After 4 Years Of Fighting - C&RT Settle Out Of Court

[Arthur Marshall]

 

 

Surely of the LCF was aft of midships, raising the bow 1" would result in the stern dropping proportionally less than 1".

 

For the stern to drop more than the amount the bow is raised, the LCF would need to be forward of midships.

I think you're right. When my water tank is empty, the front end rises by about four inches and the back only goes down an inch or two at most. I was confusing the LCF with where the boat pivots when you turn it... I think. There again, I may just be confused. Wouldn't be the first time.

[mross]

I think you're right. When my water tank is empty, the front end rises by about four inches and the back only goes down an inch or two at most. I was confusing the LCF with where the boat pivots when you turn it... I think. There again, I may just be confused. Wouldn't be the first time.

You and MtB are correct and I was wrong.

[magictime]

Not getting very far are we?

 

OK, how about this: the water level drops by a few inches. The boat heels to starboard by a few inches and the stern drops in the water by a few inches. The boat is now wedged against the bottom gate, the lock wall and the other boat in the lock in such a way that when Ken opens the top gate and the water level starts to rise again, the boat doesn't rise with it along its full length as it should. Maybe only the bow is able to rise, and as it does the stern is forced even further down into still-rising water.

 

Any takers?

[Mike Todd]

Agreed. LCF (longitudinal centre of flotation) is going to be well aft of midships due to weight of engine, batteries and fuel. More so if the fresh-water tank happened to be low. I can see the stern dropping more that 2" at the starboard gunwhale if port bow is raised 1".

But the LCF is only relevant as you are using it when the vessel is floating free. The issue here is that one end of the vessel was kept in a fixed position regardless of flotation considerations.

 

reductio ad absurdum argument: if you tied one end of a boat firmly to the land and then let sufficient water out (supposing that the lock was deep enough) so that the boat was entirely suspended out of the water, then there would be no sinking in the manner described.

 

The actual situation was probably somewhere in between and it ought to be a relatively simple task to estimate the likely behaviour - but I'm too much past A level to bother!

[Paul C]

The maths of the situation aren't massively complicated. When a boat is free floating, there is a bouyancy force acting upwards throuh the LCF (which is very similar to the LCB, for an unloaded boat) equivalent to the weight of the boat - its static, not a dynamic situation.

 

When a boat gets hung up at the front centre, some of the weight is taken by the point it hangs up, and some is supported by buoyancy. So, the amount by buoyancy reduces. Furthermore, since the boat is tilting more and more (the front is out the water, the back's in it more) the LCF also moves (because the LCF is a function of the underwater shape of the hull) backwards. Both the amount of weight supported by buoyancy, and the place it exerts this, change (for the better - ie to lessen loss of freeboard at the back (low) end). Imagine an extreme example, a crane picks up a boat at the front and suspends it 89.9deg vertical and its back end just touches the waterline. Just as it starts floating, a small force is applied at the back, but almost all the weight of the boat is still supported by the crane at the front.

 

Regarding the fact it was hung up at the side, its a similar bunch of calculations but for the side-to-side geometry.

 

It may indeed be that the CRT man applied some kind of (erroneous) rule of thumb, that a 10" lift at the front results in a 1" loss of freeboard at the rear. Basic science tells us it can't be a 10" loss of freeboard at the rear, when the front is jammed and the water level under it drops a mere 1".

 

Also remember that as the boat tilts more and more, eventually there's a point where the friction forces which are supporting the jammed end/side are overcome, partially because its angle is changing, and partly because more and more weight is being applied. It might be that on a particularly hard obstacle such as a large, strong bolt, or metal cill buffer or other metalwork, the weight does not deform the supporting area and the angle increases to allow it to slip off. But on a softer obstcle such as a wooden structure or certain types of masonry or a foreign object such as a tree branch etc, this deforms as more weight is applied, it might slip off due to the amount of deformation or deform in such a way as to 'lock' or 'stick' the boat on it more and more - becoming wedged in a certain position. And of course, the contact points of the boat are going to have some bearing (excuse the pun) on how this occurs and resolves itself.

[IanD]

But the LCF is only relevant as you are using it when the vessel is floating free. The issue here is that one end of the vessel was kept in a fixed position regardless of flotation considerations.

 

reductio ad absurdum argument: if you tied one end of a boat firmly to the land and then let sufficient water out (supposing that the lock was deep enough) so that the boat was entirely suspended out of the water, then there would be no sinking in the manner described.

 

The actual situation was probably somewhere in between and it ought to be a relatively simple task to estimate the likely behaviour - but I'm too much past A level to bother!

LCF determines where you need ballast to float the boat level, allowing for where the heavy bits are and how much displacement (especially at the stern) is removed by the swim.

 

Once you've done this and you start to apply force to the boat (e.g. a hangup), the amount of bow rise/stern sink is only set by the area of the hull at the water surface at bow and stern because this sets how much water is displaced -- good old Archimedes again. Since the counter will normally be under water by an inch or two, this means the change in displacement by lifting the bow and sinking the stern by the same amount are very similar since they are similar shapes, LCF or LCB has no influence on this. With a typical narrowboat draught the rise and fall at the two ends will be similar to each other, even allowing for differences in hull shape.

 

Exactly the same applies for lateral tipping, the distance one side goes up and the other side goes down will be almost the same. If you work out the numbers, the "stiffness" in the sense of amount of vertical movement for a given amount of force is the same for both lengthwise and widthwise tipping -- if compared to standing in the middle of the boat you stand on one gunwale and it drops by X (and the other one rises by X), standing on the stern will also drop that by X (and raise the bows the same amount). If you move to one side of the stern it will drop by 1.4X and the opposite bow corner will rise by 1.4X -- the distance is bigger because the "stiffness" is less, but the up and down movements still balance.

 

(actually these numbers ignore the overall vertical movement of the hull, adding a weight pushes it down -- but a hangup pulls it up...)

 

So it comes back to the fact that if a boat gets hung up at one corner and the water level drops a given amount, the loss in freeboard at the opposite corner will be somewhat less than this -- not almost double as I said before, that was wrong :-)

 

If you don't believe me get a rectangular piece of wood, float it in the bath, draw a line round it at the water surface, and have a play by lifting one corner. I guarantee that after all these years Mr. Archimedes is still right ;-)

Edited January 18, 2017 by IanD

[mross]

"the "stiffness" in the sense of amount of vertical movement for a given amount of force is the same for both lengthwise and widthwise tipping"

 

Do you really mean 'vertical'? Tipping is not a vertical movement, it is rotational and more horizontal than vertical.

Edited January 18, 2017 by mross

[rowland al]

Did you have fenders deployed Rod.

Many people have asked you this.....

GOT IT????

Obviously it must have been something far more complex.

[Rod Bender]

Quote Arthur "It might be worth asking for something sensible instead, like "make sure lock chambers & gates are free of protrusions/indentations that HAVE snagged boats".

 

Indeed that is exactly what I have asked members to do - "ask CRT tp make lock 40 and all Bank Newton locks safe."

These locks alone.

 

Bank Newton flight have twenty two incidents on CRT records involving chamber wall hang-ups, including 8 at Lock 38 and 7 at Lock 40.

 

On three occasions when I have returned to Bank Newton since my boat sinking I have taken photos of fresh chamber wall scars. Evidence of boat hang-ups and the probability of another serious incident, or death.

[Arthur Marshall]

Quote Arthur "It might be worth asking for something sensible instead, like "make sure lock chambers & gates are free of protrusions/indentations that HAVE snagged boats".

 

Indeed that is exactly what I have asked members to do - "ask CRT tp make lock 40 and all Bank Newton locks safe."

These locks alone.

 

Bank Newton flight have twenty two incidents on CRT records involving chamber wall hang-ups, including 8 at Lock 38 and 7 at Lock 40.

 

On three occasions when I have returned to Bank Newton since my boat sinking I have taken photos of fresh chamber wall scars. Evidence of boat hang-ups and the probability of another serious incident, or death.

They'll be on a list. The trouble is, the list is probably huge and getting longer every year. Maintenance in general is now so poor you can see the end of the system as a decent navigation is in sight. It's not just the locks, it the towpath near collapses and breaches, lack of maintenance at most locks and just about all the sluice gizmos rusting solid. Isn't the term "managed decline"?

[Mike Todd]

LCF determines where you need ballast to float the boat level, allowing for where the heavy bits are and how much displacement (especially at the stern) is removed by the swim.

 

Once you've done this and you start to apply force to the boat (e.g. a hangup), the amount of bow rise/stern sink is only set by the area of the hull at the water surface at bow and stern because this sets how much water is displaced -- good old Archimedes again. Since the counter will normally be under water by an inch or two, this means the change in displacement by lifting the bow and sinking the stern by the same amount are very similar since they are similar shapes, LCF or LCB has no influence on this. With a typical narrowboat draught the rise and fall at the two ends will be similar to each other, even allowing for differences in hull shape.

 

Exactly the same applies for lateral tipping, the distance one side goes up and the other side goes down will be almost the same. If you work out the numbers, the "stiffness" in the sense of amount of vertical movement for a given amount of force is the same for both lengthwise and widthwise tipping -- if compared to standing in the middle of the boat you stand on one gunwale and it drops by X (and the other one rises by X), standing on the stern will also drop that by X (and raise the bows the same amount). If you move to one side of the stern it will drop by 1.4X and the opposite bow corner will rise by 1.4X -- the distance is bigger because the "stiffness" is less, but the up and down movements still balance.

 

(actually these numbers ignore the overall vertical movement of the hull, adding a weight pushes it down -- but a hangup pulls it up...)

 

So it comes back to the fact that if a boat gets hung up at one corner and the water level drops a given amount, the loss in freeboard at the opposite corner will be somewhat less than this -- not almost double as I said before, that was wrong :-)

 

If you don't believe me get a rectangular piece of wood, float it in the bath, draw a line round it at the water surface, and have a play by lifting one corner. I guarantee that after all these years Mr. Archimedes is still right ;-)

To some extent your analysis assumes that the boat is wholly supported by flotation (such that Archimedes applies) However, in a hang up (of whatever kind) some of the weight (the down force from the mass) is taken by the point of hanging leaving a very different moment supported by flotation.

[Sbg]

That was point I was trying to get at, and have been talking to some physicist friends about. The hangup creates a second point, carrying some of the weight (apparently up to 33% if the bow is completely lifted out and the snag is 10' back on a 60' boat), with the buoyancy point (I think this is the metacentre) taking up the rest. As the water level drops, the metacentre moves rearwards but remains the fulcrum around which the boat rotates in pitch. Buoyancy is reduced as a greater proportion of the hull remaining in the water is taken up by the swim, but this is balanced (maybe (probably?) not entirely) by the amount of weight being carried by the snag.

 

If I have this right, the lever length is from the snag to the metacentre, and the stern will drop by a corresponding ratio dependant upon the lever length from the metacentre to the stern. The heel effect will also be apparent as the hull will rotate to the right, hence the water coming in over the starboard cant. Despite earlier statements to the contrary I think the formulae and maths are actually pretty complicated and I'm trying to correlate the calculations needed against Ken's observations of the event i.e. a snag 10' from the bow combined with a water level drop of a few inches conspired to swamp the starboard stern cant. Still too many unknown factors to try to make sense of it though - at least to this bear of little brain.

Edited January 19, 2017 by Sbg

[Rojo]

After taking into account the formulae stated, I am still puzzled by the fact that 'flood water was pouring through the legs of the owner of the second boat'.

[WotEver]

After taking into account the formulae stated, I am still puzzled by the fact that 'flood water was pouring through the legs of the owner of the second boat'.

I guess it depends on where said owner was standing. On the stern of the sinking boat?

[mross]

We need a naval architect! They know how to calculate the stability of a vessel when it is not floating free. If a ship goes aground on rocks, it very quickly becomes unstable if the tide goes out and can capsize. Less so on mud.

[IanD]

Instead of all the arguing about formulae, I'm still waiting for somebody to float a length of wood in their bath to prove it. I'd do it myself but it's so expensive heating up all that water when you live in a house ;-)

 

[and it really isn't that complicated, all you need is Archimedes' principle and Newton's laws...]

[Rod Bender]

After taking into account the formulae stated, I am still puzzled by the fact that 'flood water was pouring through the legs of the owner of the second boat'.

His right foot on his boat, left foot on my boat,, water pouring over the starboard cant, down the steps, through the Starboard access door and into my boat.

 

My thanks to members post regarding boat hang-ups, interesting and informative. The post below appears to be 'spot on'.

 

That was point I was trying to get at, and have been talking to some physicist friends about. The hangup creates a second point, carrying some of the weight (apparently up to 33% if the bow is completely lifted out and the snag is 10' back on a 60' boat), with the buoyancy point (I think this is the metacentre) taking up the rest. As the water level drops, the metacentre moves rearwards but remains the fulcrum around which the boat rotates in pitch. Buoyancy is reduced as a greater proportion of the hull remaining in the water is taken up by the swim, but this is balanced (maybe (probably?) not entirely) by the amount of weight being carried by the snag.

 

If I have this right, the lever length is from the snag to the metacentre, and the stern will drop by a corresponding ratio dependant upon the lever length from the metacentre to the stern. The heel effect will also be apparent as the hull will rotate to the right, hence the water coming in over the starboard cant. Despite earlier statements to the contrary I think the formulae and maths are actually pretty complicated and I'm trying to correlate the calculations needed against Ken's observations of the event i.e. a snag 10' from the bow combined with a water level drop of a few inches conspired to swamp the starboard stern cant. Still too many unknown factors to try to make sense of it though - at least to this bear of little brain.

[Mike Todd]

That was point I was trying to get at, and have been talking to some physicist friends about. The hangup creates a second point, carrying some of the weight (apparently up to 33% if the bow is completely lifted out and the snag is 10' back on a 60' boat), with the buoyancy point (I think this is the metacentre) taking up the rest. As the water level drops, the metacentre moves rearwards but remains the fulcrum around which the boat rotates in pitch. Buoyancy is reduced as a greater proportion of the hull remaining in the water is taken up by the swim, but this is balanced (maybe (probably?) not entirely) by the amount of weight being carried by the snag.

 

If I have this right, the lever length is from the snag to the metacentre, and the stern will drop by a corresponding ratio dependant upon the lever length from the metacentre to the stern. The heel effect will also be apparent as the hull will rotate to the right, hence the water coming in over the starboard cant. Despite earlier statements to the contrary I think the formulae and maths are actually pretty complicated and I'm trying to correlate the calculations needed against Ken's observations of the event i.e. a snag 10' from the bow combined with a water level drop of a few inches conspired to swamp the starboard stern cant. Still too many unknown factors to try to make sense of it though - at least to this bear of little brain.

You are not quite correct: once one end is suspended then it becomes the centre of rotation. Whilst he behaviour of the unsupported end is a matter for Archimedes, it does depend hugely on the shape. The distribution of the onboard load (ballast, cargo people etc) is not especially important as the hangup is creating a fulcrum.

[Horace42]

I put a long shallow plastic dish (the sort shops use to supply perishable goods in) in a sink of water and with a couple of coins in one end to represent the weight of the engine, and held the front to represent the snag.

Slowly drained some water from the sink and watched the dish gradually tilt until the stern gunnel reached water level - then suddenly water rushed in the dish and it sank pronto - albeit the plastic was buoyant and almost floated to the surface after it sank.

I guess the same would happen to a narrowboat - when water flows over the gunnel - the boat will sink in seconds - and being metal will stay sunk.

[Horace42]

As a general observation all the videos I have seen of ships sinking at sea - the stern drops slowly (relatively speaking) until the gunnell is under water. Then they go down vertically and sink very rapidly.