A shameless bump post so it appears on my helpful fellow bloggers blog rolls...;-)
She is up photographed and has her own brochure now....
http://narrowboats.apolloduck.co.uk/feature.phtml?id=284510
Please pass this info around as you see fit.... blog about it would be even nicer ;-)
Thanks
Nev & Rachel
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12 hours ago
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2 comments:
Nev - I tried to leave this comment on your latest post on Percy, but it doesn't seem to have worked:
Nev, there are only two circumstances where the voltage at the batteries would be exactly the same as the voltage at the alternator. One: no current is flowing; two: there is zero resistance between the batteries and the alternator.
Case one could theoretically arise if the batteries just happened to be at the same voltage as the alternator - unlikely if you are measuring 14.5V.
Case two would imply that the connections and the cables connecting the batteries and alternator together have zero resistance, which is impossible at normal temperatures.
Ohm's Law is essential: V = IR (voltage equals current times resistance). The voltage here is that across the resistance in question, i.e. the cables/connectors. If we assume a reasonable alternator output of 20A, then a resistance as low as 0.01 ohms would give a voltage drop of 20 x 0.01 = 0.2V. A measurable voltage drop. If the resistance of the cables/connections were to rise to 0.05 ohms then you'd get a significant drop of one volt.
As the batteries charge up, the current flowing from the alternator falls, so that the effect of the resistance diminishes. In the example above with resistance of 0.01 ohms and a reduced current of 2A, the voltage drop is now only 2 x 0.01 = 0.02V, barely noticeable.
Introducing a forward-biassed diode, as in a split-diode-charging system, into the circuit immediately gives you an extra voltage drop of 0.5V - 0.7V, reasonably irrespective of the amount of current flowing.
This gives an indication of the importance of good clean connections, and as thick and short a cable as practicable (the thicker the cable the less resistance; the shorter the cable the less resistance too).
I hope this is helpful, if not to you, Nev, then to anyone else who might chance upon this!
Thanks Halfie, I have posted it up with an explanation. Always learning ...thanks for taking the time to explain,
Nev
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