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  #31   Spotlight this post!  
Unread 04-06-2018, 04:14 PM
bt20304 bt20304 is offline
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Re: Possible Issues with Feeding Voltage to a 775 Constantly?

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Originally Posted by nuclearnerd View Post
By slip I think you mean the elevator drops (back drives) on its own when power is removed right?

If so, yes you can hold position by applying voltage to the motor. There are issues you should be aware of doing this. Without careful control, a "stalled" motor has a very low resistance, and zero air cooling, so it will pass a lot of current and heat up very quickly to internal-wire-melting temperature. Look carefully at the last graph on this page "Locked Rotor Stall": http://motors.vex.com/vexpro-motors/775pro You'll see that applying more than 3 or 4 volts to a stalled 775 pro will kill it pretty quickly.

That's fine though - if you program it to never go above 4 volts (or current limit to maybe 20A). Just be aware that will also reduce the motor torque to less than 33% of it's rated value (and the power would be only ~10% of its rated value), so double check your calculations to make sure the elevator will still hold position if you limit the voltage or current this way.


This answer deserves a "kudos". I couldn't have said it any better.

A note i would add though. A motor is just a converter, from electrical to rotational. For every watt of input you'll get ~.7 watt of power. simple right. so keeping with electrical principles. If you give it steady voltage, you would get steady speed. but you don't want speed from the motor. I would try giving it constant current, which should give you constant torque to prevent back drive, also easier to do calculation on force it will exert in "hold" mode.

just food for thought. best of luck and let CD know how it works
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Unread 04-08-2018, 08:06 AM
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Re: Possible Issues with Feeding Voltage to a 775 Constantly?

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Originally Posted by bt20304 View Post
For every watt of input you'll get ~.7 watt of power. simple right.
If you were thinking that for every watt of input you'll get ~.7 watt of mechanical power output, that would be simple wrong. The mechanical output power depends on the motor's operating point.


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Unread 04-08-2018, 11:13 PM
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Re: Possible Issues with Feeding Voltage to a 775 Constantly?

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Originally Posted by Ether View Post
.... The mechanical output power depends on the motor's operating point.


Please show how to calculate that.
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Unread 04-09-2018, 06:56 AM
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Re: Possible Issues with Feeding Voltage to a 775 Constantly?

There are measurements of the CIM power curve here. At zero speed and free speed, the CIM (as with any other motor) is producing to mechanical power, because power = torque * angular speed, and in each case, one of those is zero. Vex also has power curves for most of the other FRC-legal motors linked from here.

When running at full throttle, a good approximation to these curves is to model the motor as a resistor and a generator which produces voltage proportional to rotational speed, in series with each other. You can get the resistance from the stall current: R = V/I = 12V / 131A = .092Ω. Then, you can calculate the generator ratio from that and the free speed. As the motor is drawing 2.7A, the voltage across the resistor is V = IR = 2.7A * 0.092Ω = .248 V. The generator ratio is therefore V/ω = (12V - 0.248V) / 5330 rpm = 2.2mV/rpm.

Note that due to the relatively flat efficiency in the upper third of the range, if running pretty fast but not free (3500-5100 rpm), 0.60 times the electrical power is a decent approximation to the mechanical power produced by a CIM - good to within 10%.
Added: The BAG is the only motor listed there which reaches a 70% efficiency as claimed above.
When doing output power computations, assuming that torque is proportional to current is a good approximation.
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Last edited by GeeTwo : 04-09-2018 at 07:04 AM.
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Unread 04-09-2018, 07:42 AM
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Re: Possible Issues with Feeding Voltage to a 775 Constantly?

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Originally Posted by frank-g-e View Post
Please show how to calculate that.
Sure.

Here are the equations, if you want to learn to do it manually.



Here is a simple motor calculator that has the above equations built-in. Just select a motor from the main menu, then select an operating point from the next menu.



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Unread 04-10-2018, 08:39 PM
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Re: Possible Issues with Feeding Voltage to a 775 Constantly?

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Originally Posted by Ether View Post

Sure.

Here are the equations, if you want to learn to do it manually.



Here is a simple motor calculator that has the above equations built-in. Just select a motor from the main menu, then select an operating point from the next menu.



I just uploaded a ZIP file containing specs for multi-motor CIM, miniCIM, and 775Pro for use with the motor calculator.



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Unread 04-16-2018, 02:10 PM
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Re: Possible Issues with Feeding Voltage to a 775 Constantly?

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Originally Posted by bt20304 View Post
This answer deserves a "kudos". I couldn't have said it any better.

A note i would add though. A motor is just a converter, from electrical to rotational. For every watt of input you'll get ~.7 watt of power. simple right. so keeping with electrical principles. If you give it steady voltage, you would get steady speed. but you don't want speed from the motor. I would try giving it constant current, which should give you constant torque to prevent back drive, also easier to do calculation on force it will exert in "hold" mode.

just food for thought. best of luck and let CD know how it works
Ran the method of applying constant voltage throughout the entire Alpena district & MSC. Did not run into any noticeable issues.

For worlds I will test with the application of current )) Thanks
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Unread 04-16-2018, 02:11 PM
Lesafian Lesafian is offline
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Re: Possible Issues with Feeding Voltage to a 775 Constantly?

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Originally Posted by Ether View Post



I just uploaded a ZIP file containing specs for multi-motor CIM, miniCIM, and 775Pro for use with the motor calculator.



Thanks for providing this.

Perhaps I can use this to do actual calculations instead of taking the risk of damaging a motor.
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Last edited by Lesafian : 04-16-2018 at 02:13 PM.
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