The grains of sedimentary rocks come in every size, from clay too small to see to boulders bigger than a car. Size matters because it records how much energy moved the grains. Geologists sort sizes into classes on the Wentworth scale, where each class limit doubles or halves: 2 mm, 4 mm, 1/16 mm, and so on. That makes it a natural fit for logarithms.
| Class | Diameter (mm) | φ |
|---|---|---|
| Boulders | over 256 | below −8 |
| Cobbles | 64 to 256 | −6 to −8 |
| Pebbles | 4 to 64 | −2 to −6 |
| Granules | 2 to 4 | −1 to −2 |
| Sand | 1/16 to 2 | 4 to −1 |
| Silt | 1/256 to 1/16 | 8 to 4 |
| Clay | under 1/256 | above 8 |
Why it works
Because the class limits are powers of 2, a base-2 logarithm turns them into whole numbers. The phi scale is $\varphi = -\log_2 d$, with $d$ in millimeters. A 4 mm pebble is $2^2$ mm, so $\varphi = -2$; fine sand of $\tfrac18 = 2^{-3}$ mm has $\varphi = 3$. The minus sign was chosen so that the sand, silt and clay geologists measure most often get positive numbers. The log also makes it easy to average and compare sizes that differ a thousandfold.
Key idea
$\varphi = -\log_2 d$ ($d$ in mm), and $d = 2^{-\varphi}$. Each step of 1 in phi halves or doubles the size. Coarse grains have small or negative phi; fine grains have large phi.
Worked examples
Worked example: Size to phi
What is $\varphi$ for a grain $\tfrac{1}{16}$ mm across?
- $\tfrac{1}{16} = 2^{-4}$.
- $\varphi = -\log_2 2^{-4} = -(-4) = 4$: the boundary between sand and silt.
Answer: $\varphi = 4$.
Worked example: Phi to size, with a calculator
What is $\varphi$ for a grain 0.3 mm across?
- $\log_2 0.3 = \dfrac{\ln 0.3}{\ln 2} = \dfrac{-1.204}{0.693} = -1.74$.
- $\varphi = 1.74$: medium sand.
Answer: $\varphi \approx 1.74$.
Common mistake
It is natural to expect phi to grow with grain size, the way most measurements do. Because of the minus sign, it does the opposite: a boulder has $\varphi$ around $-9$ and clay around $9$. Check: a 1 mm grain has $\varphi = 0$; anything bigger is negative.