我将以八度为单位创建的代码移植到pylab中.其中一个移植方程在python中产生的结果与在八度音程中产生的结果截然不同.
解释的最佳方式是显示由八度和pylab从同一方程生成的图.
这是八度音阶中原始方程的简化片段.在这个小测试脚本中,phi保持为零的函数结果从〜( - pi,pi)绘制:
clear
clc
close all
L1 = 4.25; % left servo arm length
L2 = 5.75; % left linkage length
L3 = 5.75; % right linkage length
L4 = 4.25; % right servo arm length
L5 = 11/2; % distance from origin to left servo
L6 = 11/2; % distance from origin to right servo
theta_array = [-pi+0.1:0.01:pi-0.1];
phi = 0/180*pi;
for i = 1 : length(theta_array)
theta = theta_array(i);
A(i) = -L3*(-((2*cos(theta)*L1*(sin(phi)*L4-sin(theta)*L1)-2*sin(theta)*L1*(L6+L5-cos(phi)*L4-cos(theta)*L1))/(2*L3*sqrt((L6+L5-cos(phi)*L4-cos(theta)*L1)^2+(sin(phi)*L4-sin(theta)*L1)^2))-((2*sin(theta)*L1*(L6+L5-cos(phi)*L4-cos(theta)*L1)-2*cos(theta)*L1*(sin(phi)*L4-sin(theta)*L1))*(-(L6+L5-cos(phi)*L4-cos(theta)*L1)^2-(sin(phi)*L4-sin(theta)*L1)^2-L3^2+L2^2))/(4*L3*((L6+L5-cos(phi)*L4-cos(theta)*L1)^2+(sin(phi)*L4-sin(theta)*L1)^2)^(3/2)))/sqrt(1-(-(L6+L5-cos(phi)*L4-cos(theta)*L1)^2-(sin(phi)*L4-sin(theta)*L1)^2-L3^2+L2^2)^2/(4*L3^2*((L6+L5-cos(phi)*L4-cos(theta)*L1)^2+(sin(phi)*L4-sin(theta)*L1)^2)))-((cos(theta)*L1)/sqrt((L6+L5-cos(phi)*L4-cos(theta)*L1)^2+(sin(phi)*L4-sin(theta)*L1)^2)-((sin(theta)*L1-sin(phi)*L4)*(2*sin(theta)*L1*(L6+L5-cos(phi)*L4-cos(theta)*L1)-2*cos(theta)*L1*(sin(phi)*L4-sin(theta)*L1)))/(2*((L6+L5-cos(phi)*L4-cos(theta)*L1)^2+(sin(phi)*L4-sin(theta)*L1)^2)^(3/2)))/sqrt(1-(sin(theta)*L1-sin(phi)*L4)^2/((L6+L5-cos(phi)*L4-cos(theta)*L1)^2+(sin(phi)*L4-sin(theta)*L1)^2)))*sin(acos((-(L6+L5-cos(phi)*L4-cos(theta)*L1)^2-(sin(phi)*L4-sin(theta)*L1)^2-L3^2+L2^2)/(2*L3*sqrt((L6+L5-cos(phi)*L4-cos(theta)*L1)^2+(sin(phi)*L4-sin(theta)*L1)^2)))-asin((sin(theta)*L1-sin(phi)*L4)/sqrt((L6+L5-cos(phi)*L4-cos(theta)*L1)^2+(sin(phi)*L4-sin(theta)*L1)^2)));
end
plot(theta_array,A)
由此产生的八度音阶图如下所示: …