Linear Interpolation

How to use it

Review the methodology below to make sure it aligns with your project's requirements, then:

1. Enter the two known points, (x₁, y₁) and (x₂, y₂).
2. Choose whether to find y at a given x, or x at a given y.
3. Enter that x (or y). The result and a graph of the line appear as you type. If the value is outside the two known points, the calculator says so: that is an extrapolation, not an interpolation.

Read more in our blog post, Unlock the power of precision with our linear interpolation calculator.

Methodology, equations and examples

Linear interpolation is a numerical method for estimating values between two known data points by assuming a straight line between them. It is a basic form of interpolation: when the data between two points is close to a straight line, linear interpolation fills in the values in between.

Linear interpolation is used in mathematics, computer graphics, data analysis and engineering. It gives a simple, quick approximation of missing or intermediate values, especially when the data lies close to a straight line. For more complex curves, or where a smoother fit is needed, other methods such as spline interpolation may be more appropriate.

To perform linear interpolation, we need two adjacent data points, (x₁, y₁) and (x₂, y₂). The goal is to estimate the value of an unknown point (x, y) on the line between them:

y = y₁(x₂ − x)/(x₂ − x₁) + y₂(x − x₁)/(x₂ − x₁)

Here (x − x₁) is the distance from the first known point to the unknown point, (x₂ − x) the distance from the unknown point to the second, and (x₂ − x₁) the distance between the two known points. Their ratios place the unknown point along the line joining the two known points. Solving the same line for x gives the x at a given y:

x = x₁(y₂ − y)/(y₂ − y₁) + x₂(y − y₁)/(y₂ − y₁)

Linear interpolation assumes a straight line between the points, which is not always accurate, particularly if the data follows a more complex pattern. In such cases spline interpolation or another method may give better results. Within a small range, though, linear interpolation is a simple and quick estimate.

Example

Suppose you want the total cooling capacity of a Bryant® packaged air conditioning unit from its performance data table, at these design conditions:

• Condenser entering air temperature: 79 °F
• Evaporator entering wet bulb temperature: 63 °F

From the manufacturer's table, at 63 °F entering wet bulb:

• x₁ = 75 °F, y₁ = 55.04 MBtuh (total capacity at 75 °F condenser entering air)
• x₂ = 85 °F, y₂ = 52.59 MBtuh (total capacity at 85 °F)
• x = 79 °F

Solution 1, graphically. Plot the two known points, (75 °F, 55.04 MBtuh) and (85 °F, 52.59 MBtuh), and draw a straight line between them. Draw a vertical line at x = 79 °F, and from where it meets the line, a horizontal line to the y-axis. It reads about 54 MBtuh.

Solution 2, by the equation.

y = 55.04 × (85 − 79)/(85 − 75) + 52.59 × (79 − 75)/(85 − 75) = 33.024 + 21.036 = 54.06 MBtuh

Solution 3, with the calculator. Enter the same values. The result is 54.06 MBtuh, the calculator's starting example.

So at 79 °F condenser entering air, the total cooling capacity is about 54.06 MBtuh.

Why use linear interpolation

• Data approximation: estimate values between known data points, such as between the rows and columns of an equipment performance table.
• Data visualization: fill in missing or incomplete points to plot a continuous series.
• Function approximation: approximate a function from a set of discrete data points by joining them with straight lines, where the function itself is not known or is hard to compute.
• Time series: fill in missing values in a sequence from the values either side.
• Numerical methods: linear interpolation is a building block for more complex schemes, such as cubic spline interpolation.

To interpolate between four points, in two directions at once, use the bilinear interpolation calculator.