pantograph mechanism
gearhead
http://www.broan.com/ImageLibrary/broan/pdf/servicemanuals/93043470.pdf
and started looking at its geometry.
I diagrammed it with all the long bars the same length, middle pivots
dead center on them (short bars half the length of the long bars, and
the anchor pins the same distance apart as the long bar end pivots),
with the mechanism partly expanded.
Of course as the mechanism expands, the ram platen stays parallel to
the upper rail because the mechanism has symmetry. But with the
dimensions I chose, the ram platen moves sideways as it advances.
So I took another look at the drawing. I can see that on the long
bars with middle pivots, those pivots are not dead center between the
ends -- they are positioned at about a third of the way.
Apparently somebody found just the right bar lengths (i.e. pivot
distances) so that not only does the ram stay parallel, but also
advances dead straight without translating sideways in the process.
I've spent hours on this and haven't come up with a solution. Once
you start allowing the dimensions to vary without any restraints, the
problem has so many degrees of freedom that it's hard to know how to
approach it.
I've tried searching the web for information on that mechanism but
haven't come up with anything.
This is a challenge for you geometry enthusiasts.
Philippe 92
> I found this scissor mechanism for a trash compactor interesting
> http://www.broan.com/ImageLibrary/broan/pdf/servicemanuals/93043470.pdf
> and started looking at its geometry.
> So I took another look at the drawing. I can see that on the long
> bars with middle pivots, those pivots are not dead center between the
> ends -- they are positioned at about a third of the way.
> Apparently somebody found just the right bar lengths (i.e. pivot
> distances) so that not only does the ram stay parallel, but also
> advances dead straight without translating sideways in the process.
Ji,
The horizontal displacement is never = 0
It is **best** approximated to 0 when the ratio is not 1/3
but ~ 0.35
You may relate this to well known Tchebichef's linkage in which the
optimum is CD = AD = DM = (3 + sqrt(7))/2, BC = (4 + sqrt(7))/3
The Tchebichef's linkage is as follows :
Let B and C fixed points.
A is linked to B by a rod with unit length AB
a rod AM is linked at A, and let D the midpoint of AM
this rod is linked to C by a rod CD
See attached figure http://cjoint.com/?1dDqvX6TTUE
When A rotates around B, M describes a curve with an "optimally
flat" part.
The condition being that the curvature radius at lower point with
horizontal tangent is infinite.
that is there is a point with y' = y" = y"' = ... = 0
(as many null derivatives as possible)
Your linkage is a bit different, but results into the same kind
of curve, with an "optimally flat" vertical part.
I didn't wrote the exact equations for this curve, resulting into
an exact condition on the linkage ratio to get infinite cuvature
radius, I just played and adjusted the ratio with geometer's
sketchpad.
Best Regards.
--
Philippe C., mail : chephip, with domain free.fr
site : http://mathafou.free.fr/ (mathematical recreations)
Michael Robinson
"Philippe 92" <nos...@free.invalid> wrote in message
news:mn.ebda7db3a...@free.fr...