Estimate a tabulated value that varies along two independent axes. Enter two x bounds, two y bounds, the value at each of the four corners, and your target coordinates. The calculator interpolates along x at each y bound, then interpolates those results along y.

Use distinct bounds on each axis. Match each corner value to its labeled coordinate: f00 belongs to (x0, y0), f10 to (x1, y0), f01 to (x0, y1), and f11 to (x1, y1).

Use consistent units for all x coordinates, consistent units for all y coordinates, and one common unit for the four values. The two axes may represent different quantities. Enter plain numbers; the estimated value has the same unit as the corner values.

Bounds may be entered in either order if the corner assignments follow that order. No target clamping is applied.

\[t=\frac{x-x_0}{x_1-x_0},\qquad u=\frac{y-y_0}{y_1-y_0},\qquad f(x,y)=(1-t)(1-u)f_{00}+t(1-u)f_{10}+(1-t)uf_{01}+tuf_{11}\]

With x bounds 0 and 10, y bounds 0 and 20, and corner values f00 = 100, f10 = 120, f01 = 140, and f11 = 180, the target (5, 10) gives t = 0.5 and u = 0.5. Interpolation along x gives 110 at y0 and 160 at y1. Halfway between those values, the estimate is 135.

At the rectangle’s center, each corner has weight 0.25. At any corner, the formula returns that corner’s supplied value.

What happens outside the rectangle?

If t or u falls outside the interval from 0 to 1, the result is an extrapolation. The same formula is used, but some weights become negative and the estimate may lie outside the range of the corner values. Accuracy beyond the supplied grid depends on how the underlying quantity behaves.

Why is a result unavailable when bounds coincide?

If x0 = x1 or y0 = y1, a normalized coordinate requires division by zero. That coordinate and the final estimate are unavailable, even if the target lies on the collapsed axis. Supply two distinct bounds on both axes.

When is bilinear interpolation appropriate?

Use it for four values at the corners of a rectangle in two independent coordinates. It assumes linear variation along each axis while the other coordinate is held fixed. It reproduces the supplied corners but does not establish the accuracy of estimates between them.

How does this differ from other interpolation methods?

Bilinear interpolation uses four rectangular grid corners. Triangular barycentric interpolation uses three triangle vertices. This calculator does not fit a regression model, evaluate a separately defined curve, or transform coordinate systems.