Mathematics Update

We are indeed using some maths in our design. The servos provide 44 oz*in of torque. We are using three. One of them opens and closes the golf ball grabber. Because the golf ball will slide up a shallow ramp of card stock and the radius is quite small indeed between the servo and the golf ball, it is reasonable to simply assume that the servo will indeed lift the ball.

The second servo is positioned on the end of one of the grabber arms and actuates the draw bridge down which the balls will roll into the bucket. This draw bridge had a weight of 1.2 oz in the preliminary stages and its estimated current weight is 3 oz. Weight is distributed approximately evenly along the draw bridge and it has a total length of about 8 inches. It will be resting on the bucket while in operation so it is not necessary to include the weight of the golf ball in the torque calculations. Nor will the servo ever have to lift it entirely of its own power if all goes according to plan. However, assuming the worst case scenario, that the servo does in fact have to live the draw bridge, the torque on it would be 3 oz * 4 in (average length) = 12 oz*in, well under the rating of 42 at 6V.

The third servo is the problematic one. This one rotates the entire grabber assembly, which weighs quite a bit indeed. The unextended draw bridge is about 4 inches from this servo for a torque of 12 oz*in (when extended, it has an average distance of about 7 inches, but again, this servo will not have to hold that weight, for the extended draw bridge will rest on the bucket). The other two servos are placed at approximately 4 inches each from this third servo and at a 90 degree angle with each other, so the worst possible torque will occur when one is straight up and the other is straight out with a torque of 6.4 oz*in. The weight of a golf ball at 3 inches from the servo is 1.6*3 = 4.8 oz*in. The grabber assembly itself weighs about 6 oz and has a skewed weight distribution; the average distance of that thing from the servo is, according to the two-pencil method of finding the center of mass of an object, about 1 inch. The total torque we are putting on this thing so far is... 31.2 oz*in. Additionally, in order to get the 180 degrees of rotation that we need, we must gear the servo in a 3:2 ratio, giving it a torque of merely 28 oz*in. We will add more supports to the assembly, and the servo is clearly not powerful enough to lift it.

Therefore!!

To the servo that lifts the assembly, we attach a second rod in the opposite direction from the assembly with sufficient weight to offset the assembly and allow the servo to do its job.

The motors!

We need as much torque as we can get. When I built a gearbox in quarter 2 to help Mr. Podmers test stuff for this class, eyeballing the axle indicates that with such huge wheels as we are using, speed will not be problematic. Our design requires as much torque as possible, so we can not afford to engineer around speed in the first place. Knowing that the robot will move reasonably fast is quite sufficient for our purposes.

The previous torque calculations hold for the motors that drive the wheels, with the ratio of the radius with respect to the third servo to the radius with respect to the motors conservatively considered to be 8.5/6. Each small motor has a torque of 6.82 g*cm; conversion to oz*in, sending the torque through the gearbox, and multiplying by 2 (because we have 2 motors) gives us a total torque of 65 oz*in, more than enough to lift both the assembly and the counterweight.