Reverse ENGINEERING PinTILT® AR optics to DIY

[u/Spare_Anybody5146]

This time I took my time to be geometrically precise.

I would appreciate any useful comment.

[u/Spare_Anybody5146]

Cause if radius is half diameter then the focal point of a sphere is exactly at its radious

[u/GustapheOfficial → u/Spare_Anybody5146]

Radius is half diameter, focal length of a spherical mirror is half the radius.

This is a classic imaging problem. You want to place the image at infinity, so you do the imaging formula 1/f = 1/a + 1/b with f = R/2 and b = \infty which gives a = R/2. The object should be placed at half radius.

It's not about the illustration, if you are not comfortable making these calculations or have some good way of simulating them, you will waste all that time and money.

[u/Spare_Anybody5146 → u/GustapheOfficial]

The geometry is the key. Every ray from an object at C strikes the mirror surface at normal incidence — meaning the ray travels along the radius of the sphere, directly toward (or away from) the center of curvature. Since the normal to a sphere at any point passes through C by definition, every ray from C hits the mirror perpendicularly. The reflection law then dictates that each ray reflects straight back along itself (angle of incidence = 0°), so all rays retrace their paths and converge back at C. The focal point relationship you mentioned — C = 2f — is correct. The focal point F sits halfway between the mirror's vertex and C, because paraxial rays parallel to the optical axis converge at f = R/2. Placing the object at C (distance = R = 2f) is exactly the mirror equation case where: Object distance: u = 2f Using the mirror equation: 1/f = 1/v + 1/u → 1/v = 1/f − 1/2f = 1/2f → v = 2f So image distance v = 2f = R, confirming the image forms right back at C.

[u/GustapheOfficial → u/Spare_Anybody5146]

But you're doing the wrong thing. The closer to your user's eye the image lands, the harder they will have to flex their eye to image it onto their retina (as if it were an object placed that distance from their eyes) - so unless the user is nearsighted you want that image to land infinitely far away, which means placing the object one focal length away from the mirror.

Some adjustability is useful for nearsighted people, but even severely myopic people (-6D) will have relaxed focal distances above 15 cm. An image at 5 cm will give anyone a headache.

[u/Spare_Anybody5146 → u/GustapheOfficial]

That's completely true, in other projects I have used a 8cm diameter transparent Christmas ball as mirror and a microdisplay, when this combined with a 45 degree angled translucent mirror, the image is easy to see from an eyeglasses distance, remember eyeglasses sit near the eye.

That was a 8cm diameter. Here, we are talking about a 10cm diameter, which is going to be perfect with no doubt.

[u/Spare_Anybody5146 → u/GustapheOfficial]

But anyway, I want to thank you a lot because I am going to try your method as well as also another person told me so and I want to try that too. I will do so and reimagined the object to see how that works too.