In 2020 I suggested that adding 1 mm of sand (see conversions to common units here) for every 1 L/m2 of clipping volume might be a topdressing rate that would keep OM246 constant.
Since then, I’ve studied—at multiple golf courses—how much sand has been applied, how many clippings have been collected, and how the OM246 test results have changed. It turns out that the 1:1 ratio of sand added for clippings removed isn’t enough, at most places, to keep OM246 constant. If you add the 1:1 ratio, OM246 will tend to increase.
It finally occurred to me why this is. The amount of organic material we start with in the rootzone plays a significant role in how much sand is required to keep the same organic material percentage for every increment of added organic material.
Here’s an equation you can use to find a sand topdressing estimate based on your clipping volume and total organic material.
Estimated sand amount to hold organic matter constant
Here’s an equation you can use to find sand depth in mm. Then use the OM246 calculator unit conversions to express sand in your favorite unit.
$$
d =\ K \cdot V \cdot \frac{100 – \mathrm{OM}%}{\mathrm{OM}%}
$$
where
- $d$ is the depth of sand to apply over the period, in mm.
- $V$ is the fresh clipping volume harvested over the same period, in L m$^{-2}$.
- $\mathrm{OM}%$ is the starting total organic material percentage in the sampled layer, measured by loss on ignition.
- $K$ is an empirical coefficient, in mm L$^{-1}$ m$^{2}$, that converts clipping volume into the sand depth required at a given OM%. My current best estimate of $K$ is 0.106,
- The factor $\frac{100 – \mathrm{OM}}{\mathrm{OM}}$ is the ratio of mineral mass to organic matter mass in the layer. Maintaining a constant OM% requires sand mass and new OM mass to enter the layer in this same ratio, so the sand depth scales with both the clipping volume and the existing mineral-to-OM ratio.