# Binocular Disparity

Hello,
I have a problem with calculating distance while receiving depth perception. I hope somebady could help me with this problem.
My fixation point (3 D TV) is located 5 m away from me. How do I calculate the distance of objects just visible before or after the fixation point? For example with a stereo acuity of 15 arcsec and an interocular distance of approx. 6.3 cm? So, for example, can I still see an object at a distance of 4.8 m? How would this be calculated?

“see”? sure. everything’s visible.

to distinguish distances? different question. if it’s two points on a flat (but angled) surface, there’s a lot more going on than just trigonometry. relative differences are probably resolvable a lot better. perhaps talk to a medical researcher.

given z=5~m distance, d=63~mm IPD, and and ordinarily parallel optical axes, that is 0.361 degrees off axis for each eye because \tan(\alpha) \cdot z = {d \over 2}

now \tan(\alpha \pm 15 ~\text{arcsec}) \cdot z' = {d \over 2} (both eyes turn that much)

• both eyes -15 arcsec: z’ = 5.058 m
• both eyes +15 arcsec: z’ = 4.943 m

so you might expect about 2.9 cm of depth resolution at 5 meters. that’s because we have a difference of four times 15 arcsec in these numbers and the difference is about 11.5 cm. don’t make me use derivatives.

that’s absolute depth, in the absence of any other depth cues.

15 arcsec at 5 meters is about 0.36 mm, so your display should be better than that.

yes I found some publications for your name