8XL on 12 cells, as powerful as 8L?

[GriffinRU]

Originally Posted by maxxdude1234

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

[maxxdude1234]

Originally Posted by GriffinRU

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.

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

[GriffinRU]

Originally Posted by maxxdude1234

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.

[GriffinRU]

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.