Sunday, 26 August 2012

t = s/v

There's a boat coming the other way.


Thing is, which of us is going get to the bridge first? The farther away you both are, the harder it is to tell. Who's nearest? How fast are they going? Are they passing moored boats? Am I passing moored boats? Is that sparkle on the water 400 yards ahead a bow wave, or propellor turbulence, or a flotilla of ducklings?


You take these decisions all the time at 70 mph on the motorway. The brain is fantastic at making almost unconscious judgements about the relative motion of vehicles. At 3 mph on the cut you'd think everything would be even easier, but I've found it more tricky. Distances are deceptive through a bridge, but there's a lot hanging on it. Your nautical elegance is partly a function of how you manage such meetings!

I can see some other decisions coming up. Will Rich be able to do the paintwork? How long will it take? If he can, should we stay on board to learn, or go home and let him get on with it? What if it's raining…  Decisions! It's crunch-time. No – unfortunate expression, especially if the boat coming the other way is a GRP.
















The equation in the title is: time = distance / velocity. It's the particular variation of this equation that applies here (who's going to get there first?). I don't remember why "s" stands for distance, but I'm sure it used to when I was at school. And I think velocity is a better concept here than speed because it involves direction as well as speed, and direction sure is important when you're heading for a bridge.


Meanwhile, the heron just got on with applying the physics of motion to the art of fishing for dinner, without even thinking about it.

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