Daisy Chain vs ?
 

I see on most websites that everyone daisy chains.
What's the difference between going with a daisy chain layout and going with a matrix(?) layout?

http://img204.imageshack.us/img204/9206/34367001bv8.jpg

If this is concerning electronics (batteries) one is in series and one is parallel; the difference being voltages adding or current adding.

Thus my next question: where's a good tutorial for newbies on volts, currents, etc?

This is extremely basic: http://www.kpsec.freeuk.com/voltage.htm
This seems to have information on general components involved in electronics: http://www.design-technology.info/electronics/
Wikipedia has some good stuff to say on series and parallel: http://en.wikipedia.org/wiki/Series_...allel_circuits

These were based on some quick searches. If you have any questions feel free to ask and we should be able to help.

And remember, the game is MUCH easier with DC power. AC will take a lot of studying to understand fully... things just don't work the same.

Thanks GenKreton, i'll get straight to studying. :)

Ok, i think i'm getting a grasp here.

So 2 automotive batteries in a series circuit with 12V/500A = 24V/500A
but
2 automotive batteries in a parallel circuit with 12V/500A = 12V/1000A

The A will stay the same across a circut but the V will not. So if one component needs 1V/1A and another component also needs 1V/1A then the total need for the circut's power supply is 2V/1A

Alright, so now i pick a random motor off of some website just to see how much i can understand:

Motor:
Universal AC/DC Motor, Open Enclosure, 1 HP, 10,000 RPM, 115 Volts, Non Standard NEMA Frame, Service Factor 1.00, Ambient 40 C, 60 Hz, Full Load Amps 12.1, Insulation Class A, Thermal Protection None, Rotation CCW, Face with Four Tapped Size 10-32 Holes, 3 1/4 Inch OC 90 Degrees Apart Mounting, Ball Bearings, Shaft Dia 7/16 In, Shaft Length 1 1/4 In DrillSpot Price: $248.19
(like i said, just a random motor.)

Now i take a generic imaginary Auto battary of 12V/500A DC.
9.58 batteries - or 10 batteries - are needed to meet the required volts of the motor. However, that will be @ 500A (serial connection) which is too much for the motor and will mostlikely put on a pretty smoke effect and maybe some sparks. so i'll have to cut the 500A down to no more than the 12.1 amps. (currently +5v and +487.9A excess power)
To get the power down then i will need a type of resistor.
but to calculate what size then i would need apparently alot of math and a handy calculator.

But so far, I'm getting it, right?

Current is dependent on the motor draw. That is, voltage NEEDS to be around the rated value but current capabilities need to be higher to handle overload conditions.

With that said, all motors can run well above their ratings, up to 2x, for short periods of time without doing damage.

Also, pulse width modulation is definitely worth looking up on wikipedia. All DC conversions use PWM to control the applied voltage to the motor. 120 V on a 115/120 V motor will always use maximum power, which is not desirable in a car. Pulse width modulation turns the signal on and off so it looks like a step function and the average applied voltage is lower.

All conversions use the matrix method because most DC motors run off of 120 V or 144 V motors and use 12 V deep cycle lead acid batteries, generally from the fork lift industry.

I hope this helps. Feel free to ask more :)

"Current is dependent on the motor draw. That is, voltage NEEDS to be around the rated value but current capabilities
current capabilities of the circut?
need to be higher to handle overload conditions.
so if it asks for 100A then you should be prepared to provide 200A if need be?

With that said, all motors can run well above their ratings, up to 2x, for short periods of time without doing damage.

Also, pulse width modulation is definitely worth looking up on wikipedia.
Sorry, Wiki is speaking in greek on this toppic. it SOUNDS like you pulse the electricity into the motor (rather than a constant current) as a method of controlling speed without going below the recommended electrical input for the motor. (instead of sending 1/2 juice, send full juice only 1/2 the time) All DC conversions use PWM to control the applied voltage to the motor. 120 V on a 115/120 V motor will always use maximum power, which is not desirable in a car. Pulse width modulation turns the signal on and off so it looks like a step function and the average applied voltage is lower.

All conversions use the matrix method because most DC motors run off of 120 V or 144 V motors and use 12 V deep cycle lead acid batteries, generally from the fork lift industry.

I hope this helps. Feel free to ask more :)"

Careful, i'm the kinda guy that can ask alot of questions when i want to comprehend something.

The voltage a motor requires to run won't vary too much, ideally not at all. The current draw varies with the load placed on the motor. Your batteries should be prepared to deliver over the current listed for the motor, so you can overdrive it for short periods of time. To simplify, the batteries will supply (ideally) a constant voltage (that will see drops during high load, or in other words current) and the current will vary with the power demand placed on the motor.

You are correct in your observation about PWM. I'm sorry for referencing the article before observing how overly complicated. to get 25% of the power out of your motor, you essentially close the circuit 25% of the time. This is by far the most efficient way to simulate lower voltages.