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Ohm's Law
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Jeff Crosson
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Ohm's Law - 05.17.2008, 04:24 PM

I looked up things about this, and here's something useful: Here.

The main equations people have used are: Amps = Voltage / Resistance (Ohms) and Power (Watts) = Voltage x Amps

I've read that people say the when the voltage is doubled the amperage is doubled also. So when you double the voltage you quadruple the power (watts). Also, slower motors have higher resistance (Ohms). See Neu's website to find these.

So how this deals with us:
A lot of people run truggies with a Neu 1515 1.5D or 1Y on 3S or 4S. I'm going to compare 1.5 (2700 kv) on 3S vs the 3D (1360kv) on 6S. These are not real world rating, just for comparison's sake. 11.1 / 0.004 = 2775 amps. And then 22.2 / 0.012 = 1850 amps.

What I've always heard around here is that when you double the voltage you want to get a motor that is about half the kv, and then it will be half the amps. But that's not how it is, the doubled voltage doubles the amps also, and the only thing that's helping keep the amps down is the higher resistance (Ohms) that slower motors have.

What I've read elsewhere is that you should get a motor that has four times the resistance if you want to double the voltage and want to keep the same amount of power (watts) and half the amps. It always seems like people who want to have higher voltage setups just multiply the kv times the voltage and match it up the rpms with the lower voltage steups and call it done, because we assume that the half kv means half the amps which doesn't seem true

I am missing something? Can some people who know about this type of stuff explain this?
   
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sleebus.jones
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05.17.2008, 04:57 PM

I think the basic idea is that to do the same amount of work, if you raise the voltage, you will need less amps to perform said work. Generally speaking, I usually pick the highest cell count I can run on an esc, then pick a motor with a KV that gives me a RPM that I'm comfortable with. Since I use mainly fegaios, that is in the 35,000 rpm range. I don't worry about much else, because I know that higher voltage is the most efficient way to run.

Methinks you overengineer this a bit. ;) Asking questions is good tho! :)
   
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BrianG
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05.17.2008, 05:15 PM

Yeah, you kinda have to think of a motor as a variable resistor which is dependant on mechanical load.

When a motor is running, it's impedance is X, so Y amount of current is pulled with Z voltage.

If you decrease or increase the mechanical load, the impedance will change, so less or more current will flow with the same voltage.

This impedance comes from the coil wires, which has two components: DC resistance and inductance. DC resistance is the pure resistive component and is very small since a coil is simply a length of wire. The inductance is resistance that changes with frequency and called inductive reactance (formula is XL = 2 x Pi x frequency x inductance). Increase in frequency means an increase of inductive reactance. The total impedance is a vector sum of the resistance (0* phase angle) and the inductive reactance (-90* phase angle).

Now a coil with AC voltage running through it creates an electromagnetic field. This field expands and contracts, and while doing this, induces a counter voltage in the adjacent coil windings. This counter voltage is impeding current flow, which in effect, looks like added resistance.

Ok, so now you have a motor which can be mechanically loaded. At no load, a certain amount of back-EMF is generated with a given amount of voltage. This produces a current value. When you mechanically load a motor, you are still applying the same voltage, but since the motor is spinning slower (because of the load), less back-EMF is generated, which appears as less resistance, so more current flows.

Load a motor too much and the it stalls. This results in NO back-EMF and therefore the current is limited only by the wires/battery/ESC and the DC resistance of the coil wires (which you recall is VERY low). So, if a motor has 0.01 ohms of coil resistance, at 14.4v (4s), the current theoretically could be 1,440A (14.4v / 0.01 ohms)! Of course, niether your batteries or ESC or motor can handle this for more than a few milliseconfds without extreme heat (otherwise known as fire ).

So, it's not as simple as pure ohms law, but with a little study, it DOES all tie together.
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Jeff Crosson
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05.17.2008, 05:24 PM

I see that a lot of people use 1515 1.5d or 1y on 4S and use 1y as a moderate racing motor. But I want to use 6S, so I ordered the 1515 3D to get about the same rpms the the 1y 4S combo, so will I have about the same speed but about double as much runtime?

Last edited by Jeff Crosson; 05.17.2008 at 05:25 PM.
   
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azjc
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05.17.2008, 05:48 PM

Quote:
Originally Posted by Jeff Crosson View Post
I see that a lot of people use 1515 1.5d or 1y on 4S and use 1y as a moderate racing motor. But I want to use 6S, so I ordered the 1515 3D to get about the same rpms the the 1y 4S combo, so will I have about the same speed but about double as much runtime?
the 3d is a bit slow on 6s the motor has a KV rating of 1360 which will give you a RPM total of about 30,000...I like to keep the total @ 35,000-40, 000, I have the 2.5d on 6s and I am happy with the results
   
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Jeff Crosson
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05.17.2008, 05:56 PM

Most people use the 1515 1y 2200 kv on 4S and that's 30,000 rpm. The 1515 3D 1360 kv on 6S in 30,000 rpm. So I'm right on. I'm racing it on a medium-sized dry track where too much motor isn't good.
   
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azjc
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05.17.2008, 06:02 PM

Quote:
Originally Posted by Jeff Crosson View Post
Most people use the 1515 1y 2200 kv on 4S and that's 30,000 rpm. The 1515 3D 1360 kv on 6S in 30,000 rpm. So I'm right on. I'm racing it on a medium-sized dry track where too much motor isn't good.

I didnt know you are racing on a smaller track, I assumme gearing it for 35mph would be right....I am a basher so my mind was there....
   
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azjc
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05.17.2008, 05:23 PM

Quote:
Originally Posted by Jeff Crosson View Post
I looked up things about this, and here's something useful: Here.

The main equations people have used are: Amps = Voltage / Resistance (Ohms) and Power (Watts) = Voltage x Amps

I've read that people say the when the voltage is doubled the amperage is doubled also. So when you double the voltage you quadruple the power (watts). Also, slower motors have higher resistance (Ohms). See Neu's website to find these.

So how this deals with us:
A lot of people run truggies with a Neu 1515 1.5D or 1Y on 3S or 4S. I'm going to compare 1.5 (2700 kv) on 3S vs the 3D (1360kv) on 6S. These are not real world rating, just for comparison's sake. 11.1 / 0.004 = 2775 amps. And then 22.2 / 0.012 = 1850 amps.

What I've always heard around here is that when you double the voltage you want to get a motor that is about half the kv, and then it will be half the amps. But that's not how it is, the doubled voltage doubles the amps also, and the only thing that's helping keep the amps down is the higher resistance (Ohms) that slower motors have.

What I've read elsewhere is that you should get a motor that has four times the resistance if you want to double the voltage and want to keep the same amount of power (watts) and half the amps. It always seems like people who want to have higher voltage setups just multiply the kv times the voltage and match it up the rpms with the lower voltage steups and call it done, because we assume that the half kv means half the amps which doesn't seem true

I am missing something? Can some people who know about this type of stuff explain this?
Man.....that was a whole lot of technobabble(something I get criticized for)...I am with Sleebus about trying to keep things a bit more simple, and going with as much voltage as youe ESC/motor can handle and mating a motor that keeps you in the 35,000rpm range. You can always gear up with a quality motor (Neu) if you want more top speed, also gear your rig to go 40-45mph you dont want to go WOT than lay off and do the yo-yo thing its harder on the ESC
   
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BrianG
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05.17.2008, 05:28 PM

Generally, if you stay at around the same total rpm with different motor/battery, runtime will be about the same. The same amount of work is being done.

If you increase voltage and motor windings and use the same capacity battery, runtime will increase. Just keep in mind added voltage usually makes for heavier batteries, so that will affect runtime a little. But the reduced current from a higher voltage setup makes everything run cooler and more efficient, which could counter the added battery weight.
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