Angle of pitch and heel for a boat snagged on the side wall

[sbg]

Hi everyone, this is a real world problem following the sinking of a narrowboat in 2014. The boat caught on the side of the lock as the water level dropped, and the resultant combination of pitch and heel resulted in the boat taking on water at the stern and sinking - luckily with no loss of life or injury. Without giving too much away (as the owner has settled the case with the navigation authority) I can offer the following information:

- boat length 60'
- boat displacement 15 tons (steel construction)
- The snag occurred on the port side, 10' from the bow
- Narrowboats tend to sit lower at the stern than at the bow so the CoG is probably aft of the centrepoint (Assumption - is this true?)

The boater stated that the water level dropped "a few inches" then later stated that the sinking actually occurred when the level had dropped by 2'. A few maths wizards have suggested that a 2' drop at the stern would equate to 2 degrees of pitch which seems a very small amount to me, and I'm pretty sure my boat (about the same size) rose by a couple of degrees at the bow just by having an empty water tank... I also think that my boat going up the slipway took on more than a 2 degree pitch

I'm not trying to make the boater look bad, and am in no way connected with the event or the owner (or the Authority) but am trying to make sense of the physics of such an event and what a 2' drop would look like in terms of the angle. Of course, there is the heel effect to consider due to the boat snagging on one side.

Grateful for any help!

 

[BvU]

Hello sbg,

A few maths wizards have suggested

If they are any good they have told you (not just suggested). 2' on 60' or on 50' IS two degrees, yes. Do you want he formula ?

rose by a couple of degrees at the bow just by having an empty water tank

You mean it bent ? Did the stern stay horizontal ? or did the whole boat (as a reasonably stiff object) pitch ?

I also think that my boat going up the slipway took on more than a 2 degree pitch

I suppose so: a 2 degree slipway would have to be very long to be useful ??!

trying to make sense of the physics

No physics so far !

 

[anorlunda]

The COG is certainly behind the snag point, 1/6 hull length from the bow.

The boat would experience both pitch and heel causing the stern corner away from the wall to dip most and be the point of entry of the water.You ask about calculating the pitch, but you also need the angle of heel and (most important) the free board at the stern quarter, to make a conclusion. Free board is the height of the hull above the water line when level. Free board in the stern is usually less than at the bow.

I'm not familiar with the term narrowboat. Do you mean something like this?

That type of construction would be highly likely to get swamped if it snagged on the lock wall while descending. I would not trust feet or angle estimates by eye witnesses, so calculations would be pointless.

 

[jim hardy]

Stability depends on heights of centers of gravity and bouyancy., too
CG is usually above CB so there is some angle beyond which the vessel will try to capsize.
The upward force of snag makes a torque that opposes restoring moment , effectively moving center of bouyancy toward snag,
and that greatly reduces the angle where you lose stability..
so i expect it sank from excessive roll rather than pitch. Did it capsize ?

http://www.learneasy.info/MDME/MEMmods/MEM23006A/fluid_mech/fluid_statics.html

@anorlunda I'm no naval architect , but do remember basics of Statics course from 1965... is above plausible ?

old jim

 

[anorlunda]

@anorlunda I'm no naval architect , but do remember basics of Statics course from 1965... is above plausible ?

Yes that's valid @jim hardy , but first you have to modify the numbers by taking on some tons of water first. Many boats capsize as they are sinking. In addition, assume that the front of the boat is snagged on the lock wall; the boat is not floating free. (By the way, the angle where you lose stability on my boat is 120 degrees. That is the effect of having a heavy keel low down. It is like a rolly-polly toy.).

Here is a video of a ship capsizing because of stability. Start at 1:10.


What I think happened in this case is more like in the following videos. Lifting the bow makes the stern go under water.

Start this video at 3:07.


And this video (only 39 seconds long)


Most dramatic is the first 7 seconds of this video. (the next 50 seconds is another stability failure)

 

[Baluncore]

Pitch and Heel can be treated separately here, as heel is not the major problem. The fundamental problem arises from several factors but the most significant is that the snag or hang-up is not at the end of the boat but at 10' along the 60' hull. That is only 20' from the mid-point of the hull.
The centre of mass of the hull is effectively a fulcrum. The point that is lifted by 2' is 20' from that fulcrum, but the stern is 30' beyond the fulcrum. So freeboard at the stern will actually be reduced by about 3'. That was more than sufficient to cause the flooding.
The pitch angle was then close to Pa = Asin( 2' / 20' ) = 5.7°

I have ignored the issue that lifting one point on the hull reduces the buoyancy required to support the hull. It is important in the detail but not in the understanding of the lever and fulcrum model.

The situation is aggravated when;
1. Most of the buoyancy of a hull is in the middle of the vessel. The ends of the hull are tapered underwater to reduce drag. This significantly reduces buoyancy at the end corners of the hull. It is more pronounced at the pointed stern than at the more rounded bow.
2. The closer the hang-up point is to the centre of mass the greater will be the loss of freeboard at the far end of the other side of the hull.
3. The direction of water flow in the canal is from the low end of the hull. A “bow” wave reduces freeboard. Also, flow along the rising bottom of the hull draws the hull downwards.

 

[sbg]

Thanks for the replies so far folks - I appreciate it. Apologies if the initial statement was a little vague - I now have more information from the owner.

Anorlunda - yes, that is the correct type of boat. 60' long, 6'10" wide and (mostly) flat bottomed. It 's a common inland waterways vessel in the UK and although most have no keel, the baseplates are usually 10mm steel and they are remarkably stable. A few intrepid people have even motored across the English Channel - properly prepared and accompanied by a pilot of course!

The boat sank at the stern rather than capsizing; according to the owner the induced heel was a factor but the greater issue was the change in pitch i.e as the bow stayed snagged and the water dropped, the stern was swamped.

BvU - thanks for the welcome. I understand that 2' on a 60' length rotating about a single end pivot point would be 2 degrees, but I think this is more complex. I think Baluncore has hit the nail on the head as the snag was 20' from the boat centre of mass, and the fulcrum effect would be to drop the stern by 3' However, would there not be a further complicating factor in that the vessel continued to have buoyancy, albeit reduced? It seems to me that with the bow lifting out of the water, the apparent centre of buoyancy (is this the metacentre Jim?) would move rearwards but would continue to keep the boat afloat until the water came over the cant. In answer to your other questions, our water tank was in the bow, and when empty, caused the bow to rise noticeably from the fully loaded mark - the whole boat pitched. As our boat was constructed in the same fashion (all steel) as the narrowboat pictured, there was no bending involved! However, on reflection, this could have been no more than a couple of inches... so I am incorrect in stating two degrees here.

I'm trying to understand the forces in action in this case and the maths involved - I've completed a diploma in marine survey and although we dealt with stability, the metacentre calculations focused on loading and jib effect with cranes. I suspect a similar set of equations are appropriate but am struggling to figure out various rotational effects, countered by the buoyancy.

Edited to add that the owner now states that the freeboard at the stern was 16" but - frustratingly - the water level changes stated was "a few inches". I have no idea why the water level drop quoted has been changed but there you are.

 

[jim hardy]

Roll, pitch, or both ? An eyewitness or cellphone video would sure be nice.

Lifting bow moves CB aft

water into stern moves CG aft,
water goes to low side of vessel moving CG that way

did water first come over stern at a corner or clear across it ?

Look at your boat and work it in your head.

 

[Baluncore]

I think Baluncore has hit the nail on the head as the snag was 20' from the boat centre of mass, and the fulcrum effect would be to drop the stern by 3' However, would there not be a further complicating factor in that the vessel continued to have buoyancy, albeit reduced?

There are several complicating factors that slightly reduce loss of freeboard, but not enough to prevent the sinking. The mass that must be supported by hull buoyancy is less because the snag is supporting the forward 33% of the boat mass. But the remaining 66% centre of mass has moved aft from the 30' section to the 40' section. The buoyancy needed to support the remaining mass is slightly less because of the lever length as the centre of buoyancy moves towards the stern. But the boat had a motor and propeller, so would have had a more tapered hull at the stern. The reduced stern buoyancy limits how far the centre of buoyancy can migrate aft as the pitch increases.

But the overriding effect is the great length of the rectangular section hull and the immediate loss of a significant part of the normal buoyancy when one end of the hull is slightly raised. That effect is magnified for a long hull with a flat bottom and shallow draft.

 

[anorlunda]

the owner now states that the freeboard at the stern was 16"

The ratio of length to stern freeboard is 45:1. That same ratio on my boat is 8:1. That is why I say the narrowboat type of construction is very vulnerable to this sinking by the stern problem. @Baluncore said the same thing in different words.

But the overriding effect is the great length of the rectangular section hull and the immediate loss of a significant part of the normal buoyancy when one end of the hull is slightly raised. That effect is magnified for a long hull with a flat bottom and shallow draft

But it seems that nobody else saw the connection between that ratio and the three videos in #5 showing boats sinking at the stern at boat ramps. Backing a boat down the ramp is very analogous to this case of the boat getting snagged on the lock wall, simply because the bow is lifted relative to the stern.

A boat with 45:1 or higher ratio can not be backed down a ramp into the water without sinking the stern, whereas a boat with 8:1 ratio can do that successfully. Of course ramp angle also plays a role, but it does not change the principle. Most boats snagged on the lock wall will not sink, but this one did. The different outcome is dictated by the dimensions.

 

[CWatters]

I've also experienced a hang up as described in the OP. Can confirm that the roll was much more of a concern than the pitch. Fortunately the rope causing the hang up broke before we took on water.

 

[sbg]

Roll, pitch, or both ? An eyewitness or cellphone video would sure be nice.

Lifting bow moves CB aft

water into stern moves CG aft,
water goes to low side of vessel moving CG that way

did water first come over stern at a corner or clear across it ?

Look at your boat and work it in your head.

No videos available and the only pictures are after the event and the subsequent draining of the lock - not helpful in this case.
The water came over the starboard stern cant, primarily due to the induced pitch / lowering of the stern but exacerbated by the roll.

 

[sbg]

I've also experienced a hang up as described in the OP. Can confirm that the roll was much more of a concern than the pitch. Fortunately the rope causing the hang up broke before we took on water.

Was the boat roped to the centreline t-stud, forward or stern points?

 

[CWatters]

Was the boat roped to the centreline t-stud, forward or stern points?

In our case it was one side of the transom/stern. We were in a lock and the person on the bank put a full turn around a post. As the lock emptied they found the rope had jammed and they were unable to pay out rope.

 

[sbg]

In our case it was one side of the transom/stern. We were in a lock and the person on the bank put a full turn around a post. As the lock emptied they found the rope had jammed and they were unable to pay out rope.

Ah, that makes sense.

 

[Baluncore]

It seems rare that a boat might snag on one side of the lock, so I wonder if the report has become confused in the telling. It seems much more likely for such a hang-up to occur on the cill. A leaking gate during low water levels can catch a narrow boat on the cill, 10' from the bow.
https://en.wikipedia.org/wiki/Lock_(water_navigation)#Cill

I googled images 'narrowboat sinks in lock' and got plenty of hits.
http://l7.alamy.com/zooms/a9d3f840f...ign-on-a-lock-gate-warning-of-cill-bwmwy7.jpg
http://waterwaynews.blogspot.com.au/2013/08/photograph-of-latest-huddersfield.html
http://www.dailymail.co.uk/news/art...rowboat-started-sink-getting-wedged-lock.html
http://www.alamy.com/stock-photo-ba...ow-boat-stuck-on-the-stone-cill-67798693.html

 

[CWatters]

Yes hang ups on the Cill are a well known problem. Many locks have a warning sign in addition to the white line. That could well have been the cause.

Now that I think about it I've also a small cabin cruiser hang up on the edge of a lock. The cruiser was made from two fibre glass mouldings (a top and a bottom). and where the two parts join there was quite a large inverted lip fitted with a rubbing strip. However most canal boats don't normally have such a wide rubbing strips.

 

[sbg]

It seems rare that a boat might snag on one side of the lock, so I wonder if the report has become confused in the telling. It seems much more likely for such a hang-up to occur on the cill. A leaking gate during low water levels can catch a narrow boat on the cill, 10' from the bow.
https://en.wikipedia.org/wiki/Lock_(water_navigation)#Cill

I googled images 'narrowboat sinks in lock' and got plenty of hits.
http://l7.alamy.com/zooms/a9d3f840f...ign-on-a-lock-gate-warning-of-cill-bwmwy7.jpg
http://waterwaynews.blogspot.com.au/2013/08/photograph-of-latest-huddersfield.html
http://www.dailymail.co.uk/news/art...rowboat-started-sink-getting-wedged-lock.html
http://www.alamy.com/stock-photo-ba...ow-boat-stuck-on-the-stone-cill-67798693.html

No - this was not a cilling event although you are right in that the majority of lock sinkings are down to the boat either being on the cill (usually at the stern) or a bow / stern fender becoming wedged in the gates and hanging up. I am an experienced boater and want to understand more of what happened as by all accounts this sinking occurred very quickly.
The lock in question has stone sides and has a bulge below the high water line - this is the subject of a legal challenge as the owners of the boat have settled out of court but still believe there is a Health and Safety issue i.e. the Canal and River Trust have not maintained the lock to the minimum safety standards. I'll try to upload the pictures from the original website.

Most modern narrowboats have a 10mm steel baseplate which usually protrudes a few mm beyond the junction with the hull sides. This chine protects the welded join from damage from underwater objects, canal pilings, lock sides etc but the downside is that the protrusion along the length of the hull increases the possibility of being hung up. There were two boats in the wide lock and the pictures from the event clearly show the witness marks on the lock side bulge where the boat was snagged, eventually slipping off as the boat took on water.

I'm trying to stay away from the legal stuff to focus on the events and dynamics of the boat becoming swamped and sinking.

Edited for typos.

 

[sbg]

Here's what I think happened:

Under normal circumstances the boat has a single centre of gravity, with a centre of buoyancy. 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. I think the formulae and maths are actually pretty complicated and I'm trying to correlate the calculations needed against the owner'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. I think the key is how much the metacentre moves.

 

[Baluncore]

Thank you sbg for confirming the situation.

Most modern narrowboats have a 10mm steel baseplate which usually protrudes a few mm beyond the junction with the hull sides.

It would not take much heel for the 10mm plate to slide off the snag. There must have been something preventing the hull from moving away from the wall, maybe the other hull in the lock?

The heel required would be the arctangent of the coefficient of friction of hull against lock. Likewise a hull would not slide lengthways off a snag (or a cill) until the pitch angle exceeded the arctangent of the coefficient of friction of hull on lock.

What was the beam of the boat in question ?

 

[sbg]

Yes, the other boat in the lock could have prevented any great sideways movement although I had a narrowboat briefly hang in a single-width lock a few years ago.
The boat that sank is a standard width narrowboat so should be 6' 10" wide along most of it's length; the baseplate would be slightly less wide than this as the max width is usually at the point the lower hull ends and rubbing strake is placed to protest the glossy paintwork above (as can be seen in Anorunda's post #3 above)

 

[sbg]

Found them. The first image shows the alignment of the lock wall and the bulge in the stonework. The second shows the scarring / witness marks from the hang up, and the third (very sad) shows the boat once the lock had been drained.

 

[CWatters]

The witness mark appears to be close to the high water point so that says something about the height of the hang up point on the boat.

 

[anorlunda]

@sgb, now I've lost track of what question you are asking. Please restate the question.

 

[Baluncore]

It is reassuring that the inward opening, high lock gates in the picture indicate that the boat snagged at the down-hill end of the lock where there is no high cill to complicate the issue.

The green slime in the picture indicates that the high water level in that lock is close to or just above the base of the top layer of stone. It appears that weathering of the upper two layers of stonework permitted the hull to move closer than usual to the wall. The loss of some mortar below the second layer of stone provided the shelf for the hull base plate to snag.

Now we just need to do the numbers to find the water level drop necessary to flood the far end of a snagged hull.

 

[sbg]

I'm trying to understand the angles of pitch and roll induced by the boat hanging up, and the water level drop that would be needed to swamp the stern starboard cant. Without carrying out any photogrammetry, I think we have to assume that that hang up was (as stated) about 10' from the bow. the boat is 60' long, with an estimate weight of 15t. The key question is "at the moment the water overtopped the cant, what were the angles of pitch and roll of the boat. I don't think we can tell how much the water level dropped without knowing the measurement from the snag to the high water mark, and the measurement of how much the water dropped before the boat hung up - unless anyone knows better?

 

[Baluncore]

Maybe an initial heel increased the effective width of the increasingly diagonal hull and so cammed the two boats firmly between the walls. If the victim had a beam of 6' 10”, with a hull height of 3' 10”, then it would widen to a maximum of the diagonal = 7' 10”, when heeled at 29.3°.

We do not know the width of the second boat, nor the width of the lock, so we cannot tell how far the victim heeled before the cam locked at the pinch point 10' from the bow. There has been no report of the cabins touching so there may not have been much heel before it locked.

I have been thinking about a computation strategy. The snag is so high on the wall that we can assume the snag is a fixed point from when the water started to fall. The contact of the victim on the second boat falls with the second boat and the water. Now we must work backwards by assuming the stern on the starboard side is at the critical depth, on the very edge of flooding. In that position we must find the volume and centroid of the submerged part of the hull. That gives us the moment of buoyancy about the snag. Likewise we know the moment of the unsupported mass. At some point, as the water falls, the buoyancy moment will become insufficient to counter the mass moment. That is when the boat would actually start to take on water and sink rapidly.

I am looking now for an efficient way to calculate the submerged hull volume, centroid and moment about the snag, when the hull is both heeled and pitched.

 

[CWatters]

Sounds horribly complicated to me. Perhaps building a physical scale model would be quicker?

Edit: Actually I suppose there might be some software around in the ship design world that could do this.

 

[jim hardy]

Narrowboats have pretty flat bottoms ?
http://narrowboatsale.com/narrow-boat.htm

The majority of can boats have a flat bottom - some with a slight cut up at the forward end - although some boats - notably those built by Springer Engineering Ltd have a slight V to the bottom. Modern boats with a flat bottom have the bottom plate extended beyond the sides by up to one inch (25 mm) to form a so-called wearing chine. This construction also allows a better external weld between the side and bottom plates.

Just assuming the hull to be a rectangular prism with CG and CB dead center would be a starting point to figure out an algorithm.
Given width and length and weight, draft is easy from that...Free body diagram with point of snag fixed, ...

Do we know the initial freeboard at stern ?

 

[anorlunda]

I fail to see how the angle is interesting. Suppose it has to be 10 degrees to swamp? or 5 degrees? or 20 degrees? What significance does that number have? Nevertheless, the calculation sounds like fun, so here is a partial solution.

I began with a cross sectional rectangle 32 inches high by 720 inches (60 feet) long. Forget the beam. We model the boat as a squared off barge with 16 inches of freeboard and (assumed) 16 inches of draft, so 16*720 is the cross section of the submerged area..

To swamp the barge without tilting it, you need to double the displacement, so 200% load at zero angle will swamp it.

We tip the boat because of the snag and look for the exact moment when water first comes in over the stern. Note two things. First, that only a small triangle at the stern is submerged. Second, that the weight of the boat is supported by two points, the snag pivot, and the COB of the submerged triangle. This is depicted in the picture.

So now, using only geometry, I calculated the volume of the submerged triangle as a function of the angle of tilt. It is scaled as a fraction of the 16*720 rectangle defined as 100%. Below is a table of my results. Note that beam does not come into the calculation because the answer is expressed as a fraction of the triangle to the rectangle. The ratio of submerged cross sections is the same as the ratio of submerged volumes for any value of beam B of a rectangular barge.

angle in degrees, submerged fraction needed to swamp.
0, 200.00%
5, 51%
10, 26%
15, 17%
20, 13%
25, 11%
30, 9%

The part I didn't solve is the division of total vessel weight between the submerged triangle and the pivot point. Obviously, when the angle was 0 degrees, the boat is not snagged and the pivot carries zero. Also, at angle 90 degrees, the pivot carries zero of the weight. Maybe someone else could do that. When we have both, we can solve for the critical angle where the boat gets swamped from the stern.

There is no estimate of roll (heel). But if there was moderate heel, it might be approximated by reducing the effective freeboard. For severe heel, my 2D calculation would need to be made 3D to calculate the submerged fraction of volume.

 

[Baluncore]

Sounds horribly complicated to me. Perhaps building a physical scale model would be quicker?

The experiment has already been done with a full size model, see pictures in post #22. We need to replicate that experiment at lower cost. I think a numerical model of the hull could be adjusted more easily than carving a scale model replica. The process may sound complicated but it can actually be broken down into neat modules. It would also be interesting to pivot a virtual narrow-boat on a virtual lock cill.

Just assuming the hull to be a rectangular prism with CG and CB dead center would be a starting point to figure out an algorithm.

I tried that simplistic approach with a calculator. Those assumptions hide the unexpected early sinking. Since the hull pivots on the snag fulcrum, it is the moment of mass versus the moment of buoyancy that must be solved. That situation is aggravated because there is little buoyancy at the stern with the narrow engine room, propeller and rudder.

Do we know the initial freeboard at stern ?

Yes, about 18". There are several plans for narrow-boats on the web. They conform closely with each other because they are constrained in design by the locks and bridges.

I fail to see how the angle is interesting.

The particular numerical angles of pitch and heel are not interesting. But those two angles are critical state variables in the model.

The part I didn't solve is the division of total vessel weight between the submerged triangle and the pivot point. Obviously, when the angle was 0 degrees, the boat is not snagged and the pivot carries zero. Also, at angle 90 degrees, the pivot carries zero of the weight. Maybe someone else could do that. When we have both, we can solve for the critical angle where the boat gets swamped from the stern.

You understand the problem. But once the water has fallen about 18”, the pivot carries 1/3 of the mass of the boat, a load that then does not change. That is not a simple cosine function. The mass centroid does not migrate, the buoyancy moment is dynamic.

Nevertheless, the calculation sounds like fun,

I agree. I am not going to do it the difficult way, I want a neat solution applicable to other hulls. Existing software could solve the problem but it is more fun to gain an understanding of the situation while reinventing the wheel.

The prismatic hull of a narrow-boat is a trivial problem. It is the lump of engine mass and the subtle lack of buoyancy at the stern that makes it more complex. The trapezoid rule for calculating the area of a polygon defined by points is brilliant in 2D, but in 3D it is certainly not. Consider a surface defined by shrinking a film onto a constellation of many points.
https://en.wikipedia.org/wiki/Convex_hull#Convex_hull_of_a_finite_point_set
The volume of that “convex hull” can be computed. The buoyancy modification to include a changing water level is more complex as it involves interpolation along the lines between the active 3D vertex points.

I had forgotten how foul and murky the waters of the British canals really were. This seems like a good time to take a cruise on the mathematical sea of convex hulls. I will think about hull computation as I drift across the crystal clear waters of the wilderness in my back garden. I am in no particular hurry to end the multidimensional exploration.

 

[anorlunda]

once the water has fallen about 18”, the pivot carries 1/3 of the mass of the boat, a load that then does not change.

I don't see how you say that. Especially up to 90 degrees. But if that is correct, then my answer is less than 10 degrees. My table in #30 says that at 10 degrees, the stern can only support 26% of the displacement before swamping (with zero heel).

the lump of engine mass

The engine mass is about the same as two crew members. It is possible that four or more crew might be in the cockpit during locking. . Fuel and water tanks can also exceed the mass of the engine. Tanks are located where?

With a total displacement of 17 tonnes, I wouldn't worry about engines crew and tanks. Ballast dominates all that.

 

[Baluncore]

I don't see how you say that. Especially up to 90 degrees.

Imagine a crane attached to the mid-line, 10' from the bow. Slowly lift the vessel, once the bow comes out of the water, about one third of the mass is being supported by the crane, that is the front 20' of the 60' hull. At some point the stern will flood, if you continue to 90° the entire mass will be supported by the crane. The hull will slide off the snag long before it reaches 45°. That was explained in post #20.

The mass that must be supported by hull buoyancy is less because the snag is supporting the forward 33% of the boat mass.

But if that is correct, then my answer is less than 10 degrees. My table in #30 says that at 10 degrees, the stern can only support 26% of the displacement before swamping (with zero heel).

The pitch angle was then close to Pa = Asin( 2' / 20' ) = 5.7°

Post #6 and #9 explains why 1/3 of the mass is carried by the snag.

The engine mass is about the same as two crew members. It is possible that four or more crew might be in the cockpit during locking.

Narrow-boats that are not carrying a full load appear to float with the stern trimmed low. That is partly because of the engine mass, but mainly due to the significant lack of buoyancy under the stern.

 

[sbg]

Do we know the initial freeboard at stern ?

16"

 

[sbg]

This feels like we are getting somewhere - thank you fellows for your contributions thus far. After reading the comments above, there are a couple of other things to note.
Narrow boats tend to be ballasted 'stern down' i.e. the draft at the stern is greater than at the bow as this has been found to enable the boat to be better under control, 'swim' better with reduced wake and so on.
Water tanks tend to be located in the bow under the foredeck and can often be in excess 400 litres when full.
When operating in locks, most of the crew tend to be ashore working the lock gear but in this case in addition to the steerer, there was a female adult and a child aboard - thankfully rescued prior to sinking and uninjured.
The snag point must have been under water when the lock was full, as the baseplate was reported to snag on it as the water level dropped - therefore the snag must have been more than the depth of hull draught at 10' rear of the bow. This could be 12" or less.

Dependant upon the hull fabricator, the 'swim' of the hull (the sweep of hull curve from the widest point to the point at which the prop exits the stern) can be of differing lengths and shapes, but as an example, on my boat the swim started about 8' fore of the propshaft.