48V in a 36v system

[OneEye]

I imagine one would usually use R=V/I to measure the resistance per length of material based on a known voltage and measured current.

In its basic form, Ohm's law is only directly relevant to a purely resistive load.  Add inductance or capacitance and you need to expand on Ohm with concepts like
"reactance".  It's the inductive nature of motor windings that delay the current from instantly reaching the value predicted by a straight V/R=I equation.  That's the ploy these folks are using when trying to implement a high "V" limited "I" motor that will not melt into a pile of slag.  I suppose the theory is to cut off the voltage when the current has reached a limiting value that would overheat the wiring in the motor.  That's where things get wierd, as the inductive nature of the winding will create a voltage spike as it tries to maintain the same current by expending the energy it has moved into a magnetic field.  All in all I'm not sure it's completely relevant beyond say 48V unless someone is designing an affordable after-market controller for the product on hand.  Even then there are the practical limits of how many batteries in series are required to achieve a high voltage.  The theory is interesting at least.  I'm glad Pete has his 48V system up and running with few problems.  

Did we ever figure out what the differences are (controller? motor?) that keep melting myelectricbike's wiring and haven't seemed to bother pdonahue?  I think there was a corroborating report about frying things at higher voltages from an Endless Sphere user, but I don't recall the specifics.

[myelectricbike]

I'm aware of AC values having different equations and of inductive and capacitative resistance as well as the concept of voltage spikes (negative) from dumping EMF stored in the windings when the applied power is removed and of stable CEMF being generated as possible contributing factors to my power phase lead insulation melting, especially where contact is made with other leads and metal.

My calculations show it to be a planned designed feature of motors limited to 750 watts to assure compliance with the law owing to the size of the axle hollow and the insulation melting temp of 60c. Move to 14AWG from 16AWG, add 2 more parallel strands to the motor windings and it all goes away. Maybe the motor Pete has is an earlier model made prior to the 750 limit ban.

[pdonahue]

What equation did you use to obtain a resistance of .0014 per 3 ft of 16AWG and what were the values for voltage, current and power?

R=P/I²
R=V/I
R=V²/P

 

The resistance of the wire came from a AWG vs resistance table.  I actually just noticed that I added an extra zero after the decimal point so that means that the power dissapation above would need to be multiplied by a factor of 10.  (The chart states that 16AWG has a resistance of 0.00475 ohms per foot so for 3 feet that would give 0.014, not 0.0014)  That merely makes the problem worse than I had stated above.

Other than that, I used only P=I²R  and V=IR.   The three formulas you list above are just permutations of these two identities.

Good bye for today, I'm about to head home on the 48 volter again.

Pete

[OneEye]

I always thought it was a positive voltage spike in the direction of current.  Ah well, perhaps a difference in terminology/sign convention.  I was never very good at either.

[myelectricbike]

Pete

Okay then at line voltage of 120 volts AC (but lets say DC) we are told 16AWG has a current rating of 10 amps.10 amps at 120 volts gives us a wattage rating of 1200.

My heat gun is 1200 watts and if I use a 16AWG extension cord you can feel it getting warm with prolonged use and the heat gun seeming to not have as much heat or power.

If we use the formula R=V²/P or the R=P/I² formula to calculate resistance then R= 12 ohms, so I think the 10 amp rating refers to the wattage of the appliance rather than to the wattage a shorted wire can handle.

I think this is where you misunderstand. I am not talking about a shorted wire load but rather the load of the appliance, which a 16AWG wire with only a .014 resistance can carry.

So how does this apply to a 1 volt, 1200 amp load?

Again we are not talking about a shorted 3 ft. length of 16AWG but rather a 1200 watt load.

Imagine we have a 1200 watt resistance load designed not to operate at line voltage but rather at exactly one volt and 1200 amps. To simulate this load let us use our equations to see just what the results will be.

Starting with P=IV we get R=V/I or 1/1200=0.000833333

0.000833333 ohms then is the resistance of our appliance load.

Lets select a 1200 watt, 0.000833333 ohm wire wound ceramic resistor to simulate this load.

Now we can cut our 3 ft piece of 16AWG in half, attach it to each end of out appliance load and operate our 1200 watt appliance all day long at 1 volt and 1200 amps using our 3 ft length of 16AWG, 10 amp, 120 volt, .014 ohm piece of wire.

Here is the full set of equations for 10 amps at 120 volts and 1200 amps at 1 volt.

	P=watts	I=amps	V=volts	R=ohms
P=I2R	1200	1200	1	0.000833333
P=IV	1200	1200	1	0.000833333
P=V2/R	1200	1200	1	0.000833333
V=P/I	1200	1200	1	0.000833333
V=IR	1200	1200	1	0.000833333
V=(PR).5	1200	1200	1	0.000833333
I=(P/R).5	1200	1200	1	0.000833333
I=V/R	1200	1200	1	0.000833333
I=P/V	1200	1200	1	0.000833333
R=P/I2	1200	1200	1	0.000833333
R=V/I	1200	1200	1	0.000833333
R=V2/P	1200	1200	1	0.000833333
	P=watts	I=amps	V=volts	R=ohms
P=I2R	1200	10	120	12
P=IV	1200	10	120	12
P=V2/R	1200	10	120	12
V=P/I	1200	10	120	12
V=IR	1200	10	120	12
V=(PR).5	1200	10	120	12
I=(P/R).5	1200	10	120	12
I=V/R	1200	10	120	12
I=P/V	1200	10	120	12
R=P/I2	1200	10	120	12
R=V/I	1200	10	120	12
R=V2/P	1200	10	120	12

[myelectricbike]

Mike

We might not be talking about the exact same thing. I'm talking about the stored inductive charge in the stator windings at the moment the applied voltage is disconnected and which must be dumped before CEMF becomes stable. Under no load conditions (freewheel) its just a mild millisecond negative spike with a fast positive recovery to stable CEMF but for a load condition it is a more pronounced negative spike and a longer positive recovery to the point of stable CEMF.

[OneEye]

Fair enough.  I guess I've already stated this is mostly just an academic amusement until a tested circuit is published and avaiable to build.  Until then I'll just grab the marshmallows and roast them over the flame war the boards have devolved to of late.

[myelectricbike]

I know Mike, but since this whole forum thing was my idea I feel responsible for what goes on here and try to keep my responses civil. As time moves on I am able to step back and let more and more of it go so that eventually one day it will just run itself without a moderator or without any help from me. I can dream but its better people who want a fight take it out on me than on other owners. I can handle it (I think) especially with the good, honest and intelligent help I am receiving from owners (or future owners) such as yourself. Were it up to me you would be in charge yesterday. No fooling.

[cadstarsucks]

Pete

Okay then at line voltage of 120 volts AC (but lets say DC) we are told 16AWG has a current rating of 10 amps.10 amps at 120 volts gives us a wattage rating of 1200.

My heat gun is 1200 watts and if I use a 16AWG extension cord you can feel it getting warm with prolonged use and the heat gun seeming to not have as much heat or power.

If we use the formula R=V²/P or the R=P/I² formula to calculate resistance then R= 12 ohms, so I think the 10 amp rating refers to the wattage of the appliance rather than to the wattage a shorted wire can handle.

I think this is where you misunderstand. I am not talking about a shorted wire load but rather the load of the appliance, which a 16AWG wire with only a .014 resistance can carry.

But that is NOT how it works!  The resistance of the motor is more like 0.5 ohm.  You MUST take out the motor voltage BEFORE you can calculate the resistance.  Energy can neither be created nor destroyed.  If the resistance were 12 ohms you would NOT have and energy left to propel the bike as it would all be lost in the winding resistance.

[myelectricbike]

Click here Dan.

Aside from this I have not yet addressed winding resistance other than to mention briefly the dumping of stored inductive charge when PWM power to a BLDC motor is turned off under no load and load conditions.

As for "winding resistance" itself, which is inductive and reactive in proportion to BLDC motor RPM with a smaller resistive and unreactive component, again you need to present a non-confrontational, clear, detailed and step by step demonstration of calculation if you want your spin on how BLDC motor winding resistance is computed to be read and considered by me. We have plenty of other sources for such descriptions and so far nothing you have said offers any improvement over them.

[cadstarsucks]

Click here Dan.

Aside from this I have not yet addressed winding resistance other than to mention briefly the dumping of stored inductive charge when PWM power to a BLDC motor is turned off under no load and load conditions.

Actually you have regularly implied that the entire voltage drop and all the energy delivered to the motor is dissipated by motor resistance.

I patiently, at first, pointed out that if all the power was dissipated I the winding that there would be nothing left to power the bike.

[myelectricbike]

Dan, you keep making this same confrontational claim, which is based on your imagination.

[pdonahue]

When you are running a 1200 watt heat gun and drawing 10A of current from a 120V source, the voltage across the wire will be:
V= IR = 10 * 0.014 = 0.14 Volts

The voltage across the plug terminals of the heat gun will be 120V - 0.14 = 119.86v.
From this we can calculate the resistance of the heat gun:
R= V/I = 11.986 ohms

The power dissapated by the heat gun is:
P= I²R = 10 * 10 * 11.986 = 1198.6 watts

The power dissapated by the wire is:
P= I²R = 10 * 10 * 0.014 = 1.4 watts

The total power used is 1200 watts, but the total power given off as heat in the wire is 1.4 watts.  That is why I would rather touch the wire than the tip of the heat gun.

Pete

Pete

Okay then at line voltage of 120 volts AC (but lets say DC) we are told 16AWG has a current rating of 10 amps.10 amps at 120 volts gives us a wattage rating of 1200.

My heat gun is 1200 watts and if I use a 16AWG extension cord you can feel it getting warm with prolonged use and the heat gun seeming to not have as much heat or power.

[myelectricbike]

Thanks for providing greater detail as to the dissipation of wattage and heat. Now on to inductive reactance and "winding resistance..."

[pdonahue]

If you apply 1 volt to a piece of wire and a 1200 watt resistor of value 0.0008333, you will have the following:

     R_total= R_{1/2 wire} + R_resistor + R_{1/2 wire}

                            = 0.014/2  + 0.0008333 + 0.014/2

                            = 0.0148333

The total current that will run through the wires and the resistor will be:

I=V/R = 1 / 0.0148333 = 67.4A

Each half of the wire will be dissapating:
P=I²R = 67.4 * 67.4 * 0.007 = 31.8 watts
That will certainly be enough to melt the insulation in a VERY short time.

The actual resistor would only be dissapating:
P=I²R = 67.4 * 67.4 * 0.0008333 = 3.79 watts
It would still get quite hot to the touch, but should be able to easily handle the heat.

So how does this apply to a 1 volt, 1200 amp load?

Again we are not talking about a shorted 3 ft. length of 16AWG but rather a 1200 watt load.

Imagine we have a 1200 watt resistance load designed not to operate at line voltage but rather at exactly one volt and 1200 amps. To simulate this load let us use our equations to see just what the results will be.

Starting with P=IV we get R=V/I or 1/1200=0.000833333

0.000833333 ohms then is the resistance of our appliance load.

Lets select a 1200 watt, 0.000833333 ohm wire wound ceramic resistor to simulate this load.

Now we can cut our 3 ft piece of 16AWG in half, attach it to each end of out appliance load and operate our 1200 watt appliance all day long at 1 volt and 1200 amps using our 3 ft length of 16AWG, 10 amp, 120 volt, .014 ohm piece of wire.

Here is the full set of equations for 10 amps at 120 volts and 1200 amps at 1 volt.

	P=watts	I=amps	V=volts	R=ohms
P=I2R	1200	1200	1	0.000833333
P=IV	1200	1200	1	0.000833333
P=V2/R	1200	1200	1	0.000833333
V=P/I	1200	1200	1	0.000833333
V=IR	1200	1200	1	0.000833333
V=(PR).5	1200	1200	1	0.000833333
I=(P/R).5	1200	1200	1	0.000833333
I=V/R	1200	1200	1	0.000833333
I=P/V	1200	1200	1	0.000833333
R=P/I2	1200	1200	1	0.000833333
R=V/I	1200	1200	1	0.000833333
R=V2/P	1200	1200	1	0.000833333
	P=watts	I=amps	V=volts	R=ohms
P=I2R	1200	10	120	12
P=IV	1200	10	120	12
P=V2/R	1200	10	120	12
V=P/I	1200	10	120	12
V=IR	1200	10	120	12
V=(PR).5	1200	10	120	12
I=(P/R).5	1200	10	120	12
I=V/R	1200	10	120	12
I=P/V	1200	10	120	12
R=P/I2	1200	10	120	12
R=V/I	1200	10	120	12
R=V2/P	1200	10	120	12