Typical distance between Smartphone and eyes is approximatly l=40cm.
Model | x | y | x:y | d[“] | l[cm] | Res(l=40cm) |
---|---|---|---|---|---|---|
Samsung Galaxy A40 | 1080 | 2340 | 6:13 | 5.9 | 19.98 | 540×1170 |
Nokia 1 | 480 | 854 | 9:16 | 4.5 | 40.11 | 450×800 |
Calculate l in bash using gnuplot:
echo 'x=1080; y=2340; d=5.9; print "l = ", (2*d*2.54) / (sqrt(x**2+y**2)*tan( (2/60.0) * pi/180.0) )," cm" ;' | gnuplot 2>&1 echo 'x=480; y=854; d=4.5; print "l = ", (2*d*2.54) / (sqrt(x**2+y**2)*tan( (2/60.0) * pi/180.0) )," cm" ;' | gnuplot 2>&1
Calculate x:y in bash using gnuplot:
echo 'x=1080.0;y=2340.0; do for [c=1:30] { print c,":",c*y/x ;}' | gnuplot 2>&1 echo 'x=480.0;y=854.0; do for [c=1:30] { print c,":",c*y/x ;}' | gnuplot 2>&1
What is the minimal distance l on wich the human eye can separeate each pixel?
The resolution of human eyes is alpha=2”. The diagonal of the display is d.
Display: _____ |\ | | \ | | \D | y | \ | |____\| x
It applies for the pixel diagonal D: D²=x²+y² ⇒ D=sqrt(x²+y²) The metric size of a pixel p is : p=d/D ⇒ p=d/sqrt(x²+y²)
human resolution: | pixel 2 p __alpha___________________| pixel 1 p l
We know: tan(alpha)= 2p / l. So it applies for l: l= 2p / tan(alpha)
It follows: l = 2d/( sqrt(x²+y²) * tan(alpha) )
Please note that 2“ is a small angle, so it can be linearized well by small angle approximation.