Force — Question 4

Question 4

A force sensor can require a zero correction

A uniaxial sensor reads 0.4 divisions at zero force and 2.4 divisions at a known positive 10 N load. Assume a linear calibration. A new reading is 1.6 divisions.

1. Find force as a function of reading R.

2. Infer the new load.

3. Why is multiplying R directly by 5 N per division incorrect?

Question 4 - Solution

A force sensor can require a zero correction

(a) Use the change from zero.

$$k=\frac{10-0}{2.4-0.4}=5\,\mathrm{N/div},\quad\boxed{F=k(R-R_0)=5(R-0.4)\,\mathrm N}$$ The slope is $10/(2.4-0.4)=5$ N per division, giving $F=5(R-0.4)$ N for numerical R in divisions.

(b) Apply the offset.

$$F(1.6)=5(1.6-0.4)=6\,\mathrm N$$ At R=1.6, $F=5(1.2)=6$ N.

(c) Keep the intercept.

$$F_{\mathrm{naive}}-F=5R-5(R-0.4)=2\,\mathrm N$$ Using $5R$ gives 8 N and predicts a nonzero 2 N load at the zero-force reading. A calibration slope alone is insufficient when the sensor has an offset.