Getting down to the last bits on my 43/44 WM20 recomissioning (yeah... I know you've heard that one before). I'm looking at a battery at this point. I want a sealed type to fit a hollowed out Lucas battery case. I've got a tight space because of the battery box wall thickness. I'm looking at a 2.5 Amp-Hour and think that should be adequate. Anyone have an opinion if something like this should work:
And a related question: what size fuse would be adequate? I'm going to install two fuses, one for the main wiring harness and another for my brakelight. All my tailights are LED, so I'll be using a 1 amp there. What would be a good amp rating for a standard main bulb system with horn? I'm thinking 10 amp.
I didn't realize that they made those Cyclon batteries as small as 2.5 amp. I usually fit the 5 amp version. There is room to fit 2 x 5 amp Cyclons in a normal dummy battery box. Ron
Fuse rating should also take account of any recommendations from the supplier of an electronic regulator if you're using one. (Dynotec suggest 15 amp on a 6v system).
I have no experience of a hollowed out battery case but can squeeze a 13a/h AGM in one of Ian's dummy boxes.
The greater the capacity, the longer you can run with lights if there is a problem and here in Belgium we have compulsory daytime headlights.
If I am not mistaken, I fitted 2 units of 6V 7 or 7.5Ah (each) in mine, in parallel connection,
In order to get 14 or 15Hh (Amp/Hour) capacity.
A 2.5Ah will do OK, and also let your dynamo “Rest” allot, at it would be quick to fully charge, but as Rik mentioned,
Would be quicker to drain when you run the regular lamp with no charge.
When only using the LED, it would last for hours.
I used only one 10A fuse on the main (+) batt lead for the entire system.
My back light is LED and so is the sidecar’s lamp. I usually use the low light at the headlamp, which is also LED,
And only use the main "H" lamp (which is Halogen) when it is really dark and no street lights around.
If you want to calculate,
The horn draws up to about 4-5 Amp, and for the headlight it is 1A for every 6Watts on a 6V system.
P(Output in Watts)=I (Current in Amps) x V (Volts) so I=P/V
