Re: [CAD_CAM_EDM_DRO] Hexapod question - rigidity

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.

RayHex