Math Bridge: Barometric Altitude Resolution

A field team faces an unresolved physical question: How one pressure count becomes centimetres They must answer it before changing barometric pressure in pascals on the real device. Predict the direction first.

See the relationship before changing it

The figure reads from left to right. The blue card is barometric pressure in pascals. The middle card applies this page's relationship. The green card is relative altitude. Walk the arrows once: 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.

Walk the arrows. The same pressure count represents slightly more height as pressure falls because H/P grows.

Derive the baseline in four named moves

1
Name the input. The chapter baseline for barometric pressure in pascals is 101325.
2
Name the relationship. H=RT/(Mg); h=-H ln(P/P0); |dh/dP|=H/P
3
Substitute the chapter fixture. Set barometric pressure in pascals to 101325. The page ledger gives relative altitude as 0.00 m.
4
Read the result. Keep m beside the value. Use it only inside the technical boundary on this page.

Predict, then change barometric pressure in pascals

Try Predict the direction of relative altitude. Move one control, calculate, then check your prediction.

Observe The same pressure count represents slightly more height as pressure falls because H/P grows. Reset the control to 101325 and compare relative altitude.

Explain Only barometric pressure in pascals 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 barometric pressure in pascals moves. Field effects named in the page's technical boundary stay fixed.

1. Pressure holds up the air above

A thin slice of air loses pressure with height because pressure supports the slice's weight. The ideal gas law links that air density to pressure and temperature.

Phoebe: Less air overhead means lower pressure, but weather can also change the air column without moving the sensor.

2. Separate and integrate

Hydrostatic balancedP=−ρg dh.

Ideal gas densityρ=PM/(RT), so dP/P=−Mg dh/(RT).

Integrateln(P/P0)=−Mgh/(RT), hence h=−RT ln(P/P0)/(Mg).

3. Find centimetres per pascal

Differentiate h with respect to pressure. The minus sign says height rises when pressure falls; resolution uses the magnitude.

|dh/dP|=RT/(MgP)=scale height/P

Multiply this cm/Pa value by pressure noise in Pa to estimate noise-limited altitude resolution.

4. Try the pressure

H=RT/(Mg); h=−H ln(P/P0); |dh/dP|=H/P

TryLower pressure from the sea-level baseline toward the chapter's 700 m example.

ObserveAt 101,325 Pa, H≈8.43 km, |dh/dP|=8.32 cm/Pa, and 0.12 Pa maps to about 1.00 cm.

ExplainThe same pressure count represents slightly more height as pressure falls because H/P grows.

Technical boundaries.

This isothermal ideal-gas model assumes fixed temperature, molar mass, and gravity.

temperature compensation: Needs separate evidence
local pressure baselines: Needs separate evidence
weather tracking: Needs separate evidence
sensor bias/drift: Needs separate evidence
mounting effects: Needs separate evidence
filtering latency: Needs separate evidence
an external height reference: Needs separate evidence

Use field evidence or a deeper model before release.

5. Work the sea-level count

H=RT/(Mg)=8.43 km

H/P0=8.32 cm/Pa; 8.32×0.12=1.00 cm

This is a noise-limited resolution estimate, not absolute altitude accuracy.

6. Move to the chapter's 700 m case

P≈93,255 Pa → |dh/dP|≈9.04 cm/Pa

9.04×0.12≈1.09 cm

A weather-driven baseline change produces the same mathematical height change, so the chapter correctly treats the output as relative.

7. Check yourself

Why does pressure fall as height rises?

Answer: There is less air above the sensor, so less air-column weight must be supported per unit area.

What does 0.12 Pa mean near sea level in this model?

Answer: About 1.00 cm of noise-limited relative-altitude resolution.

Can this equation distinguish weather from climbing stairs?

Answer: No. Both change pressure, so a fresh baseline or independent reference is required.

Honesty boundary.

These are the chapter inputs, worked results, and named teaching assumptions.

R=8.314 J/(mol K): Charge or energy value
T=288.15 K: Chapter input or worked result
M=0.028964 kg/mol: Distance, wavelength, or size
g=9.80665 m/s²: Time, interval, or service-life value
P0=101,325 Pa: Sensor scale, pressure, or digital result
8.43 km scale height: Distance, wavelength, or size
8.32 cm/Pa: Distance, wavelength, or size
catalog-typical 0.12 Pa noise: Named teaching assumption
about 1.00 cm sea-level result: Distance, wavelength, or size
93,255 Pa example: Named teaching assumption
9.04 cm/Pa: Distance, wavelength, or size
about 1.09 cm result come from the chapter: Distance, wavelength, or size

Relative pressure is not surveyed altitude.