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Methods for determining fertilizer doses

All methods of determining fertilizer doses are based on data from long-term or episodic field and production experiments, and differ in the completeness and accuracy of reflecting the patterns of relationships of plants, soils and fertilizers.

All existing methods and their modifications for determining fertilizer doses can be divided into:

  • methods of generalizing the results of experiments with empirical fertilizer doses;
  • methods of generalizing the results of experiments with the help of balances of nutrients.

All of the above methods of optimizing fertilizer doses can fairly objectively predict the value of crop yields. But despite this, they require improvement in terms of a comprehensive approach, taking into account the conditions of growing crops and economic payback of fertilizers.

Methods based on generalization of data with empirical fertilizer doses

Generalization carried out under the methodological guidance of the Geographical Network of experiments All-Russian Institute of Fertilizers and Agrochemistry (VIUA) in all soil and climate zones with different crops, the results of field experiments allowed to determine the effectiveness of certain types of fertilizers on different types of soils and doses of organic and mineral fertilizers for major crops on different types and subtypes of soils. Subsequently, differentiation of doses within the varieties (subtypes) of soils, taking into account the provision of nutrients to preceding crops and varietal characteristics was carried out.

Based on the generalized results of the experiments doses, optimum timing and methods of fertilizing before sowing, during sowing and after sowing for the main crops in all soil and climatic zones were also developed.

According to the Geographic Network of Experiments VIUA and agrochemical service CINAO, for the main soil and climatic zones of Russia on the prevailing types of soils with an average content of mobile phosphorus and exchangeable potassium recommended optimum doses of macrofertilizers for major crops, as well as doses and methods of making microfertilizers.

Table. Optimal doses of mineral fertilizers (kg/ha) for main crops (generalization by Litvak, 1990)

Crop
Zone
N
P2O5
K2O
Winter wheatNon-Black Earth
100
90
90
Forest-steppe
85
80
65
Steppe
75
70
50
CornForest-steppe
100
80
70
Steppe
80
70
60
PotatoesNon-Black Earth
95
90
110
Forest-steppe
90
90
90
Steppe
85
80
70
Silage cropsNon-Black Earth
100
80
105
Forest-steppe
100
75
80
Steppe
65
60
55
Sugar beetNon-Black Earth
145
135
175
Forest-steppe
135
140
150
Steppe
120
120
105

Table. Doses and methods of microfertilizer application for major crops (summarized by Litvak, 1990)

Crop
Element
Soil application, kg/ha a.s.
Seed treatment, g/t a.s.
Top dressing, g/ha a.s.
before sowing
at sowing
Cereals
В
-
0,2
30-40
20-30
Cu
0,5-1,0
0,2
170-180
20-30
Mn
1,5-3,0
1,5
80-100
15-25
Zn
1,2-3,0
-
100-150
20-25
Мо
0,6
0,2
50-60
100-150
Beets (all types)
В
0,5-0,8
0,15
120-160
25-35
Cu
0,8-1,5
0,3
80-120
70
Mn
2-5
0,5
90-100
20-25
Zn
1,2-3,0
0,5
140-150
55-65
Мо
0,5
0,15
100-150
100-200
Leguminous
В
0,3-0,5
-
20-40
15-20
Cu
-
-
120-160
20-25
Mn
1,5-3,0
-
100-120
-
Zn
2,5
0,5
80-100
17-22
Мо
0,3-0,5
0,06
150-160
25-30
Vegetables and potatoes
В
0,4-0,8
-
100-150
-
Cu
0,8-1,5
-
-
20-25
Mn
2-5
-
100-150
-
Zn
0,7-1,2
-
-
-
Мо
-
-
80-100
30-150
Flax
В
0,3-0,5
0,1
50-60
5-10
Cu
1-6
-
100-120
-
Mn
3,0
-
80-100
30
Zn
3,5
-
-
-
Мо
3,0
-
150-160
150-250
Leguminous grasses
В
0,5-0,6
-
20-40
25-35
Cu
3,0
1,5
150-160
20-35
Mn
1,5-3,0
-
50-70
-
Zn
1,3
-
100-120
55-65
Мо
0,2-0,3
-
100-120
150-250
Cereal grasses
В
0,5-0,6
-
-
25-35
Cu
0,8-1,5
-
-
25-35
Zn
0,7-1,2
-
100-120
55-65
Мо
0,2-0,3
-
150-200
150-250

Regional research institutions offer more specific recommendations for crops, types, subtypes and varieties of soils, indicating the levels of planned yields, cultivation of soils and in combination with doses of organic fertilizers.

In each set of specific natural and economic conditions of territories based on the results of at least 7-10 reproducible experiments with one crop or variety the regional institutions of the Geographic Network of Experiments and Agrochemistry Service determine the quantitative indicators of fertilizer efficiency:

  • yield increase from the optimal dose;
  • element removal per unit of main and by-products and coefficients of soil element and fertilizer use;
  • coefficients of return or intensity of element balance;
  • correction coefficients to doses depending on soil class;
  • rates of mineral fertilizer inputs to produce a unit of increase and yield as a whole;
  • optimum levels of nutrients in soil;
  • standard costs of fertilizers per unit of change in the content of mobile forms of elements in the soil;
  • key indicators of product quality;
  • economic indicators of fertilizer efficiency;
  • mathematical models characterizing the relationship between crop productivity, soil fertility, fertilizer doses, weather and agrotechnical factors;
  • levels of environmental constraints in fertilizer application.

Based on the results, specific recommendations for doses and ratios of fertilizers are developed, but even in this case it is necessary to correct the doses in relation to a specific enterprise, agrocenosis and field.

This group of methods also includes calculations of doses according to the standards of mineral fertilizer inputs for the whole crop according to the formula:

D = Y⋅H1⋅Kn,

or an increase in the harvest:

D = ΔY⋅H2⋅Kn,

where D – the dose of N, P2O5, K2O for the desired yield or gain, kg/ha a.s.; Y and ΔY – desired yield or gain respectively, t/ha; H1 and H2 – rates of fertilizer costs per unit yield and gain, kg a.s.; Kn – correction factor for soil class of phosphorus and potassium availability; when calculating nitrogen doses Kn = 1.

Norms of fertilizer application and correction factors to fertilizer doses are specified in the regional recommendations of research institutions, agricultural experimental stations, centers and stations of Agrochemical Service.

The third direction of the group of methods based on the generalization of data with empirical doses of fertilizers is the search for mathematical dependencies of yield on doses of fertilizers. The first such attempt was made in 1905 by German scientist E.A. Mitscherlich, who proposed the following equation:

lg(A – Y) = lgA – K⋅x,

where A – the maximum possible yield; Y – the actual yield; K – the coefficient of proportionality, which characterizes the relationship between yield and fertilizer dose; x – fertilizer dose.

The fourth direction of the group of methods is the development of regression models based on the results of planning, carrying out and statistical evaluation of the results of multifactor experiments with empirical doses of fertilizers. Equation with powers of 0.5 and 1 for factors and 0.5 for pairwise interactions proved to be the best mathematical model for determining quantitative dependence between yield and fertilizer doses:

Y = а0 + а1N0,5 + a2N + a3P0,5 + a4P + a5K0,5 + a6K + a7(NP)0,5 + a8(NK)0,5 + a9(PK)0,5,

where Y is yield; a0 is a free term of the equation; a1, a2, …, a9 are terms of the equation characterizing the effect and interaction of factors; N, P, K are fertilizer doses.

The fifth direction of this group of methods is the development of mathematical models with the use of computer technology to determine the optimum doses of fertilizers for crops, taking into account the functional dependence on many environmental factors:

Y = f(xn),

where Y is the yield; xn are the variables affecting the yield, such as doses and ratios of fertilizers, soil class and granulometric composition, weather conditions, varietal characteristics, predecessors, etc.

Various research institutions, based on the generalized results of field experiments, analyses and observations, have developed software packages for determining fertilizer doses. Thus, CINAO developed the RADOZ (abbreviation for “rational doses”) software complex, which was upgraded to RADOZ-2 and later to RADOZ-3. The modernization is related to the increase in the number of factors affecting crop yields.

Practical application of any of these methods and modifications makes it possible to avoid gross errors in the application of fertilizers. However, they are determined empirically without taking into account the biological needs of crops in nutrients and do not provide an answer to the question of soil conditions; according to them, despite the correction factors, it is impossible to quantify the balance of elements without special calculations.

Methods based on the generalization of data using balance sheet calculations

In this group of methods the biological characteristics of crops and varieties in the consumption of nutrients to create the planned high quality yields with simultaneous regulation of soil fertility (class, cultivation) for specific natural-economic conditions are the basis for determining the optimal doses of fertilizers. Consumption of soil nutrients and fertilizers by crops is determined by the results of field and production experiments, which transforms the field method from empirical to analytical, which allows to pass from the statement of yield increments to the prediction of their efficiency.

This group of methods is promising first of all for regions with sufficient moisture and irrigated agriculture, where the limiting factor for high and sustainable yields is the lack of nutrients, and fertilizer availability is high enough – not less than 100 kg/ha a.s.

Detailed soil characteristics are available in soil and agrochemical maps, which should be in each farm. The use of soil nutrients by specific crops is determined by coefficients of utilization (CUE) or by correction factors to the doses depending on the fertility of a particular soil.

Differences in effective fertility and cultivation of soils can also be taken into account through differentiated balance coefficients of use of mineral and organic fertilizers of relative balance indicators, i.e. return coefficients, intensity of balance and difference coefficients of fertilizer use.

There are many methods and modifications of balance calculations to determine optimal fertilizer doses.

Table. Differentiated by soil fertility differential coefficients of the use of nutrient elements of organic and mineral fertilizers in Non-Black Soil Zone (average for the rotation of crop rotations), %[1] Yagodin B.A., Zhukov Y.P., Kobzarenko V.I. Agrochemistry / Ed. by B.A. Yagodin. - Moscow: Kolos, 2002. - 584 p.: ill.

Fertility (class) of soil
N
P2O5
K2O
organic
mineral
organic
mineral
organic
1
40-50
55-65
45—55
35-45
75-85
2
45-55
60-70
50-60
40-50
80-90
3
50-60
65-75
55-65
45-55
85-95
4
55—65
70-80
60-70
50-60
90-100
5
60-70
75-85
65-75
60-70
95-105
6
70-80
80-90
70-80
70-80
100-110

Example. It is necessary to determine optimal doses of mineral fertilizers in combination with 20 tons/ha of half-decomposed manure containing 0.4% N, 0.2% P2O5 and 0.5% K2O to obtain 4.0 t/ha of grain with a grain:straw ratio of 1:1.5, winter wheat variety Mironovskaya 808 on sod-podzolic medium loamy calcareous soil with phosphorus and potassium (according to Kirsanov) respectively 70 and 100 mg/kg (3rd class) and salt extract pH 6.2 (6th class), precursor – vetch-oat mixture, under which N60P60K60 was introduced.

Maps of soil hydrolysable and mineral nitrogen supply are usually not made due to the high variability of these indicators even during one month, so the provision of soil with hydrolysable nitrogen is determined analytically or tentatively by the content of organic matter, total nitrogen or by phosphorus or potassium, which are at a minimum. Since the content of nitrogen in humus is an average of 4%, and, according to generalized information of the All-Russian Institute of Fertilizers and Agrochemistry, its hydrolysable forms of 4-7%, then at 2.5% of soil humus content of total nitrogen will be 0.1%, and hydrolysable – 0.004%, or 40 mg/kg. In order to determine nitrogen availability for the element that is in the minimum according to the adopted soil classification, its content corresponding to the same class as the element that is in the minimum is used.

In all methods and modifications determine the economic removal of a crop or variety in nutrients to create the planned yield at the cost per unit of the main and the corresponding amount of by-products from zonal, regional directories and recommendations. Or select a field in the farm, where a close level of yield of a variety is already achieved, take samples of grain and straw and subject them to chemical analysis. At the content of N, P2O5 and K2O in grain, respectively, 2.5; 0.8 and 0.6% and in straw 0.5; 0.2 and 1.2%, the economic yield with the planned yield will be:

N 130 (2,5-40 + 0,5-60) kg,

Р2O5 46 (0,8-40 + 0,2-60) kg,

К2O 96 (0,6 -40 + 1,2-60) kg,

and the cost of 1 ton of grain with the corresponding amount of straw, respectively, N 130:4 = 33 kg, P2O5 46:4 = 12 kg, K2O 96:4 = 24 kg.

Then the main methods are calculations of the elementary balance, on the surplus, on the relative indicators of the balance, based on one or a combination of several methods.

Elementary balance method

The method of elementary balance is the most common and the least accurate method, because it uses highly variable under the influence of many factors coefficients of use of soil elements (CUE) and more stable differential coefficients of fertilizer use. The calculations are carried out according to the formula:

where D – dose of N, P2O5 and K2O, kg/ha a.s.; RY – economic removal of an element with the planned yield, kg/ha; S – stock (content) of mobile forms of an element in the soil, kg/ha; KU – coefficient of element use from soil, fractions of one (at 10% – 0.1; 20% – 0.2, etc.); W – amount of an element in organic fertilizer, kg/ha; KW – difference coefficient of organic fertilizer element use, fractions of one; P – amount of an element in forecrop fertilizer and/or in post-harvest forecrop residues, kg/ha; K1 – difference coefficient of fertilizer and/or forecrop residues use, fractions of one; E – pre-sowing (row) fertilizer, kg/ha a.kg/ha; KE – difference coefficient of utilization of pre-sowing fertilizer, fractions of one; K2 – difference coefficient of fertilizer utilization at pre-sowing application, fractions of one.

For this example, optimal doses according to this method on the background of 20 t/ha of manure are as follows:

This method is widespread because it accounts for all nutrient inputs and outputs.

The elementary balance method takes into account:

  • nutrient removal by the crop;
  • content of mobile nutrients in the soil;
  • coefficient of utilization of nutrients from soil;
  • coefficient of nutrient use from fertilizers;
  • mass of arable soil layer or soil layer for which the calculation is made.

Agrochemical indicators of soil supply maps of nitrogen, phosphorus and potassium in mg per 100 g of soil are translated into kg/ha by multiplying by the coefficient corresponding to the soil difference and the depth of the calculated arable layer. For example, for the 0-22 cm tilled layer of sod-podzolic soils it is 30, that is the mass of 1 ha of tilled layer of sod-podzolic soil is considered equal to 3000 t, for a layer up to 30 cm – the coefficient is 40.

This balance method is also applied with clarifications and modifications. Objectivity of the method depends on the reliability of the listed data, which can vary significantly depending on soil properties, weather conditions, doses and forms of fertilizer, term and method of application and other factors.

Method of calculations on the planned yield

The method of calculations on the planned yield increment is more accurate method compared to the previous one, as it takes into account the provision of soil with nutrients with the help of correction factors to doses, which are less dependent on various factors than the coefficients of useful use. However, for this method it is necessary to know the yield without fertilizers, which is best determined by data from experiments with fertilizers, based on which in the case under consideration it is equal to 2.0 t/ha. The yield can also be determined by the element that is in the minimum, using the coefficient of its use (CUE). The calculations are carried out according to the formula:

where D – the dose of N, P2O5 and K2O, kg/ha a.s.; Rp – element removal with the planned yield increase, kg; W – amount of element in organic fertilizer, kg/ha; KW – difference coefficient of organic fertilizer element use, fractions of one; P – amount of element in the predecessor fertilizer and/or post-harvest residues of the preceding crop, kg/ha; K1 – difference coefficient of fertilizer and/or residues of the preceding crop, fractions of one; E – pre-sowing (row) fertilizer, kg/ha a.s.; KE – difference coefficient of use of pre-sowing fertilizer, fractions of one; K2 – difference coefficient of use of fertilizer at pre-sowing application, fractions of one; K3 – correction coefficient to dose depending on soil class, (for the 3rd class for cereals, legumes and grasses equals 1).

For the above example according to this method optimal doses of mineral fertilizers on the background of 20 t/ha of manure will be:

In the above methods, when calculating fertilizer doses for the planned yield or increment, the fertilization of the preceding crop must be taken into account. If the preceding crops were grown on fertilized soils, to the soil nutrients calculated from the yield of the crop in the current year, add the effect of fertilizers applied at a rate of 10-15% of the original amount of the active substance in them.

For example, on unfertilized soil you get 20 tons/ha of green mass of corn, which yields 50 kg N, 20 kg P2O5 and 70 kg K2O. Corn is placed after the sugar beet, under which 150 kg N, 80 kg P2O5 and 150 kg K2O were brought; 15% of this amount will be 22,5 kg N, 12 kg P2O5 and 22,5 kg K2O. Thus, placing corn after the sugar beet, you can harvest about 30 tons/ha of green matter without additional fertilization. With a planned yield of 50 t/ha, the estimated dose for the formation of an additional 20 tons of green mass will require an additional 50 kg N, 20 kg P2O5 and 70 kg K2O.

Calculation of optimum doses with the help of balance coefficients of fertilizer use, differentiated by soil fertility

Calculation of optimum doses with the help of balance coefficients of fertilizer use, differentiated by soil fertility is the best method, because it allows you to simultaneously regulate the provision of soil with nutrients. Calculations to obtain the planned yield are carried out according to the formula:

where D – the dose of N, P2O5 and K2O, kg/ha a.s.; RY – economic removal of the element with the planned yield, kg/ha; W – the amount of element in the organic fertilizer, kg/ha; K1 – the difference factor of fertilizer and/or residues of the previous crop, fractions of one; K2 – the difference factor of fertilizer use at preplant application, fractions of one.

On the background of 20 t/ha of manure, mineral fertilizer doses are as follows:

Calculation of optimum fertilizer doses with return rates or balance intensities

The calculation of optimum fertilizer doses with return rates or balance intensities is more complicated, because these indicators are difficult to take into account the effect of fertilizers by years: for this purpose additional indicators are introduced, such as distribution coefficients of fertilizer action by years, which are derivatives of difference coefficients and have the same disadvantages.

Methods for determining doses of fertilizers, taking into account the annual increase in soil fertility and the removal of a nutrient element by the crop

Example. D = 80 kg/ha – amount of fertilizer (a.s.) to be applied annually to the soil; L = 4 years – number of years after the survey; R = 30 kg/ha – removal of nutrients on average per year, kg/ha; C = 50% (0.5) – proportion of the nutrient element going to replenish soil nutrients in the arable layer, from the value characterizing the positive balance; I – planned increase of the nutrient element, mg per 100 g of soil in the arable layer.

Content of nutrition elements in arable layer equivalent to 1 mg per 100 g of soil equivalent to 30 kg of phosphorus in arable layer is equal:

that is in 4 years phosphorus reserves in the soil increases by 3.3 mg per 100 g of soil, or annually by 0.8 mg per 100 g of soil. Using this formula determine the dose of phosphorus fertilizer (kg/ha), assuming that the average yield of phosphorus will be 30 kg/ha per year and its content in the soil should increase in 5 years by 4 mg/100 g (I), for example:

This method allows you to determine the dose of fertilizer for the planned yield, as well as the rate of growth of nutrients in the soil and its state of cultivation.

According to N.N. Mikhailov’s method, fertilizer doses for cereal crops on soils with low nutrient content, and for row crops – with average content are calculated for the planned yield, taking into account the increase of soil fertility.

Table. Possible removal of phosphorus and potassium from the soil[2]Agrochemistry. Textbook / V.G. Mineev, V.G. Sychev, G.P. Gamzikov et al; ed. by V.G. Mineev. - M.: Publishing house of the All-Russian Scientific Research Institute named after D.N. Pryanishnikov, … Continue reading

Content of mobile Р2O5 and K2O in soil
Nutrients assimilated by plants from soil, kg/ha
Р2O5
K2O
Very low
0-10
до 45
Low
10-20
45-90
Medium
20-40
90-180
High
40-80
180-360
Very high
over 80
over 360

Calculations of phosphorus and potassium requirements for the planned yield of winter rye are presented in the table.

Table. Demand and supply of phosphorus and potassium in winter rye yield planning 4 t/ha[3]Agrochemistry. Textbook / V.G. Mineev, V.G. Sychev, G.P. Gamzikov et al; ed. by V.G. Mineev. - M.: Publishing house of the All-Russian Scientific Research Institute named after D.N. Pryanishnikov, … Continue reading

Indicators
Р2O5
K2O
Soil supply
low
low
needed to form a harvest, kg
48
112
Possible removal from the soil, kg
10
45
is required to provide at the expense of fertilizers
38
67
Supply with 20 tons of manure
25
72
Mineral fertilizers are required
13
secured

Nitrogen in this case is better optimized using the NMIH method (see “Nitrogen fertilizers“).

Although this method takes soil fertility into account when calculating nutrient requirements for the planned yield, it is also not highly accurate, as it uses too wide ranges of plant use of phosphorus and potassium from the soil. Also rather relative coefficients of nutrient use from organic and mineral fertilizers are taken.

Over a rotation, mineral fertilizer nitrogen is used by an average of 60%, phosphorus by 35% and potassium by 75%. Nitrogen and phosphorus of organic fertilizers is used by 50%, potassium – by 75%. Systematic determination of nitrogen, phosphorus and potassium content in the crop allows for annual adjustment of utilization rates and compilation of more reliable balances of nutrients.

The balance data allows you to more accurately calculate the dose of mineral fertilizers, depending on the specific conditions and the goal. Doses of fertilizer calculated to obtain the planned yield and a given content of nutrients in the soil are determined by the formula:

where D – nutrient dose, kg/ha; R – nutrient removal by the planned harvest, kg/ha; K1 – coefficient of nutrient use with regard to aftereffect; SS – the specified nutrient content in the soil, mg/100 g; SF – actual nutrient content in the soil, mg/100 g; K2 – coefficient of conversion mg/100 g to kg/ha; K3 – coefficient of fertilizer consumption to increase nutrient content in the soil; t – time for which the target nutrient content in the soil is to be obtained.

Example. It is necessary to obtain 4 t/ha of grain of winter wheat and achieve in 10 years, the actual content of phosphorus (SF) 10 mg per 100 g of soil. With a grain yield of 4 t/ha, winter wheat takes out 48 kg/ha of P2O5 (R). To determine the dose of phosphorus, removal (48 kg/ha) divided by the coefficient of its use by plants from fertilizers, taking into account the effects, provided that 2/3 of phosphorus is made with mineral and 1/3 with organic fertilizers, the coefficient of use is 0.4 (K1). In this case you need to apply 120 kg/ha of P2O5.

The average content of mobile phosphorus for 10 years (t) increase to 10 mg, or 5 mg per 100 g of soil (SS), which corresponds to 150 kg/ha (5 mg⋅30).

According to long-term studies, about 0.4 of the amount of phosphorus (K3) applied in excess of the dose for the planned yield goes to increase its assimilable forms in the soil. To achieve a given level of content of mobile forms of phosphorus in the soil over 10 years will require 375 kg/ha of P2O5 (150 kg: 0.4), or an average of 37.5 kg/ha for the year while maintaining the level of planned yields. Given this amount, the required dose calculated to obtain the planned yield and a given nutrient content will be 157.5 kg/ha (120 + 37.5):

Calculation of fertilizer doses according to soil evaluation

T.N. Kulikovskaya recommends to calculate fertilizer doses according to the point assessment of soils. On the basis of experimental data the price of arable land score, kg of production per score was developed.

Table. Price point of arable land, kg of products per one point[4]Agrochemistry. Textbook / V.G. Mineev, V.G. Sychev, G.P. Gamzikov et al; ed. by V.G. Mineev. - M.: Publishing house of the All-Russian Scientific Research Institute named after D.N. Pryanishnikov, … Continue reading

Crop
Sod-podzolic soils
Peat-bogs
sandy loam
sandy loam, underlain by moraine
sandy loam, underlain by sand
sandy
Winter rye
33
33
36
30
44
Winter wheat
36
34
28
25
36
Barley
39
38
35
25
43
Oats
33
30
30
28
35
Potatoes
260
250
245
240
262
Flax (fiber)
7,8
7,0
Sugar beets
290
330

Example. Planned yield of winter wheat 5 t/ha of grain. Soil is sandy loamy, arable score Sa = 58, price score for winter wheat PS = 34 kg.

Agrochemical properties of the soil: pH 6.0; humus – 1.8%; P2O5 – 14 and K2O – 12 mg per 100 g of soil; bulk weight 1.3 g/cm3; weight of 20 cm of arable layer 2600 t. Correction factor for agrochemical properties K = 1.23.

First, determine the amount of yield that can be obtained at the expense of effective soil fertility:

Y = Sa⋅PS⋅К = 58⋅34⋅1,23 = 2.42 t/ha.

Consequently, when fertilizer is applied, grain gain will be 2.58 t/ha (5.0-2.42) (table).

Table. Calculation of fertilizer doses for the planned yield of winter wheat 5 t/ha dry matter from 1 ha[5]Agrochemistry. Textbook / V.G. Mineev, V.G. Sychev, G.P. Gamzikov et al; ed. by V.G. Mineev. - M.: Publishing house of the All-Russian Scientific Research Institute named after D.N. Pryanishnikov, … Continue reading

Indicators
N
P2O5
K2O
Planned increase in grain yield, t/ha
2,58
Removed from 100 kg of grain, kg
3,5
1,2
2,5
Total yield per increment, kg/ha
90,3
30,9
64,5
Utilization factor from mineral fertilizers, %
60
25
45
Required to contribute with regard to the utilization factor, kg/ha
150,5
123,6
143,3

Thus, to get 5 tons of grain we put N151P124K143, or a total of 418 kg/ha NPK.

According to studies of the Belarusian Research Institute of Soil Science and Agrochemistry, for different soils determined the payback of mineral and organic fertilizers.

Table. Recoupment of fertilizers by yield, kg production per 1 kg NPK and 1 ton of organic fertilizers[6]Agrochemistry. Textbook / V.G. Mineev, V.G. Sychev, G.P. Gamzikov et al; ed. by V.G. Mineev. - M.: Publishing house of the All-Russian Scientific Research Institute named after D.N. Pryanishnikov, … Continue reading

Crop
1 kg of NPK on soils:
1 ton of organic fertilizer
loamy
sandy loam
sandy
peat-bogs
Winter rye
6,3
6,0
5,0
5,9
10-14
Winter wheat
7,2
6,0
6,0
12-18
Barley
6,5
6,2
4,5
6,0
7-12
Oats
5,7
5,4
4,5
5,8
10-12
Potatoes
30
30
28
35
100
Flax (fiber)
1,4
1,3
-
-
-
Sugar beet
35
33
-
30
120

Using rates of recoupment of a unit of nutrients, calculate the dose of fertilizer for an additional yield of 2580 kg of grain of winter wheat. So, for 1 kg of NPK applied to loamy soil, 6 kg of grain, 2580 kg will require 430 kg/ha (2580:6). Knowing the optimum ratio of nutrients, which for winter wheat is N:P:K = 1,0:0,9:1,2, respectively – for 5 tons/ha of grain N139P125K166.

The calculation method is based on a large number of experimental data, and the dose is largely consistent with the biological characteristics of the crop.

Method for determining the really possible yield

The method for determining the really possible yield (RPY) by the content of nutrients in the soil was proposed by Ermokhin, Neklyudov, and Krasnitsky in 2000. The method is based on the fact that the content of nutrients in the soil is a limiting factor and determines the really possible yield.

The approach of the method allows us to estimate the natural fertility of the soil and determine the possible yield without fertilizers and then predict the effectiveness of fertilizers.

The authors propose a formula for the calculation:

where RPYNS – really possible yield due to soil nutrients (without fertilizers), t/ha; m – nutrient content in the soil, which is in the minimum, mg/100 g; h – depth of arable layer, cm; d – volume mass of arable layer, g/cm3; Ku – coefficient of plant nutrient use from soil; H – nutrient consumption by plants to create a unit of main production, including the side, kg/t.

In addition to the content of nutrients in the soil it is necessary to know the ratio of elements that take part in the formation of yield and their availability to plants.

For the conditions of Western Siberia the ratio of nutrients in the soil, which characterizes the balanced nutrition and allows to determine which of the elements is in the first minimum, was determined.

The optimum ratio of these elements in the soil layer 0-30 and 0-40 cm is equal:

P2O5 mg/100 g ≈ 10⋅NO3 mg/100 g ≈ K2O mg/100 g.

The balanced ratio of P2O5:NO3 is 10, P2O5:K2O is 1. If the P2O5:NO3 ratio is less than 10, it indicates a phosphorus deficiency, if more than 10, the soil contains nitrogen in the minimum. The ratio of K2O to NO3 is similarly characterized.

To determine which element is at a minimum, the action factor of the element to be applied as a fertilizer is established. The highest action coefficient (Kd) indicates that this element is at a minimum and will limit the yield of the cultivated crop.

Example. When the content in the soil N-NO3 – 0.7 mg/100 g; P2O5 – 10.5 mg/100 g; K2O – 10.0 mg/100 g; the ratio of nutrients will be: P2O5:N-NO3 = 10.5:0.7 = 15; K2O:N-NO3 = 10.0:0.7 = 14.3. Consequently, the limiting crop yield on this soil is nitrogen.

When using fertilizers the authors recommend for the soils of Western Siberia to use the optimal levels of nutrients established for grain crops A.E. Kochergin (mg/100 g): N-NO3 – 1,5; P2O5 – 15,0; K2O – 15,0.

The coefficient of action of a nutrient element can also be determined by another method: by the ratio of the optimal and actual content of the nutrient element in the soil:

From these data, it follows that the limiting factors limiting yield in this soil are all three nutrients, most of all nitrogen, as the Kd of nitrogen has the highest value of 2.14.

For Siberia Y.I. Ermokhin with co-authors (2000) give the following coefficients of utilization of nutrients by plants from soil reserves: nitrate nitrogen – 0,6-0,8 (60-80%), mobile phosphorus – 0,1 (10%), exchangeable potassium – 0,2-0,3 (20-30%). Knowing these indicators, determine the really possible yield (for example, barley) without the use of fertilizers.

Example.

N-NO3 = 0,7 mg/100 g

P2O5 = 10,5 mg/100 g

K2O = 10,0 mg/100 g

h = 30 sm

d = 1,2 g/sm3

HN = 35,6 kg/t

HP = 12,1 kg/t

HK = 25,1 kg/t

КП N = 0,6

КП P = 0,1

КП К = 0,3

Determine the RPY, t/ha = ?

Solution.

Taking into account the current nitrification (NT = 70 kg/ha) RPY will be in nitrogen:

on phosphorus:

on potassium:

Consequently, with this characteristic of the soil, the possible barley yield would be 1.6 t/ha.

Methods of calculating fertilizer doses for the planned yield (N.K. Boldyrev)

N.K. Boldyrev in 1962 on the basis of complex methods of leaf and soil diagnostics proposed methods of calculation of fertilizer doses for the planned yield.

A simplified method of calculating fertilizer doses according to the chemical composition of leaves and mobile nutrients of soil is based on establishing the degree of need (DN) for a nutrient element by equation:

DN = OC : AC,

where OC is the optimal content of the element, AC is the actual content of the element.

The degree of need is specified by another element in relative abundance, or by the optimal ratio between the norms of elements in the leaves, given the equality:

%N (L) = 12% Р (L) = 1,2% К (L) or %N (L) = 5,2% P2O5 (L) = %К2O (L).

Equation:

%N = 12⋅%P = 1,2⋅%K = 12⋅%S = 12⋅%Mg = 6⋅%Ca.

N.K. Boldyrev called the equation of the optimal balance of elements in the leaves of cereal crops during the flowering phase. The optimal ratios of elements for other crops have been developed.

Table. Indicators of normal levels of elements and the optimal ratio between them in the leaves of some crops[7]Agrochemistry. Textbook / V.G. Mineev, V.G. Sychev, G.P. Gamzikov et al; ed. by V.G. Mineev. - M.: Publishing house of the All-Russian Scientific Research Institute named after D.N. Pryanishnikov, … Continue reading

Crops
Yield level, 100 kg/ha, at "B" value
Sampling time for analysis (growth phase)
The organ of the plant, the tier of leaves
Content in % on absolutely dry matter
Optimal ratio between the elements
N
Р
К
N/P
N/K
Р/К
Spring and winter wheat
40 - 45 (0,8)*
tillering
aboveground
5
0,43
4,2
12
1,2
10
Barley
50-60 (1,00-1,25)
tubing
all leaves
4
0,33
3,3
12
1,2
10
end of flowering (6-8 days after full appearance of spikelets)
3-4 leaves, counting from the spike
3
0,25
2,5
12
1,2
10
aboveground
2,1
0,25
2,1
8
1,0
8
Corn for silage
500 (0,8) 800-1000 (1-1,25)
6-8 leaves
aboveground
4
0,34
3,4
12
1,2
10
Corn on grain
45-50 (0,8)
cob flowering
2 cobbler leaves
3,2
0,27
2,7
12
1,2
10
80-100(1 -1,25)
aboveground
2,5
0,21
2,0
12
1,2
10
Sunflowers for grain
35-40(1,0)
Head formation before flowering
all leaves
3,1
0,25
2,8
12
1,1
11
Cereal perennial grasses
120-140 for 2 mowings
Beginning of flowering
aboveground
2,6
0,26
2,5
10
1,0
10

*In parentheses is the value of the coefficient of action of the balanced element in the leaves (CABEL, or “B”), corresponding to a certain level of plant yield

N.K. Boldyrev gives three tables with the ratio of NPK in leaves in three growth phases: tillering, piping, and the end of flowering. In the center of each table the optimal ratio of elements (NPK in leaves in the three growth phases) is given. Up and down from the center grows the imbalance associated with element deficiency (degree of need < 1) or excess (degree of need > 1).

Table. Ratio between nitrogen and phosphorus (% P) in leaves (tillering, tubing and late flowering phases) as a basis for assessing nutritional conditions and determining the degree of need (DN) and fertilizer rates for cereal crops (by N.K. Boldyrev)

Number of the reference point from the center of the optimum (1-CO)
N/P ratio
DN
Nitrogen and phosphorus nutritional conditions and their equilibrium
N
P
7
22,5-24
0,5
2,0
strong phosphorus deficiency with a large excess of nitrogen
6
20
0,6
1,7
5
18
0,7
1,5
4
16
0,75
1,3
average phosphorus deficiency with an average nitrogen excess
3
14
0,8
1,2
2
13
0,9
1,1
ratio close to normal
1-CO
12
1
1
balanced N and P supply
2
11,2
1,1
0,9
ratio close to normal
3
10
1,2
0,8
average nitrogen deficiency with an average phosphorus excess
4
9,0
1,3
0,75
5
8
1,5
0,66
strong nitrogen deficiency with a small excess of phosphorus
6
7,1
1,7
0,6
7
6,0
2
0,5

Table. Ratio between nitrogen and potassium (% K) in leaves as a basis for assessing nutritional conditions, determining the degree of need (DN) and fertilizer rates for cereal crops (by N.K. Boldyrev)

Number of the reference point from the center of the optimum (1-CO)
Ratio N/K
DN
Nitrogen and potassium feeding conditions and their equilibrium
K
N
7
2,25-2,4
2,0
0,5
severe potassium deficiency with a large excess of nitrogen
6
2,0
1,7
0,6
5
1,8
1,5
0,7
4
1,6
1,3
0,75
average potassium deficiency with an average nitrogen excess
3
1,4
1,2
0,8
2
1,3
1,1
0,9
ratio close to normal
1-CO
1,2
1,0
1,0
balanced N and K power supply
2
1,12
0,9
1,1
ratio close to normal
3
1,0
0,8
1,2
average nitrogen deficiency with an average potassium excess
4
0,9
0,75
1,3
5
0,8
0,66
1,5
severe nitrogen deficiency with a slight excess of potassium
6
0,71
0,6
1,7
7
0,6
0,5
2

Table. Ratio between potassium (% K) and phosphorus (% P) in leaves as a basis for assessing nutritional conditions, determining the degree of need (DN) and fertilizer rates for cereal crops (by N.K. Boldyrev)

Number of the reference point from the center of the optimum (1-CO)
K/P ratio
DN
Phosphorus and potassium nutritional conditions and their equilibrium
P
K
7
20
2,0
0,5
severe phosphorus deficiency with a large excess of potassium
6
18
1,8
0,56
5
16
1,6
0,63
4
14
1,4
0,7
average phosphorus deficiency with an average potassium excess
3
12,5
1,2
0,8
2
11,2
1,1
0,9
ratio close to normal
1-CO
10
1
1
balanced K and P power supply
3
8
0,8
1,25
average potassium deficiency with an average phosphorus excess
4
7,1
0,7
1,4
5
6,3
0,6
1,6
severe potassium deficiency with a slight excess of phosphorus
6
5,6
0,56
1,8
7
5
0,5
2

Knowing the Needs Degree (DN) values, determined by the equation DN = OC : AS or tables, you can calculate the fertilizer dose.

If DN < 1, the plants do not need this element and the calculation of a dose is not carried out. If the value of DN is from 1.1 to 3-4, the indicator is included in the formula:

D (kg/ha) = DN ⋅ MN,

where D is the dose of the active substance (a.s.); MN is the minimum dose (kg a.s.) used in the main fertilizer, the value of which is set in field experiments.

The value of MN in the basic fertilizer for cereals, corn and peas is usually 30 kg/ha a.s., for potatoes and vegetables – 45 kg/ha a.s. under normal growing conditions to get the yield to 4 t/ha of grain, 25 t/ha of potatoes and 50 t/ha of cabbage. Under irrigation conditions the minimum dose (MN) can be 1.5-2 times more. In the complex method of leaf diagnostics the dose adjustment of the missing element by other basic elements in some excess or deficiency is given.

Example. Calculation of nitrogen rate to produce 4 t/ha of spring wheat grain.

For the tillering phase optimum content in leaves N = 5%, P = 0.43%, K = 4.2%. Equilibrated element (nitrogen) action coefficient in leaves B = 0.5 for 3.2-4.0 t/ha yield, 0.63 for 4.1-5.0 t/ha yield, and 0.8 for 5.5-6.0 t/ha yield. Nitrogen requirement of spring wheat per 100 kg of grain with 4 kg of straw. The coefficient of use from mineral fertilizers N – 63%, P2O5 – 20% and K2O – 63%. Minimum nitrogen rate NMIN = 45 kg.

For the phase of the end of flowering the optimal content of elements in the leaves N = 3%, P = 0.25%, K = 2.5%. The coefficient of action of elements in leaves (B) for yields of 4.0 t/ha is 0.8, for yields up to 5.0 t/ha of grain is 1.0, for yields of 5.1-6.0 t/ha is 1.25. Other indicators are the same as for the tillering phase.

The actual indicators of the chemical composition of leaves of the unfertilized variant in the phase of tillering are N = 3.92%, P = 0.46%, K = 4.3%, in the phase of the end of flowering N = 2.10%, P = 0.29%, K = 3.74%.

Then, for the tillering phase the degree of need for nitrogen is:

For the late blooming phase, the degree of nitrogen requirement will be:

This degree of wheat’s need for nitrogen must be corrected for the lack of phosphorus in the leaves or in the soil if this deficiency cannot be overcome. Considering that the degree of need N = 1.7,

From there, the nitrogen dose would be N kg/ha = DN ⋅ 45 = 1.02 ⋅ 45 = 46 kg/ha instead of 80 kg/ha. This correction is important when determining nitrogen doses for winter and spring crops whose fields have less than the norm content of mobile phosphorus or potassium.

Thus, the dose of nitrogen by leaf chemistry in the phase of tillering is N kg/ha = DN ⋅ MN = 1.4 ⋅ 45 = 63 kg/ha, in the phase of end flowering N kg/ha = DN ⋅ MN = 1.7 ⋅ 45 = 76 kg/ha, the average – 70 kg/ha.

If we take into account the excessive content of potassium in leaves compared with the optimal content, then

and the nitrogen dose will be N = 2.1 ⋅ 45 = 94 kg/ha.

According to the sum of the three calculations, adjusted for some excess of phosphorus and potassium in the two phases of growth, the nitrogen rate will average (63 + 76 + 94) : 3 = 78 kg/ha.

By analogy with the complex method of analytical leaf diagnosis N.K. Boldyrev recommends to calculate fertilizer doses by the content of mobile nutrients in the soil, i.e. to apply a complex method of soil diagnosis. The method is based on determining the normal nutrient composition of the “soil”, which provides a high yield of grain, for example, 4 t/ha of grain of spring wheat. They can be called the optimal parameters of soil fertility by mobile forms of nutrients.

Table. Indicators of the normal nutrient composition of different types of soils, ensuring the receipt of 4 tons of spring wheat and the corresponding level of yield of other crops[8]Agrochemistry. Textbook / V.G. Mineev, V.G. Sychev, G.P. Gamzikov et al. - M.: Publishing house of the All-Russian Scientific Research Institute named after D.N. Pryanishnikov, 2017. - 854 p.

Soil type
Content in soil before sowing (7-10 days before sowing), mg/kg
N-NO3
P2O5 по
K2O по
by Chirikov
by Machigin
by Kirsanov
by Truog
by Chirikov
by Machigin
by Kirsanov
by Maslova
Black Earth:
common
25
180
-
-
200
180
-
-
320
leached
25
180
-
-
200
180
-
-
-
carbonate
25
-
25*
-
-
-
250
-
-
Chestnut
25
-
25
-
-
-
250
-
-
Sod-podzolic with medium granular composition
25
-
-
200
-
-
-
200
-
Peat-bog
125
-
-
1000
-
-
-
1000
-

*For yield levels of 5.0-5.5 t/ha of winter wheat grain, the content corresponds to 35 mg/kg of soil

The table is supplemented by indicators of the optimum ratio between the elements in the soil, which are necessary for subsequent calculated adjustments of fertilizer doses by ratios between mobile nutrients.

Table. Relationship between mobile nutrients in the soil as a basis for assessing nutritional conditions and determining the degree of need for fertilizers cereal crops[9]Agrochemistry. Textbook / V.G. Mineev, V.G. Sychev, G.P. Gamzikov et al; ed. by V.G. Mineev. - M.: Publishing house of the All-Russian Scientific Research Institute named after D.N. Pryanishnikov, … Continue reading

Number of the reference point from the center of the optimum
Р2O5:(N-NO3) mg/kg
Degree of need DN
Characteristics of nitrogen and phosphorus nutritional conditions, their equilibrium
for N
for Р2O5
7
14,4
2,0
0,5
severe nitrogen deficiency with a large excess of phosphorus in the soil
6
12,5
1,8
0,56
5
11,2
1,56
0,63
4
10,0
1,4
0,72
average nitrogen deficiency with an average phosphorus excess
3
9
1,25
0,8
2
8
1,11
0,9
ratio close to normal
1 NO*
7,2
1,0
1,0
balanced supply of N and P2O5
2
6,3
0,9
1,14
ratio close to normal
3
5,6
0,8
1,28
average phosphorus deficiency with an average nitrogen excess
4
5
0,72
1,44
5
4,5
0,63
1,6
severe phosphorus deficiency with a large excess of nitrogen
6
4
0,56
1,8
7
3,6
0,5
2,0

*NO – nutritional optimum in terms of nutrient ratios. Designations: 1 NO – balanced nutrition, defined by the equation: mg/kg P2O5 – 7,2⋅Nnit mg/kg K2O; 2 – nutrition imbalance is weak; 3-4 – medium; 5-7 – strong.

Table. Relationship between potassium and nitrate nitrogen in the soil as a basis for assessing nutritional conditions, determining the degree of need DN and norms of the missing element in fertilizer[10]Agrochemistry. Textbook / V.G. Mineev, V.G. Sychev, G.P. Gamzikov et al; ed. by V.G. Mineev. - M.: Publishing house of the All-Russian Scientific Research Institute named after D.N. Pryanishnikov, … Continue reading

Number of the reference point from the center of the optimum
K2O:(N-NO3) mg/kg
Degree of need DN
Characteristics of potassium and nitrogen nutrition conditions, its equilibrium
for N
for K2O
7
14,4
2,0
0,5
severe nitrogen deficiency with a large excess of phosphorus in the soil
6
12,5
1,3
0,56
5
11,2
1,6
0,63
4
10,0
1,4
0,72
average nitrogen deficiency with an average phosphorus excess
3
9,0
1,25
0,80
2
8,0
1,12
0,9
ratio close to normal
1 NO*
7,2
1,0
1,0
balanced nutrition K2O
2
6,3
0,9
1,14
ratio close to normal
3
5,6
0,8
1,28
average phosphorus deficiency with an average nitrogen excess
4
5,0
0,72
1,44
5
4,5
0,63
1,60
severe phosphorus deficiency with a large excess of nitrogen
6
4,0
0,56
1,80
7
3,6
0,5
2,0

Table. Ratio between mobile phosphorus and exchangeable potassium (according to Chirikov) as a basis for assessing nutritional conditions and determining the degree of need of DN in the missing element[11]Agrochemistry. Textbook / V.G. Mineev, V.G. Sychev, G.P. Gamzikov et al. - M.: Publishing house of the All-Russian Scientific Research Institute named after D.N. Pryanishnikov, 2017. - 854 p.

Number of the reference point from the center of the optimum
P2O5:K2O mg/kg
Degree of need DN
Characteristics of potassium and nitrogen nutrition conditions, its equilibrium
for K2O
for P2O5
7
2,0
2,0
0,5
severe nitrogen deficiency with a large excess of phosphorus in the soil
6
1,8
1,3
0,56
5
1,6
1,6
0,63
4
1,4
1,4
0,72
average nitrogen deficiency with an average phosphorus excess
3
1,25
1,25
0,80
2
1,12
1,12
0,90
ratio close to normal
1 NO*
1,0
1,0
1,0
balanced nutrition K and P
2
0,9
0,9
1,12
ratio close to normal
3
0,8
0,8
1,25
average phosphorus deficiency with an average nitrogen excess
4
0,72
0,72
1,4
5
0,63
0,63
1,6
severe phosphorus deficiency with a large excess of nitrogen
6
0,56
0,56
1,8
7
0,5
0,5
2,0

N:P2O5:K2O ratios in soil characterize the qualitative side of plant nutrition and its balance. The ratio between the optimal and actual content of Nnit and P2O5, between Nnit and K2O is expressed by quantitative indices of the degree of need for the missing element, which are used as correction factors for adjusting the fertilizer rate.

Optimal ratios between mobile nutrients of the soil are set in field experiments with varying doses of fertilizers on the factorial or conventional schemes by determining the correlation relations graphically or by grouping between the indicators of relations of pairs of elements and the amount of yield. Optimal ratios between mobile nutrients of soil for grain crops on some types of soils are expressed by the equations of nutrient balance.

For ordinary and leached chernozems (Chirikov’s method for phosphorus and exchangeable potassium) equality:

P2O5 mg/kg of soil = 7,2⋅N-NO3 mg/kg = K2O mg/kg.

For carbonate chernozems and chestnut soils (Machigin’s method for phosphorus and potassium):

Р2O5 mg/kg of soil = N-NO3 = K2O : 10.

Indicators of optimum soil nutrient composition and optimum ratio between mobile nutrients are used to determine fertilizer doses in other calculation methods as well. Calculation of fertilizer doses according to agrochemical soil analysis is carried out by analogy with the method of leaf diagnostics.

The norm of the missing element is determined by multiplying by the minimum dose of MN element, equal to 30.

For nitrogen under irrigation conditions Mn is differentiated taking into account the planned yield and is: MN = 30 for yields up to 3.6-3.8 t/ha; MN = 45 for yields 4.0-5.0 t/ha and MN = 60 for yields above 6.0 t/ha. For phosphorus MH under irrigation is 45 kg P2O5.

Initial data of soil analysis are taken for ordinary chernozem and getting the planned yield of spring wheat grain at irrigation 4.0 t/ha. Humus content in the layer 0-30 cm is 6.0%, the content 7-10 days before sowing nitrate nitrogen (N-NO3) – 14.0 mg/kg; P2O5 and K2O (by Chirikov) – 182 and 1176 mg/kg respectively. The volume mass of 1 cm3 of the analyzed layer d = 1.05, the depth of the analyzed layer h = 30 cm. Mass of the analyzed soil layer in million kg per 1 ha to convert nutrients from mg/kg to kg/ha, i.e. the conversion factor for the mass of the layer Ml, is determined by the formula:

or

For chestnut soil and irrigation for nitrate nitrogen using 0-60 cm definition layer at d = 1.2 g/cm3, Ml will be 7.2.

Procedure for determining the fertilizer dose

1. Comparison of actual data on N-NO3, P2O5 and K2O with optimum values – 25, 180, 180 mg/kg respectively indicates that wheat plants for 4.0 t/ha yield does not need phosphorus and potassium fertilizers and only need nitrogen.

2. The degree of plants’ need for nitrogen is determined using the equation

correction for phosphorus

The degree of needing N adjusted for the ratio of K2O and N-NO3 because of the excess of exchangeable potassium reaches a very large value.

Correction for potassium

Therefore, the equilibration of nitrogen nutrition is an excess of potassium in the soil, when the correction for potassium exceeds 3, DN for phosphorus is (180 : 182) = 0.99, and for potassium DN = (180 : 1176) = 0.15, that is, the need for these elements is absent.

3. The dose of nitrogen in the basic fertilizer is determined by the formula

DN kg/ha = DN ⋅ MN,

where DN = 1.8, the minimum dose (MN) is 30 kg/ha or 45 kg/ha of nitrogen, N kg/ha = 1.8 ⋅ 30 = 54, N kg/ha = 1.8 ⋅ 45 = 81.

These calculated doses equal to 54 and 81 kg/ha of nitrogen almost coincide with the optimal doses identified in field experiments. There was no need to apply phosphorus and potassium fertilizers.

Based on many years of research of Kazakh Agrotechnical University (KATU) made by V.G. Chernenok, gradations of soils by P2O5 content were corrected for conditions of Northern Kazakhstan.

Table. Gradation of dark chestnut and chernozem soils of Northern Kazakhstan on the content of P2O5 and the effectiveness of phosphorus fertilizers[12]Agrochemistry. Textbook / V.G. Mineev, V.G. Sychev, G.P. Gamzikov et al. - M.: Publishing house of the All-Russian Scientific Research Institute named after D.N. Pryanishnikov, 2017. - 854 p.

Availability class
Availability index
Content Р2O5 in the 0-20 cm layer, mg/kg
Fertilizer efficiency
Actual effect of P60, %, average for 12 field-rotations
I
Very low
Up to 15
Very high (30-50%)
-
II
Low
15-25
High (20-30%)
24
III
Medium
25-35
Medium (10-20%)
13
IV
Increased
35-45
Low (5-10%)
6
V
High
Over 45
Not available
0

Application of balance methods of calculations

The disadvantage of all balance methods is that they do not take into account the previous crops, agrochemical indicators, the state of soil cultivation and other indicators affecting the coefficient of use of nutrients by plants from soil and fertilizer. Therefore, they are considered as indicative, especially if the data for calculations are taken from reference materials. In practice, they give satisfactory results and streamline the application of fertilizers.

The objective of the above calculation methods is to obtain a yield in the current year using the natural fertility of the soil. Fertilizers compensate for the amount of nutrients that cannot be obtained from the soil. In this case there is no systematic improvement of fertility and replenishment of nutrients that are used to form the crop.

Balance calculations of fertilizer doses for the planned yield, taking into account the increase in fertility, have options:

  1. Obtaining high yields by applying small doses with simultaneous depletion of soil nutrients.
  2. Obtaining high yields while maintaining effective fertility at the initial level.
  3. Obtaining the maximum possible yields of the given crop with simultaneous increase of effective fertility.

Determination of phosphorus fertilizer doses

Based on the correlation analysis of experimental data was established quantitative relationship between the content of nutrients in the soil and crop productivity, defined the optimal parameters for different crops and proposed a way to achieve them.

Table. Optimal levels of nutrients in the soil for different crops (V.G. Chernenok, 1993, 2009)

Crop
Content, mg/kg
N-NO3
Р2О5
K2О
Wheat
12-15
35
400
Barley
12-15
35
400
Oats
10-12
28-30
400
Corn
10-12
40
500
Millet
10-12
40
360
Buckwheat
10-12
30-32
400
Chickpeas, peas
12-15
28-30
440
Rapeseed
15-18
30-32
-

This allowed us to propose a new and more accurate way to determine the deficiency of phosphorus in the soil and fertilizer doses for crops by the optimization formula:

DP (kg a.s./ha) = (Poptimal – Pactual) ⋅ K,

where the difference (Poptimal – Pactual) shows the deficit of phosphorus in the field in mg/kg of soil, the coefficient of K = 10 in zonal soils, DP – the amount of phosphate fertilizers, which should be made to eliminate the deficit and create optimal conditions for nutrition phosphorus for crops in this field.

When calculating fertilizer doses, the lower indicator of the optimal level should be included.

If the phosphorus deficiency is very high and it is not possible to bring its content to the optimal level in one application, it can be done in several applications, bringing in the first stage to the average level – 25 mg/kg of soil.

Creating an optimal level of phosphorus in the soil allows the crop to realize the potential of forming the maximum yield in all conditions of moisture.

The optimum regime of phosphorus nutrition contributes to the effective consumption of moisture. For example, at a low level of nutrition on average over 20 years, the coefficient of water consumption was 20 mm, at an average – 12 mm, the optimum – 8 mm.

If there is a large deficit of phosphorus in the soil, when it is not possible to bring the content to the optimal level in one go, you can determine the dose of fertilizer for a certain increase in yield

Example. To increase the yield by 0,5 t it is necessary to increase the content of phosphorus in the soil by 6 mg (5 t⋅1,2 mg P2O5), for this purpose 60 kg a.s. of fertilizer (6⋅10) should be applied.

Table. Relation of spring wheat yield to P2О5 content in soil (KATU, Chernenok, 1970-1990).

Content Р2О5, mg/kg soil
Yield, 100 kg/ha in years:
very dry
medium
humid
10
3-4
7-8
10-12
15
5-6
10-12
17-20
20
6-7
14-16
22-25
25
8-9
17-20
28-30
30
10
21-25
33-37
35
11-12
25-30
38-40
Consumption mg Р2О5 per 100 kg of harvest
3
1,2-1,4
0,8-1,2

Determination of nitrogen fertilizer doses

Thanks to many years of research (Chernenok 1993,1997), 4 main factors of effectiveness of nitrogen fertilizers: nitrate content in the layer 0-40 cm, the content of mobile phosphorus, their ratio and moisture conditions were determined. A quantitative relationship between these factors has been established, which allows us to determine the needs of crops in fertilizers and ensure their effectiveness.

Taking into account these factors a zonal scale of soil nitrogen and phosphorus availability was developed.

Table. Gradation of grain crops nitrogen availability, according to the content of N-NO3 mg/kg of soil in the 0-40 cm layer (Chernenok V.G.)

Availability class
Availability of Р2O5
Nitrogen fertilizer requirements
Recommended dose of N, kg a.s./ha
Normative increase
Very low - medium
Medium - high
100 kg/ha
%
Very low
Up to 4
Up to 6
Very high
60
3-5
30 и >
Low
4-8
6-9
High
45
2-3
20-30
Medium
8-12
9-12
Medium
30
1 -2
10-20
Increased (optimum)
12-15
12-15
Low
-
<1
<10
High
>15
>15
Not available
-
-
0

To calculate the dose of nitrogen fertilizers, you need to determine the content of nitrate nitrogen in the soil in the 0-40 cm layer. If the content of hydrolysable nitrogen is known (according to Tyurin-Kononova), the index should be converted to nitrate nitrogen (N-NO3) by multiplying by the factor 0.26. For subsequent crops placed in the rotation after the fallow, the N-NO3 content according to the average annual data is reduced by 30%.

More accurately the nitrogen deficit in the soil and the need for nitrogen fertilizers is calculated by a formula which takes into account the biological characteristics of the crop, the requirements of the conditions of nitrogen nutrition and the nitrogen content in the soil N-NO3 mg/kg in the layer 0-40 cm (Chernenok V.G.):

DN = (Noptimal – Nactual) ⋅ K ⋅ Kmoisture,

where: DN – dose of nitrogen fertilizers, kg/ha a.s.; Noptimal – the optimum content of nitrate nitrogen in the soil, mg/kg in the layer 0-40 cm; Nactual – the actual content of N-NO3, mg/kg in the layer 0-40 cm; K – the equivalent of kg a.s. fertilizers 1 mg N-NO3 soil, equal to 7.5 kg (that amount of N fertilizers, which should be made to increase the N-NO3 in soil by 1 mg/kg); Kmoisture – moisture correction factor.

The formula assumes bringing the phosphorus content to an optimum level. If the P2O5 content in the soil even after the application of phosphate fertilizers remains below the optimal, which can be determined by dividing the dose of fertilizer applied by 10 (the equivalent cost of kg a.s. fertilizer to increase phosphorus in the soil by 1 mg/kg given the actual content of soil phosphorus):

P (mg/kg) = Pactual + DP/10,

where DP is the applied dose of phosphorus.

In this case, the dose of nitrogen fertilizer is calculated by the formula:

DN = (1/3Pactual – Nactual) ⋅ К ⋅ Kmoisture,

where (1/3Pactual) is an indicator of the level of N-NO3 mg/kg to which nitrogen should be brought, providing the optimum ratio of P : N, equal to 2.5-3.

Example. The soil contains 15 mg P2O5/kg of soil. We applied 120 kg a.s., the content of P2O5 increased to 27 mg (15 + 120/10). Optimal levels of phosphorus have not been reached, so you should not bring the nitrogen to 12 or 15 mg. For 27 mg of P2O5, the optimal N-NO3 content would be 27:3 = 9 mg.

This allows you to maintain the optimal phosphorus to nitrogen ratio for plants in phosphorus deficient soils. Which saves the amount of fertilizer and funds, as phosphorus deficiency will not allow to realize the full dose of nitrogen, designed to bring it to the optimum.

Kmoisture – correction factor for moisture, calculated by the formula:

Kmoisture = Aactual/275,

where: Aactual – actual (predicted) annual precipitation; 275 – normative precipitation, a constant value equal to the mean annual precipitation for the study period.

Actual (predicted) precipitation is calculated based on actual precipitation for September-May plus predicted precipitation for vegetation period. If in June the amount of precipitation is expected within the norm – the long-term average is added, if higher – 1.5 norm, lower – 0.5 norm.

Kmoisture allows to correct normative indicators and calculate dose and increment from nitrogen fertilizers for any wetting year within zonal variation of precipitation. These tables are applicable for all soils.

Table. Doses of nitrogen fertilizers and yield gains (100 kg/ha) depending on N content in soil and Kmoisture

Precipitation per agricultural year, mm
Kmoisture
Nitrogen availability
Very low
Low
Medium
N dose, kg a.s./ha
Yield increase
N dose, kg a.s./ha
Yield increase
N dose, kg a.s./ha
Yield increase
200
0,7
42
2,1-3,5
32
1,4-2,1
21
0,7-1,4
225
0,8
48
2,4-4,0
36
1,6-2,4
24
0,8-1,6
250
0,9
54
2,7-4,5
40
1,8-2,7
27
0,9-1,8
275
1,0
60
3,0-5,0
45
2,0-3,0
30
1,0-2,0
300
1,1
66
3,3-5,5
50
2,2-3,3
33
1,1-2,2
325
1,2
72
3,6-6,0
54
2,4-3,6
36
1,2-2,4
350
1,3
78
3,9-6,5
58
2,6-3,9
40
1,3-2,8
375
1,36
82
4,2-6,8
61
2,7-4,1
41
1,4-2,7

More precisely the increase is determined by the formula:

IN = 1,24 – 0,14N-NO3 + 1,62 Kmoisture + 0,06P/N,

IN – increment from nitrogen fertilizers; N-NO3 – content in soil, mg/kg in 0-40 cm layer; P/N – ratio of actual P2O5 content, mg/kg of soil in 0-20 cm layer to N-NO3, mg/kg in 0-40 cm layer.

In market conditions it is important to know the possible yield increase, so that even before fertilizing, based on the prevailing prices, to determine how this method will be economically justified.

Correlation analysis showed high reliability (r = 0.93) forecast efficiency of nitrogen fertilizers, calculated by the method given, taking into account all 4 factors determining the efficiency.

Sources

Yagodin B.A., Zhukov Y.P., Kobzarenko V.I. Agrochemistry / Edited by B.A. Yagodin. – Moscow: Kolos, 2002. – 584 p.: ill.

Agrochemistry. Textbook / V.G. Mineev, V.G. Sychev, G.P. Gamzikov et al. – M.: Publishing house of the All-Russian Scientific Research Institute named after D.N. Pryanishnikov, 2017. – 854 с.