Quaternion Analysis of Momenta and Forces Acting Upon a RigidBody

We will start by considering the movements of a massparticle moving with respect to a central location and how systems of forcesmay be recast in a number of alternative formulations, leading up to a wrench,which is a special instance of a quaternion.

The basic arrangement is a reference point () and a moving mass (), which is displaced from the reference point by a spatialinterval (). Any locationwill do for a reference point, but there are some locations that make sensebecause they simplify the description or they represent anatomically relevantpoints, which may be on the axis of rotation of a joint or some othersignificant location. Since we areinterested in anatomical movement, we are interested in movements ofarticulated rigid bodies. Thatmeans that there is a physical connection between the reference point and themoving mass, but the distance between the two need not be constant, especiallyif there is an intervening joint. We are interested in situations when the direction of changes withtime, that is, where there is a rotation component.

The ratio of the direction of the velocity to the direction of the radialvector

An object with a mass of lies at a pointthat is at a location relative to areference point, , and it is moving with a velocity . We start bycomputing the unit vector and magnitude of each parameter.

We may compute the ratio of the direction of the velocity tothe direction of the radius, which is . It has avector and an angle .

The directions and define aplane. The vector of , , is a perpendicular to that plane and is the angularexcursion that turns into .

It is important for understanding what follows todifferentiate between quaternions, which are written bolded and in italics,vectors, which are just bolded, and scalars, which are neither bolded nor initalics. So, is the vector ofthe quaternion and is its angle,which is a scalar. Of course,scalars and vectors are special instances of quaternions, but since theycombine in different ways, it is convenient to differentiate them.  In addition, the practice ofsymbolically differentiating between the types of entities forces a morecareful consideration of what is being said. Being forced to consider what formthe elements of an expression take prevents many potential logical errors.

Ratio of the velocity to the radial vector from the origin

Note that the ratio quaternion may also be the unitquaternion of the ratio of the velocity to the radial displacement, , which may be the simpler, more direct, way of calculatingit.

The tangential and centrifugal components of the velocity

From the unit vector, , we can construct two rotation quaternions, one with anangle of 90° ( radians) and one with an angle of 0° (0 radians).

The rotation, , gives the tangential component of the velocity and therotation, , gives the centrifugal component. From an examination of the above illustration, one caneasily write down the expressions for the two velocities.

Linear momenta

If the mass of the object is , then the tangential momentum of the object relative to thecenter of rotation is the mass times the tangential velocity and thecentrifugal momentum is the mass times the centrifugal velocity.  The linear momentum is the mass timesthe velocity. Note that the linearmomentum is the sum of the tangential and centrifugal momenta.

Although the linear momentum is the product of a quaternionand a vector, the result is a vector, because the vector of the quaternion isperpendicular to the rotating vector. Consequently, the linear momentum is a non-orientable vector if theradial vector is a non-orientable vector. The radial vector is the difference between two locations, therefore isnot intrinsically orientable. Linear momentum has the same direction as its velocity and it isproportional to the mass and speed of the moving object.  We will find that angular momentum is aquaternion vector that does have an intrinsic orientation.

A simple example

This leads to a counter-intuitive, but observationallysubstantiated situation. Assumethat the mass is rotating on a circular trajectory about a reference point.

Linear momentum is an expression of the effort that it wouldtake to stop the movement, its inertia. Inertia is proportional to the amount of material present and to thespeed at which it is moving, therefore to momentum. The tangential momentum is how much of that momentum isdirected along a trajectory that is perpendicular to the radial vector from thepoint of reference to the moving mass. The centrifugal momentum is the effort that would be required to keepthe mass from moving radially. Ifwe were to instantaneously break the physical connection between the mass andthe reference point, it would move away from the reference point with a momentumequal to the centrifugal momentum. Note that the total linear momentum is a conserved quantity in systemsthat are not subject to external forces, but, neither of the component momentaare conserved and their values are functions of the location of the referencepoint.

There are circumstances in which the component linearmomenta may be informative. If thereference point is on the axis of rotation for a bone, then the centrifugalmomentum is an expression of how much the rotation tends to distract or compressthe joint and the circumferential momentum is an expression of the energy ofthe rotation. Time differentialsof linear momenta are forces.

This implies Newton’s first law of mechanics.  If is zero, thatis, there is no external force, then, there can be no change in momentum,meaning that the object must continue in the same direction at the samespeed. That includes the situationwhen the object is not moving ().

Let us consider a system in which a mass is rotating about afixed reference point at a constant velocity. Then the position of the point can be written as a functionof time and its velocity readily computed, by taking the derivative of itslocation.

It follows that rate of change of the velocity is directedin the opposite direction to the radial vector or the tether.

In a frictionless system, the mass will continue to spinindefinitely, unless the tether is cut, in which case, it will travel away fromthe central fixation, while continuing in the direction that it had at theinstant of the disconnection. Inthis case, the linear momentum, which is equal to the tangential momentum,changes periodically. There is noexternal application of force, but the momentum is periodic.  There is a periodic internalapplication of force that moves the circling mass out of a straight trajectory.

Angular momentum

The angular momentum is the vector interval from thereference point to the point mass times the tangential momentum.

Since the angular momentum is a product of vectors, it is aquaternion. However, since thetangential velocity is by definition perpendicular to the radial vector, thequaternion is a vector, a quaternion vector, which means that it isorientable. It not only has adirection and magnitude, but also a sense, that is, a direction of rotationabout its axis. Angular momentum, , is a quaternion vector, times times the mass of the moving particle, where and may bequaternion vectors or standard non-orientable vectors. It is perpendicular tothe plane defined by the radial and velocity vectors and turns to . It isproportional to the mass of the particle, the distance from the origin to themass particle, and the velocity of the particle.

Angular momentum versus linear momentum and their conservation

Angular momentum is a different type of entity than a linearmomentum. We can see that in thefact that a non-orientable vector may represent linear momentum and angularmomentum requires an orientable vector. Linear momentum is an expression of the inertia of an object.  It is the effort required to bring amoving mass to rest in the assumed frame of reference.  It expresses the amount of effortrequired and the direction in which it must be applied. The change in linearmomentum that an object experiences is the effort that must be done to bringabout the change. Angular momentumalso embodies an effort, but it is the effort to rotate a body.  It is in essence the law oflevers. The longer the lever arm,the greater the mass, or the greater the velocity, the more effort it takes tostop the rotation. When we uselevers, we try to match two lever arms so that the impressed force is such thattheir angular momenta are equal.

The angular momentum vector is perpendicular to the plane inwhich the movement is occurring and its magnitude is proportional to the effortto the effort needed to stop the rotation. It also specifies the direction in which the effort must beexerted by its direction. It is avector that has direction, magnitude and sense. That is to say it is orientable. Linear momentum is expressed in kilogram-meters per secondand angular momentum is expressed in kilogram-meters squared per second.

A reason that we care about both linear and angular momentumis that they are both conserved in a closed system. If there are no external forces, then the velocities may beshifted between the components of the system, but the sums of both types ofmomentum over all of the components must remain the same magnitude and in thesame direction. If the connectionbetween the circling mass and the center of rotation were cut instantaneously,the mass would continue indefinitely in the direction that it is going themoment of the cut. That wouldsatisfy both conservation laws because the velocity remains the same and theperpendicular to the velocity is constant, therefore the product of the radialvector and the velocity will be constant.

The two conservation laws set constraints upon the possibleoutcomes of events in closed systems. They also allow us to shift our point of view in useful ways that oftenclarify a situation or allow us to express the description of an event indifferent ways that are equivalent, but often give insight into the nature ofthe event. We will spend most ofthe remainder of this essay considering such transformations.

Angular momentum versus linear momentum and their conservation

Angular momentum is a different type of entity than a linearmomentum. We can see that in thefact that a non-orientable vector may represent linear momentum and angularmomentum requires an orientable vector. Linear momentum is an expression of the inertia of an object.  It is the effort required to bring amoving mass to rest in the assumed frame of reference.  It expresses the amount of effortrequired and the direction in which it must be applied. The change in linearmomentum that an object experiences is the effort that must be done to bringabout the change. Angular momentumalso embodies an effort, but it is the effort to rotate a body.  It is in essence the law oflevers. The longer the lever arm,the greater the mass, or the greater the velocity, the more effort it takes tostop the rotation. When we uselevers, we try to match two lever arms so that the impressed force is such thattheir angular momenta are equal. The angular momentum vector is perpendicular to the plane in which themovement is occurring and its magnitude is proportional to the effort to the effortneeded to stop the rotation. Italso specifies the direction in which the effort must be exerted by itsdirection. It is a vector that hasdirection, magnitude and sense. That is to say it is orientable. Linear momentum is expressed in kilogram meters per second and angularmomentum is expressed in kilogram meters squared per second.

A reason that we care about both linear and angular momentumis that they are both conserved in a closed system. If there are no external forces, then the velocities may beshifted between the components of the system, but the sums of both types ofmomentum over all of the components must remain the same magnitude.  If the connection between the circlingmass and the center of rotation were cut instantaneously, the mass wouldcontinue indefinitely in the direction that it is going the moment of thecut. That would satisfy bothconservation laws because the velocity remains the same and the perpendicularto the velocity is constant, therefore the product of the radial vector and thevelocity will be constant.

The two conservation laws set constraints upon the possibleoutcomes of events in closed systems. They also allow us to shift our point of view in useful ways that oftenclarify a situation or allow us to express the description of an event indifferent ways that are equivalent, but often give insight into the nature ofthe event. We will spend most ofthe remainder of this essay considering such transformations.

A simple example

Consider a simple example, where a mass is rotating along acircular trajectory about the reference point. We can write the description of the trajectory as a functionof time.

From these relations we can readily compute the ratio of thevelocity to the location.

We can see that this is a correct expression for the ratiojust by inspection. It followsthat the tangential velocity is and thecentrifugal velocity is 0. Thelinear momentum is , which is also the tangential momentum.

The angular momentum follows from its definition.

The angular momentum is perpendicular to the plane of thelocation and velocity and it is proportional to the mass, the angular velocity,and the square of the distance from the reference point to the mass point.  Doubling the radial arm quadruples theangular momentum.

The rates of change of linear and angular momentum

It is an experimental observation that a force applied to arigid body may be moved along its line of action without changing it rotatory action on the rigid body.  That means that there is a radialvector, , that is perpendicular to the line of action for a velocity,unless the line of action passes through the origin of the radial vector.  Consequently, we can compute the ratioof to , , and the expression for the angular momentum becomessimpler.

That leaves us with the problem of finding . It lies in theplane of and and it isrotated relative to by the negativeof minus the anglebetween and , .

Then, the radius is obtained by rotating through and taking thevector of the quaternion.

We can see that moving the force along its line of actionwill not change the moment about the point because the is the scalar factor in both forms of the expression.

In both cases we are effectively using a cross product andthe magnitude of the cross product may be interpreted as the area of theparallelogram that has the two vectors as its sides. In moving the force vector along its line of action we areeffectively shifting a parallelogram that lies between two lines so that it isa rectangle. The two crossproducts are represented in the above figure as oriented areas.  It is easily appreciated that the areasof the two cross products are the same. The two oriented areas are in the same plane, so, the perpendiculars tothe areas are in the same direction.

The move to forces and torques

If we return to our simple example, it is straightforward tocompute the force that is acting on the mass to hold it in a circularorbit. Force is the timederivative of the linear momentum. We have computed the linear momentum and we assume that the mass isconstant, so the derivative of the velocity is the critical parameter.

The force is directed in the direction opposite the radialvector and it is proportional to the mass, the distance between the centralreference point and the mass point, and the square of the angular velocity.  The force is entirely centripetal, thatis pulling the mass towards the central reference point, and no force isrequired to move the mass along its circular course.

If the trajectory of the mass is not circular, then therewill be a tangential force. We cansee that because the ratio vector would have a scalar component since the anglebetween the radial vector and the velocity vector would not be a right angle (), therefore the ratio of the velocity to the radial vectorwould be a quaternion with both a vector part and a scalar part.  The velocity will have componentsparallel and perpendicular to the radial vector.

Another way to view the situation is to consider a force orcollection of forces that act upon a rigid object. Most placements of the force(s) will cause the object torotate and translate to a new location. We must choose a reference point.  It may be a pivot point in a joint, acenter of mass or an arbitrary point. Its location is denoted by O.  The point of application of the force F is rfrom the reference point. Theforce may be resolved into a force that is perpendicular to the radial vectorin the plane defined by the radial and force vectors, the tangential force (), and a force that is in the direction of the radial vector,the centrifugal force (). The force will rotate therigid body about O and theforce will draw thebody in the direction of the radial vector. We have effectively computed these vectors in the sectiondealing with the resolution of velocities.

Torque Vectors and Torque Quaternions

Torque is the rotatory effort being applied relative to areference point in a rigid body. It resembles angular momentum in that it is avector quantity that is proportional to the displacement from the referencepoint to the point of application of the force and to the magnitude of theforce. In vector analysis it is expressedas the cross-product of the radial displacement times the force.  It may also be expressed as thedifferential of the angular momentum.

Actually, those two definitions are not actually the samequantity if is a function of time. Therefore let us consider the definition of torque in some moredetail. We will start with someexamples to get a feel for the way forces and torques behave.

An example: uniform rotatory movement

For uniform rotatory movement about a fixed point, we canwrite the expression for the radial vector as a function of time and thevelocity and acceleration follow directly from that description.

It follows that the ratio quaternion is the mass times theacceleration, divided by the radial vector.

In this instance, the velocity is always perpendicular tothe radial vector, so and we can writethe expression for the angular momentum and differentiate it to obtain thetorque.

The angular momentum is constant, therefore, the torque iszero. It does not take anyadditional effort to keep the mass circling once it has been placed in orbitabout the center of rotation.

An example: non-uniform rotatory movement

Now that we have set out the form of the analysis, let usconsider a slightly more complex situation, where the movement is not circularor of uniform speed. The locationslie on an elliptical trajectory that may move out of a plane.  It may rise and fall relative to theplane as one goes around the ellipse. The location, velocity and acceleration are as follows.

It follows that the ratio quaternion is the mass times theacceleration, divided by the radial vector.

If we multiply this ratio times the radial vector the resultis the total force vector.

If movement is entirely in a single plane and circular , then the force is the same as was computed for the uniformcircular movement.

If the movement is in a single plane, but elliptical, thenthe total force is somewhat more complex. It is a little counter-intuitive that despite the elliptical orbit,where the moving mass slows and speeds up at different parts of the orbit, theforce that holds it on trajectory is a constant central force.  Of course that is true because planetsin the solar system follow elliptical orbits and they are held by a constantcentral force, namely the gravitational force between the mass of the sun andtheir mass.

For both of these the force is directed in the oppositedirection to the radial vector. The situation for the full example is not as simple, but itis similar in that the force is in the same general direction.  If the angular velocity of the thirdcomponent, , is the same as the angular velocity of the other directions,, then the force vector would be in the direction opposite tothe radial vector. The ratio ofthe direction of the force to the direction of the radius is given by thefollowing expression.

The relationship between the force vector and the radialvector is somewhat complex. Thevector that turns the radial vector into the force vector is always in the i,j-plane and it is perpendicular to the radiusvector. We know that the vectorcomponent is perpendicular to the radius because of exchanging of the sine andcosine functions in the vector. This relationship is illustrated in the following figure.  The horizontal ellipse is the excursionof the radius vectors and the vertical ellipse is the excursion of itsperpendicular. In the ratio of theforce to the radius, the ellipse is multiplied by a constant () and a variable function of time (), but that does not change the direction of the turningquaternion’s vector. It does affect the angular excursion of the radial vectornecessary to align it with the force vector.

Calculation of the torques for the second example

The torques are computed as follows for each case.  When the trajectory is circular, thenthe torque is zero, as was computed.

When the trajectory is elliptical and confined to a singleplane then the angular momentum is the product of the mass times the vectorpart of the product of the radial vector times the velocity vector.

As stated above for the central force that holds planets ontheir elliptical orbits, there is no torque for an elliptical orbit, becausethe angular momentum is a constant vector, even though the mass changes speedas it moves around its orbit.

Clearly the torque moves in a circle in the plane of theelliptical movement, but it cycles at a rate that is the sum of the rates forthe elliptical component and the oscillation above and below the plane.  This arrangement is too complicated tobe readily visualized. Fortunately, this complexity is not typical of anatomicalmovements.

An example that is more representative of an anatomicalmovement is a pendulum. Let usconsider a periodic pendular movement as a third example.

Pendular example

Consider a joint in which the armature swings back and forthalong a circular trajectory, but at a rate that varies sinusoidally.  In particular, the pendulum swings inthe i,j – plane about the origin,with lever arm of one unit. It swings through a maximal angularexcursion of with an angularvelocity of . Consequentlyit swings from to and back in oneunit of time. We can write down ittemporal course fairly easily and all else follows from that decription.

The angular momentum is in-phase with the pendulum and thetorque is shifted 90° out of phase. The angular momentum is maximal at and , but in opposite directions. It is minimal at and , when it switches polarity. The torque is proportional to the mass, the maximal excursionand the rate of the swinging and it points in a direction perpendicular to theplane of the pendular movement.

Force couples

As with velocity, a force can be moved along its line ofaction without fundamentally changing its effect on the rotation of the rigidbody, therefore we can rewrite the expression for angular momentum.

The radial component becomes shorter but the force becomesproportionately greater so the product is same. The angular moment is the turning capacity of the force atits current point of application relative to the reference point.  We can make the concept more flexibleby introducing the concept of a force couple. If two equal, but opposite forces are applied at points separatedby an interval d, then themoment of the force couple is the moment.

As discussed above, the moment is the product of twovectors, therefore a quaternion, but since the vectors are perpendicular, it isa vector, so it is a quaternion vector.

Force couples are essentially equivalent to the laws oflevers. If we have a referencepoint, , then the force couple is the turning effort upon a rigid body relative to the referencepoint. Three situations areillustrated in the following figure. In each case the net angular momentum about the reference point is thesame. In the first situation, onthe left, the forces are applied symmetrically to either side of the origin, sothat is each is appliedway from the origin and both forces will tend to rotate therigid body counterclockwise. Theresultant of the two forces is the sum of the two torques, . In the secondsituation, in the middle, one of the forces is applied to the origin, thereforedoes not produce any torque. Theother force is displaced from thereference point so the torque is the same, . In the thirdsituation, both forces are applied to the same side of the reference point,therefore the rigid body would rotate about a different point if it were ableto do so. If the pivot point is at, then the two forces tend to move the rigid body in oppositedirections. However, the neteffect is the same, . In addition tothese types of variation, the force couple is unchanged if the distance betweenthe points of force application is reduced or expanded as long as the magnitudeof the force in concurrently raised or lowered by the same proportion.

Force couples rotate the object but do not translateit. Consequently, the moment ofthe force couple is the amount of turning capacity due to the force.  A useful feature of force couples isthat their moments are free vectors, that is, vectors that do not have aparticular location. So we canmove them as necessary as long as we keep the same moment.  The moment does define the plane anddirection of the rotation, that is, the plane and sense of the quaternionvector.

Equivalent systems of forces

Since equal and opposite forces applied to the same pointwill have a net force of zero, we can apply the force F and its negative to the reference point withoutchanging the mechanical situation for the rigid body. That was done in panel B for the figure above.  However, we can rearrange ourinterpretation and consider the negative force vector as the other half of aforce couple that has one force vector displaced dfrom the reference point. Theother force vector applied at Owill now translate the rigid body in the direction of F but not rotate it.

So, it is possible to replace a single force acting at apoint on a rigid body with a force couple, which rotates the rigid body, and aforce acting at the reference point, which translates the rigid body withoutrotating it. The torque of theforce couple and the translating force are always perpendicular for any singleforce.

The advantage of this arrangement becomes apparent when wewish to examine the consequences of applying several forces to different partsof the rigid body. A torque/forcepair can be generated for each force so that the torques all reference the samepoint in the body and all the forces pull from that point.  The forces and torques add vectoriallywith the others of the same type so that we obtain a force acting at thereference point and a torque for a force couple acting about that point.  However, unlike the case for individualforces, the force vector and the torque are generally not perpendicular whencombining a number of forces acting at different points on the rigid body.  Still, we have reduced a set of forcesto a single force and a torque.

Forces are vectors and torques are quaternions.  Both may be added to others of the samekind, but they may not be added to each other. The differences are subtle, but important.  A force can be represented by a rightquaternion. That is a quaternionthat may be expressed as a ratio of two vectors at right angles to each other,but a force is indistinguishable from its reflection. Torques are also right quaternions, the ratios of twovectors separated by an angular excursion of radians, but their sense of rotation is quite specific, sothat the reflection of a torque is different from the original torque.  One is a right-handed rotation and theother is a left-handed rotation. Still,both forces and torques add vectorially.

If multiple forces act together at a point, then the netforce is the vector sum of the component forces, which is a force.  This rule is well established byexperimental observation.

If we have two torques acting at the same point, then thenet result is the vector sum of their two quaternions.  That is obviously true when they areacting about the same or opposite axes of rotation. The total turning effort is the scalar sum of the magnitudesof the turning efforts about that axis of rotation. It is less clear that when the two turning efforts areacting in different directions that the combined effort is about an axisbetween the two component axes. Since an object can more in only one direction at any given time, theremust be a single axis of rotation and it is not in the direction of either ofthe two components.

The above figure is a demonstration that torques to addvectorially. We have two torques, and , which are not aligned. Since torque is a radial displacement times a force, we canexpress the torque as a force acting in the plane perpendicular to the torquevector times a force in that plane. Each torque has its own plane and the two planes intersect in a linethat is the ratio of the two planes. Therefore, we can take our radial displacement to be a unit vector inthe direction of the intersection and the force to be a force of appropriatemagnitude in each plane. Theresult is two forces acting at a point and we can sum those forces to obtain anequivalent force that acts in an intermediate direction.  This is just the run for addingforces. But that combinationforce, acting at the radial displacement of the component torques, will generatea torque that is the vector sum of the component torques.  By this means, we can sum any twotorques with a common reference point and by induction any set of torquesacting at a common locus. Consequently, torques sum vectorially.

It should be stressed that this argument depends on the twotorques having a common reference point, on being a unitvector, and upon the force vectors being perpendicular to the unit radialdisplacement. We have constructedan equivalent system that has nice properties. Torques do not add vectorially if the do not have a commonreference point, but we can compute an equivalent system in which they do havea common reference point, therefore it is always possible to add torques.

Computing torques at other locations, moving the point of reference

When two torques are not referenced to the same point thenwe can move them to a common reference point. The torques added are not those given, but their equivalentswhen referenced to the common locus. If we know the value of a torque referenced to a point and we wish tocompute its value referenced to point , we subtract the torque referenced to that would beobtained by applying the force at from its torquereferenced at . That isillustrated in the above figure when we have a force applied at thepoint . By definitionthe torque relative to point is , but we know the torque at and thedisplacement from to , which is .

More generally, if we have a number of forces that areapplied to various points on a rigid body, then we can extend the aboveargument to the equivalent force and torque.

The final relationship says that if we know the equivalent forceand torque at a point, we may compute the force and torque at another pointthat lie from thereference point. That is aremarkable simplification of the system.

The force/torque pair of a combination of forces will generally not bemutually perpendicular

If all the forces are applied at a single point, then therewill be no torque, since they are all pulling radially. If all of the forcevectors lie in a single plane, then they will all have torques in the samedirection or its negative, which are both perpendicular to the plane thatcontains the force vectors. Consequently, the force and torque are perpendicular and the pair can bereduced to single force in the plane, but displaced from the reference point(see the next section). Finally, if all the applied forces are applied in a single direction orits negative, then they will all have their torques in the same plane, which isperpendicular to the directions of the force vectors. The sum of the torques will be in the same plane, therefore,it must be perpendicular to the sum of the force vectors.  Otherwise, the torque and force of acombination of forces applied to a rigid body will not be mutuallyperpendicular without very special balancing of their magnitudes to make it so.

The next step would to be to try to convert the force/torquepair to a single force applied at some point on the rigid body.  To do that, we need one more tool, theconsequences of moving the reference point.

Reducing a force/force couple pair to a single force

We start with a set of forces applied at thepoints and referenced to the point. Afterprocessing as described above, we will have a mutually orthogonal force, , and force couple torque, , at point that is formallyequivalent to the set of applied forces. We can compute the couple that would result if we shift the reference pointto a different location displaced from . We did thatabove. The linear force is notchanged by displacement, therefore, it remains . The torque isthe torque at plus the torquefrom applying at from .

It follows that if we choose the offset so that , then the force/torque pair is replaced by an arrangementwhere there is only a force and that force negates the action of theforce/torque at . That is, theforce produces the same amount of torque as exists at . The rotationabout by the force applied at is equivalent tothe sum of all the forces applied to the rigid body. So the system of forces is reduced to a single equivalentforce applied to a particular point on the rigid body.

We still need to compute the value of . Clearly, the displacements are in the plane perpendicularto the torque’s vector. Otherwise,the torque vectors that cancel cannot point in opposite directions.  Secondly, the displacement is mostusefully chosen to be perpendicular to the force vector because then . That directioncan be readily computed by computing the ratio of the plane of the torque tothe plane of the force. The ratioof those planes is their intersection. We can determine the magnitude of the displacement, , by taking the ratio of the torque to the force vector.

This is a remarkably simple result.  The system reduced to a single forcethat is equivalent to the force/torque pair is the system that displaces theforce from the reference point by the negative of the ratio of the torque tothe force. Note that we still arereferenced to a point.

There is nothing in this argument that depends upon theorthogonality of the torque and the force so we can extend it to the generalcase where the torque vector is not perpendicular to the force vector.  Then the displacement is chosen to liein the plane that is perpendicular to the force vector and in the plane that isthe perpendicular to the torque vector, which means that it lies in theintersection of the two planes and the ratio of two planes is theirintersection. So the displacementis the ratio of the two vectors.

The force applied at is formally andphysically equivalent to the force/torque pair applied at , when referenced to . The referenceto the origin of the system, , is critical to the interpretation of torque, which isalways in reference to a location.

Wrenches and their Pitches

When a number of forces are applied to a rigid body at anumber of points that are not coplanar, then the net result is almost always aforce/force couple pair in which the force and torque vectors are not mutuallyorthogonal. The object is movinglinearly in one direction and rotating in a plane that is not perpendicular tothat direction of movement. Whenthat occurs, it is not possible to reduce the system to a single force actingat a point. There is anirreducible torque. However, wecan simplify the system to a force and a torque that is in the samedirection. Such a combined vectorpair is called a wrench. For anexcellent introduction to this material in a vector analysis framework see Beerand Johnston (Beer and Johnston Jr1990).

We do not see a great deal about wrenches these days, butthey were of great interest in the late 1800’s and early 1900’s.  They turn up in the advanced analysisof mechanical systems. They arerelated to screws, which are the movement version of the same process, aconcurrent rotation about an axis of advancement. We consider that type of movement elsewhere.

Let us now consider how one obtains a wrench as asimplification of a system of forces acting on an object.  The mechanical situation is a rigidbody that is being moved by multiple forces that do not act in a commondirection or in a common plane. Wehave shown that such a system may be simplified to a force, , and a torque, , acting at a reference point, . As statedabove, the force vector and the torque vector will not generally be mutuallyperpendicular. We have also seenthat we can effectively eliminate a torque perpendicular to the force vector bymoving the point of application of the force some distance along a vector thatis perpendicular to both the force and torque vectors.  The applied force generates the sametorque about the reference point.

We cannot remove torque that is not perpendicular to theforce vector by this means. However, we can split the torque into a component that is perpendicularto the force vector, , and a torque that is in the same direction, . The torquethat is perpendicular, , can be embedded in a displacement of the point ofapplication of the force, , leaving the torque in the direction of the force vector, .

Let us start with the force, , and the torque . We can computethe unit vector of the ratio of to , . Let the angleof the ratio be . Then we canwrite the expressions for the components of as a function ofthe ratio of to , , which is a quaternion.

A knowledge of the force and the components of the torqueperpendicular to it allows one to compute the distance that one needs to goalong to apply theforce so as to generate the same torque.

The parallel torque can be moved with the force.  It is the turning effort in the planeof the force, that is in the plan perpendicular to the force.

Although wrenches do not figure prominently in moderntreatments of mechanics, they were well known in the late 1800’s (Joly 1905). At that time it was common practice to speak of the pitch ofthe screw or the wrench. As ithappens the pitch of the wrench is the amount of turning in the planeperpendicular to the force vector (parallel torque) for a given amount offorce. If we express the pitch bythe symbol , then the scalar is the ratio ofthe magnitude of the torque parallel to the force to the magnitude of theforce. Since the torque and theforce are in the same direction, their ratio is a scalar and that scalar is thepitch of the wrench.

This leads to an interesting relationship.  The ratio of the torque to the forcefor a system of forces acting on a rigid object is the sum of the offset orlever arm and the pitch, which is a quaternion.

Recapitulation and a More General Solution

We started with a very general situation, where we had arigid body that is being acted upon by a number of forces that are applied atvarious point on or within the object. For instance, the object might be a bone that is being pulled on by aset of muscles attached at several insertions, plus the ligaments that bind thebone to another bone, the force of gravity acting at the center of mass for thebone, and the reaction forces in the joint that resist its compression ordistraction.

Forces add vectorially so one can compute a resultant forcethat may formally replace the various forces acting on the bone.  Torques also add vectorially.  Consequently, we can choose areasonable reference point, such as a point on the axis of rotation for thejoint, and compute the torque generated by each force in reference to thatreference point. While the torquefor any force is always perpendicular to the plane defined by the force vectorand the radial vector from the reference point to the point of application ofthe force, when we add the forces and the torques, the resultant force, , and torque, , will not generally be perpendicular.

We have seen that it is possible to resolve the torque intotwo components, one perpendicular to the resultant force vector () and one parallel to it (). Let thosetorque vectors be called the perpendicular torque and the parallel torque,respectively. We can create anequivalent mechanical system with the torque being generated by the force () displaced from thereference point. It follows fromthe definition of torque that the torque is the radial displacement times theforce vector in such a system, that is the lever arm times the force acting onthe lever arm. Therefore, thelever arm for rotation in the plane of the torque perpendicular to the forcevector is the ratio of the perpendicular torque to the force.

In this system, the force causes the rigid body to rotate inthe plane that contains the force vector, , and the displacement vector, , while being translated in the direction of the forcevector.

However, we still have a torque that is parallel to theresultant force vector, the twist that accompanies the translation, thewrench. The wrench rotates therigid body about the axis of the force vector as it translates the object.  The relative proportions of rotationand translation are expressed in the pitch of the wrench.

Let us consider how we might further generalize thisanalysis and perhaps simplify it in the process by replacing the force/torquepair with a single force. Tostart, we resolve the torque into two components, as has already beendone. But, in addition, weconsider the implications of the parallel torque. As a consequence we will be able to move directly to anequivalent mechanical system that has a single force.

The following figure illustrates the situation where thetorque and force are not mutually orthogonal. A force is combined witha torque that lies at anangle of to the forcevector. The torque is resolvedinto a perpendicular torque, , and a parallel torque, . Theperpendicular torque can be replaced with a displacement of the force vector from thereference point for the force torque pair. Consequently, the displaced force has the same vector valueas the original force, . This isessentially the analysis that is spelled out above. The new step is to note that a force vector at the samelocation may also replace the parallel torque. The new force vector is perpendicular to the plane definedby and and itsmagnitude is to the magnitude of the parallel force as the magnitude of is to themagnitude of .

The parallel torque force does not advance the object in thedirection of the force vector, but it does rotate it in the plane perpendicularto the force vector. We have twoforces acting at a common point, therefore we can add them vectorially.  The result is the force . However, it isapparent that the force is simply the force in the plane of the torque thatsatisfies the relationship implicit in the definition of a torque.

We have taken advantage of the fact that is in fact aquaternion, although we have treated it as a vector up to this point.  It rotates the torque vector through90° and re-scales it to the appropriate length. This expression is in fact more general and simpler than theanalysis that led to a wrench. Itmoves directly from the force and torque to the radial displacement and forcethat will give the same mechanical outcome. It is more general in that is not a uniquevalue. We can redefine it asfollows.

The product in the numerator rotates the torque vectorthrough 90° in the direction opposite to the rotation from the force vector tothe torque vector and the scalar in the denominator scales the force to be theinverse of the ratio of the magnitude of the tensor to the magnitude of theforce, that is the inverse of the generalized pitch.

An example of distributed forces that create a wrench

Consider a set of four forces that are applied at where each forceis perpendicular to the coordinate axis and tilted in the direction of . The net effectof such an arrangement is a torque about the origin and a force in thedirection of .

In the above illustration, the vectors are 0.2 units long inthe -plane and one unit longin the direction of . Consequently, the resultant force is0.4 units in the direction of and the resultant torque is 0.8 units,also in the direction of . This arrangement is clearly a wrench inthat the torque and force vectors are in the same direction.  Also, because of the symmetry of thearrangement it is clear that one can replace the wrench with a single vector ata unit distance from the origin that is directed 0.8 units tangent to a unitcircle in the -plane and 0.4 units inthe direction of , if the rotation is constrainedto occur about the origin. We arenot really replacing the four vectors with a single force vector, but with aforce couple where one of the vectors is at the origin. The simplification is not as intuitiveas the wrench interpretation, which leads directly to a force pulling theobject in the direction of the axis while the object is being spunabout that axis. A point on theobject will follow a screw trajectory.

Joints generally resist substantial translations, becausethey are an inefficient way of achieving movement, therefore, the forcecomponent is apt to be the component that is resisted by compression ordistraction of the structures of the joint. Compression is absorbed by cartilage and distraction is usuallyabsorbed by ligaments or muscles. Rotation may also be resisted by joint structures, so that the joint isconstrained to move in a particular plane.

References

Beer, F. P. and E. Johnston Jr,Russell (1990). Vector Mechanics for Engineers. Statics, Dynamics. NewYork, McGraw-Hill Book Company.

Joly, C. J.(1905). A Manual of Quaternions. London, Macmillan and Co. Ltd.{available in facsimile reprint from Kessinger Publishing (www.kessinger.net)}.