# Inverse Kinematics 5DOF - general approach - Wrong solution MATLAB

Can anyone name a good source for a general approach (cookbook-like) for the inverse kinematics regarding a 5-DOF?

The paper by De Xu et al. is aiming for a general approach, but it doesn't work for my code. Maybe you could check that one as well?!

I am new to MATLAB, the code is no masterpiece:

Parameter used:

a=[0 82 93 55 95 0];

d=[47 0 0 0 0 10];

alpha=[90 0 0 0 0 0];

theta=[15 30 40 -90 -15 0];

Result for Theta 1-5:

1. 5,080
2. 38,029
3. 75,947
4. -90,247
5. 20,719

EDIT: The results 1-5 mentioned above are the one the code gave me. But those are wrong, because:

Position of TCP (taken form the Forward Kinematics):

px= -5.7853

py= 5.7606

pz= 25.908

Position of TCP (using the calculating values for Theta 1-5):

px= 47.0

py= -151.0

pz= 43.9430

All values in degrees. I am using 5 servos by Dynamixel ax-12a.

a=[0 82 93 55 95 0];
d=[47 0 0 0 0 10];
alpha=[90 0 0 0 0 0];
theta=[15 30 -90 0 45 0];

T01=[cos(theta(1)),-sin(theta(1))*cos(alpha(1)),sin(theta(1))*sin(alpha(1)),a(1)*cos(theta(1));
sin(theta(1)),cos(theta(1))*cos(alpha(1)),-cos(theta(1))*sin(alpha(1)),a(1)*sin(theta(1));
0,sin(alpha(1)),cos(alpha(1)),d(1);
0,0,0,1];
T12=[cos(theta(2)),-sin(theta(2))*cos(alpha(2)),sin(theta(2))*sin(alpha(2)),a(2)*cos(theta(2));
sin(theta(2)),cos(theta(2))*cos(alpha(2)),-cos(theta(2))*sin(alpha(2)),a(2)*sin(theta(2));
0,sin(alpha(2)),cos(alpha(2)),d(2);
0,0,0,1];
T23=[cos(theta(3)),-sin(theta(3))*cos(alpha(3)),sin(theta(3))*sin(alpha(3)),a(3)*cos(theta(3));
sin(theta(3)),cos(theta(3))*cos(alpha(3)),-cos(theta(3))*sin(alpha(3)),a(3)*sin(theta(3));
0,sin(alpha(3)),cos(alpha(3)),d(3);
0,0,0,1];
T34=[cos(theta(4)),-sin(theta(4))*cos(alpha(4)),sin(theta(4))*sin(alpha(4)),a(4)*cos(theta(4));
sin(theta(4)),cos(theta(4))*cos(alpha(4)),-cos(theta(4))*sin(alpha(4)),a(4)*sin(theta(4));
0,sin(alpha(4)),cos(alpha(4)),d(4);
0,0,0,1];

T45=[cos(theta(5)),-sin(theta(5))*cos(alpha(5)),sin(theta(5))*sin(alpha(5)),a(5)*cos(theta(5));
sin(theta(5)),cos(theta(5))*cos(alpha(5)),-cos(theta(5))*sin(alpha(5)),a(5)*sin(theta(5));
0,sin(alpha(5)),cos(alpha(5)),d(5);
0,0,0,1];

T56=[cos(theta(6)),-sin(theta(6))*cos(alpha(6)),sin(theta(6))*sin(alpha(6)),a(6)*cos(theta(6));
sin(theta(6)),cos(theta(6))*cos(alpha(6)),-cos(theta(6))*sin(alpha(6)),a(6)*sin(theta(6));
0,sin(alpha(6)),cos(alpha(6)),d(6);
0,0,0,1];

T06=T01*T12*T23*T34*T45*T56;        %transformation matrix
inv_T01=inv(T01);
inv_T12=inv(T12);
inv_T23=inv(T23);
inv_T56=inv(T56);
T05=T06*inv_T56;
%needed for theta4 and theta5
Inverse=inv_T23*inv_T12*inv_T01*T06*inv_T56;
T_4_5=T34*T45;

N=T06(1:3,1);
nx=N(1,1);
ny=N(2,1);
nz=N(3,1);
O=T06(1:3,2);
ox=O(1,1);
oy=O(2,1);
oz=O(3,1);
A=T06(1:3,3);
ax=A(1,1);
ay=A(2,1);
az=A(3,1);
P=T06(1:3,4); %comapre to (17) in paper
px=P(1,1);
py=P(2,1);
pz=P(3,1);
px_2=T05(1,4);
py_2=T05(2,4);

%theta1
help1=px_2;               %not sure
help2=py_2;                 %not sure
help3=(px_2*(-1));         %not sure
help4=(py_2*(-1));         %not sure

theta11=atan2(help2,help1);
theta12=atan2(help4,help3);     %using this
theta1=atan2(py,px);            %most common way

%Theta2
B1=(az*d(6)-pz+d(1));
B2=(sqrt((-ax*d(6)+px).^2+(-ay*d(6)+py).^2))-a(1); %taking only positive value, alternative: [1 -1]*sqrt
B3=((B1.^2+B2.^2+a(2).^2-d(4).^2));
%auxiliary angle
beta=(atan2(B2,B1));
root=((a(2)*a(2))*sqrt(B1.^2+B2.^2));

theta21=(asin(B3/root)-beta);%works best
theta22=pi-theta21;

%theta3

theta3=atan2(B1-a(2)*sin(theta21),B2-a(2)*cos(theta21))-theta21; %using theta21 is close to determined angle BUT still far away from acceptable

%theta4
%auxiliary
OX=(ox*cos(theta1).*sin(theta21+theta3));
OY=(oy*sin(theta1).*sin(theta21+theta3));
OZ=(oz*cos(theta21+theta3));
theta4=atan2(-OX-OY-OZ,ox.*sin(theta1)*oy.*cos(theta1));%as in the case

%theta5
%auxiliary
NX=(nx*cos(theta1).*cos(theta21+theta3));
NY=(ny*sin(theta1).*cos(theta21+theta3));
NZ=(nz*sin(theta21+theta3));
AX=(ax*cos(theta1).*cos(theta21+theta3));
AY=(ay*sin(theta1).*sin(theta21+theta3));

theta5=atan2(NX+NY-NZ,-AX+AY-az*sin(theta21+theta3));%as in the case

enter code here

• Apart from theta2 it seems to work. Someone ones how to calculate theta2? thanks
– Max
Dec 25, 2017 at 10:53
• would you please send the paper link?(it is broken).and did using radians instead if degree solved the problem?
– alex
Jul 24, 2018 at 14:25

I never saw a conversion between degrees in radians in your code, so that would be the number one problem to fix. All of your transform matrices are using sin and cos terms, which only accept values in radians, and using them with values in degrees. I'm a bit busy at the moment but I'll go over the code in more detail later today.