The Bennett linkage 101
.
(5.66)
Similar to relationship between θ 1e and θ 1f , there are two values of δ for
each θ 1f .
The flattened configuration typically has a rhombus shape. Among all the
rhombuses with the same side length the square has the largest area, i.e.
.
(5.67)
Substituting Eq. (5.66) into Eq. (5.67) gives
.
(5.68)
Considering Eq. (5.61), Eq. (5.68) can be simplified as
.
(5.69)
So when θ 1e and θ 1f satisfy both of Eqs (5.63) and (5.69), the linkage based
on the alternative form expands to a square. Solving both equations simultaneously, we obtain,
,
(5.70)
.
(5.71)
Moreover, for any square fully deployed configuration, we always have
,
(5.72)
due to Eqs (5.61), (5.65) and (5.70).
.
(5.66)
Similar to relationship between θ 1e and θ 1f , there are two values of δ for
each θ 1f .
The flattened configuration typically has a rhombus shape. Among all the
rhombuses with the same side length the square has the largest area, i.e.
.
(5.67)
Substituting Eq. (5.66) into Eq. (5.67) gives
.
(5.68)
Considering Eq. (5.61), Eq. (5.68) can be simplified as
.
(5.69)
So when θ 1e and θ 1f satisfy both of Eqs (5.63) and (5.69), the linkage based
on the alternative form expands to a square. Solving both equations simultaneously, we obtain,
,
(5.70)
.
(5.71)
Moreover, for any square fully deployed configuration, we always have
,
(5.72)
due to Eqs (5.61), (5.65) and (5.70).