Storyboard

The main components of soil include both coarse and fine sand, silt, and clay. The proportion of these three components varies depending on the type of soil. Due to significant differences in particle size among these components, spaces known as pores are formed. These pores can accommodate or facilitate the flow of water, which in turn affects the moisture content of the unsaturated part of the soil.

>Model

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Iframe

>Top

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Image

>Top

Soils have a basic structure composed of different layers that result from their initial formation and how they develop, influenced by factors such as erosion and vegetation growth. While the thickness of these layers can vary significantly, they exhibit clear patterns in terms of their organic matter content:

Soil profile (commons.wikimedia.org - translated)

O: Organic Surface Layer

This layer originates from the accumulation of organic matter, primarily consisting of decomposing plant material and humus. The amount of organic material can reach up to 100%, depending on the type of vegetation and its stage of development.

A: Surface Soil

The A horizon is formed through a combination of mineral erosion, the addition of organic matter from the upper horizon, and the activities of soil organisms. Most of this layer is composed of soil's inherent mineral components, with an organic matter content that can reach up to 10%.

B: Subsoil

The B horizon accumulates minerals and leached materials that have moved downward from the A horizon. This accumulation may also result from the arrival of minerals transported from other areas and chemical processes occurring within the horizon. In this layer, the majority of components are soil minerals, with an organic matter content that can reach up to 5%.

C: Substratum

The C horizon is primarily formed by geological processes and contributes to the overall terrain configuration. It provides materials that interact with the upper layers. In this layer, most components consist of soil minerals, with an organic matter content that can reach up to 1%.

The thickness of the first two layers can vary from a few centimeters to several tens of centimeters, while the latter layers can span from tens of centimeters to meters. It is important to note that the thickness values mentioned in the Wikipedia image are provided solely as illustrative examples and may vary widely in reality.

ID:(15066, 0)

Concept

>Top

Soil is composed of varying proportions of sand, silt, and clay. Therefore, to model soil behavior accurately, it is necessary to represent its structure as a mixture of these components.

The simplest way to determine the composition of a particular sample is by calculating the mass of each component in relation to the total mass. To do this, physically separating each sample into its individual components is required, taking advantage of the differences in particle size. After drying and grinding the soil, a sieving process is performed using multiple sieves of different sizes, chosen based on the required precision:

The image illustrates a scheme that separates three components, which could correspond to measurements of sand, silt, and clay. However, it is possible to use multiple sieves to differentiate, for example, coarse sand, medium sand, and fine sand, as needed.

The main masses are as follows:

• the dry mass of sand in the sample ($M_a$): Mass of all sand grains.
• the dry mass of silt in the sample ($M_i$): Mass of all silt grains.
• the dry mass of clay in the sample ($M_c$): Mass of all clay grains.

ID:(2067, 0)

Concept

>Top

A ternary diagram is a visual tool used to represent the relative proportions of three components, such as silt, sand, and clay. In this diagram, each corner is dedicated to one of these components, and various points within the triangular space signify unique combinations of these constituents.

To interpret a ternary diagram for silt, sand, and clay:

1. Identify the components' locations:
- Sand typically occupies the bottom-left corner.
- Silt is found at the bottom-right corner.
- Clay is situated at the top corner.

2. Analyze the mixtures:
- Points located inside the triangle denote diverse mixtures of these three components.

3. Understand lines and zones:
- Lines traversing the triangle signify consistent ratios of the three components.
- Different regions within the triangle correspond to distinct soil classifications.

In particular, for the specified zones, the following ranges can be estimated:

where any point has to satisfy the condition

ID:(2070, 0)

Top

>Top

Parameters

$\rho_c$

rho_c

Density of a grain of clay

kg/m^3

$\rho_a$

rho_a

Density of a grain of sand

kg/m^3

$\rho_i$

rho_i

Density of a grain of silt

kg/m^3

$m_c$

m_c

Mass of a clay plate

kg

$m_a$

m_a

Mass of a grain of sand

kg

$m_i$

m_i

Mass of a grain of silt

kg

$\rho_p$

rho_p

Particle density

kg/m^3

$\rho_s$

rho_s

Solid Density

kg/m^3

$v_a$

v_a

Volume grain of sand

m^3

$v_c$

v_c

Volume of a grain of clay

m^3

$v_i$

v_i

Volume of a grain of silt

m^3

Variables

$M_c$

M_c

Dry mass of clay in the sample

kg

$M_a$

M_a

Dry mass of sand in the sample

kg

$M_i$

M_i

Dry mass of silt in the sample

kg

$g_c$

g_c

Mass fraction of clay in the sample

-

$g_a$

g_a

Mass fraction of sand in the sample

-

$g_i$

g_i

Mass fraction of silt in the sample

-

$N_c$

N_c

Number of clay grains in the sample

-

$N_a$

N_a

Number of sand grains in the sample

-

$N_i$

N_i

Number of silt grains in the sample

-

$V_s$

V_s

Solid volume of a component

m^3

$V_c$

V_c

Solid volume of clay

m^3

$V_a$

V_a

Solid volume of sand

m^3

$V_i$

V_i

Solid volume of silt

m^3

$M_s$

M_s

Total Dry Mass of Sample

kg

Calculations

• Alternative method
• Traditional method
• Strategy
• 2D
• 3D

Equation

Number

First, select the equation: to , then, select the variable: to

---

Calculations

Symbol

Equation

Solved

Translated

Calculations

Symbol

Equation

Solved

Translated

Variable Given Calculate Target : Equation To be used

Equations

ID:(15216, 0)

Equation

>Top, >Model

During the analysis of a sample, the first step involves removing the contained water to prevent its influence. In this way, we obtain the total Dry Mass of Sample ($M_s$), which corresponds to the sum of the dry mass of sand in the sample ($M_a$), the dry mass of silt in the sample ($M_i$), and the dry mass of clay in the sample ($M_c$):

$M_s = M_a + M_l + M_c$

Conditions

Variables

$M_c$

Dry mass of clay in the sample

$kg$

6035

$M_a$

Dry mass of sand in the sample

$kg$

6033

$M_i$

Dry mass of silt in the sample

$kg$

6034

$M_s$

Total Dry Mass of Sample

$kg$

5987

Relation

It's important to emphasize that this mass must be completely dry, as moisture can distort the weight of each component. Additionally, any components such as rocks and organic material other than sand, silt, or clay must be removed from the sample.

ID:(4729, 0)

Equation

>Top, >Model

To describe the soil model, we must first understand its composition. To do this, we introduce the variable representing the mass fraction of sand in the sample ($g_a$). This fraction is calculated from the dry mass of sand in the sample ($M_a$) and the total Dry Mass of Sample ($M_s$), using the following relationship:

This relationship is expressed as follows:

$g_a =\displaystyle\frac{ M_a }{ M_s }$

Conditions

Variables

$M_a$

Dry mass of sand in the sample

$kg$

6033

$g_a$

Mass fraction of sand in the sample

$-$

5797

$M_s$

Total Dry Mass of Sample

$kg$

5987

Relation

ID:(4716, 0)

Equation

>Top, >Model

To describe the soil model, we must first understand its composition. To do this, we introduce the variable representing the mass fraction of silt in the sample ($g_i$). This fraction is calculated from the dry mass of silt in the sample ($M_i$) and the total Dry Mass of Sample ($M_s$), using the following relationship:

This relationship is expressed as follows:

$g_i =\displaystyle\frac{ M_i }{ M_s }$

Conditions

Variables

$M_i$

Dry mass of silt in the sample

$kg$

6034

$g_i$

Mass fraction of silt in the sample

$-$

10098

$M_s$

Total Dry Mass of Sample

$kg$

5987

Relation

ID:(15064, 0)

Equation

>Top, >Model

To describe the soil model, we must first understand its composition. To do this, we introduce the variable representing the mass fraction of clay in the sample ($g_c$). This fraction is calculated from the dry mass of clay in the sample ($M_c$) and the total Dry Mass of Sample ($M_s$), using the following relationship:

This relationship is expressed as follows:

$g_c =\displaystyle\frac{ M_c }{ M_s }$

Conditions

Variables

$M_c$

Dry mass of clay in the sample

$kg$

6035

$g_c$

Mass fraction of clay in the sample

$-$

10099

$M_s$

Total Dry Mass of Sample

$kg$

5987

Relation

ID:(15065, 0)

Equation

>Top, >Model

Since the mass fraction of sand in the sample ($g_a$), the mass fraction of silt in the sample ($g_i$), and the mass fraction of clay in the sample ($g_c$) are calculated based on their proportion in relation to the respective masses of the dry mass of sand in the sample ($M_a$), the dry mass of silt in the sample ($M_i$), and the dry mass of clay in the sample ($M_c$) in the sample, which sum up to the total Mass ($M_t$) for normalization:

We obtain a normalization condition:

$g_a + g_i + g_c = 1$

Conditions

Variables

$g_c$

Mass fraction of clay in the sample

$-$

10099

$g_a$

Mass fraction of sand in the sample

$-$

5797

$g_i$

Mass fraction of silt in the sample

$-$

10098

Relation

Development

When we add the mass fraction of sand in the sample ($g_a$) in the sample:

$g_a =\displaystyle\frac{ M_a }{ M_s }$

along with the mass fraction of silt in the sample ($g_i$) in the sample:

$g_i =\displaystyle\frac{ M_i }{ M_s }$

and the mass fraction of clay in the sample ($g_c$) in the sample:

$g_c =\displaystyle\frac{ M_c }{ M_s }$

we obtain:

$g_a + g_i + g_c = \displaystyle\frac{M_a}{M_t} + \displaystyle\frac{M_i}{M_t} + \displaystyle\frac{M_c}{M_t} = \displaystyle\frac{M_a + M_i + M_c}{M_t}$

Given that $M_t$ represents the total mass, as per the condition:

$M_s = M_a + M_l + M_c$

this results in:

$g_a + g_i + g_c = 1$

ID:(15072, 0)

Equation

>Top, >Model

To model the soil, it is important to introduce the variable corresponding to the number of sand grains in the sample ($N_a$), which can be calculated by dividing the dry mass of sand in the sample ($M_a$) by the mass of a grain of sand ($m_a$):

$N_a = \displaystyle\frac{ M_a }{ m_a }$

Conditions

Variables

$M_a$

Dry mass of sand in the sample

$kg$

6033

$m_a$

Mass of a grain of sand

$kg$

4942

$N_a$

Number of sand grains in the sample

$-$

4941

Relation

ID:(1539, 0)

Equation

>Top, >Model

To model the soil, it is crucial to introduce the variable corresponding to the number of silt grains in the sample ($N_i$), which can be calculated by dividing the dry mass of silt in the sample ($M_i$) by the mass of a grain of silt ($m_i$):

$N_i = \displaystyle\frac{ M_i }{ m_i }$

Conditions

Variables

$M_i$

Dry mass of silt in the sample

$kg$

6034

$m_i$

Mass of a grain of silt

$kg$

5988

$N_i$

Number of silt grains in the sample

$-$

10100

Relation

ID:(15067, 0)

Equation

>Top, >Model

To model the soil, it is crucial to introduce the variable corresponding to the number of clay grains in the sample ($N_c$), which can be calculated by dividing the dry mass of clay in the sample ($M_c$) by the mass of a clay plate ($m_c$):

$N_c = \displaystyle\frac{ M_c }{ m_c }$

Conditions

Variables

$M_c$

Dry mass of clay in the sample

$kg$

6035

$m_c$

Mass of a clay plate

$kg$

6278

$N_c$

Number of clay grains in the sample

$-$

10101

Relation

ID:(15068, 0)

Equation

>Top, >Model

To model the soil's behavior, it is necessary to introduce the variable corresponding to the solid volume of sand ($V_a$). This variable can be calculated from the number of sand grains in the sample ($N_a$) and the volume grain of sand ($v_a$) using the following equation:

$V_a = N_a v_a$

Conditions

Variables

$N_a$

Number of sand grains in the sample

$-$

4941

$V_a$

Solid volume of sand

$m^3$

6554

$v_a$

Volume grain of sand

$m^3$

5798

Relation

Development

If we model the grain as a sphere with a radius of $r_a$, we can calculate its volume using the formula:

$v_a =\displaystyle\frac{4 \pi }{3} r_a ^3$

With the density of the sand denoted as $\rho_a$, we can determine the mass of a single sand grain using the equation:

$m_a = \rho_a v_a$

By dividing the mass of the sample's sand component $M_a$ by the mass of a single grain, we can find the number of grains:

$N_a = \displaystyle\frac{ M_a }{ m_a }$

This allows us to calculate the volume of the sand component:

$V_a = N_a v_a$

ID:(10366, 0)

Equation

>Top, >Model

To model the soil's behavior, it is necessary to introduce the variable corresponding to the solid volume of silt ($V_i$). This variable can be calculated from the number of silt grains in the sample ($N_i$) and the volume of a grain of silt ($v_i$) using the following equation:

$V_i = N_i v_i$

Conditions

Variables

$N_i$

Number of silt grains in the sample

$-$

10100

$V_i$

Solid volume of silt

$m^3$

6555

$v_i$

Volume of a grain of silt

$m^3$

4943

Relation

Development

If we model the grain as a sphere with a radius of $r_i$, we can calculate its volume using the formula:

$v_i = a_i ^3$

With the density of silt denoted as $\rho_i$, we can determine the mass of a single silt grain using the equation:

$m_i = \rho_i v_i$

By dividing the mass of the silt sample $M_i$ by the mass of a single grain, we can obtain the number of grains:

$V_c = N_c v_c$

This allows us to calculate the volume of the silt component:

$V_i = N_i v_i$

ID:(10365, 0)

Equation

>Top, >Model

To model the soil's behavior, it is necessary to introduce the variable corresponding to the solid volume of clay ($V_c$). This variable can be calculated from the number of clay grains in the sample ($N_c$) and the volume of a grain of clay ($v_c$) using the following equation:

$V_c = N_c v_c$

Conditions

Variables

$N_c$

Number of clay grains in the sample

$-$

10101

$V_c$

Solid volume of clay

$m^3$

6556

$v_c$

Volume of a grain of clay

$m^3$

5799

Relation

Development

If we model the grain as a thin plate with length and width $l_c$ and height $w_c$, we can calculate its volume using the formula:

$v_c = w_c l_c ^2$

With the density of clay denoted as $\rho_c$, we can determine the mass of a single clay grain using the equation:

$m_c = \rho_c v_c$

By dividing the mass of the clay sample $M_c$ by the mass of a single grain, we can obtain the number of grains:

$N_c = \displaystyle\frac{ M_c }{ m_c }$

This allows us to calculate the volume of the clay component:

$V_c = N_c v_c$

ID:(15069, 0)

Equation

>Top, >Model

Using the definition of density, it is possible to calculate the solid volume of sand ($V_a$) directly from the dry mass of sand in the sample ($M_a$) and the density of a grain of sand ($\rho_a$) using the following formula:

$V_a =\displaystyle\frac{ M_a }{ \rho_a }$

Conditions

Variables

$\rho_a$

Density of a grain of sand

2.63e+3

$kg/m^3$

10095

$M_a$

Dry mass of sand in the sample

$kg$

6033

$V_a$

Solid volume of sand

$m^3$

6554

Relation

The direct calculation has the advantage that it does not depend on the shape of the grains:

If we assume that density is homogeneous, we can calculate the total volume independently of the number and shape of the grains.

ID:(3168, 0)

Equation

>Top, >Model

Using the definition of density, it is possible to calculate the solid volume of silt ($V_i$) directly from the dry mass of silt in the sample ($M_i$) and the density of a grain of silt ($\rho_i$) using the following formula:

$V_i =\displaystyle\frac{ M_i }{ \rho_i }$

Conditions

Variables

$\rho_i$

Density of a grain of silt

2.65e+3

$kg/m^3$

10094

$M_i$

Dry mass of silt in the sample

$kg$

6034

$V_i$

Solid volume of silt

$m^3$

6555

Relation

This direct calculation has the advantage that it does not depend on the shape of the grains:

If we assume that density is homogeneous, we can calculate the total volume independently of the number and shape of the grains.

ID:(15070, 0)

Equation

>Top, >Model

Using the definition of density, it is possible to calculate the solid volume of clay ($V_c$) directly from the dry mass of clay in the sample ($M_c$) and the density of a grain of clay ($\rho_c$) using the following formula:

$V_c =\displaystyle\frac{ M_c }{ \rho_c }$

Conditions

Variables

$\rho_c$

Density of a grain of clay

2.8e+3

$kg/m^3$

10093

$M_c$

Dry mass of clay in the sample

$kg$

6035

$V_c$

Solid volume of clay

$m^3$

6556

Relation

This direct calculation has the advantage that it does not depend on the shape of the grains:

If we assume that density is homogeneous, we can calculate the total volume independently of the number and shape of the grains.

ID:(15071, 0)

Equation

>Top, >Model

Once we know electron Energy ($E$), susceptible ($S_t$), and infected ($I_t$), we can determine the solid volume of a component ($V_s$) by summing the various components, as described in the following equation:

$V_s = V_a + V_l + V_c$

Conditions

Variables

$V_s$

Solid volume of a component

$m^3$

6038

$V_c$

Solid volume of clay

$m^3$

6556

$V_a$

Solid volume of sand

$m^3$

6554

$V_i$

Solid volume of silt

$m^3$

6555

Relation

ID:(4734, 0)

Equation

>Top, >Model

Since we already know the total Dry Mass of Sample ($M_s$) and the solid volume ($V_s$) from the sample, we can introduce the solid Density ($\rho_s$) and calculate it using the following equation:

$\rho_s = \displaystyle\frac{ M_s }{ V_s }$

Conditions

Variables

$\rho_s$

Solid Density

$kg/m^3$

4944

$V_s$

Solid volume of a component

$m^3$

6038

$M_s$

Total Dry Mass of Sample

$kg$

5987

Relation

ID:(15073, 0)

Equation

>Top, >Model

The solid Density ($\rho_s$) can be calculated from the density of a grain of sand ($\rho_a$), the density of a grain of silt ($\rho_i$), and the density of a grain of clay ($\rho_c$), as well as from the factors describing its composition: the mass fraction of sand in the sample ($g_a$), the mass fraction of silt in the sample ($g_i$), and the mass fraction of clay in the sample ($g_c$). This leads us to the following relationship:

$\displaystyle\frac{1}{ \rho_s }=\displaystyle\frac{ g_a }{ \rho_a }+\displaystyle\frac{ g_i }{ \rho_i }+\displaystyle\frac{ g_c }{ \rho_c }$

Conditions

Variables

$\rho_c$

Density of a grain of clay

2.8e+3

$kg/m^3$

10093

$\rho_a$

Density of a grain of sand

2.63e+3

$kg/m^3$

10095

$\rho_i$

Density of a grain of silt

2.65e+3

$kg/m^3$

10094

$g_c$

Mass fraction of clay in the sample

$-$

10099

$g_a$

Mass fraction of sand in the sample

$-$

5797

$g_i$

Mass fraction of silt in the sample

$-$

10098

$\rho_s$

Solid Density

$kg/m^3$

4944

Relation

Development

If we express the inverse of the solid Density ($\rho_s$) defined in the total Dry Mass of Sample ($M_s$) and the solid volume ($V_s$) using the equation:

$\rho_s = \displaystyle\frac{ M_s }{ V_s }$

And considering that the solid volume is the sum of the solid volume of sand ($V_a$), the solid volume of silt ($V_i$), and the solid volume of clay ($V_c$):

$V_s = V_a + V_l + V_c$

We obtain:

$\displaystyle\frac{1}{\rho_s}=\displaystyle\frac{V_s}{M_s}=\displaystyle\frac{V_a+V_i+V_c}{M_s}$

Substituting the volumes with the relationships for the density of a grain of sand ($\rho_a$)

$V_a =\displaystyle\frac{ M_a }{ \rho_a }$

for the density of a grain of silt ($\rho_i$)

$V_i =\displaystyle\frac{ M_i }{ \rho_i }$

and the density of a grain of clay ($\rho_c$)

$V_c =\displaystyle\frac{ M_c }{ \rho_c }$

We get:

$\displaystyle\frac{1}{\rho_s}=\displaystyle\frac{1}{\rho_a}\displaystyle\frac{M_a}{M_s}+\displaystyle\frac{1}{\rho_i}\displaystyle\frac{M_i}{M_s}+\displaystyle\frac{1}{\rho_c}\displaystyle\frac{M_c}{M_s}$

With the mass fraction of sand in the sample ($g_a$)

$g_a =\displaystyle\frac{ M_a }{ M_s }$

for the mass fraction of silt in the sample ($g_i$)

$g_i =\displaystyle\frac{ M_i }{ M_s }$

and the mass fraction of clay in the sample ($g_c$)

$g_c =\displaystyle\frac{ M_c }{ M_s }$

We obtain:

$\displaystyle\frac{1}{ \rho_s }=\displaystyle\frac{ g_a }{ \rho_a }+\displaystyle\frac{ g_i }{ \rho_i }+\displaystyle\frac{ g_c }{ \rho_c }$

ID:(15074, 0)

Equation

>Top, >Model

The particle density ($\rho_p$) can be calculated as the average density where the weighting factors are the components. Therefore, with the mass fraction of sand in the sample ($g_a$), the mass fraction of silt in the sample ($g_i$), the mass fraction of clay in the sample ($g_c$), and the density of a grain of sand ($\rho_a$), the density of a grain of silt ($\rho_i$), the density of a grain of clay ($\rho_c$), it can be defined as follows:

$\rho_p = \rho_a g_a + \rho_i g_i + \rho_c g_c$

Conditions

Variables

$\rho_c$

Density of a grain of clay

2.8e+3

$kg/m^3$

10093

$\rho_a$

Density of a grain of sand

2.63e+3

$kg/m^3$

10095

$\rho_i$

Density of a grain of silt

2.65e+3

$kg/m^3$

10094

$g_c$

Mass fraction of clay in the sample

$-$

10099

$g_a$

Mass fraction of sand in the sample

$-$

5797

$g_i$

Mass fraction of silt in the sample

$-$

10098

$\rho_p$

Particle density

$kg/m^3$

10168

Relation

ID:(15127, 0)