Linear Interpolation

Linear Interpolation

Linear Interpolation is the simplest form of interpolation used to estimate a value between two known data points. It assumes that the change between the points follows a straight line. While :math:Polyfit seeks to find a single curve for an entire dataset, linear interpolation works "locally" between adjacent coordinates.

1. The Mathematical Formula

Given two points $(x_0, y_0)$ and $(x_1, y_1)$, the value :math:y for any :math:x in the interval $[x_0, x_1]$ is calculated as:

$y = y_0 + (x - x_0) \cfrac{y_1 - y_0}{x_1 - x_0}$

This formula represents the weighted average of the two endpoints based on how close :math:x is to either side.

2. Implementation in SepalSolver

In SepalSolver, the Interp1 method performs 1D linear interpolation. It first searches the input array to find the correct "bin" (the two closest :math:x values) and then applies the linear formula.

double[] xData = [0.0, 10.0, 20.0];
double[] yData = [0.0, 100.0, 400.0];
​
// Estimate the value at x = 5.0 (midway between 0 and 10)
double queryX = 5.0;
double result = Interp1(xData, yData, queryX);
​
// Result: 50.0
Console.WriteLine($"Interpolated value at {queryX} is {result}");

Interpolated value at 5 is 50

Examples

.. Admonition:: Example 1 : Reading Steam Tables

Engineers often use tables of data where values are only provided at fixed intervals (e.g., every 5 degrees). Linear interpolation allows for the estimation of properties at specific, non-tabulated temperatures.

double[] temps = [100, 105, 110];
double[] pressure = [101.3, 120.8, 143.3];
​
double currentTemp = 107.5;
double pEst = Interp1(temps, pressure, currentTemp);
​
Console.WriteLine($"Estimated pressure at {currentTemp}C: :math:{pEst} kPa");

Estimated pressure at 107.5C: :math:132.05 kPa

.. Admonition:: Example 2 : Lookup Tables in Control Systems

In real-time control, complex mathematical functions are often replaced by "lookup tables" to save processing power. Linear interpolation provides a fast way to get reasonably accurate intermediate values from these tables.

double[] rpm = [0, 1000, 2000, 3000, 4000];
double[] torque = [0, 150, 280, 310, 290];
​
double currentRPM = 2500;
double tEst = Interp1(rpm, torque, currentRPM);
​
Console.WriteLine($"Torque at {currentRPM} RPM: :math:`{tEst}` Nm");

Torque at 2500 RPM: :math:`295` Nm

Exercise: Manual Implementation

Task: Implement the core logic of a linear interpolator between two specific points.

double x0 = 2.0, y0 = 10.0;
double x1 = 4.0, y1 = 20.0;
double targetX = 3.0;
​
// Calculate the slope (m)
double slope = (____ - ____) / (____ - ____);
​
// Calculate interpolated Y
double interpolatedY = y0 + slope * (targetX - x0);
​
// Expected: 15.0
Console.WriteLine($"Value at 3.0: :math:{interpolatedY}");

2D Linear Interpolation (Interp2)

2D Interpolation, or Bilinear Interpolation, is used to estimate values of a function $z = f(x, y)$ when the data is provided as a grid. This is essential in fields like meteorology (mapping temperatures across a geographic area) or structural engineering (calculating stress over a surface).

1. How it Works

Bilinear interpolation is essentially a two-step linear interpolation. To find the value at a point $(x, y)$ inside a rectangular cell defined by four points:

1. Interpolate along X: Perform linear interpolation between the points in the $x$-direction to find two intermediate values at the target $x$ coordinate.

1. Interpolate along Y: Perform a final linear interpolation between those two intermediate values in the $y$-direction to find the result $z$.

2. Data Structure

In SepalSolver, Interp2 requires: double[] X: A vector of coordinates for the columns. double[] Y: A vector of coordinates for the rows. * double[,] Z: A 2D matrix where Z[i, j] is the value at (X[j], Y[i]).

double[] xCoords = [0.0, 1.0];
double[] yCoords = [0.0, 1.0];
// Define Z values at (0,0), (1,0), (0,1), (1,1)
double[,] zGrid = {
    { 10.0, 20.0 }, // y = 0
    { 30.0, 40.0 }  // y = 1
};
​
double qX = 0.5, qY = 0.5;
double result = Interp2(xCoords, yCoords, zGrid, qX, qY);
​
// Result: 25.0 (the center of the square)
Console.WriteLine($"Value at (0.5, 0.5): {result}");

Value at (0.5, 0.5): 25

Examples

.. Admonition:: Example 1 : Topographic Height Estimation

Imagine you have a grid of elevation data from a survey. You want to know the height at a specific coordinate that wasn't directly measured.

double[] longitude = [100.0, 101.0, 102.0];
double[] latitude = [50.0, 51.0, 52.0];
double[,] elevation = {
    { 500, 550, 600 },
    { 520, 580, 630 },
    { 540, 610, 650 }
};
​
double lon = 100.5, lat = 50.5;
double height = Interp2(longitude, latitude, elevation, lon, lat);
Console.WriteLine($"Height at ({lon}, {lat}): :math:{height} m");

Height at (100.5, 50.5): :math:537.5 m

.. Admonition:: Example 2 : Pressure Mapping

In aerodynamics, sensors on a wing provide pressure data at specific $(x, y)$ points. Interp2 allows the solver to reconstruct the pressure distribution across the entire surface for lift calculations.

double[] x = [0, 10];
double[] y = [0, 10];
double[,] z = { { 0, 100 }, { 100, 200 } };
​
// Target: Exactly 25% into the grid from the origin (2.5, 2.5)
double targetX = 2.5;
double targetY = 2.5;
​
double val = Interp2(x, y, z, targetX, targetY);
​
Console.WriteLine($"Interpolated Z: {val}");

Interpolated Z: 50

.. Admonition:: Example 3 : Reading steam table

In thermodynamics, the properties of water and steam (such as Enthalpy $h$, Entropy $s$, or Specific Volume $v$) are determined by two independent state variables, typically Pressure ($P$) and Temperature ($T$).

these properties are governed by complex non-linear physics, they are often provided in massive, discrete grids known as Steam Tables.When an engineer needs the enthalpy at a specific state that isn't exactly in the table, they must use Bilinear Interpolation (Interp2).

double[] Temperature = [1000, 1100, 1200, 1300, 1400, 1500, 1600, 1800, 2000],
    Pressure = [0.01, 0.02, 0.03, 0.04, 0.05, 0.06];
double[,] SpecificVolume = new double[,]
{
    {58.758, 29.379, 19.586, 14.689, 11.751, 9.7927},
    {63.373, 31.686, 21.124, 15.843, 12.674, 10.562},
    {67.988, 33.994, 22.663, 16.997, 13.598, 11.331},
    {72.604, 36.302, 24.201, 18.151, 14.521, 12.101},
    {77.219, 38.61,  25.74,  19.305, 15.444, 12.87},
    {81.834, 40.917, 27.278, 20.459, 16.367, 13.639},
    {86.45,  43.225, 28.817, 21.613, 17.29,  14.409},
    {95.68,  47.84,  31.894, 23.92,  19.136, 15.947},
    {104.91, 52.455, 34.97,  26.228, 20.982, 17.485}
};
​
double T = 1350, P = 0.0373;
double sv = Interp2(Temperature, Pressure, SpecificVolume, T, P);
Console.WriteLine($"Specific Volume of superheated water at T = {T}, P = {P} is {sv}");

Specific Volume of superheated water at T = 1350, P = 0.0373 is 20.413475