Re: Re: Hexapod question - rigidity

> From: "Raymond Heckert" <jnr@...>
>
> The ideal configuration of a Stewart Platform would consist of six
> struts; A-B, C-D, E-F, G-H, I-J, and K-L; a 'base-triangle' " M-N-O",
> and a 'table-triangle' "P-Q-R", where the "A" end of strut A-B is
> 'joint-connected' to the "C" end of the C-D strut at the point "M".
> Likewise, the "E" end of the E-F strut is 'joint-connected' to the "G"
> end of the G-H strut at point "N". And, the "I" end of the I-J strut is
> 'joint connected' to the "K" end of the K-L strut at point "O". The
> forgoing comprises the 'base-linkage'. Now, the 'table linkage'
> consists of ends "D" & "F" 'joint-connected' at point "P", and the ends
> "H" & "J" 'joint-connected' at point "Q". Finishing the table, we
> joint-connect' "L" & "B" together at point "R". As you will see, if
> you make a drawing of this, the unit is made of triangles
> interconnected in such a way as to make for a 'geometrically-rigid'
> structure. Note that the lengths of each strut is variable, and
> dependent on the length of at least one other strut, and depending on
> the complexity of the move, dependent upon the other five struts! I
> hope this can give everybody a picture of the unit. Hint: the
> 'table-triangle' is initially oriented 180° around from the
> 'base-triangle'. i.e they 'point' in opposite directions.

Hi RayHex.

How'd you like that last blast of snow?

Your design here is real close to what was done with the NIST cable
hexapod. It is also about what was done with the Ingersoll at NIST. You
run into some practical limits when you try to co-locate the ends of the
triangles with rigid struts and still have them slide as the platform
moves.

A prof, John W. Sutherland, at Michigan Tech had a nice page on Stewart
geometry but I couldn't find it the last time I looked. Looks like they
put all of his stuff behind a password.

Ray Henry