question and answer thread.

This problem is a bit unclear. First, I do not understand what is meant by the distances between endpoints being constant, since I do not see this as a dynamic problem. And it cannot mean that all the distances are equal, since at most four points can have this property in 3-space. So I will assume that is superfluous information. Also, I do not understand the restriction on lengths of the vectors, since according to the problem, one of the informations given is the length of the vectors, so these are not variable or dynamic either... Again, I will assume this is superfluous information. So basically, I have transformed this problem into:

1) You are given six points in the x-y plane. In fact, they lie on the regular hexagon x=cos(n p / 3), y=sin(n p / 3) for n = [0, 5] e Z.

2) You are given six distances that represent the lengths of vectors from these points. Additionally, all points must lie in the positive-z subspace.

3) You know that the points of these vectors are coplanar.

Q) Determine the equation that these vectors must obey to satisfy these conditions.

Now, I still am confused as to whether you are given in 2) distances that are assigned to a particular collection of vertices (distance 1 to vertex 1, etc.) or just a collection of distances in which assignment must be made. First it is important to realize that there may be none, one, many, or infinite solutions to a particular set of distances, so the equation one arrives at may not be constructive. But a general brute force attack would be to enumerate all 15 groups of four vertices. The coplanarity condition can be seen as
| x1 x2 x3 x4 |
| y1 y2 y3 y4 |
| z1 z2 z3 z4 | = 0
| 1 1 1 1 |
since the determinant is proportional to the volume of the parallelipiped described by the four points (which would be zero if they are coplanar). This will give 15 equations on the points. Additionally you get the six distance equations (of the form (xp - xv)^2 + (yp - yv)^2 + (zp - zv)^2 = d^2 where the p's refer to point coordinates and the v's refer to vertex coordinates on the hexagon). This gives 21 equations for 18 coordinates, and thus is over determined (thus allowing all possibilities mentioned above).

From here I would try to simplify somewhat, but I find it difficult to proceed without knowing whether the distances are tied to a particular vertex or not...

I would consider that a sufficient answer.

But even if it's not what souldog was looking for, it would be tedious to describe exactly what it was, since he doesn't have a solution anyway.

I am pretty bad at asking these question I realize. Let me not try and phrase this without some technical jargon.

What I had in mind here was the forward static analysis of the nearly general stuart platform. This is a problem from robotics.

First some background.
Let the X-Y plane represent the ground. Then you have six actuators (A device which can generate motion in One dimension)
connected to the Ground on the vertices of a Hexagon. These actuators are the vectors in the problem. The other end of the actuators are connected to another plane (Call this the platform). The connections are fixed, only the angles the actuators (vectors) make with the ground and the platform can change as the length of the actuators change, the connections are made with universal joints, The connections of the actuators on the platform again lie on the vertices of a hexagon, but one of a much smaller size than the ground (In reality the Heaxgons are not regular, but it does not matter) These devices are used to move an object connected to the platform to simualte motion. As the length of the six actuators changes the platform can acheive any orientation and position within the range of motion. Think of this as a solid body problem. There are six degrees of freedom for a solid body moving in three space. The position of its center of mass and the orientation of a coordinate system connected to the center of mass relative to a fixed reference frame. You need six linear actuators to generate all six degrees of freedom. Devices like this are used for flight simulators and earthquake simulators (though often a different configuration is used for earthquake simulation).

The device I have described here is often called a Stuart platform (Gough actually used it first).

As in any such problem you can perform a dynamic or static analysis. I was mixing in dynamic features which do not matter.

In the real world you usually measure the lengths of the actuators and send them to a computer which generates command signals based on these lengths (The control is actually through a servo mechanism, but that is another story). There are two problems related to this for a static analysis.

1. Inverse Static analysis : Given that you know where the center of the platform is (this is the center of the hexagon determined by the connection points for the actuators) what are the length of the actuators. Here I assume you have a fixed reference frame and you know the coordinates of the center of the platform as well as the Eular angles for the plane.
This is easy.

2. Forward Static Analysis: Given the length of the six actuators, where is the center of the platform and what is its orientation?

This is the question I wanted to ask.

Let me now start by saying that I do know the solution to this. It runs along the methods Galathaea was outlining and results in solving a 21st order polynomial by numerical techniques. This method uses no restriction on the lengths of the actuators and results in many solutions which are not meaningful. Notice Galathaea you had not used in any way that the vectors were in the positive Z subspace.

So is it possible to make the above problem determined (One solution) if you further introduce that the lengths you are given satisfy 8<=Length(i)<=10 where i = 1,2,3,4,5,6.


I am sorry you find this tedious SolarFlare, but you were not asking a question.

Quote:

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Originally posted by souldog
I am sorry you find this tedious SolarFlare, but you were not asking a question.

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I'm terribly sorry. I didn't mean that at all. If I actually thought the questions were tedious, I wouldn't post in this thread, would I? I guess I was just tired when I made that post, and couldn't express myself well. What I basically meant is that I expected there might be a long argument as to what the problem specifically meant, when it didn't really matter that much about the specifics. But you actually cleared it up quite nicely.

Sorry i didn't respond earlier, but I've been under the weather the past few days (I hope none of you catch anything from me!). Anyways, looking at this I have a clearer idea how to proceed, but just wanted one more clarification. I think the important part here is found in the statement

Quote:

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Originally posted by souldog
The distance between the free ends of any two of the vectors is constant

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From the description of the problem, it seems like what is really meant here is that the distance between the free ends of any two vectors whose origins are adjacent vertices of the base hexagon are all equal. But before I proceed on something like this, I'd like to know if this is the correct interpretation.

PS: souldog, I actually enjoy working out problems like these, so don't get discouraged. It is sometimes just as difficult to ask a question correctly as it is to answer it, as the numerous errors that have popped up in the popular mathematics competitions show.

Well let me say this in a different way. The vectors connect to the platform on a hexagon and you know the coordinates of the vertices of that hexagon relative to a reference fixed to the center of the platform (note: in a dynamics problem this would be the moving reference frame, but again this is not dynamics). So you can take the distance between any two ends of the vectors on the plane determined by the platform as known constants. (dynamically the ends of the vectors on the platform form a hexagon which moves around as the lengths of the vectors change, but does not change shape. Rigid body motion)

Note: The solution of this problem is not necessary for the control of a stuart platform. Numeric algorithms can be found which work fine. Why this problem interests me is: Why is it hard? All solutions that I have seen have not taken constraints on the lengths of the actuators into account. These constraints, by the way, can also be interpreted as a limitation on the angles the vectors form with the ground and platform (i.e. the scalor product) Physically if you tell the platformto go some where (i.e. specify the six lengths of the actuators), then it goes to one place.
With the constriants the problem is determined.

To me the question is how to introduce the constraints into the problem tomake it determined. If this turns out not to be easy then to determine why.

A solution to this problem would almost certainly be publishable, by the way.

Galathaea, why don't you post another problem. Maybe something a little more reasonable

Although I am still messing around with the vector problem, I will post a new question that I though might be enjoyable for some to work on. It is kind of a turn of pace.

Some here may know that the torus may be topologically represented as a square, where opposite sides are specified as having been glued. As well, th Klein bottle may be represented as a square with one pair of opposite edges glued directly, and one pair glued with "opposite orientation", ie. if we say the square's left and right edges have this distinction, the left edge's points going up correspond with right edge's points going down.

Now, let us look at a similar 3-dimensional space problem and the topology induced by a gluing. Take a cube. Each face viewed from above is glued to the opposite face turned one-quarter clockwise (in the view). A picture is attached to illustrate, which I hope is sufficient. Call the strange manifold built up from this gluing a Q-manifold.

In normal 3-space, normal cubes will fill space and have corresponding faces and edges when 8 cubes meet at a corner. The question I have is this: how many of the Q-manifold's cubes meet at a corner in the gluing? If you wish, you can also say what type of geometry this imposes on the Q-manifold, as it follows fairly quickly, but it is not necessary. One hint on the last optional part is that of the eight homogeneous geometries in 3-space, the 5 "lesser" ones are immediately ruled out in this case, leaving the decision betwen the 3 more popular geometries: euclidian, elliptic, or hyperbolic.

2

(since nobody else seems to know - or hasnt read the question, more likely - i'll take my best unfounded guess ;))

Incorrect ;)

infinite

Nope. That's not it either. :)

It looks like there might be some guessing go on here...:p
Until now, I guess the answer 1 would have been one technically correct answer, since only one box comprises this space, but that's not what I meant. Maybe a better way to look at this problem is: corners become identified by the gluing. Start with one corner and find out how many other corners coincide with it because of the gluing. Add them up (including the original).

If a correct answer is not given by tonight, I'll post the solution and let the next question be asked.

1?

I guess I misunderstood this question. I assumed you were asking: Well the manifold is Euclidean and R^3 is the covering space and we are acting on the boundaries of the fundamental polyhedron with one of the crystalligraphic groups (translation by one and rotation by 90 degress clockwise about each axis). If I take any corner on any cube in a tessellation of the cube over R^3 then it is identified with at least one corner on every other cube. Therefore, all the cubes meet at a corner, ie an infinite number.

To the question: It seems to me that four corners are identified as a single point. If I take any corner then the the other three corners not adjacent to it are identified.

Quote:

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Originally posted by souldog
To the question: It seems to me that four corners are identified as a single point. If I take any corner then the the other three corners not adjacent to it are identified.

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Yes! That's the one! The three corners diagonally across from a chosen corner are identified with it, making the total 4 corners which meet. Similarly, three edges coincide in the gluing. Because each corner has a solid angle of p / 2, the sum is 2p < 4p and so the geometry is elliptic. It couldn't have been hyperbolic at all since the cube has only 8 corners...

Quote:

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Originally posted by souldog
I guess I misunderstood this question. I assumed you were asking: Well the manifold is Euclidean and R^3 is the covering space and we are acting on the boundaries of the fundamental polyhedron with one of the crystalligraphic groups (translation by one and rotation by 90 degress clockwise about each axis). If I take any corner on any cube in a tessellation of the cube over R^3 then it is identified with at least one corner on every other cube. Therefore, all the cubes meet at a corner, ie an infinite number.

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Aha! I see where you were coming from here, but the other corners in the tesselation aren't actually coincident in the covering space. In this approach, there are then 8 cubes adjoining a particular vertex and they are identified in pairs by their particular symmetry orientation, so again the answer is four.

By the way, that view of the manifold is why it is commonly called the quaternionic manifold (my "Q-manifold"), because its symmetry group is the quaternion group.

Okay! Your up, souldog!