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

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.