On Thu, 10 Jun 2004 01:45:59 GMT, otnmbrd wrote:



Steven Shelikoff wrote:

But the force generated is actually pretty high.

Not really. If it was, it would overcome the ahead or astern component.

I guess it's all relative. I think it's pretty high because the side
force from the prop walk is greater than what I can do by pushing the
stern sideways from the dock. It's not as high as the ahead component
because the prop is pretty efficient in that direction. However, I
wouldn't be surprised if, for some props, it were somewhere near as high
as the astern thrust.

For instance, say there was no way of controlling the direction of the
boat. No rudder, no keel, the form of the hull in the water is such
that it can move equally well in all directions and pivot just as easily
also. I.e., if the same force applied in any direction at the stern
will move the stern in that direction by the same amount. Something
like a beachball with a prop sticking out of one side.

I wouldn't be surprised if, for some props, when you put the prop in
reverse the "stern" of the boat (meaningless wrt hull form, but in this
case the stern is simply where the prop is sticking off of) moves more
sideways than backward, spinning the boat more than it pulls it
backwards.

For instance, due to
the full keel of my boat it's much easier to move it fore and aft vs.
spin it sideways. Yet when I throw the thing in reverse, all that
happens at first is prop walk.

Because of propwalk and the fact that your prop is in the stern and
everything forward of that follows your prop, much like the "tail of a dog".

Or course the boat is all attached to itself and what the stern does the
rest has to also. But what I'm talking about is that the stern moves
sideways and the boat pivots around a point that is somewhere within the
outline of the boat. That's different than the stern pulling the boat
backwards and the rest of it following.

By keeping the rudder fully to port I
can, by alternating forward and reverse power every few seconds, spin my
boat completely around in very little more than it's LOA. I use that
"feature" to get into very tight spaces.

If I remember correctly you have a LH prop which means this is totally
understandable and expected.

No, that wash is coming from the upward and to the right (for a right
handed prop).

No, only some of it. G it's amazing what you can see and watch over a
period of years. Because of the proximity to the surface, the blade,
starting at 000*, pushes water right, then down, BUT, again, because of
the proximity to the surface (no real column of water above that
directly in line with the pitch of the blade) it pushes water up into
the air, thus not having full efficiency.

You're being fooled by what you see though. Because the water being
pushed up into the air on the right side of the prop is actually due to
the blades coming up on the left side as the blade travels to the right
on it's way from somewhere past 270 up to 360.

Once again, my feeling is that
it doesn't reach full efficiency until @ 045*, and moving the prop
further underwater doesn't totally negate this fact/effect, just reduce it.

For example, say the prop is 2 feet deep. When the blade
is at 315 degrees, that's when it's pushing water out 45 degrees to the
right,
Maybe, depends on the pitch, but for our discussion, ok. (as long as you
realize the direction is also UP toward the surface .....

Of course. That's the point I'm making, that when the blade is at 315
degrees it's pushing water UP toward the surface at a 45 degree angle
from horizontal (or vertical for that matter) to the *right*. The water
being thrown up in the air on the *right* side of the prop is actually
from the blade on the left side of the shaft coming up and over to the
right.

which works out to be 2 feet to the right. By the time the blade
goes past 0 degrees, it's only pushing water sideways and then down.

Understood, but I feel the column of water above this is not "solid" ...
when you push on something, if it can't go straight, it goes to the
side, in this case up and to the surface .... path of least
resistance..... less efficiency.

Ok, the water above is not solid. But the prop is only pushing up
against this non-solid water when there is an upward component to it's
travel around the circle. Once the prop reaches 0 degrees and until it
reaches 180 degrees, there is absolutly no upward component to the
travel of the blade. The vertical component is all down. So how is a
blade with only a downward component to it's motion going to push the
column of water up?

The upward wash to the left of the boat is from the 180 to 270 part of
the rotation and the upward wash to the right is from the 270 to 0 part
of the rotation.

The main left component is from 090* to 180* during maximum prop
efficiency. From 180* - 270* the "left" component starts at maximum and
constantly decreases, as the upward component increases (nothing happens
individually, everything happens concurrently).

Exactly. And the main right component is from 0 to 90 during maximum
prop efficiency. From 270 to 0 the right component starts at minimum
and constantly increases (if the prop is not right at the surface). The
closer it is to the surface, the further around the clock the minimum
efficiency point is, both for the decreasing and increasing part.

3. Ignore the quadrants 000-090, and 180-270 ( My sense - the prop
efficiency is in transition during these portions of rotation. In one
the efficiency is increasing and the other it is decreasing ... 0% net


Separating out all the other effects and only discussing the efficiency
of the effect we're talking about here, it's not increasing from 0 to
90. It's at maximum at 0 and stays there until 180.

NO

According to your theory, YES it is. Because you have yet to explain
how a prop blade with absolutely no upward component (true all the way
from 0 to 180 degrees) can push the water column up.

It's decreasing as
you go from 180 to 270 and then increasing again as you go from 270 to
360.

No. The "left" component is decreasing as you go from 180-270, but the
right component (from 270-000) is staying relatively low (compared to
the left component from 090 t0 180) because the direction is up toward
the surface at the same time as it is to the right, rather than
(090-180) down toward "solid" water and to the left.

That's just not true if the blade is not breaking the surface. The
minimum efficiency occurs when the water column the blade is pushing
against is at a minimum. The length of the water column is that of a
line drawn perpendicular to the blade from the blade to the surface.

Just for the hell of it, I came up with a formula for the length of the
column of water the blade is pushing against when given the angle of
it's rotation during the portion from 270 to 360 degrees. It's:

column length = (D-0.5B*sin(theta-270))/cos(theta-270)

where
theta = the angle of rotation from 270 to 360 degrees.
D = depth of the center of the prop,
B = length of the blade. I'm drawing the perpendicular line from the
middle of the blade, which is where the 0.5 multiplier for B comes from.
If you want to draw it from the tip of the blade, take out the 0.5.

I'll leave it to you to verify that the formula is correct. Now, what
you have to do is put in the constants D and B and find the value of
theta that gives the minimum column and you'll see where the efficiency
reaches the minimum value on the way from 180 to theta and then start to
gain efficiency on the way from theta to back 0.

As an example, I'll use a prop depth of 3 feet and a prop diameter of 1
foot. So, D (depth) = 3 feet, B (blade length) = 0.5 feet and solve for
the theta that gives the minimum column. The answer is 274.7 degrees.
So for that configuration, the efficiency decreases from 180 to 274.7
degrees where it reaches a minimum. Then it starts to rise on it's way
from 274.7 degrees up to 360 where it's at a maximum again, and equal to
what it was at 180 but in the opposite direction.

Using the same blade diameter, if the prop is 1 foot deep, the
efficiency reaches a minumum at 284.4 degrees. If the prop is 10 feet
deep, the efficiency reaches a minimum at 271.4 degrees.

For a prop that's not right at the surface the decreasing force to
the right as it goes from 180 to 270 balances out the increasing force
to the left as it goes from 270 to 360 because the balance point (i.e.,
where the force is at a minimum due to minimum efficiency) is right at
or very near 270 degrees. I.e., the size of the water column when it's
at 280 degrees is the same as when it's at 350, just in the other
direction. 290=340, etc.

Rather than going into detail .... NO. You cannot compare a force from
180-270, to a force from 270-360. you MUST compare a force from 180-270,
to 000-090, and a force from 270-360, to 090-180.

Um, yes you can. In fact, you MUST. You cannot just look at certain
parts of the rotation while ignoring other parts. You have to look at
the balancing forces the entire way around. You MUST balance all of the
horizontal components against eachother and see what the resulting
horizontal force component is. There are horizontal components all the
way around except for when the blade is exactly at 90 or 270 degrees.
So you must consider what's happening all the way around and find what
portions balance out the other portions and what's left over that
doesn't balance out.

The "length" of the water column from 090-180, is far greater than the
"length" of the water column from 270-360. The body (propellor pitch) is
constantly changing direction of "push".

While that's true, it's also true that the "length" of the water column
from 0-90 is far greater than the "length of the water column from 270
to 0. So what?

But for a prop that's very near or at the surface, those forces don't
balance out. That's because as the prop gets closer to the surface the
balance point (where the force is at a minimum due to minimum efficiency
because of the smallest water column before you get to air) move further
around the rotation. In this case, the force at 280 /= the force at 350
and you have a net sideways force.

True, but that sideways force does NOT equal the force at 100-170,
because the column of water above it is less than the column of water
below 100-170.

Which is completely irrevelant because the sideways force from 100-170
is balanced by the equal and opposite sideways force from 10-80. Again,
you're not considering the entire rotation and what parts of it balance
out what other parts and what's left over after the balancing act.

difference.)the blades are still pushing back, but there is no net
effect (arguably) which we can readily apply to "propwalk". Instead .....

4. Concentrate on the quadrants 090-180 and 270-000. From 090 to 180 the
blade is pushing back against a relatively solid column of water, down
against a relatively solid column of water and increasingly LEFT against
a relatively solid column of water. During this entire quadrant of
rotation, the blade is at maximum efficiency...... BUT, from 270-000 the
blade is pushing back relatively nearer the surface, up toward the
surface, and RIGHT towards and relatively close to the surface, where it
can and does break the surface or at least bulge the surface. So.....


And the quadrant from 0 to 90 exactly balances out the quadrant from 90
to 180. And the quadrant from 180 to 270 "almost" exactly balances out
the quadrant from 270 to 0. The smaller the ratio of prop size/prop
depth, the closer those quadrants (180-270 and 270-0) will balance out.

NO,NO,NO You cannot compare 000-090, to 090-180 or 180-270, to 270-360.
You Must compare opposites !! 000-090 and 180-270 or 090-180 and 270-360.

YES, YES, YES you can compare 0-90 to 90-180 because they ARE opposites
to each other in the sideways direction. In fact, they completely
cancel each other in the sideways direction. The easiest way to balance
the forces is to just exactly what you think you cannot do. So,
covering the angle of rotation from 0 to 180, the net sideways force is
ZERO. Now, you have left to find the net sideways force from 180 to
360. This net force is NOT zero *because* of what I described above,
i.e., the minimum force is not right at 270 degres. And the closer to
the surface you get, the more the sideways force is out of balance. So,
for the first example above (3' depth 1' prop) the in forward with a RH
prop, for the portion of the trip from 180-360, the blade is pushing
right for 94.7 degrees but is pushing left for only 85.3 degrees. THERE
is the out of balance force that your "water column" theory says will
cause prop walk.

5. My sense from this. The blade, in these two all important quadrants,
is more efficient between 090-180, than it is between 270-000. The


Most definitely. But you're ignoring the other two all important
quadrants.

Nope, I'm narrowing down the important quadrants of push, germane to the
discussion.

But you can't do that and still come up with a correct answer. You have
to balance out the sideways force for the entire rotation. And the
easiest way to do that is to first take out the parts that completely
balance each other and then look at what's left.

differences if you add in depth of the hub of the prop, may diminish,
but for a boat floating on the water surface, the efficiency will never
be equal...... VBG ..... propwalk.

I disagree that there is no such thing as a "more solid" column of water.


That doesn't really matter since it's not important as long as you
realize that the "solidity" of the column of water (if there is such a
thing) is the same for the same angle to the right vs. to the left of
the prop.

No. The column of water is important to the "net" right and left forces.

That's not what I said. What I said is that what you have to realize is
that the column of water is the same for the same angle right vs. left
of the prop.

[...]
I disagree that by moving the blade a few feet down in the water, you
will totally negate the effects or differences in blade efficiency.


Not totally. But it doesn't take going very far down before everything
*nearly* balances out left and right due to the effect we're talking
about.

No, if that were the case, then your boat wouldn't experience propwalk.

Yes it would. Just not totally from the effect you're postulating.
But it would contribute.

Steve

On 10 Jun 2004 02:08:17 GMT, (Gould 0738) wrote:

I have it but it's on the boat, not available right now. I read it but
don't remember a discussion about what exactly causes prop walk. What
does it say?

That unequal blade thrust is the result of the difference in effective pitch
between the descending and ascending blade when a propeller is rotated on an
angle, (90 degrees from the angle at which the shaft exits the shaft log)
though a horizontally advancing medium (water under the boat).

(Page 143 in my 1985 edition, may be different in later versions)

That would make sense if prop walk only occured when the the blade is
moving through a horizontally advancing medium (water under the boat).
That effect I'm sure is a contributing factor though. But it's 0 when
the prop is not advancing through the medium, i.e, when you first throw
the boat in reverse from a standstill, the stern kicks out before the
boat starts moving significantly backwards. In fact, if you're moving
forward slightly and kick it in reverse, the stern keeps moving sideways
all the way from the slight forward motion through reverse motion,
including when there's absolutely no horizontal advancing medium. So
there must be some other cause(s) of prop walk as well.

Steve

On Thu, 10 Jun 2004 00:05:11 GMT, (Steven Shelikoff) wrote:

On Wed, 09 Jun 2004 17:57:52 GMT, otnmbrd wrote:

I've always wondered if a submarine, 1,000 feet underwater, experiences
"propwalk" .... I'm betting it doesn't G.
At any rate, to save some unnecessary reading, by the numbers on some
points at issue:

1. We are not talking about some highly visible, easily measured,
variations in efficiency. These are very subtle, but sufficient to cause
the condition.

But the force generated is actually pretty high. For instance, due to
the full keel of my boat it's much easier to move it fore and aft vs.
spin it sideways. Yet when I throw the thing in reverse, all that
happens at first is prop walk. By keeping the rudder fully to port I
can, by alternating forward and reverse power every few seconds, spin my
boat completely around in very little more than it's LOA. I use that
"feature" to get into very tight spaces.

2. Due to it's relative proximity to the surface, I do not feel that a
blade is at maximum efficiency at this angle of rotation. My
sense/guess/feel, is that isn't doesn't reach maximum until about 045*.
Why? Sit on a tug, tied to the stern of a ship and watch "propwash" for
awhile. The wash is pushed to the side, breaking the surface during the
initial part of it's rotation. However ....

No, that wash is coming from the upward and to the right (for a right
handed prop). For example, say the prop is 2 feet deep. When the blade
is at 315 degrees, that's when it's pushing water out 45 degrees to the
right, which works out to be 2 feet to the right. By the time the blade
goes past 0 degrees, it's only pushing water sideways and then down.
The upward wash to the left of the boat is from the 180 to 270 part of
the rotation and the upward wash to the right is from the 270 to 0 part
of the rotation.

3. Ignore the quadrants 000-090, and 180-270 ( My sense - the prop
efficiency is in transition during these portions of rotation. In one
the efficiency is increasing and the other it is decreasing ... 0% net

Separating out all the other effects and only discussing the efficiency
of the effect we're talking about here, it's not increasing from 0 to
90. It's at maximum at 0 and stays there until 180. It's decreasing as
you go from 180 to 270 and then increasing again as you go from 270 to
360. For a prop that's not right at the surface the decreasing force to
the right as it goes from 180 to 270 balances out the increasing force
to the left as it goes from 270 to 360 because the balance point (i.e.,
where the force is at a minimum due to minimum efficiency) is right at
or very near 270 degrees. I.e., the size of the water column when it's
at 280 degrees is the same as when it's at 350, just in the other
direction. 290=340, etc.

But for a prop that's very near or at the surface, those forces don't
balance out. That's because as the prop gets closer to the surface the
balance point (where the force is at a minimum due to minimum efficiency
because of the smallest water column before you get to air) move further
around the rotation. In this case, the force at 280 /= the force at 350
and you have a net sideways force.

difference.)the blades are still pushing back, but there is no net
effect (arguably) which we can readily apply to "propwalk". Instead .....

4. Concentrate on the quadrants 090-180 and 270-000. From 090 to 180 the
blade is pushing back against a relatively solid column of water, down
against a relatively solid column of water and increasingly LEFT against
a relatively solid column of water. During this entire quadrant of
rotation, the blade is at maximum efficiency...... BUT, from 270-000 the
blade is pushing back relatively nearer the surface, up toward the
surface, and RIGHT towards and relatively close to the surface, where it
can and does break the surface or at least bulge the surface. So.....

And the quadrant from 0 to 90 exactly balances out the quadrant from 90
to 180. And the quadrant from 180 to 270 "almost" exactly balances out
the quadrant from 270 to 0. The smaller the ratio of prop size/prop
depth, the closer those quadrants (180-270 and 270-0) will balance out.

Of course, all this assumes no hull overhang.

5. My sense from this. The blade, in these two all important quadrants,
is more efficient between 090-180, than it is between 270-000. The

Most definitely. But you're ignoring the other two all important
quadrants.

differences if you add in depth of the hub of the prop, may diminish,
but for a boat floating on the water surface, the efficiency will never
be equal...... VBG ..... propwalk.

I disagree that there is no such thing as a "more solid" column of water.

That doesn't really matter since it's not important as long as you
realize that the "solidity" of the column of water (if there is such a
thing) is the same for the same angle to the right vs. to the left of
the prop.

I disagree that a propellor is at maximum efficiency at 000* (on a boat
floating on the surface of the water).

If it's not then there's something else at work then the effect we're
talking about here. Because at 0 degrees the blade is pushing against
an infinite column of water directly to the right (for a RH prop).
Well, maybe not infinite because of the curvature of the earth and
because there may be a shoreline before the curvature of the earth comes
into play.

I disagree that by moving the blade a few feet down in the water, you
will totally negate the effects or differences in blade efficiency.

Not totally. But it doesn't take going very far down before everything
*nearly* balances out left and right due to the effect we're talking
about.

I think we've probably hashed, thrashed, and rehashed this enough in the
NG, Steve. However, feel free to continue via E-mail if you wish.

Nah. I'd rather continue it here. Finally a civil boating related
thread in the midst of a sea of bickering and political crap.

Steve

Please continue it here! It's quite interesting, although I think you're going
to end up with something like, "God created prop walk to **** off people too
cheap to buy two engines."

John H

On the 'Poco Loco' out of Deale, MD
on the beautiful Chesapeake Bay!

John H wrote:
On Thu, 10 Jun 2004 00:05:11 GMT, (Steven Shelikoff) wrote:


On Wed, 09 Jun 2004 17:57:52 GMT, otnmbrd wrote:




I think we've probably hashed, thrashed, and rehashed this enough in the
NG, Steve. However, feel free to continue via E-mail if you wish.

Nah. I'd rather continue it here. Finally a civil boating related
thread in the midst of a sea of bickering and political crap.

Steve


Please continue it here! It's quite interesting, although I think you're going
to end up with something like, "God created prop walk to **** off people too
cheap to buy two engines."

John H

On the 'Poco Loco' out of Deale, MD
on the beautiful Chesapeake Bay!

EG Actually, the effects of "propwalk" are just as important, useful
and noticeable, with two engines, as they are with one.

otn

Please continue it here! It's quite interesting, although I think you're
going
to end up with something like, "God created prop walk to **** off people too
cheap to buy two engines."

John H

Actually, prop walk is a gift, not a curse.
Once you learn how to use it, it makes it much easier to handle a single screw.

Maybe God created twin engines for people incapable of handling singles?
And in that case it's a good thing. What a shame it would be to have you all
shorebound, watching the real he-men having fun on the water. (if you didn't
notice, that was a troll!)

On Thu, 10 Jun 2004 05:07:43 GMT, (Steven
Shelikoff) wrote:

So
there must be some other cause(s) of prop walk as well.

==================================

Absolutely, because we all seem to agree that prop walk occurs even
with horizontal prop shafts.

G And, awaaaaaaay we go .......

Steven Shelikoff wrote:


I guess it's all relative. I think it's pretty high because the side
force from the prop walk is greater than what I can do by pushing the
stern sideways from the dock. It's not as high as the ahead component
because the prop is pretty efficient in that direction. However, I
wouldn't be surprised if, for some props, it were somewhere near as high
as the astern thrust.

I can't really argue pro or con with that, but I think/sense you are
high on your propwalk numbers versus astern.


For instance, say there was no way of controlling the direction of the
boat. No rudder, no keel, the form of the hull in the water is such
that it can move equally well in all directions and pivot just as easily
also. I.e., if the same force applied in any direction at the stern
will move the stern in that direction by the same amount. Something
like a beachball with a prop sticking out of one side.

I wouldn't be surprised if, for some props, when you put the prop in
reverse the "stern" of the boat (meaningless wrt hull form, but in this
case the stern is simply where the prop is sticking off of) moves more
sideways than backward, spinning the boat more than it pulls it
backwards.

I would picture an ever increasing spiral as you gained speed.



For instance, due to

the full keel of my boat it's much easier to move it fore and aft vs.
spin it sideways. Yet when I throw the thing in reverse, all that
happens at first is prop walk.

Because of propwalk and the fact that your prop is in the stern and
everything forward of that follows your prop, much like the "tail of a dog".


Or course the boat is all attached to itself and what the stern does the
rest has to also. But what I'm talking about is that the stern moves
sideways and the boat pivots around a point that is somewhere within the
outline of the boat. That's different than the stern pulling the boat
backwards and the rest of it following.

The pivot point has changed position. Now, rather than being @1/3rd of
the way from the bow, it's @1/3rd of the way forward of the stern.
As a possible explanation for this .... You're DIW, single screw RH
prop. Put your rudder hard left, then put your engine ahead. Generally,
the first thing you will notice is the beginning of a turn to port.
My sense on this .... the hard left rudder with propwash going over it
can more easily overcome the mass of the boats tendency to not turn than
it's tendency to want to remain DIW, so you notice the turn first, then
see ahead motion.

No, that wash is coming from the upward and to the right (for a right
handed prop).

No, only some of it. G it's amazing what you can see and watch over a
period of years. Because of the proximity to the surface, the blade,
starting at 000*, pushes water right, then down, BUT, again, because of
the proximity to the surface (no real column of water above that
directly in line with the pitch of the blade) it pushes water up into
the air, thus not having full efficiency.


You're being fooled by what you see though. Because the water being
pushed up into the air on the right side of the prop is actually due to
the blades coming up on the left side as the blade travels to the right
on it's way from somewhere past 270 up to 360.

No. With these direct drive diesels, frequently when they stop the
engine, one blade will stop at TDC. When they restart the engine, you
can watch that blade and what it does, before the effects of the one
behind it become apparent.
In nature, there are many things which take the path of least resistance
.....water is one of them.
You are correct that at 000* the blade is pushing against an infinite
column of water (facing 090), and to the right of that column is solid
water toward and to the ocean floor, but .... uhoh, to the left, is a
short column and air.
If the blade pushing right into that infinite column could hold that
water in line, then the blade would be at maximum efficiency. However,
it can't. The water wants to take the path of least resistance, so it
"leaks" off, up into the air. This is why I don't feel the blade is at
maximum efficiency during the entire rotation from 000-090. My own sense
is that it doesn't start approaching maximum until it reaches @ 045* of
rotation..... until then, some of of that "push" is being lost upward.


Maybe, depends on the pitch, but for our discussion, ok. (as long as you
realize the direction is also UP toward the surface .....


Of course. That's the point I'm making, that when the blade is at 315
degrees it's pushing water UP toward the surface at a 45 degree angle
from horizontal (or vertical for that matter) to the *right*. The water
being thrown up in the air on the *right* side of the prop is actually
from the blade on the left side of the shaft coming up and over to the
right.

No real argument, once the prop is in full rotation, but underlying that
is also, what I've stated earlier, above.


Ok, the water above is not solid. But the prop is only pushing up
against this non-solid water when there is an upward component to it's
travel around the circle. Once the prop reaches 0 degrees and until it
reaches 180 degrees, there is absolutly no upward component to the
travel of the blade. The vertical component is all down. So how is a
blade with only a downward component to it's motion going to push the
column of water up?

See above


The upward wash to the left of the boat is from the 180 to 270 part of
the rotation and the upward wash to the right is from the 270 to 0 part
of the rotation.

The main left component is from 090* to 180* during maximum prop
efficiency. From 180* - 270* the "left" component starts at maximum and
constantly decreases, as the upward component increases (nothing happens
individually, everything happens concurrently).


Exactly. And the main right component is from 0 to 90 during maximum
prop efficiency. From 270 to 0 the right component starts at minimum
and constantly increases (if the prop is not right at the surface). The
closer it is to the surface, the further around the clock the minimum
efficiency point is, both for the decreasing and increasing part.

Let me say this another way.
The "right" component starts just after 270* when the efficiency of the
prop is down. The right component continues through 000* (during the
entire time of which the efficiency of the blade is down. As the blade
passes 000* the right component is directly right, but because the push
can "leak" off (as I've stated above) the efficiency of the blade is
still down, but beginning to increase, just at the point that the
"right" component is beginning to decrease and head down. This reduced
efficiency continues to increase to a point @045* at the same time that
the right component is rapidly approaching zero.... i.e. the "right
component occurs during the period, the majority of which the propellor
efficiency is down.
On the other side, the left component begins just after 090* when the
prop is at maximum efficiency and continues to increase as the blade
approaches 180* (during this entire time, the blade is at maximum
efficiency). After 180, the blade is still pushing left, but it's also
beginning to push up which means that the closer it gets to 270* the
left component is decreasing as is the efficiency.
Now, if I compare the first paragraph to the second, at no time during
the rotation, during the time the "right" component could be said to
dominate, is the propellor ALSO at maximum efficiency. Whereas at the
same time on the other side when the "left" component could be said to
be dominate (is that the word I want?) the blade is a maximum efficiency
all the way to 180* and then begins to decrease as the "left" component
decreases ....... LOL PROPWALK.


According to your theory, YES it is. Because you have yet to explain
how a prop blade with absolutely no upward component (true all the way
from 0 to 180 degrees) can push the water column up.

See above above.

No. The "left" component is decreasing as you go from 180-270, but the
right component (from 270-000) is staying relatively low (compared to
the left component from 090 t0 180) because the direction is up toward
the surface at the same time as it is to the right, rather than
(090-180) down toward "solid" water and to the left.


That's just not true if the blade is not breaking the surface. The
minimum efficiency occurs when the water column the blade is pushing
against is at a minimum. The length of the water column is that of a
line drawn perpendicular to the blade from the blade to the surface.

The blade/ water column does not have to break the surface, it just has
to be able to "bulge" it.

Just for the hell of it, I came up with a formula for the length of the
column of water the blade is pushing against when given the angle of
it's rotation during the portion from 270 to 360 degrees.

Ouch, math. I've hated math since they made me do celestial and Great
circles, long hand .... no slide rule, even.
At any rate, I'm going to have to print this out and study it, but on
first read I have a problem with your conclusions.
Being just a simple sailor I decided to draw a picture and try and look
at the water column lengths that way.
I started with a circle to represent the path of the blade and at 000* I
drew a horizontal line to represent the surface.
I then cut out a "U" shape where the curved part was the same diameter
as my circle and at 270* and 090* I drew the vertical lines of the "U"
until they intersected the horizontal surface line on the full circle.
(in retrospect I should have only drawn the line up from 270*, but
that's all right). Trough the center of the "U" where the curve met the
verticals I drew another horizontal line to represent the blade and
along this line, in a number of places, additional vertical lines which
extended to the surface.
I then overlaid this "U" onto the circle and began rotating it within
the circle. At the outer limits of the "U" (tip of the blade) the water
column line immediately became shorter as I rotated, and continued to do
so until 000 where it immediately return to infinity.
At the same time, looking at the other lines between the tip and
midpoint of the blade, the column initially became shorter and at a
certain point of rotation (roughly 315*) for the center one the column
began to increase. Naturally, as you got close to the hub the column
resumed increasing sooner.
The problem with this from my standpoint is that at no point is the
column as large for the majority of the blade during this entire (for
the most part) as it is between 090-180 which would be it's opposite
balancing blade.


Um, yes you can. In fact, you MUST. You cannot just look at certain
parts of the rotation while ignoring other parts. You have to look at
the balancing forces the entire way around. You MUST balance all of the
horizontal components against eachother and see what the resulting
horizontal force component is. There are horizontal components all the
way around except for when the blade is exactly at 90 or 270 degrees.
So you must consider what's happening all the way around and find what
portions balance out the other portions and what's left over that
doesn't balance out.

OK let me rewrite: We are discussing a rotating machine which creates
thrust. At a minimum this is being done with 2 blades for balance, and
in this case you would have to compare what is happening at one blade
with exactly what's happening with the one 180* opposite to understand
the net results.
However, boats can have 2,3,4,5,6 bladed props, in which case you would
need to compare the total constant effect as one sum of effect for the
entire rotation, but not side by side effects.


The "length" of the water column from 090-180, is far greater than the
"length" of the water column from 270-360. The body (propellor pitch) is
constantly changing direction of "push".


While that's true, it's also true that the "length" of the water column
from 0-90 is far greater than the "length of the water column from 270
to 0. So what?

Actually, the length of the water column is greater from 000-180 than it
is from 180 to 000, but that puts us back to the beginning and ignores
all the variables of direction, push, etc., we've introduced.


But for a prop that's very near or at the surface, those forces don't
balance out. That's because as the prop gets closer to the surface the
balance point (where the force is at a minimum due to minimum efficiency
because of the smallest water column before you get to air) move further
around the rotation. In this case, the force at 280 /= the force at 350
and you have a net sideways force.

True, but that sideways force does NOT equal the force at 100-170,
because the column of water above it is less than the column of water
below 100-170.


Which is completely irrevelant because the sideways force from 100-170
is balanced by the equal and opposite sideways force from 10-80. Again,
you're not considering the entire rotation and what parts of it balance
out what other parts and what's left over after the balancing act.

No, I'm considering the entire rotation, but for the basic cause I'm
sticking with even number blades such as two/four where you must compare
the opposites: Two bladed - compare 10-80 to 190-260, four bladed same
as two, but add, 100-170 compared to 280-350.
G now make odd number blades and my feeling is you have to compare all
at the same time for wherever they are in rotation, but obviously the
effect (net) come out the same.


difference.)the blades are still pushing back, but there is no net
effect (arguably) which we can readily apply to "propwalk". Instead .....

4. Concentrate on the quadrants 090-180 and 270-000. From 090 to 180 the
blade is pushing back against a relatively solid column of water, down
against a relatively solid column of water and increasingly LEFT against
a relatively solid column of water. During this entire quadrant of
rotation, the blade is at maximum efficiency...... BUT, from 270-000 the
blade is pushing back relatively nearer the surface, up toward the
surface, and RIGHT towards and relatively close to the surface, where it
can and does break the surface or at least bulge the surface. So.....


And the quadrant from 0 to 90 exactly balances out the quadrant from 90
to 180. And the quadrant from 180 to 270 "almost" exactly balances out
the quadrant from 270 to 0. The smaller the ratio of prop size/prop
depth, the closer those quadrants (180-270 and 270-0) will balance out.

NO,NO,NO You cannot compare 000-090, to 090-180 or 180-270, to 270-360.
You Must compare opposites !! 000-090 and 180-270 or 090-180 and 270-360.


YES, YES, YES you can compare 0-90 to 90-180 because they ARE opposites
to each other in the sideways direction.

No, because they are not happening at the same time.... do that and
that boat will bounce left and right


But you can't do that and still come up with a correct answer. You have
to balance out the sideways force for the entire rotation. And the
easiest way to do that is to first take out the parts that completely
balance each other and then look at what's left.

If you view it from your point of view, correct. However, from mine, the
net "right" never exceeds the net "left" because the net "right"
occurs when the propellor blade is at reduced efficiency while the net
"left occurs during maximum efficiency.

That doesn't really matter since it's not important as long as you
realize that the "solidity" of the column of water (if there is such a
thing) is the same for the same angle to the right vs. to the left of
the prop.

G Sorry, but as you can see, I totally disagree with that.

Not totally. But it doesn't take going very far down before everything
*nearly* balances out left and right due to the effect we're talking
about.

No, if that were the case, then your boat wouldn't experience propwalk.


Yes it would. Just not totally from the effect you're postulating.
But it would contribute.

Steve

I don't think you will ever reach that balance point on a small boat
floating on the surface,or even a ship .... goes back to my question
about subs. Do they experience propwalk at a depth of 1,000 feet? If
not, at what depth does it stop ... sure would confirm or deny our
points, wouldn't it?

otn (phew)

Seems to me that the major bone of contention is the efficiency of the prop
between 270 and 90 degrees of rotation.
Personally, I don't think it's all that good .
Maybe that's why Mr. Arneson developed his drive unit.

Shen

On Thu, 10 Jun 2004 20:16:24 GMT, otnmbrd wrote:

G And, awaaaaaaay we go .......

Steven Shelikoff wrote:


I guess it's all relative. I think it's pretty high because the side
force from the prop walk is greater than what I can do by pushing the
stern sideways from the dock. It's not as high as the ahead component
because the prop is pretty efficient in that direction. However, I
wouldn't be surprised if, for some props, it were somewhere near as high
as the astern thrust.

I can't really argue pro or con with that, but I think/sense you are
high on your propwalk numbers versus astern.

It all depends on the prop. One with big flat paddle blades might be
pretty efficient in reverse and have much more thruse than prop walk.
Something like a fixed sailboat prop with 2 tiny blades might not be,
and the small amount of reverse thrust could be overwhelmed by prop
walk.

For instance, say there was no way of controlling the direction of the
boat. No rudder, no keel, the form of the hull in the water is such
that it can move equally well in all directions and pivot just as easily
also. I.e., if the same force applied in any direction at the stern
will move the stern in that direction by the same amount. Something
like a beachball with a prop sticking out of one side.

I wouldn't be surprised if, for some props, when you put the prop in
reverse the "stern" of the boat (meaningless wrt hull form, but in this
case the stern is simply where the prop is sticking off of) moves more
sideways than backward, spinning the boat more than it pulls it
backwards.

I would picture an ever increasing spiral as you gained speed.

Me too. And how tight the spiral is dependent on how much prop walk
there is relative to reverse thrust.

For instance, due to

the full keel of my boat it's much easier to move it fore and aft vs.
spin it sideways. Yet when I throw the thing in reverse, all that
happens at first is prop walk.

Because of propwalk and the fact that your prop is in the stern and
everything forward of that follows your prop, much like the "tail of a dog".


Or course the boat is all attached to itself and what the stern does the
rest has to also. But what I'm talking about is that the stern moves
sideways and the boat pivots around a point that is somewhere within the
outline of the boat. That's different than the stern pulling the boat
backwards and the rest of it following.

The pivot point has changed position. Now, rather than being @1/3rd of
the way from the bow, it's @1/3rd of the way forward of the stern.

Doesn't matter. It's still within the confines of the boat.

As a possible explanation for this .... You're DIW, single screw RH
prop. Put your rudder hard left, then put your engine ahead. Generally,
the first thing you will notice is the beginning of a turn to port.
My sense on this .... the hard left rudder with propwash going over it
can more easily overcome the mass of the boats tendency to not turn than
it's tendency to want to remain DIW, so you notice the turn first, then
see ahead motion.

But this is not true in reverse since you don't have the propwash going
over the rudder.

No, that wash is coming from the upward and to the right (for a right
handed prop).

No, only some of it. G it's amazing what you can see and watch over a
period of years. Because of the proximity to the surface, the blade,
starting at 000*, pushes water right, then down, BUT, again, because of
the proximity to the surface (no real column of water above that
directly in line with the pitch of the blade) it pushes water up into
the air, thus not having full efficiency.


You're being fooled by what you see though. Because the water being
pushed up into the air on the right side of the prop is actually due to
the blades coming up on the left side as the blade travels to the right
on it's way from somewhere past 270 up to 360.

No. With these direct drive diesels, frequently when they stop the
engine, one blade will stop at TDC. When they restart the engine, you
can watch that blade and what it does, before the effects of the one
behind it become apparent.

How many blades does the prop you're watching have? If it's one or two
then maybe you're not being fooled. If it's 3 or more, especially if
it's 4 or more, then you can't separate what you think you may be seeing
one blade doing from what the rest are doing because they are too close
to each other in angle. For instance, if you're watching a 4 blade
prop, what "splashing" you think you're seeing from the blade at TDC
starting to move over and down could just as easily be from the blade at
270 and starting to come up and over.

In nature, there are many things which take the path of least resistance
....water is one of them.
You are correct that at 000* the blade is pushing against an infinite
column of water (facing 090), and to the right of that column is solid
water toward and to the ocean floor, but .... uhoh, to the left, is a
short column and air.
If the blade pushing right into that infinite column could hold that
water in line, then the blade would be at maximum efficiency. However,
it can't. The water wants to take the path of least resistance, so it
"leaks" off, up into the air. This is why I don't feel the blade is at
maximum efficiency during the entire rotation from 000-090. My own sense
is that it doesn't start approaching maximum until it reaches @ 045* of
rotation..... until then, some of of that "push" is being lost upward.

*If* that's true, that the blade can be moving down but some of the
water it's pushing against moves up because it's the path of least
resistance, *then* it also must be true that the blade starts loosing
efficiency well before it's reaches 180 on it's way from 90 to 180 for
the same reason.

If the prop is deep enough that the blade diameter to prop depth ratio
is small, then those efficiency losses will almost exactly balance out.
It's the same effect on efficiency as what's happening from 180 to 360
but due to something else. In the 180 to 360 case it's due to the blade
directly pushing against a varying sized water column. From 0 to 180
it's from the wash of a blade taking the path of least resistance, i.e.,
to the surface, even if the blade isn't traveling in that direction.

So, if you think there is less efficiency at 0 and it doesn't reach
maximum until 45 (well beyond where I think by we'll go with it) then
it's also true that the blade starts to loose efficiency at around 135
and continues to loose it until at 180 it's about where it was at 0.

Of couse, the exact numbers depend on the shape of the blade and the
depth to diameter ratio. The effect of the wash due to shape seems hard
to predict. But the effect of depth to diameter ratio is the same as
the formula I gave for the other side of the rotation. I.e., if a 1'
blade is 3' deep then there's about 8-9 degrees more push one way then
the other way.

[...]
Ok, the water above is not solid. But the prop is only pushing up
against this non-solid water when there is an upward component to it's
travel around the circle. Once the prop reaches 0 degrees and until it
reaches 180 degrees, there is absolutly no upward component to the
travel of the blade. The vertical component is all down. So how is a
blade with only a downward component to it's motion going to push the
column of water up?

See above


The upward wash to the left of the boat is from the 180 to 270 part of
the rotation and the upward wash to the right is from the 270 to 0 part
of the rotation.

The main left component is from 090* to 180* during maximum prop
efficiency. From 180* - 270* the "left" component starts at maximum and
constantly decreases, as the upward component increases (nothing happens
individually, everything happens concurrently).


Exactly. And the main right component is from 0 to 90 during maximum
prop efficiency. From 270 to 0 the right component starts at minimum
and constantly increases (if the prop is not right at the surface). The
closer it is to the surface, the further around the clock the minimum
efficiency point is, both for the decreasing and increasing part.

Let me say this another way.
The "right" component starts just after 270* when the efficiency of the
prop is down. The right component continues through 000* (during the
entire time of which the efficiency of the blade is down. As the blade

It's down, but increasing on it's way to 0, just like it was up but
decreasing on it's way from 180 to 270.

passes 000* the right component is directly right, but because the push
can "leak" off (as I've stated above) the efficiency of the blade is
still down, but beginning to increase, just at the point that the
"right" component is beginning to decrease and head down. This reduced
efficiency continues to increase to a point @045* at the same time that
the right component is rapidly approaching zero.... i.e. the "right
component occurs during the period, the majority of which the propellor
efficiency is down.

While I can agree with that, the effect is nowhere near as much as
you're making it out to be. The efficiency during the right motion is
not constantly low. It's varying for the entire trip right. Same thing
as for it's trip left. There are many angles for the trip right where
the efficiency is higher then the efficiency of the opposing angle left.

On the other side, the left component begins just after 090* when the
prop is at maximum efficiency and continues to increase as the blade
approaches 180* (during this entire time, the blade is at maximum
efficiency). After 180, the blade is still pushing left, but it's also

Um, no. Not according to your "leakage" theory. Unless for some reason
you want to argue that the water can only follow the path of least
resistance in one direction ... Which I think you'll have a hard time
convincing anyone of.

I think you're problem is that you're looking at 2 different things at
once ("leakage" and "column length") and not applying the effects both
of them all the way around the prop rotation. It's easy to see why
there's a net force in some direction if you can, in your mind, ignore a
force that exists in some quadrants yet count it in other quadrants.

beginning to push up which means that the closer it gets to 270* the
left component is decreasing as is the efficiency.
Now, if I compare the first paragraph to the second, at no time during
the rotation, during the time the "right" component could be said to
dominate, is the propellor ALSO at maximum efficiency. Whereas at the
same time on the other side when the "left" component could be said to
be dominate (is that the word I want?) the blade is a maximum efficiency
all the way to 180* and then begins to decrease as the "left" component
decreases ....... LOL PROPWALK.

Sure, if you say it that way it's easy. But the underlying basis of the
paragraph above is wrong. While it can contribute to prop walk, the
difference between the left and right components during the entire 360
degree trip is significantly less then one might assume if they took the
paragraph above at face value.

No. The "left" component is decreasing as you go from 180-270, but the
right component (from 270-000) is staying relatively low (compared to
the left component from 090 t0 180) because the direction is up toward
the surface at the same time as it is to the right, rather than
(090-180) down toward "solid" water and to the left.


That's just not true if the blade is not breaking the surface. The
minimum efficiency occurs when the water column the blade is pushing
against is at a minimum. The length of the water column is that of a
line drawn perpendicular to the blade from the blade to the surface.

The blade/ water column does not have to break the surface, it just has
to be able to "bulge" it.

True. But the only time what you said above is true is if the blade
*is* breaking the surface. If not, then the minimum efficiency point is
beyond 270 degrees.

Just for the hell of it, I came up with a formula for the length of the
column of water the blade is pushing against when given the angle of
it's rotation during the portion from 270 to 360 degrees.

Ouch, math. I've hated math since they made me do celestial and Great
circles, long hand .... no slide rule, even.
At any rate, I'm going to have to print this out and study it, but on
[...]
began to increase. Naturally, as you got close to the hub the column
resumed increasing sooner.
The problem with this from my standpoint is that at no point is the
column as large for the majority of the blade during this entire (for
the most part) as it is between 090-180 which would be it's opposite
balancing blade.

You have to open up your U so that it's more like a V if you want to see
the effects of prop wash taking the path of least resistance. If you
do, you'll see that the efficiency begins to drop before the blade gets
to 180, just like it wasn't at max until some angle past 0.

Um, yes you can. In fact, you MUST. You cannot just look at certain
parts of the rotation while ignoring other parts. You have to look at
the balancing forces the entire way around. You MUST balance all of the
horizontal components against eachother and see what the resulting
horizontal force component is. There are horizontal components all the
way around except for when the blade is exactly at 90 or 270 degrees.
So you must consider what's happening all the way around and find what
portions balance out the other portions and what's left over that
doesn't balance out.

OK let me rewrite: We are discussing a rotating machine which creates
thrust. At a minimum this is being done with 2 blades for balance, and

Actually, it can be done with 1 blade if you balance the blade with a
"blob" of heavy material on the other side. A 1 blade prop is the most
efficient since it's not in the wash of any other blades.

in this case you would have to compare what is happening at one blade
with exactly what's happening with the one 180* opposite to understand
the net results.
However, boats can have 2,3,4,5,6 bladed props, in which case you would
need to compare the total constant effect as one sum of effect for the
entire rotation, but not side by side effects.

The easiest way to do it is to just look at 1 blade and sum all of the
horizontal components for it's entire rotation. You can't just look at
what's going on in the "quadrant of interest." To do the summing of
components, it's easiest to do a comparison of equal and opposite forces
and cancel them out and see what's left that's not equal and opposite.
I.e., look at the quadrants where there's only a motion right and see
how much of an opposing force you can cancel out from a corresponding
quadrant where there's a motion left.

The quadrants that correspond most are those from 0-90 vs. 90-180 where
the vast majority of the horizontal forces (but not all) cancel out and
from 180-270 vs. 270-360, where again, the vast majority of the
horizontal forces (but not all) cancel out.

The "length" of the water column from 090-180, is far greater than the
"length" of the water column from 270-360. The body (propellor pitch) is
constantly changing direction of "push".


While that's true, it's also true that the "length" of the water column
from 0-90 is far greater than the "length of the water column from 270
to 0. So what?

Actually, the length of the water column is greater from 000-180 than it
is from 180 to 000, but that puts us back to the beginning and ignores

Well of course. But again, so what. You haven't broken it down far
enough to get any significant information if you look at those large
spans of angle because there's a cancelling right and left component in
both 0-180 and 180-0.

all the variables of direction, push, etc., we've introduced.

Yup.

But for a prop that's very near or at the surface, those forces don't
balance out. That's because as the prop gets closer to the surface the
balance point (where the force is at a minimum due to minimum efficiency
because of the smallest water column before you get to air) move further
around the rotation. In this case, the force at 280 /= the force at 350
and you have a net sideways force.

True, but that sideways force does NOT equal the force at 100-170,
because the column of water above it is less than the column of water
below 100-170.


Which is completely irrevelant because the sideways force from 100-170
is balanced by the equal and opposite sideways force from 10-80. Again,
you're not considering the entire rotation and what parts of it balance
out what other parts and what's left over after the balancing act.

No, I'm considering the entire rotation, but for the basic cause I'm
sticking with even number blades such as two/four where you must compare
the opposites: Two bladed - compare 10-80 to 190-260, four bladed same
as two, but add, 100-170 compared to 280-350.
G now make odd number blades and my feeling is you have to compare all
at the same time for wherever they are in rotation, but obviously the
effect (net) come out the same.

No, you don't really have to compare blades at all as long as you
believe the same thing is happening to each blade. The only thing you
might have to worry about the number of blades for is where the wash of
one affects the one behind it (but again, that should be the same for
all blades as long as the prop is symmetrical, which it likely is) and
because the total prop walk will be higher the more blades you have.
That's because, as you add more blades you're not adding more efficiency
and the marginal increase in thrust from each additional blade is
smaller the more blades you have. But the prop walk effects we're
talking about here seem like they would be cumulative and not have a
lower marginal effect for each additional blade.

difference.)the blades are still pushing back, but there is no net
effect (arguably) which we can readily apply to "propwalk". Instead .....

4. Concentrate on the quadrants 090-180 and 270-000. From 090 to 180 the
blade is pushing back against a relatively solid column of water, down
against a relatively solid column of water and increasingly LEFT against
a relatively solid column of water. During this entire quadrant of
rotation, the blade is at maximum efficiency...... BUT, from 270-000 the
blade is pushing back relatively nearer the surface, up toward the
surface, and RIGHT towards and relatively close to the surface, where it
can and does break the surface or at least bulge the surface. So.....


And the quadrant from 0 to 90 exactly balances out the quadrant from 90
to 180. And the quadrant from 180 to 270 "almost" exactly balances out
the quadrant from 270 to 0. The smaller the ratio of prop size/prop
depth, the closer those quadrants (180-270 and 270-0) will balance out.

NO,NO,NO You cannot compare 000-090, to 090-180 or 180-270, to 270-360.
You Must compare opposites !! 000-090 and 180-270 or 090-180 and 270-360.


YES, YES, YES you can compare 0-90 to 90-180 because they ARE opposites
to each other in the sideways direction.

No, because they are not happening at the same time.... do that and
that boat will bounce left and right

Yes, because we're only looking at the quadrants for a force
calculation. In trying to solve the problem, there's no difference in
the result if you look at one blade at a time and sum it's forces for
the entire rotation vs. trying to see what's happening at any given prop
angle and worrying about what's happening to all the blades. It's just
much easier to concentrate on one blade the whole way around if you want
to come up with a correct answer. It may be the fact that you're trying
to think about what's happening to all the blades simultaneously that's
leading you astray.

Oh, and if your theory is correct, the boat will feel like it's bouncing
left and right because the prop walk force will be pulsating.

But you can't do that and still come up with a correct answer. You have
to balance out the sideways force for the entire rotation. And the
easiest way to do that is to first take out the parts that completely
balance each other and then look at what's left.

If you view it from your point of view, correct. However, from mine, the
net "right" never exceeds the net "left" because the net "right"
occurs when the propellor blade is at reduced efficiency while the net
"left occurs during maximum efficiency.

That may not be true once you're underlying foundation is butressed.
I.e., once you consider the decreasing efficiency as the blade travels
from 90 to 180 due to the "leakage" part of your theory. Once you
consider that the min/max efficiency points are not at 270/90 and will
change where they are at different depths.

Even just applying your theory of "leakage" and "column length" is more
complicated then you're making it out to be. And while I can agree that
there is a net sideways force from your theory, it's much smaller than
you're envisioning and does nothing to explain prop walk once a prop
gets far enough away from the surface that the depth to diameter ratio
is very large.

That doesn't really matter since it's not important as long as you
realize that the "solidity" of the column of water (if there is such a
thing) is the same for the same angle to the right vs. to the left of
the prop.

G Sorry, but as you can see, I totally disagree with that.

I'm not sure how you can. How is it possible that, from the same point
under the water, the "solidity" is different if you go up 45 degrees to
the right vs. going up 45 degrees to the left? If you can disagree with
that underlying foundation then all bets are off and anything is
possibly.

Not totally. But it doesn't take going very far down before everything
*nearly* balances out left and right due to the effect we're talking
about.

No, if that were the case, then your boat wouldn't experience propwalk.


Yes it would. Just not totally from the effect you're postulating.
But it would contribute.

I don't think you will ever reach that balance point on a small boat
floating on the surface,or even a ship .... goes back to my question

I don't think so either. In fact, it will never reach a balance point
no matter how low you go. But it doesn't take going very deep (i.e.,
only a few feet with a 1 foot prop) for the effect to balance out enough
that the reverse thrust should easily be able to overcome it. It
doesn't though. Which means there must be some other, or many other,
causes of prop walk that all combine.

about subs. Do they experience propwalk at a depth of 1,000 feet? If
not, at what depth does it stop ... sure would confirm or deny our
points, wouldn't it?

It would help.

Steve

For the first part of this response, I'll comment on some specific
items, as they are side issues.
For the part on "propwalk theory", I'm stepping back and taking a look,
because as Shen has said our major stumbling block revolves around "prop
efficiency" at certain angles and our individual visualization of what's
occurring.

Steven Shelikoff wrote:

The pivot point has changed position. Now, rather than being @1/3rd of
the way from the bow, it's @1/3rd of the way forward of the stern.


Doesn't matter. It's still within the confines of the boat.

It mainly matters in a sense of maneuvering to visualize the point about
which the boat is turning, so you can understand what is happening at
each end. I find a number of people who understand the 1/3 aft for
ahead, but forget to change this to 1/3 forward, for astern.


But this is not true in reverse since you don't have the propwash going
over the rudder.

You don't have the propwash over the rudder, but you still have the
"propwalk steering component" which can turn the boat more easily and
sooner than the astern thrust can start moving the mass of the boat astern.


How many blades does the prop you're watching have? If it's one or two
then maybe you're not being fooled. If it's 3 or more, especially if
it's 4 or more, then you can't separate what you think you may be seeing
one blade doing from what the rest are doing because they are too close
to each other in angle. For instance, if you're watching a 4 blade
prop, what "splashing" you think you're seeing from the blade at TDC
starting to move over and down could just as easily be from the blade at
270 and starting to come up and over.

They're apt to have anywhere from 3-6. This view has occurred over many
years, watching many props, some on diesels some on steam turbine
(sitting on a dock watch the warm up spinning of the prop fwd and astern).
You need to remember, these are props with @ 25' diameters turning,
initially at maybe 15-30 RPM.
G Call me an oddball for standing there watching/studying this, but
then again, I like watch porpoise swim in the bow wave and seeing how
they move with it.

On to propwalk ....

In reading the last few post, it's obvious we are now just chasing each
other in circles and resolving nothing.
A main issue is prop efficiency at various angles of rotation.
My opinion:
For a boat floating on the surface, when discussing prop efficiency we
must consider that the prop is affecting two mediums -water and air.
The prop is at maximum efficiency when it is affecting water alone and
at less than maximum efficiency when it affects air and water.
Is the depth of the hub of the prop important? To a degree, yes, but for
the boats we are discussing, the effect is constant, it's degree varies.
For instance, hub at surface half of prop out of water (Arneson drive);
blade tip at surface, hub submerged; blade totally submerged..... the
blade is more efficient, overall, in one semicircle, than the other, in
all of these conditions, in my opinion.
During the rotation of the blade, it is my opinion that the blade is at
the most maximum efficiency between the angles 045* and 225* of rotation
because it is acting against only the single not compressible medium of
water.
However, between the angles 225* and 045* the blade is operating at LESS
than maximum efficiency because it is acting upon both water AND air.

BG Until we resolve this issue between us, we are just butting
heads.....and I'm running low on paper ....

otn