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Frost Point Temperature
-7.58
°C (over ice)
Intermediate γ (gamma) -0.6428

What Is the Frost Point?

The frost point is the temperature to which air must be cooled, at constant pressure and water-vapor content, for water vapor to deposit directly as frost (ice) rather than condense as liquid dew. It is the sub-freezing analogue of the dew point and uses saturation values calculated over ice instead of over liquid water. The frost point is essential in meteorology, aviation icing forecasts, cold-storage engineering, and frost-protection planning for agriculture.

Diagram showing water vapor depositing as frost crystals on a cold surface below freezing
The frost point is the temperature at which air becomes saturated with respect to ice, forming frost.

How to Use This Calculator

Enter the current air temperature in degrees Celsius and the relative humidity as a percentage (0.01–100). The calculator returns the frost point temperature in °C using the Magnus-Tetens approximation with coefficients tuned for saturation over ice. Frost point is most physically meaningful when the air temperature is below 0 °C.

The Formula Explained

We first compute an intermediate term γ that blends the humidity ratio with the temperature:

$$\gamma = \ln\!\left(\frac{\text{RH}}{100}\right) + \frac{22.46 \times T}{272.62 + T}$$

Then the frost point is:

$$T_f = \frac{272.62 \times \gamma}{22.46 - \gamma}$$

The constants 22.46 and 272.62 °C are the Magnus coefficients for the ice phase (Sonntag/Alduchov-style fits). At 100% relative humidity, \(\ln(1) = 0\) and \(T_f\) collapses back toward \(T\), as expected.

Graph of saturation vapor pressure curves over water and over ice meeting at zero degrees
Below 0 °C, saturation over ice is lower than over water, so the frost point differs from the dew point.

Worked Example

Suppose T = −5 °C and RH = 80%. First, $$\gamma = \ln(0.80) + \frac{22.46 \times -5}{272.62 - 5} = -0.22314 + \frac{-112.3}{267.62} = -0.22314 - 0.41962 = -0.64276.$$ Then $$T_f = \frac{272.62 \times -0.64276}{22.46 + 0.64276} = \frac{-175.25}{23.10276} \approx -7.59 \text{ °C}.$$ So frost would form on surfaces cooled to about −7.6 °C.

FAQ

Frost point vs dew point — what's the difference? Below 0 °C, vapor can deposit as ice. The frost point uses ice-saturation coefficients and is slightly higher than the liquid-water dew point at the same conditions.

Can I use this above freezing? The math runs, but the result is only physically meaningful for sub-freezing air where ice saturation applies. Above 0 °C use a dew point calculator.

Why must humidity be above 0%? The formula takes the natural log of RH/100, which is undefined at 0; we clamp tiny values to avoid errors.

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