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[longshanks]

Hi Folks,

I'm after some advice re power for the latest build.

Modern chine tug hull 30" x 10". Motors 2 x MFA 919D 6:1 driving 50mm dia 4 bladed props + Raboesch (Race 400 7.2v) bow thruster.

Using Mtroniks Marine 15 speed controllers for all.

Looking to use SLA batteries - always used NiMh packs before.

Due to hull dimensions/openings I need to use 2 x 6volt 7amp batteries wired in series to give me 12 volts.

1. Can I take off from one battery to supply 6 volts to bow thruster?

2. Can I treat the batteries as 1 x 12 volt when it comes to charging or do I need to disconnect and charge separately?

3. Can you get a reasonably priced (what I would call ) a peak charger for SLA? Using a 350/500 mAh charger how long do you charge for?

Please keep any replies fairly basic, I'm only a simple guy 

Regards
longshanks

[craftysod]

Q1 if two batteries are wired together for 12v you cannot get 6v from just placing the wires to 1 battery
Q2 if the batteries are connected,treat it as a 12v batt or you can disconect them and charge as 6v,recommended occasionely
I stand to be corrected
Mark

[Rex Hunt]

Q1 if two batteries are wired together for 12v you cannot get 6v from just placing the wires to 1 battery
Q2 if the batteries are connected,treat it as a 12v batt or you can disconect them and charge as 6v,recommended occasionely
I stand to be corrected
Mark

Corrected you will be!

If you connect the 2 batteries in series to give 12V you can take a feed across the one that is attached to your negative return to provide 6v.

Recharging is best done separately otherwis one battery may be overcharged and the other undercharged.


Rex

[malcolmfrary]

Adding to Rex, if you do use the tap method, you will get the negativemost battery discharging soonerr, because it is driving the thruster.  If you the re-charge the pair as if it was a 12 volt battery, you run the risk of not fully charging the tapped battery.
Running the thruster from the 12v supply is possible.  It will not be in continuous use, and it is being run through a controller, so does not have to be presented with excess voltage unless you want it that way.  You can also get a resistor to limit current through the thruster motor.

[stallspeed]

Longshanks,it would be easier to run everything off twelve volts and either...

• use a bow thruster esc that has the facility to limit the maximum power OR
• fit a volt drop resistor in the bow thruster feed.

[malcolmfrary]

While one big battery might be electrically ideal, it might not be physically possible due to the space available or the access.

[stallspeed]

Longshanks made that perfectly clear.
I'm suggesting two ways that allow the 7.2v bow thruster to run off both batteries.

[alan colson]

Answer to question 3 is 1 hour for every 1/2 amp, so a 7amp battery should take 14 hours although if you run a battery that flat you are likely to do it more harm than good. Also you should be aware that as the name suggests these batteries are sealed, do not try to charge at higher than the directions printed on the battery as it may explode because the gases have nowhere to escape. As for a cheap charger, try ModelPower.
Alan

[DickyD]

Cheap charger  http://www.component-shop.co.uk/html/body_lead_acid.html

Excellent service, get all my chargers and batteries there.

[longshanks]

Thanks guys for all your replies.

Looks like I can do what I thought.

Stallspeed, could you explain a bit further about a voltage drop resistor. Seems if I can do that it will simplify things re charging.

Regards
longshanks

[craftysod]

Rex
Thank you for correcting me,
I learn something new every day, then forget something i already new
Mark

[stallspeed]

Stallspeed, could you explain a bit further about a voltage drop resistor. Seems if I can do that it will simplify things re charging.

The ohmic value is R(Ω)= (12- 7)/I where I is the thruster consumption (I) at 6 or 7.2 volt.

The power rating of the resistor should exceed the value I²R.

[malcolmfrary]

The ohmic value is R(Ω)= (12- 7)/I where I is the thruster consumption (I) at 6 or 7.2 volt.

The power rating of the resistor should exceed the value I²R.

Or, a simple practical way, measure the resistance of the motor.  Use a resistor of as near that value as you can get.  This will approximately halve the voltage applied to the motor (a motor is not pure resistance, but this is near enough for our purposes).  As Stallspeed says, you then need the appropriate power rating of the resistor to avoid burning the thing out.  I would calculate on the worst case voltage and current (12 volts at whatever current the resistor would pass at that to give the wattage rating, then pick the next higher rating.)  This builds in a safety factor.  Or I might pillage a dead toaster or hairdryer for the element, and experiment with various lengths of the element.

[FullLeatherJacket]

Working on a max efficiency current of 3.3A (Herr Graupner's data) I reckon about 1.5R and 16W. In practical terms that's series 1.0R and 0.47R, both wire-wound. Craplins do them at 10W but you'd be better off with ali-cased 25W ones.....or a P93 
FLJ

[sheerline]

If 20w resistors are unavailable, what about 2off  .. 6.8 ohms 10w in parallel, that should do it ok.