How does more voltage make a car faster?

[dure16]

Why does a car feel so much more powerful with more voltage?

I’ve been wrestling with this question and hope someone here can enlighten me.

Watts move the car and the ESC can deliver enough current for the motor to hit 1200w in either scenario. Both cars are exactly the same except for battery and pinion. The motor stays cool with either setup.

Both Cars
6S ESC
2200kv motor (1200 watts)

3S battery
Larger pinion geared for 50mph
Estimated current = 95a (1200w/12.6v)

4S battery
Smaller pinion geared for 50mph
Estimated current = 71a (1200w/14.8v)

 

[fish]

MAX RMP Potential.

on one of my rigs, for example, with a 20p/50s, 3.3 ration on 6S RPM potential is 51k/102mph. on 3S its 26krpm's/51mph.

If you use a gear calculator, you will see how the pinion, spur and tire diameters change the MPH potential but not the RPMs.

 

[Sandracer_NL]

Power is indeed Watts. I think you feel the speed.
kv=rpm x voltage.
Higher voltage -> more rpm

 

[dure16]

MAX RMP Potential.

on one of my rigs, for example, with a 20p/50s, 3.3 ration on 6S RPM potential is 51k/102mph. on 3S its 26krpm's/51mph.

If you use a gear calculator, you will see how the pinion, spur and tire diameters change the MPH potential but not the RPMs.

I’ve taken that into account with pinion changes. Both cars are geared for the same speed.

 

[fish]

I’ve taken that into account with pinion changes. Both cars are geared for the same speed.

Like this?
One is working very hard the other not so much.

 

[dure16]

Like this?

Yes like that. A little less different though since I'm only talking 3S and 4S.

Are you saying that motors perform better at higher RPMs?

 

[LibertyMKiii]

It is made up of two things, power and RPMs.

For example say you have a 6s Arrma that comes with a 150a esc and 2050kv motor. (they actually are closer to 1910kv on a tester)

3s setup
150a * 11.1v = 1,665 watts

6s setup
150a * 22.2v = 3,330 watts

The other aspect is motor RPMs. The motor's KV is how many RPMs it will turn unloaded per volt.

3s on a BLX2050 motor
11.1v * 2050kv = 22,755 rpms

6s on the same BLX2050 motor
22.2v = 2050kv = 45,510 rpms

_________________________________________________________________________

I have taken a 1/7 scale to 99mph on 2s but it took 600+ amps to generate the watts needed with the XLX2
126 mph on 3s and same story. It took 600+ amps to get there.

You can make watts with volts or amps, but its always safer to do it with higher voltage assuming the motor is the correct KV to match that.

 

[Hector_Fisher]

Unfortunately as the famous saying goes: "one does not just increase amperage".

Amps are a byproduct of voltage and resistance. So calculating your amperage is a little more complicated than just using the motor's max power rating and the battery voltage. For example, batteries are prone to voltage sag under load, and the load that the motor draws from the battery is dependent on what gearing you use, what size the motor is, and how heavy the car is. Also, batteries have various internal resistance, electrical wire size can vary, etc. You get the idea.

In additional to Liberty's comments, it's good to remember that a motors rating power (in your example 1200 watts) is a number provided by the manufacturer and based a certain set of circumstances they used during their testing. Motors are only restricted by how fast you can spin them before they fly apart and how much heat they can handle before disintegrating, but under the right conditions, a motor can put out more than it's rated power.

Conclusion: If you're good at understanding math, the simplest way to understand why voltage affects power is with the power equation:
Power = Voltage x Amperage.

In this equation, if you increase your voltage, then power (on the other side of the equation) will also increase.

While this is a simplified way of looking at it, overall if you increase voltage, you're power will increase.

 

[SrC]

The short abreviated answer is that more volts makes more RPM's.
Like stated above, Lipo Volts x Motor KV = RPM's.
Just at some point, you have to consider the motor's max sustained RPM rating and Heat ouput that will result.

 

[dure16]

Unfortunately as the famous saying goes: "one does not just increase amperage".
View attachment 248515
Amps are a byproduct of voltage and resistance. So calculating your amperage is a little more complicated than just using the motor's max power rating and the battery voltage. For example, batteries are prone to voltage sag under load, and the load that the motor draws from the battery is dependent on what gearing you use, what size the motor is, and how heavy the car is. Also, batteries have various internal resistance, electrical wire size can vary, etc. You get the idea.

In additional to Liberty's comments, it's good to remember that a motors rating power (in your example 1200 watts) is a number provided by the manufacturer and based a certain set of circumstances they used during their testing. Motors are only restricted by how fast you can spin them before they fly apart and how much heat they can handle before disintegrating, but under the right conditions, a motor can put out more than it's rated power.

Conclusion: If you're good at understanding math, the simplest way to understand why voltage affects power is with the power equation:
Power = Voltage x Amperage.

In this equation, if you increase your voltage, then power (on the other side of the equation) will also increase.

While this is a simplified way of looking at it, overall if you increase voltage, you're power will increase.

Thank you! This is the piece I was missing.

I assumed that increasing amps (within reason) was easy and immediate. And so I thought similar motor wattage could be achieved by simply increasing volts or amps (either side of the equation).

It didn't occur to me that increasing amps wasn't linear and came with other challenges like resistance and voltage sag.

 

[Hector_Fisher]

Thank you! This is the piece I was missing.

I assumed that increasing amps (within reason) was easy and immediate. And so I thought similar motor wattage could be achieved by simply increasing volts or amps (either side of the equation).

It didn't occur to me that increasing amps wasn't linear and came with other challenges like resistance and voltage sag.

One way to look at and understand amperage is to understand the similarities between the flow of electricity and the flow of water. In that comparison voltage is like water pressure, electrical resistance is like flow resistance in the pipes (smaller/larger pipes, pipe bends or straight, blockages, etc), and amperage is like flow rate (how much water is moving through). You can increase pressure on the water through a pump or change flow resistance by changing pipe size, but you can't directly "increase flow". Any way you try to interact with water (moving it faster or getting stuff out of its way) is done through pressure and resistance. Flow is a byproduct of what pressure and resistance you put on the water. Similarly, you can change electrical "flow" by increasing/decreasing voltage and lowering or raising resistance, and amperage is the measurement of how much flow you've created.

In reality, unless you have an ESC capable of measuring amperage, or an ammeter in the circuit, the only way you'll be able to tell whether there is an increase or decrease of amperage in your system is through heat. More amperage = more heat. Less amperage = less heat.

 

[N.Mango]

 

[SirCrashALot]

Haha!
This is such basic knowledge, that i knew this when i was 5, playing with Lego trains and electric motors.

 

[mat60]

Haha!
This is such basic knowledge, that i knew this when i was 5, playing with Lego trains and electric motors.

I understood amps as far as amps in household wiring...breakers and overloading them and the heat you get when you do but when I was 5 I was playing with other things.

 

[N.Mango]

None of us knew this when we were 5yo

 

[SirCrashALot]

None of us knew this when we were 5yo

View attachment 248687

Look at the title, though.