1
$\begingroup$

I am tracking eye movements in mice passively viewing naturalistic stimuli or experimental stimuli meant to elicit reflexive eye movements. I would like to define a transformation function that translates displacement in pixels (x, y) to degrees of visual angle (azimuth, elevation). Here is an example image:

Example Image

I have seen this paper referenced several times in other papers that track eye movements in mice, but it seems like the position of my cameras relative to the eye is different than what is shown in the paper and I worry this will affect the validity of the transformation.

I have the ability to track any feature of the eye that is useful. Right now I am tracking the center of the pupil, the corners of the eye, and the center of the top and bottom eyelid.

EDIT #1

Here is a description of the camera's position relative to the animal. The solid black object represents the animal (A=anterior, P=posterior, D=dorsal, V=ventral). The angle from each orientation represents the angle from the animal to the camera. enter image description here

$\endgroup$
1
$\begingroup$

In theory you can use the geometry of your setup to transform your data to the orientation reffered in the paper you mentioned using Scheimpflug Principle and then use the methods provided in it.

Practically, you can only approximate the geometry of your setup. If you have some idea about angles in something like this

enter image description here

please let me know, I might be able to do the math for you.

EDIT: The lens plane is parallel to image plane in cameras so now it is just the question of orientation of subject plane and lens plane(camera plane).

$\endgroup$
1
  • $\begingroup$ Thank you for your comment! I'm not sure I totally understand how to fill in the value of the angles in the figure you attached to your response, but I updated my original post to include a diagram of the position of the camera relative to the animal/eye. Is this information sufficient? $\endgroup$
    – jbhunt
    Jan 2 at 23:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.