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A discussion mostly about the principles of turfgrass nutrition, with a focus on soil nutrient analyses and their use in modern turf management

July 16, 2015

Transcript

1. A discussion mostly about the principles of turfgrass nutrition, with

a focus on soil nutrient analyses and their use in modern turf management Micah Woods Chief Scientist Asian Turfgrass Center www.asianturfgrass.com 16 July 2015
2. The main principle To have good turf, we need to

make sure that the grass is supplied with each nutrient in adequate amounts.
3. The main principle If the grass already has enough of

an element, adding more will provide no benefit.
4. The main principle Grass obtains nutrients from the soil, or

from fertilizer. If the soil contains more of an element than the grass requires, that element is not required as fertilizer. If the soil doesn’t have enough, that element should be applied as fertilizer.
5. There are only 2 fundamental questions 1. Is this element

required as fertilizer? 2. If it is required, how much? I’ll explain three ways to answer those questions – in words, as an equation, and in graphical form.
6. In words The quantity of an element required as fertilizer

is the diﬀerence between the amount the grass requires and the amount present.
7. As an equation One can express the quantity of an

element required as fertilizer as Q. a + b − c = Q where, a is the quantity of the element used by the grass b is the quantity of the element required in the soil c is the quantity of the element present in the soil Q is the quantity of the element required as fertilizer
8. As an equation amount needed a + b − amount

present c = fertilizer requirement Q a is a site-specific use estimate, b is the MLSN guideline, and c is the soil test result.

10. In graphical form 0 25 50 75 Soil Test Annual

Plant Uptake Fertilizer Applied Remaining in Soil K (ppm)
11. 0 5 10 Soil Test Annual Plant Uptake Fertilizer Applied

Remaining in Soil K(g m2)
12. 13 -8 0 5.3 0 5 10 Soil Test Annual

Plant Uptake Fertilizer Applied Remaining in Soil K(g m2)
13. 104 -2 0 102 0 25 50 75 100 Soil

Test Annual Plant Uptake Fertilizer Applied Remaining in Soil Ca(g m2)
14. Beware! These topics are misleading and irrelevant ▶ The concept

of locked-up nutrients ▶ Focusing on what an element does, rather than how much is present ▶ Looking at percentages of elements rather than quantities ▶ Water or saturated paste extracts to look at availability ▶ The idea that an element can be exchangeable but not available

element

25. Example: how to apply this approach Let’s look first at

what the grass uses. That is, how much of each element is harvested by the grass.
26. 0 10 20 30 40 N K P Ca Mg

S Element Amount in bentgrass leaves (g/kg)
27. Represent elements in proportion to N For every 1 gram

of N, bentgrass will use ▶ 0.5 g K ▶ 0.125 g P ▶ 0.125 g Ca ▶ 0.05 g Mg ▶ 0.05 g S
28. Amount of elements to apply If you don’t do soil

testing, apply each element in proportion to N. This ensures that the grass is supplied with all of each element that it will use. But it is much more eﬀicient to do soil testing.
29. As an example for K Let’s say the soil test

K is 50 ppm and the annual N rate is 16 g m-2.
30. As an example for K Let’s say the soil test

K is 50 ppm and the annual N rate is 16 g m-2. a + b − c = Q
31. As an example for K Let’s say the soil test

K is 50 ppm and the annual N rate is 16 g m-2. a + b − c = Q In g m-2, a = 8, b = 5.5, c = 7.5.
32. As an example for K Let’s say the soil test

K is 50 ppm and the annual N rate is 16 g m-2. a + b − c = Q In g m-2, a = 8, b = 5.5, c = 7.5. In ppm, a = 54, b = 37, c = 50.
33. As an example for K Let’s say the soil test

K is 50 ppm and the annual N rate is 16 g m-2. a + b − c = Q In g m-2, a = 8, b = 5.5, c = 7.5. In ppm, a = 54, b = 37, c = 50. amount needed 54 + 37 − amount present 50 = fertilizer requirement 41
34. As an example for K Let’s say the soil test

K is 50 ppm and the annual N rate is 16 g m-2. a + b − c = Q In g m-2, a = 8, b = 5.5, c = 7.5. In ppm, a = 54, b = 37, c = 50. amount needed 54 + 37 − amount present 50 = fertilizer requirement 41 Convert between 3-dimensional (ppm) and 2-dimensional (g m-2) based on rootzone depth and soil bulk density. For a 10 cm deep rootzone with bulk density of 1.5 g cm-3, 1 g m-2 equals 6.7 ppm. Using that conversion, Q of 41 ppm is a K requirement of 6.1 g m-2.