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Discussion Starter · #1 ·
Hi,

It may be a dumb question but I just want to be sure I have understanden how power supply works for slot car racing.
1/ The voltage is used to set the maximum speed of the engine. More volts you have more r.p.m you get.
2/ The amp is used to set the degree of acceleration of the engine (most of the time I hear people talking about power but I guess it's all about acceleration?). More amps you have the quickest an engine can reach its maximum speed.

Am I right?

I am using a 2 lines Sport Scalex 14 meters track with a variable output power supply and most of the time I am racing by myself (I know it's sad!). I used 17V (the Scalex Power base lower down the voltage so actually the track is polarized with only about 16V) and 2A. What about you? Any advice?

By the way if you want to buy a power supply I have found a pretty cheap and good one. The brand (distributor) is Velleman. The power supply has a size of a laptop one and does have a set of changeable slots to change the ouput voltage (5V, 6V, 7V, 8V, ... up to 24V) with 4A at 5V and 1.5A at 24V. And it comes only for 39€. So far it seems to works perfectly.

Thanks.

Regards,
Maverick.
 

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DT
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I can't remember my school science. Even though I was good at it, I must have been looking out the window that day. This is what I found on the net:
QUOTE Comparing electrical wires with plumbing pipes works as a good analogy in most, but not all, cases. Voltage is the pressure forcing the electrical current through the wire. Amperage is the measure of the speed of flow through the wire, while watts would be the measure equivalent to gallons per minute. The neutral line is the equivalent of the drain pipe.

Just as a larger pipe can carry more water at the same pressure, a larger wire can carry more electrical current, with a big exception. If a pipe is too small, the only negative result is that less than the necessary amount of water will trickle through. With wiring, if the wire is too small, it begins to function the same way the heating element does and gets hot. This heat can quickly build up enough to melt the insulation, set fires, and electrify objects including people and pets.

QUOTE (Maverick @ 9 Jan 2004, 04:22 AM)most of the time I am racing by myself
Why don't you come and have a go on my track sometime, you're less than an hour away by car or train. Send me an email.
 

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Rich Dumas
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Maverick, I am afraid that I will have to give you a rather complicated answer to what seems to be a simple question. Remember that you are not just wanting the motor to turn, it must also do work, that means horsepower (very little horses!). With electric motors we are talking about watts or volt amperes (VA). Voltage is a measure of electrical potential, it is like water pressure. Amperage is a measure of the amount of electricity and you need both voltage and amperage to do work. Your power supply will, of course, set the maximum voltage that your motor can see. All that you are able to control directly is the voltage. In any case the motor will draw the current that it needs, unless you power supply is not able to provide it. Your strategy of choosing a voltage that you feel comfortable with is sound. Raising the voltage will probably require modifications to your cars. If you intend to actually race you need to determine the voltage that will be used and adapt both yourself and the cars to that condition.
 

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Discussion Starter · #4 ·
Thanks,

I am little bit ashamed
since I have an engineer degree (mechanical one). I mean I understand pretty well what voltage and amperage are. I just have some issue to understand what is the impact on the engine when you play with those parameters.
But I do have more elements now to understand. I didn't take in account the notion of work.
Work and power are related, if I am right Power = Work / Time. The work for a linear movement is define by W = Force (Newton) * moving (Meter). For a rotative movement it's different W = Radius * Force.
The engine force will have to be bigger than the the weight force, the magnet force and friction force between the tires and the track (and other friction forces in the whole mechanical system like in gears for example) in order to move the car.
So because Power = Voltage * Amperage we have P = UI = W/T = Rd*F/T, so F = UIT/Rd. So if I do it simply the moving force is proportional to U and I. Because U is a fixed value and so doesn't vary, the only variable parameter here is the amperage and the engine will drain as much as amperage it needs according the weight of the car, the magnet force and the friction forces to move the car.

It's still a little bit confusing since I guess both voltage and amperage can't be dissociated to explain how an engine works. I am wondering what happens if you power a track with enough volts (such 24V for example) but with a small amperage such 100mA. Because the Force required to move the car is related to the power you apply to the engine (U*I) in theory if U is big enough you can have a very small amperage but I am pretty sure that the car won't move if there is not enough amps...hmmmm weird and you might as well fry the engine!

Sorry if I have looked as a nerd in this post
. I jus want to understand.

Regards,
Maverick.
 

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Scott Brownlee
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QUOTE I am racing by myself (I know it's sad!)

It's not sad - most of us do. Enjoy it!

Scott
 

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Rich Dumas
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Just to add to the confusion there are a few more factors to consider with respect to DC motors. First off you could measure the resistance of the motor windings and then calculate the current draw using Ohm's Law. For a stock HO car with a 6 ohm armature and 18 volts you get three amps. For a restricted open car with a 1.2 ohm armature you get 15 amps! Fortunatly a DC motor does not act like a simple resistor once it begins to turn. A spinning DC motor acts like a generator, it produces a back EMF that pushes against the applied voltage and has the effect of increasing the resistance of the arm. More spin, more back EMF. Because of this effect you can't get unlimited power from a motor by increasing the voltage. It is interesting to note that, if you are free running a motor and it is working properly, the amperage that it draws is just about constant as you vary the voltage. The strength of the motor magnets play a big part in the power that the motor puts out. While you put volts and amps into the motor you get RPMs, torque and also some heat out. The magnets mostly contribute to the torque. A motor with weak magnets is not going to have a lot of punch, but magnets that are too strong for the armature wind will actually limit the top speed. As a general observation 1/32nd racers do not seem to fuss with their motors like HO and 1/24th racers do. I suppose that is because 1/32nd motors are not really designed to be taken apart.
 

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Hi Maverick
Without going too deeply into it, the answers above are about right. Volts will increase your revs up to a point. Amps will determine torque (pulling or turning power) If you have a limited output unregulated supply then the more amps a motor draws, the less volts are available. Revs will drop but acceleration rises.
Two things to watch out for. Modern Scalextric powerunits are AC and require some kind of rectifier in the system. this rectifier will shed 2 to 2.5 volts across it.
A lot of power supplies on the market are relatively cheap. A lot of them also have poor rectifiers. When measured, they may put out 15 volts DC but when tested with a multi meter they can put out as much as 30+volts AC!! (Admitedly at very low current). It is this AC ripple that will cause your motors to run hot and loose performance. Whenever buying a non specific power supply always attempt to find out what the AC leakage is at the power out terminals
 

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"Two things to watch out for. Modern Scalextric powerunits are AC and require some kind of rectifier in the system. this rectifier will shed 2 to 2.5 volts across it."


Scaley sport power bases are AC? Oh no, I'm lost again!


I was under the impression that you could hook up ANY aftermarket supply to a scaley power base (via the correct harness) and just go. Do you have to put a rectifier in the system or doesit come built in to the power supply?

The wiring harness that Prof Motor sells are alleged to be able to be used with any aftermarket supply (this is what I was planning).

Help!

Frosty

 

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Scaly sport powerbase - the track bit where u plug in ur power supply to - has a rectifier built in, which consumes a bit of power. You can plug in AC or DC to this. DC will come out on the rails. But you cant reverse the polarity of the DC to drive the cars the other way.
 

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Thanks Astro

That one had me worried for a second there!


The next question that springs to mind is the grade of wires inside the powerbase, should you not pull the thing apart and re-solder higher grade wire in place of the stock wiring?


Frosty (clearer than before)

 

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Al Schwartz
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Perhaps this will clarify things a bit:

Ohm's Law holds that E (voltage)= I (current) X R (resistance) Thus you can set two, but not 3 of the parameters.

Consider two limiting cases:

Open circuit (i.e., no connection) R is infinite thus, for any value of V, I must = 0

Short circuit R= 0, therefore for any value of V, I must be infinite = clearly impossble so that the only real solution is that V=0

Real life is somewhere in between. If you were to connect a 25 volt, 10 milliamp supply to a car, the voltage would drop to balance the equation (observing, of course the issues raised above about back EMF etc. It is not simply a question of the static resistance of the armature windings).

well, one might ask - Where does the voltage "go"? One way to look at it is this: We have to consider a complete circuit - controller, track connections, track, car and the power supply! This is a series circuit and the voltage drop across each element will be proportional to its resistance. The internal resistance of an automobile battery is very low so that no matter how much current a car draws, the effective resistance of the battery as a component of the circuit is very small compared to the resistance of the motor. If, however, we try to power a car with a supply that has a very low current capacity, e.g. 10 milliamps, the effective resistance of the supply will become very large compared to that of the motor, thus the voltage drop across the motor will be small.

As a graphic example of series resistances consider this: A water tap, a length of hose and a hose nozzle. If the water tap is fully open and you have 100' of 5/8" hose, you will get a good spray of water from the nozzle. If you substitute a 3/4" hose, the volume of water delivered by the nozzle will be eve greater but, if the wall tap is barely cracked open, not matter what hose you use, all that will come from the nozzle is a trickle.

EM
 

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Umm....with the sport power bases there was a problem reported on another forum about them frying if DC power packs were used. There was an excellent article on how to rewire them written by "Jimmy in Greece" but I just can't recall the link at the moment. I did copy the info into a doc so I could do the mod (which I did and it works perfectly) so here is the exact text as written by Jimmy.

Jimmy, if you read this I hope it is ok with you to post this here.

"Here's a mod for sports power bases that allows the use of DC power supplies without running the power through the bridge rectifier circuit. Follow the instructions with drawings.

There are 2 bridge rectifier circuits, one for each power supply circuit. I modified both of them so that you won't have problems if you plug into the either one.

1. Remove the 8 diodes I've marked on drawing spwrbas2.jpg with "circle X". The other 2 diodes just give power to the red LED so they should be no problem.
2. Next add the 2 blue wires and the 2 red wires (spwrbas1.jpg) as shown.
3. Your replacement power supply should use the same type of connector with the center pin wired for positive (+)

If you just remove the diodes, you have an open circuit between the power plug and the controller/track circuit.

The reason one should remove the diodes is that the remaining circuit with the jumpers installed would allow 2 of the diodes to act like fuses under certain situations and since they are only about 1 amp from what I can see this could cause problems like smoke and possibly a safety issue if one diode happened (however unlikely) to short out."

These are the pics referred to above
spwrbas2.jpg

spwrbas1.jpg
 

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Discussion Starter · #13 ·
I didn't notice before that the Sport power base requires an AC power supply. But yes near the input plug jack it's written 16V ~.
So now I am starting to understand why when I am using my variable DC power supply the power base drop the voltage.
I guess the 4 diodes associated to each input are in fact what we call a Graetz bridge whom the role is to make linear the alternative current. Since the current is already linear with a DC power supply, the current will pass only through 2 diodes and theses diodes will drop the voltage of 0.6V each, so a total of 1.2V. That about the voltage drop I have notice on my track when I am mesuring the voltage with a multimeter.
Could somebody confirm that?

I did break the wire of my sport scalex power supply. Can somebody tell what is the voltage on the track when you are mesuring it with a multimeter. It is less than 16V? Under each track section it's written 12V DC.

Cheers,
Maverick.
 

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Discussion Starter · #14 ·
QUOTE (wixwacing @ 10 Jan 2004, 04:13 PM)When measured, they may put out 15 volts DC but when tested with a multi meter they can put out as much as 30+volts AC!! (Admitedly at very low current).
I am a little be lost here
. How a DC output power supply could put out an AC current?

Cheers,
Maverick.
 

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QUOTE How a DC output power supply could put out an AC current?
Because the output is rectified (converted) from an AC input.
If this conversion is not complete, then leakage of AC takes place together with the DC.
ie, both AC and DC are present at the same time in the output.
Electro-boffins take over here, for more detailed explanation, please!
 

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Rich Dumas
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The inexpensive power supplies tat come with any slot car set produce a DC voltage with a fair amount of AC "ripple" riding on it. The rectifier part of the power supply uses either four discrete diodes or a single package that has the diodes inside of it. One pair of diodes will pass one half of the AC sine wave, the other pair passes and inverts the other half of the sine wave. This configuration is called a bridge or full-wave rectifier. The output is now all the same polarity, but it is varying in amplitude. To get rid of most of the ripple you need a fairly large capacitor or two.
 

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Discussion Starter · #17 ·
QUOTE (Tropi @ 17 Jan 2004, 02:54 PM)QUOTE How a DC output power supply could put out an AC current?
Because the output is rectified (converted) from an AC input.
If this conversion is not complete, then leakage of AC takes place together with the DC.
ie, both AC and DC are present at the same time in the output.
Electro-boffins take over here, for more detailed explanation, please!

Thanks. I didn't know that. I will check out that with my multimeter.
 

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Although I am no expert, I have been reading about model power and its supply for many years. What many genuine experts are saying is that the 12 volts that used to be the standard nominal figure for powering cars and trains is absolutely ample,
on condition that there is enough amperage.

Enough amperage?
Minimum 1 amp per car/lane for standard home-set motors.
Double that for hotter motors.
A 12 volt 10 amp supply will cover most any home requirement.
Double the amperage again (to 20 amps) for heavy club use.

It appears that, some years ago, when 12 volts @ 1 amp or more was the standard spec for home-set power supplies, the odd kid would lick the track or shove the wires in his mouth - typical child curiosity. This was capable of hurting them and the manfacturers reduced the amperage to puny levels to avoid potential legal problems from irresponsible parents. It was found that a few more volts could raise the motor speed, though it still left the cars under powered overall.
That's the background and it makes sense to me.

That nominal 12 volt figure is how car batteries are nominally classified, though it appears that those batteries are technically 13.2 volts, based on their make-up of 6 cells of 2.2 volts apiece, making 13.2 volts in total. I gather that a freshly charged battery CAN display 13.2 volts but that it rapidly drops to 12 volts after a tiny bit of use. It then maintains that 12 volt figure until it is fairly close to discharge (flattened).
This is where the 12 volt standard originated and it is a GOOD STANDARD.
 

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Good call Mecoprop. Indeed, looking at the array of transformers and wall warts that fill up many a nook and cranny in my shed, it seems that somewhere in the 80's that the voltage and amperage of Scalex power supplies got reduced.

I've got some that are 10v 1.1amp, others 12v 1a and some old style transformers, one that has two outputs, one 12v the other 15v at 2a. I even found a 16v 0.5a one that came genuine with a Scalextric set
. Maybe Australia has odd electricity...

I recall reading somewhere that the change from DC to AC on Sport was a "safety thing".

As for kid's licking the track
....perhaps it was because Scaley track tastes like a Big Mac!
 

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Beppe Giannini
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Meco,
it's the Amps that give you an "electric shock" - and you need very little to feel it (I forgot how much, just look at the differential protection relay on the home mains, should be well below 0.1 A).
Therefore :
- any wall wart will be able to supply so small a current
- raising the voltage (i.e. what "drives" the current, for a given resistance) would only make things worse
- for humans, the resistance is nearly all contact (wet feet vs. rubber shoes), not internal to the body (we are mainly water, right?)

Let's face it : if a kid insists on licking the rails, it's probably that Darwin didn't mean him to become a great kisser

Ciao
 
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