# Cartesian movement of the TCP frame relative to the base frame

Hello !

Here are my tcp frame and my base frame of my UR3 robot :

I want to move my tcp frame so that the XtcpYtcp plan is parallel to the XbaseYbase plan.

I know I have to change Rx,Ry and Rz orientations of my tcp but I dont know the exact values ​​to have a perfect parallelism.

Is there a way to orient the tcp frame in relation to the base frame with polyscope ?

Thank you

One way is to convert rotation vector to rpy, then set x to 3.14 and y to 0, convert back to rotation vector and assign to pose.
Try rotvec2rpy() and rpy2rotvec() script functions.

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Thank you for the answer but I just realize that I made a mistake in my explanation…
From the begining I was meaning the tcp frame and not the tool frame.
modified the post

i dont fuly understand your question but i hope this helps

i thougt UR calculation was RPY ( so first move RZ then RY then RX still didnt figured this fully out so i can be wrong)

for this i used this site

so if u want to go to your tcp plane relative to use

x= something
y=something
z= something
var1 = P[x,y,z,0,0,0]
Var1=pose_trans(Var1,p[0,0,0,d2r(180),0,0])
moveJ to var1

to move relative in your plane.

x2= something
y2=something
z2= something
var2 = P[x2,y2,z2,0,0,0]
Var2=pose_trans(Var1,var2)

this is solution by heart so i can be wrong.

kind regards,

Corné