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[–] 44 points 8 months ago (4 children)

I assumed he'd estimated it based on how distorted the face appears behind the glasses. I do that all the time.

At this angle it's hard for me to do that, since I usually use the edges of the face to estimate it. negative glasses pull the line inwards, positive outwards. I can reliably tell when someone is wearing fake glasses (0 strength) for example, and probably estimate strength within 30% of the actual value.

If the image was higher res maybe I could estimate this case too. Or this professional optometrist is just a lot better at it than I am.


Strong negative glasses: (Note the faces contours in the glasses appearing well inside the faces contours around the glassed)

Fake glasses:

Positive glasses:


PS: Searching for generic terms yields 100% fake glasses, so I took a specific person I remember having strong glasses for myopia.

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  • [–] 7 points 8 months ago

    This is so cool. Thank you for sharing how to do that.

    At this angle it's hard for me to do that, since I usually use the edges of the face to estimate it

    This is a screenshot of a photo, but you can still see the left side offset if you zoom in. The original photo likely has much better quality to see it with.

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