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**Arctic sea ice / Roundhouse punches from the ApocalyptoKraken**

« **on:**

**Today**at 04:51:50 AM »

These Total Precipitable water maps look like theres seven water Giants wading Around the equator swinging tentacled fists under the cover of the jetsteam. In Nth and Southern Hemispheres each, three in the pacific, two per Atlantic and Indian Oceans. Every time the God Coriolis tries to deflect them, the following Kraken punches it back on course for the poles. The circulation has gone Feral! (or should that be Ferral?). Back flows seem to be at altitude. And Hadley seems history.

Quickstab at what this setup might mean numerically. Peer review and alternative approaches most welcome:

1,680,000,000,000,000J

18.748 kg/sqm

200km x 50kmph (ballpark flow estimate) x 24hr = 240 000 sqkm = 240 000 000 000 sqm

240 000 000 000 sqm x 19 kg x 4200J = 19,152,000,000,000,000 Joules per day

=19.52 petajoules per day

Quote wikipedia:

"The petajoule (PJ) is equal to one quadrillion (10^15) joules. 210 PJ is equivalent to about 50 megatons of TNT. This is the amount of energy released by the Tsar Bomba, the largest man-made nuclear explosion ever."

"The gigajoule (GJ) is equal to one billion (10^9) joules. 6 GJ is about the amount of potential chemical energy in 160 L (approximately one US standard barrel) of oil, when combusted."

So about ten days of ApocalyptoKraken suckerpunches = 1 Tsar Bomba = 35 million Barrels of oil burned.

North pole at summer solstice gets 12.64 kWhrs per sqm. = 12.64 x 60mx 60s kJ / sqm = 45,504,000 J per sqm per day = 45 504 000 000 000J per sqkm per day.

19 152 000 000 000 000 / 45 504 000 000 000 = 420 sqkm of full midsummer insolation. About 4200sqkm of 10% absorbed as per bright white ice.

If we were to anticipate that in a few months time there might be twice as much water vapour per sqm incoming and 4000 km front of it crossing for flux calculation purposes, then it would be 80x this. So simular to 336 000 sqkm of normal midsummer insolation.

(Neglecting other energy transported in the humid air of course. This probably of a far larger magnitude. Anyone want to do an estimate for the specific heat transport capacity of moist air column incoming, say flux area 1000km wide by 5km deep, velocity 50kmph?).

Quickstab at what this setup might mean numerically. Peer review and alternative approaches most welcome:

1,680,000,000,000,000J

18.748 kg/sqm

200km x 50kmph (ballpark flow estimate) x 24hr = 240 000 sqkm = 240 000 000 000 sqm

240 000 000 000 sqm x 19 kg x 4200J = 19,152,000,000,000,000 Joules per day

=19.52 petajoules per day

Quote wikipedia:

"The petajoule (PJ) is equal to one quadrillion (10^15) joules. 210 PJ is equivalent to about 50 megatons of TNT. This is the amount of energy released by the Tsar Bomba, the largest man-made nuclear explosion ever."

"The gigajoule (GJ) is equal to one billion (10^9) joules. 6 GJ is about the amount of potential chemical energy in 160 L (approximately one US standard barrel) of oil, when combusted."

So about ten days of ApocalyptoKraken suckerpunches = 1 Tsar Bomba = 35 million Barrels of oil burned.

North pole at summer solstice gets 12.64 kWhrs per sqm. = 12.64 x 60mx 60s kJ / sqm = 45,504,000 J per sqm per day = 45 504 000 000 000J per sqkm per day.

19 152 000 000 000 000 / 45 504 000 000 000 = 420 sqkm of full midsummer insolation. About 4200sqkm of 10% absorbed as per bright white ice.

If we were to anticipate that in a few months time there might be twice as much water vapour per sqm incoming and 4000 km front of it crossing for flux calculation purposes, then it would be 80x this. So simular to 336 000 sqkm of normal midsummer insolation.

(Neglecting other energy transported in the humid air of course. This probably of a far larger magnitude. Anyone want to do an estimate for the specific heat transport capacity of moist air column incoming, say flux area 1000km wide by 5km deep, velocity 50kmph?).