Need to know more about disparity formula about sphericalviewsynthesis HOT 4 CLOSED

Hi Jarrome,

indeed the notation is confusing.
To set it straight:

• The angular disparity gamma, corresponds to the displacement of the phi and theta angles, i.e. gamma = (phi_displacement, theta_displacement).
• The partial derivatives relate the angular displacements to the cartesian displacements.
• For the vertical disparity case, we only have displacement along the y-axis.
• In addition, the longitudinal partial derivative w.r.t the displacement along the y-axis is zero, thus the displacement of phi is zero, i.e. phi_displacement = 0.

We are then left with angular disparity being (0, theta_displacement), with the latter being theta_displacement = -sin(theta) * y_displacement / r.

from sphericalviewsynthesis.

Hi, zuru,

Thank you. But that's not what I'm confusing.

Actually, the y_displacement is the baseline on y-axis, r is the depth.
So my understanding is, if we follow the Law of sine, the formula should be like

sin(theta_displacement) = -sin(theta) * y_displacement / r instead.

Maybe I miss something? Or the ignoring of sine is just for the convenience of deep learning?

from sphericalviewsynthesis.

Hi Jarrome,

can you please check the law of sines document in #2 can then update the formula using the notation (angles and segments) in that document?

I think the confusion arises from the frame of reference, i.e. which r and theta are used, and the fact that they are not aligned (i.e. bottom/top and top/bottom).

from sphericalviewsynthesis.

Hi, zuru,

Yes, zuru, the sin(theta_displacement) = -sin(theta) * y_displacement / r might be inaccurate but my idea is theta_displacement should be solved with arcsin or in your law of sines document Eq. (7) a arctan to solve the \gamma.

While the paper is with theta_displacement = -sin(theta) * y_displacement / r, that's what I'm confusing.

from sphericalviewsynthesis.