How much does my boat weigh?

[Machpoint005]

1 hour ago, TheBiscuits said:

Not done the Rochdale recently then? 

 

No, and point taken ?, but any lock, anywhere is "full" when the water level matches the level outside the top gates, whatever that level is.

[Mike Todd]

54 minutes ago, Machpoint005 said:

No, and point taken ?, but any lock, anywhere is "full" when the water level matches the level outside the top gates, whatever that level is.

I think the point being made is that there are some locks on some canals (fewer than than certain folk would have you imagine) that do not make a level for one of several reasons. The bottom may be leaking faster than the top can supply or the top may be supplying faster than the bottom can empty. Or the upper pound may be overflowing the top gates which are higher than the bottom gates. In any case it may be necessary to resort to traditional boater techniques (LTRU) to persuade the gates to open.

[zenataomm]

I give up ….. how much does your boat weigh?

[TheBiscuits]

2 hours ago, zenataomm said:

I give up ….. how much does your boat weigh?

Does the nearest guess win it?

[Iain_S]

11 hours ago, WotEver said:

I’ve noticed that a 1.5l bottle of pop weighs a fair bit more than a 1l bottle. Does that help?

And it weighs a fair bit more if it comes in a glass bottle.

[Peter X]

A tonne per metre length is going to give you a bit of an over estimate for most canal boats. This is because one cubic metre of water weighs one tonne, so from Archimedes we know the weight of a boat in tonnes is the number of cubic metres of water it displaces when floating. If the portion of a boat below the water was a box 2 metres wide and half a metre deep (i.e. about 6'8" wide and 1'8" deep) the formula would be exact. In practice the depth varies, and a typical canal boat is tapered at each end, so Tony Brooks' formula is better, and that 36' Springer (if it has a one square metre cross section below the water in the middle) is going to be quite a bit less than 12 tonnes.

 

Us mathematicians have formulae for working out the volumes of all sorts of odd shapes, and given equations to define a shape can generally use calculus to work out its volume, but boat hulls have some pretty complex curves and designs vary (e.g. the V shaped bottom of a Springer), so no simple formula can give an exact answer. 

[WotEver]

7 hours ago, Iain_S said:

And it weighs a fair bit more if it comes in a glass bottle.

Yes that’s true...

[Alan de Enfield]

Question : Does a litre of 'still' water weigh more than a litre of 'fizzy' water ?

(Logic being that the gas content weighs less than water.)

[howardang]

7 hours ago, Peter X said:

A tonne per metre length is going to give you a bit of an over estimate for most canal boats. This is because one cubic metre of water weighs one tonne, so from Archimedes we know the weight of a boat in tonnes is the number of cubic metres of water it displaces when floating. If the portion of a boat below the water was a box 2 metres wide and half a metre deep (i.e. about 6'8" wide and 1'8" deep) the formula would be exact. In practice the depth varies, and a typical canal boat is tapered at each end, so Tony Brooks' formula is better, and that 36' Springer (if it has a one square metre cross section below the water in the middle) is going to be quite a bit less than 12 tonnes.

 

Us mathematicians have formulae for working out the volumes of all sorts of odd shapes, and given equations to define a shape can generally use calculus to work out its volume, but boat hulls have some pretty complex curves and designs vary (e.g. the V shaped bottom of a Springer), so no simple formula can give an exact answer. 

A useful method which has been used for many years in stability and tonnage calculations in the marine industry is Simpsons rules (1st, 2nd & 3rd Rules)  which allow a good approximation of areas and volumes of ship shaped areas and volumes, sufficient so that they  can then be applied in many stability and tonnage calcs for ships and boats.  For anyone who may be interested they are worth having a look at this brief description on Wikipedia, 

https://en.wikipedia.org/wiki/Simpson's_rules_(ship_stability)

 

but if you want a better description get hold of a copy of Derrets "Ship Stability" which is one of the standard works on ships stability calculations.

Howard

[Mike Todd]

2 hours ago, howardang said:

A useful method which has been used for many years in stability and tonnage calculations in the marine industry is Simpsons rules (1st, 2nd & 3rd Rules)  which allow a good approximation of areas and volumes of ship shaped areas and volumes, sufficient so that they  can then be applied in many stability and tonnage calcs for ships and boats.  For anyone who may be interested they are worth having a look at this brief description on Wikipedia, 

https://en.wikipedia.org/wiki/Simpson's_rules_(ship_stability)

 

but if you want a better description get hold of a copy of Derrets "Ship Stability" which is one of the standard works on ships stability calculations.

Howard

I'm very rusty now (nearly 40 years since I had anything to do with this) Simpsons only calculates area (usually the ship's cross section - the problem being considered in the latter posts above is how to do a similar (but different) calculation for volume, taking into account that the cross section area is not a constant.

[Tony Brooks]

10 minutes ago, Mike Todd said:

I'm very rusty now (nearly 40 years since I had anything to do with this) Simpsons only calculates area (usually the ship's cross section - the problem being considered in the latter posts above is how to do a similar (but different) calculation for volume, taking into account that the cross section area is not a constant.

but if you did enough cross sections at say 6" intervals and then subdivide each cross section with vertical lines the sum of the volumes of each 6" slice of structure would be a fair approximation of the volume. However for weight calculations you would only use the submerged part of the hull.  Too much faffing about for little real gain for inland boats I would say.

[Mike Todd]

3 minutes ago, Tony Brooks said:

but if you did enough cross sections at say 6" intervals and then subdivide each cross section with vertical lines the sum of the volumes of each 6" slice of structure would be a fair approximation of the volume. However for weight calculations you would only use the submerged part of the hull.  Too much faffing about for little real gain for inland boats I would say.

Just so - when our new boat was launched (fully fitted) the crane weighed it for us!

[Alan de Enfield]

1 minute ago, Mike Todd said:

Just so - when our new boat was launched (fully fitted) the crane weighed it for us!

Eggs Actly.

I cannot imagine any reason for wanting to know the weight of your boat, except for booking a crane.

If you work on 1 tonne per yard then it will be an overestimate and you will be on the 'safe-side'.

It may mean you book an 'oversize' crane and it'll cost you a bit more, but generally the crane size is determined by the horizontal distance (boom length) required to lift the boat from the water as much as the boat weight.

You may need a 50 tonne crane to lift a 20 tonne boat because the boat is (say) 15 metres away from the nearest point that the crane can access.

A cranes boom is rated at lower and lower figures as the 'lift' distance increases.

[howardang]

3 hours ago, Mike Todd said:

I'm very rusty now (nearly 40 years since I had anything to do with this) Simpsons only calculates area (usually the ship's cross section - the problem being considered in the latter posts above is how to do a similar (but different) calculation for volume, taking into account that the cross section area is not a constant.

Like you, it is some time since I used Simpsons Rules in anger,  which as you say are used to calculate areas, but they can (and still are) used to find volumes both of regular and irregular ship/boat shaped objects by setting suitably spaced ordinates between water-planes and  summing the total volumes so found to arrive at the total volume of the vessel. If one of the water-planes is the waterline of the vessel, the total underwater volume can be calculated and hence the displacement, which will give the weight of the vessel, using the correct density for the water the vessel is floating in at the time of the calculation.

It is too tedious to go into details of how to cope with irregular shaped water-planes but this can be done by dividing the water-plane into  two or more separate parts and adding the results.

Using the rules it can also help to arrive at the block coefficient of the vessel which would also allow a quick calculation to find the displacement. 

 

Howard

[cuthound]

6 hours ago, Alan de Enfield said:

Eggs Actly.

I cannot imagine any reason for wanting to know the weight of your boat, except for booking a crane.

If you work on 1 tonne per yard then it will be an overestimate and you will be on the 'safe-side'.

It may mean you book an 'oversize' crane and it'll cost you a bit more, but generally the crane size is determined by the horizontal distance (boom length) required to lift the boat from the water as much as the boat weight.

You may need a 50 tonne crane to lift a 20 tonne boat because the boat is (say) 15 metres away from the nearest point that the crane can access.

A cranes boom is rated at lower and lower figures as the 'lift' distance increases.

 

Exactly, I once had to hire a 1000 tonne crane to lift a 40 tonne containerised generator set onto the roof of a 14 storey building in Docklands because of the limited weight rating of the paving around the building and the need for the generator to be close to the centre of the building.

[GUMPY]

Wheres @Albion when you need him!

He used to have a simple excel file that asked you for various parameters Length, beam, swim lengths etc that gave out an answer in tons.

I will look and see if i have a copy somewhere but it was a while ago.

Strangely I still have it. File below

 

Displacement.xls

Edited September 19, 2018 by Loddon

[Tim Lewis]

22 hours ago, Alan de Enfield said:

Eggs Actly.

I cannot imagine any reason for wanting to know the weight of your boat, except for booking a crane.

If you work on 1 tonne per yard then it will be an overestimate and you will be on the 'safe-side'.

It may mean you book an 'oversize' crane and it'll cost you a bit more, but generally the crane size is determined by the horizontal distance (boom length) required to lift the boat from the water as much as the boat weight.

You may need a 50 tonne crane to lift a 20 tonne boat because the boat is (say) 15 metres away from the nearest point that the crane can access.

A cranes boom is rated at lower and lower figures as the 'lift' distance increases.

 

The size of the crane used is normally determined by the size of cranes in the crane companies yard that morning :-)

 

Edited September 20, 2018 by Tim Lewis

[TheBiscuits]

1 hour ago, Tim Lewis said:

 

The size of the crane used is normally determined by the size of cranes in the crane companies yard that morning :-)

 

 

[Alan de Enfield]

And these :

 

 

 

 

 

[NickF]

On 19/09/2018 at 09:18, Alan de Enfield said:

Question : Does a litre of 'still' water weigh more than a litre of 'fizzy' water ?

(Logic being that the gas content weighs less than water.)

Suerley that depends on whether you have taken the top off the bottle!