PD-area change due to distortion caused by off-center gaze¶

We assume that the plane of the pupil is located parallel to the camera plane and when the eye is fixating the middle of the screen $(0,0)$, then the pupil is a perfect circle with radius $r$. In that case, the recorded pupil signal will be the area of a circle

\[ A=\pi r^2. \]

However, when fixating another point on the screen $(x,y)$, there is a displacement where $\alpha$ is the displacement angle in x-direction and $\beta$ in y-direction. Because the pupil is a circle, we can reparametrize $(x,y)$ to $(x',y')=(\sqrt{x^2+y^2},0)=(d,0)$ so that the displacement has only one dimension parametrized by angle $\theta$.

When $\theta>0$ (i.e., the eye is not perfectly fixating $(0,0)$), the pupil plane is tilted relative to the camera plane. When projecting the tilted pupil onto the camera plane, the radius in this direction is shortened to $a=r\cos \theta$. Therefore, the projection of the pupil is an ellipse with parameters $a$ and $r$ and the actual area of this ellipse is

\[ A'=\pi a r=\pi\cos\theta r. \]

When fixating point $(x,y)$, the distance of this point from the center is $d=\sqrt{ x^2+y^2}$. When the eye is a distance of $h$ away from the screen, then the displacement angle is

\[\theta=\arctan\frac{d}{h}\]

such that

\[a=r\cos\arctan\frac{d}{h}\]

which wolframalpha assures us is equal to

\[a=\frac{r}{\sqrt{1+\frac{d^2}{h^2}}}\]

and therefore

\[A=\frac{\pi r^2}{\sqrt{1+\frac{d^2}{h^2}}}\]

Therefore, we can correct the raw signal $ which we assume is the area of a circle with radius $r$.