breaking the minibot/ controled descent

[Daniel_LaFleur]

Originally Posted by ADTRavenger3585

what is the best way to safely stop the minibot at the top of the pole or put it in a controled descent. please give me a good explanation. danke.
-ethan and 3585

The best way, we found, for a controlled descent, is to make the motor act as a brake on the way down.

To do this you need to connect the positive pole of the battery to both electrical connections of the motor, during descent. Use a light switch that gets switched when the minibot hits the top to do this.

Basically, have the lightswitch change the connections on the motor from + and - to + and +.

[fox46]

Actually, it has nothing to do with connecting the motor to (+) but rather, shorting the leads to the motor so that as it backdrives down the pole it acts as a generator. With the dead short on the "generator output" it resists rotation.

[TEE]

haha, the title of this thread confused me at first... you said "breaking" instead of "braking"

[tdramble]

We implemented a circuit that switched the motors from driven to "braking" via a short when it reaches the top, here is a video: http://www.youtube.com/watch?v=KEXraq_Tmy0

[Mr V]

Originally Posted by fox46

Actually, it has nothing to do with connecting the motor to (+) but rather, shorting the leads to the motor so that as it backdrives down the pole it acts as a generator. With the dead short on the "generator output" it resists rotation.

Correct it is turning the motor into a generator that creates the dynamic braking effect.


Most teams short it via the (-) side of the circuit rather than the (+) side, though both ways will do the same thing.

[Daniel_LaFleur]

Originally Posted by Mr V

Correct it is turning the motor into a generator that creates the dynamic braking effect.


Most teams short it via the (-) side of the circuit rather than the (+) side, though both ways will do the same thing.

My understanding (and I may be wrong, it's been a while since I'ce studies regenerative braking) was that connecting the (-) sides together dissipated the energy as heat in the motor while connecting the (+) to both sides would dissipate the energy by recharging the battery. Am I incorrect in this? and if I'm not wouldn't it be better not to heat an already heated motor?

[Vikesrock]

Originally Posted by Daniel_LaFleur

My understanding (and I may be wrong, it's been a while since I'ce studies regenerative braking) was that connecting the (-) sides together dissipated the energy as heat in the motor while connecting the (+) to both sides would dissipate the energy by recharging the battery. Am I incorrect in this? and if I'm not wouldn't it be better not to heat an already heated motor?

Regardless of which side you connect, the battery is not part of the completed circuit and thus no current flows in or out of it. All energy dissipation occurs in the motor and the wiring/components that are shorting the leads.

[Al Skierkiewicz]

Daniel,
By shorting the motor leads together, you cause all generated currents to flow in the short and no where else. It does not matter whether the the positive or negative lead of the battery is connected at all. No current will flow through the battery when the motor leads are shorted. This method is identical to the the "brake" mode on either the Victor or Jag. The amount of current generated and the amount of braking action is RPM dependent. The faster the motor is spinning the more braking is evident.