Transform the Wrench to the Body Frame -- a more general case.

Due to the symmetry of the helical shape, the external wrench in the body frame will be the same.

In the following, we can verify that the wrench is independent of

thus independent of s

% Derive the representation of force in the body frame. 
clc; clear;
syms alpha r F_e M_e l_t GJ EI GA EA M_y M_z F_y F_z phi s r theta rot_n;
assume([alpha r F_e M_e l_t GJ EI GA EA M_y M_z F_y F_z phi s r theta rot_n], 'real')
pi = sym(pi);

is the reference postion vector

p_star = [ r*cos(-phi); r*sin(-phi); s*sin(-alpha); ]

p_star =

R_star = Rz(pi - phi)*Rx(-pi/2 - alpha)

R_star =

Ad = simplify( [R_star zeros(3); skew(p_star)*R_star R_star])

Ad =

Notice that the

and

are downward in the global brame,

Fe = [0;0;F_e]; 
Me = [0;0;M_e];
We = [Me; Fe]; % 

Using Equation (20)

W = simplify(Ad'*We) % this is the wrench in the body frame

W =

This is corresponding to Eq. 22 in the paper.