View Full Version : How many MP (GP) is an 8X10 equivalent to?
The recent post about the scanning back for 8X10 and the 1 gigapixel Bryce Canyon image got me wondering how many maga pixels a 4X5 and 8X10 transparency would be equivalent to. I know that film grains are not equivalent to pixels, so it is not directly comparable, and it would obviously vary depending on film. Let's just say, how many mega pixels would a camera have to deliver to match the resolution of 4X5 and 8X10 Velvia(the old iso 50)? I had heard one statistic of 400 MP to match an 8X10 transparency. Any one have any ideas? Thanks!
In general, you can retrieve just about all the information captured by a photographic film by scanning around 4000 ppi. Clearly, this number must vary based on the film and it's internal structure. You can get into religious wars debating this stuff, and whether negative film is sharper than tranny film, etc. and I'm not going there with anyone.
Basically, pick a number you like. Then do the math:
(4000 ppi)(4 in)(4000 ppi)(5 in) = 320 Mpixels.
(4000 ppi)(8 in)(4000 ppi)(10 in) = 1280 Mpixels.
A low grain high resolution film like Velvia probably can be scanned at even higher scanner resolution to good effect. Therefore, IMHO, your estimate of 400 Mp for an 8x10 tranny is quite low. Of course, YMMV.
Great question. Now, let's see if anyone actually gives a straight answer.
I would think that to capture all the information on film, you would need 5000dpi. This is the standard scanning resolution offered by Wci for scanning 35mm.
One has to keep in mind that the pixels obtained from scanning film are quite grainy/noisy compared to those obtained with digital capture. Because of that, while a 4000dpi scan from 35mm has more than 20 Mp, almost everybody agrees that a 11 Mp image direct from a 1Ds camera is superior. Apparently a lot of people even prefer a direct 6 Mp image, although clearly the resolution is less.
You can scan as high as your scanner's resolution will go. A Linocolor Tango will easily produce a 2 gig file from an 8x10, and I don't see why you couldn't go to 4 gig with Mac OSX. Whether you could do anything practical with a 4 gig image file is quite another thing. FWIW, I just produced a 14 x 44 foot billboard - they print these at 9 dpi (not a typo) - so the stuff you see on the highway is never more than 50 mb. There isn't much point in going for higher and higher resolutions if you know what your final output is going to be. Even if you're Thomas Struth or Richard Avedon, and want to print expensive 8 foot tall custom prints, a 4 gig image would be way overkill (and quite a slog to RIP.)
Straight answer - a gig scan of an 8x10 should cover all your potential needs and then some. Most people agree that 35mm scans of 50-60 mb are the max, and 8x10 is simply 20 35mm slides tightly packed. 20 x 50mb = 1 gig.
The fact that scanning with 4000 dpi captures all information that is present on the film doesn't mean that so many information is really there, especially as you go to greater formats. A 35mm slide may contain as much as 3000 dpi, but a MF slide will much probably contain only 2500 dpi or less, a 4x5 would be 1800 dpi and 8x10 1200 dpi (don't pay too much attention to the values, this is just to illustrate my argument).
Hence, I would suggest another approach. The question is: how many line pairs/mm (lpm) do we have on our slides? Let us say 30 lpm for 8x10 (and I'm not even sure that we always get so many). One line pair is equivalent to two pixels (one black, one white in a simplified model). Hence, the total number of pixels: (pixel count in the length x pixel count in the height of the slide) (one inch=25,4 mm) (30x2x10x25,4)x(30x2x8x25,4) = 185 MegaPixels
Take any value you want for the resolution on your particular film and you will obtain a different count. As for myself, I would say that a standard 8x10 is around 100 MPix. By the way, I recently read an ad' for a 120 MPix scanning digital back that says that the resulting image contains more information that a 8x10 slide...
How much detail one can get from a scanned transparency is avoiding the point. The question is how many megabites from a digital camera does it take to show the same amount of detail that one can see (with suitable magnification) on a 4x5 or 8x10 Velvia transparency.
Bill, multiply the ratio of surface areas to 35mm (which is about 15 for 4x5 and 60 for 8x10) by the number of pixels in a digital camera that you think matches 35mm film (something between 5mp and 11mp). For 8x10, by this calculation, the low estimate would be 300 megapixels. Color file size is 3x number of pixels (one byte each for RGB). This doesn't take into account possible loss in the 8x10 due to lens diffraction limits, film imperfect flatness, etc...
What size are you going to print to? Up to at least 16"x20" a 16 to 22 MP ( kodak, leaf valeo, imacon) back will at least equal the amount of detail you can get out of 4x5 transparency like Astia 100F, which has more resolution than RVP. If you think I'm wrong, prove it and not with calucations but with images.
Ellis, up to at least 0.1" x 0.15" a desktop webcam (logitech, apple, ...) will equal the amount of detail you can get out of 4x5 transparency like Astia 100F, which has more resolution than RVP :-).
QT; laughing out loud but thank you for reinforcing my point -- although you go in the opposite direction. Is there a webcam that is 12 14 or 16 bit per channel capable?
I can resolve lettering that is less than 1/10 of a mm tall on a 4x5 negative, close to the edge of the image circle. That is with an old camera and a single coated lens. If you assume that it takes at least 10 rows of pixels to resolve a letter, then a 35mm negative is about 10 meg. That would put 8x10 over 500.
Research done at Cornell would tend to support this.
I just played with shooting a full unfolded sheet of newspaper with a Argus C3 and resolved lettering (on a drug store print) that was about .09mm tall on the negative.
I think using legibility as a benchmark is better than debating the validity of MTF.
Most average people would tend to agree that a medium that retained legibility at a given reproduction ratio contained more information than one that didn't.
"FWIW, I just produced a 14 x 44 foot billboard - they print these at 9 dpi"
What, you mean people don't climb up to the billboard with a loupe, to examine the details and check for grain?
(4000 ppi)(4 in)(4000 ppi)(5 in) = 320 Mpixels
(4000 ppi)(8 in)(4000 ppi)(10 in) = 1280 Mpixels
To make such a high-res scan may be a little overkill, since you can get a very sharp image by printing out at 300-400 dpi. I doubt that most of us could tell the difference, once a print gets up past 360 dpi.
So let's say you scan at 4000 ppi, and print at 400 dpi. That means your print will be a 10x enlargement, an 80x100 inch image. That may be larger than you can output to any affordable printer, or hang on the wall in your home, for that matter.
If you only want a 16x20 print, that's a 2x enlargement: you could scan at 800 ppi, and print out at 400 dpi. The 800 ppi file would be 146 megabytes for color, 49 megabytes for B&W.
If you scan at 1600 ppi and crop a little, you still have plenty of pixels to toss out, and still make a very fine 16x20 image. The size of that 1600 ppi file would be 586 Megabytes for color, 195 for B&W.
A 1200 ppi file, allowing a 3x enlargment, would be 330 megabytes for color, and 110 megabytes for B&W.
You can print out to fine art paper, or print a negative onto a transparency material, and make Platinum/Palladium prints, or other alternative process. You can also remove blemishes & dust spots, and control contrast & brightness to an almost infinite degree, without using a single sheet of paper.
"FWIW, I just produced a 14 x 44 foot billboard - they print these at 9 dpi"
Lear: O reason not the need: our basest Beggers Are in the poorest thing superfluous. Allow not Nature, more then Nature needs: Mans life is cheape as Beastes.
is that what they mean by "beaste boyes"?
There are in fact three questions within this question, depending on what you are trying to do:
1) How many pixels in a digital camera would give an image of the same apparent quality as a LF print.
2) How many pixels must we scan a LF neg. so that the digital print looks subjectively "as good as" the chemical print.
3) How many pixels are required to capture every detail on the LF neg.
In recent times I have done some investgation on 1). The figures quoted for the digital resolution of a camera to match 35mm seem to lie around 10 Mpixel. So if we expand this to LF sizes
4x5 = approx 15 times 35mm size = 150 Mpixel
8x10 = approx 60 times 35mm size = 600 Mpixel
Of course one could argue that the relationship of LF and 35mm is not a simple linear one, so that the above figures are optimistic or pessimistic.
You know that, and I know that. But the poster asked how many mega pixels would a camera have to deliver to match the resolution of 4X5 and 8X10 Velvia(the old iso 50)?
The point is, there's a hell of a lot of information on a piece of 8x10 film. Whether or not it is all usable information is where the religious arguments start. Personally, I agree with you - I've never scanned anything to the file sizes I'm talking about (at 16bit, grayscale, that's 640MB for the 4x5, and 2.5GB for 8x10. It would take my drum scanner over 5 hours to scan an 8x10 at that resolution!)
Personally, I have to have a really first class image, both aesthetically and technically, before I'll go higher than a 10x enlargement on any film, 35mm to 4x5 (sorry, I'm too old and weak to haul 8x10 equipment up and down the mountains anymore ;-).
hmmm...someone asked if anyone would post a straigt answer... I'll try.
The question was pretty straight forward: "how many mega pixels would a camera have to deliver to match the resolution of 4X5 and 8X10 Velvia(the old iso 50)?"
The answer is equally straight forward...
PPI = 2R*25.4 (ok 50.8R for the sticklers out there).
Where R is the resolving power of the film expressed in line-pairs per mm.
Its a simple application of the Nyquist Theorem... paraphrased, in a digital system, the sampling rate needed to reproduce an analog signal is 2 times the frequency of that signal. In this case the resolving power of the film is the frequency.
I don't know the resolving power of Velvia 50, but I would guess its somewhere between 100-150 lpmm, so a digital sensor that exactly replicated its resolving power would need between 5000 and 7550 ppi... so, at least 500MP for 4x5.
Of course, the real goal is to produce the same results in the final product. The question becomes, in a particular digital system, how much resolution is required in the capture media to produce results with the same apparent resolution in the final product as a film based system that uses film X as the capture media?
To answer this, you need to also know the impacts of the print production methods on resolution. In a traditionaly enlarged photo print, the paper and the enlarging system (and even the chemistry) have resultion factors that are part of the overall equation. Digital printing is completely different, but each technique has its own set of resolution impacting attributes. Thus, to get the same resolution in the final print, the resolution required of the capture media will be different between film and digital systems.
Hogarth - You are right ! I missed the question itself. Oops ! :-)
Scott, you're right, but way too optimistic, too! The limiting factor in 8x10 is rather the lens, not the film. Take a look at the Schneider's MTF curves for similar lenses of different focal lengths, e.g. Apo Symmar 150 and 300. You will see how much coverage goes to the expense of resolving power. By the way, at f/32, diffraction only limits the resolving power of the lens to about 45 lpm. Velvia can resolve about 60 lpm with real life, low-contrast subjects. Combine both with the (1/D)2 rule and you get 36 lpm as a resultant resolution on the film! And we don't even take care of depth of field, film planeness and other real life effects...
Powered by vBulletin® Version 4.2.0 Copyright © 2013 vBulletin Solutions, Inc. All rights reserved.