Working on the steering cables for Rutu and have run up against a math
problem. The steering lines come off the drum down through the pedestal to
a pair of sheaves. To avoid wear the lines have to be aligned very
accurately to the sheave grooves. In order for the boat to turn the same
direction as the wheel the lines have to cross inside the pedestal so the
sheaves have to be canted outwards from vertical about 6 degrees to make the
lines run fair. BUT the lines lead off from the pedestal base about 15
degrees to the left. It looks like when I angle the sheaves in the
horizontal plane the angle of the cant changes. .

How do I calculate what the actual cant angle should be before I start
welding this thing together?

--
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

Lemme see if I can put this in a text book format. The coordinates of 3
points in 3 dimensions where X is width, Y is height and Z is depth:

Point A: X=0 Y=0 Z=0
Point B: X=7 Y=24 Z=36
Point C. X=5.5 Y=50 Z=37

One line runs from point A to Point B and a second from Point B to point C?
What are the horizontal and vertical angles of intercept of the two lines?

--
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

"Ignoramus15109" wrote in message
...
On Wed, 22 Feb 2006 15:34:43 -0500, Glenn Ashmore
wrote:
Working on the steering cables for Rutu and have run up against a math
problem. The steering lines come off the drum down through the pedestal
to
a pair of sheaves. To avoid wear the lines have to be aligned very
accurately to the sheave grooves. In order for the boat to turn the
same
direction as the wheel the lines have to cross inside the pedestal so the
sheaves have to be canted outwards from vertical about 6 degrees to make
the
lines run fair. BUT the lines lead off from the pedestal base about 15
degrees to the left. It looks like when I angle the sheaves in the
horizontal plane the angle of the cant changes. .

How do I calculate what the actual cant angle should be before I start
welding this thing together?


Um, can you simply write down coordinates of the relevant points?

If so, then the answer is that it is very simple to figure out all
angles when you know coordinates.

i

A sketch would be helpful. It's not clear what "outward from
vertical" means -- which way is out?

A sketch from two views would help a lot.

Don't think I could do it in ASCII art.

Picture two ropes coming down from either side of the steering drum that
cross each other as they lead to the sheaves. The sheaves have to lean
inwards toward each other to line up with the ropes. That would be fairly
easy to draw out and measure the angle that the sheaves have to lean if the
ropes went round the sheaves then straight back but they don't. They have
to go to another pair of sheaves that are 24" back, 36" down and 7" to one
side.

The 7" to the side makes it complicated.

--
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

I think I see where you are headed and I see that ABG is a leaning right
triangle but where did the 2 in (2,24,0) come from?

--
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

"Ignoramus15109" wrote in message
...
On Wed, 22 Feb 2006 16:14:40 -0500, Glenn Ashmore
wrote:
Lemme see if I can put this in a text book format. The coordinates of 3
points in 3 dimensions where X is width, Y is height and Z is depth:

Point A: X=0 Y=0 Z=0
Point B: X=7 Y=24 Z=36
Point C. X=5.5 Y=50 Z=37

One line runs from point A to Point B and a second from Point B to point
C?
What are the horizontal and vertical angles of intercept of the two
lines?


you have a vector AB (A to B)

AB = (Xb-Xa, Yb-Ya, Zb-Za) = (7-0, 24-0, 36-0) = (7, 24, 36)

consider vector AG = (2, 24, 0) -- it is a projection of vector A to
the horizontal plane (Z being the vertical dimension).

AB and AG are the hypotenuse and leg of a right triangle, see

http://www.algebra.com/calculators/g...ythagorean.mpl

length of AG divided by length of AB is the cosine of the "horizontal
angle" of vector AB. arccosine of this value os the angle.

You can continue in the same style and soon you wil have all the
angles.

i

Glenn Ashmore wrote:

A sketch would be helpful. It's not clear what "outward from
vertical" means -- which way is out?

A sketch from two views would help a lot.

Don't think I could do it in ASCII art.

Picture two ropes coming down from either side of the steering drum that
cross each other as they lead to the sheaves. The sheaves have to lean
inwards toward each other to line up with the ropes. That would be fairly
easy to draw out and measure the angle that the sheaves have to lean if the
ropes went round the sheaves then straight back but they don't. They have
to go to another pair of sheaves that are 24" back, 36" down and 7" to one
side.

The 7" to the side makes it complicated.

--
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

Wouldn't mounting the pulleys with a swiveling mount allow them to self
align?

Pete C.

Wouldn't mounting the pulleys with a swiveling mount allow them to self
align?

If I could do it with the mounts in tension that would work but the sheaves
will be bolted to the underside of the cockpit floor and have to be rigidly
mounted. The commercial units made by Edson are adjustable in all axies
but they cost big bucks.

That does give me an idea to mock it up with the sheaves free to move as
they want and then measure the angles.

--
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 writes:

How do I calculate what the actual cant angle should be before I start
welding this thing together?

Draw it in a CAD program. It reports the results. Dunno how you can be
attempting your ambitious design project without this, at least in these
days.

Glenn Ashmore wrote:

A sketch would be helpful. It's not clear what "outward from
vertical" means -- which way is out?

A sketch from two views would help a lot.



Don't think I could do it in ASCII art.

Picture two ropes coming down from either side of the steering drum that
cross each other as they lead to the sheaves. The sheaves have to lean
inwards toward each other to line up with the ropes. That would be fairly
easy to draw out and measure the angle that the sheaves have to lean if the
ropes went round the sheaves then straight back but they don't. They have
to go to another pair of sheaves that are 24" back, 36" down and 7" to one
side.

The 7" to the side makes it complicated.



I started doing this problem on the cad system but I think more info is
needed.
Maybe the easiest way without going overboard G is to mock it up with
string and cardboard.

Fred

" I started doing this problem on the cad system but I think more info is
needed.
Maybe the easiest way without going overboard G is to mock it up with
string and cardboard.

Fred

That is the project for this evening. :-)

--
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

On Thu, 23 Feb 2006 17:42:07 -0500, "Glenn Ashmore"
wrote:


" I started doing this problem on the cad system but I think more info is
needed.
Maybe the easiest way without going overboard G is to mock it up with
string and cardboard.

Fred

That is the project for this evening. :-)

Hey Glenn,

I'm surprised you are not using chain & cable, like most this size
But even so, surely, on the boat you are building, wouldn't the
crossover be done between the quadrant and the blocks, rather than
between the pedestal and the blocks?
I you must have "fixed" turning blocks, then why not set up a jury-rig
with a gimbaled or floating block on each line, with one fixed and
one adjustable fleet guide on each one. Or like this:

http://www.edsonmarine.com/sailboat/sailboat_toc.html#sswc
and pick "Articu-Lock™ Sheaves"

Take care. Having fun yet??

Brian Lawson,
Bothwell, Ontario.

On Wed, 22 Feb 2006 16:14:40 -0500, "Glenn Ashmore"
wrote:

Lemme see if I can put this in a text book format. The coordinates of 3
points in 3 dimensions where X is width, Y is height and Z is depth:

Point A: X=0 Y=0 Z=0
Point B: X=7 Y=24 Z=36
Point C. X=5.5 Y=50 Z=37

One line runs from point A to Point B and a second from Point B to point C?
What are the horizontal and vertical angles of intercept of the two lines?

I get a tilt angle of plane ABC ( the plane of the pulley) from Y
axis of 13.42 degrees.

The azimuth or bearing angle of plane ABC is 3.72 degrees from the
roll axis (Z) of the boat.

The included angle between AB and BC is 124.65 degrees.

I don't know what "angle of intercept" means. Intercept with what?

Hey Glenn,

I'm surprised you are not using chain & cable, like most this size
But even so, surely, on the boat you are building, wouldn't the
crossover be done between the quadrant and the blocks, rather than
between the pedestal and the blocks?
I you must have "fixed" turning blocks, then why not set up a jury-rig
with a gimbaled or floating block on each line, with one fixed and
one adjustable fleet guide on each one. Or like this:

http://www.edsonmarine.com/sailboat/sailboat_toc.html#sswc
and pick "Articu-LockT Sheaves"

That would have ben my first choice but Edson wants an arm and a leg for
those things.

What I have done is knock off an Edson rope steerer that they designed for
the Open 60's. Spiraled Delrin drum running on Torlon rollers in bronze
races. Running 8mm Amsteel lines to a composite radial. All shop built.
Lucked up on 4 custom made 8" Harken steering quadrants in their close outs
and that is what I am trying to mount. If I had bought it all out right it
would have been a little over $15K not counting the 52" carbon wheel. Got a
little over $500 and many hours in it right now but it looks like it will be
strong, light and quiet.

After playing with a mock up I think I am going to come straight out of the
pedistal and make the cross over in the long run to the radial. The
geometry will be a lot simpler and a bit more tollerant of my gorilla
welding. With all the crap that has to go aft of the pedistal it has been
a real challange finding a clear route to the rudder. That is why I have
the off center routing down close to the hull.

--
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.

That is amazingly close to what I measured on the mock up a few minutes ago.
I got 13.5 and 4 degrees. Close enough! Now if I can just keep my welds
somewhere close to that. :-)

--
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

"Don Foreman" wrote in message
...
On Wed, 22 Feb 2006 16:14:40 -0500, "Glenn Ashmore"
wrote:

Lemme see if I can put this in a text book format. The coordinates of 3
points in 3 dimensions where X is width, Y is height and Z is depth:

Point A: X=0 Y=0 Z=0
Point B: X=7 Y=24 Z=36
Point C. X=5.5 Y=50 Z=37

One line runs from point A to Point B and a second from Point B to point
C?
What are the horizontal and vertical angles of intercept of the two lines?

I get a tilt angle of plane ABC ( the plane of the pulley) from Y
axis of 13.42 degrees.

The azimuth or bearing angle of plane ABC is 3.72 degrees from the
roll axis (Z) of the boat.

The included angle between AB and BC is 124.65 degrees.

I don't know what "angle of intercept" means. Intercept with what?

On Wed, 22 Feb 2006 21:29:50 GMT, Ignoramus15109
wrote:

On Wed, 22 Feb 2006 16:14:40 -0500, Glenn Ashmore wrote:
Lemme see if I can put this in a text book format. The coordinates of 3
points in 3 dimensions where X is width, Y is height and Z is depth:

Point A: X=0 Y=0 Z=0
Point B: X=7 Y=24 Z=36
Point C. X=5.5 Y=50 Z=37

One line runs from point A to Point B and a second from Point B to point C?
What are the horizontal and vertical angles of intercept of the two lines?


you have a vector AB (A to B)

AB = (Xb-Xa, Yb-Ya, Zb-Za) = (7-0, 24-0, 36-0) = (7, 24, 36)

consider vector AG = (2, 24, 0) -- it is a projection of vector A to
the horizontal plane (Z being the vertical dimension).

AB and AG are the hypotenuse and leg of a right triangle, see

http://www.algebra.com/calculators/g...ythagorean.mpl

length of AG divided by length of AB is the cosine of the "horizontal
angle" of vector AB. arccosine of this value os the angle.

You can continue in the same style and soon you wil have all the
angles.

i

Not quite, Iggy.

First, you have interposed Z and Y from Glenn's definition.

Even with that corrected, your procedure might get the traces of the
cables, but not the orientation of the sheave.

The plane of the sheave can be non-ambiguously defined as a vector
aligned with the axle which is normal to the plane of the sheave.
This would be the vector cross-product of any two vectors in the
plane which are AB, BC and AC. If you convert this to a scalar
magnitude times a unit vector, then the unit vector will be the
direction cosines of the resultant.

You can then find elevation or zenith angle (tilt) which is
independent of azimuth in ship's coordinates, and the projection
of this axle vector in the XZ (horizontal) plane to get azimuth
angle.

The results, from just a few lines on a MathCAD sheet a tilt of
3.722 degrees and azimuth of 13.443 degrees.