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Replacing motors


david_denham

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 Hi David

Have to agree with WTD, just service them. The issue with putting in a new motor there are no new motors that fit with out having to make motor housing adaptors, then there is the issue of getting the gearing to mesh as the gears will be imperial on the old models where today it is metric, in fact the whole chassis is made with imperial measurments so it not just the case of changing the central gear to a new one.

Clean the commutator, oil the two bearings front and back, fit new brushes if required, if the magnet is very week either replace the old for a new or just add two small neo magnets either side of the old one. Oil the axels on pony main chassis and bogie, spot of oil on every rivet and the three screws and or rivets on coupling rod, motion gear and on piston slider and support arms. Good for 250 hours continuas running.

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These old motors will for decades. Kept in good maintenance and cleaned they will live on.  Remagentise the magnet now and then they get weak, clean the commutator, clean and lubricate the bearings, most people forget to add a little lubrication to the back bearing surface behind the magnet and the motor then runs noisy. Why replace something that is fully repairable, and runs forever if looked after.

 

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Totally agree, these old motors will run for decades. Kept in good maintenance and cleaned they will live on.  Remagentise the magnet now and then they get weak, clean the commutator, clean and lubricate the bearings, most people forget to add a little lubrication to the back bearing surface behind the magnet and the motor then runs noisy. Why replace something that is fully repairable, and runs forever if looked after.

 Today's modern non-servicable motors fitted to most modern locos won't last that long

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Hi Tony. Thanks for your concise summary re motor servicing/replacement. This leads me to the most depressing conclusion

that replacing the burnt-out motor of a Lima GWR Railcar (near mint, incidentally) is just not on. A great pity. So it's either sell on eBay

for Spares or keep to adorn the layout. Am I right ?   Cheers

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There was an 'argument' running a few weeks back from when I suggested brushless motors, and someone said they don't make them small enough -

Sorry, they do! -quote -

Brushless motors like this one are the perfect motor for small models and micro models. This motor has the advantage of being able to fit in a GWS IPS gearbox, so there are many models already out there 'crying' for this motor. A motor like this will transform most models in terms of performance if they use the Ips system already.As with all Feigao motors the quality is superb and the product is amazing value.RPM-V 4100Size : 12mm x 30mmTurns 36 Shaft Diameter: 1.5mm Weight: 17g - 0.60ozResistance: 0.6ohmAmp No Load : 0.2A Feigao Motor Code : 1208436

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As has been said, with the Lima there are many many options to keep it running. A good service may fix it. Otherwise you could replace it with a donor motor bought second hand. A Hornby donor chassis would provide a better runner than the Lima original. You can also use a motor conversion kit to a CD type motor.

Worst case scenario you can bodge a CD motor into the existing housing yourself, it's relatively easy to do.

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  • 2 years later...

 I've just used one of those CD remotoring kit in my LIMA 94XX and what a transformation!

 

However being designed for a lower voltage the loco goes much too fast, Can someone give me a good guess at a resistor I can put in series to slow things down a bit (or even a lot!)

 

I say guess since I have no way of determining the resistance of the motor.

 

As an aside I was not confident at gluing the motor in and I wanted to be able to replace it if necessary so I 3D printed an adaptor ring. Not sure if it will work in any other LIMA locos but it is fine for the 94XX

 

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Not going to stick my neck out and suggest a resistor value for you, but I will describe a process by which you will be able to determine a best guess estimate for yourself.

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Now you say the CD motor is rated at a lower voltage than a normal loco motor, but you haven't stated any values. So for the purpose of this testing process let us assume it is 6 volts DC.

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You haven't stated what the current is that the motor is rated at for full speed. So for the purpose of this testing process let us assume 100mA.

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Let us also assume that the maximum output voltage of your Analogue controller is 12 volts when under load.

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You will of course need to substitute your own measured current and voltage values in the following formulas in place of the assumptions above.

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So, what you want in this assumed example is a resistor that drops 6 volts across it (dropping the 12 volt controller down to 6 volts) when drawing 100mA (the current drawn when the loco is at the desired maximum speed).

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What you need in the way of tools to perform this analysis:

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  • A variable DC voltage source (your controller in this case) capable of providing a voltage in the range of 0-12 volts. Your controller may be higher than 12 volts....see comment further below.
  • A multi-meter to measure DC voltage.
  • A multi-meter to measure DC current.

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Using the multi-meter to measure DC voltage. Connect your CD motor to your power source and crank it up until the voltage reads 6 volts DC across the CD motor [or in your case the rated voltage of the CD motor if not 6 volts].

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Now use the multi-meter to measure the DC current being drawn by the motor at 6 volts [or its rated voltage] . Whatever this current value is, lets call it 'A'.

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The resistor value from Ohms Law is R=V/A

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Now using the assumed example values from the formulas above [you will need to substitute your own measured values in the formulas].

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Thus the value of R in Ohms = 6 volts / A amps

Now if the value of A was indeed 100mA, then R = 6V / 0.1A = 60 ohms

If the current A was actually 50mA, then R would equal 6 volts / 0.05A = 120 ohms and so on.

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Now you need the wattage rating of the resistor. This resistor is going to get hot and will need to be kept well clear of plastic etc.

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Watts = V x A

So using the example above, Watts = 6V x 0.1A = 0.6W

Where A is the value you measure in the test above and the V value in the formula is the voltage that you need to drop across the resistor to obtain the CD motor rated voltage when your controller is set to max speed.

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The 50 & 100mA values stated above are plucked from the air purely for the purpose of example demonstration. You will need to measure the current for your particular motor. You will also need to measure the maximum 'on load' voltage output of your controller as it may be higher than 12 volts. It is important to measure the controller maximum voltage when 'on-load' because it may be higher 'off-load' if not regulated. Just run one of your normal locos at maximum speed and then measure the voltage across the track from the controller. This will be the 'on-load' voltage.

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When sourcing the resistor I would double the Wattage rating of the resistor you source. So for 0.6 Watts I would get a 2W resistor (nearest standard value as 1W is less than double).

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So for example, if your controller measured 14 volts DC on load when at maximum speed setting, the voltage you want to drop across the resistor when the loco is at maximum speed will be 8 volts and not 6 volts as used in the example.

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So as you can see, you can only use my formulas as a process guide and will need to input your own values depending upon what you measure in practice in the real world, to get a more accurate resistor specification.

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I should point out, that using a resistor IMO to slow down the motor is not ideal, but without a sophisticated electronic control circuit custom designed, built and added to the input of the motor within the loco chassis then there really isn't any other reasonable alternative.

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