More or less amps?
 

I have played with brushless some. I love it.:043:
I am a little confused in some areas.
It takes amps to create torque.
Would more volts make more torque as well? :032:
That would make more scence to me for you would have more wattage at same amp draw.
My question is. If you put the same volts and amp draw from two different motors. Would one create more torque than the other?
I was wondering if I get a xl1200 and wanderer 12xl. Would the wanderer put out more torque for it has a larger rotor on the same amp draw?

Torque is force. Horsepower is the rate of torque ("time" is a factor here).

Horsepower can also be expressed in watts. Voltage generates an amount of current flow across a given resistance. The product of voltage and current generates a certain wattage. So, for a given voltage and resistance, the current will be X. If you double the voltage on the same resistance, you double the current, which quadruples the wattage. So, if you have a motor that pulls 30A on 14.4v, you know it will pull more current with higher voltage, and less current at lower voltage. We are assuming that resistance is constant here, but BL motors essentially run on PWM AC so you have to take inductive reactance into account, but to make things simpler (if a little inaccurate) we don't usually worry about it.

To find HP, you can use the following equation:

HP=( engine_speed * engine_torque) / 5252

Engine speed is in rpm.
Engine torque is in foot-pounds.

Since you know the HP rating of a motor (watts) depending on how many cells you use, you can figure out torque by messing with the equation to get:

engine_torque = ( HP * 5252 ) / engine_speed

That's how I see it anyway...

Tahnks brian. I wanted to know if a motor like the xl1200 and xl12 turn both spin about 1200 kv give or take 51. :D One motor is rated at 55 amp the other at 26. I was wonder if they would produce the same amount of torque on the same amount of amps or would the xl12turn produce motor for it has more rotor area. If that is the case I should possible get more run time out the xl12turn than xl1200.
By looking at what you explained. If i go up on voltage the watts will go up and the torque as well. So If I pack a 8s lipo on instead of 6s with a lower kv( higher resistance) I can get the sam torque with the same rpm with lower amp draw. Am I on the right track?

CHC, I'm not really sure about the motors themselves, but don't forget that the watts (or HP) you get is a function of torque and motor speed. You could probably plot a horsepower and torque curve vs motor speed for each motor based on the formulas I supplied. It won't be "exact" because of some losses, but it should give you a rough idea.

You'll need a voltmeter and a high current (~100A) meter, or I suppose you could use a watt meter. If a watt meter is unavailable to you, measure the battery voltage and current draw for different battery voltages. From that, calculate wattage. You know a motor has a kv rating, so you can calculate motor rpm based on the voltage you are using. Once you know the power and rpm, you can calculate torque using the second formula from my previous post. You can do this for different cells counts from 6v up to whatever you want to get a plot. Do this for both motors.

Note: Since you are measuring input power, there will be some inaccuracy because of controller inefficiency, but as long as you use the same controller for each motor, the inaccuracy will at least be consistent. Also, you'll have to provide a consistent load for each motor.

Oh, and don't forget that torque is ft-lbs. You can convert that to oz-in if you want.

Once you get the plots done, it should be intersting to see if the sweet spot is truly around 35k-40k rpm.

My plan is to run them on same truck with same esc with same batteries. I was happy with apeed so same gearing as well. I am not sure how to plot a map or a power curve. I don't have a watt meter and stuff either. I guess like an eagle tree would tell me. I was accually wondering before I bought the motor(12xl). I have ran the truck with xl1200. Thanks for you help brian. I am not maxing the watts the motors are rated for. I am under voltage and gearing lower for trying to get maximum run time. I was thinking of getting a 22 series for if it would have more torque with the same amp draw. I let you know as i test more

lol, looks like I didn't really answer your question CHC. Sounds like someone with experience with those two motors will have to help you out. I will say that for a given wind, a motor with a longer can will have more torque, but lower speed, but I'm sure you've long since figured that out! Of course, the build quality of the motor does help too. An XL and a Lehner of the same winds are a different story. :)

Yeah I got you. I guess it is one way to find out. I have to get it and see. If I get a longer run time out of it I guess the amp draw will be less and the torque would have to be close to the same. I may get it to try.

If you multiply kv x kt, you get the same value for all brushless motors.

Unfortunately, some BL motors don't supply the KT value. It would be nice though...

kt is kv/1352 I think.

Edit: It's 1352/kv (oz.in/amp)

So, what you are saying then is that torque = kv^2/1352 ? If so, then a motor with lets say 1500 rpm/v will have a torque of a little more than 1664. What unit is this? Does this also mean that a S, L, and XL motors have the same torque if their kv are all the same?

Could you please elaborate on this some?
Cause I don't think all the motors are equal. Now if you put kv x amp draw = kt I would go along with that. And what is the same value you get for all brushless motors.

Sorry, it's actually 1352/kv in (oz.in/amp). Here are some equations to get you to that number.

Power = Volts x Amps = Velocity (m/s) x Force (Newtons)

Lets say you have a motor with X kv (rpm/volt). Volts = V, Amps = I. For now, the efficiency of this motor will be 100%

With V voltage the motor is spinning at VX rpm = 2(pi)(V)(X)/60 radians per second (r/s).

If radius = 1 meter, r/s = meters per second (m/s) away from the center of the shaft.

Efficiency = 100%
Power in = VI = Power out = (m/s)(Newtons) = 2(pi)(V)(X)(Newtons)/60

If Power out (VI) = 2(pi)(V)(X)(Newtons)/60, then
The torque in Newtons = 60VI/(2(pi)(V)(X)) newtons a meter away from the center of the shaft.

To convert to oz.in/amp,
39.37 inches/meter and 3.6 oz/newton

oz.in = 39.37(3.6)(60)(V)(I)/(2(pi)(V)(X))
oz.in = 1353(I)/(X) the V's cancel out.
oz.in/amp = 1353/X = 1353/kv

Put in any value for kv, amps, voltage and you will get that number every time. The reason why I used 100% efficiency is so I don't need to factor in resistance of the motor. 100% efficiency assumes the motor has a constant kv at any amp rate and has 0 ohms resitance, but you can find efficiency just by finding the rpm drop constant. With the I = V/r equation, you can find the max amp rate where the motor is spinnin at 0 rpm and is 0% efficient. Basically the rpm drops linearly until it reaches that point. rpm drop = kv(I)(r). Rpm = (kv)(volts) - (kv)(amps)(resistance). Efficiency = (V-rI)/V.