mlines
Well-Known Member
What stops a submarines propellor from staying still and the whole submarine rotating instead (is it just the shear mass and the "fin" providing stability)
Prop walk. Pysical reason why does this happen
- Thread starter Uricanejack
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What stops a submarines propellor from staying still and the whole submarine rotating instead (is it just the shear mass and the "fin" providing stability)
JayBee
Well-Known Member
What is the difference in water density across the diameter of a yacht's fully submerged propeller disc? Not much. It is probably safe to assume that propwalk is not caused by differing water densities, or precession.
Prop walk or transverse propeller thrust is caused by the helical discharge from the propeller and its interaction with the rudder and hull.
With an ahead movement of a right handed propeller:
Transverse thrust is of much greater significance when using an astern movement.
The helical discharge from a right handed propeller working astern splits and passes forward towards either side of the hull. In doing so it behaves quite differently. On the port quarter it is inclined down and away from the hull whilst on the starboard quarter it is directed up and on to the hull. This flow of water striking the starboard quarter can be a substantial force capable of swinging the stern to port, giving the classic kick of the bow to starboard.
(With acknowledgements to The Shiphandler's Guide, by Captain RW Rowe, FNI, published by The Nautical Institute)
This was extracted in part from a book written for big ship handlers, but the dynamics are about the same for most yachts, in my experience.
Prop walk or transverse propeller thrust is caused by the helical discharge from the propeller and its interaction with the rudder and hull.
With an ahead movement of a right handed propeller:
- The helical discharge from the propeller creates a larger pressure on the port side of the rudder.
- A slight upward flow from the hull into the propeller area puts slightly more pressure onto the down sweeping propeller blades.
- The net result is a tendency for a right handed propeller to give a small swing to port when running ahead.
Transverse thrust is of much greater significance when using an astern movement.
The helical discharge from a right handed propeller working astern splits and passes forward towards either side of the hull. In doing so it behaves quite differently. On the port quarter it is inclined down and away from the hull whilst on the starboard quarter it is directed up and on to the hull. This flow of water striking the starboard quarter can be a substantial force capable of swinging the stern to port, giving the classic kick of the bow to starboard.
(With acknowledgements to The Shiphandler's Guide, by Captain RW Rowe, FNI, published by The Nautical Institute)
This was extracted in part from a book written for big ship handlers, but the dynamics are about the same for most yachts, in my experience.
nigelmercier
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I didn't say it was what causes tides, it what causes the second bulge to be on the far side. Brian Cox explained it like this, so I'm happy to take his word for it.
marklucas
Well-Known Member
This is what I have always taught. And once the boat starts moving backwards flow is established over the rudder and so control over the prop-kick can be gained.
charles_reed
Well-Known Member
If you want, you can induce propwalk, in reverse. Just use frequent bursts of forward and reverse, dwelling more in reverse, and you can actually move the boat sideways.
I would agree, however that the Autoprop exhibits far less prop-walk than a fixed two-blade prop.
Personally I find prop-walk a very useful adjunct for berthing.
marklucas
Well-Known Member
Tides
As posted last June:
Hopefully this will clarify things. Blame my brother for asking me to explain why, his pulling holes in the basic explanation (he's got an engineering PhD) and a desire to finally get to the bottom of it all.
So we know that tides can happen twice a day (but not always). And that Newton realised it was the rate of change of gravity that caused the tide, not absolute gravity itself. But how do we put all this together and make an argument that holds water.
Now, the simple, reasonably correct explanation of the tides is as follows:
The diagram below represents the earth-moon system. The shaded area around the earth represents the earth's oceans (we assume for now a uniform ocean with no continents). Point A is at the surface of the ocean nearest the moon. Point C is at the surface of the ocean farthest from the moon. Point B is the centre of the earth.
The arrows in the top diagram pointing toward the moon represent the force of the moon's gravity at these three points. Since the force of gravity depends on distance, point A is attracted to the moon most strongly, point C least strongly, and point B at intermediate strength.
The lines leading downward from each point show how each point would move under the influence of the moon's gravity. Point A moves farthest toward the moon, B, not as far, and C least of all. The result is as if the ocean were bulging out toward and away from the moon. Thus, as the earth rotates under this bulge, any point on earth has two high tides and two low tides each day.
This is (or is similar to) the explanation you would be given on a theory course and is quite adequate for the Yachtmaster exam (practical or theory).
Now, we will consider certain facts that will gradually make our understanding more accurate.
First, it turns out that the direct vertical pull of the moon on the ocean (as at point A) is nowhere near strong enough to raise the tides we observe. The forces that actually move the oceans are the forces acting at points D and E in the diagram below. These act horizontally, not vertically, and moving water horizontally is a lot easier than pulling it up vertically.
But, where is the actual centre of the system from which the Earth is experiencing the rotation causing the force?
It’s tempting to think that this is the moon, but the moon and the Earth exist in a steady state – the centre of which is the centre of mass of the two – known as the barycentre. To find the location of the barycentre is quite simple: imagine that the Earth and the moon are joined by a rigid pole – where would you position a support to balance the pole? This is just a simple weight balance / leverage problem, the distance of the barycentre from the centre of mass of the smallest object, simply being the ratio of the largest mass to the total mass of the two objects multiplied by the distance between the two centres of mass.
For the Earth-Moon system, this happens to be inside the Earth. So we can draw a plan diagram of the Earth and moon and their barycentre that looks like this:
The lines from E1 and E3 are the pole we mentioned previously. As you would expect we can see that the centre of gravity of the Earth (E 1,2,3) rotates around the barycentre, and this gives the eccentric movement of point X at the Earth’s surface. The Earth however still revolves around the Earth’s centre every 24 hours (giving us a clue that the Earth’s revolution does not cause the tide – merely the location of the highs and low tides).
Now we can see that the earth rotates around the barycentre we can expect there to be a centrifugal force, Fc, acting on the centre of the Earth (just like on a fairground ride). This is the force that the centre of mass of the Earth experiences as it rotates around the barycentre. However, as we have derived a centre of gravity for the Earth, any force acting on the centre of gravity of the Earth will act on the whole of the Earth and be the same at any point.
Taking a cross-section view and drawing in the moon’s force of gravity Fg we get:
The key thing to notice here is that at X the forces Fg and Fc are equal – they must be otherwise the Earth-moon system would not be stable (a bit like two ice-skaters holding hands and spinning – if the forces from each partner weren’t’ the same then they would wobble and then fall over). We know (intuitively and experimentally) that as you get nearer to the moon, gravity gets stronger, so at B, Fg must be greater than Fc creating a resultant force towards the moon. At A the opposite is true - Fg is less than Fc and so there is a force away from the moon.
This means that the force producing the tides is created by the rate of change of the moon’s gravity, not the absolute strength of it. Thus if the Earth were closer to the moon, the tides would be stronger (because gravity is proportional to the inverse square of distance, so the rate of change with distance will be greater closer to an object). This is shown in an extreme case between Jupiter and its innermost moon, Io, where the tides create earthquakes on Io. For the Earth’s moon, the effect is less marked, but it is this force that means the same side of the moon always points towards the Earth (called tidal locking).
As the force of gravity is proportional to the inverse square of distance, we can show by differentiation that the tidal force created by the rate of change of gravity is proportional to the inverse cube of distance.
Now the lunar day (also known as a tidal day) at 24.84 hrs, exceeds the solar day (24 h) since the moon is revolving around the earth with a period of 27.32 days. This adds 50 minutes a day to the time taken for the same point on Earth to come directly under the moon again, as shown below:
As we have two tidal bulges, then the tides should be 12 hours and 25 minutes apart, and for those parts of the world that have semi-diurnal tides, this is indeed the case. It is worth noting that 27.3 days is not the time between new moons: due to the Earth’s own movement along its orbit around the sun – the moon’s state is caused by the relative positions of the moon and sun as viewed from the Earth, not where the moon is around the Earth – this has a period of 29.5 days.
We can repeat this exercise with every other body in the solar system and so calculate the tidal force created by it, as well as the effects caused by the elliptical orbits of planets, moons etc. However, it is fairly obvious that after the moon the next major body creating tides will be the sun, due to its mass – which after taking into account its distance creates tides 44% of the combined tidal effect of the sun and moon. Clearly, when they line up we will get the two forces acting together giving us Spring tides. But always remember there are many more factors involved, not least the elliptical nature of many orbits and that many planets / moons exist at an angle to the plane of the overall solar system.
OK, so that’s the force and timing that creates the tides, but what are its effects?
At any point on the earth's surface, the tidal force produced by the moon's gravitational attraction may be separated or "resolved" into two components of force - one in the vertical, or perpendicular to the earth's surface - the other horizontal or tangent to the earth's surface. This second component, know as the tractive ("drawing") component of force Ftr is the actual mechanism for producing the tides, shown at C.
The force is zero at the points on the earth's surface directly beneath and on the opposite side of the earth from the moon A and B (since in these positions, the lunar gravitational force is exerted in the vertical - i.e., opposed to, and in the direction of the earth-gravity, respectively).
Any water accumulated in these locations by tractive flow from other points on the earth's surface tends to remain in a stable configuration, or tidal "bulge”. Indeed, in the open ocean, the actual rise of the tidally induced wave crest is only one to a few feet. It is only when the tidal crests and troughs move into shallow water, against land masses, and into confining channels, that noticeable variations in the height of sea level can be detected.
But why do Spring tides in Europe arrive 2 days after full / new moon?
Well, only in the Southern Ocean can the tidal bulge move unhindered by bits of land and so it builds up a single continuously moving wave. It takes two days for this wave to travel up the Atlantic and reach Europe, giving additional emphasis to the locally created tides, but which are not able to develop a continuous wave.
Indeed, land can create basins in which the sea moves. Now each basin will have a natural period based on its size and depth (just like carrying a washing-up bowl full of water). If the period of the basin were to be around 12 hours 25 minutes then it would synchronise with the semi-diurnal tides and the results would be large as each tide would reinforce the existing wave. This is indeed what happens in the Bay of Fundy in Nova Scotia which experiences the world’s largest tides of 16m as it has a period of 13 hours.
Conversely, the Pacific basin has a natural frequency of 30 hours, meaning that the natural period and the tide are not in synch and so most of the Pacific only has one tide a day, or at best mixed tides (one tide is much larger than the other).
As posted last June:
Hopefully this will clarify things. Blame my brother for asking me to explain why, his pulling holes in the basic explanation (he's got an engineering PhD) and a desire to finally get to the bottom of it all.
So we know that tides can happen twice a day (but not always). And that Newton realised it was the rate of change of gravity that caused the tide, not absolute gravity itself. But how do we put all this together and make an argument that holds water.
Now, the simple, reasonably correct explanation of the tides is as follows:
The diagram below represents the earth-moon system. The shaded area around the earth represents the earth's oceans (we assume for now a uniform ocean with no continents). Point A is at the surface of the ocean nearest the moon. Point C is at the surface of the ocean farthest from the moon. Point B is the centre of the earth.
The arrows in the top diagram pointing toward the moon represent the force of the moon's gravity at these three points. Since the force of gravity depends on distance, point A is attracted to the moon most strongly, point C least strongly, and point B at intermediate strength.
The lines leading downward from each point show how each point would move under the influence of the moon's gravity. Point A moves farthest toward the moon, B, not as far, and C least of all. The result is as if the ocean were bulging out toward and away from the moon. Thus, as the earth rotates under this bulge, any point on earth has two high tides and two low tides each day.
This is (or is similar to) the explanation you would be given on a theory course and is quite adequate for the Yachtmaster exam (practical or theory).
Now, we will consider certain facts that will gradually make our understanding more accurate.
First, it turns out that the direct vertical pull of the moon on the ocean (as at point A) is nowhere near strong enough to raise the tides we observe. The forces that actually move the oceans are the forces acting at points D and E in the diagram below. These act horizontally, not vertically, and moving water horizontally is a lot easier than pulling it up vertically.
But, where is the actual centre of the system from which the Earth is experiencing the rotation causing the force?
It’s tempting to think that this is the moon, but the moon and the Earth exist in a steady state – the centre of which is the centre of mass of the two – known as the barycentre. To find the location of the barycentre is quite simple: imagine that the Earth and the moon are joined by a rigid pole – where would you position a support to balance the pole? This is just a simple weight balance / leverage problem, the distance of the barycentre from the centre of mass of the smallest object, simply being the ratio of the largest mass to the total mass of the two objects multiplied by the distance between the two centres of mass.
For the Earth-Moon system, this happens to be inside the Earth. So we can draw a plan diagram of the Earth and moon and their barycentre that looks like this:
The lines from E1 and E3 are the pole we mentioned previously. As you would expect we can see that the centre of gravity of the Earth (E 1,2,3) rotates around the barycentre, and this gives the eccentric movement of point X at the Earth’s surface. The Earth however still revolves around the Earth’s centre every 24 hours (giving us a clue that the Earth’s revolution does not cause the tide – merely the location of the highs and low tides).
Now we can see that the earth rotates around the barycentre we can expect there to be a centrifugal force, Fc, acting on the centre of the Earth (just like on a fairground ride). This is the force that the centre of mass of the Earth experiences as it rotates around the barycentre. However, as we have derived a centre of gravity for the Earth, any force acting on the centre of gravity of the Earth will act on the whole of the Earth and be the same at any point.
Taking a cross-section view and drawing in the moon’s force of gravity Fg we get:
The key thing to notice here is that at X the forces Fg and Fc are equal – they must be otherwise the Earth-moon system would not be stable (a bit like two ice-skaters holding hands and spinning – if the forces from each partner weren’t’ the same then they would wobble and then fall over). We know (intuitively and experimentally) that as you get nearer to the moon, gravity gets stronger, so at B, Fg must be greater than Fc creating a resultant force towards the moon. At A the opposite is true - Fg is less than Fc and so there is a force away from the moon.
This means that the force producing the tides is created by the rate of change of the moon’s gravity, not the absolute strength of it. Thus if the Earth were closer to the moon, the tides would be stronger (because gravity is proportional to the inverse square of distance, so the rate of change with distance will be greater closer to an object). This is shown in an extreme case between Jupiter and its innermost moon, Io, where the tides create earthquakes on Io. For the Earth’s moon, the effect is less marked, but it is this force that means the same side of the moon always points towards the Earth (called tidal locking).
As the force of gravity is proportional to the inverse square of distance, we can show by differentiation that the tidal force created by the rate of change of gravity is proportional to the inverse cube of distance.
Now the lunar day (also known as a tidal day) at 24.84 hrs, exceeds the solar day (24 h) since the moon is revolving around the earth with a period of 27.32 days. This adds 50 minutes a day to the time taken for the same point on Earth to come directly under the moon again, as shown below:
As we have two tidal bulges, then the tides should be 12 hours and 25 minutes apart, and for those parts of the world that have semi-diurnal tides, this is indeed the case. It is worth noting that 27.3 days is not the time between new moons: due to the Earth’s own movement along its orbit around the sun – the moon’s state is caused by the relative positions of the moon and sun as viewed from the Earth, not where the moon is around the Earth – this has a period of 29.5 days.
We can repeat this exercise with every other body in the solar system and so calculate the tidal force created by it, as well as the effects caused by the elliptical orbits of planets, moons etc. However, it is fairly obvious that after the moon the next major body creating tides will be the sun, due to its mass – which after taking into account its distance creates tides 44% of the combined tidal effect of the sun and moon. Clearly, when they line up we will get the two forces acting together giving us Spring tides. But always remember there are many more factors involved, not least the elliptical nature of many orbits and that many planets / moons exist at an angle to the plane of the overall solar system.
OK, so that’s the force and timing that creates the tides, but what are its effects?
At any point on the earth's surface, the tidal force produced by the moon's gravitational attraction may be separated or "resolved" into two components of force - one in the vertical, or perpendicular to the earth's surface - the other horizontal or tangent to the earth's surface. This second component, know as the tractive ("drawing") component of force Ftr is the actual mechanism for producing the tides, shown at C.
The force is zero at the points on the earth's surface directly beneath and on the opposite side of the earth from the moon A and B (since in these positions, the lunar gravitational force is exerted in the vertical - i.e., opposed to, and in the direction of the earth-gravity, respectively).
Any water accumulated in these locations by tractive flow from other points on the earth's surface tends to remain in a stable configuration, or tidal "bulge”. Indeed, in the open ocean, the actual rise of the tidally induced wave crest is only one to a few feet. It is only when the tidal crests and troughs move into shallow water, against land masses, and into confining channels, that noticeable variations in the height of sea level can be detected.
But why do Spring tides in Europe arrive 2 days after full / new moon?
Well, only in the Southern Ocean can the tidal bulge move unhindered by bits of land and so it builds up a single continuously moving wave. It takes two days for this wave to travel up the Atlantic and reach Europe, giving additional emphasis to the locally created tides, but which are not able to develop a continuous wave.
Indeed, land can create basins in which the sea moves. Now each basin will have a natural period based on its size and depth (just like carrying a washing-up bowl full of water). If the period of the basin were to be around 12 hours 25 minutes then it would synchronise with the semi-diurnal tides and the results would be large as each tide would reinforce the existing wave. This is indeed what happens in the Bay of Fundy in Nova Scotia which experiences the world’s largest tides of 16m as it has a period of 13 hours.
Conversely, the Pacific basin has a natural frequency of 30 hours, meaning that the natural period and the tide are not in synch and so most of the Pacific only has one tide a day, or at best mixed tides (one tide is much larger than the other).
charles_reed
Well-Known Member
Nah! Nah! Its the effect of Saturn's contra-rotating rings
nigelmercier
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I retract my earlier suggested link, I recall seeing the water doing exactly this. Thank you.
Last edited:
nigelmercier
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Great explanation, and good to know that Brian Cox is still awesome.
AngusMcDoon
Well-Known Member
Density of water increases with depth (assuming constant temperature) at the rate of 0.46% per....kilometer!
Your statements above are true as density does increase with depth, but it's so small that it isn't the cause of propwalk.
Coaster
Well-Known Member
Thanks for this. I will give your method a try next year.
Uricanejack
Well-Known Member
Thanks for all your thoughts. I got a couple of excellent explanations of tides, which I quite enjoyed by the way.
Still not a lot wiser about prop walk.
Though some of you have some very intriguing ideas. My personal theory based on absolutely nothing. Is the difference in pressure depth increases? But it seams to small.
I liked the gyroscopic precession but not quite convinced.
The asymmetrical propeller diagram was very good and I liked that. I think it is very likely a contributing factor on many boats but it doesn’t explain pop walk when the shaft is level. You still get prop walk with a level shaft.
The water against the hull I’m not so sure.
The quoted paper from nautical institute is probably the nearest thing to an engineering or scientific explanation I have seen.
I suspect it’s a combination of more than one of these.
Still not a lot wiser about prop walk.
Though some of you have some very intriguing ideas. My personal theory based on absolutely nothing. Is the difference in pressure depth increases? But it seams to small.
I liked the gyroscopic precession but not quite convinced.
The asymmetrical propeller diagram was very good and I liked that. I think it is very likely a contributing factor on many boats but it doesn’t explain pop walk when the shaft is level. You still get prop walk with a level shaft.
The water against the hull I’m not so sure.
The quoted paper from nautical institute is probably the nearest thing to an engineering or scientific explanation I have seen.
I suspect it’s a combination of more than one of these.
Mel
Well-Known Member
I always understood that the prop walk was all about drag along the hull at low speed - we tend to only consider this when manoeuvring around tight spaces !
So as we all know there is no such thing as perfection in human construction technology, which can be seen if you measure the water line length of your hull – take a tape measure and get in your dingy measure the length at water line, bow to stern and you always find there is some difference between Port and Starboard ! Also consider that we tend to berth our boats in a normal orientation relative to the sun and we get different algae or other growth side to side which also has its effect.
Putting all this together provides you with the most obvious answer for this phenomena.
Phew that took some thinking about – my brain cell is really tired now !
So as we all know there is no such thing as perfection in human construction technology, which can be seen if you measure the water line length of your hull – take a tape measure and get in your dingy measure the length at water line, bow to stern and you always find there is some difference between Port and Starboard ! Also consider that we tend to berth our boats in a normal orientation relative to the sun and we get different algae or other growth side to side which also has its effect.
Putting all this together provides you with the most obvious answer for this phenomena.
Phew that took some thinking about – my brain cell is really tired now !
alan_d
Well-Known Member
I think your brain cell should recruit some helpers - nothing you have said explains why the direction of prop-walk is determined by the direction of rotation of the propellor.
SHUG
Well-Known Member
The explanation is completely correct. The Eart-Moon system rotates about the common centre of mass which is 1/3 of the radius from the surface of the Earth.Do not know if its true but I like that explanation, some how it works in my head. Although that is mainly holow so is not affect much by gravity
As for prop. walk I thought I understood it until I drew a diagram and found that on paper anyway , everything balances out. Mmmmmm!
DaveS
Well-Known Member
I was reading a concise and convincing explanation of prop walk the other week but now can't remember where it came from. Basically it was as some have said here: with a prop on a conventional inclined shaft running astern the water flows hitting the quarters of the hull are quite different, hence the sideways thrust. Going ahead, or in either direction with a horizontal shaft like a sail drive, most flow misses the hull and the effect is greatly reduced. All seems quite simple.
I do recall "paddle wheel effect" explanations being given on RYA courses many years ago which to my shame I accepted uncritically at the time. A bit more rational thought as displayed by some on this thread makes it obvious that that explanation is incorrect since (a) it does not explain why prop walk is much greater going astern, and (b) the water density difference between top and bottom of the prop is negligible. (A real paddle wheel works because it is only in water for half or less of its depth.)
I do recall "paddle wheel effect" explanations being given on RYA courses many years ago which to my shame I accepted uncritically at the time. A bit more rational thought as displayed by some on this thread makes it obvious that that explanation is incorrect since (a) it does not explain why prop walk is much greater going astern, and (b) the water density difference between top and bottom of the prop is negligible. (A real paddle wheel works because it is only in water for half or less of its depth.)
Jaguar 25
Well-Known Member
Water density - prop walk
Could it be that there is an effective difference in water density above and below the prop centreline related to the amount of air entrained in the water due to the effect of the prop????? So, more air entrained nearer the surface so the prop pushes harder against the denser water/air mixture below the prop centreline.
Could it be that there is an effective difference in water density above and below the prop centreline related to the amount of air entrained in the water due to the effect of the prop????? So, more air entrained nearer the surface so the prop pushes harder against the denser water/air mixture below the prop centreline.
charles_reed
Well-Known Member
It's misleading to call it prop-walk (though we do and understand the phenomenon).
It's really hull movement caused by unequal forces from the prop stream.
It's really hull movement caused by unequal forces from the prop stream.
DaveS
Well-Known Member
If that was the correct explanation then you would expect a similar effect going ahead and astern - and that's not what happens.
marklucas
Well-Known Member
Because:
- Forward thrust immediately creates flow over the rudder and so provides steering. In reverse, the water is sucked from everywhere behind the boat - so very little flows over the rudder, until the boat actually starts moving. With your boat tied up see how quickly you get steerage in forward and reverse.
- In forward the upward flow of water has less hull to act on and it is shallower, so further away.
