September 28, 2016

For this week, most of my time was spent preparing for the first presentation, which included calculating out a few numbers as well as coming up with some concept designs. One of the first calculations that I made this week was finding the stall torque of the motor that we already have at our disposal: A Crystalyte SAW_408 motor paired with a 48 volt, 25 amp motor driver. Through looking into the superway archives, I found a data sheet on this motor, giving the potential torque and rotations per second at various amperage, given it had a constant voltage of 48 volts. Which has a stall torque of about 105 Nm. Now given that last years design used this one motor as a direct drive system, I used the outer diameter of the hub (about .406 m) to find the force that the direct drive motor could produce; approximately 519 N of force without accounting for friction. Yet, the bogie, which we are weighing at 600 lbs or 272 kg, would have a force of 780N going down a 17 degree slope, so using one motor as a direct drive would not be a viable option.

From there, I went into determining a more realistic weight distribution between the two bogies as it rests in the middle of a 17 degree incline. Surprising, it appears that the wheel that is higher on the track supports more of the systems weight compared to the lower wheel, due to carriage being underneath the bogies and causing the rear to swing up. For the calculations, I based it on the idea that the carriage was 400 lbs hanging from the cross member in the middle, much like the summer team developed. Each of the bogies therefore weight about 100lbs and were support (vertically) by one wheel on the track. With the wheel base of the bogies being 26 inches and the pivot point of the  carriage hanging about 10 inches below the cross member. Upon running through the calculations for weight distribution, I discovered that both wheels would experience the same force sliding down the track 390N, but in terms of normal force, the front wheel would support 1583 N compared to the back wheel's 1015 N. What makes this information so important is that it give me a rough estimate of what size motor I would need for each bogie. And more importantly an idea of about the contact friction between the wheels and the metal rail.

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.

Last Week's Post

Sorry, due to the tardiness of this post.
Last week, I did some simply calculation to the amount of normal force of the bogie to remain steady on the upward slope, basing it off of the an estimated weight of 300 lbs.