Just to be clear... Power = Torque*RPM

Both motors will have the same RPM limit, based on ESC limits/ball bearings/magnet/ballance...

So, do you still consider L an XL have the same output?

Later we can talk about cells, turns and gear ratios.

Artur

Thats incorrect, let me explain. Power is r.p.m and torque. I'm sure you know that Watts = current x voltage. In electrical motors, torque is proportional to current, and rpm is proprtional to voltage. Therefore power is a product of torque and rpm.
I think you've got a slight misunderstanding there. The 'watt' is a unit for measuring power. Hope I explained this well.

At the same voltage, and geared for the same speed, yes. However, at their maximum r.p.m, the XL will have the potential to produce more power (if geared to do so).

Then next question :)

How is Current (which is torque*koeff) and RPM curves look for 8XL and 8L if you plot it vs voltage?
Efficiency curve would be nice too ... but later

Artur

The current that either motor draws will depend on the load put on the motor. For a given load the L motor will pull more current than the XL, because it has a lower torque constant (kt rating). At a given rpm, the L motor will need require a lower voltage to achieve that rpm than the XL will need.

The point I was initially trying to make (all those posts back) is that at a given speed both motors will give out the same power. The XL will be more torquey, and the L more revvy. The XL might be a tiny bit more efficient, and hence draw less power from the batteries

Well, I cannot say so because motor output (Motor Output = Total power * Efficiency) won't be the same across entire RPM range. And both motors will have similar (not the same though) current-torque curves with load or without (as long as it would be inside motor powerband, based on size). But with the same voltage one motor can be at pick efficiency while another not.

Artur

er.....please tell me what part of my post was incorrect? Your post, although correct, had nothing to do with mine. Please, if something I posted was incorrect then explain in what way it was incorrect - don't just say its wrong and don't explain why. Thanks

*edit* when I said "at a given speed both motors will put out the same power", I am refering to the speed of the vehicle they are moving. This might be where the confusion lies

I like your *edit* :)

And then talking about what current which motor will draw at fixed load is less important, right?

Just to get back, where we started, you said:
And before that by (Rtsbasic) :
So, to correct these statements I start my posts. Bottom line - 8L and 8XL cannot produce the same power at given voltage because they have different Kv number based on motor construction. And do not look on resistance because inductance is the parameter what plays the major role in motor efficiency. When you deal with DC motors you look for active and reactive resistance, the last one is dominant and primary depends on motor size, RPM and any parameter responsible for change in frequency applied to motor phases.

So can you get the same power on the wheels from 8L and 8XL with 12 cells, yes you can, but one will be more efficient then another. And both can be tweaked to be at top efficiency, but then one will have higher output then another. That is what I am trying to say in my posts.

I can see how my last paragraph can confuse, but it is all about balance and tweaks.

Artur

P.S. 8L will spin up faster and will have less rotation mass plus weights less then XL.

Apart from this statement, I fully agree with you last post. The 8L and 8XL can produce the same power at a given voltage, if they are both geared to achieve the same speed. However, if they are both geared optimally, then they will achieve different speeds and hence have different power output

Well, if we clear with the rest, then let’s think about both systems.

At given voltage (12 cells) 8L with fixed load will draw current I(8L) which generates torque T(8L), based on motors magnet size and windings; 8XL - I(8XL) and T(8XL).

If we try to normalize torque values from 8L and 8XL we need to introduce coefficient responsible for size difference (magnet, core, windings…). This coefficient will be RPM dependant or more precisely efficiency curve mirror.

So Motor torque for 8L = n(8L)(RPM)*T(8L) and 8XL = n(8XL)(RPM)*T(8XL).

Now virtual experiment:

8L (Kv 3079) on 12 cells with fixed load spins at 30000 RPM’s and draws I(8L)A
8XL (Kv 2084) on 12 cells with the same load spins at 23000 RPM and draws I(8XL)A

If I(8L) equals I(8XL) then Total power equals, but shaft power differs.
And based on motor size and guys feedback I will say that I(8L) should be higher.

So what we have then? 8L has higher current meaning higher torque, but because of smaller footstep has lower coefficient “n”. While 8XL has good size and great “n” but barely has current to create that torque.

Without good motor I-V-RPM curves we can discuss that forever.

Artur

What do you mean by shaft power? I'm guessing you mean power output, as opposed to power input. With the same power input, both the 8L and 8XL will have very similar power outputs. This will depend on how efficient either one is. Without specific data it is hard to say which will be more efficeint (on 12 cells) - but personally I would say that the 8L will be more efficient at 30,000rpm than the 8XL at 23,000rpm. The difference will be fairly small though.

I think/hope we are both in agreement now

Shaft power and power Output are the same in my previous post.

Power input for 8L and 8XL will be the same only if current will be the same, see post above. If that true then we have unique situation, where 8L runs in RPM range pass peak efficiency and 8XL before, but thats not true, because DC motors have close to linear torque curve (current draw) vs voltage, which should put 8XL current draw to lower value then 8L.

Why do you think one motor would be more efficient on 30000 then another at 23000?

Artur

P.S. Lets clean this up in PM and put clear answer for other guys.

Finally...
 

Answer comes from http://www.aveox.com/technical/dc.html

Calculating Motor Performance
Use these handy equations to calculate steady state motor performance. A spread sheet will help in visually graphing motor parameters. If the Torque constant is not supplied by the motor manufacturer, you can measure the motors no-load RPM/Volt and use the following equations to calculate the torque constant.
Torque constant: Kt=Kb x 1.345
Current draw of motor: I = [V-(Kb x kRPM)]/Rm
Torque output of motor: J = (Kt x I) - (Kt x Inl)
RPM of motor: kRPM = (V - RmI) / Kb
Power output of motor: Po = (J x RPM)/1345
Power input: Pi = V x I
Motor efficiency: Eff = (Po/Pi) x 100
Current at peak motor efficiency: Ie max = Sqrt [(V x Inl)/Rm]

Symbol Definitions:
Eff = Efficiency
I = Current
Iemax=Most efficient current
Inl = No load current
J = Torque (oz-in/A)
Kb = Voltage constant (Volt/1000 RPM)
Kt = Torque constant (oz-In/A)
Pi = Power input (Watts)
Po = Mechanical power output (Watts)
Rm = Terminal resistance
RPM = Revolutions/minute
V = Voltage

Based on this motors with the same Kv numbers have the same Kt numbers, but RPM and Amps will be different. Meaning that power output of both motors won't be the same.

Artur

P.S. Apply the same math for this http://www.rc-monster.com/forum/showthread.php?t=2775 as well.

very interesting?

That makes interesting reading artur. Now I see that two different sized motors with the same kv and kt ratings can have different power outputs. Thanks for researching that!