Forum Discussion
Pixel Convergence Math
Given IPD, and horizontal screen resolution (2160), how might one calculate the distance at which pixels converge and stereoscopy is lost?
This is complicated by the differing angular resolutions from the center to the edge of the screen. With anti aliasing and head wobulation you also get a bit of information from beyond the angular resolution of the pixel.
But if we simply devide 108 degrees by 1080 to get an angular resolution of 10 pixels per degree. do a simple trigonometry calculation with https://www.carbidedepot.com/formulas-trigright.asp a tringle with a .1 degree arc(angle A) and a width of 64mm(average IPD(use side a)) is 36 meters long, so that should be about the distance where objects are lined up with the same pixels in both eyes.
But if we simply devide 108 degrees by 1080 to get an angular resolution of 10 pixels per degree. do a simple trigonometry calculation with https://www.carbidedepot.com/formulas-trigright.asp a tringle with a .1 degree arc(angle A) and a width of 64mm(average IPD(use side a)) is 36 meters long, so that should be about the distance where objects are lined up with the same pixels in both eyes.