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Re: paper: 4 CIM versus 6 CIM theoretical calculations
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Re: paper: 4 CIM versus 6 CIM theoretical calculations
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Last edited by cbale2000 : 09-12-2016 at 13:56. |
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Re: paper: 4 CIM versus 6 CIM theoretical calculations
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The closer you are to stall, the more the CIM:MiniCIM heat ratio goes down (which is bad), but not by a large amount. The ratio gets closer to 1:1 as you go nearer to free speed as well, but by the time you are running that fast the heat generated isn't too much anyway. It's possible to gear the MiniCIM such that it's always running at a higher RPM than the CIM, perhaps by using 11t and 12t pinions, but it probably isn't necessary. Maybe a team that has done CIM + MiniCIM combos can chime in here? I wonder if using higher-resistance cables to increase your resistance would be a valid strategy to help prevent brownout of 6+ CIM drivetrains. EDIT: The numbers for a 2 CIM + 2 MiniCIM drivetrain are as follows: Approximated as 3.33 CIMs 2,931 theoretical rpm 333 W/CIM 1,108w total, compared to 1,280w for a 4-CIM drive, or about 86% as efficient. Not too great of a drop if you're running a significantly lighter robot and don't need the power. Last edited by asid61 : 09-12-2016 at 14:21. |
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Re: paper: 4 CIM versus 6 CIM theoretical calculations
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Re: paper: 4 CIM versus 6 CIM theoretical calculations
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Typically you design your drive system around 80% of free speed which is more realistic. AndyMark always used CIMS at 4455 rpms when providing gearbox performance data in Feet per Second. A CIM + Mini CIM per side would take about 125% of the time to speed, and distance traveled compared to a two CIM drive. Last edited by InFlight : 09-12-2016 at 15:10. |
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Re: paper: 4 CIM versus 6 CIM theoretical calculations
If you didn't have motor controllers, or (illegally) put both motors on the same controller, this would certainly be the case. Are motor controllers designed to handle the reverse bias? - Oh, right. If you decide to change from 100% speed to 50% speed in a short period of time, you'll create the same situation, so they have to be designed that way. Never mind.
Last edited by GeeTwo : 09-12-2016 at 19:01. |
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Re: paper: 4 CIM versus 6 CIM theoretical calculations
Electrical Power and Mechanical Power are not the same.
The Electrical Power into the a DC Motor is Volts x Amps (Watts) The Mechanical Power Out is: Power Out = Torque*(0.112985) * Speed * (2*pi/60) in Watts Note: 1 lb-in = 0.112985 Nm, and radian/sec 2*pi/60 = rpm The Motor Efficiency is η=(Power Out) /( Power in) At near free speed, the torque and Mechanical Power approach zero. |
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Re: paper: 4 CIM versus 6 CIM theoretical calculations
For any of the DC motors used in First Robotics, we can also use the following equations to determine their performance:
Current (Amps) = Torque Load * ((Stall Current- Free Current)/Stall Torque) + Free Current Torque Load = (Current-Free Current) * Stall Torque/(Stall Current-Free Current) Speed = Free Speed - (Free Speed / Stall Torque) * Torque Load Last edited by InFlight : 09-12-2016 at 18:51. |
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