<s>[INST] Moscow papyrus area of hemisphere I found this interesting text on wikipedia that explains how egyptians calculated the area of (supposedly) hemisphere, referred to as "basket". It is contained in the Moscow Papyrus. Here is the text:

Example of calculating a basket. You are given a basket with a mouth of $4 \frac{1}{2}$. What is its surface? Take $\frac{1}{9}$ of $9$ (since) the basket is half an egg-shell. You get $1$. Calculate the remainder which is $8$. Calculate $\frac{1}{9}$ of $8$. You get $\frac{2}{3} + \frac{1}{6} + \frac{1}{18}$. Find the remainder of this $8$ after subtracting $\frac{2}{3} + \frac{1}{6} + \frac{1}{18}$. You get $7 + \frac{1}{9}$. Multiply $7 + \frac{1}{9}$ by $4 + \frac{1}{2}$. You get $32$. Behold this is its area. You have found it correctly.

Now I am trying to understand what all these numbers represent by imagining this "basket" and I have been unsuccessful. Maybe someone understands the logic behind all of this to explain in more detail?
EDIT
So to be more specific. The algorithm obviously works ignoring the small error compared to our current formula. They were very practical and didn't have sophisticated mathematical knowledge. So this is why I essentially want that whoever answers this forgets about our modern mathematics and thinks exclusively in the context of this algorithm and its operations to somehow make sense of this.
 [/INST] If the diameter of the hemisphere is $d=4.5$  then


*

*Take $\frac19$ of $2d$ to give $\frac{2}{9}d$ leaving $\frac89$ of $2d$, i.e. $2d - \frac{2}{9}d = \frac{16}{9}d$

*Take $\frac19$ of $\frac{16}{9}d$ to give $\frac{16}{81}d$ leaving $\frac89$ of $\frac{16}{9}d$, i.e. $\frac{16}{9}d - \frac{16}{81}d = \frac{128}{81}d$

*Multiply $\frac{128}{81}d$ by $d$ to give $\frac{128}{81}d^2$
The suggestion is that the curved surface area of a hemisphere is about $\frac{128}{81}$ times the square of the diameter
We know the surface area of a sphere is  $4\pi r^2 = \pi d^2$, implying the curved surface area of a hemisphere is $2\pi r^2 = \frac{\pi}{2} d^2$
So this suggests that $\frac{\pi}{2} \approx \frac{8}{9} \times \frac{8}{9} \times 2 = \frac{128}{81}$ and thus $\pi \approx \frac{256}{81}$
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