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Angle of prop shaft - theoretical question.
On Wed, 09 Jun 2004 04:56:51 GMT, otnmbrd wrote:
Steven Shelikoff wrote: *The* root cause is a net force sideways. G read the next part carefully [...] Let's now look at prop efficiency (these are MY views based on what I see, read and feel). If we follow the rotation of a RH fixed pitch prop, looking at it from astern starting at top dead center (ooo*) the particular blade is pushing water to the right (and back ...always back, but we will ignore that component for this discussion) at minimal efficiency. this efficiency, however, is increasing as the propellor turns towards 45* and the direction is changing to a increasingly down direction. As the blade reaches 45*, efficiency is close to maximum and from here the direction is more down than to the side. Someplace just prior to 90* the efficiency becomes maximum and as the blade rotates toward 135* the angle of push changes to the left You haven't explained why the prop is less efficient at 0 and gains efficiency on it's way from 0 to 90. Imagine the blade is just beneath the surface Again, ignoring the after component, up until the blade reached o, coming from 315, the pitch of the blade was not pushing towards a solid wall of water, it was pushing the water up and to the right into air ... it was less efficient, compared to it's opposite blade which was rotating from 135 to 180 which was pushing down and to the left against solid incompressible water. As the blade starts rotating to 090 it is pushing to the right and as it rotates, also begins to push down(into more solid water) , so that it's efficiency begins to increase as the angle it's pushing down, increases and it stops pushing to the right. There's no such thing as "more solid water" unless you're talking about ice. Once the blade goes past 0 degrees, all the way from there until 180 degrees it's pushing against the same thing. The water may be minutely more dense and under a tiny bit of less pressure for the 0 to 90 part of the trip than from the 90 to 180 part of the trip. But that has absolutely nothing to do with your theory of the water column being backed by air vs. something else. Your theory gives no reason why there's less efficiency for the 0 to 90 part of the rototation than for the 90 to 180 part of the rotation. It also gives no reason why there's a difference in efficiency between the 180 to 270 part of the rotation than from the 270 to 360 part since it's backed by air for that entire trip (unless there's hull overhang). To explain that, you have to start looking at the amount of water before you reach air. There's also one more thing you haven't thought about and that's the fact that we're dealing with 2 interfaces on the way down as well as on the way up. On the way up, the water column being pushed by the prop can bulge the air at the surface. On the way down, the water column being pushed by the prop can bulge the bottom, especially if the bottom is made of something like soft mud. It may provide more support than air at the surface, but the difference isn't as large as if there truly was a single interface on the way down. I can think of some reasons why that may be correct. But the reason given having to do with a column of water only backed by air and a bulge at the surface isn't it. How do you know that? Because there's no difference in the "backing" of the column of water for the entire travel of the prop from 0 to 180 degrees. It's not backed by air for any portion of that rotation. It's backed by practically non-compressable water against the bottom of the sea for that entire period of the rotation. So that theory provides no explanation for why there would be a difference in efficiency from 0 to 90 degrees vs. 90 to 180 degrees. That's because the whole way from 0 to 180 degrees there is either an infinite column of water (right a 0 and 180) or the column of water is supported by the sea floor (everywhere else between 0 and 180) Not initially, but shortly after 000* it begins to be and increases. Yes, initially and for the entire trip it doesn't change. It doesn't increase at all beyond 0. It's the same the whole way to 180. Why aren't you considering 180 - 000? Looking at the prop just beneath the surface, are you saying that the pitched blade is pushing against a solid column of water as it goes from 180* and approaches the surface at 000*? Your theory that the difference in efficiency is due to the column of water being backed by air vs. not backed by air doesn't explain any difference during the way up either since it' backed by air the whole way from 180 up to 0. You have to start looking at the amount of water before you get to air. [...] Also, so there's no misunderstanding, remember, my feeling about props efficiency decrease on it's upward rotation is about the prop pushing water up and into air, losing efficiency. .... and most importantly, these are my views/visualizations .... G yours may vary. I realize that. It just doesn't support your discussion about what happens to the prop efficiency from 0 to 180 degrees. All it does is explain why the prop is less efficient on the upward part of it's trip vs. the downward part of it's trip which creates a net upward force off the centerline, which lifts the stern and lists the boat. It does nothing to explain why there's a net sideways force (if there's no overhang, which I think we've agree is the root cause of prop walk that must be explained. If you can see a net upwards force, I'm halfway there. I'm agreeing that if your theory is correct, i.e., that there is a difference in efficiency whether the column of water being pushed by the prop is backed up by air vs. something other than air, then there will be a net upwards force. You agree that the blade is more efficient from 000-180 than from 180-000? If so, forget those numbers, consider the prop just beneath the surface (to help the visualization) and look at the blade rotation from 090-270 and 270-090, considering a 12 inch dia prop on a boat in 3,000 feet of water. At 090 the blade is pushing directly down. as the blade rotates past 090, it continues to push down, but also begins to push to the left. As the blade rotates toward 180 the downward push decreases as the sideways push (to the left) increases,until you reach 180 where the blade is pushing directly left. During this time, the blade has been pushing against a solid column of water, 3,000 feet deep (maximum efficiency). As the blade passes 180, it continues to push left, but also begins to push up (against a 12" column of water), efficiency decreases and as the blade approaches 270, the upward component increases as the left component decreases until you reach 270* where the blade is now pushing directly up against 6" of water. From 090 to 180 the blade was pushing at maximum efficiency in an ever increasing left component. From 180 to 270 the efficiency was decreasing at the same time as the left component was decreasing. Now, at 270, the blade is pushing directly up against 6" of water and as soon as it passes 270 begins to push to the right in a decreasing column of water. As before, as the right component increases the up component decreases, until it disappears at 000* in zero inches of water....the Move the blade down in the water a few feet and this is no longer true. I.e., if the blade is 12" and it's 3 feet below the surface, it's pushing against the smallest column of water just past 270 degrees. As it rotates from just beyond 270 until it reaches 0 degrees, the size of the water column before it reaches air is ever increasing, thus increasing it's efficiency during the trip from 270 back to 0. prop is relatively inefficient during this whole period or arc of rotation, compared to it's opposite 090-180 As the blade passes 000* it is pushing directly right and as it rotates past 000* begins to push downward (and decrease pushing right) and consequently begins to increase in efficiency as it gets further down and into more solid water, until we once again reach 090*. Again, not true. Once it passes 0 degrees, there's no difference in what it's pushing against all the way until 180 degrees. So you can forget that portion of the trip giving any net force in the sideways direction. If we consider that we've got a 2 bladed prop, blade A passing between 090 and 270 started out at maximum efficiency and continued at that to 180* where it's efficiency began to decrease. Blade B passing between 270 and 090, on the other hand started out at poor efficiency, which it maintained until 000, where it started to pick up efficiency, going to maximum at 090. No. Actually, if the blade is at the surface, the efficiency decreased from 270 to 0 (if we're now talking about the amount of water before air) and then got to maxumum right at 0 and stayed there until 180. If the blade is a few feet down, it was at minimum efficiency just past 270 and then gains efficiency on it's way to 0 degrees. The net greater push is to the left (hull goes right) .... propwalk G I've exaggerated numbers, for clarity (at least, for my attempt at it) and don't really know how I could explain this in another way to make the point I'm trying to get across, clearer. The problem is that geometry doesn't support what you're trying to show. To support your theory, you're stating assumptions about what the prop is doing during it's rotation and for several parts of the rotation, those assumptions are just wrong. Also, you're assuming more than just a difference in efficiency in water being backed by air vs. non-air. You're also assuming that the efficiency changes due to how much water there is before you get to air. Except for a very tiny amount of water (i.e., right at the surface) that's not much of a factor. That's because right at the surface, the size of the prop is huge relative to the depth of the prop. So there's a large relative difference between the amount of water above the blade when it's at 180 degrees vs. 0 degrees. But move a small prop down a few feet and that becomes less and less of a factor. Steve |
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