque at the pintle you have to divide the total force on
the rudder by the distance from the pivot point to the center of force
on the rudder, not the length of the tiller arm. The ceenter of
pressure is usually about 1/3 to 2/5 of the width of the rudder.
Without knowing the shape of the rudder it is impossible to know what
that this is but it is bound to be more than 20cm.

Once you get the required torque you can divide by the tiller arm to get
the force required to turn the rudder hard over. Normally you don't
have to do this because the vendors give the turning force based on a
particular length of tiller but you are going outside the norm so you
need to carry it further.

The MTC72 delivers 434 ft. lb on a 7.72" tiller. That works out to
about 675 pounds of force. As Meindert pointed out with an 8" bracket
you will have a "virtual tiller" length of about 11.3". That will
require more stroke than the MTC72 has so you will need to shorten the
brackets to about 7" (180mm) to stay in the stroke range. That will
give you an effective tiller arm of about 9.9". (250mm) Applying 675
pounds force at 90º to a tiller .825 ft. long produces a torque of 556
ft.lb. (76KgM)

That is the SIMPLE part. Steering rams and tillers convert linear
motion to circular motion. Exccept when the ram is perpendicular to the
tiller some portion of the force goes into compression or tension on the
tiller arm itself. The amount of linear force converted to circular
force varies as the Sine of the angle between the tiller and the ram.
In a "normal" arangement with the ram perpindicular to the tiller in the
neutral position, 82% of the force goes to turning at 35º hard over.
You are starting out at 45º so the turning force is only 71%. At hard
over away from the ram the angle is about 15º and the turning force is
about 26%. Conversely hard over towards the ram the angle is about 105º
and the turning force is about 96%.

With two rams working oposite each other you will get about 60% of the
total force converted to torque. SOOOOOO, two rams developing 76KgM
each times 60% works out to about 91 KgM. If your rudder's center of
pressure actually is close to 20cm from the pintle you have lucked up.

OTOH, you might wonder where the other 40% of the force went. It is
pulling against your pintle so you might better give the whole idea a
bit more thought.
--
Glenn Ashmore

I'm building a 45' cutter in strip/composite. Watch my progress (or lack
there of) at: http://www.rutuonline.com
Shameless Commercial Division: http://www.spade-anchor-us.com

"Glenn Ashmore" wrote in message
news:Uu%Db.5870$JD6.4178@lakeread04...


garry crothers wrote:


Perhaps I am missing something in my calculation, taken from Vetus
calalogue
I had contemplated using approx 20cm brackets set back from the pintels
by
same 20cm distance.
(using Vetus MTC72 ram with a stroke of 225mmm I would need 196mm lever
to
give me 2 x 35 degree rotation of rudder)

Area of rudder 0.69 m2
Max Speed 16 Kmh
Force on rudder = 23.3 X 0.69 x (16 x 16)
approx 4000N

Torque = Force on Rudder x Lever.
4000 x 0.2
800Nm


This figure is just outside of the rated capacity of the MTC72, thats
why I
was asking if I used two Rams , one on each side of the rudder, can I
divide
the torque required by 2.


Took me a little while to figure this out. Being over 50 I have trouble
working with all these newfangled French measurements. I work in feet
knots and pounds. To me KmH is that other dial on the speedometer and a
Newton is a cookie with figs in the middle. :-)

Other than the physical arangement of the rams I see a problem with your
figures.

To get the torque at the pintle you have to divide the total force on
the rudder by the distance from the pivot point to the center of force
on the rudder, not the length of the tiller arm.



AAhhh!!!
I new I was missing something in that calculation

snip

OTOH, you might wonder where the other 40% of the force went. It is
pulling against your pintle so you might better give the whole idea a
bit more thought.
--
Glenn Ashmore



o.k
I'm convinced, I'll lose the 2nd ram idea, and go back to the drawing
board and redo the figures.

Thanks for all your help Glenn

garry