Re: kinematics answer

[robin szemeti <robin-at-rszemeti.demon.co.uk>]

On Tuesday 10 September 2002 08:13, Till Franitza wrote:
> Hi Robin!
> It is as trivial as you supposed.
> This is the code of the trivial kinematics. Do not bother about a,b and c.
> These are the (for example rotational) axes you do not need.

ahh .. thats what a,b and c are ....

> pos is the cartesian position that you refer to in g-code and that you see
> in the display of the NC normally. In EMC you can switch between worls
> coordinates and joint coordinates. the inverse kinematics and the forward
> kins have to fit together. 

yes that makes sense .. I had figured that forward kinematics was 
controller->machine and inverse was machine feedback > display which was 
quite wrong .. infact quite the opposite of what happens.

so forward kinematics maps joints to cartesian coords and inverse does 
*exactly* the opposite ... no wonder I was having difficulty understanding 
... I was trying to work out how forward kinematics did what really inverse 
kinematics is doing.

> So, for example you find out that when your y-axis moves 100mm,
> the x-Position of the TCP moves 1 mm down. To compensate this you figure
> out you have to make the y-Axis go 1% of the x motion up:

so inverse kinematics 

> joint 0 = pos x+ pos y * 0.01
> joint 1 = pos y
> joint 2 = pos z
>

> so the forward kinematics:
>
> pos x = joint 0 - joint 1 *0.01
> pos y = joint 1
> pos z = joint 2

yes .. simple once you know what the variables represent .. all I needed was 
joints being the 'machine' and pos being our desired cartesian position and 
that bit about world and machine views and I was away ... no problem.

Thanks .. that will definitely do it ... I'll let you know how it allworks 
out in the end.

-- 
Robin Szemeti