website/docs/library/restable.md
We list the number of hexagons and pentagons at each H3 resolution. There are always exactly $12$ pentagons at every resolution.
| Res | Total number of cells | Number of hexagons | Number of pentagons |
|---|---|---|---|
| 0 | 122 | 110 | 12 |
| 1 | 842 | 830 | 12 |
| 2 | 5,882 | 5,870 | 12 |
| 3 | 41,162 | 41,150 | 12 |
| 4 | 288,122 | 288,110 | 12 |
| 5 | 2,016,842 | 2,016,830 | 12 |
| 6 | 14,117,882 | 14,117,870 | 12 |
| 7 | 98,825,162 | 98,825,150 | 12 |
| 8 | 691,776,122 | 691,776,110 | 12 |
| 9 | 4,842,432,842 | 4,842,432,830 | 12 |
| 10 | 33,897,029,882 | 33,897,029,870 | 12 |
| 11 | 237,279,209,162 | 237,279,209,150 | 12 |
| 12 | 1,660,954,464,122 | 1,660,954,464,110 | 12 |
| 13 | 11,626,681,248,842 | 11,626,681,248,830 | 12 |
| 14 | 81,386,768,741,882 | 81,386,768,741,870 | 12 |
| 15 | 569,707,381,193,162 | 569,707,381,193,150 | 12 |
:::caution Cell areas are computed with a spherical model of the earth using the authalic radius given by WGS84/EPSG:4326. :::
The area of an H3 cell varies based on its position relative to the icosahedron vertices. We show the average hexagon areas for each resolution. All pentagons within a resolution have the same area.
| Res | Average <ins>Hexagon</ins> Area (km<sup>2</sup>) | Pentagon Area* (km<sup>2</sup>) | Ratio (P/H) |
|---|---|---|---|
| 0 | 4,357,449.416078381 | 2,562,182.162955496 | 0.5880 |
| 1 | 609,788.441794133 | 328,434.586246469 | 0.5386 |
| 2 | 86,801.780398997 | 44,930.898497879 | 0.5176 |
| 3 | 12,393.434655088 | 6,315.472267516 | 0.5096 |
| 4 | 1,770.347654491 | 896.582383141 | 0.5064 |
| 5 | 252.903858182 | 127.785583023 | 0.5053 |
| 6 | 36.129062164 | 18.238749548 | 0.5048 |
| 7 | 5.161293360 | 2.604669397 | 0.5047 |
| 8 | 0.737327598 | 0.372048038 | 0.5046 |
| 9 | 0.105332513 | 0.053147195 | 0.5046 |
| 10 | 0.015047502 | 0.007592318 | 0.5046 |
| 11 | 0.002149643 | 0.001084609 | 0.5046 |
| 12 | 0.000307092 | 0.000154944 | 0.5046 |
| 13 | 0.000043870 | 0.000022135 | 0.5046 |
| 14 | 0.000006267 | 0.000003162 | 0.5046 |
| 15 | 0.000000895 | 0.000000452 | 0.5046 |
*: Within a given resolution, all pentagons have the same area.
Here are the same areas, but in m<sup>2</sup>.
| Res | Average <ins>Hexagon</ins> Area (m<sup>2</sup>) | Pentagon Area* (m<sup>2</sup>) |
|---|---|---|
| 0 | 4,357,449,416,078.392 | 2,562,182,162,955.496 |
| 1 | 609,788,441,794.134 | 328,434,586,246.469 |
| 2 | 86,801,780,398.997 | 44,930,898,497.879 |
| 3 | 12,393,434,655.088 | 6,315,472,267.516 |
| 4 | 1,770,347,654.491 | 896,582,383.141 |
| 5 | 252,903,858.182 | 127,785,583.023 |
| 6 | 36,129,062.164 | 18,238,749.548 |
| 7 | 5,161,293.360 | 2,604,669.397 |
| 8 | 737,327.598 | 372,048.038 |
| 9 | 105,332.513 | 53,147.195 |
| 10 | 15,047.502 | 7,592.318 |
| 11 | 2,149.643 | 1,084.609 |
| 12 | 307.092 | 154.944 |
| 13 | 43.870 | 22.135 |
| 14 | 6.267 | 3.162 |
| 15 | 0.895 | 0.452 |
*: Within a given resolution, all pentagons have the same area.
The area of an H3 cell varies based on its position relative to the icosahedron vertices. We compute the minimum and maximum values for the hexagon areas (excluding the pentagons) at each resolution, and show their ratio.
| Res | Min <ins>Hexagon</ins> Area (km^2) | Max <ins>Hexagon</ins> Area (km^2) | Ratio (max/min) |
|---|---|---|---|
| 0 | 4,106,166.334463915 | 4,977,807.027442012 | 1.212276 |
| 1 | 447,684.201817940 | 729,486.875275344 | 1.629468 |
| 2 | 56,786.622889474 | 104,599.807218925 | 1.841980 |
| 3 | 7,725.505769639 | 14,950.773301379 | 1.935248 |
| 4 | 1,084.005635363 | 2,135.986983965 | 1.970457 |
| 5 | 153.766244448 | 305.144308779 | 1.984469 |
| 6 | 21.910021013 | 43.592111685 | 1.989597 |
| 7 | 3.126836030 | 6.227445905 | 1.991613 |
| 8 | 0.446526174 | 0.889635157 | 1.992347 |
| 9 | 0.063780227 | 0.127090737 | 1.992635 |
| 10 | 0.009110981 | 0.018155820 | 1.992740 |
| 11 | 0.001301542 | 0.002593689 | 1.992782 |
| 12 | 0.000185933 | 0.000370527 | 1.992797 |
| 13 | 0.000026562 | 0.000052932 | 1.992802 |
| 14 | 0.000003795 | 0.000007562 | 1.992805 |
| 15 | 0.000000542 | 0.000001080 | 1.992805 |
:::caution Edge lengths are computed with a spherical model of the earth using the authalic radius given by WGS84/EPSG:4326. Average edge lengths were calculated exactly for resolutions 0 through 6 and extrapolated for finer resolutions. :::
| Res | Average edge length (Km) |
|---|---|
| 0 | 1281.256011 |
| 1 | 483.0568391 |
| 2 | 182.5129565 |
| 3 | 68.97922179 |
| 4 | 26.07175968 |
| 5 | 9.854090990 |
| 6 | 3.724532667 |
| 7 | 1.406475763 |
| 8 | 0.531414010 |
| 9 | 0.200786148 |
| 10 | 0.075863783 |
| 11 | 0.028663897 |
| 12 | 0.010830188 |
| 13 | 0.004092010 |
| 14 | 0.001546100 |
| 15 | 0.000584169 |
<i>Hexagons have 7 hexagon children. Pentagons have 6 children: 5 hexagons and 1 pentagon.</i>
</div>By definition, resolution 0 has $110$
hexagons and $12$ pentagons, for a total of $122$ cells.
In fact, every H3 resolution has exactly $12$ pentagons, which are always centered at the icosahedron vertices; the number of hexagons increases with each resolution.
:::tip Formula Accounting for both hexagons and pentagons, the total number of cells at resolution $r$ is
$$ c(r) = 2 + 120 \cdot 7^r. $$ :::
We can derive the formula above with the following steps.
First, let $h(n)$ be the number of children $n \geq 0$ resolution levels below any single hexagaon. Any hexagon has $7$ immediate children, so recursion gives us that
$$ h(n) = 7^n. $$
Next, let $p(n)$ be the number of children $n \geq 0$ resolution levels below any single pentagon. Any pentagon has $5$ hexagonal immediate children and $1$ pentagonal immediate child. Thus, $p(0) = 1$ and $p(1) = 6$.
For $n \geq 1$, we get the general recurrence relation
$$ \begin{aligned} p(n) &= 5 \cdot h(n-1) + p(n-1) \ &= 5 \cdot 7^{n-1} + p(n-1). \end{aligned} $$
For $n \geq 0$, after working through the recurrence, we get that
$$ \begin{aligned} p(n) &= 1 + 5 \cdot \sum_{k=1}^n\ 7^{k-1} \ &= 1 + 5 \cdot \frac{7^n - 1}{6}, \end{aligned} $$
using the closed form for a geometric series.
Finally, using the closed forms for $h(n)$ and $p(n)$,
and the fact that (by definition)
resolution 0 has
$12$ pentagons and $110$ hexagons,
we get the closed form for the total number of cells
at resolution $r$ as
$$ \begin{aligned} c(r) &= 12 \cdot p(r) + 110 \cdot h(r) \ &= 2 + 120 \cdot 7^r. \end{aligned} $$
A notebook to produce the cell count table above can be found here.
Cell areas are computed with a spherical model of the earth using the authalic radius given by WGS84/EPSG:4326.
The h3-py-notebooks repo
has notebooks for producing the
average cell area table
and the
min/max area table.