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

[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.

 

[anorlunda]

@sgb

You have not answered why knowing the angle of pitch is so significant as to make it worth all this trouble.