# How many Mega-Pixel are Necessary in Digital Photography?¶

**The optimum size of a digital image in mega-pixel**

Author: | D.Thaler |
---|---|

Created: | Feb 2014 |

Last changed: | 2014-03-17 |

## Summary¶

Modern digital photo cameras still compete in ever higher resolutions and therefore output images of ever increasing pixel size. Full Frame Digital SLR now (2014) stop at about 36 Mpx [1] (e.g. Nikon D800). But how many million image pixels are actually useful?

- The visual acuity of humans varies between 0.5 arc-minutes angular resolution for younger people up to 1.0 and more for older people. [2]
- Consider a photo print. This print can be reasonably watched as a whole (without pixel peeping) from a distance not closer than the image diagonal, more comfortable even more distant.

Both statements imply a certain optimum image resolution [1] measured in mega-pixel.

Detailed calculation gives following result for \(M\) in mega-pixel (valid only for distances longer than one image diagonal because of a simplifying approximation; see Calculation):

Let \(\epsilon\) be the angle (*in radiant*) that can be resolved
by human eyes (visual acuity), \(k\) be the multiples of the image
diagonal as a measure for the watching-distance and
\(a=\frac{x}{y}\) be the aspect ratio of the image (width divided
by height), and \(N_{\mathrm{Nyquist}}=2\;..\;4\) the necessary
factor for increasing the pixel-number because of the Nyquist-Shannon
theorem [3].

The result can be displayed as graph for different values of acuity:

Depending on the visual acuity there are optimum resolutions reaching from

- estimated
**60 Mpx**for very sharp eyes (acuity of 0.5’) and a close watching distance of about the image diagonal or even closer - about
**14 Mpx**for good eyes (0.75’) and a moderate watching distance of 1.5 times the image-diagonal **less than 6 Mpx**for standard eyes** (1.0’) and a comfortable watching distance of 1.5 to 2 times the image-diagonal.

Note

**6 to 10 mega-pixel** is a sufficient image resolution for
photo prints watched in standard situations. - Virtually all
computer screens cannot meet even this moderate condition.

**Very high end performance** has room for improvements up to about **60-80
Mpx**, provided there is no blur due to diffraction because of small
apertures and other optical restrictions. [4] [5]

Footnotes and References

[1] | (1, 2) The number of effective sensor pixels or image pixels
differs from the nominal number of sensor pixels due to a
variety of reasons and is usually significantly
lower. Compare: Understanding Digital Camera Resolution |

[2] | Welche Sehschärfe ist normal? |

[3] | (1, 2) en.wikipedia.org/wiki/Nyquist-Shannon_sampling_theorem |

[4] | http://en.wikipedia.org/wiki/Depth_of_field#Diffraction_and_DOF . |

[5] | http://www.cambridgeincolour.com/tutorials/diffraction-photography.htm |

## Calculation¶

Given an image with the relative size *width* x *height* (\(x
\cdot y\)) the diagonal is \(d=\sqrt{x^2+y^2}\).

With *Pythagoras* and the relation for the aspect ratio \(a=x/y\)
the image diagonal can be written as \(d^2 = x^2 + y^2 = x^2 +
x^2/a^2 = x^2\left(1+\frac{1}{a^2}\right)\)

leading to

A similar calculation \(d^2 = x^2 + y^2 = y^2 + y^2 a^2 = y^2\left(1+{a^2}\right)\)

yields

Generally the image-diagonal \(d\) is watched from a distance \(l=kd\) (with \(k\) as a multiple of the diagonal \(d\)) with the angle \(\alpha\).

With the trigonometrical relation

a relation for the watching angle \(\alpha\) can be derived:

With the visual acuity \(\epsilon\) the number of pixels that can be resolved along the diagonal is approxymately

provided the distance is not too small so that (see figure above) roughly \(c \approx l\) is valid. Anyway the approximation poses a more severe condition on pixel resolution than the exact calculation. The exact solution would take into account that the resolved pixel size at the image edges is coarser than at the center.

With the help of (1) and (2) and using the similarity \(N_x = N_d \cdot x/d,\;\mathrm{etc...}\) one gets for the image resolution in pixel

Further refinement yields

One might expect equation (5) is the final result. But to resolve
a pixel pattern like *“0,1,0,1,0,1,0”* in one dimension at least a
doubled sampling rate is necessary because one cannot rely that the
samples exactly fit the patterns. If the sample falls in between, the
result will be series like *“0.5,0.5,0.5,0.5,0.5,0.5,0.5”* and so no
structure is resolved. Mathematically this is due to the so called
**Nyquist-Shannon-Theorem** [3]

The factor 2 is true for 1-dimensional data f(x). Having 2-dimensional data f(x,y) one might expect the square \(2^2=4\). This is not necessarily true. Most structures in 2-D images are rather lines and filaments than “pepper and salt”-structures and together with sophisticated 2-D sensor grid designs this is likely to reduce the factor.

So let us call the factor necessary to for correct sampling the
*Nyquist-factor* \(N_{\mathrm{Nyquist}}\) and give it a value
somewhere between 2 and 4. Thus the final result shown to us:

Note

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