Interpret a temperature-only stability comparison

Compare the parcel with its surroundings

The plotted contrast is ΔT = T_parcel − T_environment at the same height. Positive values mean a warmer parcel; negative values mean a cooler parcel. At equal pressure and comparable composition, temperature differences help interpret density differences. But this model omits virtual-temperature corrections for moisture, mixing with surrounding air and condensate loading.

The contrast is therefore a temperature-only buoyancy proxy. It is not actual buoyancy, vertical acceleration or convective available potential energy, CAPE. No equation of motion is solved. The profiles are supplied hypothetical states rather than a simulation showing that the parcel reaches each height.

Saturation and a later crossing are different events

At the default surface gap of 8°C, the approximate LCL is 1 km. With environmental decrease b = 7°C/km, the parcel is 20.2°C and the environment 23°C there. Saturation has been reached, but the parcel is still cooler by 2.8°C.

At 3 km, parcel temperature is 8.2°C and environment 9°C, leaving ΔT = −0.8°C. The parcel approaches the environment above the LCL because its prescribed 6°C/km cooling is slower than the environmental 7°C/km decrease. It does not cross within the displayed layer.

Change b to 8°C/km while keeping the surface gap fixed. At the LCL the parcel is still cooler, now by 1.8°C. At 3 km it is warmer by 2.2°C. A crossing lies between those levels. Capture the comparison and keep the LCL marker separate from the temperature-equality point.

Derive the post-saturation crossing

Let L denote the approximate LCL in kilometers. Above L, parcel temperature simplifies to 30 − 6z − 3.8L. The environment is 30 − bz. Subtracting gives ΔT = (b − 6)z − 3.8L.

For b = 8 and L = 1, equality requires 2z − 3.8 = 0, so z = 1.9 km. Check by substitution: parcel and environment are both 14.8°C there. The browser’s crossing is a temperature-equality diagnostic within its approximations, not an actual level of free convection.

For b = 7 and L = 1, the same algebra gives 3.8 km. That lies beyond the 3 km plotting range; it must not be displayed as a crossing inside the viewed layer. An algebraic result also needs to lie on the branch for which it was derived.

If b = 6, division by b − 6 is not allowed. For a positive L, the post-saturation contrast is the constant −3.8L, so no later crossing occurs. If L = 0 and b = 6, the two prescribed profiles coincide throughout instead of having one isolated crossing.

Interpret rates with their assumptions

A surrounding temperature that falls more rapidly with height can make the lifted parcel warmer relative to its environment. The result depends on whether the parcel is on its unsaturated or saturated branch. A lapse-rate comparison therefore needs the parcel’s moisture state and starting conditions, not just one environmental number.

NOAA uses parcel comparisons to introduce atmospheric stability, while also stating assumptions about heat exchange and parcel behavior. NOAA parcel-theory discussion. The browser deliberately retains a simpler temperature diagnostic rather than presenting a full sounding analysis.

Compare the idealized lapse-rate regimes

For this rate comparison, imagine a parcel initially matched to its environment at a level and displaced slightly upward. Compare the environmental decrease b with the model’s saturated rate 6 and unsaturated rate 9.8°C/km:

• Below 6°C/km, either parcel cools faster than its surroundings: the temperature-only response is stable for both unsaturated and saturated displacements.
• At 6°C/km, the saturated response is neutral, while the unsaturated response remains stable.
• Between 6 and 9.8°C/km, the unsaturated response is stable but the saturated response is unstable: this is conditional instability in the simplified rate comparison.
• At exactly 9.8°C/km, the unsaturated response is neutral and the saturated response is unstable. This exact value is a paper comparison because the control uses whole-number steps.
• Above 9.8°C/km, both parcel types cool more slowly than the surroundings: both responses are unstable in this temperature-only classification.

Here neutral means that an initially matched parcel’s temperature difference does not change under the corresponding displacement rule. It does not erase an existing offset. At b = 6, the plotted parcel with a positive initial dry segment retains its negative post-saturation contrast of −3.8L; it is not suddenly equal in temperature to the environment. These regime labels describe idealized slope responses, not guaranteed motion or a complete moisture-corrected buoyancy calculation.

No cloud or thunderstorm is guaranteed by a positive contrast. A lifting mechanism, the ability to traverse earlier layers, humidity-related density effects and evolving cloud processes are absent. Likewise, a negative contrast at one level does not provide a complete forecast for the whole atmosphere.

Build a diagnostic record

On paper list LCL, parcel and environmental temperatures at 1 and 3 km, and whether a post-saturation crossing lies inside the plotted range. Add an assumptions column. Separate what follows from the displayed equations from what would require measurements or more complete physics.

Transfer: check branch and range

Take surface gap 16°C, so L = 2 km, and b = 10°C/km. The post-saturation equality formula gives 1.9 km. Is that a valid post-saturation crossing?

Check the atmospheric-stability transfer

No. The formula applies only at or above L = 2 km, but 1.9 km is below it. Before saturation, ΔT = (10 − 9.8)z = 0.2z, already positive for z above zero. At the LCL the contrast is +0.4°C and remains positive above it. The algebraic root belongs outside the branch and must not be labeled a post-saturation crossing.