I have been aware of the general consensus of needing between 250 and 300
pixels-per-inch when printing photos to get a good quality print. What I'd never
bothered figuring out was *why?*

This page describes my investigations and turns them into a few simple formulas that help indicate what resolutions, in relation to printing and capture (camera), are needed in various circumstances. I've been pleasantly surprised by how close the formulas match the general newsgroup recommendations.

I hope this information is found to be useful.

The Page Summary at the end of this page includes a form that automates all the calculations shown on this page.

How much resolution is enough? This is a common question in newsgroups. The answer is simple. It depends!

It depends on several things:

- Your visual acuity (the quality of your eyesight)
- The quality of light under which you view the subject (photo)
- The distance at which you view the subject (photo)

According to the 15th Edition of the Encyclopædia Britannica (1977):

The power of distinguishing detail is essentially the power to resolve two stimuli separated in space, so that, if a grating of black lines on a white background is moved farther and farther away from an observer, a point is reached when he will be unable to distinguish this stimulus pattern from a uniformly gray sheet of paper. The angle subtended at the eye by the spacing between the lines at the point where they are just resolvable is called the resolving power of the eye; the reciprocal of the angle, in minutes of arc, is called the visual acuity. Thus, a visual acuity of unity indicates a power of resolving detail subtending one minute of arc at the eye; a visual acuity of two indicates a resolution of one-half minute, or 30 seconds of arc. The visual acuity depends strongly on the illumination of the test target, ... thus, with a brightly illuminated target, with the surroundings equally brightly illuminated (the ideal condition), the visual acuity may be as high as two. When the illumination is reduced, the acuity falls so that, under ordinary conditions of daylight viewing, visual acuity is not much better than unity.

- Macropædia, Volume 7, page 104.

The article goes on to relate the 30 seconds of arc to the diameter of the light-sensing cones and rods on the retina, establishing 30 seconds of arc (a visual acuity of 2) to be the absolute maximum resolution possible in the human visual system.

Clearly, as you view subjects from greater and greater distances, the level of detail you can see diminishes. The distance between the details is best measured by the angle between them as you view them. As they move further away, the angle diminishes, and thus the detail also diminishes.

The minimum angle between the details was established above, under Visual Acuity, as 30 seconds of arc under ideal conditions and one minute of arc under ordinary conditions.

The relationship between angle, distance and detail separation is given by the formula (see Math Forum: Ask Dr. Math FAQ: Circle Formulas):

c = 2 × d × tan( θ ÷ 2 )

where c (the chord length) is the detail separation distance, d (the distance to the chord midpoint) is the viewing distance, and θ (the angle in radians) is the visual acuity angle.

To aid the maths, the angles given above in degrees have been converted to radians (see Dr. Math item):

Angle | ... in Radians |
---|---|

30 seconds of arc | 0.000145444 |

1 minute of arc | 0.000290888 |

To show how this works in practice, consider a standard 6 by 4 inch print. It depends on the person, but these might typically be viewed at a distance of 15 inches.

Basing the calculation on ordinary (typical) viewing conditions where the visual acuity angle is 0.000290888 radians (θ), the distance (D) between two details such that they are just distinguishable when viewed from a distance of 15 inches (V) is:

D = 2 × V × tan( θ ÷ 2 )

D = 2 × 15 × tan( 0.000290888 ÷ 2)

D = 0.004363323

The reciprocal of that is the "details per inch" = 229. This would be the minimum "pixels per inch" required for printed details to be at the limit of visual resolution under ordinary conditions.

A larger print, intended to be viewed at a distance of, say 30 inches, would require a minimum PPI of 115.

A quick way to determine the minimum PPI resolution to print your photos is to divide 3438 by the intended viewing distance in inches. This is for ordinary conditions; double the number for ideal conditions.

Where did "3438" come from? Given the visual acuity angle for ordinary viewing, the equation can be written as:

1 ÷ PPI =
2 × V × tan( 0.000290888 ÷ 2)

1 ÷ PPI =
V × 0.000290888

PPI = 3437.746747 ÷ V

Pixels in your photo image are not the same as dots printed on your photo paper. Each pixel on your screen can be one of 16 million different colours. Printers have a much more limited range of colours per dot. At minimum it is the number of different colour ink tanks installed in the printer, plus one (for white - the blank paper). Some printers are capable of overprinting their colours, giving their printed dots a wider range of colours.

For example, my HP Officejet G95 uses the HP PhotoRet III "colour layering technology". The printer has four inks - black, cyan, magenta, and yellow. According to HP technical documentation, the PhotoRet III technology can produce over 3500 different colours per dot. Different printers with more ink tanks (the additional colours usually being light cyan, light magenta, and light black) will be able to produce more colours per dot.

In any case, no matter the technology or the number of ink colours, printers
simply *cannot* come anywhere near the 16 million different colours per
pixel your images are capable of.

Printers address this problem by printing smaller dots and more of them. The same principle that causes a fine weave of black and white threads to appear gray from a distance is used to make the small number of colours available to a printer appear to be a much wider range of colours when viewed from a distance, even if that distance is only a few inches.

There is no simple method to determine the
DPI to print a photo at. Clearly, it's
related to the viewing distance and, by extension, the
PPI. The
DPI value *must* exceed the
PPI value by a significant amount in
order for the colours of the printed dots to blend imperceptibly.

The amount by which the DPI must exceed the PPI depends on how the printer makes up the extra colours. The fewer colours the printer is able to produce per dot, the more dots are required to spread the average colour around.

The short answer is, print at the highest DPI your printer allows. For printers that allow a trade-off between resolution and colours-per-dot (like my OfficeJet), the only way to determine which is best is to try the different configurations and see for yourself which is better. I can't tell the difference in quality, but the "lower-resolution-with-more-colours-per-dot" configuration prints faster for me.

If the assumption was made that digital cameras can record the full 16 million different colours per pixel, the calculation to determine the required resolution would be simple. Take your required print size in inches and the PPI you calculated for the required viewing distance and multiply them togther.

For example, a 6 by 4 inch print to be viewed at 15 inches would require a minimum (6 × 229) × (4 × 229) = 1374 × 916 = 1258584 pixels = 1.2 megapixels.

However, digital cameras are not capable of capturing 16 million different colours per pixel. Instead they use a colour filter (called a Bayer filter) made up of red, green, and blue and measure 256 different shades of each. My Inside my QV-3000 web page has close-up photos of a CCD where you can see the filter. How Stuff Works has a page describing the process.

(Note that recent innovations, in particular the Foveon X3, are expected to remove this limitation and result in cameras capable of measuring the full 16 million colours at each pixel.)

The process to convert these separate colours reduces the effective resolution of the camera by at least half. Therefore, the required megapixels calculated above should be doubled to 2.4 megapixels.

Rearranging the maths will result in a simpler formula. All dimensions in inches or square inches as appropriate.

Megapixels (MP) = (Bayer filter
correction) × (print width ×
PPI × print height ×
PPI) ÷ 1000000

MP = 2 × (print area ×
PPI²) ÷ 1000000

PPI = (3437.746747 ÷ viewing
distance (V) )

PPI² = (11818102.69 ÷
V²)

MP = 2 × (print area ×
11818102.69 ÷ V²) ÷ 1000000

MP = (print area × 23.63620539
÷ V²)

Using the previous example, the 6 by 4 inch print has an area of 24in². The square of the viewing distance is 225. The required megapixels are (24 × 23.6) ÷ 225 = 2.5.

What these formulas are calculating is the resolution beyond which any higher resolution is pointless. Your personal preferences may allow the resolution to be lower. Note also that where your printers DPI does not greatly exceed the PPI of your prints, the quality will be poorer and will need to be viewed at distances greater than those indicated here.

These numbers are for high quality under 'ordinary' conditions. Double the PPI and quadruple the MP for 'ideal' conditions.

For highest quality prints, scale your photos so that the PPI is no smaller than the value indicated by the formula:

PPI = 3438 ÷ (viewing distance in inches)

Print your photos at the highest dots-per-inch (DPI) your printer allows. Preferably, the printer DPI should triple (or more) the image PPI.

When performing the rescaling in your photo editor, *do not* select
the resample option. You will either have more pixels than is strictly required,
in which case the printer driver will happily 'down-sample' your photo
automatically. The extra pixels may even help it produce a slightly better
print. If you don't have enough pixels than the maths would indicate, then there
isn't anything you can do about it. Pixels cannot be invented and you'll have to
make the best of those you have.

For example, if you're printing a photo intended for viewing at a distance of 15 inches, the minimum PPI to aim for is 3438 ÷ 15 = 229, and your printer DPI should exceed 687.

To determine the required megapixel resolution of the digital camera, use this formula:

MP = (print area in square inches) × 23.6 ÷ (viewing distance in inches)²

For example, a 6 by 4 inch print to be viewed at 15 inches will require a minimum (6 × 4 × 23.6) ÷ 15² = 566.4 ÷ 225 = 2.5 megapixels.

NOTE: Units are not specified since they do not affect the numbers. The numbers you enter must all use the same units, of course! If the width and height are inches, then the viewing distance must also be entered in inches (and vice versa), and the calculated PPU will be pixels per inch.

For a bit of a background on resolution, see Scanning Basics 101 - All about digital images by Wayne Fulton.

The Luminous Landscape has several tutorials on Resolution and Sharpness that may be of interest.

Terry Dawson has some relevant pages too.