Week 3

For this week, I started some basic calculations for the braking force needed for a 600 lb object to be going down a slope of 17 degrees at a speed of 20 mph. Say we wanted it to stop in 5 seconds, 284.8 lb-force would be needed in order to brake at that rate. Or for stopping in three seconds down the same slope and the same speed a total of 358 lb-force would be needed. Now to calculate the force needed to be applied for the actual brake, I would need to know a bit more about the caliper and rotor compared to the wheel, so for now, I just assumed the caliper to be the same size as the rotor at 8 inches and a brake pad of 1.5 inches wide with a friction coefficient between the brake and brake pads being .3. The result is 1253 lb-f force needed to clamp onto the brake rotor for stopping in 5 seconds and 1574 lb-f for braking in 3 seconds. Which, I will have to see in the caliper that we currently possess can handle that about of clamping force. Also, I am not currently sure it that will cause the rotor to lock up and skid the wheel as apposed to slow it down at a steady rate.
As far as the design of the motor and braking arrangement are concerned, I figure that we could put the motor and brake on a bracket in front of the bogie, which a chain that will drive the wheels that are in contact with the rail supporting the bogie's weight. Similarly, there will also be a fail-safe in place in front of the bogie acting as a physical brake between the rail and the bogie instead of using the wheel. This fail safe is basically a rubber stopper on an actuator that retracts under power and clamps into place if the bogie loses power.
The motor is still up in the air at this point, I need to do calculations for going up a slope of 17 degrees at 600lbs to determine the torque, voltage, and amperage that would be ideal.