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## The Moon’s Orbit is Wrong, It Can Change a Lot, And Tides Will Too

The basic “issue” here is simple: We don’t really know what the moon has done in the past, because we don’t know what it is doing now; but we do know it could have made much stronger tides in the past, so could do that again. We also know that present tides are about 1/2 the total overturning force bringing cold deep water to the surface, so we also know that changes in tide forces could and would have major impacts on how cold it gets, and / or when ice sheets break up. So much for “settled science”…

Before anyone gets up in arms over my saying the moon’s orbit is wrong: It wasn’t me! See:

https://duckduckgo.com/?q=moon+orbit+wrong

Moon Orbit Wrong Cornell University – YouTube
Moon Orbit Wrong Cornell University associate Lorenzo Lorio, Has Researched i’m sure because of public outcry & observations made by You, my Friends, Visitors & others on the Internet and Concluded that Indeed there Is Something Wrong With The Sun Moon & Earth.

Cornell University – Moons Orbit is Wrong |
“collected” The Moons orbit Wrong according to Cornell University associate Lorenzo Lorio. He has researched the mystery and he is sure! Because of public outcry & observations made by myself, my friends, Visitors & others on the Internet and concluded that indeed there Is …

reinep.wordpress.com/2011/11/16/cornell-university-moons-orb… More from reinep.wordpress.com
Moon Orbit Wrong Cornell University Says. , page 1
Frankly I don’t care at this point what is causing it, the fact is something is not right with the moon. Thats all I need to know at this point.

Just read on and if it still bugs you, take it up with Cornell! ;-)

OK, this will mostly be a “follow the links” and “toss stuff at the walls” posting. I’ve had a 10 hour day at work, it’s a bit late, I need to get up and do it all again in about 7.5 hours, so every hour here is one less hour of sleep. That means things will be a bit terse without a lot of “connect the dots” prose.

First up, the genesis: I’ve been following this lunar connection for quite some time. The “tipping point” for me was a paper that pretty much shows how lunar tide forces can drive the 1470 year cycle AND explains why it wanders around that average (it is more of a 1200 to 1800 year variable pseudo cycle). From this article: https://chiefio.wordpress.com/2011/11/03/lunar-resonance-and-taurid-storms/

This paper: http://www.pnas.org/content/97/8/3814.full

The 1,800-year oceanic tidal cycle: A possible cause of rapid climate change

Charles D. Keeling* and
Timothy P. Whorf

Pretty much lays it all out, IMHO. (Graphs from it are in the various articles). That paper ASSUMES certain periods and orbital cycles for the moon, in order to make the math more tractable. It basically sets the orbital parameters to what they are now, and changes at only what is known and assumed cyclical. Essentially, a lunar model orbit. From that it calculates tide forces and what they would do.

I went on to explore some of the history of what that cyclicality has done with an eye to what it might do in:

Along the way also looking at a variety of orbital cycles and things, and how one of them might explain the 54-56 ish year cycle of PDO and similar weather cycles (when an 18 year lunar cycle repeats at the same spot over the globe) in these postings:

https://chiefio.wordpress.com/2013/01/04/lunar-cycles-more-than-one/

https://chiefio.wordpress.com/2013/01/24/why-weather-has-a-60-year-lunar-beat/

All of this, more or less, based on the notion that the orbit of the Moon is known, and fairly constant.

Well, it isn’t, really. It’s not totally wonky, just not as stable as everyone assumes and prone to ‘wander’ some. Wandering in ways that can increase tides. Perhaps quite a lot. Perhaps enough to cause major shifts of weather and climate. That’s what this posting is about. Those missing bits.

What started me down this particular path was a posting at WUWT. It looked to me like a follow on to some discussions of ice age glaciations over at Tallbloke’s Talk Shop about what causes glacials to end. I’d commented on one or two of the threads, but can’t find just which one now. At any rate, here’s a couple that are of interest.

https://tallbloke.wordpress.com/2012/09/29/a-better-explanation-for-the-end-of-ice-ages/

https://tallbloke.wordpress.com/2014/01/10/clive-best-does-the-moon-trigger-interglacials/

This all ends up in a posting at WUWT, which gets me looking into lunar orbital mechanics a bit more.

http://wattsupwiththat.com/2014/01/14/do-super-tides-kick-start-interglacials/

So that’s the “set up”. A key point to keep in mind is that the natural state of the planet is frozen into a glacial period. We are presently in an Ice Age. They last for millions of years. What folks often call an Ice Age is really just one big glacial episode in that bigger Ice Age. Each glaciation runs about 100,000 years. Yet inside those glaciations can be interstadials where things get nearly as warm as now for a few hundred to a couple of thousand years, or interglacials where the warmth lasts a few thousand to 10,000 years.

So what hooks all this together? Surely not just a simple repeating pattern of predictable lunar orbits? That’s close, but there just doesn’t seem to be quite enough power to make a full on glacial just go away. But, what if the lunar forces changed not just in time, but in size and scale as well? What if that were happening now?

First, back at Clive’s posting, he shows a pattern of how certain orbital alignments can cause “super tides”, when the lunar orbit and the earth orbit both align with closest approaches in sync. This can cause some very large tides. I’m going to borrow a graph from his posting for here:

Change in lunar tide force with eccentricity

Below it the text:

Fig 5: Variation in strength of lunar tides with orbital eccentricity relative to today.

We see that spring lunar tides for a lunar orbit twice the current eccentricity would be about 60% higher than they are today. Lunar tides are about twice the strength of solar tides so overall spring tides would have been at least 50% stronger than they are today, and Super Perigean tides would have been 20% stronger again.

Are these super-tides the catalyst to break up the large northern ice sheets and exit ice ages once every 100,000 years ?

Note those numbers. 60% larger and then 20% more on top of that. Call it

But what if tides might be even larger? He, too, assumes that the orbit is regular and predictable and about like it is now. Then puts in some of the expected eccentricity changes. But is that the only thing going on?

First up, NASA has a nice page on the expected changes of the lunar orbit. Well worth a read:

http://eclipse.gsfc.nasa.gov/SEhelp/moonorbit.html

It also has some very nice graphs. I’ll only put one or two here for illustration of what’s going on. They show that the length of month wanders around rather a lot (lunar month) and that the tide forces ought to wander a lot with that.

Length of Lunation Variation

So several hours more or less on any given lunation (month). Here’s another graph, showing different “bins” of length:

5000 years of lunations, binned by length

And one showing how any given anomalistic month can vary:

Length of Anomalistic Month variation.

Similarly eccentricity and even inclination wobble about (in addition to the earth doing the same…)

Eccentricity changes for Lunar Orbits

Lunar Orbital Inclination Changes

Note that changes by about 0.3 degrees out of 5.x in just a couple of years. So how much the moon is pulling water north or south in tides changes from this and from eccentricity changes. Now set the whole thing in motion and stir (or rather, let it stir the oceans of the earth…)

To quote that NASA site:

The mutual gravitational force between the Sun and Moon is over twice as large as between the Moon and Earth. For this reason, the Sun plays a dominant role in perturbing the Moon’s motion. The ever changing distances and relative positions between the Sun, Moon, and Earth, the inclination of the Moon’s orbit, the oblateness of Earth, and (to a lesser extent) the gravitational attraction of the other planets all act to throw the Moon’s orbital parameters into a constant state of change. Although the Moon’s position and velocity can be described by the classic Keplerian orbital elements, such osculating elements are only valid for a single instant in time (Chapront-Touze’ and Chapront, 1991). Nevertheless, these instantaneous parameters are of value in understanding the Moon’s complex motions particularly with respect to the three major orbital cycles that govern eclipses.

So the Sun stirs the Lunar Orbit and the Moon stirs the oceans. And it does this in a quasi periodic way that has some surprising excursions from time to time… I note in passing that changes to the “oblateness” of the earth is likely to cause some earthquake and volcano changes, IMHO.

OK, so all these things can mix to cause some significant changes in where the cold water is on this planet, how much mixes into surface waters, and just how big the tides are. Anything else?

How about the lunar eccentricity is increasing and we don’t know why?

http://arxiv.org/abs/1102.0212

On the anomalous secular increase of the eccentricity of the orbit of the Moon
Lorenzo Iorio
(Submitted on 1 Feb 2011 (v1), last revised 22 Apr 2011 (this version, v6))

A recent analysis of a Lunar Laser Ranging (LLR) data record spanning 38.7 yr revealed an anomalous increase of the eccentricity of the lunar orbit amounting to de/dt_meas = (9 +/- 3) 10^-12 yr^-1. The present-day models of the dissipative phenomena occurring in the interiors of both the Earth and the Moon are not able to explain it. We examine several dynamical effects, not modeled in the data analysis, in the framework of long-range modified models of gravity and of the standard Newtonian/Einsteinian paradigm. It turns out that none of them can accommodate de/dt_meas. Many of them do not even induce long-term changes in e; other models do, instead, yield such an effect, but the resulting magnitudes are in disagreement with de/dt_meas. In particular, the general relativistic gravitomagnetic acceleration of the Moon due to the Earth’s angular momentum has the right order of magnitude, but the resulting Lense-Thirring secular effect for the eccentricity vanishes. A potentially viable Newtonian candidate would be a trans-Plutonian massive object (Planet X/Nemesis/Tyche) since it, actually, would affect e with a non-vanishing long-term variation. On the other hand, the values for the physical and orbital parameters of such a hypothetical body required to obtain the right order of magnitude for de/dt are completely unrealistic. Moreover, they are in neat disagreement with both the most recent theoretical scenarios envisaging the existence of a distant, planetary-sized body and with the model-independent constraints on them dynamically inferred from planetary motions. Thus, the issue of finding a satisfactorily explanation for the anomalous behavior of the Moon’s eccentricity remains open.

So “something is going on” that is increasing orbital eccentricity, but it is unexplained by all that we know. This link to a pdf does a good job of stepping through what was looked at, and what is still stuck.

http://moriond.in2p3.fr/J11/transparents/iorio.pdf

Oddly, Germany seems to get a lava flow just about the start of each Interglacial. Almost as though there was some tidal force working on the magma and crust.

Rieden volcanic complex – 380,000 to 430,000 years old in the WEVF. It consists of plagioclase-free fallout, flow deposits, and scoria cones in various compositional types (leucite-phonolite, nephelinite, leucitite).

Southeastern sector of EEVF – around 215,000 to 225,000 years old. Basaltic (basanite and tephrite) scoria cones.

Niedermendiger lava flow – between 100,000 to 150,000 years old. Major basaltic (tephrite) lava flow along with lesser alkalic (phonolite) eruptions in the EEVF.

Laacher See eruption – about 13,000 years ago. Major eruption of alkalic (phonolite) tephra and pumice at the Laacher See volcano in the EEVF. Volume of erupted magma was approximately 5 km³, which equals or exceeds all mafic eruptions in the WEFV.

What’s a person to do?…

### But Wait, There’s More!

Turns out that in the past, tides were quite different from today. These two articles are mostly looking at things like the Hudson Bay acting as a tide damper for the North Atlantic, so during glacial times it can’t do that and tides are much higher.

http://www.sciencedaily.com/releases/2011/08/110801095116.htm

Ancient Tides Quite Different from Today — Some Dramatically Higher, Some Lower

Aug. 2, 2011 — The ebb and flow of the ocean tides, generally thought to be one of the most predictable forces on Earth, are actually quite variable over long time periods, in ways that have not been adequately accounted for in most evaluations of prehistoric sea level changes.
[…]
Some tides on the East Coast of the United States, for instance, may at times in the past have been enormously higher than they are today — a difference between low and high tide of 10-20 feet, instead of the current 3-6 foot range.

And tides in the Bay of Fundy, which today are among the most extreme in the world and have a range up to 55 feet, didn’t amount to much at all about 5,000 years ago. But around that same time, tides on the southern U.S. Atlantic coast, from North Carolina to Florida, were about 75 percent higher.

The findings were just published in the Journal of Geophysical Research. The work was done with computer simulations at a high resolution, and supported by the National Science Foundation and other agencies.

Yes, models… but I trust models of water flow in a basin rather a bit more than some hypothetical gas calculations… we’ve been doing tide prediction and modeling for a long while now AND testing the results…

This part in particular was interesting:

One of the most interesting findings of the study, Hill said, was that around 9,000 years ago, as Earth was emerging from its most recent ice age, there was a huge amplification in tides of the western Atlantic Ocean. The tidal ranges were up to three times more extreme than those that exist today, and water would have surged up and down on the East Coast.

One of the major variables in ancient tides, of course, was sea level changes that were caused by previous ice ages. When massive amounts of ice piled miles thick in the Northern Hemisphere 15,000 to 20,000 years ago, for instance, sea levels were more than 300 feet lower.

But it’s not that simple, Hill said.

“Part of what we found was that there are certain places on Earth where tidal energy gets dissipated at a disproportionately high rate, real hot spots of tidal action,” Hill said. “One of these today is Hudson Bay, and it’s helping to reduce tidal energies all over the rest of the Atlantic Ocean. But during the last ice age Hudson Bay was closed down and buried in ice, and that caused more extreme tides elsewhere.”

Now again, note those sizes. Up to 3x as large as now. I think that is likely to have some effect on ice sheets and cold water / surface water mixing…

The Oregon State University version of the news release:

http://oregonstate.edu/ua/ncs/archives/2011/jul/ancient-tides-different-today-%E2%80%93-some-dramatically-higher

OK, so is there any other reason to think tides might have been higher in the past? Perhaps even during recorded history? How about a Greek observer?

https://en.wikipedia.org/wiki/Pytheas_of_Massilia#Pytheas_on_the_tides

Pytheas on the tides

Pliny reports that “Pytheas of Massalia informs us, that in Britain the tide rises 80 cubits.” The passage does not give enough information to determine which cubit Pliny meant; however, any cubit gives the same general result. If he was reading an early source, the cubit may have been the Cyrenaic cubit, an early Greek cubit, of 463.1 mm, in which case the distance was 37 metres (121 ft). This number far exceeds any modern known tides. The National Oceanography Centre, which records tides at tidal gauges placed in about 55 ports of the UK Tide Gauge Network on an ongoing basis, records the highest mean tidal change between 1987 and 2007 at Avonmouth in the Severn Estuary of 6.955 m (22.82 ft). The highest predicted spring tide between 2008 and 2026 at that location will be 14.64 m (48.0 ft) on 29 September 2015. Even allowing for geologic and climate change, Pytheas’ 80 cubits far exceeds any known tides around Britain. One well-circulated but unevidenced answer to the paradox is that Pytheas is referring to a storm surge.

Matching fragments of Aëtius in pseudo-Plutarch and Stobaeus attribute the flood tides (πλήμμυραι plēmmurai) to the “filling of the moon” (πλήρωσις τῆς σελήνης plērōsis tēs sēlēnēs) and the ebb tides (ἀμπώτιδες ampōtides) to the “lessening” (μείωσις meiōsis). The words are too ambiguous to make an exact determination of Pytheas’ meaning, whether diurnal or spring and neap tides are meant, or whether full and new moons or the half-cycles in which they occur. Different translators take different views.

That daily tides should be caused by full moons and new moons is manifestly wrong, which would be a surprising view in a Greek astronomer and mathematician of the times. He could have meant that spring and neap tides were caused by new and full moons, which is partially correct in that spring tides occur at those times. A gravitational theory (objects fall to the center) existed at the time but Pytheas appears to have meant that the phases themselves were the causes (αἰτίαι aitiai). However imperfect or imperfectly related the viewpoint, Pytheas was the first to associate the tides to the phases of the moon.

OK, maybe the guy over estimated. Maybe he used a strange “cubit”. There are some who assert him a charlatan (but reason to think they are not all that accurate either…). But what if he did see rather large tides? Maybe not 3x as high as now, but just double? Or even 1.5x the tide. Do we really know what the lunar orbital elements were in 400 BC? We are not even sure why they are changing today. Just how eccentric can the lunar orbit get? How much elevation change? Do we really know?

We are, in fact, a binary planet system. The moon is always in a “concave to the sun” orbit. It does not make curlicues in space and does not go retrograde relative to the sun. The moon and earth share a common orbit about the sun, slowly changing who is in the lead, and who has the inside or outside track. The sun stirs this pot, as do the gas giants (and who knows what else).

In the end, all we can really say is that it has changed, is changing, and will change. The best guesses we have include saying that tides were much higher a few thousand years ago. So why not two thousand years ago? And why not 2000 years from now?

If we can’t define the range of the tides, how can we define the range of ocean overturning force and cooling, or the range of tide induced cooling or warming that is normal? Yes, we can figure tide tables for a few years. With reduced accuracy even longer. But thousands? History and what we know of modeling tides says “things change”, and sometimes quite a lot.

Is it enough to end glacials and break up ice sheets? To perhaps lead to warmer than normal times when tides slack off? Will increasing lunar eccentricity cause more cooling? How much is driven by the sun? And how much by the other planets?

There’s an awful lot of loose ends and unknowns in this “settle science”…

### A Close

With that, I close this posting. I get 6 hours sleep tonight. Hopefully more tomorrow night ;-) But this collection of lunar links and observations was nagging at me and would not let go until posted. I’ll just add one more minor point here: Stonehenge has many aspects that track and predict lunar motions and positions. Perhaps 4000 years ago there was even more reason to watch and track the moon. Perhaps we ought to do a bit more of that today.

To me, it is pretty clear that there are a lot of things we don’t know. Many things about CO2 that are flat out wrong. And a great deal of historical change being ignored by Grant Seekers with the CO2 boogyman. For my money, it’s the moon that matters most, not a few parts per million of CO2 in a water laden sky. Certainly the moon matters more on a water world with oceans and giant tides.

Post Script: I’m skipping my usual proof read / link QA in the interest of being awake at my desk tomorrow (today?). Please note any “issues” and I’ll fix them later. For now, sleep demands its due.

A technical managerial sort interested in things from Stonehenge to computer science. My present "hot buttons' are the mythology of Climate Change and ancient metrology; but things change...
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### 23 Responses to The Moon’s Orbit is Wrong, It Can Change a Lot, And Tides Will Too

1. R. de Haan says:

“So the Sun stirs the Lunar Orbit and the Moon stirs the oceans”.

Not only that. The shift in jet stream patterns going further south (and North on the SH) in combination with blocking triggers a shift in wind patterns and wind strengths.
Watch the news about 50 ft waves hitting Hawaii, the coast of Portugal and the British coast recently: http://www.telegraph.co.uk/news/worldnews/europe/portugal/10555954/At-least-four-injured-as-freak-wave-hits-Portugal-coast.html

Almost all this winter season Europe is in an airflow that originates from Central America, hence the relative warm Western European winter. This wind causes the big waves that hit the Portugal and British coast.
What we see here is a massive destruction of heat and the prelude to much colder times.

Watch this video posted by Adolfo in a response on the previous CET article, suppress the sound, forget all about the niro BS and watch the air flows and the typed comments: https://www.youtube.com/watch?v=E-HPxYYUqG4#t=164

2. R. de Haan says:
3. omanuel says:

Thanks for this intriguing post. If the Moon was produced by fission of the proto-Earth, as I believe, then Earth probably also suffered a traumatic event.

4. R. de Haan says:

E. M. Smith, why don’t we introduce the next crypto currency and really get something done?

5. philjourdan says:

While some think the “science is settled”, it is work like this that raises the questions that are unanswered (and to some – refused to answer). These questions make the issue of how man evolved and is now dominant seem to be the more pressing ones. For what was done can be undone. And there is nothing we can do to stop it.

Fascinating read. Thanks for the sacrifice of sleep for bringing it to us.

6. p.g.sharrow says:

Sun moon earth tides brain dump! Lol Sometimes you just have to unload before you can rest. Good job.
The earth moon pair dances around the sun as they have done for 4.5 billion years. One thing for sure is the dance has not been boring. The benign period of the last few thousand years is not normal. The ancients knew this and watched the skies for hints of changes over a much shorter time then we now expect. Any expectation that the present conditions are for all time is foolish. Gravity is not as constant as most believe. The earths’ sea level? has as much as 600feet of variation from theoretical level due to local mantel density. If my concept is correct, local Aether density has an effect on the manifestation of gravity within the solar system. Lots of moving parts. The best we can do is adapt. Some things can’t be changed but it would be nice to know cause and effect. At least put an end to “sacrificing Virgins” to placate the gods. pg

7. Bloke down the pub says:

Amusing to think that, some time in the future, the Apollo missions may be seen to have been worth the expense if only for the lunar ranging reflectors that were left on the surface.

8. clivebest says:

Fascinating and thought provoking!
I have been trying to use the JPL ephemeris to calculate the moon’s orbital parameters 17,000 years ago. I beginning to conclude that the Milankovitch cycles really only apply to the earth-moon barycentre. The effective moon-earth orbit is far more complex as you have described. Therefore it seems likely that larger excursions in the gravitational coupling between the two must result as the overall eccentricity increases.

How could we prove this hypothesis ?

9. E.M.Smith says:

@R. de Haan:

Well, for starters, the only reason a crypto currency gets acceptance is due to it being promoted, and I’m not a big promoter. Then there is the problem that anyone can launch a crypto currency, so it is inherently prone to “competition”. Oh, and the U.S. Feds have just started to play in the crypto currency playground, so it’s stepping into the bull pen just at the wrong time.

Opposed to that is that anyone can issue a currency. ( Just an I.O.U., really) So crypto currencies are not all that different. All gain acceptance based on others using it.

I don’t know. Somebody will make money off of it..

@PhilJourdan:

You are welcome. Catching up this afternoon ;-)

IMHO we became dominant due to a willingness to eat most anything and an interest in weapons…

Hand skills and brains came later…

@P.G. Sharrow:

Yup. Need to unload “stuff” then sleep can come…

I’m pretty sure that as the whole shebang goes to extremes, tides and waves in the UK / Portugal / etc. all go much larger. The historical reports were not fanciful…

Don’t know what happens in the rest of the world, but there are large waves recorded in the California cliffs…

Rather like those waves in R. de Haan’s link

@Bloke Down The Pub:

Surprising, but likely to be correct. So many things you can do with a precision reflector…

10. R. de Haan says:

E.M.Smith says:
25 January 2014 at 9:50 pm

@R. de Haan:

“Well, for starters, the only reason a crypto currency gets acceptance is due to it being promoted, and I’m not a big promoter. Then there is the problem that anyone can launch a crypto currency, so it is inherently prone to “competition”. Oh, and the U.S. Feds have just started to play in the crypto currency playground, so it’s stepping into the bull pen just at the wrong time.

Opposed to that is that anyone can issue a currency. ( Just an I.O.U., really) So crypto currencies are not all that different. All gain acceptance based on others using it.

I don’t know. Somebody will make money off of it..

11. A C Osborn says:

Chiefio, can you have a look at why Clive Best’s comments have got stuck in moderation please?
I am not sure which of the 2 moon threads he posted on, so I commenting on both.
Thanks

12. E.M.Smith says:

@R. de Haan:

I’ll contact you. Give it a day or two…

@A C Osborn:

The first time someone comments here it goes to moderation. How long it stays there depends on how hectic it is at work… Once approved, future comments are only “moderated” after the fact, or if they trigger some key word that WordPress is hung up on at the moment.

I’ve approved both of his comments, so they are now visible. Also, future comments from Clive ought to show up immediately.

One is upthread here, so the other is likely on the other thread.

@Clive:

Sorry for the slow moderation. I have a “day job” now and it takes a lot of my time. (Oh for some of that “Oil Money” that is supposed to flow like honey to skeptics /sarc ;-)

If the JPL ephemeris is the same as NASA (which I would expect) it gets off kilter in as little as a 1000 years per that Indian data / paper. I think you will need to make a new one that does match the recorded eclipses to have any hope of doing 17,000 years.

Per the Earth Moon gravitational range being larger with eccentricity: Same thing I was pondering. Once you get the idea that the moon is a distinct planet orbiting the sun, it suddenly frees the orbital ranges some… I’ve even found myself wondering, with a 5 degree tilt to the ecliptic, if anything really prevent the moon from going straight over the north or south pole? A comparison graph I say put the Earth as a basketball under a regulation basket, then the Moon would be a tennis ball at the 3 point line and the moon could range up to one tennis RACKET above the earth in the orbit… An interesting visual…

So just how extreme can the eccentricity get? We don’t know. We know what we claim now is wrong. We know it gets larger (and maybe quite large given the ancient claims of giant tides). Since the “eccentricity” relative to Earth is really a mathematical artifact of the lunar orbit relative to the Sun, what prevents the lunar / solar eccentricity from getting nearly the same as the earth / solar and having the moon do “laps” nearly overhead? I don’t know…

13. Wayne Job says:

Classic and Einstein gravity do not explain the tides, they are forecast only by precedent.
Some thing else causes them?

14. craigm350 says:

Reblogged this on CraigM350 and commented:
A great thought provoking read. Thank you.

15. Power Grab says:

It occurred to me yesterday (sorry if I’m extremely late on this topic) that TPTB might not necessarily be trying to destroy this country, per se, but “merely” its currency. Is that why there is so much interest in new currencies? I read that the Chinese want to use their currency for oil trading. Will it instead end up being some kind of virtual currency?

Re the moon: the crescent shapes seem to be less east-and-west oriented, and more north-and-south oriented. I read somewhere that a person had noticed the change and wondered about it. I thought they were talking about a photoshopped image. The next time I had a chance, I noticed that it indeed does have north-and-south (top-and-bottom) crescents rather than east-and-west crescents now. Is that the biggest clue that something is off about the orbits of the sun, moon, and earth?

Sorry to be so dense and late to the party, but I’ve had lots of diversions lately. It’s like I’m mostly under water, but once in a while come up for air and have a burst of female intuition.

16. Harris Keillar says:

Apparently Newton said that calculating the moon’s orbit was the only thing that gave him a headache!

17. E.M.Smith says:

@Power Grab:

Currency and currencies are interesting topics, but not for this thread. I’ll put someting up (maybe tonight). The Chinese want to use their currency for oil as then THEIR buys do not have to clear through a US Bank. (Last time I looked, major US $currency conversions had to pass through US banks, so US visibility and ot some extent control. Part of why the Arabs started looking into trading oil in Euro when the USA started “freezing assets”. If it doesn’t land on a US bank, it can’t be US Froze…) As per the crecents: Good point. As the ecliptic excursion takes it above / below the ecliptic, the crecents will wander more up / down too… Nice quick way to figure out where it is relative to the solar / ecliptic orbit… As per “Late to the party”: I’ve been that way myself lately, so I understand ;-) BTW, going to be at Grandma’s Kitchen about 1 PM today Feb 2… 18. Chris in Calgary says: A very recent publication by the same author sheds little further light on the matter: http://arxiv.org/abs/1404.6537 _The lingering anomalous secular increase of the eccentricity of the orbit of the Moon: further attempts of explanation of cosmological origin Lorenzo Iorio (Submitted on 26 Apr 2014) A new analysis of extended data records collected with the Lunar Laser Ranging (LLR) technique performed with improved tidal models was not able to resolve the issue of the anomalous rate$\dot e$of the eccentricity$e$of the orbit of the Moon, which is still in place with a magnitude of$\dot e=(5\pm 2)\times 10^{-12}$yr$^{-1}$. Some possible cosmological explanations are offered in terms of the post-Newtonian effects of the cosmological expansion, and of the slow temporal variation of the relative acceleration rate$\ddot{S} S^{-1}$of the cosmic scale factor$S\$. None of them is successful since their predicted secular rates of the lunar eccentricity are too small by several orders of magnitude. _

19. geran says:

Bloke down the pub says:
25 January 2014 at 2:40 pm
Amusing to think that, some time in the future, the Apollo missions may be seen to have been worth the expense if only for the lunar ranging reflectors that were left on the surface.
>>>>>>>>
Yes, and they are using those reflectors to claim that the Moon is moving away from Earth at about 4 cm/yr. And, the “science” they use to explain this is almost as funny as climate “science”.

20. Mike Mellor says:

Re moon the cause of seismic and volcanic activity. As the Earth-Moon barycentre is well below the Earth’s crust, the moon is surely the best candidate for the cause of continental drift?


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# Convergence of a sequence

Problem statement:

Determine the limit of the following sequence:

$\sqrt{a},\sqrt{1+\sqrt{a}}, \sqrt{1+\sqrt{1+\sqrt{a}}},...$

My progress:

Let´s begin by introducing some notation. Let $a_{n}$ denote the nth term of the sequence. We have $a_{1}=\sqrt{a}$ and $a_{n}=\sqrt{1+a_{n-1}}$. My instinct tells me now to rewrite as $a_{n}^2-a_{n-1}-1=0$ which has a root $\frac{1+\sqrt{5}}{2}$ (neglect the negative root for obvious reasons).

However: My friend told me this is only an eventually value of the sequence and not necessarily. I have to determine that this sequence converges before i can conclude this. How can I do this? And what does it actually mean when I solve the quadratic(because that is only an instinct of mine)?

-
Correct, I´ll change that. – EricAm Dec 29 '12 at 1:20

If you know that the sequence is convergent with limit $l$, then making $n \to \infty$ in $a_{n}^2-a_{n-1}-1=0$ yields

$$l^2-l-1=0 \,.$$

But, as your friend told you, this helps you only if the sequence is convergent.

To prove convergence, you'll see that the root of the quadratic play a huge role. You need to distinguish among three cases:

Case 1:

$a= \frac{1+\sqrt{5}}{2}$. Then you can prove by induction that $a_n= \frac{1+\sqrt{5}}{2}$.

Case 2:

$a< \frac{1+\sqrt{5}}{2}$. Then you can prove by induction that $a_n< \frac{1+\sqrt{5}}{2}$ and $a_n <a_{n+1}$. Conclude that $a_n$ is convergent.

Case 3:

$a>\frac{1+\sqrt{5}}{2}$. Then you can prove by induction that $a_n> \frac{1+\sqrt{5}}{2}$ and $a_n >a_{n+1}$. Conclude that $a_n$ is convergent.

-
Or you can just use the fact that $f(x)=\sqrt{1+x}$ is a contraction in $(1,\infty)$ and thus must have a fixed point. Specifically $|f(x)-f(y)| < \frac{1}{\sqrt{2}}|x-y|$ – Thomas Andrews Dec 29 '12 at 1:33
If $a_n\to l$ then $a_n^2\to l^2$ and $a_{n-1}\to l$ so $a_n^2-a_{n-1}-1\to l^2-l-1$ @AdamYac – Thomas Andrews Dec 29 '12 at 1:35
Thank you Thomas, I fully understand. – EricAm Dec 29 '12 at 1:39

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×

# Bonse's Identity! Interesting!

Prove that $$\displaystyle {p}_{n}\# >p_{n+1}^2$$ for $$n \geq 4$$, where $$p_k$$ is the $$k$$th prime number and the primorial is defined as $$\displaystyle {p}_{n}\# = \prod_{k=1}^n p_k$$.

Note by Kartik Sharma
1 year, 6 months ago

Sort by:

Assume $$p_n \# > p_{n+1}^2$$ for some $$n$$; this will be the inductive hypothesis. Then $$p_{n+1} \# > p_{n+1}^3$$, and so we seek to prove that $$p_{n+1}^3 \geq p_{n+2}^2$$.

It is known that for $$n \geq 0$$, $$2p_{n+1} > p_{n+2}$$ (alternate form of Bertrand's postulate). As such, cubing both sides produces $$p_{n+1}^3 > \dfrac{p_{n+2}}{8} p_{n+2}^2$$, which implies that $$p_{n+1}^3 > p_{n+2}^2$$ if $$p_{n+2} > 8$$. For $$n \geq 3$$, this is obviously true, so $$p_{n+1}^3 > p_{n+2}^2$$ (stronger result than required), and thus $$p_{n+1} \# > p_{n+2}^2$$.

The hypothesis is verifiably true for $$n = 4$$. By induction, because $$p_n \# > p_{n+1}^2 \longrightarrow p_{n+1} \# > p_{n+2}^2$$, we have that the hypothesis is true for all $$n \geq 4$$. · 1 year, 6 months ago

Clever way to go about it, did not the alternate form. · 1 year, 6 months ago

It's an inequality, by the way. An identity is an exact equivalence. · 1 year, 6 months ago


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# sampling from normal distribution using MCMC doesn't seem correct [closed]

I'm trying to sample from the normal distribution with a mean of .5 (like flipping a coin) and an arbitrary standard deviation using a markov chain (sequence of numbers where each number is dependent on the previous number of the sequence).

since this is a normal distribution under a markov chain, I expect the sampled values to fluctuate between 0 and 1 but mostly around .5 (like a random walk). I expect a result similar to this:

but when I try to do it my result looks like this:

I don't know why in that example the MCMC iteration never gets negative thetas or thetas greater than 1. all my plots are never contained between 0 and 1. if Normal(proposed_theta) outputs a value greater than 1 or is negative that is very problematic.

## closed as off-topic by Peter Flom♦Apr 22 at 11:16

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question appears to be off-topic because EITHER it is not about statistics, machine learning, data analysis, data mining, or data visualization, OR it focuses on programming, debugging, or performing routine operations within a statistical computing platform. If the latter, you could try the support links we maintain." – Peter Flom
If this question can be reworded to fit the rules in the help center, please edit the question.

• This is going to require people to watch the video. Can you make this question self-contained? I suspect there would be an on-topic question here. – gung Apr 22 at 14:49

What's missing in what you are doing is an acceptance step that accepts a proposal based on the normal density $$f(x_i)$$ at the current state of the chain $$x_i$$ and that at the proposed value $$x_{i+1}'$$. This could be based on using an acceptance probability $$\min(1, f(x_{i+1}')/f(x_{i}))$$ (move to the proposal that probability, make the previous value the next one otherwise), if you want to use Metropolis Hastings.


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# Billboarded trees

## Recommended Posts

daniel_i_l    295
I'm making a map with billboarded trees, it workes great with one tree but with a few trees it looks like this: I'm using tga files as the trees with an alpha channal and rendering the trees in blend mode. Why is there that weird color were the trees overlap? Is it a problem with the pictures, the order that I'm drawing, the blendFunc..? Thanks.

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soconne    105
You either need to sort your tree billboards or using alpha testing.

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bpoint    464
The best solution is to sort the tree polygons with relation to the camera, and draw from back-to-front. It's obviously more involved and requires more processing power, though...

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Glaiel_Gamer    106
cant you just disable depth testing before drawing the trees?

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bpoint    464
Quote:
 Original post by Glaiel_Gamercant you just disable depth testing before drawing the trees?

Disabling Z compares would actually make the problem worse.

A tree with a farther Z could be drawn on top of a tree with a closer Z. You could potentially see a base of a tree appear in front of tree leaves. :)

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Dorvo    272
Quote:
 Original post by bpointThe best solution is to sort the tree polygons with relation to the camera, and draw from back-to-front. It's obviously more involved and requires more processing power, though...

Why not sort front-to-back? Wouldn't that speed up the rendering and depth testing?

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bpoint    464
Quote:
 Original post by DorvoWhy not sort front-to-back? Wouldn't that speed up the rendering and depth testing?

Front-to-back drawing wouldn't give the right results. The pixels on the outside of the tree edges would be blended with the terrain, and not other (potential) trees in the background.

Hence, back-to-front drawing is required.

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hplus0603    11356
Front-to-back won't work if you actually use blending.

The traditional way to do billboards is to sort back-to-front, turn off Z write, and keep Z-test on. Then draw them with blending enabled.

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Dorvo    272
Quote:
 Original post by hplus0603Front-to-back won't work if you actually use blending.The traditional way to do billboards is to sort back-to-front, turn off Z write, and keep Z-test on. Then draw them with blending enabled.

Understood. See? You learn something new every day. [grin]

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daniel_i_l    295
I sorted back to front like this:
struct STree{ VERTEX position; double rotation; double height,width; GLuint tID; STree(double x, double y, double z, double h, double w, GLuint t); void Render();  //overload '<' used for sorting	    friend bool operator<(const STree& lhs, const STree& rhs)	{		return ( (lhs.position.x - Camera.PositionX())*(lhs.position.x - Camera.PositionX())               + (lhs.position.z - Camera.PositionZ())*(lhs.position.z - Camera.PositionZ())               >  (rhs.position.x - Camera.PositionX())*(rhs.position.x - Camera.PositionX())               + (rhs.position.z - Camera.PositionZ())*(rhs.position.z - Camera.PositionZ()) );	}};...sort(Trees.begin(), Trees.end())

every time a drew the trees.
Or should I've sorted it in one of these ways?
friend bool operator<(const STree& lhs, const STree& rhs)	{		return ( (lhs.position.x * Camera.PositionX())               + (lhs.position.z * Camera.PositionZ())               >  (rhs.position.x * Camera.PositionX())               + (rhs.position.z * Camera.PositionZ()) );	}		friend bool operator<(const STree& lhs, const STree& rhs)	{		return ( lhs.position.z               <  rhs.position.z );	}

But it looked like this(no matter how I sorted them):

The same problem?
What did I do wrong?
Thanks.

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Aragon    144
i am an opengl newbie but i remeber something from
directx where i got the same problem

directx:
g_pd3dDevice->SetRenderState( D3DRS_ALPHAREF, DWORD(wert) ); // wert = 0<->255

thoose command changes the edge of colorkey effect
at edges..and it looks like same at your screens

but i dont know the openGL command for this

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grayrobertos    122
I had a similar problem, it is to do with the linear filtering on the textures, blending the white under the alpha channel of the TGA. Simply change the background colour to black or something for a quick fix.

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rapunzel    170
grayrobertos is right. It definitely looks like the linear filtering is disturbing here. An easy way to solve this is to use alpha-testing like this I think:

glEnable(GL_ALPHA_TEST);
glAlphaFunc(GL_GREATER, 0.99);

drawyourtrees();

glDisable(GL_ALPHA_TEST);

You ensure, that only pixels with alpha of 1.0 (or very very close to 1.0) in you billboard texture are drawn and all others are skipped. So you also skip the fragments that result from a linear interpolation of an alpha=0.0 and alpha=1.0 (which creates some alpha between 0.0 and 1.0) ...

greetz

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daniel_i_l    295
I did the glAlphaFunc(GL_GREATER, 0.99); and I coud barely see the trees!
I changed it to 0.5 and even then there was a small problem?
Also the picture of the tree is one layer, the tree on a transparent background, how do I change the background color to black? (gimp)
Thanks.

[Edited by - daniel_i_l on March 24, 2006 2:54:22 AM]

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alfith    2337
With The Gimp just add a black layer below your tree layer (if you can't move the new layer below, duplicate your tree layer and delete the original one)
Instead of black, you can use a color that is close to the one used for your tree because if you use mipmapping, you will probably get dark borders around far distance trees.

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daniel_i_l    295
I added a layer under the tree layer and now I get a lot of black squares (all of the black showed up in the game) gimp said that tga can't handle layers and it merged everything together?
Thanks.

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alfith    2337
I forgot a step sorry (I don't have gimp here).
then you can merge your two layers (it must be merged but not flattened ! if you flatten, you will get the alpha removed). And try using a green color instead of black.

what you should get is one layer (the tree on a black or better green background) and its mask.
hope it is ok now...

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daniel_i_l    295
I created a mask for the first layer, then I merged the layers, and I got a tree on a green background, but again I saw green squares? It seens like it should do that cause all I have is one layer of a tree on a green BG?
Thanks? (I'm a total newb to gimp and 2D art in general)

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alfith    2337
I took gimp...
so:
you have a tree on a transparent background
right clic on the layer
"Alpha to selection"
choose your foreground color (a green)
Ctrl+, (or "Edit\Fill with FG color")

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daniel_i_l    295
Thanks, one question, when I save the gimp says that tga cant handle layers and masks and asks if I wan't to export, do I say yes?

Also, after doing that there still was some of the problem that I mentioned above, will fixing the pic as you said work without sorting and disableing z-write?
Thanks.

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alfith    2337
fixing the pic will provide you with a "good" texture well suited for what you want to achieve in all conditions (with alpha test or blending, with or without mipmapping...)

to render, as others said, you can do
- alpha test with Z write on. the "problem" is that you will get "blocky" trees as the opacity of your trees will be "full" or "none"
- blending with sorting + Z write off. here, you need to sort the trees from back to front and render them after the other objects.
- blending + alpha test + Z write on. you will get artifacts depending of the glAlphaFunc value you use...

I am using the last option for my trees... with a not too bad (and not too small) texture artifacts can be really small

Hope this helps...


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# Math Help - Spanning tree question.

1. ## Spanning tree question.

If $|Sp(G)| = 1$, plz prove that $G$ is a tree.
where $Sp(G)$ = {spanning trees of $G$}

2. Please take no offence at this comment. But from the tenor of your general questions I have to wonder how careful you are in trying to ingest the meaning of the definitions on Graph Theory.
Let me give you a website that may help you: Graph Theory.
To you current question: If one edge is removed from a tree, is it still connected?
What does it mean to span?
What is a minimal span?

3. ## Thanks.

Originally Posted by Plato
Please take no offence at this comment. But from the tenor of your general questions I have to wonder how careful you are in trying to ingest the meaning of the definitions on Graph Theory.
Let me give you a website that may help you: Graph Theory.
To you current question: If one edge is removed from a tree, is it still connected?
What does it mean to span?
What is a minimal span?


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# Thread: Independence and Multiplication rule help

1. ## Independence and Multiplication rule help

Hi guys, I'm back again with another question hopefully you guys can help me.

[Question]
We have a fuse box with 20 fuses, 5 of which are defective.

If 2 fuses are selected at random, in succession, and without replacement, what is
the probability that both fuses are defective?

[SOLN]
A = "1st fuse is defective", P(A) = 5/20 = 1/4
B = "2nd fuse is defective", P(B) = 4/19

P(A n B) = (1/4) * (4/19) = 1/19

? - My confusion has to do with independence, since you can't calculate P(B|A) without having P(A n B)
and since the above solution only seems to be calculable if you know the events are independent. Since then P(B|A) = P(B).

But it seems to me that the events would be dependent since there is no replacement. For instance if the first fuse chosen
was not defective then wouldn't the probability of choosing a defective fuse on the 2nd try actually be 5/19?

I guess then my actual question is how can I figure out whether or not the events are dependent if (A n B) is not known?
I hope I am making some sort of sense.

Thanks for your help guys, I sure need it.

2. ## Re: Independence and Multiplication rule help

The events certainly are dependent.

The probability the first fuse is defective is $\dfrac {5}{20}=\dfrac 1 4$

This leaves 4 defective fuses in the 19 total remaining.

So the probability that the 2nd fuse is defective, given the fact that the first one is, is $\dfrac {4}{19}$

And the probability of these two events

$P[\mbox{fuse 1 and fuse 2 were defective}] = P[\mbox{fuse 2 is defective |fuse 1 was defective}]P[\mbox{fuse 1 was defective}] = \dfrac{4}{19} \dfrac {1}{4} = \dfrac {1}{19}$

3. ## Re: Independence and Multiplication rule help

P(B|A) = P(B), IF the events are independent
P(B|A) != P(B), IF the events are dependent

But in this solution we say they are dependent, but then calculate as if it were independent?

4. ## Re: Independence and Multiplication rule help

Originally Posted by mathgex

P(B|A) = P(B), IF the events are independent
P(B|A) != P(B), IF the events are dependent

But in this solution we say they are dependent, but then calculate as if it were independent?
there's nothing about independence in the solution.

$P[A \cap B] = P[A | B]~P[B]$ regardless of whether $A$ and $B$ are independent or not.

maybe a better way to look at it is

$P[A \cap B] = P[A | B]~P[B]$

if $A$ and $B$ are independent then $P[A | B] = P[A]$ and we get the well known

$P[A \cap B] = P[A]~P[B]$ for independent events.

5. ## Re: Independence and Multiplication rule help

So are we are saying then that P(B) is not actually 4/19 but P(B|A) is 4/19..
And we simply just aren't considering P(B) in this case?

P(B), in my mind, is probability that 2nd fuse is defective regardless of what was chose first, I guess.
P(B) = ?? and P(B|A) = 4/19

Changing the solution to:

A = "1st fuse is defective", P(A) = 5/20 = 1/4
(B|A) = "2nd fuse is defective also", P(B|A) = 4/19

P(A n B) = P(A) * P(B|A) = (1/4) * (4/19) = 1/19

6. ## Re: Independence and Multiplication rule help

Originally Posted by mathgex
So are we are saying then that P(B) is not actually 4/19 but P(B|A) is 4/19..
And we simply just aren't considering P(B) in this case?
Yes. Let's take a look at what P[B] would be.

$P[B] = P[B | A]~P[A] + P[B | !A]~P[!A]$

Just to be clear
!A = good first fuse

as before

$P[B | A ] = \dfrac 4 {19}, ~~P[A]=\dfrac 1 4$

however

$P[B | !A ] = \dfrac 5 {19}$

and $P[!A] = \dfrac {15}{20} = \dfrac 3 4$

so $P[B] = \dfrac {4}{19} \cdot \dfrac {1}{4} + \dfrac {5}{19} \cdot \dfrac{3}{4} = \dfrac {1}{19}+\dfrac{15}{76} = \dfrac {19}{76}= \dfrac {1}{4}$

P(B), in my mind, is probability that 2nd fuse is defective regardless of what was chose first, I guess.
exactly

7. ## Re: Independence and Multiplication rule help

Ahhh excellent, thank you so much for your help on this Romsek, clear as a bell now.
You rock!


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JH

# Show that if we want to approximate the sum of the series $\sum_{n = 1}^{\infty} n^{-1.001}$ so that the error is less than 5 in the ninth decimal place, then we need to add more than $10^{11,301}$ terms!

## $1.07 \times 10^{11,301}$

Sequences

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##### Kristen K.

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Idaho State University

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### Video Transcript

this problem will show that if we want to find a partial some to approximate this infinite sum here such that the ear is less than this decimal here that will be eight zeros and then a five in the ninth spot. Then we'LL show that we'LL need to take and larger than this quantity over here. So this is our am So that means we should find off of X by just replacing and with X And then here X is bigger than or equal to one. And the reason for pointing that out is if you use in terms then from the section we know that the air is bounded above and we want thiss which is equal to one. You can go ahead and replace F with the formula up here and we would like this here to be less than the given quantity. So instead of writing this each time, let me just go ahead and denote this by some water, eh? And now this will go ahead and saw friend. So that means that we should just go ahead and use the power rule here and integrate this and that inequality is equivalent to this inequality here and we could just keep rewriting this and then this is equivalent to. So now, finally, to solve this for end will raise both side to the one thousand power that'LL cancel out this here and that'LL leave us with n larger than one thousand over, eh? All to the one thousand color and going to the calculator we get and we can see here that will need more than this many terms because this larger this numbered out here is larger and that completes the proof.

JH

#### Topics

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# Question about nodal analysis with an ideal op amp circuit

Iam trying to do nodal analysis of this op-amp circuit, but iam not quite sure if I have done it correctly or if I have missed something. The op-amps are considered being ideal so there is no current flowing in or out. I hope the picture is clear enough.

• If they are ideal opamps, then their inputs do not accept or inject current. Without current, R3 and R6 cannot drop a voltage. So the output of their respective opamps must appear at their input. So, I'm not at all sure where you got your "0" values at the opamp inputs. Must be an error in your perspectives (or a unique value for Vin.) – jonk Nov 8 '20 at 21:15

Your node A equation is incorrect because the voltage across R2 is $$\V_A\$$, not $$\V_B\$$. (It will turn out in this circuit that these voltages are the same, but for the purpose of doing nodal analysis, you can't assume this a priori)

Your node B equation is incorrect because you didn't include current coming from the op-amp output pin.

Your equation set is incorrect because you didn't include an equation for the output node.

The fact that it's difficult to write the equation for node B without knowing the internals of the op-amp, is why for hand calculation we don't usually use the nodal analysis. Instead we use equations derived from the knowledge that in negative feedback the two input terminals of the op-amp will be driven to the same voltage.

• Alright, because Iam thinking that the equation for node B is maybe not necessary because of the negative feedback of the op-amp. – Håkan Mjölk Nov 8 '20 at 18:16

Assuming ideal op-amp for analysing.

At Node A: $$\frac{(V_A-V_{in})}{R_1}+\frac{V_A}{R_2}=0$$

$$\V_A\$$ is the voltage at non-inverting input. Voltage at inverting and non-inverting inputs should be equal for negative feedback. Therefore:

At Node B: $$\frac{(V_B-V_{A})}{R_3}+\frac{(V_B -V_C)}{R_4}-i_0=0$$

$$\i_0\$$ is assumed to be the current flowing out of op-amp output to node B.

Current flowing into inverting input should be zero, means $$\\frac{(V_B-V_{A})}{R_3}=0\$$, so $$\V_B=V_A\$$.

At Node C: $$\frac{(V_C-V_{B})}{R_4}+\frac{V_C}{R_5}=0$$

and finally $$\V_{out} = V_C\$$ just like we saw in the case of $$\V_B = V_A\$$.

If $$\V_{in}\$$ and all resistor values are known then every equations can be solved.

• On Node C, do you mean VC/R5 ? – Håkan Mjölk Nov 8 '20 at 18:23
• Oh yes. Thanks for correcting! – Meenie Leis Nov 8 '20 at 18:25

Following Meenie's explanation, since ideal opamp has no input current in either inverting/non-inverting inputs the current throughout $$\R_3\$$ and $$\R_6\$$ can be considered as zero so $$\V_b=V_a\$$ and $$\V_{out}=V_c\$$, then reaching to $$V_{out}=V_{in}\frac{R_2}{R_1+R_2}\frac{R_5}{R_4+R_5}$$ $$\R_3\$$ and $$\R_6\$$ are not necessary in the ideal case. You always can simulate the circuit with this fantastic tool provided in here, I made it for you so you can confirm by simulating the circuit below.

simulate this circuit – Schematic created using CircuitLab


text:
### Home > CALC > Chapter 9 > Lesson 9.3.1 > Problem9-80

9-80.
1. Write the integral that calculates the volume generated when the region bounded by y = x2 and x = y4 is rotated: Homework Help ✎


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# Indicator 74. Consumption of ozone-depleting substances (MDG Indicator)

## Rationale and definition:

This indicator measures the consumption trends for ozone-depleting substances (ODS) controlled under the Montreal Protocol on Substances that Deplete the Ozone Layer, thereby allowing inference of the amounts of ODS being eliminated as a result of the protocol. It is expressed in ODP Tons, which is defined as the Metric Tons of ODSs weighted by their Ozone Depletion Potential (ODP).1

The Montreal and the Vienna Convention for the Protection of the Ozone Layer target the complete phase-out of use of ODS.

A


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# Is sine of angles greater than 90 degrees a convention?

The sine function is defined as the opposite side of the angle in question over the hypotenuse of the $90^\circ$ triangle.

$$\sin(â) = \frac{\text{opposite side}}{\text{hypotenuse}} \tag{0^\circ<â<90^\circ}$$

Therefore, I cannot see why this is valid for angles equal or greater than $90^\circ$. I know that in the trigonometric cirle, we can just look for the simetric angle in the other side and then the sine value will be the same, since the radius of the circle is $1$ (then the sine is just the height of the point, or the $y$ value). But I cannot understand how they became with this definition. For me, there's no sense to admit that the sine of an angle greater than $90^\circ$ will be symmetric to a triangle on the other side.

I know that the Taylor expansion for sine can converge for any angle (in radians) but this expansion is derived from the derivative of the sine function, which should be only defined for $0^\circ<x<90^\circ$.

-
We defined $\sin x := \sum_{i=0}^\infty {(-1)^n\cdot\frac{x^{2n+1}}{(2n+1)!}}\ (x\in\mathbb{R})$ –  moose Sep 24 '13 at 5:41
More or less everything is a convention. Definitions are arbitrary, although they are subject to "natural selection". Useful ones carry on, others are forgotten. –  Dan Shved Sep 24 '13 at 6:30

When $\sin$ is defined geometrically, it is typically as the vertical coordinate on the unit circle. Certainly this is the only sensible extension from $[0^\circ, 90^\circ]$ if you want to retain any of the nice properties like analyticity, angle addition formulas, Euler's formula, etc.


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# Recursion Warmup - Tower Numbers

Probability Level 1

What is the next number in this sequence? (There are no hidden cubes.)

$0, 1, 3, 6, 10, 15...$

×


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Volume 317 - The 9th International workshop on Chiral Dynamics (CD2018) - Session: Plenary Program
Experimental studies at lowQ2 of the spin structure of the nucleon at Jefferson Lab
A. Deur
Full text: pdf
Published on: February 28, 2020
Abstract
We summarize the experimental program of Jefferson Lab that studies the nucleon spin structure at low $Q^2$. This program completes the precise experimental mapping of the nucleon spin structure functions $g_1(\nu, Q^2)$ and $g_2(\nu, Q^2)$ and their moments started at SLAC, CERN and DESY at high $Q^2$, and continued at Jefferson Lab at intermediate $Q^2$.
The results presented cover the domain where Chiral Effective Field Theory ($\chi$EFT) should describe the strong interaction. They provide a comprehensive set of benchmark measurements for $\chi$EFT.
The preliminary conclusion is that nucleon spin structure data are still challenging for $\chi$EFT in spite of the notable improvements in these calculations.
DOI: https://doi.org/10.22323/1.317.0005
How to cite

Metadata are provided both in "article" format (very similar to INSPIRE) as this helps creating very compact bibliographies which can be beneficial to authors and readers, and in "proceeding" format which is more detailed and complete.

Open Access


text:
How close to escape velocity are most Oort-cloud comets?

User @antlersoft wrote a nice answer to my question on the difference between barycentric and heliocentric models of the solar system when applied to comets (edge cases of the systems). In a comment, @DavidHammen mentions that

Planetary perturbations near perihelion will change the aphelion for most highly-eccentric objects, regardless of the coordinates used.

Because most comets have inclinations different than that of the ecliptic plane and do not intersect the orbit of a planet, with the famous exception of Comet Bowell in a close swingby of Jupiter, with an inclination of only 1.66 degrees. As a result, why are such small perturbations of planets several AU away enough to take a comet from a closed to an open orbit? How close to escape velocity are most Oort-cloud comets?

It's a consequence of the reversibility of orbits that anything that free-falls from a near-interstellar distance arrives at nearly escape velocity, or faster. Take the Vis-Viva equation, the formula for the velocity of objects in Keplerian orbits/trajectories:

$$v=\sqrt{\mu\left(\frac{2}{r}-\frac{1}{a}\right)}$$

Where $$r$$ is the radial distance, $$a$$ is the semi-major axis, and $$\mu$$ is the standard gravitational parameter.

When $$a$$ is very, very large compared to $$r$$, things start to look like this:

$$v=\sqrt{\mu\left(\frac{2}{r}\right)}$$

Which is the formula for escape velocity.

So basically, almost anything falling from the Oort cloud to the inner solar system is going to be scraping the underside of escape velocity in the inner system, at minimum.

For an example, the inner edge of the Oort cloud is believed to start at about $$2\,000\, \mathrm{AU}$$ from the Sun, according to NASA's Solar System Overview. As such an object in a highly-elliptical orbit that takes it out that far would have a semimajor axis of about $$1\,000\,\mathrm{AU}$$. Jupiter's Semimajor axis is about $$5 \,\mathrm{AU}$$.

The Vis-viva calculation puts an object orbiting the Sun with a semi-major axis of $$1000\, \mathrm{AU}$$ moving at $$18\,814\, \mathrm{m/s}$$ when it reaches $$5\, \mathrm{AU}$$. Solar Escape velocity at $$5\, \mathrm{AU}$$ is $$18\,837\,\mathrm{m/s}$$, a difference of $$23 \,\mathrm{m/s}$$ that would be swallowed whole if I'd used proper significant figures in the calculation.

• Very nice explanation. In essence: it's just jumping a frog off the comet to trigger it :) Dec 17, 2021 at 18:58
• Any object obeying the vis-viva equation is bound to the star, anything bound has a factor of $1/\sqrt(2)$ between its perihelion velocity and the escape velocity at perihelion... Dec 17, 2021 at 19:46
• @AtmosphericPrisonEscape Both of your statements are incorrect. The Vis-viva equation works for all Keplerian orbits and trajectories, as along as you're using negative semi-major axis for hyperbolic trajectories. Escape velocity /$\sqrt{2}$ = the circular orbit velocity for the chosen distance, only. Dec 17, 2021 at 19:55
• Just that velocity at perihelion is not a radial one, but a tangential one, as the object has to orbit. Furthermore, the approximation for large a is only valid in the approximation of infinite eccentricity, which is not given for all small solar system bodies. Dec 17, 2021 at 20:52
• @AtmosphericPrisonEscape If an object at perihelion $r_p$ is moving at $v_p=\sqrt{\frac{\mu}{r_p}}$, then that orbit must be circular. For an elliptical orbit, it will always be moving faster than that speed at perihelion, which invalidates your $1/\sqrt{2}$ ratio claim. Under the Keplerian-Newtonian two-body simplifications, any velocity of magnitude less than the local escape velocity, in any direction results, in an elliptical orbit. Dec 17, 2021 at 21:15


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## Discussion Forum

### Daily Magic Spell Discussion - January 9, 2017

Daily Magic Spell Discussion - January 9, 2017

For today's Daily Magic Spell Discussion, we will look at the problem given on January 9, 2017.

Problem: Let $N$ be the largest seven-digit palindrome that can be constructed from the digits $1,2,3,4,5,6,7$ (repeats are allowed) such that the sum of every pair of consecutive digits is a prime number. What is the number $N$?

Solution: The largest number is $7676767$ since $7+6 = 6+7=13$ is the largest prime you can get.

To approach this question, note that the largest prime number that you can add together by two digits is $7$ and $6$. Some students may have attempted to try and brute forced methods so that two consecutive digits add to a prime number. Some students may not realize that this can be done by swapping the digits. Because of this, their answers are marked incorrect.

Some students may also approach the question without realizing that they can repeat digits.Because of this, students would respond $7652341$. This is achieved by starting with $7$, asking if adding $6$ will yield a prime number. If so, move onto the next digit. If not, ask if $4$ adds up to a prime number.

If you have any other approaches to tackling this question, please do not hesitate to respond to the forum discussion. We would love to offer you feedback!


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## The method of Frobenius to Fuchsian partial differential equations.(English)Zbl 0993.35004

Let $$t\in \mathbb C$$, $$x=(x_1,\ldots, x_n) \in \mathbb C^n$$. The following Fuchsian differential operator is considered: $\begin{gathered} P=t^m D_t ^m+ P_1(t,x,D_x) t^{m-1}D_t ^{m-1}+ \ldots + P_m(t,x,D_x), \\ P_j(t,x,D_x)=\sum_{\alpha\leq j} a_{j,\alpha}(t,x) D_x ^{\alpha}, \quad (1\leq j \leq m), \\ {\text{ord}}_{D_x}P_j(0,x,D_x) \leq 0, \quad (1\leq j \leq m), \end{gathered}$ where $$m$$ is a positive integer, and $$D_t=\partial/\partial t$$, $$D_{x_j}=\partial/\partial x_j$$. The coefficients $$a_{j,\alpha}$$ are supposed to be holomorphic in a neighborhood of $$(t,x)=(0,0)$$. Let $$a_j(x)=P_j(0,x,D_x)$$. The polynomial $C(x,\lambda)= (\lambda)_m +a_1(x) (\lambda)_{m-1}+ \ldots +a_m (x)$ is called a indicial polynomial, and its roots are called the characteristic exponents of $$P$$ in $$x$$. H. Tahara investigated the solutions of such equations under the assumption that the characterictic exponents $$\lambda_k(0)$$, $$k=1,\ldots, m$$, do not differ by an integer [J. Math. Soc. Japan 36, 449-473 (1984; Zbl 0545.35015)]. The aim of the paper under review is to construct the solutions of the Fuchsian equation without any assumption on the characteristic exponents, and using the same Frobenius method originally used for ordinary differential equations with regular singular points and than used by H. Tahara.
The structure of the paper: in section 2 the review and modification of the Frobenius method are given, in section 3 the statement of a main theorem is presented, sections 4-8 contain its proof. In conclusion some variants of the result are discussed. For example, its Nagumo-type version, that is, the Fuchsian operators with not only holomorphic coefficients but also with $$C^{\infty}$$ coefficients. Besides, the well-posedness of the characteristic Cauchy problem is considered.

### MSC:

 35A07 Local existence and uniqueness theorems (PDE) (MSC2000) 35C20 Asymptotic expansions of solutions to PDEs 35B40 Asymptotic behavior of solutions to PDEs

### Keywords:

regular singularity; characteristic index

Zbl 0545.35015
Full Text:


text:
## Tuesday, November 13, 2012

### Fearing the Bond Viligantes

Paul Krugman posted on his blog an interesting write-up of a model which proposes that an "attack" by what he has called the "invisible bond vigilantes" is expansionary. In other words, an increase in the lending premium on sovereign debts should increase aggregate demand and thereby real GDP.

Krugman's mechanism makes a great deal of sense: (1) the increase in the risk premium generates depreciation of the currency for all domestic interest rates; (2) the lower exchange rate of the dollar causes an increase in net exports; and (3) thus we have a shifting to the left of the IS curve in the IS-LM model for all interest rates.

I agree with Krugman on several points, but in this post I want to propose a simple model which might make the best form of the counter-argument -- i.e. that such an increase in the risk premium would be contractionary. I also note a few areas in which his model and also my own make some assumptions which I, well, wouldn't bet very much on at all.

(Nick Rowe also has some brief comments on Krugman's model here, and so do Tyler Cowen, Brad deLong, David Beckworth and Scott Sumner.)

Let me first note the general areas of agreement. It's pretty obvious that a country with a floating exchange rate and an independent currency (like the U.S. or U.K.) is in a totally different situation than a country without either (like Greece or Spain) insofar as debt can create default risk. I also agree with some key components of the model -- you'll see me re-use much of his framework in a moment -- and with the insight that an increase in the risk premium should generate higher net exports and thereby, cet. par., higher real GDP in the short run.

But here's how I thought about this. In terms of intuition, an increase in the risk premium is a supply shock. A central bank which follows some sort of rule regarding price stability should be somewhat constrained as to how much devaluation they can allow. From this, one can reason that the sign on the effect to real GDP should always be negative, given that the central bank will split the impact between a decline in real GDP and an increase in the price level according to its preferences.

And that's what one sees come through in this little model.

First, start with the linearized demand function y = -ar + be, where r is the real interest rate, e is the nominal exchange rate in terms of the price of foreign currency, and a and b are constants. Note the signs of the terms mean that an increase in the real interest rate decreases real GDP and an increase in exchange rate increases real GDP -- i.e. when foreign currency is more valuable, our net exports increase.

Second, assume a real domestic interest rate determined by r = r* + p, where r* is the real risk-free rate of return and p is the risk premium. An increase in the risk premium increases the interest rate.

Third, assume that purchasing power parity (PPP) holds to some reasonable extent in some reasonably short time frame. That is to say, an identical tradable goods and services should cost roughly the same whether I'm buying them in dollars, yen, etc. We can write this out as P = ePf, where P is the domestic price level, e is (again) the exchange rate, and Pf is the foreign price level.

Fourth, assume some monetary policy rule. Here are three which I think characterize the broad swath of options, which I proceed to consider individually.

(1) price-level targeting: P = P*
(2) NGDP targeting: Py = N
(3) a Taylor-type rule: P + Ty = k, T > 0

To consider (1) with the assumptions explained above:

e = P* / Pf

y = -a(r* + p) + b(P* / Pf)

∂y/∂p = -a.

This means that real GDP will fall in response to an increase in the risk premium, due to its effect on real domestic investment, when the central bank is targeting the price level. (This should also be the response of an inflation-targeting central bank.) The strength of this effect will depend on the "a" parameter, which is the responsiveness of domestic investment to changes in the real interest rate.

To consider (2):

$e = \frac{N}{yP_f} \\ y = -a ( r^* + \rho ) + b\frac{N}{y P_f }\\ y^2 + ay ( r^* + \rho ) + b\frac{N}{y P_f } = 0 \\ \frac{\partial y}{\partial \rho} = \frac{-ya}{2y + a ( r^* + \rho)} \\ \frac{\partial y}{\partial \rho} \approx \frac{-a}{2}$

Again, this means that real GDP will fall, though less than under the previous monetary policy specification, due to the supply shock's effect on domestic investment.

To consider (3),

$e = \frac{k-Ty}{P_f} \\ y = -a ( r^* + \rho ) + b\frac{k-Ty}{P_f} \\ y = \frac{-a ( r^* + \rho ) + \frac{bk}{P_f}}{1 + \frac{bT}{P_F}}\\ \frac{\partial y}{\partial \rho} = \frac{-a}{1 + \frac{bT}{P_F}}\\ 0 < \frac{\partial y}{\partial \rho} < -a$

This finding should make a lot of sense. If T = 0, then there is no weight on output stabilization, and the Taylor-type rule becomes a price-level rule. As T rises, the effect on output is dampened -- though, naturally, by progressively larger increases in the price level.

Making some small and I believe acceptable assumptions -- the aggregate demand function, the determination of an interest rate, a soft PPP, and a monetary policy rule -- we see that an increase in the risk premium is likely contractionary. The intensity of the contraction, furthermore, depends on the willingness of the central bank to accept large increases in the price level to cushion output in the short run.

The major differences with Krugman's model in terms of structure are as follows. First, I assume that domestic investment will feel the sovereign's risk premium. My understanding is that this is consistent with the actual operation of debt markets; the debt of large companies will be knocked down in terms of credit rating when the sovereign's credit rating falls. Second, there are some important real-nominal distinctions in my model versus his; it's not clear why an independent central bank would tolerate the inflation Krugman's model builds in but never directly addresses. I think these explain the opposite findings.

Three further notes. First, I think the assumption of a fixed risk-free rate i* in the context of a run on US sovereign debt is highly strained, for the same reason that the small open economy model is not the same as the large open economy model. Second, when the risk premium rises, the increase in the real interest rate is likely not to capture the full effect on domestic investment -- there are other mechanisms, most importantly tighter lending standards, which will cause an even larger decrease in investment. Third, in the context of an increase in the risk premium on U.S. debt, the U.S. dollar is -- by our experience in the last recession -- likely to appreciate, unless global debt markets are sufficiently strong to withstand a global risk-off. Heavy capital flows into U.S. Treasuries will prevent devaluation for the time period Krugman's model expects an expansionary effect. It is worth noting that all three reasons suggest that an increase in the risk premium is likely to result in a decrease of real GDP in excess of what is predicted by my model.

Correction: I originally had "e" marked as the real exchange rate. It should have been the nominal exchange rate.

1. Evan: I disagree with your model (unless I misunderstand it).

Let me re-write your demand function as y = -ar + b(e-1). Then PPP can be interpreted as "the elasticity parameter b approaches infinity, because domestic and foreign goods become perfect substitutes."

1. Do you disagree with the assumption of PPP? If there's going to be a deviation from it, given flight-to-safety capital inflows into the US, it's going to be on the contractionary side.

2. I do disagree with PPP, but that's not the issue. Suppose I agreed with PPP. I would say PPP is a consequence of your demand function, in the limit as b approaches infinity. You cannot assert PPP as an independent equation, it is like having an overdetermined model.

Take the limit as b approaches infinity. A tiny real exchange rate depreciation would cause an infinitely big increase in output determined.

3. OK, I think I understand what went wrong; for PPP to hold, b must approach infinity, given the definitions. I'm not exactly sure if this can be fixed -- at this point, I'm in a little over my head -- but is the problem corrected by assuming e as the nominal exchange rate, rather than the real exchange rate? Or, is it just that PPP cannot hold in the context of the model? The reason I had invoked it was to connect exchange rates to the domestic price level, and then to monetary policy, which I saw as the underlying issue here. Thanks for your input.

4. The "e" in y=-ar+be should really be the real exchange rate (just as the 'r" should be the real interest rate). The equation is a simplified version of the IS curve in an open economy. Net exports depends on the relative price of foreign to domestic goods, which is the real exchange rate. For example, if the foreign price level doubled, and the exchange rate depreciated by half, net exports should stay the same, because the relative price of foreign and domestic goods would stay the same.

This is really just a specific example of a more general rule. Every equation in a model (except the central bank's reaction function/target/whatever) should be homogenous in nominal variables. If you double all the nominal variables, and leave the real variables unchanged, the equation should still be true. (Except for the central bank's reaction function/target/whatever, because otherwise the price level would be indeterminate). (I might be using the wrong math terminology when I say "homogenous. I got a 'D' in A-level math, a very long time ago.)

5. I think you need some sort of AS curve (or Phillips Curve) to pin down the price level (or inflation rate). Something like y=sP for an upward-sloping SRAS curve.

2. This comment has been removed by the author.

3. A few quick notes: this might be my mistake, but I don't quite follow how you've defined the real exchange rate. As I understand it, in your pseudo-IS function you explicitly label 'e' as the real exchange rate, but then go on to define the law of one price as being: P = P* x e, which doesn't quite make sense to me. The LOP is ordinarily written like that, but the 'e' in that equation corresponds to the 'nominal' exchange rate. That makes sense because, say, in a one good economy, if the consumption item is now more expensive (P is higher) and prices in the foreign country stay the same, the nominal exchange rate must appreciate in response so that the LOP holds.

It seems to me, then, that in case (1) you are simply keeping the real exchange rate constant, which Krugman effectively discusses in his proposed modelling. Price level targeting is, in this instance, a fixed nominal exchange rate regime.

As for (2), I don't quite understand what the monetary authority is supposed to be doing. Setting a target for nominal output is probably less informative than assuming the monetary authority simply sets a target of N for real output. In that event, the TR line in Krugman's model is a vertical line at the desired level of output, and the monetary authority will allow the interest rate to increase in the exact amount of the increase in the risk premium, which ought to have no effect on output (naturally). Setting a target in terms of nominal output, seems to me, would have, again, no effect on output: the increase in risk premium has a negative effect on investment which is perfectly offset by the expansionary effect of real depreciation.

Effectively, we need to think of Krugman's model in the following sense: the monetary authority has no control over actual interest rates, only nominal exchange rates. It uses the wedge between the long run exchange rate and the actual nominal exchange rate as the driving mechanism in the model: the CB sort of gains 'momentary' control of policy by increasing or decreasing that wedge. Thus, his model isn't actually a model of risk premia at all: note that an increase in the rate required by bond holders of the Bundesbank (the safe asset) has exactly the same implications as an increase in the risk premium, and the latter is therefore pretty irrelevant to his conclusions.

As he points out, the main result of the model is that an increase in interest (for whatever reason, I would add) implies currency depreciation via the wedge, and therefore always leads to an increase in output. The question is surely then whether this is the right way to model the relationship between the interest and nominal exchange rates.

1. Thanks -- I think I did mean the nominal exchange rate (I wrote this up in my notebook over the weekend). Your point on (1) is correct. On (2), yes regarding the real-stability-only central bank, but I think a nominal-output target isn't the same in my model. Any devaluation is going to boost the price level and real output; therefore, a return to the same Py will happen before the same y. I agree with your points about the original model, also.

4. Evan, I start from a different premise. The risk premium is currently too high and would be lowered given the appearance of "bond vigilantes". Rising yields would be a sign of recovery and return to normalcy.

1. I understand, but IIRC, didn't Krugman have a post in response to deLong re: a bond bubble, with (what I thought was) convincing evidence that bonds were appropriately priced given the mkt. forecast for future interest rates. You might respond that that path of interest rates is effectively a bet of too-low NGDP -- and I would agree -- but I don't subscribe to the reverse logic, i.e. that increasing interest rates will generate an increase in the NGDP path. Hopefully I'm not misreading you.

2. Evan,
I do not believe or advocate raising interest rates will generate an increase in NGDP path. http://macromarketmusings.blogspot.com/2012/02/can-raising-interest-rates-spark-robust.html

3. Evan,

I did a second post to clarify that I was staring from a different premise than most in this discussion.


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## anonymous 4 years ago i need help

1. anonymous

help in wat

2. anonymous

use half angle formulas to rewrite $\sqrt{1-\cos30 degrees/2}$

3. anonymous

$\sin (\theta/2) = \sqrt{(1-\cos \theta)/2}$

4. anonymous

using that formula o but..

5. anonymous

so you can rewrite it as sin(15)

6. anonymous

hwo do i get there i cantsee

7. anonymous

??

8. anonymous

what i mean the answer of the problem i sin 15 how do i get there id ont understand

9. anonymous

theta = 30 so theta/2 = 30/2 = 15

10. anonymous

can u show that

11. anonymous

show what...prove that 30/2 = 15

12. anonymous

no the whole process

13. anonymous

well i looked up the half-angle identity for sin (theta/2) **shown above** it matches your expression except you have 30 degrees for the angle

14. anonymous

ok ,..?

15. anonymous

and

16. anonymous

ok

17. anonymous

nee help trig= rewrite this using half angle formula sqrt{1= \cos 135 degress/2}\]

18. anonymous

is that a plus or minus ?

19. anonymous

sorry let me rewrite it

20. anonymous

$rewrtite using half aangle formulas \sqrt{1+ \cos 135 dgrees/2}$

21. anonymous

plus means its the cosine half angle formula take the given angle --> 135 divide it by 2 --> 135/2 = 67.5 answer = cos(67.5)

22. anonymous

so that my asnwer my teacher doesn tlike decimals

23. anonymous

then write it as a fraction...the main thing is do you get the concept? find appropriate formula , then take half the angle

24. anonymous

hello?

25. anonymous

can u show me thewhole process and ow do is hwo it in fraction

26. anonymous

hello

27. anonymous

what wre u gonnaa say

28. anonymous

are you serious...i showed you step by step...there is no other process you should have a reference in your book or notes showing the half-angle formulas the fraction would simply be 135/2 http://www.freemathhelp.com/trig-double-angles.html

29. anonymous

need help trig rewrite using half angle idnetities =1-cos210 degress/sin210 degrees

30. anonymous

need help trig rewrite using half angle idnetities =1-cos210 degress/sin210 degrees

31. anonymous

heloo

32. anonymous

$\rightarrow \frac{1-\cos 210}{\sin 210}$

33. anonymous

yes i know the formula i dont know how to use it

34. anonymous

please don;t message me the same thing you are posting, thanks and this is different it isn't any of the half-angle formulas

35. anonymous

no yes

36. anonymous

$\frac{1-\cos 210}{\sin 210} = \csc 210 - \cot 210$ not sure how else to rewrite it unless you want to use sum/difference formulas

37. anonymous

ok tanks

38. anonymous

s o how do i ewrite using half angle idnetities =1-cos210 degress/sin210 degrees

39. anonymous

by just saying it equals to 1-cos theta/sin theta

40. anonymous

???

41. anonymous

42. anonymous

no i dont grasp it

43. anonymous

i don't think you can rewrite using half-angle formulas look for yourself...see that the expression does not match any the formulas

44. anonymous

its what it says n my book

45. anonymous

yes it does tan theta /2

46. anonymous

hello

47. anonymous

yes i am wrong...my ref sheet has a different but equivalent formula for tangent sorry...hey i learned something anyway, now just take half the angle


text:
# Pololu 6V, 2.5A Step-Down Voltage Regulator D24V22F6

 Pololu item #: 2859 46 in stock Brand: Pololu Status: Active and Preferred Free add-on shipping in USA Free shipping in USA over $40 Price break Unit price (US$)
1 9.95
5 8.95
25 7.95
100 6.95

 Quantity: backorders allowed

This small synchronous switching step-down (or buck) regulator takes an input voltage of up to 36 V and efficiently reduces it to 6 V. The board measures only 0.7″ × 0.7″ yet delivers a typical continuous output current of up to 2.5 A and features reverse voltage protection. Typical efficiencies of 85% to 95% make this regulator well suited for powering moderate loads like sensors or small motors. An optional shutdown pin enables a low-power state with a current draw of around 30 μA to 350 μA, depending on the input voltage, and a power-good output indicates when the regulator cannot adequately maintain the output voltage.

Alternatives available with variations in these parameter(s): output voltage Select variant…

## Overview

The D24V22Fx family of step-down voltage regulators generates lower output voltages from input voltages as high as 36 V. They are synchronous switching regulators (also called switched-mode power supplies (SMPS) or DC-to-DC converters) with typical efficiencies of 85% to 95%, which is much more efficient than linear voltage regulators, especially when the difference between the input and output voltage is large. These regulators can typically support continuous output currents of over 2 A, though the actual available output current is a function of the input voltage and efficiency (see the Typical efficiency and output current section below). In general, the available output current is a little higher for the lower-voltage versions than it is for the higher-voltage versions, and it decreases as the input voltage increases.

These regulators have a typical quiescent (no load) current draw of around 1 mA, and an enable pin can be used to put the boards in a low-power state that reduces the quiescent current to approximately 5 µA to 10 µA per volt on VIN.

The modules have built-in reverse-voltage protection, short-circuit protection, a thermal shutdown feature that helps prevent damage from overheating, and a soft-start feature that reduces inrush current.

Several different fixed output voltages are available:

Alternatives available with variations in these parameter(s): output voltage Select variant…

The different voltage versions of this regulator all look very similar, so you should consider adding your own distinguishing marks or labels if you will be working simultaneously with multiple versions. This product page applies to all versions of the D24V22Fx family.

The D24V22Fx family is intended to replace our older D24V25Fx family of step-down voltage regulators. The two designs have the same size and similar current capabilities and input voltage ranges, but they do not have the same pinout and are based on different internal circuits, so there are fundamental differences in operation. In particular, these newer D24V22Fx regulators have much lower dropout voltages and provide a “power good” signal, and the newer design allows for higher output voltages (e.g. 12 V).

## Features

• Input voltage:
• 4 V to 36 V for the version that outputs 3.3 V
• [output voltage + dropout voltage] to 36 V for output voltages of 5 V and higher (see below for more information on dropout voltage)
• Fixed 3.3 V, 5 V, 6 V, 7.5 V, 9 V, or 12 V output (depending on regulator version) with 4% accuracy
• Typical maximum continuous output current: >2 A
• Typical efficiency of 85% to 95%, depending on input voltage, output voltage, and load
• Switching frequency: ~400 kHz
• Integrated reverse-voltage protection, over-current protection, over-temperature shutoff, and soft-start
• 1 mA typical no-load quiescent current; this can be reduced to approximately 5 µA to 10 µA per volt on VIN by disabling the board
• “Power good” output indicates when the regulator cannot adequately maintain the output voltage
• Compact size: 0.7″ × 0.7″ × 0.31″ (17.8 mm × 17.8 mm × 8 mm)
• Two 0.086″ mounting holes for #2 or M2 screws

## Using the regulator

### Connections

These buck regulators have five main connection points for five different electrical nodes: power good (PG), enable (EN), input voltage (VIN), ground (GND), and output voltage (VOUT). The board also features a second ground connection point off the main row of connections that might be convenient for applications where you are soldering wires directly to the board rather than using it in a breadboard.

The input voltage, VIN, powers the regulator. Voltages between 4 V and 36 V can be applied to VIN, but for versions of the regulator that have an output voltage higher than 4 V, the effective lower limit of VIN is VOUT plus the regulator’s dropout voltage, which varies approximately linearly with the load (see below for a graph of dropout voltages as a function of the load).

The output voltage, VOUT, is fixed and depends on the regulator version: the D24V22F3 version outputs 3.3 V, the D24V22F5 version outputs 5 V, the D24V22F6 version outputs 6 V, the D24V22F7 version outputs 7.5 V, the D24V22F9 version outputs 9 V, and the D24V22F12 version outputs 12 V.

The regulator is enabled by default: a 270 kΩ pull-up resistor on the board connects the EN pin to reverse-protected VIN. The EN pin can be driven low (under 1 V) to put the board into a low-power state. The quiescent current draw in this sleep mode is dominated by the current in the pull-up resistor from EN to VIN and by the reverse-voltage protection circuit, which altogether will draw between 5 µA and 10 µA per volt on VIN when EN is held low. If you do not need this feature, you should leave the EN pin disconnected.

The “power good” indicator, PG, is an open-drain output that goes low when the regulator’s output voltage falls below around 85% of the nominal voltage and becomes high-impedance when the output voltage rises above around 90%. An external pull-up resistor is required to use this pin.

The five main connection points are labeled on the top of the PCB and are arranged with a 0.1″ spacing for compatibility with solderless breadboards, connectors, and other prototyping arrangements that use a 0.1″ grid. Either the included 5×1 straight male header strip or the 5×1 right angle male header strip can be soldered into these holes. For the most compact installation, you can solder wires directly to the board.

The board has two 0.086″ (2.18 mm) diameter mounting holes intended for #2 or M2 screws. The mounting holes are at opposite corners of the board and are separated by 0.52″ (13.21 mm) both horizontally and vertically. For all the board dimensions, see the dimension diagram (204k pdf).

### Typical efficiency and output current

The efficiency of a voltage regulator, defined as (Power out)/(Power in), is an important measure of its performance, especially when battery life or heat are concerns. This family of switching regulators typically has an efficiency of 85% to 95%, though the actual efficiency in a given system depends on input voltage, output voltage, and output current. See the efficiency graph near the bottom of this page for more information.

The maximum achievable output current is typically over 2 A, but this depends on many factors, including the ambient temperature, air flow, heat sinking, and the input and output voltage.

### Typical dropout voltage

The dropout voltage of a step-down regulator is the minimum amount by which the input voltage must exceed the regulator’s target output voltage in order to ensure the target output can be achieved. For example, if a 5 V regulator has a 1 V dropout voltage, the input must be at least 6 V to ensure the output is the full 5 V. Generally speaking, the dropout voltage increases as the output current increases. See the “Details” section below for more information on the dropout voltage for this specific regulator version.

## Details for item #2859

The graphs below show the typical efficiency and dropout voltage of the 6 V D24V22F6 regulator as a function of the output current:

During normal operation, this product can get hot enough to burn you. Take care when handling this product or other components connected to it.

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text:
# Stationary patterns in star networks of bistable units: Theory and application to chemical reactions

 We present theoretical and experimental studies on pattern formation with bistable dynamical units coupled in a star network configuration. By applying a localized perturbation to the central or the peripheral elements, we demonstrate the subsequent spreading, pinning, or retraction of the activations; such analysis enables the characterization of the formation of stationary patterns of localized activity. The results are interpreted with a theoretical analysis of a simplified bistable reaction-diffusion model. Weak coupling results in trivial pinned states where the activation cannot propagate. At strong coupling, uniform state is expected with active or inactive elements at small or large degree networks respectively. Nontrivial stationary spatial pattern, corresponding to an activation pinning, is predicted to occur at intermediate number of peripheral elements and at intermediate coupling strengths, where the central activation of the network is pinned, but the peripheral activation propagates toward the center. The results are confirmed in experiments with star networks of bistable electrochemical reactions. The experiments confirm the existence of the stationary spatial patterns and the dependence of coupling strength on the number of peripheral elements for transitions between pinned and retreating or spreading fronts in forced network configurations (where the central or periphery elements are forced to maintain their states). An illustrative bistable systems organized on undirected networks, generally reads $\dot{u}_i = u_i(h-u_i)(u_i-a) + K \sum_{j=1}^{N}\!T_{ij}\left(u_j - u_i \right)$ where $u$ is the local activator density, function $f(u_i,h)=$ specifies local bistable dynamics and $D$ is the coupling strength. $\mathbf{T}$ is a symmetric ($T_{ij} = T_{ji}$) adjacency matrix that describes the connectivity structure of the network; its elements are $T_{ij}=1$, if there is a link connecting the nodes $i$ and $j$ $(i,j=1,…,N$), and $T_{ij}=0$ otherwise. In the star networks, because of the symmetry in the system all peripheral nodes obey the same equation and the system is reduced into a two-dimensional system of ordinary differential equations, $\dot{u} = u(h-u)(u-a) + kK(v-u) := f(u,v;k,K)\,,$ for the central hub $\dot{v} = v(h-v)(v-a) + K(u-v) := g(u,v;K)\,,$ for the peripheral nodes By solving $f(u,v;k,K)=0$ we can get an expression of $v$ in terms of $u$, and then, by substituting this expression into the equation $g(u,v;K)=0$, we obtain the sixth degree polynomial equation, $P(u) := c_0 + c_1 u + c_2 u^2 + c_3 u^3 + c_4 u^4 + c_5 u^5 + u^6 = 0\,,$ The following figure shows $P(u)$ for $k=10$ and three values of coupling strength $K$. At $K_c^{pr}$ (dashed curve) the two inner roots (shown with circle and dot) of $P(u)$ merge together and disappear through a saddle-node bifurcation. These roots correspond to the fixed points of the system. Any initial condition $(u_0,v_0)$ in the basin of attraction of a stable fixed point will eventually converge to the corresponding fixed point. Therefore, before the bifurcation, the initial center activation $(a,0)$ results in a stationary pattern (thick blue trajectory in panel (b)), and the initial periphery activation $(0,a)$ propagates towards the center (thin red trajectory in panel (b)). After the bifurcation, the center activation retreats to the passive state (thick blue trajectory in panel (c)) whereas the periphery activation propagates towards the center (thin red trajectory in panel (c)). Figures (d)–(f) and (g)–(i) illustrate similar scenarios for $k=5$ and $k=3$ respectively.   Further reading: N. E. Kouvaris, M. Sebek, A. Iribarne, A. Diaz-Guilera and I. Z. Kiss, Phys. Rev. E 95, 042203 (2017). [arXiv] N. E. Kouvaris, M. Sebek, A. S. Mikhailov and I. Z. Kiss Angew. Chem. 2016, 128, 13461-13464. [Angew. Chem.] Angew. Chem. Int. Ed. 2016, 55, 13267-13270. [Angew. Chem. Int. ed.] [arXiv]


text:
# ANOTHER GENERALIZATION OF A RAMANUJAN SUMMATION

• Lee, Seung Woo (Department of Mathematics, Wonkwang University) ;
• Lee, Chang Hyun (Department of Mathematics, Seonam University) ;
• Kim, Yong Sup (Department of Mathematics Education, Wonkwang University)
• Accepted : 2013.02.12
• Published : 2013.03.25
• 90 24

#### Abstract

The aim of this research paper is to provide certain generalizations of two well-known summations due to Ramanujan. The results are derived with the help of the generalized Dixon's theorem on the sum of $_3F_2$ and the generalized Kummer's theorem for $_2F_1$ obtained earlier by Lavoie et al. [3, 5]. As their special cases, we have obtained fifteen interesting summations which are closely related to Ramanujan's summation.

#### Keywords

Hypergeometric $_2F_1$;Dixon's summation theorem;Kummer's summation theorem;Ramanujan summation formula

#### Acknowledgement

Supported by : Wonkwang University

#### References

1. W.N Bailey, Generalized Hypergeometric Series, Cambridge University Press, Cambridge, 1935.
2. B.C. Berndt, Ramanujan's Notebooks, Part II, Springer-Verlag, New York, 1985.
3. J.L. Lavoie, F. Grondin, A.K. Rathie, and K. Arora, Generalizations of Dixon's theorem on the sum of a $_3F_2$, Math. Comp. 62(205) (1994), 267-276.
4. J.L Lavoie, F Grondin, and A.K. Rathie, Generalizations of Whipple's theorem on the sum of a $_3F_2$, J. Comput. Appl. Math. 72(2) (1996), 293-300. https://doi.org/10.1016/0377-0427(95)00279-0
5. T.K. Pogany, A.K. Rathie, and U. Pandey, Generalization of a summation due to Ramanujan, Makedon. Akad. Nauk. Umet. Oddel. Mat.-Tehn. Nauk. Prilozi XXX(1-2) (2009), 67-73.
6. A.K. Rathie, S. Malani, R. Mathur, and J. Choi, Certain summations due to Ramanujan and their generalizations, Bull. Korean Math. Soc. 42(3) (2005), 469-475. https://doi.org/10.4134/BKMS.2005.42.3.469
7. M.R. Spiegel, Mathematical Handbook, Mc Graw-Hill Book Company, New York, 1968.
8. H.M. Srivastava and J. Choi, Zeta and q-zeta functions and associated series and integrals, Elsevier Science Publishers, Amsterdam, London, and New York, 2012.