Calculating the Fertilizer Requirement for Any Turfgrass, Anywhere

This presentation builds on the fundamentals of the turfgrass nutrition talk, and the nutrient use and nutrient supply talk, by explaining a system by which a turfgrass manager can calculate the amount of any element required by any turfgrass, under any growing condition, anywhere in the world. Some common misapprehensions about turfgrass nutrition and soil testing will also be discussed. The minimum levels for sustainable nutrition (MLSN) guidelines for interpreting soil test results, and the temperature-based turfgrass growth potential (GP), which were introduced in the two previous seminars, will be discussed in even more detail.

Transcript

1. Calculating the Fertilizer Requirement for Any Turfgrass, Anywhere Micah Woods Chief Scientist Asian Turfgrass Center www.asianturfgrass.com 13 January 2016 Northern Green Expo Minneapolis, Minnesota
2. 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.
3. As an example for K a + b − c = Q
4. As an example for K a + b − c = Q In ppm, a = 67, b = 37, c = 50.
5. As an example for K a + b − c = Q In ppm, a = 67, b = 37, c = 50. amount needed 67 + 37 − amount present 50 = fertilizer requirement 54
6. As an example for K a + b − c = Q In ppm, a = 67, b = 37, c = 50. amount needed 67 + 37 − amount present 50 = fertilizer requirement 54 Convert between 3-dimensional (ppm) and 2-dimensional (ex. g m-2, or lb 1000 -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, and 1 lb 1000 -2 equals 33.5 ppm. Using that conversion, Q of 54 ppm is a K requirement of 8 g m-2 or 1.6 lb 1000 -2.
7. As an example for K a + b − c = Q
8. As an example for K a + b − c = Q In lbs/1000 2, a = 2, b = 1.1, c = 1.5.
9. As an example for K a + b − c = Q In lbs/1000 2, a = 2, b = 1.1, c = 1.5. amount needed 2 + 1.1 − amount present 1.5 = fertilizer requirement 1.6
10. As an example for K a + b − c = Q In lbs/1000 2, a = 2, b = 1.1, c = 1.5. amount needed 2 + 1.1 − amount present 1.5 = fertilizer requirement 1.6 Convert between 3-dimensional (ppm) and 2-dimensional (ex. g m-2, or lb 1000 -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, and 1 lb 1000 -2 equals 33.5 ppm. Using that conversion, Q of 54 ppm is a K requirement of 8 g m-2 or 1.6 lb 1000 -2.
11. “Precision fertilisation” from STERF
12. Fukuoka, Japan
13. Fukuoka, Japan
14. -25 0 25 50 75 100 Jan 2015 Apr 2015 Jul 2015 Oct 2015 Jan 2016 Mean daily temperature, °F Minneapolis, 2015
15. 0.00 0.25 0.50 0.75 1.00 30 40 50 60 70 80 90 100 Mean temperature, °F C3 growth potential (GP)
16. -25 0 25 50 75 100 Jan 2015 Apr 2015 Jul 2015 Oct 2015 Jan 2016 Mean daily temperature, °F Minneapolis, 2015
17. 0.00 0.25 0.50 0.75 1.00 Jan 2015 Apr 2015 Jul 2015 Oct 2015 Jan 2016 C3 growth potential (GP) Minneapolis 2015
18. 0.000 0.005 0.010 0.015 0.020 Jan 2015 Apr 2015 Jul 2015 Oct 2015 Jan 2016 Estimated N use, lbs/1000 ft2 day Minneapolis 2015
19. 0.0 0.5 1.0 1.5 2.0 2.5 3.0 Jan 2015 Apr 2015 Jul 2015 Oct 2015 Jan 2016 Cumulative N use, lbs/1000 ft2 Minneapolis, 2015
20. Represent elements in proportion to N For every 1 pound of N, expect cool-season grass to use 0.5 lb K 0.125 lb P 0.125 lb Ca 0.05 lb Mg 0.05 lb S
21. 0.0 0.5 1.0 1.5 2.0 2.5 3.0 Jan 2015 Apr 2015 Jul 2015 Oct 2015 Jan 2016 Cumulative use, lbs/1000 ft2 N K P Ca Mg S Minneapolis, 2015
22. a Estimate a from: Known amount of applied N, or Measure clippings, or Temperature-based growth potential, or PACE Turf climate appraisal form, or other method
23. b
24. c Get from a soil test
25. Beware! Common misapprehensions Locked-up nutrients are not a real thing. It doesn’t really ma er what an element does. What ma ers is that enough is available. Don’t be tricked into thinking that an element can be exchangeable but not available. Ratios or percentages of nutrients, not useful. antities of elements are useful. Stressed turf needs a reduction in stress. Adding more nutrients doesn’t solve that problem.