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