Movements of a Circular Canal

When a semicircular canal accelerates, the fluid within ittends to move in the opposite direction as it lags or leads the canal, becauseof inertial drag. This sectionconsiders how fluid moves within a ring in a number of situations where thering is rotating. The generalsituation will be described first and then a number of sample situations orincreasing complexity will be considered computationally.

A general scheme for the analysis of fluid motionin a circular ring. The ring is rotated about an axis ofrotation that does not lie in the sensory region and the inertial drag upon thefluid in the canal causes the fluid to move within the ring.

The general situation is illustrated in the followingfigure. There are two frames ofreference, one attached to the center of rotation () and one attached to a ring (). The center ofrotation has a location, , and an orientation frame of three vectors that reference itto the head or universal space or some other convenient coordinate system {}. There is anaxis of rotation for the canal, , which will be taken to be the perpendicular to the canalplane that leads to increased activity in the sensory epithelium when the canalis rotated about it. It willgenerally be assumed that the movement is in a right-handed coordinate system.  The framed vector for the canal willhave a location, , and a frame of reference, {}, where the first basis vector, , points to the direction of the optimal acceleration for thesensory epithelium, the second basis vector, , points in the direction of the axis of rotation and thethird basis vector, , will point towards the sensor. There is an extension vector from the center of the canal tothe sensor, . That extensionvector is important to what follows, because we will select points in the canalby rotating the vector around the axis of the canal. The radial vector, , extends from the center of rotation to the canal, where itmeets a vector drawn from the center of rotation to the same point in thecanal, . That point isrotated through a small angular excursion, , about the axis of rotation and it comes to a new point at theend of the vector . The linearexcursion of that point in the canal is . Thedisplacement may be resolved into three mutually orthogonal displacementvectors. Since we are interestedin the movement of fluid in the canal, the axes of the displacement are definedrelative to the canal. One axis isperpendicular to the canal in the plane of the canal, , therefore it is in the direction of the radial vector tothe point, . A second isperpendicular to the canal in the direction of the axis of the canal, , therefore in the direction of the axis of the canal, . The third, , is in the direction of the ratio of the first to the secondand therefore it is in the direction of the tangent to the canal.  Although the basis vectors are listedin this order, they will be written in the order , so that In thatorder, they form a right-handed coordinate system. Once we have these basis vectors, we can calculate thecomponents of the displacement by taking the dot products of each basis vectorwith the displacement vector. Fromthat information one can move fairly directly to the forces acting on the fluidat the selected point in the canal. The force is the second derivative of the displacement with respect totime. It is in the oppositedirection of the displacement and proportional to the rate of rotation.  We now need to get down to thenitty-gritty of the calculation.

There are several rotations that occur about vectors, so letus set up the following convention. If a rotation occurs about a vector, let the quaternion that expresses that rotation be writtenas . The italic and bold formal indicates that it is aquaternion and the bar indicates that it is a unit quaternion.  Similarly, the bar above the symbol fora vector indicates that it is aunit vector.

To simplify the calculations, let us assume that the centerof rotation is at and the basisvectors for the center of rotation frame are . Doing so doesnot affect the generalizability of the calculations and it simplifies thesymbolism. Then the center of thering is at and the radiusof the ring is . The axis ofthe ring is . If we take thecenter of the sensor to be at zero on the ring, then we can express the sampledpoint in the ring as a rotation of the extension vector from the center ofrotation to the sensor, .

The vector from the center of rotation to the selected pointis the point minus the center, which, because of the assumptions listed above,is the value of the selected point.

If the canal isrotated through a small angular excursion, , then the new locations of the point will be and we cancompute the change in location for the sampled point.

The new location of the selected point is and thedisplacement is the difference between the locations of the sampled pointbefore and after the rotation.

The components of the displacement are the dot or scalarproducts of the displacement with the local basis vectors for the canal.

It should be noted that the rotation will also change theorientation and location of the canal, therefore one must recalculate theframed vector for the canal.

However, returning to the component displacement vectors,the calculated quantity is a displacement and we are interested inacceleration. We could compute therate of change of the displacement per unit time and then compute thecomponents or we can treat each component separately. The result is the same if we stick to rotations about fixedcenters of rotation. Any of thevectors may be expressed as a vector in the plane of the change.  We parameterize the plane with twomutually orthogonal vectors, and express thedisplacement as a sum of the projections of on to and . The velocityof the displacement is perpendicular to the displacement  and proportional to the magnitude ofthe angular excursion per unit time. Then one can differentiate again to obtainthe acceleration and the result is that the acceleration is in the oppositedirection of the displacement and proportional to the square of the rate ofangular displacement per unit time.

Forces that are in the directions of the perpendicular orthe normal vectors will not affect the sensor, because they will not move fluidaround the ring of the canal. Therefore, we are most interested in the magnitudes of the forces thatare tangential to the ring, which are proportional to . While we mayeventually be interested in the actual magnitudes of the forces generated inthe canal, for the time being we are only interested in relative magnitudes.

At this point we will begin examining a number ofprogressively more complex situations to get some sense of the consequences ofrotations on the signals generated by canals. The first situation is very simple, if not very realistic inlife. It is assumed that thecenter of rotation is at the center or the circular canal and the axis ofrotation is aligned with the axis of the canal. The situation is very symmetrical so we can examine onesegment of the ring and deduce the action occurring in all segments.

F = ma

If a mass moves at a constant velocity in a straight line,then no force is required for it continue in the same manner, leaving asidefrictional forces for the time being. However, if the mass accelerates or decelerates or if it moves in acurvilinear trajectory, then there must have been a force applied to cause thechange and the mass will exert equal and opposite forces upon the agentresponsible for the change. Themagnitude of the force will be proportional to the quantity of the mass and theacceleration that it experiences.

A uniformly rotating circular ring with its axis of rotation through thecenter of the ring, aligned with the axis of the ring

If we take the origin of our coordinates to be the center ofthe ring, where the axis of rotation intersects the plane of the ring, then theposition of a segment of the ring may be described as a constant radius thatrotates at a constant angular velocity in the plane of the ring.  For simplicity, let the radius be unityand the angular velocity be one cycle per second.

The velocity vector of the moving segment is tangential tothe ring and orthogonal to the position vector.

A short time later the segment of the ring has rotated asmall distance around the axis of rotation and the velocity vector is the samemagnitude, but directed in a slightly different direction.  The change in the velocity with respectto time is the acceleration.

Notice that the acceleration vector is the same as theposition vector except for being multiplied by a negative constant.  Since the position vector isperpendicular to the ring in the plane of the ring, the acceleration vector isalso perpendicular to the ring and in the plane of the ring.  This corresponds to our physicalintuition that the uniformly spinning ring where the axis of rotation isthrough the center of rotation and perpendicular to the plane of the ring willexperience a constant centrifugal force on the outer surface of the ring.

An accelerating rotating circular ring with the axis of rotation throughthe center of the ring and aligned with the axis of the ring

If the ring in our first example is changing the speed withwhich it is rotating, then the situation is slightly more complex.  The position vector is a variablefunction of time; therefore, the derivative with respect to time is not aconstant.

The velocity vector of the moving segment is tangential tothe ring and orthogonal to the position vector.

The velocity is proportional to the derivative of theangular speed. So far, thesituation is only moderately different from that for uniform rotation of thering. The acceleration is notablydifferent.

In this situation, it is clear that the force can beresolved into two vectors. Thefirst, , is like in the first example, with constant speed ofrotation. It is directedperpendicular to the ring, in the plane of the ring and it is proportional tothe square of the angular velocity. The second vector, , is tangential to the ring, therefore is a force that wouldcause deflection of a partition transverse to the circumferential axis of thering. Consequently, a structurelike the semicircular ducts might be sensitive to changes in head velocity, butnot to uniform head rotation.

If the displacement varies sinusoidally with time (), then the outward pressure is greatest when the velocity isgreatest and the circumferential pressure is greatest when the displacement isgreatest.

We will move to a more realistic model, but first let usconsider another situation, where the rotation is not about the perpendicularaxis through the center of the ring. Let the axis of rotation be in the plane of the canal and parallel withthe sensitive axis of the sensor. In this situation the distribution of pressures is not radiallysymmetrical, so we will need to look at the distribution of pressure as afunction of angular distance from the sensor.

Let us put the situation in the format that was introducedabove to set up for the more complex situations where the mathematics is moredifficult. We start with the ringin the i,k-plane, centered on the origin,which is also the center of rotation. The axis of rotation is in the same direction as the axis of the canal,both in the direction of the jaxis. The selected point is theextension vector of the frame of the canal that extends from the center of thecanal to the sensor, , rotated through an angle of and it isdenoted by . The rotation ofthe canal about the axis of rotation is and it moves theselected point to . The excursionof the selected point is .

We can write down the expression for the various elements ofthe situation by inspection of the diagram. The new location follows from the anatomicaldescription of the rotation. Then it is a simple matter to compute thedisplacement of the selected point in the ring and derive an expression for thedisplacement that simplifies to a description in accord with our intuitionabout the situation.

The final result says that the displacement is approximatelyorthogonal to the radial vector, which means that the acceleration term isalmost totally circumferential around the ring, independent of the pointchosen. This is the same result aswe obtained by the first method.

It should be noted that what is being assessed is the changein velocity of ring rotation. Ifthe ring continues to rotate at a constant velocity, then the fluid within itwill be brought up to speed with the ring, because of friction within the fluidand between the fluid and the wall of the ring and because there is a partitionacross the ring that is somewhat distensible, but will not allow fluid to flowaround the ring. Consequently, thefluid displacement is really a change in fluid velocity and the ring isreacting to the acceleration of the fluid.

Ring spinning about an axis through the ring

Let the center of rotation be the center of the ring and letit be perpendicular to the extension vector from the center of the ring to thesensor, . Then theselected point in the ring is . Forconvenience, let the ring lie in the i,k-plane, so the quaternion of the rotation is with a vector ofk and the extension vector is i. Theaxis of the canal is . We can thenbegin calculating the excursion of the selected point.

The movement is confined to the plane of the rotation.  However, the magnitude of is assumed to be small, approaching zero, therefore, is nearly one and the movement of the fluid in the canal isessentially perpendicular to the lateral wall of the canal.  The magnitude of the force isproportional to the . So, it ismaximal in the mid-ring and small at the top and bottom of the ring.  It is in opposite directions on the twosides of the ring, as one would expect from the geometry of the situation.  There is no circumferential movement,therefore, this type of rotation would not stimulate the receptor.

Rotation about an eccentric center of rotation, in the plane of the canal

Next, consider the situation where the canal is rotatingabout a center of rotation that is outside the ring, but the axis of rotationis aligned with the axis of the canal. We would expect this situation to be quite effective in stimulating thecanal, but the forces should be a function of the point chosen in the ring.

Let the canal lie in the i,k-plane and the axis ofthe canal lie at the origin and directed along the j axis. The extension vector from the center of the canal to thesensor, , is i. The selected point in the ring is and it isrotated about the center of rotation at to obtain the new location, . The locationof the center of rotation will be three radii from the center of the ring andthe axis of rotation is parallel to the axis of the ring.  The difference in location due to therotation is . In thissituation, it will be necessary to compute the vectors for the selected point so that may be projectedon the unit vectors of the basis, to obtainthe relative amounts of acceleration along the canal and perpendicular to it.

We can see byexamining the illustration that the movement will tend to be at a substantialangle to the ring at both the top and the bottom of the ring and fairlyparallel at the two sides of the ring especially at the sensor when thearrangement is it being most distant from the center of rotation. We can also see that thecircumferential accelerations in the part of the ring more proximal to thecenter of rotation will tend to push the fluid in the opposite direction to thedirection induced in the more distal parts of the ring.  We would expect the more distalacceleration to be greater, because of the greater linear excursion with thesame angular excursion. Having made these observations, let us consider thecalculation of the displacements.

This expression for the displacement of the selected pointis sufficiently complex that one is probably far better off going withnumerical methods. First wecalculate the displacement of the selected points as a function of theirlocation on the ring. The nextfigure shows the displacement in the direction of the axis and in the direction of the axis.  The rotation is 0.05 radians which isslightly less than 2.9°. We cansee that the greatest displacements in the direction, relative to the movement of the ring as a whole,are at the most distal and most proximal parts of the ring with respect to thecenter of rotation. In the direction the greatest displacements are at the midpointsbetween those two poles and they are also about 0.05 times the radius.

We know from the expression for the displacement that thefunctions are approximately a sine and cosine function, so it is likely thatthe circumferential displacement may be almost constant. There is a phase shiftin the curve for displacement in the direction.  However, when the total displacement isplotted, we find that the amount of displacement is a cosine-like function oflocation. It is maximal at the most distal part of the ring and minimalin the location closest to the center of rotation. That is what we would expect from the structure of thesituation. Because the distalpoint is about twice as far from the center of rotation, it travels about twiceas far with the same angular excursion.

We can examine that by computing the projections of thedisplacement upon tangential and radial unit vectors. That has been done here by taking the ratio of thedisplacement to the unit vector tangential to the ring and to a radial unitvector. The tensor or magnitude ofthe ratio is the magnitude of the displacement, which is plotted immediatelyabove. The scalar of the ratioquaternions may be computed and they are the relative magnitudes of theprojections of the displacement on each reference vector for thatlocation. The sum of the squaresof the values for each location will be unity. The relative radial flow is clearly warped from a sinusoidalfunction and the relative tangential flow is similarly warped, being narrowerin the proximal part of the ring than in the distal part.

The curves for the angles between the displacement vectorand the local reference vectors is approximately saw-tooth, but with somecurvature that causes the curves to deviate from straight lines.  They are also phase-shifted relative toeach other.

When we multiply the relative amount of flow at eachlocation by the absolute amount of flow, to obtain the actual amount of flow,then the curves become symmetrical nearly sine and cosine functions.  The radial flow is nearly sinusoidaland symmetrical for flow to the exterior of the canal on one side and flowtowards the outside of the canal on the other side. That is intuitively what we would expect from the geometryof the situation. Thecircumferential flow is not symmetrical. For the distal part of the ring the flow is greater than for theproximal part and the flow is in the opposite direction for the most proximalpart of the ring. In a ring ofuniform caliber, there will be a preponderance of flow in one direction, thedirection of flow in the distal part of the ring. If one plots the function on the sameplot, there is a very small discrepancy between it and the curve of thetangential flow, but it turns out to be very good approximation to the flowdistribution. That is good becauseit implies that we can assume a fairly simple relationship between angularexcursion and net flow in a tube.

In the next figure the profiles for a series of situationsare plotted together to illustrate how the tangential and radial flows depend onthe distance from the center of rotation to the center of the ring.  The number beside the curve indicatesthe distance between the centers, in ring radii. The curves from the above calculations are indicated by thenumber 3. All the curves of a typeintersect at two points. Thatvalue is the angular excursion of the ring rotation chosen for thecalculation. When the movementexcursion is 0.05 radians, the tangential displacements have a commonintersection 0.05 radians above the zero axis. For the radial displacements, the intersects lie on thehorizontal axis through zero displacement. The curves are very close to cosine and sine waves so thatthere is a small asymmetry for tangential displacements, which means that thereis a small bias for fluid movement in the ring in a direction opposite to thedirection of ring rotation.

The geometry of the situation was chosen so that the sensorwas at the part of the ring that is maximally stimulated by the rotationbecause the force to move the fluid in the direction opposite the rotation isgreatest in the part of the ring opposite the center of rotation.  The calculations so far seem toindicate that the location of the sensor would effect the signal the sensorwould give with a given rotation. If we assume that the forces average, which is most likely since thefluid is virtually incompressible at the pressures that occur in a semicircularcanal. Then the effect of arotation would be to cause the fluid to move in one direction, the directiondictated by the pressures in the most distal part of the ring.

The net pressure in the ring may be related to the integralof the forces generated throughout the ring, which is the area between thecurve and the horizontal axis. Wehave found that the pressure profile for circumferential movement is of theform , therefore the integral will be depend only on the offset, . The k is aconstant, which would be the sensor output when there is no movement.  In the vestibular system there is abaseline discharge rate about which the discharge rate varies, so both positiveand negative displacements a will register.

The overall signal that is sensed by the ring is the angularexcursion. Theoretically, therewould be more distortion if we use large angular excursions, but we can view alarge angular excursion as a series of small excursions, each of which is quitesmall. In that case there is littledeviation from a sine or cosine function.

If the movement is happening at a particular speed, then theangular excursion per unit time will be translated into a displacement per unittime or an angular velocity. It isthe change in velocity that generates forces that tend to displace the fluid.  If a ring travels in a straight linewith constant speed, the fluid in the ring travels with the ring and there isno force. If the speed of the ringchanges or the direction of its movement changes, then there is a change invelocity and the fluid has an inertial drag, which generates reaction forces offluid movement. Therefore, the pressure on the cupula will be a linear functionof the change in the head’s angular velocity.

The average circumferential displacement is a linearfunction of the magnitude of the rotational excursion if the ring is rotatingabout an axis that is parallel with the axis of the ring.  That is remarkable, because it saysthat the push on the cupular partition is simply related to the head movement,independent of the center of rotation. Even though the semicircular canals move through much more distance whenthe head is moved from the waist than from the neck, the canals will sense bothmovements as being the same if the angular excursion is the same and the timeto complete the movement is the same. The local forces at points in the ring will be much greater in the firstsituation, moving from the waist, but the average force moving fluid around thering is the same.

Rotations about the axes of rotation near the axis of the ring

The next few section it will be shown that rotations inplanes perpendicular to the plane of the ring cause no overall circumferentialfluid movement, because the average displacement around the ring is zero.  So, circumferential movement in a ringis due to rotation in a plane near that of the ring. We will briefly consider that relationship in the nextcouple sections.

Now, we turn to rotations about axes that are tilted withrespect to the axis of the ring. In the following figure, the center of rotation is displaced three ringradii in the direction of the axis and three radii in the direction of the axis. The axisof rotation is initially parallel to the axis of the ring (0°).  Then the direction of the axis ofrotation is moved in five steps to point in the direction of the center of thering. In the most tilted instance,the ring is tilted 45° with respect to the axis of rotation (45°).

A ring (green) isrotated about an eccentric center of rotation (brown circle) with a number ofdifferent axes of rotation ranging from perpendicular to the plane of the ring(0°) to a 45° angle to the plane of the ring.

When the axis of rotation is parallel to the axis of thering all of the displacement is circumferential and the average is proportionalto the angular excursion of the rotation, but the displacement variessinusoidally around the ring. Whenthe axis of rotation is a 45° angle to the ring, but through the ring thecircumferential displacement is constant around the ring, but there is asubstantial displacement towards the wall of the ring that varies sinusoidallywith location, being greatest at the poles of the ring, the leading a trailingparts of the ring. Note that sincethe lateral and radial displacements co-vary, the displacement is neitherradial nor lateral, but oblique, more radial than lateral.

It is also interesting that the tangential displacementcurves do not cross at a common point. That reflects that the middle value of the curves shifts withangle. When the axis of rotationis perpendicular to the ring, the middle value is the magnitude of the angularexcursion of the rotation and when it is a 45° angle, it appears to be aboutthe angular excursion times the cosine of 45°. Consequently, the average displacement in a rotating ring isan index of both the angular excursion and of the inclination of the ring tothe axis of rotation.

Rotations in which the ring is tilted to the axis of rotation, but there issome rotation in the plane of the ring.

The previous situation was complicated by two thingshappening at once. The axis ofrotation was moving closer to the center of the ring at the same time as it waschanging it angle with respect to the plane of the ring.  Let us now eliminate the first  factor by examining the situation whereall the axes of rotation are through the center of the ring, but they differwith respect to their tilt. Whenthe ring is perpendicular to the axis of rotation and the axis of rotation inthrough the center of the ring, we have the initial situation considered aboveand all of the displacement is circumferential. The tangential displacement is 0.05 radians, the same as thetotal displacement of the ring about its center. The displacement is the same in all parts of the ring so thecurve is a straight line or a constant value. As the axis is tilted more with respect to the plane of thecanal (10° through 90°), the amount of tangential displacement  becomes less until there is notangential displacement of average (90°), because the ring is rotating about anaxis through the center of the ring that lies in the plane of the ring.  It is spinning like a coin spinning onit edge. There is some sinusoidalvariation in the profiles for tilted axes of rotation, because some parts ofthe ring are further from the axis of rotation than other parts.  The magnitude of that variationincreases with greater tilt.

Still, the variation is much smaller than the mean value forall but the 90° tilt example. Forevery other example, there is a non-zero average for the profile and the areabetween the profile and the zero axis is greater for less tilt.  It works out that the magnitude of theaverage tangential displacement is proportional to the cosine of the angle ofthe tilt. So we can elaborate onthe expression developed from the first situation and say that the response, , is proportional to the rate of change of rotation, angularacceleration, , and the cosine of the tilt between the axis of the canaland the axis of rotation, .

That relationship indicates that we cannot tell from onesample of a response to a movement to what extent the ring is accelerating () and to what extent it is tilted relative to the axis ofrotation ().

Returning to the previous section, we can now say that thecurves for tangential displacement did not cross at a single point because theaverage values of the responses were different in response to the differenttilts with respect to the plane of the ring.

Rotations in which the ring moves in direction of its axis

More briefly, we can consider the displacements that occurwhen the ring moves about an axis that lies outside the ring in the plane ofthe ring and the ring moves in the direction of the axis of the canal.  It moves face-on.  We would expect the main displacementto be against the lateral wall of the canal. The next figure illustrates the displacements relative tothe local coordinate system, that is radially, tangentially and laterally ororthogonal to the ring plane. Thecalculations were essentially as in the last situation, except that the centerof rotation is displaced along the axis for 1, 3,5, 7, 9, and 11 radii, and the rotation is about an axis of rotation parallelto the axis.

As expected, the major displacement is in a directionorthogonal to the canal plane or against the lateral wall of the ring.  The amount of displacement isproportional to the distance from the center of rotation to the center of thering. The sinusoidal variation isdue to the more proximal locations traveling less distance than the more distallocations.

There is also some displacement in the radial direction atthe more distal and proximal locations. That is centripetal displacement due to movement in a circle, so thatthe fluid seems to be drawn towards the center of rotation because of itdeviation from a straight trajectory.

There is also some tangential displacement, for the samereason as the radial displacements, but it is greatest at the middle of thering, where the fluid tends to flow along the tube towards the distal part ofthe ring. The displacement isbalanced on the two sides, or opposite in direction, so there is no netmovement when integrated over the whole ring.

Both the radial and the tangential displacements aresinusoidal about no displacement, so that there is no net tangentialdisplacement, but there would be a force on the outer wall of the distal ringand the inner wall of the proximal ring, which would tend to push the ringdistally as one would expect of a swinging ring.

Spinning about an axis in the plane of the ring

Another possible way of rotating the ring is to spin itabout and axis that passes through the middle of the ring in the plane of thering. It would be spinning like atop or a penny spinning on its edge. The displacements would be fairly straightforward.  They would be against the lateral wallsof the ring, in opposite directions on the two sides of the axis of rotationand zero where the axis passes through the ring. There would also be a radial displacement towards the centerof the ring because the fluid would deviate from a straight trajectory.  Calculations with the model indicatethat the lateral displacements are orders of magnitude greater than the radialdisplacement for small angular excursions. There would be no tangential displacement, because the fluidat any location will be moving in a circular trajectory in a planeperpendicular to the plane of the ring.

Conical swings about an axis parallel to the plane of the ring

A more interesting excursion is when the ring is swingingabout an axis that is not through the ring, but parallel to the plane of thering, that is experiencing a conical rotation. Many of the movements of the head will produce such atrajectory for the semicircular canals. For instance, moving one’s head back and forth in a horizontal plane, aswhen shaking it to signify ‘no’ or disagreement would cause all of the canalsto experience a conical rotation. In that particular instance, the only canal that has a plane that comesclose to containing a vertical vector is the anterior canal.  The posterior canal is in roughly thesame situation. The horizontalcanal would roughly approximate another type of conical swing where the centerof rotation is not in the plane of the canal, but the axis of rotation is inthe direction of the axis of the canal, that is formally equivalent to thefirst situation considered, because we can find a point on the axis of rotationthat is in the plane of the canal.

In the next section, we will consider the special case ofthe axis of rotation parallel to the ring plane, but not in it, and then moveon to the general case of a conical rotation. That is the situation most like normal movements ofsemicircular canals.

What happens when the tube is not of uniform caliber?

The assumption of a tube of uniform caliber has given ussome insights into the movements of fluid in a ring, it is not very like theanatomy of semicircular canals. The semicircular canals are only partial rings, about two-thirds of acircle with the ampulla at one end. Both ends of the canal are joined to the utricle, which is acomparatively large amorphous chamber. The diameter of the duct part of a canal is about a quarter to a thirdthat of that of the ampulla and the ampulla is much smaller than the utricle.We would expect comparatively little resistance to fluid flow in the utricle orthe ampulla, compared to the resistance to flow in the duct.  For laminar fluid flow in a circulartube, resistance is proportional to the fourth power of the radius of the tube,which means that the resistance to fluid flow in the duct is about two ordersof magnitude greater than the resistance in the ampulla and even greaterrelative to the utricle. Thesestructural considerations mean the fluid dynamics in the vicinity of an ampullaare like that of a hypodermic syringe with a long needle on it.  The pressure needed to move the plungerin the barrel is much greater with the needle attached than it is without it.  In much the same way, a force thattries to move fluid from an ampulla into the adjacent duct will meet asubstantial resistance to flow that depends upon the viscosity of the fluid,the length of the tube, and the diameter of the tube.

The duct acts as a damper on the system.  The same process would work in theopposite direction. Fluid movementfrom the duct through the ampulla and into the utricle would encounter a dragretarding movement through the duct. The retarding force would be proportional to the velocity of fluidmovement and opposite the direction of the force trying to move the fluidthrough the ampulla. This dampingis a good feature to have in such a system because it tends to prevent largemovements of the fluid in the canal. The ampulla is sealed off by the cupula and the cupula is distensible,but optimally only within narrow range. Too much unopposed force tending to move fluid around the canal wouldtear the cupula from its attachments to the wall of the ampulla and the cristaampullae. The cupula will act as aspring in that fluid movements will cause it to billow, like a sail in thewind. However, that distortionwill resist further distortion, presumably in a manner roughly proportional tothe amount of displacement, .  Ideally, we want a brief transient delayas the canal changes the rate at which it rotates and a rapid acceleration ofthe fluid within the canal to the new velocity.

There is an additional component acting to control fluid movements.  The duct of a semicircular canal doesnot have a circular cross-section. It is elliptical, with an approximately 2:1 ratio of axes.  That means that it can accommodate abit more fluid if there is greater internal pressure and it can become flatterif the internal pressure drops. That may be an important mechanism if the movements of fluid into andout of the ampulla. As fluid is forced into the ampulla by greater pressure onthe utricular side of the cupula, the cupula billows, thereby forcing fluidfrom the duct side of the ampulla into the duct. The duct offers resistance to fluid movement, but the fluidcan be accommodated by rounding of the duct. If there is a sustained acceleration, then the duct canbleed the excess fluid through the duct and into the utricle.  With fluid movements in the oppositedirection, the duct side of the ampulla can draw fluid from the duct byallowing the duct to flatten a bit. This mechanism would also seem to work as a spring and, in the shortterm, one would expect the fluid stored in the duct distortion to beproportional to the amount of fluid that has moved into or out of theduct. More stored fluid wouldstretch the duct more and create a greater restoring force, .

In fact, the anatomy is even stranger in that the needleextends back to end in the chamber of the syringe.