10 - Kinematic Analysis
Regular Theo Jansen Mechanism Position Analysis
import numpy as np
import matplotlib.pyplot as plt
""" --- POSITION ANALYSIS --- """
# Initialize variables
# Actual CAD measurements of the linkages:
b = 124.50
c = 117.9
d = 120.3
e = 167.4
f = 118.2
g = 110.1
h = 197.1
i = 147
j = 150
k = 185.7
m = 45
x = 114
y = 23
"""
#Jansen Holy numbers:
x = 38 # x-component of ground link
y = 7.8 # y-component of ground link
b = 41.5 # Upper 4-bar Rocker
c = 39.3 # Lower 4-bar Rocker
d = 40.1
e = 55.8 # Rigid body "bde"
f = 39.4 # Parallel-like mechanism
g = 36.7
h = 65.7
i = 49 # Rigid body "ghi"
j = 50 # Upper 4-bar Coupler
k = 61.9 # Lower 4-bar Coupler
m = 15 # Crank
"""
theta_m_degrees = np.arange(360)
theta_j = []
theta_b = []
theta_k = []
theta_c = []
theta_d = []
theta_f = []
theta_g = []
theta_i = []
theta_p = []
p = []
P_p_x = []
P_p_y = []
X3 = []
Y3 = []
n = np.sqrt(x**2 + y**2) # Ground link
"""
In this mechanism we refer to FIGURE 0 for the links and linkages of the Theo Jansen Mechanism.
The Jansen mechanism is composed of
- crank (m)
- 2 oscillating rockers (b,c)
- 2 couplers (j, k)
- 2 three-bar linkages (b,d,e) and (g,h,i)
- upper four bar linkages (nm-bj)
- lower four bar linkages (nm-ck)
- open four-bar linkage/parallel lik linkage (dc-fg)
- rigid three-bar (g,h,i)
- Ground (02-04)
- Toe (P)
We let O2 be the origin of the General Coordinate System.
Note that in general four bar notiation we use FIGURE 1 as a reference for our analysis for
the four bar linkages.
For Figure 1, we can write the vector loop equation as
d - c - a- b = 0
d exp(jtheta_d) - c exp(jtheta_c) - a exp(jtheta_a) - b exp(jtheta_b) = 0
Using these we can analyze the four linkages (nm-bj), (nm-ck).
"""
for angle in theta_m_degrees:
# Compute the position of point B
X_m = x + m * np.cos(np.deg2rad(angle)) #x comp of point B from origin
Y_m = y + m * np.sin(np.deg2rad(angle)) #y comp of point B from origin
R = np.sqrt(X_m**2 + Y_m**2) #vector of Point B from origin
alpha = np.rad2deg(np.arctan2(Y_m, X_m)) #angle of point 1 wrt O2 ground/origin
"""
---- VECTOR LOOP (1) UPPER 4-BAR LINKAGE (nm-bj) ---
Referring to FIGURE 2, we can write the equation of the components of the linkages as:
n cos(theta_n) + m cos(theta_m) - b cos(theta_b) - j cos(theta_j) = 0
n sin(theta_n) + m sin(theta_m) - b sin(theta_b) - j sin(theta_j) = 0
Although a four bar linkage we use R and cos law to get theta_j and theta_b.
This is done by making a triangle for links {b,j,R}.
Let A_j be the projecton of j on R.
Let A_b be the projection of b on R.
See FIGURE 3.
"""
A_j = (j**2 - b**2 + R**2) / (2 * j) #projecton of j on R
A_b = (b**2 - j**2 + R**2) / (2 * b) #projection of b on R
theta_j.append(np.rad2deg(np.arccos(A_j / R)) + alpha) #theta_j as function of m
theta_b.append(np.rad2deg(np.arccos(A_b / R)) + alpha) #tehta_b as function of m
"""
--- VECTOR LOOP (2) LOWER 4-BAR LINAKG (nm-ck) ---
Again, we use R and cos law to get theta_k and theta_c.
Note that we are using the same R here since it is the same points for both (nm-bj) and (nm-ck).
This is done by making a triangle for links {c,k,R}.
Let A_k be the projecton of k on R.
Let A_c be the projection of c on R.
See FIGURE 4 and FIGURE 5.
"""
A_k = (k**2 - c**2 + R**2) / (2 * k) #projecton of k on R
A_c = (c**2 - k**2 + R**2) / (2 * c) #projecton of c on R
theta_k.append(np.rad2deg(np.arccos(A_k / R)) + alpha) #theta_k as function of m
theta_c.append(np.rad2deg(np.arccos(A_c / R)) + alpha) #theta_c as function of m
"""
--- PARALLEL LINKAGE (dc-fg) --
Need to find theta_d since it is the common link between triangle bde and parallel dc-fg.
Usining cosine law with bde FIGURE 6:
"""
theta_d_value = np.degrees(np.arccos((b**2 + d**2 - e**2) / (2 * b * d))) + theta_b[-1]
theta_d.append(theta_d_value)
"""
Now looking at the parallelogram (FIGURE 7), we take theta d as the new "moving ground".
We take point 4 as the input since the rotation from m is transmitted to that point.
Let Point 5 be the output since that point helps determine point P.
Just like in our analysis with the 2 four bar linkages where we used R to extend a line
from the origin/input to the output point to help with the analysis, see FIGURE 8.
We use R3 to to extend a line from point 4 to point 5.
We let X3 and Y3 be the components of R3, and alpha3 be its angle from horizontal.
"""
X3_value = d * np.cos(np.radians(theta_d_value)) + c * np.cos(np.radians(theta_c[-1]))
Y3_value = d * np.sin(np.radians(theta_d_value)) + c * np.sin(np.radians(theta_c[-1]))
X3.append(X3_value)
Y3.append(Y3_value)
R3 = np.sqrt(X3[-1]**2 + Y3[-1]**2)
alpha3 = np.rad2deg(np.arctan(Y3_value/X3_value))
"""
Now that we have alpha3 of point 5, we can use it to find theta_f and theta_g in the parallelogram.
Just as before, we use projections to help us.
Let A_f be the projection of f on R3.
Let A_g be the projection of g on R3.
We use R3 since this is the third four-bar we are analyzing,
and the first two four-bars used the same R.
"""
A_f = (f**2 - g**2 + R3**2) / (2 * f) #projection of f on R3
A_g = (g**2 - f**2 + R3**2) / (2 * g) #projection of g on R3
A_f_clipped = np.clip(A_f / R3, -1, 1) #clipped to take away rouding errors
A_g_clipped = np.clip(A_g / R3, -1, 1) #clipped to take away rouding errors
theta_f_value = np.rad2deg(np.arccos(A_f_clipped)) + alpha3
theta_g_value = np.rad2deg(np.arccos(A_g_clipped)) + alpha3
theta_f.append(theta_f_value)
theta_g.append(theta_g_value)
"""
--- (ghi) TRIANLGE + (cip) TRIANGLE ---
Referring to Figure 9:
Now we need find find point p as a function of the crank angle theta_m
First we find theta i from cosine law of triangle {ghi}
"""
theta_i_value = theta_g[-1] - np.rad2deg(np.arccos((g**2 + i**2 - h**2) / (2 * g * i)))
theta_i.append(theta_i_value)
"""
From the diagram, P is just the sum of the x and y components of link c and link i.
Therefore, since we already knowt theta_c from the second four-bar analysis and we obtained
theta i from the ghi triangle, we can find theta p.
""" # Angular Position theta_p_value = np.rad2deg(np.arctan( (c * np.sin(np.deg2rad(theta_c[-1])) + i * np.sin(np.deg2rad(theta_i[-1]))) / (c * np.cos(np.deg2rad(theta_c[-1])) + i * np.cos(np.deg2rad(theta_i[-1]))) )) theta_p.append(theta_p_value) p_x = c * np.cos(np.deg2rad(theta_c[-1])) + i * np.cos(np.deg2rad(theta_i[-1])) p_y = c * np.sin(np.deg2rad(theta_c[-1])) + i * np.sin(np.deg2rad(theta_i[-1])) p_value = np.sqrt(p_x**2 + p_y**2) p.append(p_value) # Linear position P_p_y.append(p_y) # y-component of P P_p_x.append(p_x) # x-component of P# Position Plotplt.figure(1)# Angular Position of Toe Point (P) of Walking Mechanismplt.subplot(3, 1, 1)plt.plot(theta_m_degrees, theta_p) # Plot against theta_m_degreesplt.title("Angular Position of Toe Point (P) of Walking Mechanism")plt.xlabel(r'Crank angle(deg) or $\theta_{m}(°)$')plt.xlim([0, 360])plt.ylabel(r'Angular Position (deg) or $\theta_{P}(°)$')plt.grid(True)# Position of Toe Point (P) of Walking Mechanism (x-direction)plt.subplot(3, 1, 2)plt.plot(theta_m_degrees, P_p_x) # Plot against theta_m_degreesplt.title("Position of Toe Point (P) of Walking Mechanism (x-direction)")plt.xlabel(r'Crank angle(deg) or $\theta_{m}(°)$')plt.xlim([0, 360])plt.ylabel('Toe Point Position (mm)')plt.grid(True)# Position of Toe Point (P) of Walking Mechanism (y-direction)plt.subplot(3, 1, 3)plt.plot(theta_m_degrees, P_p_y) # Plot against theta_m_degreesplt.title("Position of Toe Point (P) of Walking Mechanism (y-direction)")plt.xlabel(r'Crank angle(deg) or $\theta_{m}(°)$')plt.xlim([0, 360])plt.ylabel('Toe Point Position (mm)')plt.grid(True)# Save the figureplt.tight_layout() # Adjust subplots to fit into figure area.# For toe pathplt.figure(2)plt.plot(P_p_x, P_p_y)plt.title("Trace Path of Toe Point (P) of Walking Mechanism")plt.xlabel('Toe Point Position (x-direction)(mm)')plt.ylabel('Toe Point Position (y-direction)(mm)')plt.grid(True)# Show the plotsplt.show()