Math Bridge: Gauge Factor 2.0

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Where a metal-foil gauge factor of 2.0 comes from

One thread from R=ρL/A to geometry, material response, and the chapter's 2.50 mV bridge output.

Phoebe, the physics guidePhoebe guides
The one targetDecompose gauge factor into two physical effects.
The chapter caseGF=2.0, ν=0.30, ε=0.001, 5.0 V.
What it buys youKnow what a calibration constant represents.

A field team faces an unresolved physical question: Where a metal-foil gauge factor of 2.0 comes from They must answer it before changing poisson ratio on the real device. Predict the direction first.

See the relationship before changing it

Set the input, apply the rule, then read the result with its unit.

The retained audit below checks several chapter fixtures. This added model holds every other chapter fixture fixed, so the numeric fixture does not switch without explanation.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline for poisson ratio is 0.3.

  2. 2

    Name the relationship. geometry=1+2ν; piezoresistivity=GF-geometry; Vout≈Vs(GFε)/4

  3. 3

    Substitute the chapter fixture. Set poisson ratio to 0.3. The page ledger gives piezoresistive term as 0.40.

  4. 4

    Read the result. Keep the stated output unit beside the value. Use it only inside the technical boundary on this page.

Predict, then change poisson ratio

Try Predict the direction of piezoresistive term. Move one control, calculate, then check your prediction.

0.3
Chapter baseline
Piezoresistive term

Observe The two contributions always add to GF=2.0; the bridge output follows the total GF, not how it is split. Reset the control to 0.3 and compare piezoresistive term.

Explain Only poisson ratio moves here. The other chapter fixtures remain fixed.

Check yourself

What should you do before trusting a moved-control result?
Answer: Predict its direction, apply the shown relationship, keep the units, and reset to the worked baseline.
What does this small model leave out?
Answer: Only poisson ratio moves. Field effects named in the page's technical boundary stay fixed.

Honesty note. This teaching model changes only poisson ratio. It replays the chapter fixture; it is not a field prediction.

1. Stretching changes three things

Resistance is R=ρL/A. A stretched foil gets longer, its cross-section gets smaller, and its material resistivity may change.

Phoebe: Gauge factor is not a magic multiplier. It is a compact account of geometry plus piezoresistivity.

2. Differentiate the resistance rule

1

Fractional changesdR/R=dρ/ρ+dL/L−dA/A.

2

Name axial strainε=dL/L.

3

Use Poisson contractionFor a round-equivalent section, dA/A=−2νε.

3. Divide by strain

GF=(dR/R)/ε=(1+2ν)+(dρ/ρ)/ε

The first term is length plus thinning. The second is the material's resistivity change. Together they must equal the measured gauge factor.

4. Try Poisson's ratio

geometry=1+2ν; piezoresistivity=GF−geometry; Vout≈Vs(GFε)/4

TryMove ν while the chapter's total GF=2.0, strain, and 5.0 V excitation stay fixed.

Geometric term
Piezoresistive term
Geometry share
Fractional resistance change
Quarter-bridge output

ObserveAt ν=0.30, geometry contributes 1.60, the material remainder is 0.40, and geometry supplies 80.0% of GF.

ExplainThe two contributions always add to GF=2.0; the bridge output follows the total GF, not how it is split.

Technical boundaries.

The gauge calculation keeps total gauge factor at 2.0 and uses small strain, an equivalent transverse contraction, and a linear quarter-bridge approximation. It does not establish the alloy's certified gauge factor, temperature response, transverse sensitivity, strain transfer through the adhesive, lead compensation, bridge completion, or calibration.

the alloy's certified GF
Needs separate evidence
temperature coefficient
Needs separate evidence
transverse sensitivity
Needs separate evidence
adhesive/host strain transfer
Needs separate evidence
lead compensation
Needs separate evidence
bridge completion
Needs separate evidence
calibration
Needs separate evidence

Use field evidence or a deeper model before release.

5. Rebuild 1,000 microstrain

ΔR/R=GFε=2.0×0.001=0.00200
Vout≈5.0×0.00200/4=0.00250 V=2.50 mV

The geometric contribution is 1.60×10⁻³ and the material contribution is 0.40×10⁻³; their sum recovers the chapter's result.

6. Cross-check 500 microstrain

ΔR/R=2.0×0.0005=0.00100
Vout≈5.0×0.00100/4=1.25 mV

This matches the chapter's Practitioner check and confirms that the decomposition has not changed its bridge model.

7. Check yourself

Why does thinning increase resistance?
Answer: A smaller cross-sectional area A makes R=ρL/A larger.
What is the geometric term at ν=0.30?
Answer: 1+2(0.30)=1.60.
What fraction of GF=2.0 is piezoresistive here?
Answer: (2.00−1.60)/2.00=20.0%.
Honesty boundary.

The gauge calculation keeps total gauge factor at 2.0 and uses small strain, an equivalent transverse contraction, and a linear quarter-bridge approximation. It does not establish the alloy's certified gauge factor, temperature response, transverse sensitivity, strain transfer through the adhesive, lead compensation, bridge completion, or calibration.

GF=2.0
Total metal-foil gauge factor
ν≈0.30
Assumed Poisson ratio
1,000
Main worked strain in microstrain
500 microstrain
Cross-check applied strain
5.0 V
Quarter-bridge excitation voltage
1.60/0.40 decomposition
Geometric and resistivity gauge-factor contributions
80.0%/20.0% shares
Geometric and resistivity shares of total gauge factor
2.50 mV result
Quarter-bridge output at main worked strain
1.25 mV cross-check come from the chapter
Quarter-bridge output at cross-check strain

The page explains one nominal small-strain model, not an alloy certificate or installation calibration.

Return to the chapter with the maths unpacked →