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How much does my boat weigh?


NbPlod

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4 minutes ago, NbPlod said:

Anybody know how much an all steel 36ft Springer might weigh? Thinking of transporting my dead-engine boat to hard standing.

My guess is that, bearing in mind the length of your boat and the typical plate thickness of Springers, it would be seven or eight tons. But it IS only a guess, albeit I hope an intelligent one, and if anyone knows some sort of formula for working out boats' weights, I would be interested in reading it.

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5 minutes ago, Athy said:

My guess is that, bearing in mind the length of your boat and the typical plate thickness of Springers, it would be seven or eight tons. But it IS only a guess, albeit I hope an intelligent one, and if anyone knows some sort of formula for working out boats' weights, I would be interested in reading it.

Bit difficult if its a V bottom Springer but with a flat bottom.

 

All in meters

Half of draft x length of front swim x beam

Half of draft x length of stern swim X beam

Draft at center of hull x length of parallel section of hull x beam

 

Added together and as 1 cu M of water weight about 1 tone you have the weight of the boat give or take.

 

It is also said that typically narrowboats weight about 1 tone per meter in length but I doubt that holds true for a Springer so Athy's figure is [probably in the right ball park.

 

I suppose you could modify the calculations for a V bottom by treating the center of the hull as two parts, the V and the section above it and halving the  V's depth x beam.

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13 minutes ago, Athy said:

Thanks, Tony.

So, in English , about a ton per yard for an average boat, if such a thing exists?

Yes, with the "about" in bold text.

 

However if the boat has never been in the water so the draft can be accurately established then you need the pivot, jack and bathroom scales method that has been described on here at least twice.

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42 minutes ago, Athy said:

My guess is that, bearing in mind the length of your boat and the typical plate thickness of Springers, it would be seven or eight tons. But it IS only a guess, albeit I hope an intelligent one, and if anyone knows some sort of formula for working out boats' weights, I would be interested in reading it.

 

The thickness of the steel is not relevant. One adds more ballast to a boat with thinner steel than to another identical boat made from thicker, in order to get the weights and trims the same. 

 

The weight can be roughly calculated by measuring the volume of water the boat displaces. Archimedes tells us the weight of the boat will be equal to the weight of the volume of water displaced. And we know how much water weighs, don't we!

 

 

Edited by Mike the Boilerman
Add a bit.
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2 minutes ago, Mike the Boilerman said:

 

The thickness of the steel is not relevant. One adds more ballast to a boat with thinner steel than another with thicker, to get the weight the same. 

 

 

Good point - though I'm sure that that some "ones" do this better than others.

 

5 minutes ago, Mike the Boilerman said:

 

 

 

The weight can be roughly calculated by measuring the volume of water the boat displaces. Archimedes tells us the weight of the boat will be equal to the weight of the volume of water displaced.

I was tempted to retort "Screw Archimedes" but of course I shan't.

Yes, but gathering this water in a manner which enables it to be weighed cannot be a simple task - hence, a way of at least guesstimating the weight without doing so is useful.

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3 minutes ago, Athy said:

Yes, but gathering this water in a manner which enables it to be weighed cannot be a simple task -

 

You don't have to actually gather the water, you look at the hull shape and approximately calculate the volume below the waterline. Its the same thing. 

 

 

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29 minutes ago, Athy said:

I was tempted to retort "Screw Archimedes" but of course I shan't.

Yes, but gathering this water in a manner which enables it to be weighed cannot be a simple task - hence, a way of at least guesstimating the weight without doing so is useful.

Fill a lock, measure the height of the water from the top of the lock wall, drive boat into lock, measure the height of the water to the top of the lock wall.

 

Multiply the difference by the width and length of the lock and you will have the volume of water displaced.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

You will find the answer is zero.

  • Greenie 1
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34 minutes ago, Mike the Boilerman said:

 

You don't have to actually gather the water, you look at the hull shape and approximately calculate the volume below the waterline. Its the same thing. 

 

 

I too was puzzling over whether Athy had made a joke or had spectacularly missed the point of the Archimedes' principle. But as Athy often assures us he never misses the point of anything I assumed it was a joke.

 

It's very easy to calculate the mass of a boat as long as you know the basic dimensions of the baseplate, the depth of the side plate and how much dry side there is along the length of the boat. Vulpes is designed to be as near as damn it 10 tonnes in fully loaded trim at which it is level in the water. It's 35' feet long so it confirms the theory of approximately 1 tonne per yard/metre.

 

I know this because I have a print of a proper technical drawing of it dating from 1968 which is interesting in the context of another thread on here.

 

JP

Edited by Captain Pegg
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1 minute ago, Alan de Enfield said:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

You will find the answer is zero.

Er, I did pass O level maths but it was some time ago. How can it be zero?

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2 minutes ago, Athy said:

Er, I did pass O level maths but it was some time ago. How can it be zero?

Did you happen to pass O'Level physics ?

 

Think about it.

 

Does a lock with a boat in it use more or less water than a lock that is cycled with no boat ?

Edited by Alan de Enfield
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3 minutes ago, Alan de Enfield said:

Did you happen to pass O'Level physics ?

 

 

No, I think it was the only subject which I failed.

5 minutes ago, Alan de Enfield said:

 

 

Think about it.

 

Does a lock with a boat in it use more or less water than a lock that is cycled with no boat ?

One would think "less", though if the lock was already full when the boat entered it, I suppose it would be the same amount. It's quite a puzzle!

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1 hour ago, Captain Pegg said:

I too was puzzling over whether Athy had made a joke or had spectacularly missed the point of the Archimedes' principle. But as Athy often assures us he never misses the point of anything I assumed it was a joke.

 

It's very easy to calculate the mass of a boat as long as you know the basic dimensions of the baseplate, the depth of the side plate and how much dry side there is along the length of the boat. Vulpes is designed to be as near as damn it 10 tonnes in fully loaded trim at which it is level in the water. It's 35' feet long so it confirms the theory of approximately 1 tonne per yard/metre.

 

I know this because I have a print of a proper technical drawing of it dating from 1968 which is interesting in the context of another thread on here.

 

JP

I think the question was the weight not the mass. Hence the displacement process.

 

The weight/m will depend, at least a little, on the length as the proportion of the length that is not full beam will change.

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26 minutes ago, Mike Todd said:

I think the question was the weight not the mass. Hence the displacement process.

 

The weight/m will depend, at least a little, on the length as the proportion of the length that is not full beam will change.

The original question was in relation to lifting, the buoyancy stuff was just the method of calculation. Obviously the OP was using the term weight in its everyday form and not by it's strict scientific definition. Do you think telling him his boat weighs 80kN (on the surface of the earth) would be helpful?

 

JP

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3 hours ago, Alan de Enfield said:

Did you happen to pass O'Level physics ?

 

Think about it.

 

Does a lock with a boat in it use more or less water than a lock that is cycled with no boat ?

It's even easier than that!

Water slopping about in a full lock would make determination of the level impossible anyway, but the boat was floating at the same level when it was just outside the lock as when it was inside with the gates shut.  

1 hour ago, Mike Todd said:

I think the question was the weight not the mass. Hence the displacement process.

 

The weight/m will depend, at least a little, on the length as the proportion of the length that is not full beam will change.

The weight for transport purposes only needs to be approximate, and a calculation of the displacement volume to the nearest 0.1 tonne (100kg or 100 litres) is easily accurate enough. The low loader will have sufficient spare capacity, also known as a safety margin. 

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1 hour ago, Captain Pegg said:

The original question was in relation to lifting, the buoyancy stuff was just the method of calculation. Obviously the OP was using the term weight in its everyday form and not by it's strict scientific definition. Do you think telling him his boat weighs 80kN (on the surface of the earth) would be helpful?

 

JP

No, but someone was enquiringly about O Level Physics . . . I wouldn't want anyone to fail an exam through reading incorrect info here . . .  

 

:) in case of doubt

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