top of page

Linear Interpolation 

Instructions:

  1. Fill in values for known point (x1, y1)

  2. Fill in values for  known point (x2, y2)

  3. Fill in either x or y

  4. Click the Interpolate button

Read more about interpolation on our latest blog post

Linear Interpolation Graph
Bryant® - 3.5 Ton 14.5 SEER Packaged Air Conditioning Unit

Methodology, Equations, and Examples

Linear interpolation is a numerical method for curve fitting or estimating values between two known data points on a straight line. It is a basic form of interpolation that assumes a linear relationship between the given data points. When you have a set of data points that do not form a smooth curve but instead lie on a straight line, linear interpolation can be used to estimate the values at points between the given data points. This method assumes that the relationship between the data points is linear and fills in the missing values accordingly.

Linear interpolation is commonly used in various fields, such as mathematics, computer graphics, data analysis, and engineering. It provides a simple and quick approximation of missing or intermediate values, especially when the data points lie in a straight line. However, other interpolation methods, such as spline interpolation, may be more appropriate for more complex curves or when a smoother fit is required.

To perform linear interpolation, we need two adjacent data points, let's call them (x1, y1) and (x2, y2), with x1 < x2. The goal is to estimate the value of an unknown point (x, y). The linear interpolation equations are:

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

 

 

Note that (x - x1) represents the distance between the unknown point and the first data point, (x2-x) represents the distance between the second data point and the unknown point, and (x2 - x1) represents the distance between the two data points. By calculating the ratio of these distances, we can determine the position of the unknown point along the line connecting the two data points. The result will be an estimation of the unknown point's value based on the assumption of a straight line between the two adjacent data points.

It's important to note that linear interpolation assumes a linear relationship between data points, which may not always be accurate, particularly if the data exhibits more complex patterns. In such cases, spline interpolation or other more sophisticated methods may be used to achieve more accurate results. However, linear interpolation is a simple and quick method for estimating values within a given range based on a straight-line approximation.

Example

Suppose you want to calculate the Total Cooling Capacity of a 5-ton Bryant® package unit. A snippet of the equipment's performance data table at design conditions is provided.  For our scenario, the design conditions are: 

  • Condenser Entering Air Temperature:  79oF

  • Evaporator Entering Wet Bulb Temperature: 63oF

Solution: 

For Solutions 1, 2, and 3, use the following inputs, which are provided in the table:

y1 = 55.04 MBtuh

y2 = 52.59 MBtuh

x1 = 75oF

x2 = 85oF
 

and the value of x is given as:
 

x = 79oF

 

Solution 1: 

To graphically find the Total Cooling Capacity at 79oF Condenser Entering Air Temperature and 63oF Evaporator Entering Wet Bulb Temperature, plot the given information on a graph. 

  1. Plot (x1, y1) where, from the performance data table, x1=75oF and y1=55.04MBtuh, the Total Capacity at 75oF.

  2. Plot (x2, y2) where, from the performance data table, x2=85oF and y2= 52.59 MBtuh, the Total Capacity at 85oF. 

  3. Draw a straight line between those two points; let's call this line the "slope."  

  4. Draw a vertical line at x=79oF. 

  5. Draw a horizontal line from where the vertical line at 79oF intersects the slope to the y-axis. The point at which this line intersects is the y value.  For our example, the y value is approximately 54MBtuh.  That is, by linear interpolation, when the Condenser Entering Temperature is 79oF, the Total Cooling Capacity is approximately 54 MBtuh.

Solution 2:

Solve for y using the equation provided:

= y1(x2-x)/(x2-x1) + y2(x-x1)/(x2-x1)

= 55.04(85-79)/(85-75) + 52.59(79-75)/(85-75)

= 54.06 MBtuh

Solution 3: 

To Solve using the Linear Interpolation calculator, enter the given inputs. The result appears at the bottom of the calculator as 54.06 MBtuh.

Linear interpolation is commonly used in various fields for the following reasons:

  1. Data approximation: Linear interpolation allows us to estimate values between known data points, providing a smooth approximation of the underlying data. This is particularly useful when we have limited data points and want to infer values at intermediate positions.

  2. Data visualization: When plotting data on a graph or chart, linear interpolation can be used to fill in missing or incomplete data points. This helps create a more continuous and visually appealing representation of the data.

  3. Function approximation: Linear interpolation can be used to approximate a function based on a set of discrete data points. An estimate of the function's behavior between the data points can be obtained by connecting the points with straight lines. This is especially useful when the function is not explicitly known or difficult to compute.

  4. Time series analysis: Linear interpolation is often applied in time series analysis to fill in missing values in a sequence of data points. By estimating the missing values using adjacent known values, the time series can be more accurately represented, allowing for more reliable analysis and forecasting.

  5. Numerical methods: Linear interpolation is foundational in various numerical methods and algorithms. It is commonly used as a building block for more complex interpolation schemes, such as cubic spline interpolation.

Linear Interpolation Graph
Screenshot of interpolation calculator results
linear interpolation graph
Anchor 1
bottom of page