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


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13 minutes ago, MtB said:

 

 

 

 

 

Is it?

 

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

Not quite

wasnt aware I should have been multiplying 12 by 20

I haven’t a clue about dinosaur money. 

Edited by Goliath
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14 minutes ago, IanD said:

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.

 

Going down hill

The upper pound does not change 

The lower pound has  a locks worth of water added to it.

But a locks worth of water makes no measurable difference in a large pound.

 

36 minutes ago, David Mack said:

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.

The volume displaced by the boat is moved from the upper pound to the lower pound

If the boat was not there that volume displaced by the boat would be water.

But yes  the boat is then displacing water  in the lower pound . 

Except a locks worth of water (less the boats displacement) has moved from the lock into to the lower pound. But if the lower pound is large the change in water level is so small as make no measurable difference.

When the lock is filled  again the water level goes down but if the upper pound is large this makes no measurable difference.

In practice pumps or feeds from rivers or reservoirs maintain the levels in a canal , usually.

T

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27 minutes ago, MartynG said:

Going down hill

The upper pound does not change 

The lower pound has  a locks worth of water added to it.

But a locks worth of water makes no measurable difference in a large pound.

 

The volume displaced by the boat is moved from the upper pound to the lower pound

If the boat was not there that volume displaced by the boat would be water.

But yes  the boat is then displacing water  in the lower pound . 

Except a locks worth of water (less the boats displacement) has moved from the lock into to the lower pound. But if the lower pound is large the change in water level is so small as make no measurable difference.

When the lock is filled  again the water level goes down but if the upper pound is large this makes no measurable difference.

In practice pumps or feeds from rivers or reservoirs maintain the levels in a canal , usually.

T

You still don't get it.

 

It doesn't matter how big or small the upper and lower pounds or the lock or the boat are, the changes in water levels above and below the lock are identical regardless of the size of the boat -- all the way from no boat at all (lock 0% full) to a boat that fits like a piston (lock 100% full).

 

If you don't understand why this is so, I don't know how to make it any clearer 😞

Edited by IanD
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4 hours ago, MtB said:

My own point of view is changing. I think narrowboats are being unfairly discriminated against by the licensing system.

 

I pay almost as much license fee for my narrow boat as a widebeam pays.

 

I pay 83% of what a 13ft 8in wide boat pays, yet my boat is only half the size. Blatantly unfair and discriminatory.

Are but you have all the waterways to use so you are in reality paying less 🤣🤣

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1 hour ago, IanD said:

It doesn't matter how big or small the upper and lower pounds or the lock or the boat are, the changes in water levels above and below the lock are identical regardless of the size of the boat -- all the way from no boat at all (lock 0% full) to a boat that fits like a piston (lock 100% full).

 

If you don't understand why this is so, I don't know how to make it any clearer 😞

 

Let's get off the locks and into the bar.

Order a scotch on the rocks.

The amount of scotch in the glass doesn't change no matter how much ice you add but the level of the liquid goes up.

Drink only the scotch (you can't, some of the ice will have melted). The level of the liquid goes down. What should be left is only the ice.

 

@David Mack's explanation was pretty good.

 

The lock when 'empty' contains 75' x 7'6" x 4' of water. 22.86m x 2.286m x 1.22m = 63.75m^3

When 'full' the lock contains 75' x 7'6" x 12' of water. 22.86m x 2.286m x 3.66m = 191.25m^3 (nothing has changed except the depth of the water)

The volume used to fill the lock is 127.5m^3

The boat is (for the sake of argument) 70' x 7' and has an average draft of 2'6. That's 21.34m x 2.134m x 0.76m = 34.6 m^3. 

 

Put the boat in at the bottom, the lock now contains 63.75 - 34.6 = 29.15m^3

Fill the lock, it now contains 191.25 - 34.6 = 156.65m^2

 

How much water has been used? Hint: subtract the amount of water in the empty lock from the amount in the full lock.

Compare this with the volume used to fill the lock calculated above.

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15 minutes ago, George and Dragon said:

 

Let's get off the locks and into the bar.

Order a scotch on the rocks.

The amount of scotch in the glass doesn't change no matter how much ice you add but the level of the liquid goes up.

Drink only the scotch (you can't, some of the ice will have melted). The level of the liquid goes down. What should be left is only the ice.

 

@David Mack's explanation was pretty good.

 

The lock when 'empty' contains 75' x 7'6" x 4' of water. 22.86m x 2.286m x 1.22m = 63.75m^3

When 'full' the lock contains 75' x 7'6" x 12' of water. 22.86m x 2.286m x 3.66m = 191.25m^3 (nothing has changed except the depth of the water)

The volume used to fill the lock is 127.5m^3

The boat is (for the sake of argument) 70' x 7' and has an average draft of 2'6. That's 21.34m x 2.134m x 0.76m = 34.6 m^3. 

 

Put the boat in at the bottom, the lock now contains 63.75 - 34.6 = 29.15m^3

Fill the lock, it now contains 191.25 - 34.6 = 156.65m^2

 

How much water has been used? Hint: subtract the amount of water in the empty lock from the amount in the full lock.

Compare this with the volume used to fill the lock calculated above.

 

And you're *still* missing the point; the volume of water "used" when cycling a lock is how much has to be put back in to restore the levels, it's nothing directly to do with the volumes in the lock. We're effectively swapping a boat (displacing 20t of water) with 20t of water in a game of "find-the-lady", and the effect on levels is the same if you replace the boat with 20t of water -- in other words, cycle the lock with no boat in it.

 

If you don't believe me, lets replace the steel boat with a boat made from 20t of ice. Do we all agree that the material of the boat makes no difference? (I sincerely hope so, Archimedes would be spinning in his grave otherwise).

 

Now cycle the 20t ice-boat from above the lock to below the lock, this will cause a given change in water levels, needing replenishment from the reservoir to put them back -- which by definition is the amount of water used by the lock.

 

Agreed?

 

We don't need to know how much this is to show that the boat is not affecting the amount of water used.

 

Compare this to exactly the same lock cycle but where the ice-boat has been melted back to 20t of water, which Archimedes tells us displaces the same volume of water. Now cycle the lock. The changes in water levels will be exactly the same as when we did this with the iceboat, except we haven't moved a boat down the lock, we've cycled an empty lock.

 

You don't need to do any complicated sums about water transfer or boat or lock volumes or water levels for this to be true, in fact they're just a distraction as well as being prone to error if you don't allow for everything and make all the correct calculations.

 

Since cycling a lock with a 20t boat (made of ice or steel) has the same effect on levels as cycling a lock with 20t of water (in other words, an empty lock), then the boat makes no difference to the amount of water used.

 

Step away from the numbers and think about this before throwing out more numbers to try and disprove it --- because it's like algebra, if an equation is true then it's true for all numbers.

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Account needs to be had of any change in displacement whilst the boat in in the lock.  Wide boats will tend to have more (and maybe heavier) crew to alight or embark during the period the gates are sealed at both end - and thus are liable to a greater effect.

 

And a wide boat (or the same length) will help reduce evaporation.   And a wide boat (i.e.  greater weight) will raise the cut more when first launched.

 

Hope that helps.

 

 

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18 minutes ago, Tacet said:

Account needs to be had of any change in displacement whilst the boat in in the lock.  Wide boats will tend to have more (and maybe heavier) crew to alight or embark during the period the gates are sealed at both end - and thus are liable to a greater effect.

 

And a wide boat (or the same length) will help reduce evaporation.   And a wide boat (i.e.  greater weight) will raise the cut more when first launched.

 

Hope that helps.

 

 

Is that why you have to stay on the boat on the Falkirk wheelm to keep it all in balance which only happens with displacement

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22 hours 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 -- same water transfer between pounds regardless of size of boat (or none).

 

Agreed?

No

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18 minutes ago, Mike Todd said:

No

<sigh> I'll restate what I actually meant, following later posts...

 

The level of the water in the canal above and below the lock both before and after the lock cycles is identical regardless of the size of the boat (or no boat). Since the water levels determine how much is needed to be drawn from the reservoir, this is "the water used by the lock" -- and it only depends on the lock, not any boat in it.

 

Agreed?

 

If you don't, please show your workings -- as the examiner would say... 😉

Edited by IanD
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On 06/05/2022 at 10:59, IanD said:

<sigh> I'll restate what I actually meant, following later posts...

 

The level of the water in the canal above and below the lock both before and after the lock cycles is identical regardless of the size of the boat (or no boat). Since the water levels determine how much is needed to be drawn from the reservoir, this is "the water used by the lock" -- and it only depends on the lock, not any boat in it.

 

Agreed?

 

If you don't, please show your workings -- as the examiner would say... 😉

A boat afloat displaces an amount of water. Before the boat enters the lock this displacement is part fo the pound above and the lock is full. However, as the boat enters the lock is displaces this amount of water out of the lock back into the pound, leaving the water in the lock when the gates are closed as being a lock full minus the displacement as indicated in the traditional formulae.

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Just now, Mike Todd said:

A boat afloat displaces an amount of water. Before the boat enters the lock this displacement is part fo the pound above and the lock is full. However, as the boat enters the lock is displaces this amount of water out of the lock back into the pound, leaving the water in the lock when the gates are closed as being a lock full minus the displacement as indicated in the traditional formulae.

And when you have finished emptying the lock, but not opened the bottom gates, the lock contains an empty lock full minus the displacement. I.e exactly one lock full less, regardless of the size of the boat.

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41 minutes ago, Mike Todd said:

A boat afloat displaces an amount of water. Before the boat enters the lock this displacement is part fo the pound above and the lock is full. However, as the boat enters the lock is displaces this amount of water out of the lock back into the pound, leaving the water in the lock when the gates are closed as being a lock full minus the displacement as indicated in the traditional formulae.

You're still not reading what I said -- it's all to do with water levels, not transfer of water.

 

If a boat goes down the lock, after the transit you've swapped 20t of boat above the lock with 20t of water below the lock. If if doesn't, you've swapped 20t of water above the lock with 20t of water below the lock.

 

In both cases the water usage (amount that the upper pound has to be topped up with) is exactly the same, and it's a lockfull of water (ignoring small changes in pound levels) -- large boat or small boat or no boat makes no difference as far as the canal (and water supply) is concerned.

 

Arguing about the fact that in the first case 20t of water went up (water transfer) is irrelevant, it makes no difference to the amount of water used.

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31 minutes ago, D Ash said:

Biscuit and his crane is on the right tracks here.   Fill the locks with concrete and crane the boats over!

And then Mike *will* be right, a wideboat will use more water going down a lock than a narrow one 🙂

 

Of course this gets reversed when it comes back up, so when moving round the system there's no difference -- unless the canals are like an Escher staircase... 😉

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1 hour ago, David Mack said:

Like this at Spon Lane and Smethwick

Capture.PNG

Yeah, but they don't go up all the way back to where you started... 🙂

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10 minutes ago, IanD said:

Yeah, but they don't go up all the way back to where you started... 🙂

Up Smethwick, up Spon Lane, then no locks via the New Main Line takes you back to your start point at the bottom of Smethwick (according to the map).

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