Parsing Movement Descriptions: Equivalent Movements

When dealing with simple unitary movements, that is,movement about a fixed center of rotation where the moving arm does not changelength, it is generally straightforward to deduce a reasonable description ofthe movement from a series of measurements of its trajectory.  In light of that, one might ask why itis necessary to develop methods to generate equivalent movement descriptionsfor movements. However, if weconsider movements that involve several joints, then it becomes obvious thatone needs simpler descriptions than the full set of generative equations forthe movement.

Elsewhere, we examine movements of the cervical spine, wherethere may be seven joints working together to produce a movement of thehead. The movements of the headare not simple swings or spins, but we find that they can be expressed fairlyfaithfully as paired rotations and translations. The difference is that the effective joints, that is, thecenters of rotation, are usually not within the anatomical joints and they maynot even be within the anatomical limits of the cervical spine. These compoundmovements often appear to occur about centers of rotation at some distance fromthe nearest anatomical joints. Even though we know that the equivalent movement description is not atrue indication of the details of a movement, it is still useful, because itdescribes the movement that is occurring in a simple manner that is still agood indication of what is occurring at a particular cervical level.

In this chapter we will concentrate upon the theoreticalaspects of computing equivalent movement descriptions, that is, finding acenter of rotation for a segment of a complex movement.  Our argument follows on the analysis ofmovement in the last chapter, where it was shown that any rotatory movementcould be expressed as a combination of a cast and a translation, because anyratio of orientations can be expressed as a conical rotation about an axis ofrotation and conical rotations can always be expressed as a cast.  When the change in location dictated bythis cast does not match the change in orientation, introduce a translation tobring the final location into alignment with the measured value, becausetranslations change location without changing orientation.

The Description of Compound Movements

Let us start with a general movement.  The object that we are monitoringstarts with a particular location, , and orientation, , its initial placement, . These arepresumably functions of time. Given at time t, the orientation will change in the nextsmall interval of time according to a rotation quaternion, . At time t, therotation is about the axis of rotation through anangular excursion of , where is an angularvelocity. The new orientation issimply a conical rotation of the frame of reference.

The change in location is more complex in that there needsto be a center of rotation, , and then the rotation is applied to the vector from thecenter of rotation to the current location, .

Note that may be aconstant, such as occurs in a single bony linkage, but the center of rotationmay be changing with time as it is carried by other linkages, such as otherjoints. In this manner, a seriesof joints may be concatenated to give a compound movement that is unlike themovements in any of its constituent joints. There may also be a translation, , which does not change the orientation but does change thelocation. Pulling all thesesources of location and change of location together, we obtain an expression inthe form of the following equation.

When working with a set of concatenated joints, oneintroduces extension vectors, written in terms of the frame of referencevectors, to compute the new location of a center of rotation for a jointassociated with that section of the assemblage of bones and joints.  The joints may be expressed as thelocations of the objects and the next joint and its orientation can be writtenas extension vectors that are functions of the new orientation of the movingobject. We described a simplesystem of this nature in the chapter on swing and spin.  It took a fair bit of concentration tofollow the implications of the movements in that system with two simple joints;assemblages become much more difficult to understand with each additionaljoint.

A Simple Armature

Consider a simple armature with two links and twojoints. The first joint is at theproximal end of the first link (at on link 1). Thesecond joint is between the links ( at ). We areinterested in the movements of the distal end of the second link ( at ). P has a location, , and an orientation, . Theorientation of the first link is .

A simple armature withtwo links

The orientation is theorientation of the second link as modified by the movements of the two joints.Quaternions written in lower case are the corresponding uppercase quaternionwith half the angle, as is the usual convention.

The location of the distal end of the first link is thelocation of the proximal end plus the rotated extension between the ends of thelink.

The movement of the distal end of the second linkbecause of the movement in the first joint is similarly expressed.

However, there is also movement in the second joint thatmoves the distal end relative to the proximal end. We can write that expression in the same format.

If we condense the last three equations into a singleequation, then the description is definitely becoming complex.  While each of the component equationsis fairly easy to understand, the combination is far from obvious.  Given a few more links, thedescriptions become impenetrable to someone who does not know the components ofthe movement. Clearly, if we start writing the descriptions for an assemblageof several joined bones, the arithmetic soon becomes overwhelming and theresults far from obvious. Itbecomes essential to find a simpler alternative description of themovements. That is when oneconsiders cutting out the detail of the internal calculations and expressing thefinal movement in terms of an equivalent description. The detailed description computes the actual movements andwe try to reduce those movements to a simpler, formally equivalent,description. How one mightaccomplish that simplification will be considered now.

Computing the Cast Center of Rotation, Given the Movement

Suppose that we have computed a segment of movement for ananatomical object. We know itsorientation and location at a series of points along its trajectory.  Now, we want to find an equivalentcompound movement that will give the observed values within an arbitrarilysmall error. Suppose that we havethree samples placements along a segment of movement, . We candetermine the angular excursions and axes of rotation for the rotations betweenthe sample points by calculating the ratios of the orientations at thosetimes. The angular excursion from to is and the axis ofrotation is . Similarly forthe interval between and , and .

It may often happen that a plane of rotation derived from aratio of orientations is oriented so that there is no way for the initial andterminal sample points of the movement can both lie in the plane.  In that case, there must be aconcurrent translation. We willaddress the translation first, because it must be removed to make it possibleto find the center of rotation for the rotation, which will be called the nodeand symbolized by for theremainder of this chapter.

Computing the Translation

For the movement segment , we compute the plane of rotation from the ratio of theorientations.

The unit vector of the quaternion, , is in the same direction or the direction opposite thetranslation, because the translation is perpendicular to the plane of therotation.

We know the beginning and ending locations, and , respectively. We know that is perpendicularto the line from to . We know that . Therefore, thetranslation is one side of a right triangle with a hypotenuse equal to . The anglebetween and the translation vector is the angle of their ratio.

The magnitude of can be computed directly.

Consequently, the translation is simply expressed.

Now, we know the end points of the arc of the rotation inthe plane of the quaternion for the ratio of the orientations, . It is possibleto compute the node of the rotation, because we know the angular excursion, whichis the angle of the quaternion , . In fact, thesolution is a double arc, symmetrical about the line segment between and . These arcswere computed in the last chapter. For small angular excursions, they are nearly circular.

The problem is now selecting a particular point on thearc. We can address this selectionin a number of ways, depending on the relationship between the movement segmentrotations. If the two movementsegments have the same vector of rotation, that is, they lie in the same planeof rotation, then we can plot both arcs, suitably displaced relative to eachother at the locations of the sample points and their intersection will be anode that gives the appropriate angular excursion for each movement.  This solution is considered in detailin the next section.

Nodes for Movement Segments That Are Coplanar

Since all simple casts, coils and screws can be expressed asa cast plus a translation, many movement segments will be coplanar.  That means that the two segments havethe vector of their quaternion in the same direction. If , then we can use the derivations of the curves for nodes,from the last chapter, to find the common node for the two excursions.  As argued there, for small offsetsbetween the movement segments, the curves must intersect because they arespatially offset relative to each other. As the angular excursions become smaller, the nodal curves become morecircular and the offsets become smaller. The segments are more apt to be similar in direction and curvature, sothe intersections are more apt to be on a line orthogonal to the movement arc.  Let us now consider the details ofcomputing the common node of two movement segments.

We start with the triangle that has the node, , the initial locus of the movement, , and the final locus of the movement, , as its apices. The angle at the node will be called and the two baseangles will be and . We note thatthis movement triangle can be divided into two right triangles with the rightangles on the baseline between the initial and final loci.  The base line is divided into asegment, , which is a side of the triangle with as one of itsangles, and a segment, , which is the side of the triangle with as one of itsangles. The perpendicular to thebase through the node is called y.

From the ratio of the initial and final orientations, weknow the value of , because it is the angle of the ratio of theorientations. And, we know theinitial and final loci of the movement segment, and . All of theother variables are subject to variation so there is not a unique solution forthe node. Instead, there is acontinuous curve that contains all the possible values of , given .

We described these curves in the previous chapter.  We are able to compute a uniquesolution if we seek a common node that satisfies the constraints of twomovement segments. Two possiblearrangements are illustrated in the following figure.

We will deal with the disjoint movement segments, becausethe conjoint case may be obtained by setting . Because wewere interested only in the shapes of the arcs in the last chapter, we assumeda constant length movement segment and computed the nodes for a variety ofvalues of . Now, we needto consider expressions for the node when the location and orientation of themovement segment is pertinent. Consequently, we need to multiply the arcs by appropriate scaling androtation factors.

To simplify the language a little, let us call a curve ofpossible nodal loci a nodal loop. Such a loop is associated with a movement segment with terminal loci and that lies in aplane that has the normal vector . The linesegment between the endpoints is the difference between them, , and the direction of that line is

The perpendicular to the movement segment is the direction rotated througha right angle about the normal to the plane. Let be thequaternion that turns into .

The location of the node is the sum of a position, , along a line coincident with the movement segment and acorresponding displacement, , perpendicular to that line in the plane of themovement. It also depends on thelength of the segment and the locationof the movement segment in space, which we take to be its initial locus, . Thecoordinates for the node when the movement segment is of unit length, , are functions of and .

If we take account of the orientation and the length of themovement segment, then the components can be written as follows.

If we return tothe disjoint movement segments that are illustrated above, then we can writetwo expressions for the common node.

This equation looks complex, but it reduces to an expressionin terms of the sine and cosine of theta. That equation can be rendered into three equations, one for the i components, one of the j components, and one for the k components. There are two unknowns and three equations, so it is largely a matter ofalgebra to solve for and , and from there to a value for . That value canbe substituted in the equation for to find itslocation.

The Isosceles Solution

A second solution to finding the node of a movement is toassume an isosceles rotation and take the node that is on a line perpendicularto the difference between the loci, though the midpoint of that line.  That solution has been developedelsewhere without having to calculate the arcs of the nodes.  The solution does not require that thetwo arcs are in the same plane and it can be computed for a single movementsegment, which makes it versatile. However, the isosceles solution may not be realistic in some situations.  The isosceles solution will give atraveling center of rotation if the movement is not a unitary cast.

The isosceles solutions are illustrated by the lines to and , in the last figure. Generally the isosceles solutions will bracket the non-isoscelessolution for coplanar movement segments.

In practice, in biological systems, the movement of the nodemay be comparatively small, especially if one breaks the movement into shortsegments. Consequently, theisosceles solution is often a useful and simple solution. However, thetrajectory of the isosceles nodes may vary with the length of the movementsegments. This phenomenon isexplored in some detail in another chapter, which deals with the movements ofthe lower cervical spine.

Nodes for Movement Segments That Are Not Coplanar

Often, perhaps most of the time in real anatomical systems,two movement segments do not have a common axis of rotation and therefore donot lie in the same plane. We willnow consider the situation where the axes of rotation are different for twosegments of movement. Inparticular, we will consider two contiguous segments of movement although it isnot necessary to the argument that they be contiguous.

We will also assume that the movement segment is convex inthe sense that a line connecting to lies entirely to one side of the movement segment in theplane of rotation , and that the movement segment intersects its plane only atthe ends of the movement, unless it is entirely in the plane of therotation. We will also assume thatthis is true of any components of the movement segment.  It may be noted that we are agnosticfor any unmeasured placements, so the movement segment could be very erraticbetween measured placements. However, if we choose our measurements reasonably close together and themovement system is well behaved, then these exceptional cases are apt to beunusual and unlikely to occur.

The plane of the rotation is the vector ofthe quaternion , . We will nowuse a result that has not been derived here, but which is derived elsewhere[Quaternion Numbers]. The ratio oftwo planes is their intersection. We will actually turn this observation around and use the observationthat the intersection of two planes is their ratio. So the line that lies in both planes is their intersection.

If the node for the rotations and must lie in therotation planes for both and , then the node must lie on the intersection of those twoplanes, . So far we donot know where on the node islocated. That is the centralproblem that must be solved. So,we need to gather the facts that we have and try to compute the value of thenodes, and .

At this point, a sketch of the situation is useful.  We can draw the segment of movement,the sample points and the intersection of the planes of rotation.  We also know the angular excursions forthe two contiguous sections of the movement segment. The rest must be constructed.

A general cast

To start with, we assume that we have two movement segmentsthat have rotations in different planes. While it is not strictly necessary, we will assume that the segments arecontiguous. So, the movement isfrom an initial placement of to a placementof and then to aplacement of . For bothmovement segments, we compute the rotation quaternion for the change inorientation.

We are interested in the direction of the plane of rotation and the angularexcursion .

The computed quaternions will be unit quaternionsbecause orientations are constructed of unit vectors.

At this point it is necessary to compute the translationcomponent and the terminal loci of the movement segments.  The means for doing so were describedabove. We know the threeplacements , the initial and final loci of the two movement segments , the direction of the rotation planes , and the angular excursions for each rotation . What we needto know is the locations of the nodes for each rotation . The criterionthat will shape our solution is that if the first translation is zero, then thenodes will occupy both planes of rotation. If the curvature of the movement segment is the same, thenodes will be coincident within that line.

We have two planes, expressed by their normal vectors, and . Theintersection of the two planes is given by their ratio.

If we rotate the intersection vector about the axis of thefirst quaternion, , through an angular excursion of , then we will have a unit vector that points from the nodeto the starting location.

We do not know the location of the node, as yet.  That is the goal of this exercise.  But, we do know that if we continue inthe direction of from the node , then we will reach the initial location, .

We also know that if we move in the direction of , we will intersect the other end of the movement segment, .

If we knew the value of and , then we could compute . With somesimple algebraic juggling of the last two equations we can eliminate and have anequation in and .

If you sketch this expression it is apparent that it issimply a statement that if you pass successively around the three sides oftriangle, you return to the starting place.

One’s first impression tends to be that though this isclearly a true statement, it is a single equation with two unknowns, thereforenot adequate for finding those unknowns. However, it is actually three equations with two unknowns.  If it is written in coordinate format,then there are three equations, one for each basis vector.  All the displacements in each directionmust total to zero. The solutionfalls out of some lengthy, but straightforward, algebra.

Once we know and , we know the location of the node, . Once we know , we know the cast that rotates into and thetranslation that brings the endpoint to . Consequently,the movement segment has beenexpressed as the sum of a cast and a translation.

Now consider the second movement segment.  The analysis is basically the same, butwith a few new features.

We know the starting and finishing locations, and . We know theplane of the rotation dictated by the ratio of the orientations.

We know the direction of the node from , because it is the same as for the first movement segment, . We obtain theunit vector that points in the direction of the final orientation, , by rotating the unit vector or the intersection in thepositive direction about the normal to the plane of rotation for the secondmovement segment, , through the angular excursion .

Once again, we can solve for and by expressingthe last equation in coordinate format and solving the three simultaneousequations.

Note that, in general, the two nodes are notcoincident. They do both lie on aline in the direction of the intersection of the two planes.  If there is no translation component,then they lie on the same line.

In the Limit

As we let the movement segments become smaller and smallerparts of the movement, they will approach the infinitesimal.  It is of interest to consider whathappens to the nodes of successive movement segments as the movement segmentsapproach that limit.

If the movement occurs in a plane, then the common nodes forpairs of segments approach a smooth trajectory, perhaps a fixed point, in theplane of the movement. Assumingthat the nodes do shift along such a trajectory, we expect that the trajectoryof the nodes would be smooth as long as the movement is smooth.

The isosceles solution is essentially the radius ofcurvature for the movement segment. Consequently, the nodes will move along a trajectory as the curve flexesand extends. A flatter segmentwill have a greater radius of curvature and a more curved segment will have ashorter radius. The curve tracedby these centers of rotation is called the evolute if the trajectory lies in aplane. The result for movements inthree dimensions is much like the evolutus considered in another chapter [OnEvolutes and Frames of Curves]. Physically, the nodes are much like the center for a weight whirling onthe end of a string and the node is the momentary location of the fixationpoint of the string. As statedabove, the isosceles solution is not in the plane of the curve unless the curveis confined to a plane. That isone of its advantages, in that it works equally well for all movements.

If the movement is not confined to a plane, then theintersections between successive movement planes rotates with the change inlocus, so that it sweeps out a curvilinear surface. As the movement segments become shorter, the successivenodes move closer together, because the curvature of adjacent segments isnearly the same. In the limit, thenodes converge on a common node that shifts as the intersection moves.  Once again, the node of the movementsfollows a smooth trajectory, but in three dimensions.

In general, the last two solutions do not yield the sametrajectory. If the movement is anon-unitary cast, then the rays to the isosceles nodes will be perpendicular tothe curve of the movement segment and the rays to the non-coplanar solutionwill be at obtuse angles to the curve.

Computation Examples

Confluent Movement Surfaces

Next, consider a computational example of a movement that isnot a cast, although each segment may be expressed as a cast withouttranslation. It is a conicalrotation in which the rotation quaternion is a function of the angularexcursion. It is also an exampleof what will be called a confluent movement surface. That term will be taken to mean that the rays from thecenter of rotation node to the loci sweeps out a smooth surface and theorientation changes in the same manner as location.

The movement starts at 0 and progresses to 18, where eachsegment is a 10° excursion. Duringthe course of the movement the length of the armature increases from 1.0 to 1.5units length and the armature sweeps down and out and then back up and in, sothat it ends pointing in the opposite direction.

The actual parameters used for this illustration were astarting locus for the distal end of the armature of and anorientation aligned with the universal coordinates, . The rotationvector is a function of the angular excursion.

The locus is rotated by that quaternion and lengthenedas follows.

The orientation is rotated by the same quaternion,consequently, there will be no translation.

This is a rather simple arrangement, but it will serveto illustrate most of the necessary points.

In this numerical example, one can easily see that theangular excursions of the movement segments are the ratios of the orientationsand, because the loci are generated by the same rotation quaternion, there willbe a convergence of the rays upon a common center of rotation node.  The intersections between successiverotation planes are clearly the line that point to the common node.

What is not visually apparent is that the isosceles solutionwill give a different result. Weknow that to be the case because the rays become longer as we progress for locus0 to locus 18. Therefore, the baseangle on the lower numbered side must be slightly larger than the angle on thehigher numbered side. Since theapical angle is 10°, by design, the two base angles must be 85° in theisosceles solution. When we takethe ratios of the rays to the baseline, we find that the values are 86.1° and83.9°.

Consequently, the isosceles node will tend to lie on theside towards the higher numbered locus, relative to the common node, andslightly further away from the baseline than the common node.  The net effect will be that theisosceles nodes will follow a trajectory that circles the common node in thedirection opposite to the trajectory of the loci.

As stated at the outset, this movement is an example of aconfluent movement surface. Themovement is such that the same rotation occurs for location andorientation. All movement isattributable to that rotation. Note that the movement is not a simple conical rotation in that therotation quaternion is a function of the angular excursion, but it ismomentarily a conical rotation at each point on its excursion.  We would like to reduce all movementsto such a confluent surface plus a translation vector that may also change withtime or angular excursion. Thenext section deals with that process.

Reducing a Movement With Translation to a Confluent Surface and aTranslation Vector

If an anatomical object moves through space, it mustnecessarily experience the same rotation for both location and orientation.  However, location can also be changedby translation, without changing orientation, therefore the change in locationmay not correspond to the change in orientation. In the above example, there was no translation, thereforethere was correspondence between location and orientation and the movementcould be expressed as a confluent movement surface. Note that the frame of reference for orientation has aconstant relationship to the rays from the common node.  If a variable translation is added thenthe movement is still smooth, but there is no frame of reference that will stayaligned with the rays from the center of rotation to the loci along themovement’s trajectory. Still it isoften of interest to separate such a movement into a rotation component and atranslation component. We sketchedhow that might be done in the analysis given above, but now it is of interestto return to the process and examine it is more detail.  The basic movement description has botha rotation and a translation and it can be written in the following form.

For instance, we might add a rotating translation to themovement illustrated above so that the movement leads and lags and rises aboveand falls below that trajectory. The orientation is not changed, but the location is altered as follows.

Movement from to may viewed asthe sum of three components: a rotation, , an orthogonal translation, , and a translation in the plane of the rotation, .

Our mission is to extract R and T from the numerical description of the movement.

Although the translation component is added as a unit, itneeds to be extracted in two steps. The first step is to compute the translation perpendicular to therotation plane, , that is computed from the ratio of the orientations.  That translation is subtracted from themovement to obtain an intermediary movement curve that is the projection of themovement curve into the planes of rotation based on orientation.  To differentiate the three sets oflocations on the three trajectories different symbols will be used for eachset. In the original movement, then’th location will be , in the intermediary trajectory, it will be , and in the rotation component, it will be .

The angular changes in location, , still do not match the angular excursions based onorientation, because we still have the translation components in the plane ofrotation, . We can computethose translations by piecemeal calculation of the rotating armature forsuccessive pairs of movement segments. The translation in the plane of the rotation is the difference betweenthe intermediary location for the armature and the computed values on theassumption that the rotation has certain characteristics, . If we subtract that translation, the result will be aconfluent surface that is the rotation component of the movement.  Because of the appearance of the formsgenerated in computing the in-plane translations, the intermediary surface willbe called the patchwork surface.

Let us now consider how one actually computes theseentities. The first step is tostart with two successive placements along the movement trajectory, and . From theorientations we can compute and angular excursion and a plane ofrotation .

Given the locations and , one can construct the right triangle that has theirdifference as its hypotenuse and one side in the direction of the normal to therotation plane as one of its sides. The angle between the hypotenuse and that side is the angle of theirratio.

The length of the translation side is and the lengthof the rotation excursion side is given by the following equation.

The first term on the right side is the length of thehypotenuse and the term in the brackets is the unit vector in the direction ofthe side, so the equation is essentially the same as the previous expression.

The excursion side extends from to , which is the movement segment when the orthogonaltranslation is subtracted. If weproceed in the same way for all the movement segments, then the result is asaw-tooth trajectory, which is the basis for the next step.  The points along that trajectory are ina surface for which the movement segments are in the planes of rotation for theorientations. We can now startcomputing the rotation surface.

If we remove the orthogonal translations, the movementbecomes a continuous trajectory of linear segments that join at theintermediary locations, . The nextfigure illustrates these as the green nodes. We know that the intermediary trajectory is the combinationof a rotation and translation in the plane of the movement segment’s rotation.

The genesis of a creased confluent surface for arotation with translations in the plane of rotation

The rotation quaternion for the movement segment, , may be a non-unitary quaternion or a unitaryquaternion. If it is a non-unitaryquaternion, then the length of the moving armature will change as the armaturemoves. Here, we will use the termarmature to indicate the link between the center of rotation and the rotatinglocus and symbolize it as .

There are two basic options in how one apportions thedeviation from a unitary cast. Thefirst option is to assume that any movement in the direction of the armature ispart of the cast and any movement perpendicular to the armature is translationin the plane. This has the nicefeature that translation is always in a plane perpendicular to the armature,but at the cost of a non-unitary cast. The second option is to assume that the cast is unitary and any movementin the plane of the rotation that deviates from that value is translation.  Under the second option the rotation issimple, but translations may be in any direction.

There is obviously a third option, which is to combine thefirst two options and apportion some of the movement in the plane of rotationto a non-unitary cast and the rest to translation. However, this option depends on having some way to decidehow the apportionment is to be handled. The choice is largely a matter or interpretation and there is no apriori basis for choosing one over theother unless one has additional information. If one plots the deviations from a unitary cast, then onemay see that there is a linear or smoothly periodic pattern to the deviationsand it is natural to choose the interpretation that best incorporates thatregularity. One may also havereasons arising from the anatomy and physics of the system for choosing oneoption over another.

Setting that aside, let us consider how one might compute aset of loci for the rotation that generate a creased confluent surface thatrepresents a rotation. To start,we need to know one armature that can be used as the basis for the calculation.  Let us call that armature . The armaturestarts in a center of rotation, , and ends in a locus, .

These values constitute a boundary condition on thecalculation of the surface.

We know the angular excursion of the segment of the surfacethat is associated with movement between two loci. It is the angle of the quaternion of the ratio of theorientations in the placements that bracket the movement segment, . The plane ofthe rotation is defined by the vector of that same quaternion, . If we writethat rotation quaternion for the segment bounded by the creases and as , then we can write down a general formula for generating thenext locus in the rotation.

If , then there must be a translation .

The in-plane translation may be resolved into a componentthat lies in the direction of the armature, , and a component that lies perpendicular to it, . These areillustrated for the last segment in the above figure.

We can determine the values of and from the following relationships, where the bar over a vectorsignifies that it is the unit vector in that direction and is the tensor ofthe quaternion .