Water water, but not everywhere

It’s not often I want it to rain over the holiday period, but we could do with a lot right now.  Herbie is in danger of being isolated in a few weeks time unless the reservoirs that feed the canals around the Leicester arm get some serious topping up.  We had planned a trip down to Oxford in September but that too looks under threat as things stand.

We are getting messages from BW talking about what their hydrologist is telling them.  I used to have a friend who was a hydrologist, and his job sounded great.  Each week he would beetle off to Thetford Forest, where he would live in a hut for a few days and take regular trips round the forest in a Land Rover collecting little bottles of water from the trees and the forest floor.  They were trying to establish how much of the rain that fell over the area actually made it to the ground and then into watercourses.  I suppose a lot of it got absorbed by the leaves, a lot of it just evaporated, then much of the rest of it got drank by the tree roots.  I don’t think a lot of it got into streams and rivers.

Last week I got to thinking about how much rain we might need to rescue my canal trip plans, and did a few simple sums.  I suspect these are grossly over simplified but it gives some idea.

Take a theoretical ten mile length of canal with a dozen single boat width locks and assume that it is fed by a reservoir a quarter of a mile square.  How many boats would be able to pass through on the proceeds of one inch of rain?  Have a guess now then let’s do the sums.

We’ll forget about extra feeder streams, water lost by evaporation and all that, so this won’t be spot on but it might give us a rough idea.

As a boat passes through the area it will take with it the volume of water used by the deepest lock in the passage.  Lets say 8 ft deep .  So that’s  about 70ft x 7ft x 8ft = 3920 cu ft.

The reservoir at 1320 feet x 1320 feet (quarter of a mile each way) has a surface area of 1742400 sq ft (phew!).  Multiply by 1/12 to give us the volume added by an inch of rain falling on its surface.  That gives us  145200 cu ft.  Divide that by the lock capacity of 3920 and we get 37.  That’s 37 boats that can now pass through.  Not many on a canal like the Oxford.  Barely a day’s worth.

Of course rain falls on the canal too.  10 miles x say 30ft wide gives us 10x 5280x30 = 158400 sq ft of surface.  x 1/12 for an inch of rain gives us 132000 cu ft added.  Divide by 3920 = 33.6 lockfulls, so another 33 boats.

So altogether an inch of rain would allow the passage of 70 boats.

Put another way, if we only get an inch of rain in a fortnight over the area, that would only provide for 5 boats a day!  And that assumes every lock is set ready for the boat, or a one up one down policy strictly enforced.
Don’t quote me on any of this because this is very rough and ready and takes no account of possible feeder streams into the reservoir or canal, evaporation, back pumping and all that.  But it does show that the odd shower of rain won’t do much to help.

Am I anywhere near right?

PS after my disaster last week I am writing this in Windows Live Writer before posting.  It does have an undo button.  Thanks Skippy for the tip.  Now let’s see if it will post to the blog.