Reply to thread

Message: <blockquote data-quote="Felisek" data-source="post: 304176" data-attributes="member: 23887">Some of your statements are not entirely precise. At the end of your post you mixed up linear dimensions with an area. Let me explain.Image resolution is always considered in linear space, i.e., how many pixels you have in a given length. For example, printer resolution would be expressed in dpi - dots per inch. Screen resolution is often given in ppi - pixels per inch. There are 6000 pixels across image in both sensors, so pixel sizes are 36 mm / 6000 = 0.006 mm for D610 and 24 mm / 6000 = 0.004 mm. So each pixel in D610 is a box with a side of 0.006 mm and each pixel in D7100 is a box with a side of 0.006 mm. This is much larger than you imply. What you really quote in your post is not pixel size in mm, but its area in sq. mm. Sorry for picking on these details, I'm just a scientist and I couldn't resist... <img src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" class="smilie smilie--sprite smilie--sprite11" alt=":rolleyes:" title="Roll eyes    :rolleyes:" loading="lazy" data-shortname=":rolleyes:" />The ratio of the pixel size is 3:2, so you could say that pixel in D610 is 50% larger than in D7100, or that pixel in D7100 is about 67% of that in D610. This also means that the resolution (in pixels per inch or millimetre) in D7100 is 50% better than in D610. The actual resolutions are about 423 and 635 ppi in D610 and D7100, respectively.However, the area of each pixel is its squared linear size, so area ratio is 9:4. Which means that the D610 pixel has 225% larger area than the D7100 pixel, or that the area of the D7100 pixel is 44% of the D610 pixel. This also means that each pixel of D7100 receives 44% light of the D610 pixel.Your conclusion is absolutely right, just wanted to clarify some tiny inaccuracies.</blockquote>

Verification