SlotForum banner
1 - 4 of 4 Posts

· Registered
81 Posts
Discussion Starter · #1 ·
HI All

I have been running both Slot.It 29k and Scale Auto 30K motors at home on a Ninco track with and without power boosters on the std Ninco 0.9A power supplies. (1 per lane)

The club track has dedicated Transformers per lane, 200VA each. Smoothing Capacitors etc etc. Should be more than I can use.

These motors have been quite close in performance, in Audi Black and Porsche Kenwood cars. The 29k motor was actually slightly faster.

However, when I get to a new local wooden track, with Magnabraid, running on the same lane and voltage, I get completely different results ..?

The motor refuses to go properly.
The scaleauto motor is a demon - outrunning everything. Really comes alive.
Then I tried a TSRF chassis and motor with std gearing. Really disappointing. This was as slow as the 29k motor. I had expected more from this. Should have been fastest of all ?

Do some motors prefer voltage to amperage ?
Any other ideas as to why the motors perform so differently on home track to club track.

Best regards


· Graham Windle
4,956 Posts
To put it simply if a power suply doesent supply enough amps for a particular motor then that motor will appear slow . On a high voltage track with low amperage the motor which requres least amps will perform best on a track with limmited voltage and infinate amps the high amp motor will be best

Your Tsrf may well have a slipping spur gear as they are just a push fit a drop of super glue will cure this

· Registered
19 Posts
simplier answer (makes to me) : the slot-it motor cannot eat fairly big voltages like a scaleauto (from 15-17 volts some motors only get hot and start to burn w/out great power gain), the scaleauto can be harder at the volts (a wooden track can use 20v or more)

· Al Schwartz
3,413 Posts
"To every complex question, there is a simple answer - and it is wrong!"

In the face of those words of wisdom - I'll try anyway

First, there is THE LAW (Ohms', that is)

E = I X R where E = voltage, I = current and R = resistance

We can apply this in a variety of ways. Let's take the simplest example first:

Take a voltage source of infinite capacity that puts out 12 volts - place a 12 ohm resistor across this supply and the equation tells us that the current through the resistor is 1 amp. If we double the resistance to 24 volts, the current drops to 0.5 amp. Conversely, if we decrease the resistance to 6 ohms, the current double to 2 amps. In the limiting case, if we simply remove the resistor, the resistance becomes effectively infinite and the current drops to zero and if we short the leads together, the resistance becomes zero and the current tries to go to infinity and we have an arc welder!

Now let us consider a real life situation - a slot car track. we need to account for the entire electrical circuit which means, in this case, the whole circle including, as we looks around the "circle", the lead from the + side of the supply, through the connecting wires to the controller and then to the track, through the motor, back to the track and back to the (-) lead of the power supply and through the power supply itself back to the (+) lead. Yes, the power supply is part of the circuit.

This is a lot of stuff, so let's simplify a bit: Assume that the wiring has zero resistance and the controller is "full on" and also has zero resistance. While not entirely correct, these assumptions make little difference. we now have two resistances to consider - the motor and the power supply. Remembering that we are dealing with a circuit or "circle", it is clear that the current must be the same everywhere in the circuit. In order for Ohm's law to be obeyed, there must be voltage drops in the circuit and the sum of these voltage drops must equal the voltage of the supply and Ohm's law tells us that the voltage drops will be proportional to the resistances of the components of the circuit.

Let us consider the motor first: the effective resistance of a spinning motor is very different from the static resistance measurement one can make with a meter. Why - well, you know about braking in slot cars, where one shorts the rails and makes the motor work as a generator and try to develop a current flow into a zero resistance? - the motor is doing the same thing even when it is running under power - this is called the "back EMF" (electromotive force) of a motor and is a function of the physical characteristics of the motor (wind, magnet strength) and the speed of rotation. This can be an incredibly complex subject so why don't we simply skip to real life - a Mabuchi motor which will measure 8 ohms across the brushes will typically draw 0.150 amps running on 12 volts so, from Ohm's law, we know that the effective resistance is 12V / 0.15 A = 80 ohms!

Good power supplies, operating within their range, will typically have internal resistances of fractions of an ohm - lets take 0.10 ohms as an example.

Why is this all important? Because the power available in an electrical circuit is determined by Volts X Amps or Watts and remember that the total voltage in the circuit must be divided among the components in proportion to their resistances

So, in the example above, the votage drop across the motor is:

80 (motor resistance) / 80.1 (total circuit resistance inc. power supply) X 12

= 11.985 volts and the voltage drop in the power supply (going to heat instead of speed) is 0.1/80.1 X 12 + 0.015 volts

And our motor has 11.985 V X 0.15 A = 1.8 watts of electrical energy with very little wasted in the power supply.

A brief note on motor design (mostly a dark science to me) motors designed for higher voltages will use smaller wire and lots more of it so the effective resistance climbs and the current flow in the circuit drops.

Now what happens when we decrease the resistance of the motor ("hotter" wind) - if the voltage stays the same, the current in the circuit will increase and the motor power will increase - why - isn't it all proportional - yes and no - remember that power = V X A and Ohm's law tells us the V = A X R so we can substitute the latter term in the first equation and get: Power = ( A X R) X (A) = current squared times resistance - so - cut the resistance in half, double the current and then square it and you double the power - not so fast - there are a whole range of other issues like number of turns of wire, back EMF again, magnetic saturation, heat, etc. etc. that nibble away at this nirvana but you get the general drift.

But it is a real world out there - with real power supplies. If you are operating from a storage battery that would normally start a large diesel engine - no worries - your power supply resistance will be forever low. If , however, you are using a wall wart manufactured in ther Orient for an F.O.B. cost of $0.39 in quantities of 100,000, life is a little different. Try to pull too much current from on of these little boxes and it will literally choke. The components become saturated, the internal resistance climbs and the voltage available to your motor drops (remember that the voltage is divided among the components of the circuit according to their resistance) Worse still, your motor slows, its back EMF drops and its effective resistance drops, thus it will try to pull more current, the power supply chokes more, its resistance increases etc. etc. etc

Bottom line - motors don't "prefer" current or voltage - the two are tightly bound together and inexorably tied to the characteristics of the power supply.

I think I'm done

1 - 4 of 4 Posts
This is an older thread, you may not receive a response, and could be reviving an old thread. Please consider creating a new thread.