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

"Glenn Ashmore" wrote in message
news:ufQDb.5545$JD6.4631@lakeread04...
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.

Say you have a bracket extending 8" off either side, and this extension is
mounted, say 8" off from the pivot point of the rudder. Then the mounting
point of the ram om this bracket is at the same point as if you would have
two tillers of 11.3" (pythagoras), each at an angle of 45 degrees from the
rudder. If you mount the each ram perpendicular to this 'virtual' tiller,
wouldn't that create the same setup as one tiller and two rams parallel to
the 'transom' (which he doesn't have)?

Meindert

Meindert Sprang wrote:
"Glenn Ashmore" wrote in message
news:ufQDb.5545$JD6.4631@lakeread04...

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.


Say you have a bracket extending 8" off either side, and this extension is
mounted, say 8" off from the pivot point of the rudder. Then the mounting
point of the ram om this bracket is at the same point as if you would have
two tillers of 11.3" (pythagoras), each at an angle of 45 degrees from the
rudder. If you mount the each ram perpendicular to this 'virtual' tiller,
wouldn't that create the same setup as one tiller and two rams parallel to
the 'transom' (which he doesn't have)?

Meindert



That makes sense but he wants to mount the rams parallel to the rudder
which, if the setup is roughly 200mm square will set them at about 45º
to the "virtual tiller" in the center position. That arangement will
reduce the effective ram force by about 1/3. To complicate matters, the
Vetus MTC72 is 487mm long in the center position so the ram pivot has to
be mounted 287mm forward of the pintle. The geometry becomes a trapezoid
rather than a parallelogram and puts the action out of balance. A quick
Autocad sketch indicates 99 mm out and 140 mm in. He would have to use
two rams or the rudder would turn one way faster than the other.

--
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:isYDb.5581$JD6.2101@lakeread04...
That makes sense but he wants to mount the rams parallel to the rudder
which, if the setup is roughly 200mm square will set them at about 45º
to the "virtual tiller" in the center position. That arangement will
reduce the effective ram force by about 1/3. To complicate matters, the
Vetus MTC72 is 487mm long in the center position so the ram pivot has to
be mounted 287mm forward of the pintle. The geometry becomes a trapezoid
rather than a parallelogram and puts the action out of balance. A quick
Autocad sketch indicates 99 mm out and 140 mm in. He would have to use
two rams or the rudder would turn one way faster than the other.

Indeed!

Meindert

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