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)