3.4 Kinematic Analysis
We had 3 different bar mechanisms that each corresponded to a different note in Mary Had a Little Lamb. Our goal for the cam design was to associate each “dip“ with a note strike.
Breakdown of Each Mechanism
Cam Design
We decided to split our cam into 8 different sections since our mechanism would only play the first 8 notes of the song. As the cam spins, if a note were to be played at the given angular position, the cam would dip into its minimum radius such that the follower would move along with it.
Going off this diagram, for each note, we generated a 16-dimensional array of radii that the cam must follow at each angular position. The reason it is 16 dimensions is that each “checkpoint” corresponds to a discrete angular interval (every 22.5 degrees) around the cams' 360-degree rotation. Between these checkpoints, we used interpolation to generate the smooth profile of the cam for the follower to have a smooth and continuous motion.
E Cam
D Cam
C Cam
Position Analysis: Solving 4-bar Mechanisms
The diagrams below illustrate the sharp decline of the angles of the mechanisms. We used an identical setup as the one used in our prototype kinematic analysis, only differentiating by the input radius vector of the cams.
E Cam
D Cam
C Cam
Cam Animations:
Velocity Analysis:
Deriving Angular Velocities:
Input cam rotation: w_cam = 10 RPM = pi/3 = 1.05 rad/second
Cam displacement per note: delta_cam_angle = 45 degrees = pi/4 = 0.79 radians
Time per strike: t_strike = delta_cam_angle / w_cam = (pi/4) / (pi/3) = 3/4 = 0.75 seconds
- Per strike, 4-bar has a displacement of 46.56 degrees (40 -> 16.71 -> 40 degrees), or 0.8129 radians.
4-Bar input angular velocity = w_2 = delta_theta2 / t_strike = 0.8129 / 0.75 = 1.0839 radians/sWith w_2 = 1.0839 radians/s, solve for velocities at each strike:
* Each displacement reflects a strike *E-Cam:
E-Cam: Relevant Velocities at Each StrikeD-Cam:
D-Cam: Relevant Velocities at Each StrikeC-Cam:
C-Cam: Relevant Velocities at Each Strike
Force of Strike:
We incorporated rubber bands into our mechanism to ensure the arm strikes the key with enough force. Given this, our linkages experience force from the rubber bands similar to a spring mechanism.
Points of Reference for Rubber Band ForceWith the rubber bands being placed where the blue line is in the diagram, the total displacement delta_x can be measured by the minimum and maximum between the points B and R. Therefore:
minimum = 6.825”
maximum = 7.708”
delta_x = (7.708 - 6.825) = 0.883”
Using Hookes law, the force exerted by the rubber band at point B is expressed by:
F_rubberBand = -k * delta_x = -k * 0.883
Since k is unknown, solving for the exact magnitude is currently not practical. But since the force grows linearly with each stretch, it will produce large enough restoring forces to strike the xylophone properly.