MATLAB: Forward Kinematics and Reverse Kinematics

robotics

In Robotics F.K. we have A*Si = B*phi where A, B are known matrices, Si is a vector with x, y and z as its elements, phi is the vector containing the angular velocity of each of the 2 wheels of the robot. MATLAB says that an eqn of the form Ax=B can be solved by A\B. I am not sure if this applies in my case i.e. A*Si = B*phi. This I know has been solved and I was told that A\B is the way to do it but does not work for me. Similarly I am stuck on the R.K. case. Please help

Best Answer

So A & B are known matrices. Si is a known vector.

OF COURSE IT APPLIES!

What A*Si? A known vector. Call it b.

So then you have a problem of the form

B*phi = v

Can you solve that using backslash? Yes. B is known. v is known.

Look at the equation. How do you solve it? Substitute variables.

Related Solutions

MATLAB: How to solve equation

m_c0 = 4*((m_c_theta1+1)*sin(gamma)^2+l*gamma*(ROOT-(1/2)*cos(gamma)-1/2)*sin(gamma)+(2*ROOT*m_c_theta1+2*ROOT-2)*cos(gamma)-2*ROOT*m_c_theta1+2*ROOT-2)*m_c_theta1/((gamma^2*l^2+4*m_c_theta1^2+8*m_c_theta1+4)*sin(gamma)^2-4*l*gamma*(m_c_theta1-1)*sin(gamma)-16*m_c_theta1)
j_L0 = (1/2)*(-l*gamma*(gamma^2*l^2+4*m_c_theta1^2+4*m_c_theta1+4)*sin(gamma)^3+(2*m_c_theta1*(gamma^2*l^2+4*m_c_theta1^2+12*m_c_theta1+8)*cos(gamma)+4*l^2*(-1+(ROOT+1/2)*m_c_theta1)*gamma^2+(16*ROOT+8)*m_c_theta1^3+(16*ROOT+8)*m_c_theta1^2+16*m_c_theta1)*sin(gamma)^2-2*l*gamma*((l^2*(ROOT-1)*gamma^2+(4*ROOT+4)*m_c_theta1^2+(4*ROOT-8)*m_c_theta1+4*ROOT-4)*cos(gamma)+l^2*(ROOT-1)*gamma^2+(4*ROOT+4)*m_c_theta1^2+(-4*ROOT-8)*m_c_theta1+4*ROOT-4)*sin(gamma)-8*(l^2*(ROOT-1)*gamma^2+4*m_c_theta1^2*(ROOT+1))*(cos(gamma)+1))/(((gamma^2*l^2+4*(m_c_theta1+1)^2)*sin(gamma)^2-4*l*gamma*(m_c_theta1-1)*sin(gamma)-16*m_c_theta1)*sin(gamma))
theta1 = arctan((-2*gamma*l*sin(gamma)^2*m_c_theta1+((-l^2*(ROOT-1)*gamma^2-4*(m_c_theta1+1)*(ROOT*m_c_theta1+ROOT-1))*cos(gamma)-l^2*(ROOT-1)*gamma^2+4*ROOT*m_c_theta1^2-4*ROOT-4*m_c_theta1+4)*sin(gamma)-4*l*gamma*(cos(gamma)+1)*(ROOT-1))/((gamma^2*l^2+4*(m_c_theta1+1)^2)*sin(gamma)^2-4*l*gamma*(m_c_theta1-1)*sin(gamma)-16*m_c_theta1), ROOT)
M = ((4*m_c_theta1+4)*sin(gamma)^2+4*l*gamma*(ROOT-(1/2)*cos(gamma)-1/2)*sin(gamma)+(8*ROOT*m_c_theta1+8*ROOT-8)*cos(gamma)-8*ROOT*m_c_theta1+8*ROOT-8)/((gamma^2*l^2+4*m_c_theta1^2+8*m_c_theta1+4)*sin(gamma)^2-4*l*gamma*(m_c_theta1-1)*sin(gamma)-16*m_c_theta1)

where

ROOT =  RootOf((-4*gamma*cos(gamma)*l*sin(gamma)+4*sin(gamma)^2*m_c_theta1^2+gamma^2*l^2*sin(gamma)^2-4*sin(gamma)^2+8-8*cos(gamma)+4*gamma*l*sin(gamma))*z^2+(-cos(gamma)*l^2*gamma^2*sin(gamma)^2-gamma^2*l^2*sin(gamma)^2-2*l*sin(gamma)^3*m_c_theta1*gamma-4*l*sin(gamma)^3*gamma+4*cos(gamma)*m_c_theta1*sin(gamma)^2+4*cos(gamma)*sin(gamma)^2-4*sin(gamma)^2*m_c_theta1-4*sin(gamma)^2)*z+cos(gamma)*l^2*gamma^2*sin(gamma)^2+2*l*sin(gamma)^3*m_c_theta1*gamma+4*l*sin(gamma)^3*gamma-2*cos(gamma)*m_c_theta1^2*sin(gamma)^2+4*gamma*cos(gamma)*l*sin(gamma)-4*cos(gamma)*m_c_theta1*sin(gamma)^2-2*sin(gamma)^2*m_c_theta1^2-4*gamma*l*sin(gamma)-4*cos(gamma)*sin(gamma)^2+4*sin(gamma)^2*m_c_theta1+8*sin(gamma)^2+8*cos(gamma)-8, z)

and RootOf(f(z),z) means the values, z, such that f(z) = 0 -- the roots of the equation.

As ROOT is a quadratic, it has two exact solutions that can be substituted in to the other equations.

MATLAB: Solve an expression and plot

beta = 0.85; gamma = .15; theta = 0.1;
alpha = .303: .001: 5;
syms Alpha delta
A = -Alpha * beta + Alpha .* delta + gamma * beta + beta;
B = Alpha .* theta * (beta - delta) + gamma * beta;
rr = solve( A^2 - 4*B*beta, delta);
rrt = double(subs(rr, Alpha, alpha(1));
[maxrrt, idx] = max(rrt);
maxrr = rr(idx);
rrpos = double(subs(maxrr, Alpha, alpha));
plot(alpha, rrpos)

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