How much printed linear resolution do you need to support optimal viewing from various distance?
I think I assumed a maximum resolving power of 1 minute of arc (MOA), so we want each pixel to subtend 1 MOA at the chosen viewing distance.
First, for a viewing distance of about 50cm (0.5m):
x/50cm == tan(0.000145444) x/50cm == 0.000145444 x = 0.0072722 cm 2x = 0.0145444 cm (base of triangle, and the length subtended at 50cm == 0.145444 mm == 175 dpi for viewing at 50cm
For further confirmation, the page on MOA for firearms states 1MOA at 100m distance is 29.08mm. A simple linear scaling confirms that this is correct.
What's a simpler formula for ready reckoning?
2.54 / (A_CONST * 2 * VIEWING DISTANCE IN CM) From left to right: 2.54 / 0.000145444 / 2 / VIEWING DISTANCE
This we arrive at:
- 218 dpi at 40cm
- 291 dpi at 30cm
- 436 dpi at 20cm
The constant is: 8,731.883061522
8732 / VIEWING DISTANCE == necessary DPI
You could round this to: 9000 / VIEWING_DISTANCE_IN_CM = necessary DPI
Distance |
Necessary linear resolution |
Nearest useful number? |
10m |
9 dpi |
10 dpi |
5m |
18 dpi |
20 dpi |
3m |
29 dpi |
30 dpi |
1m |
87 dpi |
90 dpi |
50cm |
175 dpi |
180 dpi |
40cm |
218 dpi |
225 dpi |
30cm |
291 dpi |
300 dpi |
20cm |
436 dpi |
450 dpi |
10cm |
872 dpi |
900 dpi |
I dunno if these longer distances hold up to scrutiny - are they correct? They'd suggest that long-distance roadside billboards can be comfortably printed with lolhueg pixels that are like 4.36mm in diameter.