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Russian Navy vs Japanese Navy
Isos
Re: Russian Navy vs Japanese Navy
Isos on Tue Apr 21, 2020 9:15 am
What pray tell, will Japan do exactly about all their bases, naval ports, airfields, HQs being in range of thousands of Russian cruise missiles?
And how many Russian equivalents will be in range or in danger from anything Japan has?
Where would those thousands missiles come from ? They have no UKSK in the pacific fleet. The only thing they could use is the long range bombers and kh101 but they don't have thousabds of them and their destroyers have very good AD systems.
I won't even touch the anti-naval capabilities afforded by the Tu-22M3s among other things, that can hit the whole Japanese fleet in port without leaving Russian airspace. What are the F-35s and helicopter carriers going to do against that?
F35 has 1000+km combat range and AMRAAM with 100+km range against straight flying bombers. Kh-22 has a range of less than 1000km.
Do the maths.
Funny I seem to recall Turkey being successfully discouraged from making good on their threat to push the Syrian Army back to the pre-Idlib offensive lines not too long ago, and not a single threat of nuclear strike was voiced.
Well that's different. Turkey is in range of western/southern military district that are in range to invade them if they touch Hmeimim. The base in Syria can be overwhelmed easily by Turkey and they have the support of NATO.
Russia clearly said it will use nuks against an overwhelming force.
And put the nukes on what?
Enriching and building nukes is one thing; but SSBNs, ICBMs, strategic bombers take quite a while to develop.
That would be easy to develop for them while they develop the nuclear warheads.
miketheterrible
miketheterrible on Tue Apr 21, 2020 10:00 am
Tu-22M has 2000km range and Kh-32 has 1000km range. It would launch and leave before the f-35 has time to react.
Russia has enough to strike Japan from within territory. Those defenses aren't exactly up to snuff even against subsonic munitions let alone supersonic.
Japan isn't a huge threat. Threat none the less but not a huge one for Russia nor China.
F-35 are sketchy pieces of shit that once an airfield is destroyed, they can't fly unless the jump jet variant which also lacks performance compared to its vstol variant. Japan unfortunately for them is too congested so striking their military facilities would be a very easy success for nearly everyone. And judging by US lack of success in their AD systems which Japan also share and uses, I don't think they would be downing too many low flying missiles since they would need to strike at them at roughly at best 20km distance.
GarryB
Location : New Zealand
GarryB on Tue Apr 21, 2020 12:32 pm
That's what I'm saying. Frigates are not enough for operating far away and they need destroyers or cruiser with more VLS.
With VLS there is no need for a 2.0 Kiriv of 25 000t.
Good so we agree. Certainly a new from scratch design cruiser does not need to be as big as the Kirovs are, but the new modular weapon systems and much more powerful radar and sensors and of course internal fibre optic communications and satellite communications and much better communications between platforms (ships, subs, aircraft, satellites, drones etc etc) means it will be vastly more capable yet probably not as large or heavy.
Japan would make a no fly zone pretty quickly and destroy all the ground launcher in kurils pretty quickly. They have 250 f15, 100 f16, and soon 100 f35. Russia has some flankers but most of its aviation is facing nato.
What would they destroy the ground launchers with?
The air defences over their Pacific Fleet base is pretty strong even if you don't include the ships and air force aircraft...
If Japan tries to strike targets everyone acknowledge as being in Russian territory then they lose any pretence at just wanting their islands back... just assembling a landing force will be noticed, and any attempt to send it for "exercises" near the Kuriles would be an obvious provocation that would send alarm bells ringing in Russia.
Mobilising an attack force would be enough warning to mobilise a response... and that response might kill some Russian fighter pilots, but would likely kill thousands of Japanese sailors. Russian pilots don't need to fly attack missions into Japanese territory... Japanese pilots would need to deal with the SAM threat covering the entire island and any ground based anti ship missile system within 500km of the Kuriles...
MiG-31s operate there and some of them can carry Kinzhals...
How could Japan enforce a no fly zone anywhere except over Japanese airspace... they have no authority to do so over international airspace or indeed Russian airspace over the Kuriles... but also no means to do so... you have to dominate an air space before you can declare a no fly zone... how is Japan going to wipe out or deal with Russias air defence network?
The only thing they couldn't face is the long range missiles and tupolevs. But you cab't win a war with only that.
Aircraft need airfields to operate from... Kinzhals could obliterate command centres and HQs as well as ships and military bases in Japan. Kh-32 could do the same. It would not take long to move Su-35s to the region and MiG-31s would already be there equipped with 300km range R-37M air to air missiles.
Russia is a piper tiger when it comes to power projection and rely on its nuks to forbid the other nation to touch its soldiers. Without them all its forces in Syria, Lybia, cebtral Africa would have been destroyed.
Yes, they are corrupt and all you have to do is kick the door and the whole rotten edifice will collapse itself...
Except that Russian soldiers and commanders and planners seem to have done an exceptional job in Syria... they did a better job in the Crimea than any HATO country could have... the Americans would have shot thousands of people in comparison... the French and British too.... though probably not as many dead...
Japan would not risk invading the Kuriles now or in the near future... maybe in the 1990s, but certainly not now.
They are not connected to Russian systems. The best they could do is use the radio. Networkcentric warfare is systems exchanging in real time btw themselves so that if your plane sees something then your ships see that too. In the best case they can use that for missike targeting.
You mean like they did to Syrian air defence systems in Syria... so they could easily do that then... right?
Certainly if the US is being belligerent enough... which they normally are...
Exactly the only strength of Russia is strategic nuclear weapon. The US has 180 F-22 and another 500F 35. More NATO countries are buying F-35. Russia does not even have a single serial Su-57. He will buy 76 pieces in big bulbs for the next 10 years.
It is not the strategic nuclear forces holding the Kurile islands, nor is it in Syria or anywhere else the Russians operate...
You can spout off that the US has 180 F-22s but how many are actually operational?
How many are they prepared to operate near a functioning S-400 system and Su-35s let alone Su-57s?
Would love to see HATO buy 10,000 F-35s... it would destroy them...
Pacific fleet is rust. Japan navy is way more powerfull. Japanese air force is huge and could deal with russian east forces. If they destroy pacific fleets landing ships russia will be left with nothing to take them back.
They have to take them in the first place, and I rather doubt they could... any landing ship would be obliterated fairly quickly and it would quickly turn in to a debacle for the Japs... they understand island warfare and would know not to try.
Japans force is a defence force... it will take a bit of BS to get the country behind an invasion... especially an invasion of Russia...
What saved Assad are the russian nuks. If Russia didn't had nuks its syrian base would have been destroyed easily. Actually they wouldn't have had any bases there and Russia would have been cut into pieces since the 50s.
The nukes were never even threatened to be used, and it was Russia that basically reorganised and re equipped the Syrians together with the Iranians to face off against the terrorist forces arrayed against them by HATO and Saudi Arabia.
I don't remember HATO or Saudi backing off in any way... they just had their surrogate head choppers defeated on the battlefield and could either send in their own troops to escalate or do nothing and let their terrorist friends rot... they did the latter... unsurprisingly... if they had any balls it would be their people in there doing the fighting in the first place... but even their own troops don't want to fight make some billionaire family at home into trillionaires...
Japan has everything to make its own nuks. If US leave Japan they will be alone against three nuclear armed countries and it wouldn't take them long to think about starting their own program.
Doesn't matter either... even if they started production right now it would be decades before they had as many as Russia and just having a few for deterrence is likely all their population will tolerate. Having a few wont keep them safe if they invade Russia no matter how many times they say it is Japanese territory... and a few would be easier to target...
So as lsos writes, Russia is just a nuclear power and nothing else.
Yeah, being the only country currently able to wipe out the west as we know it is not really that much... especially with the efforts of western countries to do it to themselves first...
But you did just made excellent argument in favor of Russia dropping conventional military and fully switching to nuclear
Would make them too brittle... a Ukrainian attempt to take Crimea back by force would probably only need the people of Crimea to stop and push back but an attempt by Poland to take back Kaliningrad after a particularly big HATO exercise on their territory that had lots of US armoured divisions... well actually nuking the fuck out of Poland might actually be a bit satisfying rather than wasting more Russian blood on those fucking ungrateful euro cunts....
There are times and situations when conventional military force is a proportional response to get your way, where a tactical nuclear strike is going overboard.
The Russian Forces allowed to be active in Syria, to strike terrorist and to deter hostile forces (Turkey, USA, Israel!). Turkey respects the Su-35 and the S-400 in Syria for sure
They liked the Razor so much they bought one for themselves...
Maybe we will see the Super Gorshkov soon Smile as a kind of stimulus package for russian economy ;D
Their next priorities will be ramping up production of the Corvettes and Frigates that work and are useful, and then work on a scaled up design for a destroyer sized vessel because many of the cold war types probably are not worth upgrading too much.
For example...if Russia would take the Baltics, they would determine the location of fight and the time of fight, with a short chain of logistics from Mainland the Baltics couldn't withstand Russia, even with all the might of NATO to defend them. On the other hand Russia would face big problems invading Kansas City ;D
Agreed, but the Pacific Fleet is located right next to the Kurill islands and there are forces on the Islands that would give warning of an attack that could be used to mount some quick counter attacks that should blunt any invasion.
On the Falklands Islands there were some marines and they managed to destroy a few armoured landing vehicles with LAW rockets before being overwhelmed... I would suspect warning of large Japanese landing ships would enable a few anti ship batteries to launch dozens of anti ship missiles at the landing ships and ships supporting the landing within hours... it would be like the Georgian invasion of South Ossetia... the special forces would likely arrive first and their capacity to protect the airfield while reinforcements are landed would make Japans job a nightmare...
Japan has 8 guided missile destroyers and 30 ships more or less to Gorshkov-class (tonnage). Additional 6 corvettes and 4 helicopter carrier, with F-35B they are some kind of aircraft carrier. Don't underestimate the Japanese Navy, even China respects Japan in terms of naval warfare.
The Russians have put Kinzhal in service... it is effectively an anti carrier missile... wouldn't really need it in the black sea or the northern fleet due to the conditions not really being conducive to carrier warfare... only place you would probably base them is the far east near the pacific fleet... 8 guided missile destroyers you say... how long would they last?
Would be interesting to see, but how long do you think they would remain within range... especially after the first salvo...
First time someone attacks you you kill them, kill their families, kill their entire population, incinerate their country and cripple their neighbors
You do that once and others will stop having stupid ideas
Sounds brutal, but if you are soft there is no point in carrying a stick you wont ever use...
[qutoe]You nuke the ship that hijacked Russian ship[/quote]
Sadly the west has hijacked Russian ships and kidnapped Russian people around the world... I like your idea... hand them back or boom.
Azi wrote:
with this attitude, humanity doesn't deserve to survive! ;D
PD wrote:
For Russia only thing that should matter is Russia, mankind is of secondary imporance
Deserve doesn't come in to it... you either do or you don't...
For the price of 1 Leader they can build several smaller FF/DD/CGs- the time for white elephants is over.
Sorry that is bullshit.
You can't say one ship costs the same as 3 smaller ships and then say lets make three smaller ships.
With that stupid fucking logic Russia should stop developing S-500 and get rid of S-400 and S-350 and Pantsir and TOR and all the other SAM systems they developed and just make millions of MANPADS.
An accountant would say three small ships can do what one big ship can.
Someone who knows something about navies will say American spends 13 billion dollars on their super carriers because they are worth it... when they work.
Every one keeps bringing up this american ship or that chinese ship with 200 vertical missile launchers... well small ships can't carry 200 vertical missile launchers.
You can't have a navy of speed boats... the Ukrainian navy is not a model to try to emulate. You need different sized boats for different roles and purposes, but it is important to get them right to start with because once they are right the idea is to make quite a few of them so they actually become significant.
They are at the point of accepting corvettes and frigates into service and time will tell if they are worth buying in numbers, but soon they will need destroyer and cruiser sized ships when the current models that they are using can no longer be upgraded or are ready to be replaced because they need something better.
In many ways their new frigates are as good as their old destroyers, but their smaller size gives them less endurance. Scale them up a few thousand tons and they will make great replacements... but they need to work that out before laying down new destroyers and cruisers willy nilly based on some design some company posted on the internet.
[qutoe]I would like to see a proper model of the Super Gorshkov before making any White Elephant claims about the Leader, after all, if its an elephant then what the hell is the Kirov?[/quote]
The Kirov had and has some problems but was the first ship with vertical missile launchers to enter operational service.
It will be rather like the Falklands then. Well turned up to x10
It would be like the Falklands but instead of the UK being in another hemisphere, them being right next to them on the map... all well within missile range... without refuelling.
Which is what all nuclear powers say except the US who wants to be allowed to use them on bunkers their conventional bombs can't reach...
Azi on Tue Apr 21, 2020 2:09 pm
miketheterrible wrote: Tu-22M has 2000km range and Kh-32 has 1000km range. It would launch and leave before the f-35 has time to react.
Japan is no threat in reality, because pacifism is the core of their "self defense forces"!
Of course Russia has enough to strike Japan conventional...russian subs can strike Tokio with hundreds of conventional armed Kalibr in a conflict. But the unrealistic maybe conflict will be decided by the will, who wants it more. As I wrote...a conflict favors the attacker (the defender only if he is fully aware of the attack and prepared), history proves it! In the "Russo-Japanese War" Russia was far superior to Japan, but it lost because of failed tactics.
Actually, there is no point in discussing a scenario, which cannot happen now...I'm with you. Far east is absolute safe, due to many factors, not only military. Another point in the present is genius russian foreign policy. But nobody knows what the future holds, the world is now in economic trouble and will be maybe in 5 years a different one?!
The real discussion is not the Kurile Islands!!! Real discussion is about saving russian interest in overseas. And you can't save your interests only with corvettes, for that Gorshkov class is needed. And additional Super Gorshkov is a must have! Lider class is not that important now. It should be no problem to build Super Gorshkov, if there are able to build Gorshkov-class.
miketheterrible wrote: F-35 are sketchy pieces of shit that once an airfield is destroyed, they can't fly unless the jump jet variant which also lacks performance compared to its vstol variant. Japan unfortunately for them is too congested so striking their military facilities would be a very easy success for nearly everyone. And judging by US lack of success in their AD systems which Japan also share and uses, I don't think they would be downing too many low flying missiles since they would need to strike at them at roughly at best 20km distance.
Destroying airfields applies to both sides in a conflict. An attacker would try to destroy russian airfields, using saturation attacks with CM and tactical missiles. So all problems for the opponent are your own problems. The real situation would be far more complex.
And yes, the F-35 is shit and lacks a few abilities! ;D
Tsavo Lion wrote: The 4 Kirovs r already built & only 2 will be modernized & kept, as it'll cost more $ & time to build their replacements &/ smaller ships.
https://nationalinterest.org/blog/buzz/russian-navy-dismantling-two-massive-nuclear-battlecruisers-heres-why-53827
Those 2 will be enough for another 20+ years, no need for Leaders.
Only 2 cruisers for 3 fleets?
Isos on Tue Apr 21, 2020 2:30 pm
What are you smocking ? They were far worse than Nazi back in the time. And today Abe is following the same path as the ones in charge in 1940. He is pushing towards a powerfull Japan and is changing the strategy from defence to attack.
Kurils are very important for the japanese. That's a huge ressource of food and they hve a self pride that will always push them to try to get the islands back.
I wouldn't be surprise if they still think the lands of the former empire are still theirs.
Only 2 cruisers for 3 fleets? pwnd
They have 2 kirov and 3 slava with 1 in black sea doing nothing there and 8 Oscar more or less in s Modernize them all with kalibr/oniks and send half in North and half in pacific. Add another 5+5 Gorshkov and it will be good for the next 15-20 years to protect their interests near Russia.
If they plan to have interests further away switch the Gorshkov for super Gorshkov and reinstall some bases in Cuba/Venezuela, Vietnam and one or two in Africa with permanent deployement of corvettes and some il-38/Su35.
The first solution is by far cheaper and more in adequation with Russian strategy/interests.
Tsavo Lion wrote: Why the Navy refused to build super destroyers
For the price of 1 Leader they can build several smaller FF/DD/CGs- the time for white elephants is over. Even China with more $ & longer SLOCs to defend in ice-free waters doesn't plan to build similar class of ships.
https://regnum.ru/news/it/2922231.html
The article is not very good! They are writting about small ships able to strike targets with Tsirkon as far as 1000 km. That's true for sure. But the enemy will maybe have the option to strike at longer range. How? For example the combat radius of an F-35B or C onboard a carrier, part of a carrier battle group, is fully equipped with missiles and bombs less than 1000 km. But with refueling the range increase significant. So it's able to start F-35 from a carrier, refuel them in the air (via drones) and to launch an attack without being in the AD zone of a corvette. Only from a frigate on you can threaten the attacker to become the target himself. This what I mean with an umbrrella bigger ships span for smaller ships. So you can't compare navalized S-500 with Pantsir.
Tsirkon is a great weapon with nearly no defense against it...but you have to come within combat range. So it is the ultimate weapon for subs and can convert any destroyer or cruiser in a ultimate killer, able to destroy a whole carrier group. But a small ship with such a powerful weapon must have luck.
The rules of naval warfare have not changed since cold war! The missiles now are faster and have a bit longer range, but the concepts are the same. One Kirov was able to threaten a whole carrier group and in these days bigger ships are potent too. So why China is building Type 055 destroyers?
Isos wrote:
I learned for 2 1/2 years japanese language and we learned in class much about culture of Japan. Japan is now a really western country, more or less. The old traditions are broken and nothing more than rituals. Yes of course they are no europeans, their lifestyle is different but modern. Yes, maybe nationlism will rise again but not in the old fashioned style, with an god emperor.
Abe is nationalist, but inside Japan he is critized often for his aggressive foreign stance. He seeks some kind of conflict with China. Russia (Kuriles) is no priority in his politics.
Japan have never given up the right to the area! In contrast to Germany, which accepted (I think in the 70ies) the new borders after war. For Japan the Kuriles Islands are part of Japan and will ever be. But they try to reintegrate the Kurile Island only with peaceful methods. A year ago they offered Russia money for Kuriles Islands ;D LOL They try nearly every year to retake Kurile, sometimes with money, sometimes with partnership and trade and sometimes with both.
Isos wrote: have 2 kirov and 3 slava with 1 in black sea doing nothing there and 8 Oscar more or less in s Modernize them all with kalibr/oniks and send half in North and half in pacific. Add another 5+5 Gorshkov and it will be good for the next 15-20 years to protect their interests near Russia.
I'm not a fan of the Slavas! Back in time they were nice and they try to refurbish them now, but I doubt they would be so good in future. But yes, you are right...I forgot them. They are officially cruisers.
Last edited by Azi on Tue Apr 21, 2020 3:58 pm; edited 1 time in total
miketheterrible on Tue Apr 21, 2020 3:12 pm
Russian aircraft can and have been proven to take off and land on unprepared airfields, something American jets cant. F-2 which is Japans F-16 is a hog and a problem in itself. It wouldn't survive taking off or landing on destroyed airfields. Japanese airfields are very limited and in close proximity due to geography. Russian airfields are far more spread out and can survive heavy bombardment as proven in Syria where Syrian airfields built during soviet times survived as well from heavy CM strikes.
miketheterrible wrote: Russian aircraft can and have been proven to take off and land on unprepared airfields, something American jets cant. F-2 which is Japans F-16 is a hog and a problem in itself. It wouldn't survive taking off or landing on destroyed airfields. Japanese airfields are very limited and in close proximity due to geography. Russian airfields are far more spread out and can survive heavy bombardment as proven in Syria where Syrian airfields built during soviet times survived as well from heavy CM strikes.
What do you mean by destroyed ? Runways are generally 2km or more long and 50m or more wide. A f-16 needs 600-700m to take off and just 7 or so meters wide space. The debris can be cleaned fast if it gets bombed and holes can be repaired quickly too.
Bombing with dumb bombs makes you go over air defence and an airspace full of enemy fighters. Cruise missiles suck at destroying a runway because they make a single hole. Only french have a dedicated cruise missile to destroy runways and it comes with submunitions. Iskander could come with submunition but doesn't have the range to do so.
Russia doesn't have that much bases either in the east:
https://en.m.wikipedia.org/wiki/List_of_Soviet_Air_Force_bases
Tsavo Lion
Location : AZ, USA
Tsavo Lion on Tue Apr 21, 2020 5:56 pm
Sorry that is bullshit. You can't say one ship costs the same as 3 smaller ships and then say lets make three smaller ships.
In the "all things considered" context that the article discusses, bigger & better armed FFGs of the DDG size will be enough, so there's no need for direct Kirov class CGN replacements.
https://vz.ru/society/2020/4/19/1035074.print.html
Russian warships instilled fear in Japan. Russia brought down arrogance from Japan, demonstrating the strength and power of the Pacific Fleet during tactical exercises, writes the Chinese publication Sohu. Since March 25, ambitious maneuvers have been carried out in the waters of the Sea of Japan and the Sea of Okhotsk, involving 15 Russian warships and boats, 12 support vessels, including the Varyag cruiser, the Bystry destroyer, the Marshal Krylov control ship, the Loud corvettes, and "Perfect", as well as the large anti-submarine ship "Admiral Vinogradov." In addition, about 20 units of naval aviation were involved.
https://www.youtube.com/watch?v=YpwG6Y7zjyI
The Oscar SSGNs can destroy JMSDF ships in port, in home waters & open sea with their long range Granits & Onikses; besides additional MiG-31s, Su-33/34s may be deployed to the FE in a matter of hours to join Tu-22M3s/95s already there.
Japan won't risk invading the S. Kuriles for 1 more reason: China & both Koreas will support Russia &/ may invade Japanese held islands in the W. Pacific, incl. Okinawa.
The Japanese will keep their military assets intact for future confrontations with China & Korea.
Last edited by Tsavo Lion on Tue Apr 21, 2020 6:30 pm; edited 1 time in total (Reason for editing : add a quote)
First off, no, runways if hit isn't "debris" it is craters. Which the jets would have to maneuver around. Second, Russia has plenty of cruise missiles to cause such damage. And third, if Russia decided to place Iskanders could do so in Kuril Islands and be within range.
I can also state long range Russian bombers using not dumb bombs would be sufficient too. But you are too thick to understand that. You just make up scenarios that doesnt exist. No, Russia wouldn't use dumb bombs against Japan not until Japans air defense systems were dealt with entirely. Kh-22 and 32 would be ideal as its warhead is 1,000kg.
You can have make belief viewpoints about Japans capabilities. But they cant fight worth shit and have been neutered since WW2.
kvs on Tue Apr 21, 2020 7:04 pm
We see the same old nonsense in this thread about Russia being some glorified 3rd world toilet that is not even able to play
second fiddle to mighty western nations. They even managed to get Japan into this ubermenschen camp. Western congenital Russophobia
is something else.
The yanquis are scrambling to replace their dinosaur Minuteman III to offset Sarmat. But hey, Russian key assets will be taken out
quickly. LOL. The S-500 system is going to produce brown stains in the underwear of all western military planners if not their bosses.
The only thing that is going to be taken out quickly in the initial stages of a war on Russia is the windbag hubris of the aggressors.
kvs wrote: We see the same old nonsense in this thread about Russia being some glorified 3rd world toilet that is not even able to play
Not to mention when some of these ubermench nations like Japan uses a shitty F-16 variant that is way over priced, and not really all that great. Along with not having really much in the way of being able to strike back at Russia nor at far distances. Then the fact that they use shitty American AD systems that are barely functional.
But oh well. Who cares cause its not like us debating it will change anything. Russia advances while other nations are barely able to advance without either costing hundreds of billions or try to play catch up.
Hole on Tue Apr 21, 2020 8:44 pm
Weird, isn´t it, that even western "think tanks" think that the russian military is at least the second most powerful in the west and the only Thing that makes Amiland "better" are the 800+ bases around the world and the useless carriers. But in reality Russia is just a paper tiger that couldn´t defeat Poland. Sure.
PapaDragon
Location : Fort Evil, Serbia
PapaDragon on Tue Apr 21, 2020 8:52 pm
This tread needs to be locked until this project is reactivated (if ever)
No point keeping it open
George1 on Tue Apr 21, 2020 9:27 pm
PapaDragon wrote:
in some topics discussion starts with construction or no of a ship and ends with WWIII beyween USA and Russia
in some topics discussion starts with construction or no of a ship and ends with WWIII beyween USA and Russia Very Happy
Not USA but Japan. Try to follow a little bit
Tsavo Lion on Tue Apr 21, 2020 10:22 pm
If/when the 1st Leader is ready, the USN won't be able to deploy as many CSGs anyway- even now most of them r in port/yards since before the pandemic. The GWOT & what followed bankrupted the USA.
Trump is the Demagogue, sweet talker, womanizer & failing Damage Controlman in Chief who himself avoided the draft, but still somehow got out of 3 bankruptcies, w/o a single shot.
Azi on Tue Apr 21, 2020 10:30 pm
Bullshit! Who said, wrote or implied that Russia is a 3rd world toilet (except Arrow)? What applies to Russia also applies to the United States!!! Russia could easy take out us-american key assets in overseas in minutes. Before you brand everyone as trolls and make yourself the keeper of truth, I recommend modesty.
At start of WW2...wich army was the best equipped and biggest in europe? Nazi Germany? NO!!! It was France, overwhelmed within weeks by Nazi Germany. As I wrote...the attacker can determine the rules of battle and only if the defender is fully aware of the threat he is in better position.
Yes, Russia could react to an attack in Far East, but it would take days or weeks to react. Or you think Russia will move the whole army, everything and every asset from western hemisphere to far east? Maybe pay USA to protect their borders? And move troops, aircraft and ships within seconds thousands of miles? There are many good analyzes that troops in Far East are underarmed and there are simply not enough troops there. All the fancy new stuff is in the western and souther hemisphere of Russia and here is confrontation with NATZO forces, that's correct. Today it's no problem, because Russia has good connections to every country in the region, even good connections to Japan.
And nobody is thinking of an attack NOW! But this region in Asia is very labile and a conflict will for sure not start with Russia as a beginner of the conflict, but it will end with Russia fully involved. To shape the future and to stabilize this region Russia needs a modern and strong Pacific Fleet. No more or less! And you think we are antirussian, because we praise a modern navy with some big ships (Super Gorshkov)?! Or would you prefer some old rustbuckets and only a few corvettes?
George1 wrote:
As always! ;D LOL
Hole wrote: Weird, isn´t it, that even western "think tanks" think that the russian military is at least the second most powerful in the west and the only Thing that makes Amiland "better" are the 800+ bases around the world and the useless carriers. But in reality Russia is just a paper tiger that couldn´t defeat Poland. Sure.
Get down a bit from your views! Never underestimate the enemy!
Russia could react to an attack in Far East, but it would take days or weeks to react. Or you think Russia will move the whole army, everything and every asset from western hemisphere to far east? Maybe pay USA to protect their borders?
Japan, a US ally, knows what the RF mil. doctrine says: there will be nuclear retaliation in the event that the existence of the RF is threatened by nuclear armed states or states allied with nuclear armed states. The loss of S. Kuriles may lead to Pac. Fleet surface units in Vladivostok becoming like tjhe bottled up Baltic & Black Sea Fleets of 1941-44, with Japanese & US navies entering the Sea of Okhotsk at will.
That in turn may lead to the loss of the RFE, Far North, & Siberia, Ural, & fragmentation of European Russia.
Poseidon UUVs can create tsunamis a lot more deadly than the 1 at Fukushima, destroying the entire JMSDF & the US 7th fleet.
Azi on Wed Apr 22, 2020 3:35 am
Tsavo Lion wrote:
You know the military doctrine? The russian military doctrine says that there will be NO NUCLEAR FIRST strike! NEVER! Nuclear Forces are only for retaliation, defending the homeland is task of conventional forcess. Conventional forces should be able to repel all attacks.
Yes...that's in the military doctrine of Russia! This and nothing else. I don't understand your wet dreams about the nuclear holocaust for a bit dirt, a few rocks and trees? All the bullshit about the aggressive russians who use nukes if you only watch them too long is propaganda bullshit and only to demonize Russia!!!
In reality a use of tactical nukes is possible if the core of Russia is threatend (all big cities!). The magic word ist tactical nukes and not strategic! There won't be a nuclear war for Kaliningrad, Kurile, Sachalin or Semlja Franza-Iossifa! If the conventional forces are not able to defend it, they lose it. Same for USA and parts of the Bering Strait, Hawaii etc! And the USA have an aggressive first strike doctrine in contrast to Russia, who see nuclear weapons only as a tool for survival of Russia.
You are right if Wladiwostok or large parts of Siberia are threatend, but this will never ever happen! And be realistic please! The use of Poseidon UUV means the use of a strategic nuke against civilian targets, triggering an annihilation strike for Russia and that will trigger a counter attack...with at least minimum 2 billion dead people. Poseidon is only a doomsday weapon, for the big nuclear clash in a possible WW3. There is no other use for Poseidon and will never be!!!
Tsavo Lion on Wed Apr 22, 2020 4:50 am
There's no clear mention of what kind of nukes will be used in any particular set of circumsatnces: https://www.files.ethz.ch/isn/96824/00_Apr.pdf
Pl. see pgs 36-37:
https://www.clingendael.org/sites/default/files/pdfs/20110300_haas_doctrine.pdf
The Poseidons r for deterrence ashore & tactical use at sea.
Who said that their road mobile ICBMs in Urals & Siberia & SLBMs the Barentz & Okhotsk Seas can't strike Japanese bases with conventional warheads?
Unlike the Baltic Fleet in 1902, the N. Fleet doesn't need to cross 2 oceans to reinforce the Pac. Fleet during any future confrontation with Japan- it can use the NSR.
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text:
Der Eutiner Wasserturm steht an der Ecke Bismarckstraße/Wilhelmstraße. Er wurde 1909 am höchsten Punkt der Stadt errichtet. Zur gleichen Zeit entstand das Wasserwerk am Deefstieg (früher Diebsteig).

Das Bauwerk
Der 38 m hohe, runde Backsteinturm steht auf polygonalem Sockel. Fenster und Eingangsportal weisen Spitzbögen auf. Grünglasierte Dachpfannen bilden den Übergang vom Sockel zum schmaleren, sich konisch verjüngenden Schaft. Der Unterbau trägt einen genieteten Behälter der Bauart Intze 1. Er fasst 250 m³ Wasser und wird von einem Innenzylinder durchbrochen. Durch den Zylinder führt eine Wendeltreppe zur Aussichtsplattform. Zur Spitze sind es insgesamt 156 Stufen.

→ Näheres zu den Behälterformen siehe Hauptartikel Wasserturm

Im Bereich des Wasserbehälters ist das Gebäude deutlich verbreitert.
In vielen Details zeigt der Wasserturm neugotische Elemente und erinnert damit an Bauwerke der Backsteingotik: Die zylindrische Ummauerung des Behälters endet mit einem Zinnenkranz. Zahlreiche Blendnischen, von denen die kleineren verputzt sind, gliedern das Mauerwerk. Ein kupfergedecktes Kegeldach (ursprünglich zinngedeckt) schließt den Bau nach oben ab.

Heutige Nutzung
Das Fassungsvermögen des Wasserbehälters reichte für die erweiterte Versorgung der Stadt in den letzten Jahren nicht mehr aus. Der Turm diente nur noch zur Druckstabilisierung und wurde 2006 schließlich ganz vom Netz genommen.

Heute wird der Turm in den Sommermonaten zur Besichtigung freigegeben und als Aussichtsturm genutzt. Die Besucher gelangen über eine Wendeltreppe, die in einer Röhre mitten durch den Wassertank führt, auf den Umgang zwischen Dach und Zinnenkranz.
Regelmäßig finden im Turm auch Ausstellungen zeitgenössischer Kunst statt. In drei Etagen befinden sich Galerien und Ausstellungsräume. Eine permanente Fotoausstellung zeigt sämtliche Wassertürme in Schleswig-Holstein.

Bilder

Siehe auch
 Liste der Wassertürme in Schleswig-Holstein

Literatur
 Information der Stadtwerke Eutin
 Jens U. Schmidt: Wassertürme in Schleswig-Holstein. Geschichte und Geschichten um die Wasserversorgung im Norden und ihre auffälligsten Bauten. Regia-Verlag, Cottbus 2008, ISBN 978-3-939656-71-5.

Weblinks

 Stadtwerke Eutin
 Kunst im Wasserturm

Eutin
Bauwerk in Eutin
Aussichtsturm in Schleswig-Holstein
Erbaut in den 1900er Jahren
Kulturdenkmal in Eutin
Aussichtsturm in Europa

redpajama_set_name: RedPajamaWikipedia


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text:
Craig Wednesday, 29 Jun 2022 at 2:11pm
groundswell Wednesday, 29 Jun 2022 at 2:27pm
Frog in the Hole
groundswell Thursday, 30 Jun 2022 at 7:04am
Fast and loose trucks but still in control.
udo Friday, 1 Jul 2022 at 2:20pm
Check this Old Grom..............
betcha he's glad he tossed in the Bricklaying.
icandig Friday, 1 Jul 2022 at 10:31pm
udo Saturday, 2 Jul 2022 at 5:55pm
Skeeta yew
Nick Bone Sunday, 3 Jul 2022 at 7:11am
You know Skeet udo? Think they did pretty well on that trip!
udo Sunday, 3 Jul 2022 at 4:17pm
udo Monday, 4 Jul 2022 at 8:53pm
From a few yrs ago but Fkinhell
udo Wednesday, 6 Jul 2022 at 6:06pm
Well done Good work Cal !
udo Thursday, 7 Jul 2022 at 7:05am
udo Thursday, 7 Jul 2022 at 6:58pm
Ripper Tube Ocea
Ripper of a Claim Yew !
udo Friday, 8 Jul 2022 at 8:24am
Burn the Grom you C##T
Supafreak Sunday, 10 Jul 2022 at 8:00am
udo Sunday, 10 Jul 2022 at 8:10am
udo Monday, 11 Jul 2022 at 12:06pm
Wild Man Barton from Borneo he he
burleigh Monday, 11 Jul 2022 at 2:15pm
https://www.instagram.com/p/Cf0TpC7ADP5/?hl=en
But Udo, he's not vaccinated and has spoken out against it.
Surely that makes him your instant enemy?
Island Bay Monday, 11 Jul 2022 at 2:48pm
Ultimate beard farmer!
Looks healthy and well, doesn't he.
udo Tuesday, 12 Jul 2022 at 4:21pm
udo Wednesday, 13 Jul 2022 at 9:54am
udo Thursday, 14 Jul 2022 at 9:21am
udo Thursday, 14 Jul 2022 at 2:19pm
NOoooo Maurice there is no Dark Side...its Riding what ever keeps you out there having Fun
Cut yourself a 5'8" x 22 x 3 Mini Simmons as well
simba Thursday, 14 Jul 2022 at 4:44pm
quick someone help MC ..............
DudeSweetDudeSweet Thursday, 14 Jul 2022 at 5:26pm
Ride a longboard if it makes you happy but why go full log retard and confirm to the whole "dancing with Krishna" bullshit style contrivance with the raised arms? Don't do it Maurice, just ride your long board.
Don't be like those clown log women who think they've got delicate raised fingertips like a Balinese dancer but who really look like….
velocityjohnno Thursday, 14 Jul 2022 at 9:16pm
Phil Edwards might have something to say about that silly post above.
Supafreak Saturday, 16 Jul 2022 at 7:29pm
blackers Saturday, 16 Jul 2022 at 10:33pm
Fark.
udo Sunday, 17 Jul 2022 at 7:52pm
udo Monday, 18 Jul 2022 at 6:29pm
Fkn Cool Honourable Act
Island Bay Friday, 22 Jul 2022 at 1:36pm
Ace visits and gets barrelled:
udo Monday, 25 Jul 2022 at 8:11am
udo Monday, 25 Jul 2022 at 11:43am
I focus Wednesday, 27 Jul 2022 at 10:25am
Supafreak Wednesday, 27 Jul 2022 at 11:07am
https://www.instagram.com/p/CgffgdrPoqr/
Still trying to fathom the risk versus reward
udo Wednesday, 27 Jul 2022 at 6:23pm
Supafreak Thursday, 28 Jul 2022 at 6:10am
Where the fck does Jack come from ?
groundswell Thursday, 28 Jul 2022 at 5:50pm
A post shared by The Inertia (@theinertia)
How to post embeded? fucked if i know.
I focus Thursday, 28 Jul 2022 at 6:51pm
Been a bit of swell with better winds

redpajama_set_name: RedPajamaCommonCrawl


text:
\section{Introduction}

Let $P$ be a set of $m$ points, and $H$ a set of $n$ planes in
$\reals^3$. Assume that $P$ is contained in some two-dimensional nonlinear and irreducible
algebraic variety $V$ of constant degree $D$. We wish to bound the size $I(P,H)$ of
the \emph{incidence graph} $G(P,H)$, whose edges connect all pairs $(p,h)\in P\times H$
such that $p$ is incident to $h$. In general, and in this special setup too, $I(P,H)$ can be
as large as the maximum possible value $|P|\cdot|H|$, by placing all the points of $P$
on a line, and make all the planes of $H$ contain this line, in which case $G(P,H) = P\times H$.
The bound that we are going to obtain will of course acknowledge this possibility,
and will be obtained by decomposing $G(P,H)$ as the disjont union
$G_0(P,H) \cup \bigcup_i P_i\times H_i$, where, for each $i$, there exists a line
$\ell_i$ fully contained in $V$, so that $P_i = P\cap\ell_i$ and $H_i$ is the set
of planes of $H$ that contain $\ell_i$. The residue graph $G_0(P,H)$ represents the
``accidental'' incidences, those that cannot be ``explained'' in terms of lines that
are contained in $V$ and induce large complete bipartite subgraphs of $G(P,H)$, and we
provide an explicit bound for the number $I_0(P,H) = |G_0(P,H)|$ of these incidences.
The quality of the bound will be measured by $I_0(P,H)$ and by $\sum_i |P_i|$ and $\sum_i |H_i|$.

Earlier works on point-plane incidences have considered the general
setup, where the points of $P$ are arbitrarily placed in $\reals^3$.
Initial partial results have been obtained by Edelsbrunner, Guibas
and Sharir~\cite{EGS}. More recently, Apfelbaum and
Sharir~\cite{ApS} (see also Brass and Knauer~\cite{BK} and Elekes
and T{\'{o}}th~\cite{ET05}) have shown that if the incidence graph
does not contain a copy of $K_{r,s}$, for constant parameters $r$
and $s$, then $I(P,H)=O(m^{3/4}n^{3/4}+m+n)$. In more generality,
Apfelbaum and Sharir~\cite{ApS} have shown that if $I=I(P,H)$ is
significantly larger than this bound, then $G(P,H)$ contains a large
complete bipartite subgraph $P'\times H'$, such that $|P'|\cdot |H'|
= \Omega(I^2/(mn)) - O(m+n)$. Moreover, as also shown in~\cite{ApS}
(slightly improving a similar result of Brass and Knauer~\cite{BK}),
$G(P,H)$ can be expressed as the union of complete bipartite graphs
$P_i\times H_i$ so that $\sum_i (|P_i|+|H_i|) =
O(m^{3/4}n^{3/4}+m+n)$. (This is a specialization to the case $d=3$
of a similar result of \cite{ApS,BK} in any dimension $d$.) A more
recent result of Zahl~\cite{Za} gives a grand generalization of
these bounds, to incidences between points and constant-degree
algebraic surfaces for which the incidence graph does not contain
$K_{r,3}$ as a subgraph; that is, in situations where every triple
of the given surfaces intersect in a constant number of points.
Zahl's general bound is $O(m^{3/4}n^{3/4}+m+n)$, as above. We note
that Fox et al.~\cite{FPSSZ} present a more general framework that
includes incidences problems of many kinds, and yields, for the case
under consideration, almost the same bound, namely $O(m^{3/4
+\eps}n^{3/4 + \eps} + m + n)$, for any $\eps$, where the constant
of proportionality depends on $\eps$.

Nevertheless, as we show in this paper, the bound can be substantially improved
when all the points of $P$ lie on a constant-degree variety $V$. Our main result
is the following theorem.
\begin{theorem} \label{main}
For a set $P$ of $m$ points on some two-dimensional, nonlinear,
non-conical\footnote{That is, $V$ is not the union of lines through
a common point.} and irreducible algebraic variety $V$ of constant
degree $D$ in $\reals^3$, and for a set $H$ of $n$ planes in
$\reals^3$, the incidence graph $G(P,H)$ can be decomposed as
\begin{equation} \label{gph}
G(P,H) = G_0(P,H) \cup \bigcup_i (P_i\times H_i) ,
\end{equation}
such that, for each $i$, there exists a line $\ell_i \subset V$ so that
$P_i = P\cap \ell_i$ and $H_i$ is the subset of planes of $H$ that contain $\ell_i$, and we have
\begin{align} \label{eq:main}
|G_0(P,H)| & = O\left( m^{2/3}n^{2/3} + m^{6/11}n^{9/11}\log^\beta(m^3/n) + m + n \right) , \quad\text{and} \\
\sum_i |P_i| & = O(m) \quad\text{and}\quad \sum_i |H_i| = O(n) \nonumber ,
\end{align}
where the constant exponent $\beta$ and the constants of proportionality depend on the degree $D$ of $V$.

If $V$ is a (non-linear) cone, the same result holds if we exclude
the apex of $V$ from $P$; the apex adds at most $O(n)$ incidences.
\end{theorem}

We remark that the problem studied in this paper is not as specialized as it
might appear. It is a generalization of the problem of
bounding the number of incidences between points and circles of arbitrary radii
in the plane. By applying the standard lifting transformation, which maps each
point $(x,y)$ in the plane to the point $(x,y,x^2+y^2)$ on the paraboloid
$z=x^2+y^2$ in 3-space, and maps each circle $(x-a)^2+(y-b)^2=r^2$ to the plane
$z=2ax+2by+(r^2-a^2-b^2)$, the problem becomes that of bounding the number
of incidences between points on the paraboloid and arbitrary planes in three dimensions.
This transformation is used in the second step of the analysis of the point-circle
incidence problem, as in \cite{ANP+,ArS}, and the bound derived there is very similar
to the one in (\ref{eq:main}), except that, in the case of circles, $G(P,H)$ does not
contain large complete bipartite graphs (because the paraboloid does not contain any line),
and the exponent $\beta$ is equal to the sharper value $2/11$; see~\cite{ANP+,ArS,MT} for details.

As a matter of fact, we extend this observation in Section~\ref{sec:quadric},
and derive such an improved variant of Theorem~\ref{main} for any quadric $V$.
The resulting bound is the same, but the analysis is simplified, and the exponent
$\beta$ in (\ref{eq:main}) is still $2/11$; see Theorem~\ref{thm:quadric} for the precise formulation.

As an application, we note that our bound leads to a slight improvement in a recent lower bound,
due to Rudnev~\cite{Rud}, on the number of distinct cross-ratios determined by $n$ real numbers,
improving his bound from $\Omega(n^{24/11}/\log^{6/11}n)$ to $\Omega(n^{24/11})$. More details
are provided in Section~\ref{sec:cross}.

In closing the introduction, we note that the terms $P_i\times H_i$ can only arise when
$V$ fully contain the corresponding line $\ell_i$. This issue, of analyzing lines fully contained
in algebraic varieties in $\reals^3$ (or in $\cplx^3$) has received considerable attention
in recent years, starting with Guth and Katz's work~\cite{GK2}. Briefly, if $V$ is a
\emph{ruled surface},\footnote
  That is, each point on $V$ is incident to a line that is fully contained in $V$.}
it contains infinitely many lines, and then the set of lines that form the complete
bipartite decompositions might be large (up to $\Theta(\min\{m,n\})$ lines). That is,
ruled surfaces might yield a large number of complete bipartite subgraphs in $G(P,H)$
(but then these graphs are smaller on average). In contrast, as follows from the
Monge--Cayley--Salmon theorem (see, e.g., Guth and Katz~\cite{GK2}), a non-ruled
irreducible variety of degree $D$ can fully contain at most $D(11D-24)$ lines. Hence,
if $V$ is not ruled, the decomposition in (\ref{gph}) can consist of only a constant
number of complete bipartite graphs.

\section{Proof of Theorem~\ref{main}}

Let $P$, $V$, $H$, $m$, and $n$ be as above. To obtain the representation (\ref{gph})
and the corresponding bounds (\ref{eq:main}), we first derive a weaker bound, and then
improve it via a suitable decomposition of dual space, similar to the way it has been
done for circles in \cite{ANP+,ArS}, and in more generality in \cite{SZ}.

\paragraph{A weak basic bound.}
Let $L$ denote the set of lines fully contained in $V$ and containing
at least one point of $P$. For each line $\ell\in L$ we form the
bipartite subgraph $P_\ell \times H_\ell$ of $G(P,H)$, where $P_\ell = P\cap\ell$
and $H_\ell$ is the set of all the planes of $H$ that contain $\ell$.
(Recall the discussion at the end of the introduction: There are at most $O(1)$
lines $\ell$ (and corresponding complete bipartite graphs) when $V$ is not ruled.)
As shown, e.g., in Sharir and Solomon~\cite[Lemma 5]{SS3d}, except for at
most one \emph{exceptional} point, each point $p \in V$ is incident
to at most $D^2=O(1)$ lines of $L$.

If $V$ contains an exceptional point $p_0$ then it must be a cone with $p_0$ as
an apex, and (since $V$ is nonlinear) all the lines contained in $V$ pass through $p_0$.
In this case we remove $p_0$ from $P$, losing at most $n$ incidences,
and re-apply the analysis to the set $P\setminus\{p_0\}$, which we continue to denote by $P$.

Ignoring then the exceptional point, if any, the preceding property
implies that $\sum_{\ell} |P_\ell|=O(m)$. (That is, even when $V$ is ruled,
only finitely many lines $\ell$ can arise.) As $V$ is irreducible and
nonlinear, it does not contain any of the planes in $H$. Thus, for
each $h\in H$, the intersection $h\cap V$ is a (plane) algebraic curve of
degree at most $D$, and can therefore contain at most $D=O(1)$ lines
of $L$ (as follows, e.g., from the generalized version of B\'ezout's
theorem~\cite{Fu84}, or by other, more direct arguments).
This implies that $\sum_{\ell} |H_\ell|=O(n)$. Retaining only the lines that
contain at least one point of $P$ and are contained in at least one plane of $H$,
we obtain a finite decomposition $\bigcup_\ell \left( P_\ell\times H_\ell \right)$
as a portion of $G(P,H)$, and $\sum_\ell |P_\ell| = O(m)$, and $ \sum_\ell |H_\ell| = O(n)$.

For each $h\in H$, put $\gamma_h := (h\cap V) \setminus \bigcup L$.
As just noted, each $\gamma_h$ is at most one-dimensional. By
construction, it does not contain any line, and it might also be
empty (for this or for other reasons). Note that if $h\cap V$ does
contain a line $\ell$, then $\ell\in L$, and the incidences between
$h$ and the points of $P$ on $\ell$ are all already recorded in
$P_\ell\times H_\ell$. Finally, we ignore the isolated points of
$\gamma_h$. By Harnack's curve theorem~\cite{Har}, the number of
such points on a plane curve of degree at most $D$ is $O(D^2) = O(1)$,
so the isolated points contribute a total of at most $O(n)$ incidences.
Let $G_0(P,H)$ denote the remaining portion of $G(P,H)$, after
pruning away the complete bipartite graphs $P_\ell\times H_\ell$.
This gives us the decomposition in (\ref{gph}). We further remove
from $G_0(P,H)$ all the $O(n)$ incidences involving isolated points
on the curves $\gamma_h$, and the $O(n)$ incidences with the
exceptional point of $V$, if it exists and belongs to $P$, continue
to denote the resulting graph as $G_0(P,H)$, for notational
convenience, and put $I_0(P,H) = |G_0(P,H)|$.

Let $\Gamma$ denote the set of the $n$ curves $\gamma_h$, for $h\in H$. The curves of $\Gamma$
are algebraic curves of degree at most $D$, and any pair of curves
$\gamma_h$, $\gamma_{h'}\in \Gamma$ intersect in at most $D=O(1)$ points.
Indeed, any of these points is an intersection point of $V$ with the line
$\ell=h\cap h'$; if $\ell$ is contained in $V$ then, by construction, it
is removed from both curves, and if $\ell$ is not contained in $V$, it can
meet it in at most $D$ points. In particular, the curves $\gamma_h$
are distinct and non-overlapping.

Note that $I_0(P,H)$ is equal to the number of incidences $I(P,\Gamma)$
between the points of $P$ and the curves of $\Gamma$. To bound the latter
quantity we proceed as follows.

We slightly tilt the coordinate frame to make it generic, and then project the
curves of $\Gamma$ onto the $xy$-plane. A suitable choice of the tilting guarantees
that (i) no pair of intersection points or points of $P$ project to the same point;
(ii) if $p$ is not incident to $\gamma_h$ then the projections of $p$ and of
$\gamma_h$ remain non-incident; and (iii) no pair of curves in $\Gamma$ have
overlapping projections. In addition, by
construction, no curve of $\Gamma$ contains any (vertical) segment. Let $P^*$ and
$\Gamma^*$ denote, respectively, the set of projected points and the set of
projected curves; the latter is a set of $n$ plane algebraic curves of constant maximum
degree $D$ (since the projection of a plane curve onto another plane clearly cannot increase its degree).
Moreover, $I(P,\Gamma)$ is equal to the number of incidences between $P^*$ and $\Gamma^*$.

By the result of Sharir and Zahl~\cite{SZ}, applied to $\Gamma^*$,
the curves of $\Gamma^*$ can be cut into $O(n^{3/2}\log^\kappa n)$ connected Jordan subarcs,
where the constant exponent $\kappa$ and the constant of proportionality depend on $D$,
so that each pair of subarcs intersect at most once. Using standard terminology,
the subarcs form a collection of \emph{pseudo-segments}.

We can now apply the crossing-lemma technique of Sz\'ekely~\cite{Sze}, exactly as
was done in Sharir and Zahl~\cite{SZ} or in earlier works. Since the resulting subarcs
form a collection of pseudo-segments, and the number of their intersections is $O(n^2)$,
Sz\'ekely's analysis yields the bound
\begin{equation} \label{i0ph}
I_0(P,H) = I(P,\Gamma) = I(P^*,\Gamma^*) =
O\left( m^{2/3}n^{2/3} + m + n^{3/2}\log^\kappa n \right).
\end{equation}

Adding incidences recorded in the complete bipartite decomposition, as constructed
above, and those involving isolated points and the exceptional point, if any,
we get our initial (weak) bound.
\begin{equation} \label{weakinc}
I(P,H) = O\left( m^{2/3}n^{2/3} + m + n^{3/2}\log^\kappa n + \sum_{\ell} |P_{\ell}| \cdot |H_{\ell}| \right) ,
\end{equation}
where $\bigcup_{\ell} \left( P_{\ell}\times H_{\ell} \right)$ is contained in the incidence
graph $G(P,H)$, is a finite union over lines $\ell\subset V$,
and $\sum_{\ell} |P_{\ell}| = O(m)$ and $\sum_{\ell} |H_{\ell}| = O(n)$.

\paragraph{The case $m=O(n^{1/3})$.}
Before proceeding to improve the bound in (\ref{weakinc}), we first
dispose of the case $m=O(n^{1/3})$. As above, we first remove all
the complete bipartite graphs $P_\ell\times H_\ell$, for $\ell\in L$,
from $G(P,H)$.
We then proceed to estimate $I_0(P,H)$ as follows. We call a
plane $h\in H$ \emph{strongly degenerate} (or \emph{degenerate} for short)
if all the points of $P\cap h$ are collinear. We claim that the number
of incidences between points of $P$ and non-degenerate planes is
$O(m^3+n)=O(n)$. Indeed, first discard the planes $h\in H$
containing at most two points of $P$, losing at most $O(n)$ incidences.
For an incidence between a point $p\in P$ and a surviving non-degenerate
plane $h\in H$, there exist (at least) two distinct points
$q,s \in (P\setminus \{p\})\cap h$ that are not collinear with $p$.
The ordered triple $(p,q,s)$ therefore \emph{uniquely} accounts for the
incidence between $p$ and $h$, and there are $O(m^3)=O(n)$ such triples.

To bound the number of incidences between the
points of $P$ and the degenerate planes of $H$, fix a degenerate
plane $h\in H$, and assume that $m_h:=|h\cap P| \ge D+1$; the
overall number of incidences on the other planes is at most $Dn = O(n)$.
By assumption, all points of $h\cap P$ lie on a common line
$\ell$. Since $\ell$ contains at least $D+1$ points, it must be
contained in $V$, so the incidences involving $h$ are all recorded
in the complete bipartite graph $P_\ell \times H_\ell$.
In other words, we have shown, for $m = O(n^{1/3})$,
\begin{equation} \label{smallm}
I(P,H) = O\left( n + \sum_\ell |P_\ell| \cdot |H_\ell| \right) ,
\end{equation}
where, as above, $\bigcup_\ell \left( P_\ell\times H_\ell \right)$ is contained
in the incidence graph $G(P,H)$,
and $\sum_{\ell} |P_{\ell}| = O(m)$ and $\sum_{\ell} |H_{\ell}| = O(n)$.
In the rest of the analysis, we thus assume that $m=\Omega(n^{1/3})$.

\paragraph{Improving the bound.}
As in the analysis of incidences between points and circles (or pseudo-circles) in
\cite{ANP+,ArS}, and the more general treatment in~\cite{SZ}, the first two terms
in (\ref{i0ph}) dominate when $m = \Omega(n^{5/4}\log^{3\kappa/2} n)$. When $m$
is smaller, the third term, which is independent of $m$, is the one that dominates,
and we then sharpen it as follows (the same general approach is also used in
\cite{ANP+,ArS,SZ}).

We apply any standard duality to 3-space, which maps points to planes and planes to points,
and preserves point-plane incidences. We denote by $a^*$ the dual image (plane or point)
of an object $a$ (point or plane). We thus get a set $H^*$ of $n$ points, and a set $P^*$
of $m$ planes in dual 3-space.\footnote
  The planes are special, because they come from points on a variety, but we do not know how
  to exploit this property.}
We choose a parameter $r$, to be fixed later, and construct a
\emph{$(1/r)$-cutting} $\Xi$ for $P^*$ (see, e.g., \cite{Chaz}), which partitions the dual
space into $O(r^3)$ simplices, each crossed by at most $m/r$ dual planes.

\paragraph{Incidences with boundary dual points.}
Let us first bound the number of incidences between the dual points
that lie on the boundaries of the simplices and the dual planes.

\smallskip

\noindent
(i) Each dual point that lies in the relative interior of a facet of some
simplex of $\Xi$ (but not on any edge of another simplex) has one incidence with
the dual plane that contains that facet (if any), for a total of $O(n)$ such incidences.
If such a point $h^*$ is incident to another dual plane $p^*$, then
$p^*$ must cross both simplices adjacent to the facet containing $h^*$.
We then assign $h^*$ to one of these two simplices, and the relevant incidence
will then be counted in the subproblem associated with that simplex
(see below).

\smallskip

\noindent
(ii) Any incidence between a dual plane $p^*$ and a dual point $h^*$
that is a vertex of some simplices of $\Xi$ can be charged to the
crossing of $p^*$ with some simplex $\sigma$ of $\Xi$ that is adjacent
to $h^*$. It follows that the number of incidences of this kind is at
most $O(r^3)\cdot m/r = O(mr^2)$. The only case where we cannot apply
this charging is when $p^*$ supports facets of all incidence simplices,
but then the incidence is counted in (i).

\smallskip

\noindent
(iii) The remaining incidences involve points that lie on edges of the
simplices; there are $O(r^3)$ such edges. Fix one of these edges $e$.
An incidence between a point $h^*\in e$ and a plane $p^*$ that crosses
$e$ (intersects $e$ but does not contain it) can be charged to the crossing
of $p^*$ with some adjacent simplex (at most four such charges can be made to
any plane-simplex pair), and, as in (ii), the number of such incidences is only $O(mr^2)$.

\smallskip

\noindent
(iv) We are thus left
with incidences between dual points on $e$ and dual planes that
fully contain $e$. Let $\ell^*$ be the line supporting $e$. Back in
the primal space, there exists a ``pre-image'' $\ell$ of $\ell^*$
such that all the dual planes $p^*$ that contain $\ell^*$ come from
points $p$ that lie on $\ell$, and all the dual points $h^*$ that
lie on $\ell^*$ come from planes $h$ that contain $\ell$. If
$\ell\subset V$ then all these incidences are recorded in the
corresponding complete bipartite graph $P_\ell\times H_\ell$
that we have already constructed and disposed of. Otherwise, $\ell$ can contain
at most $D$ points of $P$, meaning that in dual space each point on
any of these edges $e$ is incident to at most $D$ dual planes that fully
contain $e$. Since the sets of dual points lying on the (relative interiors of the) edges $e$ are pairwise
disjoint, we get a total of at most $O(nD)=O(n)$ incidences of this latter kind.

In total, the number of incidences with boundary points is $O(mr^2+n)$.

\paragraph{Incidences within the simplices.}
We now proceed to consider incidences between dual points in the
interior of the simplices of $\Xi$. For each simplex $\sigma$ of $\Xi$, let
$n_\sigma$ denote the number of points of $H^*$ in the interior of $\sigma$,
including the points that $\sigma$ has ``inherited'' from points on its facets, in
step (i) of the preceding analysis. By duplicating cells $\sigma$
for which $n_\sigma > n/r^3$, so that in each copy we take at most $n/r^3$
of these dual points (but retain all the crossing dual planes), we obtain
a collection of $O(r^3)$ simplices, each of which is crossed by at most $m/r$
dual planes and contains at most $n/r^3$ dual points; we denote the actual
number of these planes and points as $m_\sigma$ and $n_\sigma$ (the latter
notation is slightly abused, as it now refers only to a single copy (subproblem)
of $\sigma$), respectively, for each simplex-instance $\sigma$.

For each cell $\sigma$, we apply the bound (\ref{weakinc})
to the subset $P^{(\sigma)}$ of the points of $P$ whose dual planes
cross $\sigma$, and to the subset $H^{(\sigma)}$ of the planes whose dual points
lie in $\sigma$, and note that the case $m/r = O((n/r^3)^{1/3})$ does not arise,
because then we would also have $m=O(n^{1/3})$, which we have assumed not to be
the case. That is, we get, for each $\sigma$,
$$
I(P^{(\sigma)},H^{(\sigma)}) = O\left( m_\sigma^{2/3}n_\sigma^{2/3} + m_\sigma + n_\sigma^{3/2}\log^\kappa n_\sigma
 + \sum_\ell |P^{(\sigma)}_\ell| \cdot |H^{(\sigma)}_\ell| \right) ,
$$
for a suitable complete bipartite decomposition
$\bigcup_\ell \left( P^{(\sigma)}_\ell \times H^{(\sigma)}_\ell \right)$.

We sum these bounds, over the simplices $\sigma$ of $\Xi$. We note that the same line
$\ell$ may arise in many complete bipartite graphs
$P^{(\sigma)}_\ell \times H^{(\sigma)}_\ell$, but (i) all these graphs are contained in
$P_\ell \times H_\ell$, and (ii) they are edge disjoint, because each dual point $h^*$ lies
in at most one simplex. This allows us to replace all the partial subgraphs
$P^{(\sigma)}_\ell \times H^{(\sigma)}_\ell$ by the single graph
$P_\ell \times H_\ell$, for each line $\ell$. We thus get

\begin{align*}
I(P,H) & = O\left( \sum_\sigma \Big( m_\sigma^{2/3}n_\sigma^{2/3} +
 m_\sigma + n_\sigma^{3/2}\log^\kappa n_\sigma \Big)
 + mr^2 + n + \sum_\ell |P_\ell| \cdot |H_\ell| \right) \\
& = O\left( r^3 \Big( (m/r)^{2/3}(n/r^3)^{2/3} + (n/r^3)^{3/2}\log^\kappa (n/r^3) \Big)
 + mr^2 + n + \sum_\ell |P_\ell| \cdot |H_\ell| \right) \\
& = O\left( m^{2/3}n^{2/3}r^{1/3} + \frac{n^{3/2}}{r^{3/2}}\log^\kappa (n/r^3)
 + mr^2 + n + \sum_\ell |P_\ell| \cdot |H_\ell| \right) .
\end{align*}

We now choose ${\displaystyle r = \frac{n^{5/11}\log^{6\kappa/11} (m^3/n)}{m^{4/11}}}$, to
equalize (asymptotically) the first two terms in the bound, which become
$O(m^{6/11}n^{9/11}\log^{2\kappa/11}(m^3/n))$.
(Note that $1\le r\le m$ for $n^{1/3}\le m\le n^{5/4}\log^{3\kappa/2} n$.)
The third term becomes
$mr^2 = m^{3/11}n^{10/11}\log^{12\kappa/11} (m^3/n)$, which is dominated by the preceding
bound for $m=\Omega(n^{1/3})$, as is easily checked.
The complementary case $m=O(n^{1/3})$ has been handled by (\ref{smallm}).

This completes the proof of Theorem~\ref{main}.
$\Box$

\medskip

\noindent{\bf Remarks. (i)}
In retrospect, once we have reduced the problem to that of bounding the number $I(P^*,\Gamma^*$)
on incidences between the projected points and curves on the $xy$-plane, we could have applied,
as a black-box, the analysis of Sharir and Zahl~\cite{SZ}, and get a slightly weaker bound with
an additional (arbitrarily small) $\eps$ in the exponents. Exploiting the fact that we are
dealing here with planes allowed us to obtain the finer bound in (\ref{eq:main}).

\medskip

\noindent{\bf (ii)}
Note also that the preceding analysis closely follows the one in \cite{ANP+,ArS}, or,
more precisely, its enhanced version that exploits the bound $O(n^{3/2}\log n)$
of Marcus and Tardos~\cite{MT} on the number of cuts needed to turn a collection of
circles into a collection of pseudo-segments (so $\kappa=1$ for the case of circles).

\subsection{The case of a quadric} \label{sec:quadric}

The special case where $V$ is a quadric admits a simpler analysis with a slightly
improved bound. In this case the curves $\gamma_h$, as defined earlier, are quadratic,
and by construction, do not contain any line. Moreover, each pair of curves
$\gamma_h$, $\gamma_{h'}$ intersect at most twice, at the at most two intesrection points
of the line $h\cap h'$ with $V$. Hence, these curves form a family of pseudo-circles\footnote
  Some care has to be exercised here, because the analysis in \cite{ANP+} handles
  pseudo-circles in the plane and not on a variety in 3-space. This extension, though,
  can be handled in a rather routine manner, e.g., by breaking $V$ into a constant
  number of $xy$-monotone pieces, projecting the configuration within each piece
  onto the $xy$-plane, and applying the analysis in \cite{ANP+} to each projection
  separately. We omit here the fairly easy details.}
with a $3$-parameter algebraic representation, in the terminology of \cite{ANP+};
they inherit this representation from the $3$-parameter algebraic representation
of the planes $h$ that define them. We can therefore apply the bound in \cite[Theorem 6.6]{ANP+},
enhanced by the improvement in \cite{MT}, and obtain the following improvement.
\begin{theorem} \label{thm:quadric}
For a set $P$ of $m$ points on some two-dimensional quadric $V$ in $\reals^3$, and for a set $H$
of $n$ planes in $\reals^3$, the incidence graph $G(P,H)$ can be decomposed as
\begin{equation} \label{gphq}
G(P,H) = G_0(P,H) \cup \bigcup_i (P_i\times H_i) ,
\end{equation}
such that, for each $i$, there exists a line $\ell_i\subset V$, so that
$P_i = P\cap \ell_i$ and $H_i$ is the subset of planes of $H$ that contain $\ell_i$, and we have
\begin{align} \label{eq:mainq}
|G_0(P,H)| & = O\left( m^{2/3}n^{2/3} + m^{6/11}n^{9/11}\log^{2/11}(m^3/n) + m + n \right) , \quad\text{and} \\
\sum_i |P_i| & = O(m) \quad\text{and}\quad \sum_i |H_i| = O(n) \nonumber .
\end{align}
\end{theorem}

\subsection{An application to distinct cross-ratios} \label{sec:cross}

In a recent work~\cite{Rud}, Rudnev has obtained a superquadratic lower bound on the number of distinct
cross-ratios determined by any set $A$ of $n$ real (or complex) numbers. One of the key steps in
his analysis (see \cite[Theorem 5(i)]{Rud}) is an upper bound on the number of $k$-rich planes in
$\reals^3$, with respect to a set $P$ of $m$ points on the hyperbolic paraboloid $\Pi:\; z=xy$;
that is, planes that contain at least $k$ points of $P$. In the interest of (partial) completeness,
here is a sketch of the reduction, adapted from \cite{Rud}, that leads to this problem. In what follows,
we only consider the real case.

Let $A$ be a set of $n$ real numbers, and consider the set of M\"obius transformations on $\reals$,
each of which can be written as
$$
y = \frac{\alpha x + \beta}{\gamma x + \delta} .
$$
To simplify things, and almost without loss of generality, assume that $\gamma = 1$
(we lose transformations with $\gamma=0$, but they are trivial to handle),
so we consider transformations of the form
$$
y = \frac{\alpha x + \beta}{x + \delta} .
$$
We also assume that $\beta \ne \alpha\delta$, so as to rule out constant transformations
from the analysis.

We are given a parameter $k$, and want to bound the number $N = N_{\ge k}$ of
\emph{$k$-rich transformations}, namely those that map at least $k$ points of
$A$ to $k$ other points in $A$.

For each pair of points $(a,b)\in A\times A$, the set of transformations that map $a$ to $b$ is
$$
M(a,b) = \left\{ (\alpha,\beta,\delta) \mid ab -\alpha a + \delta b - \beta = 0 \right\} .
$$
We can represent each triple $(\alpha,\beta,\delta)$ (i.e., each M\"obius tranformation) by the
plane $z -\alpha x + \delta y - \beta = 0$. Then $M(a,b)$ is the set of all planes that pass through
the point $(a,b,ab)$ on the hyperbolic paraboloid $\Pi$. A transformation is $k$-rich if its
associated plane passes through at least $k$ such points.

This leads to the following incidence question. We have the set
$P = \{ (a,b,ab) \mid (a,b)\in A\times A\}$ of $m=n^2$ points on $\Pi$, and some set $H$ of $N$ planes
of the above form, and wish to bound $I(P,H)$, the number of incidences between $P$ and $H$.

To apply Theorem~\ref{thm:quadric} to this scenario, we note that no plane in $H$ intersects $\Pi$
in a line. To see this, we note that, for each $h\in H$, given by an equation of the form
$z -\alpha x + \delta y - \beta = 0$, for a suitable triple $(\alpha,\beta,\delta)$ of coefficients,
its intersection with $\Pi$, when projected down to the $xy$-plane, is the hyperbola
$xy -\alpha x + \delta y - \beta = 0$. This hyperbola degenerates into a pair of lines
only when $\beta=\alpha\delta$, and since we have ruled out this situation, the claim follows.

It follows that, in the notation of Theorem~\ref{thm:quadric}, we have
$$
I(P,H) = |G_0(P,H)| = O\left( m^{2/3}N^{2/3} + m^{6/11}N^{9/11}\log^{2/11}(m^3/N) + m + n \right) .
$$
We can now obtain an upper bound on the number $N_{\ge k}$ of $k$-rich transformations,
by taking $H$ to be the set of planes that correspond to these transformations, and by comparing
this bound with the lower bound $kN_{\ge k}$. That is, we have
$$
kN_{\ge k} = O\left( m^{2/3}N_{\ge k}^{2/3} + m^{6/11}N_{\ge k}^{9/11}\log^{2/11}(m^3/N_{\ge k}) + m + N_{\ge k} \right) ,
$$
from which we easily obtain, substituting $m=n^2$ and assuming that $k$ is at least some
suitable sufficiently large constant $k_0$,
\begin{equation} \label{eq:ngek}
N_{\ge k} = O\left( \frac{n^4}{k^3} + \frac{n^6}{k^{11/2}}\log k + \frac{n^2}{k} \right) .
\end{equation}
(For the second term, put $Z:=m^3/N_{\ge k}$. If the second term in
the bound for $I(P,H)$ dominates, then we get $kN_{\ge k} =
O(m^{6/11}N_{\ge k}^{9/11}\log^{2/11} Z)$, or $k^{11/2} = O(Z\log
Z)$, from which one easily gets $Z = n^6/N_{\ge k} =
\Omega\left(k^{11/2}/\log k\right)$, and the bound follows.) Note
that for $3\le k<k_0$, we have the trivial bound $N_{\ge k} =
O(|P|^3) = O(n^6)$, as easily follows from the fact that none of our
planes intersect $\Pi$ in a line. Hence, (\ref{eq:ngek}) extends to
any $k\ge 3$, by suitably adjusting the constant of proportionality.

The analysis in Rudnev~\cite{Rud} uses the bound in (\ref{eq:ngek}) to derive an upper bound on
the number $Q$ of pairs of \emph{congruent pentuples} in $A$; that is, pairs of pentuples of elements of $A$
for which there exists a M\"obius transformation that maps the points in one pentuple to the respective points
in the second pentuple. Rudnev shows that $Q$ is upper bounded by
$$
Q = O\left( \sum_{j=2}^{\log n} N_{\ge 2^j}\cdot (2^j)^5 \right) .
$$
Using (\ref{eq:ngek}), we get
\begin{equation} \label{sump}
Q = O\left( \sum_{j=2}^{\log n} \left( 2^{2j}n^4 + \frac{jn^6}{2^{j/2}} + 2^{4j} n^2 \right) \right) = O(n^6) .
\end{equation}
In the original analysis, Rudnev has used, instead of (\ref{eq:ngek}), the weaker bound
$$
N_{\ge k} = O\left( \frac{n^6}{k^5} \right) .
$$
This bound follows from Solymosi and Tardos~\cite{SoTa}, and leads to the slightly inferior bound $Q= O(n^6\log n)$.

The rest of the analysis, from upper bounding $Q$ to obtaining a lower bound on the number of distinct cross-ratios,
is spelled out in \cite{Rud}. His original analysis yields the lower bound
$\Omega(n^{24/11}/\log^{6/11}n)$, and our improvement raises it to $\Omega(n^{24/11})$.

\medskip
\noindent{\bf Remark.}
For the purpose of showing that $P=O(n^6)$, we could equally use the more general bound in
Theorem~\ref{main}. The only difference is that the second term in (\ref{eq:ngek}) increases to
${\displaystyle O\left( \frac{n^6}{k^{11/2}}\log^{11\beta/2} k \right)}$, as is easily checked.
This does not affect the subsequent analysis, as the sum of the second terms in (\ref{sump})
is still the sum of a convergent series.

\paragraph{Acknowledgement.}
The authors would like to thank Misha Rudnev for valuable interaction and comments, especially
concerning the application in Section~\ref{sec:cross}.

redpajama_set_name: RedPajamaArXiv