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Unread 20-05-2014, 01:51
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Re: motor Ke and Kt Quiz

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Originally Posted by DonRotolo View Post
I think it will also help to know
Ke = Motor Voltage constant
Kt = Motor Torque constant

In theory, Kt = Ke * (some constant)
(The value of the constant depends upon the units used).
Could you (or someone else) give an example of how Ke and Kt are used? Whenever I've wants to do motor calculations, I've just assumed free speed directly varied with voltage, that current drawn=free current + (free current-stall current)*(torque load on motor/stall torque), and power=voltage*current=Speed*torque/efficiency. Those have basically gotten me where I've needed to go in terms of motor calcs. Where does Ke and Kt come into play?
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Unread 20-05-2014, 04:17
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Re: motor Ke and Kt Quiz

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Originally Posted by DampRobot View Post
Could you (or someone else) give an example of how Ke and Kt are used?
This is how I understand it (disclaimer I am just starting to learn this stuff).

Ke is the constant that relates the back emf generated by the motor to the angular velocity, Eb=Ke*omega. Kt is the constant that relates the torque output of the motor to the current drawn, Tau = Kt*I. In an ideal motor Ke and Kt are equal.

You can use these numbers to calculate the current drawn by the motor under a given load (tau) or speed (omega). Using the earlier definition of Kt, calculating the current under any torque is simple.

I = Tau/Kt

Solving for a given speed is slightly more complicated. Say you had the following circuit:


We are assuming the motor has been running long enough such that the inductive effects of the motor windings are ~0. (This is called steady state operation)

E is the applied voltage to the circuit, Ra is the resistance of the motor, Eb is the back emf generated by the motor and Ia is the current through the circuit (what we are trying to solve for).

Using Kirchhoff's loop rule we know that ΔV of a loop is 0. The voltage drop across the resistor is I*R (from ohm's law) and Eb = Ke*omega; from here we can solve for Ia.

ΔV = 0
E-Ia*Ra-Eb = 0
E-Eb = Ia*Ra
(E-Eb)/Ra = Ia
(E-Ke*omega)/Ra = Ia

If Kt = Ke the two currents should be the same

(sorry if this is poorly written, its late)
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Unread 20-05-2014, 10:15
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Re: motor Ke and Kt Quiz

I'm not really an electrical guy, but this is my understanding.

The biggest point of ambiguity is the definition of an ideal motor. I think a more accurate way to state the problem is that the motor is "perfectly efficient". Its nice because efficiency actually has a definition that relates to the performance of the motor, as opposed to the word "ideal" which can mean a variety of things (inductance of motor is disregarded? heat effects are disregarded? I've usually seen "ideal" motors modeled as resistors).


If the motor is assumed to be perfectly efficient and linear, then only one datapoint is needed to characterize the entire motor (because the power transfer would be constant, meaning that any excess power that results from a loss in velocity must be made up with an increase in torque). You provided two datapoints, two datapoints that do not describe a perfectly efficient motor. The motor described by your datapoints operates most efficiently at halfway up the speed vs. torque graph (200rad/sec, 0.2 n-m).

If I understand correctly, the quiz is basically nonsensical because conflicting information is given. The statement is basically:
"this motor is perfectly efficient and the power input is constant, when it is going fast it outputs 60 watts, when it is going slow it outputs 40 watts. why is that?"
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Unread 20-05-2014, 16:23
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Re: motor Ke and Kt Quiz

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Originally Posted by headlight View Post
If I understand correctly, the quiz is basically nonsensical because conflicting information is given. The statement is basically:
"this motor is perfectly efficient and the power input is constant, when it is going fast it outputs 60 watts, when it is going slow it outputs 40 watts. why is that?"
The prompt of the quiz is fine.

In this context ideal means linear torque-speed tradeoff, NOT a lack of resistance.
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Unread 20-05-2014, 23:28
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Re: motor Ke and Kt Quiz

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Originally Posted by AdamHeard View Post
The prompt of the quiz is fine.

In this context ideal means linear torque-speed tradeoff, NOT a lack of resistance.
I thought it meant something different because the linear properties of the motor were specified later, didn't expect redundancy.

So in the situation presented Kt and Ke aren't supposed to be equal because of motor's internal resistance/losses?

Are there other ways that theoretical motors differ from ideal motors?

Sorry, I haven't done a lot of work with electrical theory.
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Unread 21-05-2014, 09:48
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Re: motor Ke and Kt Quiz


Let's try a different question.

Here are the specs for the 2014 BAG motor:

Spec Voltage: Vspec = 12

Spec Free Speed: Wfree = 1466 radians/sec

Spec Free Current: Ifree = 1.8 A

Spec Stall Torque: Tstall = 0.403 Nm

Spec Stall Current: Istall = 41 A

Calculate Ke and Kt from the above data, and discuss possible reasons why these values are not equal.


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Unread 21-05-2014, 10:02
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Re: motor Ke and Kt Quiz

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Originally Posted by Ether View Post

Let's try a different question.

Here are the specs for the 2014 BAG motor:

Spec Voltage: Vspec = 12

Spec Free Speed: Wfree = 1466 radians/sec

Spec Free Current: Ifree = 1.8 A

Spec Stall Torque: Tstall = 0.403 Nm

Spec Stall Current: Istall = 41 A

Calculate Ke and Kt from the above data, and discuss possible reasons why these values are not equal.


Do I get to answer or is that cheating?
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Unread 21-05-2014, 10:31
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Re: motor Ke and Kt Quiz

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Originally Posted by Paul Copioli View Post
Do I get to answer or is that cheating?
By all means Paul, please do jump in. I think there may be at least a small audience of interested students following the thread.

Adam Heard and Richard Wallace, please also feel free to post. I know you both have been holding back.


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Unread 21-05-2014, 13:07
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Re: motor Ke and Kt Quiz

Russ -- I will join the discussion when I have data to share.

Paul -- this motor is your design, and I want to go on record saying you did a fine job. It is a worthy successor to the venerable Globe, and about three orders of magnitude more flexible for a range of applications powering FRC mechanisms. You may have noticed I ordered a few of them yesterday -- will share test data (and THEN my own answer to Russ's question) when I can.
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Unread 21-05-2014, 13:27
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Re: motor Ke and Kt Quiz

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Originally Posted by Richard Wallace View Post
Paul -- this motor is your design, and I want to go on record saying you did a fine job. It is a worthy successor to the venerable Globe, and about three orders of magnitude more flexible for a range of applications powering FRC mechanisms. You may have noticed I ordered a few of them yesterday -- will share test data (and THEN my own answer to Russ's question) when I can.
Paul, did you design this motor for FRC? or was it a motor already in production (or not in production but available as a configurable item) that you picked out?

If you designed it for FRC, I'd love to hear more about the process.
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Unread 21-05-2014, 15:59
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Re: motor Ke and Kt Quiz

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Originally Posted by AdamHeard View Post
Paul, did you design this motor for FRC? or was it a motor already in production (or not in production but available as a configurable item) that you picked out?

If you designed it for FRC, I'd love to hear more about the process.
A little bit of both. CCL, the manufacturer of the CIM motor, did all the detail design work. They did not have any motor like the BAG. In our VEXpro product discussions, before we even made one product, we all agreed that losing the Globe motor from the KoP was a big loss to FRC teams. We then finalized what we wanted from a power, torque at max power, and packaging perspective. I ran some quick numbers based on DC motor design formulas (I will get the exact book details that I prefer at another time b/c I loaned it out to someone recently) to get a rough idea of what was possible.

I then took the rough design to CCL and said, "construct this like the CIM motor, use all ball bearings, and hit these power and torque numbers". I was fairly certain they could do it, but haven't had to design a DC motor in a long time so left it to them.

But this definitely is a custom motor specific to the FRC application.

The MiniCIM was a similar process except we literally took a picture of a dismantled CIM and said make it only "this" long. I gave them torque and power numbers so it would match up 1:1 with a CIM and contribute the most power possible in the meat of the CIM speed profile.
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Unread 21-05-2014, 16:15
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Re: motor Ke and Kt Quiz

So, for most small, high Flux DC motors there is this phenomenon called magnetic field saturation. It happens when the permanent magnet doesn't have enough field strength to drive the high amount of Flux through the moving parts of the motor. I am simplifying here to try to keep it short.

The bottom line is that at high current (aka, near stall), the magnetic field saturates producing less actual torque than ideal and this is one of the reasons for the difference between Kt and Ke.

However, it is not the only difference. This can really only be found with motor testing.
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Unread 21-05-2014, 14:22
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Re: motor Ke and Kt Quiz

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Originally Posted by Ether View Post
By all means Paul, please do jump in. I think there may be at least a small audience of interested students following the thread.

There is.
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Unread 21-05-2014, 16:13
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Re: motor Ke and Kt Quiz

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Originally Posted by Gregor View Post
There is.
OK, I'll toss this out for discussion


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Last edited by Ether : 21-05-2014 at 18:02. Reason: corrected power equation
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