Logarithmic Interpolation Calculator
Interpolate values accurately on semi-logarithmic (log-X, log-Y) and log-log coordinate systems. Essential for geotechnical sieve curves, acoustics, Bode plots, and exponential decay.
Why Use a Logarithmic Interpolation Calculator?
In many engineering disciplines and physical systems, phenomena do not change linearly with addition; they change proportionally with multiplication across multiple orders of magnitude. Examples include the human ear's perception of sound frequencies, earthquake Richter scales, and soil permeability.
If you attempt to apply standard linear interpolation across a logarithmic span (for instance, between 10 Hz and 10,000 Hz), the arithmetic midpoint is 5,005 Hz, whereas the logarithmic center is 100 Hz (a factor of 10). Performing logarithmic interpolation with our logarithmic interpolation calculator preserves scale fidelity by converting inputs into logarithmic space before computing fractional distances via the proven logarithmic interpolation formula.
Semi-Log (Logarithmic X) Formula
- y=Interpolated value: The calculated output value on the transformed scale.
- x₀, x₁=Boundary coordinates: The known lower and upper coordinate limits (must be > 0 for log X).
- y₀, y₁=Boundary responses: The known function values at x₀ and x₁.
- x=Target query point: The evaluation position on the logarithmic axis.
Note: For Log-Log curves, both x and y are transformed into log space: log(y) = log(y₀) + t × [log(y₁) - log(y₀)], followed by taking the antilog.
Step-by-Step Worked Examples
Walk through two complete real-world calculation problems or click any example to load its numbers directly into the calculator.
Geotechnical Sieve Grain Size Analysis (D50)
In civil engineering soil testing, percent passing is plotted on a linear Y-axis against particle diameter (mm) on a logarithmic X-axis. A sample has 38% passing at 0.075 mm and 62% passing at 0.425 mm. Find the D50 grain diameter corresponding to 50% passing.
- Convert X coordinates to log₁₀: log₁₀(0.075) = -1.124939, log₁₀(0.425) = -0.371611
- Target x log: log₁₀(0.150) = -0.823909
- Compute log fraction: t = [-0.823909 - (-1.124939)] / [-0.371611 - (-1.124939)] = 0.301030 / 0.753328 = 0.39960
- Interpolate linear Y (%): y = 38 + 0.39960 × (62 - 38) = 38 + 9.5904 = 47.59%
Electronic Low-Pass Filter Frequency Attenuation (Bode Plot)
An electrical engineer tests an active filter circuit. At frequency f = 100 Hz, gain is -3.0 dB. At f = 10,000 Hz, gain drops to -40.0 dB. Estimate the filter attenuation at f = 1,000 Hz assuming semi-log response.
- Take base-10 logarithms of frequencies: log₁₀(100) = 2.0000, log₁₀(10,000) = 4.0000, log₁₀(1,000) = 3.0000
- Logarithmic span: 4.0000 - 2.0000 = 2.0000 decades
- Fraction across span: t = (3.0000 - 2.0000) / 2.0000 = 1.0000 / 2.0000 = 0.5000
- Interpolate decibel gain: y = -3.0 + 0.5000 × (-40.0 - (-3.0)) = -3.0 + 0.5000 × (-37.0) = -3.0 - 18.5 = -21.5 dB
Logarithmic Interpolation FAQs
When should I use logarithmic interpolation instead of linear?
You should use logarithmic interpolation whenever the underlying physical process spans orders of magnitude (decades) or behaves exponentially. Classic examples include frequency response in electronics and acoustics (Bode plots), particle sieve distribution in civil engineering, sound pressure in decibels (dB), financial yield curves, and radioactive decay rates.
What is the difference between semi-log and log-log interpolation?
In semi-log interpolation, only one axis is transformed with logarithms—either X is logarithmic while Y is linear (common for frequency vs. dB or grain size curves), or X is linear while Y is logarithmic (common for exponential growth and decay). In log-log interpolation, both X and Y axes are transformed logarithmically, which is ideal for power-law relationships (y = a·x^b).
Can logarithmic interpolation handle zero or negative numbers?
No. The mathematical logarithm of zero or any negative number is undefined in real numbers (ln(x) does not exist for x ≤ 0). Any coordinate axis configured for logarithmic scaling must strictly contain positive values (> 0).
Does the choice of base (ln vs log10) change the result?
No. Because the formula divides the difference of logs by another difference of logs, the change-of-base constant ln(10) cancels out in the numerator and denominator: [log₁₀(x) - log₁₀(x₀)] / [log₁₀(x₁) - log₁₀(x₀)] = [ln(x) - ln(x₀)] / [ln(x₁) - ln(x₀)]. The final numerical result is identical regardless of base.
Why does linear interpolation produce inaccurate results on logarithmic scales?
Linear interpolation assumes equal numerical increments represent equal changes. But on a logarithmic axis, equal geometric ratios (like 1 to 10 and 10 to 100) represent equal physical distances. Using standard linear interpolation over a wide decade span heavily distorts the true curve.