Three-dimensional analysis of linear vestibulo-ocular reflex in humans during eccentric rotation while facing downwards

Abstract

When participants undergo eccentric rotation (ER), i.e., they are rotated while displaced from the axis of rotation, they undergo both rotational stimulation and linear acceleration, which induces both the angular vestibulo-ocular reflex (aVOR) and linear VOR (lVOR). During ER, the lVOR induced by tangential linear acceleration enhances the eye movement induced by aVOR. In this study, we attempted to measure aVOR and lVOR separately, while participants underwent ER while facing the ground in a dark room. We analyzed three-dimensional eye movements using a video-oculography system. The participants sat on the ER chair either directly above the center of rotation, or with their head out, head in, right ear out, or left ear out against the center of rotation. Under these conditions, the rotational axis of the eye was perpendicular to the ground for rotational stimulation (aVOR), and the axis was parallel to the ground for linear stimulation (lVOR). Thus, measured eye movements could be separated into these two components. At 0.1 and 0.3 Hz rotation, we observed aVOR but not lVOR. However, when the stimulation frequency was above 0.5 Hz, we observed both aVOR and lVOR. These data indicate that lVOR is activated when the stimulation frequency is above 0.5 Hz. We conclude that it is possible to separately analyze aVOR and lVOR, and to simultaneously assess the function of aVOR and lVOR by analyzing eye movements induced when participants undergo ER above 0.5 Hz while facing the ground.

Appendices

Appendix 1: Analyzing the rotation vector of human eye movement

A circle with a radial gaze angle of 40° was placed 1.5 m from participants. Participants wore goggles equipped with an infrared CCD camera (RealEyes) on the left eye, sat on a chair, and used their right eye to look at eight targets located on the circumference of the circle in sequence, so that their gaze moved in a circular clockwise pattern. We improved the design of the RealEyes goggles, such that it was possible to remove the cover in front of the right eye. By removing the cover, participants could obtain a wide field of view in their right eye. After completing the circular eye movement, they were asked to look at the center of the circle. From the digital images of the circular eye movement, we were able to detect the edge of the pupil, and approximate the pupil using an ellipse. We then found the minor axis, major axis, and center of the ellipse (p). The center of eye rotation (o) (yc zc) on the image plane was determined as the intersection of the extensions of the minor axes (Fig. 6a, b). After determining the center of eye rotation, we calculated the radius of rotation of the center of pupil (R) using the following formula:

$$R\sqrt {1 - ({\text{the length of }}\hbox{min} {\text{ or axis}}/{\text{the length of major axis}})^{2} } = r$$

where r is the length between o and the center of the pupil ellipse, p (Fig. 6c). The average values of R were adopted.

Fig. 6

Analysis of rotation vector of human eye movement. a Center of eye rotation in a two-dimensional image plane. Ideally, multiple minor axes of a pupil ellipse intersect at a single point. The point is the center of eye rotation in a two-dimensional plane, o. b Calculation of the intersection of multiple minor axes of a pupil ellipse. The multiple minor axes of the pupil ellipse did not intersect at a single point. The intersection of the multiple minor axes was determined as the point corresponding to the minimum sum of the squares of the distances between the point and the minor axes. c Calculation of the length of the radius of rotation of the center of the pupil. θ is the eye rotation angle from the eye position during frontal vision. r is the length between o and the center of the pupil ellipse, labeled p in the image. R is the length of the radius of rotation of the center of the pupil. d Calculation of the length of the radius of rotation of an iris freckle. When an iris freckle is on the same plane as the pupil edge, the radius of rotation, R′, can be calculated. a is the distance between the center of the pupil ellipse and the center of gravity of an iris freckle in the image, b is the distance between the center of the pupil ellipse and point c in the image, and point c is the point of intersection of the edge of the pupil ellipse and the line connecting of the center of the pupil ellipse and the center of gravity of an iris freckle in the image

Next, we calculated the length of the radius of rotation of an iris freckle (R′). When an iris freckle is on the same plane as the pupil edge, the following formula is true (Fig. 6d):

$$R^{{{\prime }2}} = R^{2} + \left( {\frac{a}{b} \cdot \frac{\text{length of major axis of pupil ellipse}}{2}} \right)^{2}$$

where a is the length between the center of the pupil ellipse, p, and the center of gravity of an iris freckle, b is the distance between p and point c in the image, and point c is the point of intersection of the edge of the pupil ellipse and a line connecting p and the center of gravity of an iris freckle. The average values of R′ were adopted. From the two-dimensional coordinates of the center of the pupil (yp zp) and the center of gravity of an iris freckle (yi zi), we were able to reconstruct the three-dimensional coordinates of the two points in the head-fixed coordinate system as $\left( {\begin{array}{*{20}c} {\sqrt {R^{2} - \left( {yp - yc} \right)^{2} - \left( {zp - zc} \right)^{2} } } & {yp - yc} & {zp - zc} \\ \end{array} } \right)$ and $\left( {\begin{array}{*{20}c} {\sqrt {R^{{{\prime }2}} - (yi - yc)^{2} - (zi - zc)^{2} } } & {yi - yc} & {zi - zc} \\ \end{array} } \right)$. The three-dimensional head coordinates are illustrated in Fig. 1. First, given the 3D coordinate of the center of the pupil (xpr ypr zpr) and the 3D coordinate of the iris freckle (xir yir zir), the cross product of the two vectors is (xcr ycr zcr) when the eye is in the reference position. Similarly, if the 3D coordinate of the center of the pupil is (xpa ypa zpa) and the 3D coordinate of the iris freckle is (xia yia zia), then the cross product of the two vectors is (xca yca zca) when the eye is in a given eye position. Then, the matrix $\left( {\begin{array}{*{20}c} {R11} & {R12} & {R13} \\ {R21} & {R22} & {R23} \\ {R31} & {R32} & {R33} \\ \end{array} } \right)$, which represents the given eye position, can be calculated as follows:

$$\left( {\begin{array}{*{20}c} {R11} & {R12} & {R13} \\ {R21} & {R22} & {R23} \\ {R31} & {R32} & {R33} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} {xpa} & {xia} & {xca} \\ {ypa} & {yia} & {yca} \\ {zpa} & {zia} & {zca} \\ \end{array} } \right)\left( {\begin{array}{*{20}c} {xpr} & {xir} & {xcr} \\ {ypr} & {yir} & {ycr} \\ {zpr} & {zir} & {zcr} \\ \end{array} } \right)^{ - 1} .$$

Finally, the rotation vector r, which represents the given eye position, can be calculated as follows:

$${\mathbf{r}} = \frac{1}{{1 + \left( {R11 + R22 + R33} \right)}}\left( {\begin{array}{*{20}c} {R32 - R23} \\ {R13 - R31} \\ {R21 - R12} \\ \end{array} } \right).$$

Appendix 2: Calculation of linear acceleration in the tangential direction

For eccentric rotation, we assumed that the amplitude of rotation could be described by $\frac{v}{2\pi f}\sin (2\pi ft)$, where v is the maximum angular velocity and f is the frequency of rotation. The radius of rotation is denoted as r and acceleration due to gravity is denoted as g. The linear velocity vectors for space- and head-fixed coordinates (Fig. 7) are as follows:

Fig. 7

Space and head coordinates. Xs, Ys, and Zs are three axes in space coordinates. X, Y, and Z are three axes in head coordinates. This schema is used during RE-ER facing the ground

Linear velocity vector:

For space coordinates:

$${\mathbf{V}}_{{\mathbf{s}}} = \left( {\begin{array}{*{20}c} { - rv\cos (2\pi ft)\sin \left\{ {\frac{v}{2\pi f}\sin (2\pi ft)} \right\}} \\ {rv\cos (2\pi ft)\cos \left\{ {\frac{v}{2\pi f}\sin (2\pi ft)} \right\}} \\ 0 \\ \end{array} } \right).$$

For head coordinates:

$${\mathbf{V}}_{{\mathbf{h}}} = \left( {\begin{array}{*{20}c} 0 \\ 0 \\ { - rv\cos (2\pi ft)} \\ \end{array} } \right).$$

Linear acceleration vector:

For space coordinates:

$${\varvec{\upalpha}}_{{\mathbf{s}}} = \left( {\begin{array}{*{20}c} { - rv^{2} \cos^{2} (2\pi ft)\cos \left\{ {\frac{v}{2\pi f}\sin (2\pi ft)} \right\} + 2\pi frv\sin (2\pi ft)\sin \left\{ {\frac{v}{2\pi f}\sin (2\pi ft)} \right\}} \\ { - rv^{2} \cos^{2} (2\pi ft)\sin \left\{ {\frac{v}{2\pi f}\sin (2\pi ft)} \right\} - 2\pi frv\sin (2\pi ft)\cos \left\{ {\frac{v}{2\pi f}\sin (2\pi ft)} \right\}} \\ g \\ \end{array} } \right).$$

For head coordinates:

$${\varvec{\upalpha}} = \left( {\begin{array}{*{20}c} g \\ { - rv^{2} \cos^{2} (2\pi ft)} \\ {2\pi frv\sin (2\pi ft)} \\ \end{array} } \right) = \left( {\begin{array}{*{20}c} g \\ { - \frac{1}{2}rv^{2} \left\{ {\cos (2\pi \cdot 2f \cdot t) + 1} \right\}} \\ {2\pi frv\sin (2\pi ft)} \\ \end{array} } \right)$$

$$\therefore {\dot{\mathbf{\alpha }}} = \left( {\begin{array}{*{20}c} 0 \\ {4\pi frv^{2} \sin (2\pi ft)\cos (2\pi ft)} \\ {4\pi^{2} f^{2} rv\cos (2\pi ft)} \\ \end{array} } \right).$$

As the frequency of centripetal linear acceleration $- \frac{1}{2}rv^{2} \{ \cos (2\pi \cdot 2f \cdot t) + 1\}$ is 2f, the frequency of centripetal linear acceleration is twice the frequency of rotation.

Angular velocity vector:

For space coordinates:

$${\varvec{\upomega}} = \left( {\begin{array}{*{20}c} 0 \\ 0 \\ {v\cos (2\pi ft)} \\ \end{array} } \right).$$

For head coordinates:

$${\varvec{\upomega}}_{{{\mathbf{scc}}}} = \left( {\begin{array}{*{20}c} {v\cos (2\pi ft)} \\ 0 \\ 0 \\ \end{array} } \right).$$

Using the above vectors, translational acceleration f _h can be calculated as follows (Angelaki et al. 1999):

$${\dot{\mathbf{f}}}_{{\mathbf{h}}} = {\varvec{\upomega}}_{{{\mathbf{scc}}}} \times {\mathbf{f}}_{{\mathbf{h}}} + {\dot{\mathbf{\alpha }}} - {\varvec{\upomega}}_{{{\mathbf{scc}}}} \times {\varvec{\upalpha}} = {\dot{\mathbf{\alpha }}} + {\varvec{\upomega}}_{{{\mathbf{scc}}}} \times ({\mathbf{f}}_{{\mathbf{h}}} - {\varvec{\upalpha}})$$

${\varvec{\upomega}}_{{{\mathbf{scc}}}} \times ({\mathbf{f}}_{{\mathbf{h}}} - {\varvec{\upalpha}}) = 0$, because ${\varvec{\upomega}}_{{{\mathbf{scc}}}} \| {\mathbf{f}}_{{\mathbf{h}}} - {\mathbf{\alpha }}$

$$\therefore {\dot{\mathbf{f}}}_{{\mathbf{h}}} = {\dot{\mathbf{\alpha }}}$$

$$\therefore {\mathbf{f}}_{{\mathbf{h}}} = \int \limits_{0}^{\text{t}} {\dot{\mathbf{f}}}_{{\mathbf{h}}} {\text{dt}} = \int \limits_{0}^{\text{t}} {\dot{\mathbf{\alpha }}}{\text{dt}} = \left( {\begin{array}{*{20}c} 0 \\ { - rv^{2} \cos^{2} (2\pi ft)} \\ {2\pi frv\sin (2\pi ft)} \\ \end{array} } \right).$$

Using the above vectors, linear acceleration in the tangential direction ${\dot{\mathbf{V}}}_{{\mathbf{h}}}$ can be calculated using the formula ${\mathbf{f}}_{{\mathbf{h}}} + {\varvec{\upomega}}_{{{\mathbf{scc}}}} \times {\mathbf{V}}_{{\mathbf{h}}}$

$$\because {\mathbf{f}}_{{\mathbf{h}}} + {\varvec{\upomega}}_{{{\mathbf{scc}}}} \times {\mathbf{V}}_{{\mathbf{h}}} = \left( {\begin{array}{*{20}c} 0 \\ 0 \\ {2\pi frv\sin (2\pi ft)} \\ \end{array} } \right) = {\dot{\mathbf{V}}}_{{\mathbf{h}}} .$$