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Motor braking power
This question is simply for personal curiosity. Using 2 identical esc's having identical programming. Does a 3.5t motor have the same, weaker, or stronger brakes then a 25.5t motor?
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esc determines braking power as in amps drawn. but a 3.5T will take long in micro seconds to brake due to turn # over a 25,5T which will be instantly ..so a 25.5T will look like it has more brakes due to its slower Kv rating .
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If they had exactly the same gearing the 3.5 would have stronger brakes at higher speed and the 25.5 at lower speed.
However if you geared them for a similar top speed the 3.5 would have stronger brakes everywhere. |
Originally Posted by faqcya
(Post 15170275)
This question is simply for personal curiosity. Using 2 identical esc's having identical programming. Does a 3.5t motor have the same, weaker, or stronger brakes then a 25.5t motor?
This is directly related to the torque of the motor. At the same voltage, a lower-turn motor has more torque (and more RPM!) than a higher-turn motor. |
Originally Posted by the rc guy
(Post 15170291)
esc determines braking power as in amps drawn. but a 3.5T will take long in micro seconds to brake due to turn # over a 25,5T which will be instantly ..so a 25.5T will look like it has more brakes due to its slower Kv rating .
And tc4 with a rsx and a 17.5 both equal settings just the 17.5 has 10* And braking is 10000x harder and responsive on the 2.5 Not everything on the interweb is accurate or true. Oh and my tc6 mmp Tekin 5.5 brakes harder than my sprint2 Tekin rs pro/lrp 5.5 |
makes sense
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how the brake frecuency (PWM) can be adjusted o n the ESC to get more brake?
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Originally Posted by Byronf1
(Post 15170544)
how the brake frecuency (PWM) can be adjusted o n the ESC to get more brake?
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If the motors were out of the car and spinning at the same RPM, the 25.5 turn would provide more braking force.
If both motors are in a car and geared for the same top speed, the 3.5 would have a much larger mechanical advantage due to the gear reduction would provide more braking force at the same speed. |
These are interesting responses and I appreciate everyone taking the time to answer.
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Originally Posted by Krio
(Post 15170790)
If the motors were out of the car and spinning at the same RPM, the 25.5 turn would provide more braking force.
Originally Posted by Krio
(Post 15170790)
If both motors are in a car and geared for the same top speed, the 3.5 would have a much larger mechanical advantage due to the gear reduction would provide more braking force at the same speed.
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Originally Posted by howardcano
(Post 15170879)
That would only be true if the ESC and wiring resistance were a significant fraction of the motor's resistance. Otherwise, if the motors have the same stator and rotor design, the rotors have the same strength, and the stators are both at 100% fill, the braking torque would be the same.
Correct. I have edited my previous post to match your observation. |
Originally Posted by Krio
(Post 15170959)
You are assuming the voltage created in the motor is the same between a 25.5 turn motor and a 3.5 turn motor at the same rpm. This isn't the case. Let's say the 25.5 motor is rated as 1,500kv and the 3.5 is 9,000. This rating also tells you (roughly, not accounting for losses/inefficiencies) the voltage created by the motor when you spin it. At 3,000 rpm, the 25.5 is creating 2 volts. The 3.5 would only be creating 0.33 volts.
The back-EMF of the motor is proportional to the magnetic field strength, enclosed flux (rotor and stator size), speed, and the number of turns. At the same speed, stator and rotor sizes, and rotor field strength, the back-EMF of the 3.5T is (3.5/25.5) = 0.137 times that of the 25.5T. The resistance of the motor is proportional to the square of the number of turns (if the stator is completely filled with wire). The 3.5 has (3.5^2/25.5^2) = 0.0188 times the resistance of the 25.5T. Torque is proportional to the current, the magnetic field strength, enclosed flux (rotor and stator size), and the number of turns. At the same speed, stator and rotor sizes, and rotor field strength, current through the 3.5T is V/R = (0.137/0.0188) = (25.5/3.5) =7.29 times that of the 25.5T. Finally, the torque of the 3.5T = 7.29*(3.5/25.5)=1 times the torque of the 25.5. The torque is the same. |
Originally Posted by howardcano
(Post 15171219)
I did not assume that.
The back-EMF of the motor is proportional to the magnetic field strength, enclosed flux (rotor and stator size), speed, and the number of turns. At the same speed, stator and rotor sizes, and rotor field strength, the back-EMF of the 3.5T is (3.5/25.5) = 0.137 times that of the 25.5T. The resistance of the motor is proportional to the square of the number of turns (if the stator is completely filled with wire). The 3.5 has 1/(3.5^2/25.5^2) = 0.0188 times the resistance of the 25.5T. Torque is proportional to the current, the magnetic field strength, enclosed flux (rotor and stator size), and the number of turns. At the same speed, stator and rotor sizes, and rotor field strength, current through the 3.5T is V/R = (0.137/0.0188) = (25.5/3.5) =7.29 times that of the 25.5T. Finally, the torque of the 3.5T = 7.29*(3.5/25.5)=1 times the torque of the 25.5. The torque is the same. I won't disagree on the calculations for an ideal motor as, yes, they should be the same. My answer was based on including inefficiencies is all in addition to the data I have come across through work. I didn't mean to imply that the 25.5 turn has 10 times the braking force due to the voltage difference in my previous post, but I wasn't sure where to start with my reasoning. I just realized that the 25.5 in question is probably ROAR spec'd to a certain wire diameter, so my assumption that the 25.5 turn would have more fill was incorrect. In the end, I meant that the 25.5 will have maybe 5% more braking power at the same rpm if the wire size was optimal for each motor. |
Originally Posted by Krio
(Post 15171445)
I won't disagree on the calculations for an ideal motor as, yes, they should be the same. My answer was based on including inefficiencies is all in addition to the data I have come across through work.
Speaking of calculations, we both missed my error in the calculation for resistance. Somehow a "1/" snuck in there! I've fixed it. |
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