Articulated cube with fixed bar (I)

[color=#999999][color=#999999][color=#999999][color=#999999]This activity belongs to the [i]GeoGebra book[/i] [url=https://www.geogebra.org/m/h3gbmymu]Linkages[/url].[/color][/color][/color][/color][br][br]In the [url=https://www.geogebra.org/m/h3gbmymu#material/zeagkrck]previous construction[/url], the degrees of freedom of the cube do not coincide with the internal degrees of freedom. For them to coincide, we have to fix the cube in space. In this construction, and [url=https://www.geogebra.org/m/h3gbmymu#material/tcsy4jut]the next one[/url], we have fixed the points O and U. Also, the point E describes its path in the XY plane.[br][list][*]O = (0, 0, 0)[/*][*]U = (0, 1, 0)[/*][*]E = (E[sub]x[/sub], E[sub]y[/sub], 0)[/*][/list]In addition, we had already seen in the [url=https://www.geogebra.org/m/h3gbmymu#material/napabhky]Four-bar linkage[/url] the difficulty of handling involved in transmitting the movement from one vertex to the others. So, now, just as we did in the [url=https://www.geogebra.org/m/h3gbmymu#material/nv6vacrh]Semi-rigid cube[/url], point A can move freely on a unit sphere centered at O, independently of point E.[br][br]We will determine (except isomer) the positions of B, D and F from those of E, A and J. Note that E has 1 degree of freedom, A has 2 and J has 3, which adds up to a total of 6. (It is worth remembering here that in GeoGebra's 3D graphical view, a free point like J has two possible movements, one vertical and the other parallel to the XY plane, selectable by clicking the mouse on that point.)[br][br]The color code used for the points is:[br][list][*]Black: fixed points.[/*][*]Grey: points with 0 degrees of freedom.[/*][*]Blue: points with 1 degree of freedom.[/*][*]Green: points with 2 degrees of freedom.[/*][*]Red: points with 3 degrees of freedom.[br][/*][/list]All the positions of the cube (except those that imply the coincidence of two or more vertices) are thus determined by the positions of E, A and J. But this does not mean that all the positions that allow the freedoms of E, A and J are possible to achieve while preserving the unit length of the edges. For this, it is necessary that the radii of the circles circumscribed to the three triangles UEJ, UAJ and EAJ are not greater than 1. When one of these three vertices reaches a limit position (some radius equal to 1), we will not be able to continue moving it in the chosen direction as long as we do not reposition any of the other two points.[br][br]Finally, note that if we appropriately move J or A to allow E to rotate one full turn around O, and then return J and A to their initial positions, then the entire cube will have returned to its initial configuration, or an isomeric configuration.
[color=#999999][color=#999999][color=#999999]Author of the construction of GeoGebra: [color=#999999][url=https://www.geogebra.org/u/rafael]Rafael Losada[/url][/color][/color][/color][/color]

Information: Articulated cube with fixed bar (I)