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<blockquote data-quote="WayneF" data-source="post: 199517" data-attributes="member: 12496">I fully agree, to extent that there is no change whatsoever in the lens.&nbsp; DX simply crops away some of it, to be a more narrow view.&nbsp; But to view that image, we can only show it much larger, enlarged on a video screen or a paper print.&nbsp;&nbsp; Assuming we framed the two images the same (to be able to compare them meaningfully), the smaller cropped DX image must be enlarged 1.5x larger than the FX image (just to be the same viewed size).&nbsp; This enlargement of a narrow view is perceived as a telephoto effect, and we compare the DX view as 1.5x greater EFFECTIVE FX focal length (meaning DX must stand back 50% farther to see the same view).&nbsp; That's pretty much the entire idea of it.Pixels per sensor area is only important to noise.&nbsp;&nbsp; Pixels per viewed area is another thing, and it is very important, and it happens that DX always has to be enlarged 50% more than FX to see it anywhere.&nbsp;&nbsp; So.... any (dimensional) number you computed on the DX sensor must be divided by 50% before it is acceptable to compare to FX.&nbsp; That is simply how life is.&nbsp; <img src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" class="smilie smilie--sprite smilie--sprite1" alt=":)" title="Smile&nbsp; &nbsp; :)" loading="lazy" data-shortname=":)" /></blockquote>

[QUOTE="WayneF, post: 199517, member: 12496"] I fully agree, to extent that there is no change whatsoever in the lens. DX simply crops away some of it, to be a more narrow view. But to view that image, we can only show it much larger, enlarged on a video screen or a paper print. Assuming we framed the two images the same (to be able to compare them meaningfully), the smaller cropped DX image must be enlarged 1.5x larger than the FX image (just to be the same viewed size). This enlargement of a narrow view is perceived as a telephoto effect, and we compare the DX view as 1.5x greater EFFECTIVE FX focal length (meaning DX must stand back 50% farther to see the same view). That's pretty much the entire idea of it. Pixels per sensor area is only important to noise. Pixels per [B]viewed area[/B] is another thing, and it is very important, and it happens that DX always has to be enlarged 50% more than FX to see it anywhere. So.... any (dimensional) number you computed on the DX sensor must be divided by 50% before it is acceptable to compare to FX. That is simply how life is. :) [/QUOTE]

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