Forum Discussion
What is the angular resolution of the DK2?
I've seen people say anywhere from just under 8 to 9 pixels/degree, and that it is a mostly uniform angular resolution across the FOV. Where does this number come from?
If I try to calculate it two ways I get vastly different results. One way is to say, there is 90 degree field of view, and 960 horizontal pixels per eye. You are getting about 85% of the pixels in that FOV so
0.85*960/90 = 9.06 pixels per degree.
The screen has a PPI of 388. I know they use a 10x magnification aspheric lens.
1/388 = 0.00258 inches per pixel.
Assuming the eye is ~2" away from the screen, the angular resolution of each pixel without a lens would be 4.4301 Minutes. Magnify that by 10x and you get 44 minutes per pixel, or 1.36 pixels/degree.
Maybe my assumption about the lens isn't right. What do you think?
If I try to calculate it two ways I get vastly different results. One way is to say, there is 90 degree field of view, and 960 horizontal pixels per eye. You are getting about 85% of the pixels in that FOV so
0.85*960/90 = 9.06 pixels per degree.
The screen has a PPI of 388. I know they use a 10x magnification aspheric lens.
1/388 = 0.00258 inches per pixel.
Assuming the eye is ~2" away from the screen, the angular resolution of each pixel without a lens would be 4.4301 Minutes. Magnify that by 10x and you get 44 minutes per pixel, or 1.36 pixels/degree.
Maybe my assumption about the lens isn't right. What do you think?
7 Replies
- Don't you have to take the ppd change due to barrel distortion into account?
"jamesj" wrote:
I know they use a 10x magnification aspheric lens.
No, 10X is just a speculation considering the size of the screen and the expected FOV. Oculus VR has never advertised the magnification of the lenses."jamesj" wrote:
Assuming the eye is ~2" away from the screen, the angular resolution of each pixel without a lens would be 4.4301 Minutes. Magnify that by 10x and you get 44 minutes per pixel, or 1.36 pixels/degree.
It doesn't work like that. To calculate the FOV correctly you need to know the surface profile of the lenses - not only the magnification - and you need to use ray-tracing with the correct equations for aspheric surfaces taking into account the refractive index of the lens.
Or you can simply launch the Oculus demo, press the spacebar to see the calculated FOV, take a screenshot to calculate the percentage of the display that is used and calculate the pixels per degree with that."Fredz" wrote:
No, 10X is just a speculation considering the size of the screen and the expected FOV. Oculus VR has never advertised the magnification of the lenses.
If you take the 8-9 pixels/degree that you get doing it that way, their lens can't be magnifying it more than ~2x if the angular resolution is about constant across the field of view. Without a lens at all the AR should be ~13.54 pixels/degree at 2" away with a 51 degree FOV. That makes sense as adding the lense increases the FOV by ~1.75 and the angular size of each pixel by ~1.5-1.7. It probably isn't exactly constant across FOV, but relatively close."Fredz" wrote:
It doesn't work like that. To calculate the FOV correctly you need to know the surface profile of the lenses - not only the magnification - and you need to use ray-tracing with the correct equations for aspheric surfaces taking into account the refractive index of the lens.
I'm mostly interested in knowing if the angular resolution actually is constant across FOV."Fredz" wrote:
"jamesj" wrote:
I know they use a 10x magnification aspheric lens.
No, 10X is just a speculation considering the size of the screen and the expected FOV. Oculus VR has never advertised the magnification of the lenses.
Lens-magnification is an arbitrary term, which isn't useful in the context of the Rift. Lens-magnification effectively is the quotient of 25cm divided by the focal length of the lens."jamesj" wrote:
I'm mostly interested in knowing if the angular resolution actually is constant across FOV.
Yes, the angular resolution approximately is constant across the FOV.- It isn't constant, the lens introduces pincushion distortion, so the the pixels are squished closer together at the center of the lens.
Also it isn't really magnifying, or perhaps you could say it is magnifying by a factor of infinity. Collimated light means no focal distance, so you cant take a ratio of the focal distance on the screen side to the viewer side.
Another consequence of collimation, the pixels per degree at the center is constant as you move your eye around relative to the lens. So eye relief and IPD dont really affect it. You can confirm this by taking off the HMD and looking at the virtual image from a distance while moving the HMD around. The pixels appear just as big as when you wear it right at the center, though they distort more rapidly off center. "mrjazz" wrote:
Lens-magnification effectively is the quotient of 25cm divided by the focal length of the lens.
Actually (25 cm / f) + 1 is often used for magnifying lenses, that's the case for the DK1 lenses (7X)."owenwp" wrote:
It isn't constant, the lens introduces pincushion distortion, so the the pixels are squished closer together at the center of the lens.
The distortion leads to approximately constant pixels/angle, because spatial distances of pixels to the screen center are approximately linearly converted to angular distances. Without the distortion it wouldn't be constant, because then distances of pixels from the screen-center would be related by the tangens-function to the resulting viewing angles:
tan(alpha)=(pixel-distance to screen-center)/(distance from screen-center to lens)
