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Finite-perturbation reference sensitivity measures the change in a comparison when one shared reference value is changed by a finite amount everywhere it occurs. It checks both local sensitivity coefficients and their usefulness over the actual reference-uncertainty range. The reference-sensitivity cancellation check is incomplete if it tests only a derivative at one point.

Let $F(R)$ be the final comparison, with all other independent inputs held fixed. A central finite difference and a second difference areFor sufficiently smooth $F$, these approach $F^{\prime }(R_0)$ and $F^{\prime \prime }(R_0)$, with errors of order $h^2$ when the required derivatives exist. Their units differ: comparison per reference unit and comparison per squared reference unit. The GUM defines sensitivity coefficients as partial derivatives and also discusses determining them by varying an input experimentally.^[ref]

An original model writes $x=R-R_0$ andIt gives $C_h=ϵ+\kappa h^2$ and $Q_h=2\beta$. Comparing $C_h$ with $C_{h/2}$ can expose cubic contamination of the local slope. But if $ϵ=\kappa =0$, every central slope is exactly zero although $F(R_0+h)-F(R_0)=\beta h^2$ is nonzero. A symmetric slope check alone misses an even-in-perturbation response.

For $\beta =1\mathrm{m}{\mathrm{m}}^{-1}$ and $h=0.001\mathrm{mm}$, the one-sided change is 1 nm, while $C_h=0$ and $Q_h=2\mathrm{m}{\mathrm{m}}^{-1}$. If the shared reference error instead has a zero-mean normal distribution with standard deviation $s=h$, the quadratic contribution has mean 1 nm and standard deviation approximately 1.414 nm. These are original model calculations. They show why exact cancellation, zero local slope and negligible finite-range influence are distinct claims.

Use step-size stability of a reference perturbation to select a sensible test range. Extremely small steps can magnify numerical subtraction errors; large steps probe a different region of a nonlinear response. Test several scales near the physical standard uncertainty and inspect one-sided changes as well as central slopes. If the model is stochastic, separate simulation noise from the response to the changed reference.

A computational perturbation changes the assigned shared reference in every dependent calculation while keeping its identity shared. Independently perturbing two uses of the same reference changes the covariance model. A physical experiment must additionally hold other influences stable, because changing a standard or its temperature may change contact or instrument behaviour. NIST separates master-block uncertainty from the comparison process in its uncertainty budget.^[ref] Retain those physical distinctions when interpreting a numerical sensitivity test.

Unequal gauge-block reference sensitivities occur when two length estimates respond differently to the same uncertain reference value. A reference contribution cancels from their difference only when its coefficients agree. Sharing a calibration source is necessary information, but does not by itself guarantee cancellation.

Let estimates have measurement models $X=f_X(R,\dots )$ and $Y=f_Y(R,\dots )$. At the stated input values define sensitivity coefficients $a_X=\partial f_X/\partial R$ and $a_Y=\partial f_Y/\partial R$. The reference contribution to $D=X-Y$ is, to first order,This follows either by differentiating the contrast directly or by retaining shared gauge-block reference covariance in the two-output propagation. The general law includes both sensitivities and covariance.^[ref]

An original ratio-transfer model illustrates the difference: $X=q_{X}R+d_X$, $Y=q_{Y}R+d_Y$, with fixed known $q_X,q_Y$. If $R=100\mathrm{mm}$, $u(R)=20\mathrm{nm}$, $q_X=1$ and $q_Y=2$, their length difference retains a 20 nm reference contribution. The reference error is shared, but appears twice as strongly in $Y$. Treating the reference contributions as independent would instead give $\sqrt{20^2+40^2}\approx 44.72\mathrm{nm}$.

The contrast $2X-Y$ cancels this model's reference term because $2q_X-q_Y=0$. It is a different measurand from $X-Y$; cancellation is not permission to replace the quantity the experiment intended to determine. More generally, $c_{X}X+c_{Y}Y$ retains reference variance $(c_X{a}_X+c_Y{a}_Y{)}^2{u}^2(R)$.

Nearly matched sensitivities can provide strong suppression. With $a_X=1$, $a_Y=0.98$ and the same 20 nm reference uncertainty, the residual reference contribution is 0.4 nm. That does not imply the total uncertainty is 0.4 nm: comparison scatter, coefficient uncertainty, temperature and wringing contributions remain.

If the transfer ratios are themselves uncertain, their derivatives introduce terms proportional to the reference length, along with any shared covariance. Determine them from the actual calibration model rather than assigning ideal coefficients solely to obtain cancellation. NIST separates reference-master history from comparator-process variability in its uncertainty calculation.^[ref] A reference-sensitivity cancellation check should accompany a claimed low-uncertainty contrast.