TheClimateChanger

Down to 32 at Zelienople Municipal AP AWOS: https://forecast.weather.gov/data/obhistory/KPJC.html

For the statewide averages, I expect Pennsylvania to check in around 67.9F for the month of August, and West Virginia around 70.0F, both of which are around 50th coldest August of record (since 1895). Meteorological summer should finish around 70.5F for Pennsylvania and 72.5F for West Virginia. That would finish up as 14th warmest for the State of Pennsylvania (since 1895) and tied for 7th warmest for West Virginia (since 1895), with 1901, 1949 & 1952. If anything, I think this projection might be slightly low for August but should be within about 0.1F of the final summer tally IMO, as I would have to be more than 0.3F off for August to be more than 0.1F off for the summer as a whole. Although the statewide numbers are so crowded that even a tenth of a degree can swing things a few places as evidenced by all the multi-year ties.

After a generally hot and humid summer, August is closing out on significantly cooler than normal. August mean should drop around 0.9F with the last 3 days factored in, with the 3-month summer average dropping around 0.3F (0.9F/3). Looking at some long-term "threaded" climate sites, PIT should finish around 73.9F, which would drop us from a 3-way tie for 11th (1877, 1880) to a 2-way tie for 16th place in the threaded record (1878). At Wheeling, summer should finish around 73.4F, dropping from a 2-way tie for 5th (2010), down to 8th place. Note - lengthy gap here from the mid 50s to late 90s... not a whole lot of hot summers in that stretch but at least a couple are missing that can be seen in the PIT and MGW threads (e.g., 1991 & 1995). Morgantown should finish around 73.9F, dropping from 7th place to a 3-way tie for 11th place (2005, 1991).

I'm still under 1/2 of an inch for the entire month.

Wow! Looks like the Army Corps is going to be a lot of dredging for a fourth consecutive autumn. Hopefully, we don't see the salt intrusion problems of recent years in NOLA.

How did this work out? Akron, Ohio was supposed to get 3.5" of rain from the "biblical storm" and 4-5" over the 2 weeks. Instead, it will be the driest month on record - not just the driest August, the driest of ANY month.

On the plus side, it will be the hottest summer on record in the UK. The UK Met Office needs to brace for some unhinged comments from social media users. For some of these guys, 1976 will always be king no matter how hot recent summers are. You can already see some of the trolls from across the pond recycling the same nonsense as the trolls here about airport tarmacs and jet engine blasts: I suspect the CONUS will slot in around 12th place overall, and warmer than any pre-21st century summers except 1934 & 1936, but I guess that's a big victory for the trolls these days. I've already been called out by the usual suspects for not providing a list of record lows. In reality, I'm just trying to take a break from the trolls!

This has just been a brutal stretch though. Here is where we were for summer to date through the 15th of the month: And here is where we are now projected to finish:

IEM is maybe a little overdone here, but very impressive dry pattern. This seems to be the trend these days with dryness August-October, after spring/early summer deluges.

In 2025, one can just make something up, post it on social media, and others will believe it. Obviously, the Carr Fork Lake number is erroneous, and the 2012 occurrence at Greenup looks to be an error as well. In 1986 and 1946, clearly a number of stations observed lows in the 30s in Kentucky. In August 1986, nearly 1/4 of all NWS stations in the State had a low in the 30s. And this is just official cooperative stations, not backyard thermometers or mesonets or whereever that is sourced from. Kentucky's highest point is 4,145 feet. Those elevated valleys in the east can radiate well.

Q: How about ignoring the human aspect? Which one carried more total heat? A temp of 92F, and dewpoint of 75F, or a temp. of 104F, and dewpoint of 50F? To determine which condition carries more total heat, ignoring the human aspect, we need to evaluate the total energy content of the air, which depends on both temperature and moisture content. The enthalpy of the air, which measures the total heat (sensible heat from temperature plus latent heat from moisture), is the appropriate metric. Wet bulb temperature can help approximate this, but we’ll calculate enthalpy directly using the given temperatures and dewpoints.Given Conditions: Condition 1: Dry bulb temperature = 92°F, Dewpoint = 75°F Condition 2: Dry bulb temperature = 104°F, Dewpoint = 50°F Step 1: Understand the Components Dry bulb temperature reflects the sensible heat (energy associated with temperature). Dewpoint indicates the moisture content, which relates to latent heat (energy tied to water vapor). Higher dewpoint means more moisture, contributing to latent heat, while higher dry bulb temperature increases sensible heat. Enthalpy (in Btu/lb of dry air) combines both sensible and latent heat and is commonly used in psychrometrics to compare total heat content. Step 2: Approximate Enthalpy CalculationEnthalpy of moist air can be calculated using the formula: h=0.24⋅T+W⋅(1061+0.444⋅T)h = 0.24 \cdot T + W \cdot (1061 + 0.444 \cdot T)h = 0.24 \cdot T + W \cdot (1061 + 0.444 \cdot T) where: ( h ) is enthalpy (Btu/lb of dry air), ( T ) is dry bulb temperature (°F), ( W ) is the humidity ratio (lb of water vapor per lb of dry air), 0.24 is the specific heat of dry air (Btu/lb·°F), 1061 is the latent heat of vaporization of water (Btu/lb at 0°F), 0.444 is the specific heat of water vapor (Btu/lb·°F). The humidity ratio ( W ) depends on the dewpoint, which determines the vapor pressure of water in the air. We can estimate ( W ) using the saturation pressure at the dewpoint and the atmospheric pressure (assumed standard at 14.696 psia or 101.325 kPa).Step 2a: Calculate Humidity RatioThe humidity ratio is given by: W=0.62198⋅PvP−PvW = \frac{0.62198 \cdot P_v}{P - P_v}W = \frac{0.62198 \cdot P_v}{P - P_v} where: PvP_vP_v is the vapor pressure at the dewpoint (psia), ( P ) is atmospheric pressure (14.696 psia). The vapor pressure PvP_vP_v can be approximated using the Antoine equation or simplified correlations for saturation pressure at the dewpoint temperature. A common approximation for saturation pressure (in psia) over liquid water is: Pv=exp⁡(20.386−5132Td+459.67)P_v = \exp\left(20.386 - \frac{5132}{T_d + 459.67}\right)P_v = \exp\left(20.386 - \frac{5132}{T_d + 459.67}\right) where TdT_dT_d is the dewpoint in °F, and Td+459.67T_d + 459.67T_d + 459.67 converts to °R (Rankine).Condition 1: Dewpoint = 75°F Td=75°FT_d = 75°FT_d = 75°F , so Td+459.67=534.67°RT_d + 459.67 = 534.67°RT_d + 459.67 = 534.67°R . Pv=exp⁡(20.386−5132534.67)≈exp⁡(20.386−9.599)≈exp⁡(10.787)≈0.487 psiaP_v = \exp\left(20.386 - \frac{5132}{534.67}\right) \approx \exp(20.386 - 9.599) \approx \exp(10.787) \approx 0.487 \, \text{psia}P_v = \exp\left(20.386 - \frac{5132}{534.67}\right) \approx \exp(20.386 - 9.599) \approx \exp(10.787) \approx 0.487 \, \text{psia} . W=0.62198⋅0.48714.696−0.487≈0.302914.209≈0.0213 lb water/lb dry airW = \frac{0.62198 \cdot 0.487}{14.696 - 0.487} \approx \frac{0.3029}{14.209} \approx 0.0213 \, \text{lb water/lb dry air}W = \frac{0.62198 \cdot 0.487}{14.696 - 0.487} \approx \frac{0.3029}{14.209} \approx 0.0213 \, \text{lb water/lb dry air} . Condition 2: Dewpoint = 50°F Td=50°FT_d = 50°FT_d = 50°F , so Td+459.67=509.67°RT_d + 459.67 = 509.67°RT_d + 459.67 = 509.67°R . Pv=exp⁡(20.386−5132509.67)≈exp⁡(20.386−10.071)≈exp⁡(10.315)≈0.300 psiaP_v = \exp\left(20.386 - \frac{5132}{509.67}\right) \approx \exp(20.386 - 10.071) \approx \exp(10.315) \approx 0.300 \, \text{psia}P_v = \exp\left(20.386 - \frac{5132}{509.67}\right) \approx \exp(20.386 - 10.071) \approx \exp(10.315) \approx 0.300 \, \text{psia} . W=0.62198⋅0.30014.696−0.300≈0.186614.396≈0.0130 lb water/lb dry airW = \frac{0.62198 \cdot 0.300}{14.696 - 0.300} \approx \frac{0.1866}{14.396} \approx 0.0130 \, \text{lb water/lb dry air}W = \frac{0.62198 \cdot 0.300}{14.696 - 0.300} \approx \frac{0.1866}{14.396} \approx 0.0130 \, \text{lb water/lb dry air} . Step 2b: Calculate Enthalpy Condition 1: 92°F, Dewpoint 75°F T=92°FT = 92°FT = 92°F , W=0.0213W = 0.0213W = 0.0213 . Sensible heat: 0.24⋅92=22.08 Btu/lb0.24 \cdot 92 = 22.08 \, \text{Btu/lb}0.24 \cdot 92 = 22.08 \, \text{Btu/lb} . Latent heat term: W⋅(1061+0.444⋅92)=0.0213⋅(1061+40.848)≈0.0213⋅1101.848≈23.47 Btu/lbW \cdot (1061 + 0.444 \cdot 92) = 0.0213 \cdot (1061 + 40.848) \approx 0.0213 \cdot 1101.848 \approx 23.47 \, \text{Btu/lb}W \cdot (1061 + 0.444 \cdot 92) = 0.0213 \cdot (1061 + 40.848) \approx 0.0213 \cdot 1101.848 \approx 23.47 \, \text{Btu/lb} . Total enthalpy: h=22.08+23.47=45.55 Btu/lbh = 22.08 + 23.47 = 45.55 \, \text{Btu/lb}h = 22.08 + 23.47 = 45.55 \, \text{Btu/lb} . Condition 2: 104°F, Dewpoint 50°F T=104°FT = 104°FT = 104°F , W=0.0130W = 0.0130W = 0.0130 . Sensible heat: 0.24⋅104=24.96 Btu/lb0.24 \cdot 104 = 24.96 \, \text{Btu/lb}0.24 \cdot 104 = 24.96 \, \text{Btu/lb} . Latent heat term: W⋅(1061+0.444⋅104)=0.0130⋅(1061+46.176)≈0.0130⋅1107.176≈14.39 Btu/lbW \cdot (1061 + 0.444 \cdot 104) = 0.0130 \cdot (1061 + 46.176) \approx 0.0130 \cdot 1107.176 \approx 14.39 \, \text{Btu/lb}W \cdot (1061 + 0.444 \cdot 104) = 0.0130 \cdot (1061 + 46.176) \approx 0.0130 \cdot 1107.176 \approx 14.39 \, \text{Btu/lb} . Total enthalpy: h=24.96+14.39=39.35 Btu/lbh = 24.96 + 14.39 = 39.35 \, \text{Btu/lb}h = 24.96 + 14.39 = 39.35 \, \text{Btu/lb} . Step 3: Compare Total Heat Condition 1: 92°F, dewpoint 75°F → Enthalpy ≈ 45.55 Btu/lb. Condition 2: 104°F, dewpoint 50°F → Enthalpy ≈ 39.35 Btu/lb. ConclusionThe air at 92°F with a dewpoint of 75°F carries more total heat (enthalpy = 45.55 Btu/lb) than the air at 104°F with a dewpoint of 50°F (enthalpy = 39.35 Btu/lb). The higher moisture content (higher dewpoint) in Condition 1 contributes significantly to its latent heat, outweighing the higher sensible heat from the higher temperature in Condition 2.Thus, 92°F with a dewpoint of 75°F has more total heat.

Funny, you don't see Bastardi trot out his population-weighted metrics in recent summers. Loves to pop them out when there's a trough in the east though.

I ran this scenario through AI and it confirmed that warming hole theory is not considering latent heat effects (increased water vapor), only sensible heat (dry bulb). Total heat (enthalpy) has likely continued to increase. In fact, if you consider latent heat effects, then places in the southwest with extreme sensible heat increases may actually have less total heat increase (since latent heat has decreased in those areas, evidenced by declining dewpoints).

Ok, well, I don't believe that. It's not 10C (18F!) warmer in a sunny neighborhood versus a shady one if the thermometer is properly sheltered or aspirated. My car thermometer doesn't register anywhere near that difference and it's housed like a foot or so above the asphalt.

Speaking of lows: