viewfinder coverage, image cropping and focal length

[jacobmiles]

<p>Hi guys. The short version of my question is this:<br>

If I use a camera with less than 100% viewfinder coverage (e.g. 96%), then in order to reproduce the image I actually saw through the viewfinder, can I always crop the same number of pixels off the final image, or does the number of pixels I need to crop from the border vary depending on the focal length and focusing distance of the lens used to take the shot?<br>

The long version of the question:<br>

100% viewfinder coverage tends to cost noticeably more - only the higher-end cameras offer it - and I'm trying to decide between the Pentax K-x, which offers less than 100% coverage, and the K-7, which offers 100% coverage but if twice the price and looks bigger and heavier. The only important difference ot me is the 100% viewfinder, so that what I see at the time of the shot is the same thing I see when I look at it the next day, when I'm not in the same frame of mind and don't remember precisely where I meant to crop the image when I pressed the shutter button.<br>

However, I don't need 100% coverage at all if I can reliably and mechanically reproduce the image I really saw through the viewfinder the next day. So, if the number of pixels I need to crop off the border doesn't change with the focal length or focus of the lens, then I can do a test once by photographing graph paper or a ruler, and determine the number of pixels on each edge that the viewfinder isn't showing me. Then I can write a script that calls ImageMagick on all of my images after I've shot them, automatically cropping en masse to match the viewfinder image. Much less expensive, a lighter and smaller DSLR, and no difference in my ability to precisely compose an image in the moment.<br>

But I can't figure out whether the cropping would have anything to do with the lens, i.e. whether I would need to crop a different number of pixels off the edges depending on the focal length and focusing distance of the lens when I shot the picture. I think the answer is no - that it's always the same number of pixels - but I wasn't sure. And I don't currently have a digital camera with a zoom lens, so I can't quickly test it out. If no one knows I'll do a test with a film camera and some different length primes, but I'd appreciate any input.<br>

Thanks for any help.<br>

- Jake</p>

[Matt Laur]

<p>95% is 95%, regardless of focal length.<br /><br />But more to the point: it's not like you're recording an image with <em>less</em> than what you saw through the viewfinder, so I'm not sure how serious an issue this really is, in practice.</p>

[jacobmiles]

<p>Thanks for the response Matt. That saves me some time, money and bulk.</p>

<p>And yes, you don't lose any image data, so it wouldn't matter in most circumstances or to most people, but one piece of information I'm often interested in that I do lose is what exactly I was looking at and how I meant to frame the shot when I pressed the shutter. Now I can determine that reliably.</p>

<p>Thanks for the quick response!</p>

<p> </p>

[alec_myers]

<p>It should be the same 95% regardless of lens, as Matt says, I think. But it needn't be the centre 95% of the frame, and it may vary camera-to-camera.</p>

[jim_momary]

Jacob -

Let's look at it as if we were in Plane Geometry back in HS. Suppose you print a 8.50 x 11.00 inch image out. That is exactly 92.00 sq. inches.

 

96% of that represents 88.32 sq. inches. If we say, arbitrarily, that the 96% rediuction is symmetrical, then if we now assume the long side drops to 10.75 inches, then using 88.32 as the area, the short side would become 8.22 inches. (8.22x10.75=88.32)

 

So, our 8.500 x 11.00 >>> 8.22 x 10.75. That is a change but not that much of one unless super critical what-you-see-is-what-you-get framing is demanded. Since your viewfinder provides less than is actually recorded, you can always later crop out that darn garbage can or Aunt Mathilda.

My cornfussed 2 cents - Jim