I've read How big can I print a JPEG photo that measures 5120 x 2880 pixels? and I've read What are the minimum DPI and size (in pixels) required to print good 10x15 photos? but I wonder what happens when you go huge.

I've been experimenting with huge panoramas, typically they are about 15,000 pixels across. How big can this realistically be printed? Yes, I can divide by 300 and then by 12 and get a bit over 4 feet wide. I can also require the viewer to stand so far away and not get up to close.

But printing something like this, especially printed to metal, costs an arm and a leg and I'd like to see if anyone has had real experience printing huge. Does going huge let you bend the rules? Could my 15k image be printed 8 feet wide and the sheer mass of it negate any low resolution badness?

  • \$\begingroup\$ As a thought exercise, consider a billboard on the side of a highway. Do you think they need 300 ppi? That would be excessive and most of those pixels will be wasted. If it's to be viewed from 2 miles away, you can use 2 ppi. You could if you want calculate it based on angle of view per pixel, much like retina screens are calculated. As a (very) rough estimate if you are viewing something from 1.5 feet away you can print 200 to 300 ppi. If you are only going to be viewing it from 150 feet away you can print 25 ppi no problem. \$\endgroup\$ Commented Mar 18, 2014 at 3:46
  • \$\begingroup\$ I'm not sure photo.stackexchange.com/questions/456/… is a dupe, but drfrog's answer there may answer my question here. You see, I don't care about the general rule, I think the general rule breaks down when you go huge, but I don't know what the rule his. His "45 degree" rule could be it. \$\endgroup\$ Commented Mar 18, 2014 at 12:46
  • \$\begingroup\$ If you can get ahold of somebody at Smugmug, they regularly get prints done very very large for their office. I believe @Shizam works there. \$\endgroup\$
    – rfusca
    Commented Mar 18, 2014 at 12:56

1 Answer 1


The answer to this depends on the viewing distance.

The usual rule of 300 PPI works well for close-up viewing, but even that isn't a hard-and-fast rule. What's more important is Pixels-Per-Degree (PPD), which is more representative of our eye's ability to resolve detail, and is dependent on a specific/typical viewing distance.

Apple's Retina displays (picking theirs just because there's abundant info, there are other similar concepts around) are designed based on PPD rather than PPI. Basically, PPD refers to the number of pixels spanned by a 1-degree arc from your eye, when viewing the screen/image.

The basic formulas to convert between PPI and PPD, for a given viewing distance d are:

PPD = d * PPI * 2 * tan(pi/360)     ≈ d * PPI * 0.01745
PPI = PPD / (d * 2 * tan(pi/360))   ≈ PPD / (d * 0.01745)
  d = PPD / (PPI * 2 * tan(pi/360)) ≈ PPD / (PPI * 0.01745)

Note: Assumes tan() is working in radians.

Now Apple's threshold for 'Retina' PPD is 53, though their displays are mostly 70–85, though the iPhones with highest PPI of 326 but lowest viewing distance are about 57 PPD. This roughly matches the recommendations for small photos of ~300 PPI, which are also often viewed at around 25cm (10") distance.

Others suggest up to 100 PPD (especially for people with better than 20/20 vision), so there's a large range for what is acceptable. You may need to pick your own PPD threshold here, probably somewhere between 50 and 100.

So what does this mean?

For a large print viewed at a distance, you don't necessarily need to keep the 300 PPI rule. Viewing from 3 metres away (~118") you could get away with about 30–50 PPI for 53–100 PPD (try looking at some large street advertising up close and see how huge the 'pixels' are printed).

Incidentally (and slightly off-topic), since viewing distance often increases roughly linearly with image size (for certain applications), you can potentially do away with these PPI and PPD concepts entirely for certain applications.

Typically*, images are viewed from a distance that makes viewing them in their entirety comfortable. So if the image is 10 metres (~400") wide, you won't be standing 25cm (10") from it, you'll be standing perhaps 12 metres (~470") away (giving a viewing angle of 45º, roughly the view of a 45mm lens on full-frame camera). This would allow a printed resolution of around 6.5–12 PPI! This will scale directly with the printed size, so there's simply a minimum resolution required for this given viewing angle of around 2400–4700 pixels across. If it's only going to be viewed as a whole, with a 45-degree viewing angle across the image, then the larger you print it, the further you stand back—image resolution is all that matters. And in that sense, something like an 8MP camera (2400x3200) or 15MP (3200x4800) is always enough for such purposes, regardless of print size.

*Not all images are viewed this way, hence why there's still a push for higher resolution cameras, and higher resolution images, so I'm certainly not saying high resolution doesn't matter.

And this is especially true for huge panoramas, where the whole point is to view the fine details in a huge image. Viewers will be looking closely at small areas, and so it comes back to the simple viewing distance calculations.

Now down to printing your 15k-pixel-wide panorama...

If you print it ~2.4m (8') wide, you'll get 156 PPI, which would give a minimum viewing distance range of about 0.5 to 1 metres (19–37") for 53–100 PPD.

You then have to decide whether this is acceptable, given the intended location of the print, the room it's in, access to the image, the likely audience, etc.

  • 1
    \$\begingroup\$ This is brilliant! Raises the viewing distance but also addresses it technically. Love it! \$\endgroup\$
    – BBking
    Commented Mar 18, 2014 at 0:51
  • \$\begingroup\$ Related, with a bit more maths, @jrista's answer over here: photo.stackexchange.com/a/1716 (which I only found after writing the above) \$\endgroup\$ Commented Mar 18, 2014 at 5:51
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    \$\begingroup\$ it also depends on the quality of the image. The more blurry it is, the less data you need to present an accurate reproduction. \$\endgroup\$
    – jwenting
    Commented Mar 18, 2014 at 8:56
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    \$\begingroup\$ If it's any help, @PaulCezanne, at 5 feet wide it will look good if you folded it up and looked at it from book/paper reading distance (at 240ppi). You literally couldn't take in the whole image at a distance where you'd actually notice a deficiency. 8 feet would be no problem from a pace away (you still couldn't see the whole thing, and you're only a pace away from one small area of the image). Unless the display space is a narrow hallway, you'll be good at almost any reasonable size - unless you have weirdo friends who always carry loupes with them. \$\endgroup\$
    – user2719
    Commented Mar 18, 2014 at 11:39
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    \$\begingroup\$ @drfrogsplat - I read your answer at photo.stackexchange.com/questions/456/… and it seems the 45 degree rule could apply here. That at some point as you wider, you back up the need for resolution decreases. \$\endgroup\$ Commented Mar 18, 2014 at 12:44

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