[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][br][br]In the following construction we can choose different elements.[br][br][b]1 point[/b][br]The position of a free point in the plane is determined by 2 coordinates, since the plane has two dimensions. We say that the point has 2 [b][i]degrees of freedom[/i][/b], since each coordinate is free to take any value.[br][br]Now, in the absence of a visible reference system (as happens in the construction), since we do not know these coordinates, point O is shown to be the same as any other point on the plane. There is nothing, except its position (which we assume to be unknown), that differentiates O from any other point.[br][br][b]2 points & 1 bar[br][/b]If we place 2 points, between them they will have 4 degrees of freedom (their 4 coordinates). However, if we impose the constraint that the distance between them be fixed, the total number of degrees of freedom is reduced to 3. This is best seen if we visualize this constraint as a segment (or non-deformable bar) with the length fixed by that distance. The position of that bar in the plane is determined by the position of O and the angle it forms with the horizontal: a total of 3 parameters or degrees of freedom.[br][br]In fact, due to the construction process, while O is a free point that can move throughout the plane (dimension 2), the point U, once a position has been fixed for O, has to remain in a circle (dimension 1) with center O and fixed radius (the given distance).[br][br]It can be shown that all (non-point) configurations of points and bars have, in the plane, at least 3 degrees of freedom. Therefore, we can consider only the degrees of freedom "that add" to these three. They are called [i][b]internal degrees of freedom[/b][/i]. The two points and one bar configuration then has 0 internal degrees of freedom.[br][br]Note that if the bar were material (for example, steel) we would not hesitate to qualify it as rigid, in the sense that it can only take one shape, it cannot be continuously deformed until it reaches a different shape. In the following examples, we will verify that this [b][i]local rigidity[/i][/b] is a consequence of the number of internal degrees of freedom being 0.[br][br]Furthermore, in this case, all the bars of the same length are [b][i]congruent[/i][/b], that is, they are identical except for isometry, they have the same size regardless of their position or orientation. In such a case, we say that the figure, in addition to being locally rigid, is also [i][b]globally rigid[/b][/i].[br][br]In order to better visualize the internal degrees of freedom, [b]from now on we will assume that points O and U are fixed[/b], as well as the distance between them. In this way, when considering the configuration {O, U, OU} with 0 degrees of freedom instead of 3, the number of degrees of freedom will coincide with the number of internal degrees of freedom.[br][br][b]3 points & 2 bars[/b][br]In this case, we clearly see that by moving E, the configuration of points and bars can take on different shapes. It is a [i][b]flexible [/b][/i]configuration, we can find various non-congruent positions. Now there is 1 degree of internal freedom (the dimension of the circle that E runs) that gives the configuration that flexibility.[br][br][b]3 points & 3 bars[/b][br]We know that all triangles with the same measures are congruent. So, as long as it is possible to obtain a triangle (one bar cannot measure more than the sum of the other two), it will be rigid. Note that although there are 2 possible positions (called [i][b]isomers[/b][/i]) for E, the two corresponding triangles (OUE and OUE') are, in this case, congruent. Indeed, by construction, point E has no freedom, so the total number of internal degrees of freedom is 0. This configuration is, then, locally and globally rigid.[br][br][b]4 points & 3 bars[/b] (a)[br]This case is similar to that of 3 points and 2 bars. By adding one more point F and bar to the OUE configuration, we are adding 1 more degree of freedom. Therefore, this configuration is flexible, with 2 internal degrees of freedom. The dimensions of the circles centered at O and E show, respectively, the degree of freedom of points E and F.[br][br][b]4 points & 3 bars[/b] (b)[br]Although this configuration is different from the previous one, we can see that it also has 2 degrees of internal freedom.[br][br][b]4 points & 4 bars[/b] (a)[br]Although this configuration is closed, like the triangular one, we quickly verified that it is not rigid. Just move E to get different non-congruent shapes. Indeed, point F lacks freedom, so the total number of internal degrees of freedom is 1 (the dimension of the circle traveled by E).[br][br]Now, what happens if we fix, in addition to O and U, the position of E? We obtain two possible forms that, in general ([url=https://www.geogebra.org/m/h3gbmymu#material/fxv3jbg4]see the following activity[/url] for the analysis of a particular case), are not congruent: UOEF and UOEF', each with 0 internal degrees of freedom. However, in the plane we cannot go from one to the other in a continuous way. A "force" jump would be needed to convert F to F'. That is why we say that, with O, U, and E fixed, the UOEF configuration is locally rigid even though it is not globally.[br][br][b]4 points & 4 bars[/b] (b)[br]Although this configuration is different from the previous one, we can see that it also has 1 degree of internal freedom. Furthermore, if we fix F, again the FUOE configuration is locally rigid, although it is not globally (since FUOE and FUOE' are not, in general, congruent figures).

[color=#999999]Author of the construction of GeoGebra: [color=#999999][url=https://www.geogebra.org/u/rafael]Rafael Losada[/url][/color][/color]

[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][br][br]Thanks to the use of scripts, we can simulate the interactive behavior of the mechanisms, overcoming the limitations imposed by the dependency structure of Dynamic Geometry. In this example, the five points are free, but maintain the constant length of the bars at all times.

The scripts used are detailed below.[br][br]When moving A:[br] SetValue(B, Intersect(Ray(A,B), Circle(A,1)))[br] SetValue(C, Intersect(Ray(B,C), Circle(B,1)))[br] SetValue(D, Intersect(Ray(C,D), Circle(C,1)))[br] SetValue(E, Intersect(Ray(D,E), Circle(D,1)))[br][br]When moving B:[br] SetValue(A, Intersect(Ray(B,A), Circle(B,1)))[br] SetValue(C, Intersect(Ray(B,C), Circle(B,1)))[br] SetValue(D, Intersect(Ray(C,D), Circle(C,1)))[br] SetValue(E, Intersect(Ray(D,E), Circle(D,1)))[br][br]When moving C:[br] SetValue(B, Intersect(Ray(C,B), Circle(C,1)))[br] SetValue(D, Intersect(Ray(C,D), Circle(C,1)))[br] SetValue(A, Intersect(Ray(B,A), Circle(B,1)))[br] SetValue(E, Intersect(Ray(D,E), Circle(D,1)))[br][br]When moving D:[br] SetValue(E, Intersect(Ray(D,E), Circle(D,1)))[br] SetValue(C, Intersect(Ray(D,C), Circle(D,1)))[br] SetValue(B, Intersect(Ray(C,B), Circle(C,1)))[br] SetValue(A, Intersect(Ray(B,A), Circle(B,1)))[br][br]When moving E:[br] SetValue(D, Intersect(Ray(E,D), Circle(E,1)))[br] SetValue(C, Intersect(Ray(D,C), Circle(D,1)))[br] SetValue(B, Intersect(Ray(C,B), Circle(C,1)))[br] SetValue(A, Intersect(Ray(B,A), Circle(B,1)))

[color=#999999]Author of the construction of GeoGebra: [color=#999999][url=https://www.geogebra.org/u/rafael]Rafael Losada[/url][/color][/color]

[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][br][br]The move from planar to 3D linkages is simple. Here we see the 3D version of the chain of 5 points and 4 bars that we had already seen in its planar version. We can observe that it has been enough to replace in the scripts the circles in charge of adjusting the distances by spheres of radius unit.

The scripts used are detailed below.[br][br]When moving A:[br] SetValue(B, Intersect(Ray(A,B), Sphere(A,1)))[br] SetValue(C, Intersect(Ray(B,C), Sphere(B,1)))[br] SetValue(D, Intersect(Ray(C,D), Sphere(C,1)))[br] SetValue(E, Intersect(Ray(D,E), Sphere(D,1)))[br][br]When moving B:[br] SetValue(A, Intersect(Ray(B,A), Sphere(B,1)))[br] SetValue(C, Intersect(Ray(B,C), Sphere(B,1)))[br] SetValue(D, Intersect(Ray(C,D), Sphere(C,1)))[br] SetValue(E, Intersect(Ray(D,E), Sphere(D,1)))[br][br]When moving C:[br] SetValue(B, Intersect(Ray(C,B), Sphere(C,1)))[br] SetValue(D, Intersect(Ray(C,D), Sphere(C,1)))[br] SetValue(A, Intersect(Ray(B,A), Sphere(B,1)))[br] SetValue(E, Intersect(Ray(D,E), Sphere(D,1)))[br][br]When moving D:[br] SetValue(E, Intersect(Ray(D,E), Esfera(D,1)))[br] SetValue(C, Intersect(Ray(D,C), Esfera(D,1)))[br] SetValue(B, Intersect(Ray(C,B), Esfera(C,1)))[br] SetValue(A, Intersect(Ray(B,A), Esfera(B,1)))[br][br]When moving E:[br] SetValue(D, Intersect(Ray(E,D), Sphere(E,1)))[br] SetValue(C, Intersect(Ray(D,C), Sphere(D,1)))[br] SetValue(B, Intersect(Ray(C,B), Sphere(C,1)))[br] SetValue(A, Intersect(Ray(B,A), Sphere(B,1)))

[color=#999999]Author of the construction of GeoGebra: [color=#999999][url=https://www.geogebra.org/u/rafael]Rafael Losada[/url][/color][/color]

[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][br][br]This construction shows an articulated cube, that is, twelve bars of unit length joined to form a cube. This cube is flexible, since the angle formed by each pair of concurrent bars at a vertex does not have to be right.[br][br]This model seems easy to build. However, it requires a strong and subtle combination of [b]Geo[/b]metry and Al[b]gebra[/b].[br][list][*]Note: this construct simulates 3D virtual space, using projections, instead of directly using GeoGebra's 3D Graphical View, because this view only appeared years after this work was published [[url=https://www.geogebra.org/m/h3gbmymu#material/dcmh7amw]1[/url]].[/*][/list]

[color=#999999]Author of the construction of GeoGebra: [color=#999999][url=https://www.geogebra.org/u/rafael]Rafael Losada[/url][/color][/color]

Note: Access to all web links has been verified on March 15, 2022.[br][br][1] Arranz, J. M., Losada, R., Mora, J. A., Recio, T., & Sada, M. (2011). Modeling the cube using GeoGebra. In L. Bu & R. Schoen (Eds.), [url=https://link.springer.com/book/10.1007/978-94-6091-618-2]Model-Centered Learning: Pathways to mathematical understanding using GeoGebra[/url] (pp. 119-131). Rotterdam: Sense Publishers.[br]