Is the current infrastructure really any worse than the 70s 80s?

[TheBiscuits]

2 minutes ago, IanD said:

OK, let's compare two cases -- a wideboat and no boat going down a lock...

 

The top gates are open, water is level with upper pound. If there's a boat (wide or not) it floats into the lock and the gates close, if not the gates close anyway.

 

Lock is emptied, water level drops by the fall of the lock, volume is the same (area*fall) whether there's a boat in there or not.

 

Bottom gates open when water is level with lower pound, boat (if there is one) floats out.

 

The same volume of water has been transferred from upper pound to lower pound (area of lock * fall) whether there was a wideboat, a narrowboat or no boat in it.

 

Exactly the same happens in reverse for a boat (or none) going up the lock.

 

Agreed?

 

No.

 

The water displaced by the boat gets left in the pound above the lock and replaced from the pound below the lock when the boat exits.

 

This volume of water hasn't been "used" when the boat is present, so it's still available (uphill) for the next lock cycle.

 

Admittedly it's unlikely that the lock doesn't leak, and if the bywash is running it's completely irrelevant anyway!

[Jen-in-Wellies]

6 minutes ago, IanD said:

The same volume of water has been transferred from upper pound to lower pound (area of lock * fall) whether there was a wideboat, a narrowboat or no boat in it.

Wide boats cause extra wear and tear on aquaducts, because they are heavier! 😀

Archimedes, what did he know?

 

[TheBiscuits]

3 minutes ago, TheBiscuits said:

The same volume of water has been transferred from upper pound to lower pound (area of lock * fall) whether there was a wideboat, a narrowboat or no boat in it.

 

The volume used is actually (area of lock * fall) - displacement.

 

 

 

[IanD]

Just now, TheBiscuits said:

 

No.

 

The water displaced by the boat gets left in the pound above the lock and replaced from the pound below the lock when the boat exits.

 

This volume of water hasn't been "used" when the boat is present, so it's still available (uphill) for the next lock cycle.

 

Please explain how the boat going into a lock changes the water level in the upper pound and the open lock.

 

Clue : it doesn't.

 

If the level is the same, the lock doesn't care whether there's a boat in it or not, the same amount of water is transferred.

 

Same for the boat coming out at the bottom.

 

If the levels above and below the pound don't change when the boat enters or exits the lock -- which they can't, it's floating on the same bit of water -- then there can't be any net movement of water as a consequence. As far as the lock is concerned, it's swapped a boat-sized lump of water for a boat.

 

Oh, I really wish I'd not got involved in this, the same people are making the same errors as all the other times the same issue came up... 😞

[MtB]

1 minute ago, Jen-in-Wellies said:

Wide boats cause extra wear and tear on aquaducts, because they are heavier! 😀

Archimedes, what did he know?

 

 

This is why there are "No Mooring" signs on aquaducts sometimes. Some at CRT thinks they can't take the extra weight of the boats 

[IanD]

2 minutes ago, TheBiscuits said:

 

The volume used is actually (area of lock * fall) - displacement.

 

 

 

No it's not. Think about it again...

 

The boat stays floating, the water comes in (or goes out) at the bottom of the lock (area*fall), so the boat displacement has nothing to do with it -- the lock doesn't care if there's a boat in there, or a lump of water with the same displacement.

 

I don't see how this can be explained more clearly, it's basic hydrodynamics -- hence the Archimedes mention...

Edited May 5, 2022 by IanD

[MtB]

1 minute ago, IanD said:

Please explain how the boat going into a lock changes the water level in the upper pound and the open lock.

 

Clue : it doesn't.

 

Playing Devil's advocaat.... (  )

 

As the boat goes in, the displaced water flows out from the lock into the upper pound, raising the water level in it ever so slightly! 

 

Or so the argument went last time all this was rehearsed.

 

[IanD]

1 minute ago, MtB said:

 

Playing Devil's advocaat.... (  )

 

As the boat goes in, the displaced water flows out from the lock into the upper pound, raising the water level in it ever so slightly! 

 

Or so the argument went last time all this was rehearsed.

 

I know -- and I'm sure you do -- that all this does is swap the places of a boat with a lump of water the same size. If this changed the water level, you'd have a perpetual motion machine... 😉

[TheBiscuits]

2 minutes ago, MtB said:

Devil's advocaat

 

That's the evil yellow stuff then?

 

 

[MtB]

1 minute ago, IanD said:

I know -- and I'm sure you do -- that all this does is swap the places of a boat with a lump of water the same size. If this changed the water level, you'd have a perpetual motion machine... 😉

 

Quite.

 

And similarly with a full and empty lock, containing a boat or no boat.

Just now, TheBiscuits said:

 

That's the evil yellow stuff then?

 

 

 

Yep, that ghastly muck. I think it is a brand name so mebbe I should of used a capital initial i.e. "Advocaat"

 

Alcoholic custard!

 

https://en.wikipedia.org/wiki/Advocaat

 

[TheBiscuits]

1 hour ago, IanD said:

The boat stays floating, the water comes in (or goes out) at the bottom of the lock (area*fall), so the boat displacement has nothing to do with it -- the lock doesn't care if there's a boat in there, or a lump of water with the same displacement.

 

On further thought you are correct. 

 

If the boat was displacing enough water from the lock to affect the water usage it wouldn't be able to get in through the headgates - it would have to be deeper than the cill to alter the area x fall, which is different to the total volume of water in the lock.

 

I was mixing up the total volume of water in the lock - which does vary by boat(s) - with the change in volume of water when the lock is cycled.

 

Add:. Further further thought says you are only partially correct, see my next post.

 

 

Edited May 5, 2022 by TheBiscuits
Add a bit

[Momac]

18 minutes ago, IanD said:

 

 

The same volume of water has been transferred from upper pound to lower pound (area of lock * fall) whether there was a wideboat, a narrowboat or no boat in it.

 

No

Compared to a lock drained down with no boat in it the weight of water moved is reduced by the weight of water displaced by the weight of the boat.

eg if a boat weighs 10Tonnes the process will require 10 Tonnes less water  to be moved with the boat in the lock compared to the same process with an empty lock.

 

 

 

 

 

 

 

 

[Momac]

28 minutes ago, IanD said:

As far as the lock is concerned, it's swapped a boat-sized lump of water for a boat.

 

Yes it has . You are getting there.

[TheBiscuits]

36 minutes ago, IanD said:

I know -- and I'm sure you do -- that all this does is swap the places of a boat with a lump of water the same size. If this changed the water level, you'd have a perpetual motion machine... 😉

 

Playing Angel's crème de menthe ...

 

If you close the lock and then crane the boat out of it, the water level in the lock drops by precisely the displacement of the boat.

 

If you then drop the lock from this state, it's clear that less water is transferred from the upper pound to the lower pound.

 

Therefore cycling a lock with a boat in it uses less water from the upper pound than cycling the same lock with no boat in it, by exactly the displacement of the boat.

 

The volume required to raise or lower the lock by it's fall remains constant as you correctly argued above, but with a boat in the lock you transfer a smaller volume of water from above the lock.

 

I think it depends how we define "using" water for a lock cycle.

 

 

 

 

 

 

 

[Momac]

34 minutes ago, MtB said:

 

 

 

As the boat goes in, the displaced water flows out from the lock into the upper pound, raising the water level in it ever so slightly! 

 

 

 

No the water level stays the same .

But with the gate closed there is less water in the pen compared to a pen with no boat in it.

 

[Midnight]

1 hour ago, MtB said:

 

As the boat goes in, the displaced water flows out from the lock into the upper pound, raising the water level in it ever so slightly! 

 

 On the Rochdale it pours over the bottom gates. I bet that's buggered up some theories. 😆

[David Mack]

1 hour ago, MartynG said:

Lets say you are upstream from a lock that is empty .

You move your boat into the lock but the water level in the lock does not rise . When you enter the lock  water equivalent to the weight of your boat  is displaced  out of the lock and into the canal upstream from the lock.

So when you close the gates and open the suices the lock drains down. With your boat in the lock the amount of water drained down is less than it would be if you had drained the lock with no boat in the lock.

Do you agree ?

Think about it this way. A typical narrow lock has a fall of 8 feet and there is about 4 feet of water over the cills, so the total depth of water in the full chamber is 12 feet (and 4 feet when empty).  When the boat enters a full lock it floats within the top 2 or 3 feet of water in the chamber. Now open the bottom paddles and allow the bottom 8 feet of water to flow into the pound below. The top 4 feet, with the boat in it, simply drops 8 feet vertically, but does not otherwise move. So an 8 foot x lock length  x lock width volume of water has passed to the lower pound. And that is regardless of the size of the boat, or indeed if one is present at all. Of course when the boat entered the lock it displaced a boatsworth of water from the chamber to the upper pound, but on leaving the lock an identical quantity of water moved from the lower pound back into the chamber. So the net effect of that is zero.

[IanD]

The water levels above and below the lock are the same after the lock has been cycled, regardless of whether a boat went up or down, or no boat at all -- or if the boat was a 20-ton block of ice which you melted in the lock, so a boat went in and no boat came out.

 

It's trivial to prove this -- and if it works one way it works the other, so you can take a boat down the lock and back up again to get back to where you started, or no boat at all. You don't solve problems like this by analysing fluid flow ( @MartynG ), you use water levels, and to show something makes no difference you try two extreme cases.

 

If you do that with no boat and a boat that completely fills the lock, you find the levels above and below the lock before and after cycling are the same in both cases, and this is true no matter how big or small the two pounds are and how much their levels change by as a result.

 

Since it's water levels that matter on a canal (because these are what has to be topped up), a lock effectively uses the same amount of water with or without a boat.

 

There are problems very like this (but sometimes a lot more complex) at the end of the relevant chapter in my fluid mechanics textbook, which might be over 40 years old but is still valid -- unless the Large Hardon Collider did something nasty to physics when nobody was looking... 😉

Edited May 5, 2022 by IanD

[MtB]

2 minutes ago, IanD said:

unless the Large Hardon Collider did something nasty

 

One's mind boggles...

[TheBiscuits]

16 minutes ago, David Mack said:

Of course when the boat entered the lock it displaced a boatsworth of water from the chamber to the upper pound, but on leaving the lock an identical quantity of water moved from the lower pound back into the chamber. So the net effect of that is zero.

 

No, the net effect on the volume in the lock chamber is zero.

 

There is still the boat's displacement of water left in the upper pound.  Which is what we are really interested in as it's available for the next lock cycle.

 

 

[IanD]

4 minutes ago, TheBiscuits said:

 

No, the net effect on the volume in the lock chamber is zero.

 

There is still the boat's displacement of water left in the upper pound.  Which is what we are really interested in as it's available for the next lock cycle.

 

 

The displacement of water doesn't matter as far as the canal is concerned, only the levels... 😉

4 minutes ago, MtB said:

 

One's mind boggles...

That was intentional -- I put that in as a deliberate typo when I gave a talk there several years ago and nobody noticed... 😉

Edited May 5, 2022 by IanD

[beerbeerbeerbeerbeer]

4 hours ago, MtB said:

 

I think you're demonstrating your poor grasp of simple arithmetic again.

 

I have read the conditions and as I've already said, with reasons, I think they are unfair.

 

 

My maths is ok

Which is why I see the simplicity of the current reckoning. 
 

If you wanna make yourself appear more erudite by adding further complications then go for it, crack on. 

[MtB]

3 minutes ago, IanD said:

The displacement of water doesn't matter as far as the canal is concerned, only the levels... 😉

 

 

Or put a different way, the displacement of water is the same at the start of the process as at the end, so the two cancel each other out. It's the difference in water levels that one uses for the calc. 

 

[IanD]

3 minutes ago, MtB said:

 

 

Or put a different way, the displacement of water is the same at the start of the process as at the end, so the two cancel each other out. It's the difference in water levels that one uses for the calc. 

 

Correct -- the canal doesn't care if it starts with a boat above the lock and the same sized lump of water below and they swap places, or if you do this again and go back up, the boat (or not) makes no difference to any water levels, which is what determines how much water the lock "uses" i.e has to be topped up from the reservoir.

Edited May 5, 2022 by IanD

[MtB]

 

 

 

2 minutes ago, Goliath said:

My maths is ok

 

 

Is it?

 

Was it not you thought 12 x 20 was 144 in another thread?