Motors

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  • as of 10/19/2006

Drive Train:

Motor:

http://maxonmotorusa.com/files/catalog/2005/pdf/05_137_e.pdf - 250002

Gearhead:

http://maxonmotorusa.com/files/catalog/2005/pdf/05_235_e.pdf - 110452

Encoder:

Encoder MR, 32 ticks/rev

Explanation on selection of motor power: The design goal that we focused on was to cross 1/3rd of the distance of the field in 1 second starting from rest.

The total length of the field is 4.9m so 1/3rd of the field = 1.63m

Using equations of kinematics:

Distance = initial velocity * time + .5 * acceleration * time^2. Since we start from rest, 1.63m = .5*a*t^2 (t was assumed to be 1 second) so a = 3.26m/s^2

Also from kinematics: final velocity = initial velocity + acceleration * time (V=at) so our final velocity is 3.26m/s

From basic energy equations: Kinetic energy = .5 * mass * velocity^2 (.5*m*V^2) So our final energy is 5.33*m J (we left mass as a variable, it's expressed in kilograms)

And finally power is energy/time so th total power we need = 5.33*m Watts Because of drivetrain considerations, we will have at least 2 motors providing power at all times. Therefore, we will need at minimum 2.67*m Watts/motor to cross 1/3rd the field in 1 second.

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