Linear Interpolation Calculator

The interpolation formula has a second form that is easier to picture. Write t for the position between the points, t = (x - x1) / (x2 - x1). Then y = (1 − t) × y1 + t × y2. That is a weighted average of the two known values. At t = 0 you get y1, at t = 1 you get y2, and at t = 0.5 you get the midpoint of the two. The closer x is to one point, the more that point's value counts.

That is why the calculator shows a position output. A value near 0 or near 1 tells you the answer sits close to a known point, while 0.5 means exactly halfway. Programmers know the same idea as lerp, written lerp(a, b, t) = a + (b - a) × t, and the calculator's position is that t.

The formula also gives the straight line through your two points. The slope is (y2 - y1) / (x2 - x1) and the intercept is the y value where the line crosses x = 0. Both are shown in the results, so you can write the line as y = mx + b.

By hand, work in four small steps. First find the slope: subtract the two y values, subtract the two x values, and divide. Second find how far your x is from the first point: x - x1. Third multiply the slope by that distance. Fourth add y1. For x = 12 between (4, 0) and (20, 100), the slope is 100 / 16 = 6.25, the distance is 8, the product is 50, and adding y1 = 0 gives 50.

On a scientific calculator, including Casio and TI models, type the whole formula with parentheses so the order is right: (x - x1) × (y2 - y1) ÷ (x2 - x1) + y1. The brackets around each subtraction matter. Without them the calculator divides and multiplies the wrong pieces. If your calculator has a memory key, store the slope first and reuse it.

In a spreadsheet, put x1 in A2, y1 in B2, x2 in A3, y2 in B3, and the new x in E2, then use =B2+(E2-A2)*(B3-B2)/(A3-A2). The calculator above does the same arithmetic and also writes out each step.

Interpolation estimates a value between two known points. Extrapolation uses the same formula for an x outside them. The arithmetic does not change, but the reliability does. Between the points you at least know the true values at both ends, while outside them nothing in your data says the trend continues. The battery example shows how that fails: the straight line predicts a negative charge.

For a smooth curve, the error of linear interpolation has a known form. If f is twice differentiable, the difference between the true value and the straight-line estimate is (x - x1)(x2 − x)/2 times the second derivative at some point between them, up to sign. The largest this can be between points h apart is h² / 8 times the biggest value of |f″|. Halving the spacing cuts the worst-case error to a quarter, which is why a finer table gives better interpolated values.

In practice, check two things. First, whether the quantity really changes at a roughly steady rate between your points. Second, whether your points are close together. Look-up tables such as steam tables and calibration charts are usually printed with closely spaced values, which keeps the straight-line error small.

A straight line is a good guess only when the data is close to straight between your two points. Compound growth, decay, pH, decibels, and anything that curves sharply between table entries will be misjudged. For data like that, interpolate on a transformed scale, such as the logarithm of the values, or use a method that follows the curve, such as polynomial or spline interpolation through more than two points.

This calculator handles one straight line through two points. It does not do bilinear or trilinear interpolation, which estimate a value on a grid or a cube. You can still work a bilinear problem by hand with three linear interpolations: interpolate along x twice, once on each of the two rows, then interpolate between those two results along y.

If your known points are noisy measurements rather than exact table values, interpolation is the wrong idea altogether. A fitted line through many points, such as a least-squares line, smooths the noise instead of passing through every error.

Bilinear interpolation estimates a value inside a rectangle when you know the value at all four corners. It is nothing more than linear interpolation done three times: twice along one direction, then once along the other. This calculator handles each pass, so you can run it three times to get a bilinear answer even though it has no dedicated grid mode.

Suppose a table gives values at x = 1 and x = 2 for two rows. At y = 10 the values are 100 and 200, and at y = 20 they are 300 and 400. To estimate the value at x = 1.25, y = 12, first interpolate along x on the y = 10 row: enter x1 = 1, y1 = 100, x2 = 2, y2 = 200 and x = 1.25 to get 125. Next do the same on the y = 20 row with 300 and 400 to get 325. Finally interpolate between those two results along y: x1 = 10 with y1 = 125, x2 = 20 with y2 = 325 and x = 12, which gives 165.

The order does not matter, as long as you finish with the other direction. Doing y first and then x gives the same 165. Bilinear interpolation is only as good as the surface is flat inside the rectangle, so it works best on a fine grid, and it should not be used outside the four corner points.

Straight-line interpolation assumes equal steps in x produce equal steps in y. Many real quantities do not behave that way. Anything that grows or decays by a percentage, such as a population, a decay curve or a sound level, moves in equal ratios, and a straight line between two points sits above the true curve in between.

There are two common fixes, and the calculator handles both without a special mode. If y changes by ratios, interpolate geometrically, which means y = y1 × (y2 / y1)^p, where p is how far x is between the points (the calculator's position output). If x covers several orders of magnitude, such as frequency or concentration, interpolate against the logarithm of x instead. Take the log of x1, x2 and x, enter those, and read the answer.

Here are two examples. Between (0, 100) and (2, 400) the linear estimate at x = 1 is 250, but a growth by ratios gives 100 × 4^0.5 = 200. For x on a log scale, take (10, 2) and (1000, 6) and look up x = 100. Using log10 of each x gives the points (1, 2) and (3, 6) at x = 2, which interpolates to 4. The plain linear answer on the original x values would be about 2.36, which is noticeably different. Always pick the scale that matches how the data behaves, and check it against a known point.

Interpolation estimates values between known data points and passes through them exactly. Curve fitting works the other way around: it chooses a smooth curve that follows the overall trend of many points and usually does not pass through any of them. Both can predict a value at a new x, but they answer different questions. If your numbers are exact table values, interpolate. If they are noisy measurements, fitting a line or curve through all of them is usually better than trusting two noisy neighbors.

Forecasting goes a step further and estimates values beyond the discrete data you have, which is extrapolation, not interpolation. The method is the same formula, but the risk is much higher because nothing constrains the answer past the last point. A trend can turn, saturate or break entirely, as the battery example on this page shows with a negative percentage.

A good rule is to use the closest two points that bracket your x, keep the answer inside the data range when you can, and treat anything outside it as a rough estimate to be confirmed. If the gap between known points is wide or the curve bends sharply, add more known points or use a method that fits the shape.

In a spreadsheet, the two-point formula is =y1+(x-x1)*(y2-y1)/(x2-x1), with each name replaced by a cell reference. It is worth knowing that Excel's FORECAST.LINEAR function is not the same thing. Its documentation describes predicting a value by linear regression on all the known x and y values, so it fits one best-fit line through every point, and its answer will usually not match a two-point interpolation between your nearest points.

In Python, NumPy's interp function performs piecewise linear interpolation over a list of increasing x values. Be careful at the edges: by default, for an x below the first point or above the last point it returns the first or last y value, not an extrapolated one, unless you pass the left and right arguments. The x list must be increasing, and NumPy does not check this for you.

You can also write the interpolation yourself in one line: y1 + t × (y2 - y1), where t is the position between the points, from 0 to 1. This is the form often called lerp in graphics and game code, and it is the same formula the calculator uses. The table below summarizes how each tool behaves.