= Two-length machine compliance separation
{title2=$E=\frac{L_2-L_1}{A_0(q_2-q_1)}$}
Prepare two uniform specimens with the same material, direction and $A_0$, but different deforming lengths $L_1,L_2$. Keep grip engagement, adapters, loading window and measurement configuration equivalent. Determine $q_i=dx_i/dF$ separately by <linear regression> with its own displacement intercept. Subtracting the two series equations eliminates the common <machine compliance>:
$$
q_2-q_1=\frac{L_2-L_1}{EA_0},\qquad
E=\frac{L_2-L_1}{A_0(q_2-q_1)},\qquad
C_m=\frac{L_2q_1-L_1q_2}{L_2-L_1}.
$$
This is an <inverse problem>: one length supplies only one equation for two unknowns, while two distinct lengths separate the length-dependent specimen contribution from the constant contribution. In a graph of compliance against length, the slope is $1/(EA_0)$ and the vertical intercept is $C_m$. It is the compliance, not $k_{\mathrm{obs}}$, that is linear in $L$.
The <dimensional analysis> is revealing: $[q_2-q_1]=\mathrm{m/N}$, so $[(L_2-L_1)/(A_0(q_2-q_1))]=\mathrm{N/m^2}$. For $L_2>L_1$, the ideal model requires $q_2>q_1$. A negative fitted $C_m$ challenges the common-intercept assumptions or the measurements. Two lengths determine two parameters exactly, so they cannot test the assumed straight line; three or more lengths allow <regression residuals> to reveal departures. Nearly equal lengths amplify slope errors through the small difference $q_2-q_1$. If the actual apparatus compliances differ by $\Delta C_m$, the inferred length slope becomes $1/(EA_0)+\Delta C_m/(L_2-L_1)$.
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