Wheatstone Bridge Single-Element Varying

So, this book is saying that when V_B is 10V, R is 100ohms, and delta R is 0.1 ohms, The linearity error is 2.5mV - 2.49875mV = 0.00125mV. I don't quite understand where 2.5mV came from.

If I use the equation on top, I get 10*(100/(100+100) - 100.1/(100.1+100)) = -2.49875mV, and if I use the equation on the bottom, I get (10/4)*(0.1/(100+0.1/2)) = +2.498751mV.

Why am I getting the opposite sign? what do those four equations represent? and why is the error 0.00125mV and where did 2.5mV come from?

I know this is a dumb question, but this has been driving me nuts..

https://www.analog.com/media/en/training-seminars/design-handbooks/Op-Amp-Applications/Section4.pdf

If you take the initial slope of the curve, using Rx = 100+$$\\sigma\$$ where $$\\sigma\$$ << 100,

You can easily show that the output voltage is approximately $$\\sigma\$$/400 for $$\\sigma\$$ << 100.
(proof left as an exercise, but use the approximation $$\ \frac{1}{1 + a} \approx 1 -a \$$, and discard terms of the order of $$\\sigma^2\$$)

So looking at the initial slope, the voltage with a 0.1$$\\Omega\$$ change (Rx = 100.1$$\\Omega\$$) should be 2.500mV, however it is actually about 2.498751mV for an error of (2.498751- 2.500)/2.500 = (about) -0.05% FS (the minus sign indicating that the voltage is less than would be expected from the initial slope).

I'm not sure about their claim of 'gain error' not being included, it sounds a bit like they're mixing up definitions of linearity. If we set the error to zero at FS as well (adjust the calibration by +0.05% to compensate for the 2.498751 - 2.5 end point discrepancy), we would expect to see worst case error somewhere near the midpoint at 100.05 ohms.

That error would be 10000mV * (100.05/200.05 - 0.5) - 1.2493753123438mV ~= 0.0125% FS (where FS is 2.498751mV). So our worst-case error is now 1/4 of what it was.

Note that because we've corrected for the error of a convex curve at the endpoints, the error at midscale is now positive (the bridge output voltage is slightly higher than would be expected with a perfectly linear response to Rx resistance).

We could get a bit better again by distributing the error evenly on both sides of the ideal value (with end points exactly calibrated the error is all on the positive side), down to +/-.00625% FS over the range 100.0 ohms to 100.1 ohms.

For a 0.1% change in one R, you expect only a 0.05% change of total R+R current on that side only.

But since the voltage on either side was 50% of Vb or 5V, you might expect 0.05% of 5V = 2.50mV as the differential voltage.

But KVL proves otherwise with a linearity error of 0.5% / %ΔR full scale ;

from KVL, $$\\dfrac{R+ΔR}{2R+ΔR}=\dfrac{ΔV}{V_B}\$$