13.3 Kinematic Analysis
Kinematic Analysis:
Mobility: M = 3(8 − 1) − 2(10) − 0 = 1 DOF
Grashof condition:
shortest + longest ≤ sum of the other two links
Loop 1: m–k–c–a
m = 90.0 mm
k = 371.4 mm
c = 235.8 mm
a = 228.0 mm
Shortest = m = 90.0 mm
Longest = k = 371.4 mm
Check:
90.0 + 371.4 ≤ 235.8 + 228.0
461.4 ≤ 463.8
Result:
461.4 ≤ 463.8 → Grashof condition satisfied
Loop 2: m–j–b–a
m = 90.0 mm
j = 300.0 mm
b = 249.0 mm
a = 228.0 mm
Shortest = m = 90.0 mm
Longest = j = 300.0 mm
Check:
90.0 + 300.0 ≤ 249.0 + 228.0
390.0 ≤ 477.0
Result:
390.0 ≤ 477.0 → Grashof condition satisfied
Loop 3: d–c–g–f
d = 240.6 mm
c = 235.8 mm
g = 220.2 mm
f = 236.4 mm
Shortest = g = 220.2 mm
Longest = d = 240.6 mm
Check:
220.2 + 240.6 ≤ 235.8 + 236.4
460.8 ≤ 472.2
Result:
460.8 ≤ 472.2 → Grashof condition satisfied
Final summary:
All three four-bar loops satisfy the Grashof condition. Therefore, the mechanism allows continuous crank rotation and should not lock due to non-Grashof geometry.
Our kinematic analysis focuses on characterizing the time-dependent motion of the assembly by evaluating not only the position, but the trace path, velocity, and acceleration of the Toe Point. In this analysis, we make sure that we map out the visual trace path of the toe point from the position vector in order to make sure that the trajectory of the end avoids breaking within our workspace limits. In order to make sure that all of the actuators can actually drive the system, the velocity plots are analyzed in order to make sure the required speeds don’t exceed the motors' peak ratings. We also analyze the acceleration plots to check where exactly we need stronger structural supports to make sure the leg doesn’t vibrate a lot.
Torque Analysis:
In order to determine what kind of motor we needed to propel the skateboard, we analyzed the torque using an impulse of 50 N·s, representing the standard push of a skateboard. Using the dimensions already used to do the Kinematic Analysis, the script below uses circle-intersection algorithms in order to map both the trajectory and velocity of the foot across an entire 360-degree rotation of the crank.
The code that we wrote for the analysis of the torque was able to produce the peak normalized torque multiplier of 0.857 N·m of required crank torque per 1 Newton of a horizontal thrust. Since this multiplier is less than one, this means that the machine has a bit of a mechanical advantage in comparison to a 1:1 direct-drive system. Though this means that to have an actually meaningful thrust in the horizontal direction, we will need a very high-torque! To do this, we have two methods: either to use an extremely high-torque gear motor, or to pair a high-torque Brushless DC motor with a planetary gearbox in order to withstand the movement done by the machine.