.

Steph Dale

Steph Dale

Western Thunderer
Tricky one that.
First step would be to measure the resistance of the motor.
Have you got a reasonable multimeter?

Steph
 
mickoo

mickoo

Western Thunderer
Surely its just ohms law? Volt drop required divided by the current, I'd use max current as a basis to begin with. Though once you have a resistance value you will need to work out the wattage as it'll be a small value and with that much current could get quite warm!

I'm suprised a lamp in series stopped the motor, BTW it's 'lamp', bulbs are what you put in the ground at spring time :thumbs: and in the same vein....as was beaten into me at school, a king is a ruler and he measures with a rule.
 
Osgood

Osgood

Western Thunderer
mickoo said:
...and in the same vein....as was beaten into me at school, a king is a ruler and he measures with a rule.
Click to expand...

I have that trouble - no matter how many 5m steel rules I acquire, they are never nearby when I need one and I find myself muttering 'Where's the ...king ruler?'

I like the idea of series coupling the 2 motors (thinking 7mm Heljan 31 here) - is this a well-used slow performance enhancer? Are there any side effects as a result?
 
Dikitriki

Dikitriki

Flying Squad
Not just Heljan. I had to couple my 2 power bogies in 10001 in series to overcome DCC control issues - nothing to do with slow running capabilities though.
 
Railwaymaniac

Railwaymaniac

Western Thunderer
Ohm’s Law – yes.

V = R * I - Rearrange to give R= V / I ( - that’s Vee over Eye, not Vee over one )
R = Resistance V = Volts I = current

Assume V = 6 volts, we are on a 12 volt system, the other six volts goes to the motor
Assume I = half an Amp, that sounds about right for a modern coreless motor?

This gives R = 12 ohms.

Next, and Power = V * I - Multiplication gives three Watts here. But if you don’t want the inside of your loco to boil then chose at least five watts.

So you need to start with a 12 ohm resistor, rated at 5 watts. You could try THIS as a starter?

Ian
 
Simon

Simon

Flying Squad
Simon Dunkley said:
Resistance is futile. You will be assimilated.
Click to expand...

Comments like this polarise, have huge capacity for impedance and are farad off the mark, they have a tiny filament of utility and fuse should return them to earth, lightening (our) conductors current load.

As a consumer of these units I fiel(d) a volt(e) face or switch is required, its a good job no one (ceiling) rose to your windings and bayonet(fitting)ed you. The frequency and amplitude of such discharges have the potential to hertz you know Henry.

He's your evil twin and earth by the way:p
 
Dog Star

Dog Star

Western Thunderer
What? Oh-mm... previous posts have indicated that Simon is wary of current elektrikeri, amazing what this DCC stuff can do to you.

regards, Prof. Barking (still crazy after all these years)
 
adrian

adrian

Flying Squad
Jordan said:
I think there's some sensible advice in this thread, somewhere.... :rolleyes: :D:D:D ...telling me the complicated sums is one thing, but actually pointing at what I might need is even more helpful - so thanks in particular to Railwaymaniac.:thumbs:

What about THIS on Ebay. Drops 12v to 7v. Seems to be for PC use, to drop the speed of the fan motor. I wonder if it'd work in a loco....:confused:
Click to expand...
That doesn't look like it would take the power - it may well be running at much lower current for a fan so would probably be a 3W resistor at most. Aside from that it's expensive, 5W resistors are 20p here, although there is p&p so you either find a local electronics store, make it up to enough for free p&p or pay the ebay premium and get the "free" p&p.
 
mickoo

mickoo

Western Thunderer
adrian said:
That doesn't look like it would take the power - it may well be running at much lower current for a fan so would probably be a 3W resistor at most. Aside from that it's expensive, 5W resistors are 20p here, although there is p&p so you either find a local electronics store, make it up to enough for free p&p or pay the ebay premium and get the "free" p&p.
Click to expand...
Failing all that, once you work out what value and wattage you need I'll look in our parts bin and if we have one will send in the post FOC ;).

FWIW the formula was given back in post #4 :thumbs:
 
adrian

adrian

Flying Squad
mickoo said:
Failing all that, once you work out what value and wattage you need I'll look in our parts bin and if we have one will send in the post FOC ;).

FWIW the formula was given back in post #4 :thumbs:
Click to expand...
Likewise - if it's not too urgent I can always add a few more bits and pieces to my next order with Rapid. I usually order enough to get free postage (latest batch of MERG-CBUS kits was £500). Alternatively I live just round the corner to CPC so can collect anything from the counter if required.
 
Zebedee

Zebedee

New Member
If I remember my schoolboy electronics correctly, if you have 2 5 ohm resistors and wire them in series you get 10 ohm resistance, if you wire them in parallel you get 2.5 ohms. so 3 5 ohm resistors in parallel would give you 1.67 ohms. Also, if you need to put 5 watts into the motor, 2 resistors wired in parallel at 2.5 Watts capacity each would be sufficient.
If you have a pile of resistors available, just have a play, you don't need to get the exact one for the job. But do yourself a favour, keep the ratings of the resistors the same, or the maths gets difficult to do in your head !
 
lancer1027

lancer1027

Western Thunderer
Hi Jordan, glad you've got it sorted.:thumbs:. I was going to suggest a couple of things but elecktrickery is not my strong point ( mind you i dont think anything is ) and you would end up even less confused than when you started.:rolleyes: :oops:

Anyway off now to hide in the cupboard at the back of the class:oops::)

Rob
 
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