I am using the following equation to determine what parachute size to use and buy: $$ S = \frac{2gm}{\rho C_D V^2} $$

Once we get the S value, we put it into the following equation to get the size: $$ D = \sqrt{ \frac{4S}{\pi} } $$

And to check it out, I tried an example (from which this reference website has provided and showed steps on how to calculate: https://apogeerockets.com/education/downloads/Newsletter149.pdf [page 3-4]) I keep failing to get the same value which they get.

For example, for the 20g, they said they got 22cm, however I only get 21cm. I just wanted to understand what am I doing wrong?

The Values: S = size of parachute
g = gravity (9.81)
m = mass (20g)

$\rho$ = density of air (1225)

Cd = coefficient drag (0.75)
V = velocity of descend (3.5)

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    $\begingroup$ I also get $21~\text{cm}$: $$ S = \frac{2 \cdot 0.02 \cdot 9.81}{1.225 \cdot 0.75 \cdot 3.5^2} \approx 0.03486 $$ and then $$ D = \sqrt{\frac{4\cdot 0.03486}{\pi}}\approx 0.2107 $$ $\endgroup$ – Matti P. Jan 17 at 11:22
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    $\begingroup$ When I look at the tabulated values for all the diameters, it seems like they consistently add $1\ldots 3~\text{cm}$ to the answer and possibly round up the answer. So I think they want to add some safety margin. $\endgroup$ – Matti P. Jan 17 at 11:25

Your calculations are correct. I can think of two reasons for the discrepancy: either Apogee made a mistake, or they have rounded the value up to the next highest integer, to be on the safe side.

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