What is the point of this? Do you really want to know the answer? Are you quizzing people?
Chesapeake & Midlantic | 03/11/14 04:21pm
Well, 25.4 divided by 3 is 8.46..., so that.
IPVMU Certified | 03/11/14 05:38pm
In other news, a 2"X4" board doesn't really measure 2 inches by 4 inches. But we know it's smaller than a 4"x6", which doesn't measure that way either.
Nominal dimensions have a way of becoming a standard metric over time. Call it right, wrong, or indifferent... the metric is what it is.
I'm not seeing the point here. Maybe we should talk about effective pixels vs. actual pixels. Or why the image sensor is square when the lens is round. OR NOT.
But I cheated and read the answer because I thought I saw a snake in the grass.
Ten-gallon hat tip to Ari for having enough guts to actually give a real answer!
Rukmini, what are you using to calculate the conversion from inches to millimeters? My calculator says 1/3" (0.333") equals 8.4582 millimeters. Multiple online calculators also say 0.333" = 8.46mm. For figuring the width and height, obviously it depends on whether the sensor is 4:3 or 16:9 (or 16:10 as a few are). For that, remember that the diagonal measurement equals the square root of the sum of the squares. For 4:3 that would be 3h x 4w = 5d so divide the diagonal measurement by 5 and multiply it by 4 for width or 3 for height.
For 16:9 sensors, 16 squared + 9 squared = 337. The square root of that is 18.36 so width would be diagonal measurement divided by 18.36 and multiplied by 16 and height would be diagonal measurement divided by 18.36 and multiplied by 9.
What a great discussion! Who could have guessed that imager size would be overstated by about 50%? It totally invalidates resolution calculations that ignorantly assume stated imager size bears some relation to a physical dimension. Trying to think of charitable interpretations... Could stated size include blade kerf that arises when separating chips from a wafer? Or perhaps it refers to standard size of the chip mounting package?