SCIENCE · INVESTIGATE AND EXPLAIN

Find dew point as an inverse saturation calculation

Use the vapor-pressure curve to find a saturation threshold while distinguishing it from a weather prediction.

Learning outcome: Calculate dew point and interpret its relation to air temperature under the stated liquid-water assumptions.

Starting recap

Dew point is the temperature at which the supplied vapor pressure equals saturation pressure over liquid water. The model solves that equality rather than predicting cooling.

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Try

A vapor pressure of 20 hPa gives a dew point of about 17.51°C, whether the selected unsaturated air temperature is 20 or 30°C.

Open the experiment

Predict

Will changing only the air temperature change dew point when vapor pressure stays at 20 hPa?

Your prediction and reason

Experiment

Hold vapor pressure at 20 hPa and compare 20 and 30°C. Then change vapor pressure to 10 and 30 hPa, recording how dew point changes.

Separate humidity, temperature and dew point

Worked teaching model · synthetic values · Liquid-water saturation approximation over 0–40 °C; fixed total pressure 1000 hPa. Changing temperature at fixed vapor partial pressure preserves the vapor-to-dry-air mass ratio before saturation, not the volume of a sealed container. Supersaturated proposals remain visible; no condensation, latent heat, nuclei, ice, cloud evolution or precipitation is calculated. Synthetic settings are not weather observations.

Air temperature: 20 °C Water-vapor partial pressure: 20 hPa

Reset keeps your prediction, captures and practice history.

Temperature, vapor abundance and saturation are different quantities
Quantity	Value and interpretation
Air temperature T	20 °C
Vapor partial pressure e	20 hPa
Saturation pressure eₛ(T)	23.369 hPa, equilibrium over liquid water
Relative humidity 100e/eₛ	85.58%, retained even above 100%
Dew point Tᵈ	17.51 °C, where eₛ(Tᵈ) = e
Mixing ratio 622e/(1000 − e)	12.694 g vapor/kg dry air
Temperature minus dew point	2.49 °C; negative means the proposed vapor pressure exceeds saturation

At or below saturation. Cooling at fixed vapor pressure reaches saturation at the dew point. Continued cooling with condensation would change the vapor pressure; the controls do not calculate that process.

Warm the air at fixed vapor pressure: relative humidity falls while dew point and mixing ratio stay fixed. Add vapor at fixed temperature: all three rise. Neither comparison alone predicts rain or cloud cover. The formula is approximate, refers to liquid water, and is not an ice frost-point calculation.

Relative humidity 85.58%; dew point 17.51 °C. At or below liquid-water saturation.

eₛ = 6.112 exp(17.67T/(T + 243.5)) hPa; RH = 100e/eₛ. Vapor mixing ratio = 12.694 g/kg dry air at 1000 hPa total pressure. Raw relative humidity is not clamped to 100%.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Reverse the saturation question

Relative humidity asks how the supplied vapor pressure compares with saturation at the selected air temperature. Dew point asks a different question: at what temperature would the saturation pressure equal that supplied vapor pressure? The browser finds a temperature on the liquid-water saturation curve, not a prediction of when the atmosphere will reach it.

At e = 20 hPa, the calculated dew point is about 17.51°C. Air selected at 20°C is above that threshold and has RH below 100%. Changing the selected temperature to 30°C lowers RH but leaves this dew point unchanged, because the supplied vapor pressure has not changed.

The usual cooling interpretation holds pressure and vapor mixing ratio constant until saturation is reached. The National Weather Service defines dew point in terms of cooling to saturation at constant pressure. NWS discussion of dew point and relative humidity. The browser fixes total pressure at 1000 hPa and directly controls vapor pressure, making that constraint explicit.

Work through the inverse calculation

The saturation approximation is e_s(T) = 6.112 exp(17.67T/(T + 243.5)). To find dew point T_d, set e_s(T_d) equal to supplied e and solve for T_d. Let L = ln(e/6.112), where ln is the natural logarithm, the inverse of exp. Then T_d = 243.5L/(17.67 − L), in degrees Celsius.

For e = 20 hPa, divide 20 by 6.112, giving about 3.272. Taking its natural logarithm gives about 1.1855. Substitute that into the numerator and denominator to obtain approximately 17.51°C. The pressure ratio uses the same hPa convention as the saturation equation. NWS reference for the inverse dew-point expression.

Check the result by substitution: calculating saturation pressure at T_d should return the original e, apart from numerical rounding. This is a consistency check between a formula and its inverse, not an independent observation of condensation.

Use a graphical inverse

On paper sketch the increasing saturation-pressure curve against temperature. Draw a horizontal line at the supplied vapor pressure. Where that line intersects the curve, project down to the temperature axis. That temperature is the dew point in this model.

Move the horizontal line upward to represent a larger vapor pressure. The intersection moves to a higher temperature. Try e values of 10, 20 and 30 hPa in the browser and record the dew points before comparing the matching RH values at a chosen air temperature.

This construction makes clear why the selected air temperature does not enter the inverse calculation. It is a separate marker used to evaluate the current RH and the distance from the saturation threshold. Air temperature and dew point coincide only at saturation in the non-supersaturated interpretation.

Respect the control resolution

The air-temperature control advances in 2°C steps. The default dew point, about 17.51°C, is between selectable values of 16 and 18°C. The slider does not need to land exactly on the threshold to establish that an inverse solution lies between those values.

At e = 20 hPa, compare 18°C and 16°C. One is above the calculated dew point and the other below. The latter produces a supersaturated proposal that the model flags; it does not automatically remove vapor or calculate droplets. Do not confuse control spacing with uncertainty in a measured temperature or with the numerical precision of the inversion.

Keep the physical interpretation bounded

This is a liquid-water relation. It does not calculate a frost point over ice. It also does not include a cooling rate, surface temperature, cloud nuclei, mixing or precipitation processes. Knowing a threshold does not tell us whether dew will appear on a particular surface or whether a cloud will produce rain.

Write a two-part result: the calculated threshold under the selected vapor pressure, followed by the additional observations needed for a real condensation question. Preserve the distinction between an equilibrium state relation and a time-dependent weather process.

Transfer to a known inverse pair

A paper-only vapor pressure is exactly 6.112 hPa, outside the browser’s vapor-pressure range. What dew point does the selected expression give, and why? Do not infer a frost point from it.

Check the dew-point transfer

ln(6.112/6.112) = ln(1) = 0, so T_d = 0°C. Substituting 0°C into the saturation equation gives 6.112 hPa. This checks the liquid-water formula at its reference value; it does not calculate an ice-equilibrium frost point or predict an observed dew event.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

Explain the inverse equation, distinguish a computed threshold from an observed dew event, and identify why a two-degree slider cannot necessarily select the exact threshold.

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

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Previous: Explain relative humidity without adding water Next: Interpret saturation limits without hiding the result

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