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Does the TIOBE rating of programming language play a prominent role in the selection of programming language in enterprise computing?
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closed as primarily opinion-based by MichaelT, Ampt, Dynamic, GlenH7, Yannis♦ Jul 3 '14 at 10:45
Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise.If this question can be reworded to fit the rules in the help center, please edit the question.
No. TIOBE is not used.
The prominent role is politics.
What do influential people know. What are they comfortable with. Not much else really matters.
When it comes to programming languages, all the "major" languages are perfectly fine. They all work well. They've all got serious, large bases of committed users. Any of the languages in the top 20 (or more) on the TIOBE list are all equally usable.
In 30 years, I've never seen a decision like programming language choice made except by default. The person who's most influential makes the final decision.
TIOBE is an index of how much "gossip" there is about a language. If you or someplace you work bases technical choices on gossip then .......
You're mixing up cause and effect. TIOBE lists languages that are most popular in the industry, not other way around.
I agree with S.Lott, it's mostly politics and decisions made on management level. Although TIOBE can be used to counter arguments like "no one is using Python" (not that it'll help -- if CTO only knows Java, you'll be programming Java).
TIOBE is not a meaningful indicator of anything. Most specifically the claim that "TIOBE lists languages that are most popular in the industry" is not a conclusion that can be drawn.
All that TIOBE tells you is which programming languages have the most documents on the web written about them. It is a huge leap of logic to conclude that that is an indicator of how much the languages are used in the real world.


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Sony confirms layoffs at three British first-party studiosWritten by: / / 3 Comments
Just one month after laying off staff at Santa Monica Studio, there have been more confirmed layoffs at Sony and three of its largest UK first-party developers – Evolution Studios (DriveClub), Guerrilla Cambridge (Killzone: Mercenary) and SCE London Studio (EyePet, SingStar).
In a statement issued to VideoGamer.com, Sony Computer Entertainment Europe said that the development of DriveClub will not be affected by the latest round of layoffs despite the game's extended development cycle. Here's Sony's statement explaining the recent layoffs in more detail:
"Regular reviews take place throughout SCE Worldwide Studios, ensuring that the resources that we have in such a competitive business can create and produce high quality, innovative and commercially viable projects.
"As part of this process, we have reviewed and assessed all current projects and have decided to make some changes to some of our European Studios. As a result of this, there will be a focused restructure within London Studio, Guerrilla Cambridge Studio and Evolution Studio to ensure that the SCE WWS is in the best position to achieve their goals going forward.
"Our first party development studios are key to our creative strategy and we are very excited about the future projects being worked on."
Is this latest round of layoffs bad news for PlayStation fans or will it make Sony more competitive moving forward as the platform holder suggests? Let us know your thoughts in the comments.
Tags: europe, evolutionstudios, guerrillacambridge, layoffs, scelondonstudio, sony

subset: commoncrawl


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Some adept choices for judgeships
Has any group of judicial nominations in Connecticut ever been more politically adept than the 16 put forth last week by Gov. Dannel P. Malloy?
Superior Court Judge Christine E. Keller was nominated for the Appellate Court. The wife of former House Speaker Thomas D. Ritter, D-Hartford, now a lobbyist, she has put in 20 years as a judge and has handled difficult cases well.
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In this question $R$ is a commutative noetherian local ring with unity.
One can construct examples of rings $R$ and zerodivisors $z$ such that $\dim R/(z)=\dim R-1$, e.g., $S\colon=k[a,b,c],\ \mathfrak{m}\colon=(a,b,c),\ R\colon=S_\mathfrak{m}/(a^2,ab)S_\mathfrak{m},\ z\colon=b^2$.
One can also construct examples of rings $R$ and zerodivisors $z$ such that $\mathrm{depth}\ R/(z)=\mathrm{depth}\ R-1$, e.g., $S\colon=k[a,b,c],\ \mathfrak{m}\colon=(a,b,c),\ R\colon=S_\mathfrak{m}/(a^2)S_\mathfrak{m},\ z\colon=ab.$
What is an example of a zerodivisor that will reduce both the dimension and the depth by $1$, simultaneously? Is that possible?


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WEIRTON - Don Ogden and his Madonna girls will enter next week's Ohio Valley Athletic Conference Class 1A Tournament on a roll after disposing of a solid Toronto club 51-33 Monday.
With the win, the Lady Blue Dons improve to 9-10 and will enter OVAC play as the No. 3 seed, meaning a visit to second-seeded Bellaire St. John next week.
"We're elated," Ogden said. "One of things we wanted to do was have some momentum going into the OVAC tournament. With the win tonight and the one Saturday (a 43-29 triumph over Oak Glen), we're headed in the right direction. We still have some things to work on but we're happy.
Article Photos
GOING TO THE BASKET — Toronto's Lexie Berger drives to the basket against the defensive pressure of Weirton Madonna's Nicolette Sellitti during the fourth quarter of Monday's contest in the Bill Barrett Gymnasium, won by the Blue Dons 51-33.
-- Mike Mathison
GRABBING A REBOUND — Weirton Madonna's Monica Bragg grabs a rebound in front of Toronto's Becca Schaming during the third quarter of Monday's contest in the Bill Barrett Gymnasium, won by the Blue Dons 51-33.
-- Mike Mathison
It was senior night inside the Bill Barrett Gymnasium and Madonna's two 12th graders (Nicolette Sellitti and Amy Davis) played key roles in the victory.
Sellitti scored 12 points while Davis added nine.
"Our seniors have been there all year, sometimes the stats don't show it, but they are the core of this team," Ogden pointed out. "They represented themselves very well tonight."
Powered by five markers from freshman Monica Bragg, the hosts grabbed a 9-8 advantage after the first period. The second stanza featured huge scoring runs by both clubs.
Two Bragg free throw makes and a conventional three-point play by Sellitti fueled a 10-0 Blue Don burst to open the quarter. Isabella Rosso banked home a triple and point guard Cameron Dowiak, who recorded six of Madonna's 15 steals on the defensive end, added a deuce.
Toronto responded with a 12-0 run of its own to seize the lead at 20-19. Makayla Bauman accounted for eight of the Red Knight points.


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7 Songs From Your Grandpa's Day That Would Make Eminem Blush
Your grandparents are full of crap.
As they grumble about how rap music is destroying civilization, what they don't mention is that the blues they were listening to in the 30s and 40s could be every bit as violent, sexually explicit and sometimes just downright insane as the worst gansta rap has to offer. Compared to some of these vintage lyrics the members of N.W.A are levelheaded concerned citizens, and Eminem's a regular damned feminist.
#7. Skip James Cuts a Woman in Half ("22-20 Blues" - 1931)
"22-20 Blues" tells the tale of a woman who just won't get her act straight. See, Skip James sent for her, on several occasions, and yet she didn't show up! The brazen audacity! Of course, in the world of blues, there's only one way to deal with minor punctuality issues: brutal, brutal murder.
Big on punctuality.
A Few Choice Quotes
Sometimes she gets unruly;
An she act like she just don't wanna do;
But I get my 22-20;
I cut that woman half in two;
Your .38 Special;
Buddy, it's most too light;
But my 22-20;
Will make ev'rything, alright;
Shooting your woman with a mere .38 pistol? That's for pussies. Ironically James soon found himself humbled when Robert Johnson recorded a far more popular version of his song. The only real change Johnson made? He upped the caliber and named it the "32-20 Blues." It was all about the gun size with those boys.
#6. Lucille Bogan Fucks Corpses ("Shave 'Em Dry" - 1935)
Looking at a picture of Lucille Bogan, it's easy to imagine her as the motherly type, making breakfast and scolding you for your dirty mouth; but in reality beneath the modest exterior was the queen of the "dirty blues," and the writer of such classics as "Sloppy Drunk Blues," "Tricks Ain't Walkin' No More" and the "Bull Dyke Women's Blues."
Her most infamous song was "Shave 'Em Dry," a three-minute ode to her own humping prowess so filthy it would Lil' Kim blush.
A Few Choice Quotes
I got nipples on my titties, big as the end of my thumb;
I got somethin' between my legs'll make a dead man come.
You know it's a good song when the first two lines reference necrophilia and giant freak nipples.
Say I fucked all night, and all the night before baby;
And I feel just like I wanna fuck some more.
You know how people ask which dead celebrity you'd like to meet if you could? We submit Lucille Bogan for your consideration.
Move over Gandhi! You shithead!
Now your nuts hang down like a damn bell sapper;
And your dick stands up like a steeple;
Your goddamn asshole stands open like a church door;
And the crabs walks in like people.
Er, actually we take that back.
#5. King Solomon Hill Sentences a Woman to Devil Rape ("Whoopee Blues" - 1932)
"Whoopee Blues" is another song about a poor blues man having to deal with a mean mistreating woman. King Solomon Hill isn't one to settle for mere murder though, he wants his woman sent to hell to do it with the Devil--which strikes us as just a tad judgmental. We're no theologians, but we're pretty sure slashing your girlfriend to death with a razor is pretty much a one-way ticket to becoming Satan's bitch.
Don't worry, he's in Hell now.
A Few Choice Quotes
Tell me you been gone all day, that you may make whoopee all night;
I'm gonna take my razor and cut your late hours;
You wouldn't think I'd be servin' you right.
I said, Undertaker been here and gone, I gave him your height and size;
You be makin' whoopee with the Devil in Hell tomorrow night.
Apparently undertakers didn't ask a whole lot of questions back then.
"Height and size, check, and you're paying cash? No problem."
Baby, you done made me love you, now I got me for your slave;
From now you'll be makin' whoopee, deep in your lonesome grave.
Ahhh, now we see. He did it because he cared too much!
#4. Josie Miles Terrorizes a City ("Mad Mama Blues" - 1924)
Obviously violence against women was a bit of an unfortunate theme of early blues; but, as Josie Miles shows, female blues singers didn't shrink from a bit of insane violence either. Josie doesn't even need a reason. In "Mad Mama Blues" she's out to wreck the city like Godzilla in a cocktail dress and there's nothing you can do to stop it.
A Few Choice Quotes
Now I could see blood runnin';
Through the streets;
Could be everyone;
Layin' dead right right at my feet.
"Hello?! 9-11? Quick! You have to... oh God, she's coming!"
Give me gunpowder;
Give me dynamite;
Yes I'd wreck the city;
Wanna blow it up tonight.
"It's Josie Miles!"
I took my big Winchester;
Down off the shelf;
When I get through shootin';
There won't be nobody left.
"Send the police! The national guard! Before it's too... arrrraaghaghh!"
Actual photograph.


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WINTERSVILLE It was a picture-perfect day for area residents to get into the holiday spirit Sunday during the 2012 Wintersville Christmas parade.
The annual parade was one of the biggest the village has seen, as more than 500 participants lined up to march down Main Street from Riesbeck's to Kroger .
"The turnout today was amazing," said Marci Zangrando, Wintersville Special Events committee member and parade organizer. "We had people show up to participate who we didn't even know were coming, which I love."
Article Photos
ZOMBIE PATROL — The Wells Township Haunted House Zombie Assault vehicle made its way to the annual Wintersville Christmas parade Sunday. On hand for the event, were, from left, E.J. and Ernie Mozingo of Follansbee; Kevin Reiner and Marci Zangrando, members of the Wintersville Special Events Committee; Doug, Christine and Kayla Striker of Richmond; Christine and Hunter Ensell of Steubenville; Sean Moodie of Wintersville; Casey Nestor of Steubenville; Cody Linn of Brilliant; and Evan West of Richmond. - Jess Looman
Members of various fire departments were on hand as well as the Indian Creek High School band, members of Brenda Casey's School of Dance, Powers Martial Arts School students and Starkdale Presbyterian Church, just to name a few.
"The Wells Township Haunted House brought its zombie assault vehicle and one of our 4-H groups brought a painted miniature horse," Zangrando noted.
"I love this parade and everyone enjoys being a part of it," she continued. "We also saw a lot of spectators line the street, which is great. The weather was perfect and it was just a wonderful day for a parade."
Mayor Bob Gale echoed Zangrando's comments.
"It was a beautiful day, we saw a great turnout with lots of last-minute entries and the parade is definitely bigger than it was last year," he stated. "We saw a lot of people on the street as well. I want to wish a Merry Christmas and Happy Holidays to everyone."
Zangrando thanked members of the village for all of their hard work and cooperation.
"I want to first thank Mayor Bob Gale and my co-organizer Kevin Reiner for all of their help," she said. "Thank you to council for their support, as well as everyone who showed up to participate. We put a lot of effort into making this parade great and we hope everyone enjoyed themselves and had a great time."


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Denote Zermelo Fraenkel set theory without choice by ZF. Is the following true: In ZF, every definable non empty class A has a definable member; i.e. for every class $A = \lbrace x : \phi(x)\rbrace$ for which ZF proves "A is non empty", there is a class $a = \lbrace x : \psi(x)\rbrace$ such that ZF proves "a belongs to A"?
This is a really great question, and it connects with many issues in set theory. Some of these other topics are related to alternative readings of the question, where one wants to allow parameters in the definition, or where one is considering only definitions in the projective hierarchy, defining sets of reals.
But let me consider the most direct interpretation of the question, as you have asked it. The answer is that it is independent of ZF, in the sense that some models of set theory will have the desired property, and others will not.
First, there are models of ZFC in which every object in the universe is definable without parameters. This is true in the so-called minimal model of ZFC (the smallest Lα which models ZFC), as well as many other models. Indeed, every countable model of ZFC has an extension to a pointwise definable model. And if every object is definable, then of course it has your desired property that every definable set has definable elements.
Second, there are models without the desired property. Consider the forcing extensions L[G] obtained by adding a Cohen real over the constructible universe L. Now, consider the set of L-generic Cohen reals. This set is definable without parameters, but it can have no definable member, since no L-generic Cohen real can be definable in L[G], since the definable objects in L[G] must all be in L, as the forcing is almost homogeneous.
Thus, some models of set theory have your property, and some do not.
Finally, other considerations to think about are:
Responding to Ashutosh's first comment below. Let A(x) be the statement "x is a real number not in OD, or there or no such reals and x is a real". Thus, A is the class of reals not in OD, if this is nonempty, and the class of all reals, if they are all in OD. This class is provably nonempty, since either it consists of all reals, or a lot of reals, depending on whether the reals are all in OD or not. But A cannot provably contain any definable set, and indeed, any ordinal definable set, since it is consistent with ZFC that there are reals not in OD, and in this case A contains no ordinal definable elements. Thus, it has the properties you desired.
Nevertheless, my answer above shows that for any definable class A, it is consistent with ZFC that A has definable elements, since in the pointwise definable models, every element of A will be definable. So there cannot be a class A that is provably nonempty, such that ZF proves that every definition fails to define an element of A. Your requested property is exactly on the borderline, between ZF proving that definable sets are not in A, and ZF not proving that any particular definable set is in A.
I've realized that your property actually implies the Axiom of Choice. That is, I claim that any model of ZF with your definability property will necessarily also model AC. The reason is that if your property holds, then every set must be in OD, for if not, then we could define the set of minimal-rank sets not in OD, which would be definable, but have no minimal element. Thus, V=HOD, and this implies AC. There is always a definable well-ordering of HOD in V, where x precedes y if x is definable in a smaller Vα than y, or if they are defined in the same minimal Vα, then by an earlier definition.
Edit: The comments below, with remarks by F. G. Dorais (please vote up his answer), have now led to the realization that the requested property is equivalent to the assertion V=HOD. Namely, if a model satisfies V=HOD, then it has a definable well-ordering of the universe, and so every definable set will have a definable member (the least one in the definable order), and conversely, if the model thinks V is not HOD, then the set of minimal rank sets outside OD will be definable, but have no definable member.
In summary, the requested definability property IS first order expressible, and is equivalent to the axiom called V=HOD, introduced by Kurt Goedel.
There is a nice model-theoretic interpretation of the property you state, which I will call property (A).
By a theory of sets, I mean any theory $T$ with an extensional relation ${\in}$.
Fact 1. If $T$ is a theory of sets with property (A) then so does every extension of $T$.
Fact 2. If $T$ is a complete theory of sets with property (A) then $T$ has a unique prime model.
Fact 3. If $T$ is a complete theory of sets with a wellfounded prime model then $T$ has property (A).
The fact that ZF does not have property (A) follows from Fact 1 and the (simpler) observation that ZFC does not have property (A). However, Facts 2 and 3 give nice ways of finding extensions of ZF which do have property (A).
Without loss of generality, $T$ has enough closed terms so that if $T$ proves that the formula $\phi(x)$ has a unique solution then $T \vdash \phi(c)$ for some closed term $c$. (Otherwise, add a new constant $c$ for each such formula $\phi(x)$ together with the defining axiom $\phi(c)$. This is a conservative extension of $T$ with the desired property.) If $M$ is any model of $T$, then let $K_M$ be the substructure of $M$ whose ground set consists of all interpretations of the closed terms. Note that $K_M$ is completely determined by the complete theory of $M$.
Observe that if $\theta(z)$ is any formula, then $\psi(x) \equiv \forall z(z \in x \leftrightarrow \theta(z))$ has at most one solution by extensionality. Therefore, $T$ proves that $\phi(x) \equiv \psi(x) \lor (\forall z\lnot\psi(z) \land x = c_0)$ has exactly one solution (where $c_0$ is any fixed closed term). So there is a constant $c$ such that, in any model of $T$, if $\{z : \theta(z)\}$ is a set then $c = \{z:\theta(z)\}$. Of course, we always have $c = \{z:z\in c\}$. So in any model $M$ of $T$ the constants $c$ are precisely all the (parameter-free) instances of comprehension which exist in $M$.
In view of this, property (A) can be formulated as:
(A) For every formula $\phi(x)$, if $T \vdash \exists x\phi(x)$ then $T \vdash \phi(c)$ for some constant $c$.
Replacing $\phi(x)$ with $\phi(x) \lor \forall z\lnot\phi(z)$, we see that property (A) implies the seemingly stronger statement:
(A') For every formula $\phi(x)$ there is a constant $c$ such that $T \vdash \exists x \phi(x) \leftrightarrow \phi(c)$.
This reformulation of property (A) immediately implies Fact 1. By the Tarski-Vaught Test, property (A') says that if $M$ is any model of $T$ then the substructure $K_M$ of $M$ is an elementary submodel of $M$. Therefore, $K_M$ is the necessarily unique prime model of the complete theory of $M$, which establishes Fact 2.
For Fact 3, suppose that $T$ is complete and that $M$ is a wellfounded prime model of $T$. I will show that $M = K_M$, which immediately implies that $T$ has property (A). For the sake of contradiction, suppose that $M \neq K_M$. Since $M$ is wellfounded, there is an $a \in M \setminus K_M$ such that $a \subseteq K_M$. Since $M$ is prime, $a$ realizes a principal type $\phi(x)$. If $b$ also realizes $\phi(x)$, then $b \cap K_M = a$ since all elements of $K_M$ are definable. Therefore, $\phi(x)$ must imply $\forall z(\phi(z) \rightarrow x \subseteq z)$ over $T$. This means that $T$ proves that $\phi(x)$ has a unique solution, which contradicts the fact that $a \notin K_M$.
To partly answer Joel's comment whether wellfoundedness can be eliminated from Fact 3, Ali Enayat [Models of set theory with definable ordinals Arch. Math. Logic 44 (2005), 363-385. MR2140616] has shown that for a completion $T$ of $ZF$ the following three properties are equivalent:
A Paris model is a model where every ordinal is first-order definable. This is slightly weaker than property (A), but it is equivalent when $T \vdash V = OD$.
The question admits an answer that may be somewhat more "concrete" in the sense of having "applications", and seems somewhat relevant (correct me if I am wrong). Namely, Kanovei and Shelah have constructed definable models of the hyperreals where each of the nonstandard elements is actually undefinable. Thus the set of nonstandard elements is definable, but none of its members are.


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Carolina Chapter Annual Awards Banquet
Event Options
Event Information
Prestonwood Country Club, Cary, NC
From : Thursday, 29 March 2012 18:00
Until : Thursday, 29 March 2012 22:00
Until : Thursday, 29 March 2012 22:00
(UTC -05:00) Eastern Time (US & Canada), Bogota, Lima
Seats available:
Unlimited seats
Event Creator:
Event Details
Mark your calendars for the STC Carolina Chapter's premiere event of the year, the Awards Banquet at Prestonwood Country Club on March 29 at 6:00 p.m. We will honor the winners of the 2010-11 competitions and introduce our new Fellow and Associate Fellows.
This is a gala affair with early cocktails, elegant dinner and dessert, and tables full of great company. Don't miss this unique opportunity to see up-close some of our industry's best work. It's also a great way to support your peers.


text:
The title is somewhat deceptive: I know that $|\mathbb{Q}|=|\mathbb{Z}|.$
But suppose I wanted to compare the sets, knowing that they are of the same cardinality but still wondering if there was some sense in which Q could be larger. There are two usual ways of thinking about this:
$\mathbb{Z}\subset\mathbb{Q}$ but $\mathbb{Q}\not\subset\mathbb{Z}$
(sorry, there's no \subsetneq symbol on math.stackexachange) and
'For any distinct integers m,n there are infinitely many rationals between m and n.'
These methods both rely on the natural mapping between integers and rationals, to wit: $n\mapsto\frac n1$. What if we didn't have this correspondence, or were working with an application where it wasn't sensible?
That is, we still have the usual order < on Q and Z, and the usual operations + and *, but we no longer have a reason to privilege the usual mapping over, say, $n\mapsto\frac2n$ or a diagonal mapping.
I'm much less concerned about the 'answer' part of the answers than I am of the 'question' part: the main point is 'how could this question be made precise?' rather than answering such question.


text:
Free personalized radio that
plays the music you love
Maurice Ravel
born in Ciboure, Basses-Pyrénées, France, composed during the Modern period
Selected Discography
Track List: Daphnis & Chloe / Mother Goose Suite
Track List: First Choice: Ravel, Bolero
Track List: Maurice Ravel: The Piano Concertos; Valses nobles et sentimentales
Track List: Piano Favourites
Track List: Ravel : Piano Works - Vol. 2
Track List: Ravel Songs (Histoires Naturelles, Chanccons Madécasses)
Track List: Ravel: Bolero, Rapsodie Espagnole, Ma Mere L'oye
Track List: Ravel: Boléro
Track List: Ravel: Boléro / Ma Mère L'Oye (Complete Ballet) / Rapsodie Espagnole / Une Barque Sur L'océan / Alborada Del Gracioso - Berliner Philharmoniker / Pierre Boulez
Track List: Ravel: Complete Works For Solo Piano
Track List: Ravel: Daphnis Et Chloe
Track List: Ravel: Daphnis Et Chloé
Track List: Ravel: Daphnis Et Chloé; Rapsodie Espagnole; Pavane
Track List: Ravel: Orchestral Works
Track List: Ravel: Orchestral Works, Vol. 2 - Valses Nobles Et Sentimentales; Gaspard De La Nuit; Le Tombeau De Couperin; La Valse
Track List: Ravel: Piano Works
Track List: Ravel: Piano Works, Vol. 1
Track List: Ravel: Rapsodie espagnole, etc.
Track List: Ravel: Sonatas & Trio
Track List: Ravel: The Complete Solo Piano Music
Track List: Ravel: The Complete Solo Piano Music
Similar Composers
Don't read this because it actually works. You will get kissed on the nearest possible Friday by the love of your life. However if you do not post this comment on at-least 3 songs you will die in 2 days. Now you've started reading so don't stop. This is so scary. Put this at- least 5 songs in at-least 143 minutes when done press f6 and your lovers name will appear on the screen in big letters. This is so scary because it actually works.you will get k.
In addition to being very thin, the bio here neglects to share that in addition to his genius, Ravel was a humble, generous and kind colleague to any and all that came to him.
Also, I just caught the magical "L'enfant est les Sortileges" performed on Classical TV. Geez I love internet TV and radio!
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The venture will process 500 million cubic feet/day (MMcfd) of natural gas by the end of this year, Iraq's South Gas Co. Director said, compared with 240 MMcfd that was being captured in November 2011 when the JV was agreed. State-run South Gas Co. holds 51% of the $17.5 billion joint venture.
By HASSAN HAFIDH
JORDAN -- A joint venture (JV) between Iraq, Shell and
Mitsubishi will by the end of this year have doubled the amount
of natural gas it captures from three giant southern oil
fields, reducing waste and helping feed domestic power and
heating needs, a senior Iraqi oil official said.
The venture will process 500 million cubic feet/day (MMcfd)
of natural gas by the end of this year, Iraq's South Gas Co.
Director Ali Hussein Khudhier told Dow Jones
Newswires, compared with 240 MMcfd that was being captured
in November 2011 when the JV was agreed.
Mr. Khudhier was speaking shortly after Shell, Mitsubishi
and South Gas Co. officially launched their JV, Basra Gas.
State-run South Gas Co. holds 51% of the $17.5 billion joint
venture, while Shell has 44% and Mitsubishi 5%. The venture
will run for 25 years.
The three southern oil fields Rumaila, Zubair and West Qurna
1 produce 1.1 Bcfd of gas, but some 700 MMcfd of that is burned
off and wasted because infrastructure does not exist to harness
it, Mr. Khudhier said. "We are producing now 400 million
cubic feet a day and targeting 500 million cubic feet a day by
the end of 2013. We are expecting to reach 2 billion cubic feet
a day in 2017," as output from those fields expands, he
The Basra Gas Co. will produce natural gas for power stations,
and LPG for cooking and heating, Mr. Khudhier said. The JV will
meet all of Iraq's needs for LPG next year and expects to
export the fuel by the end of 2014. The venture will also
produce hydrocarbon liquids that will be sent to
refineries to process it into gasoline to raise its octane.
Mr. Khudhier said that the venture is expected to invest
some $1.5 billion in the three years after the signing on
upgrading existing gas facilities and training of
He expected the new venture to start exports of gas to the
international market in 2020, once it has met local needs. This
could be achieved through the construction of a FLNG plant and
terminal off the coast of the southern city of Basra in the
Arabian Gulf, to be built by Shell and Mitsubishi, he said.
The LNG facility is expected to cost around $3 billion, Mr.
Khudhier has said previously. The project would handle the export of
600 MMcfd of LNG, he said.
Iraq has estimated natural gas reserves of 112.6 Tcfd.
Dow Jones Newswires
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Is 2016 the peak for US gasoline demand?
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Priscilla Queen of the DesertSun, October 27, 2013 - 1:00 PMSegerstrom Center for the Arts Segerstrom Hall in Costa Mesa, CA
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text: Pinterest INDIAN NIGHT JASMINE FRAGRANCE MIST Select size 3.3 fl oz Quantity 1 2 3 4 5 6 7 8 9 10 Item 3.3 fl oz Price $15.00 Now $7.00 Add to bag Email When In Stock Please note that there is a limit on this product. Offers available on this product See Promotional Details Product description Escape to India with night jasmine. A light spritz infused with Indian Night Jasmine. This seductive, feminine blend of floral and oriental accords was inspired by a moonlit night in a Mughal garden. Oriental scent Light and refreshing Layer with other Indian Night Jasmine products to build the fragrance Shipping Close


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To the editor:
Attorney General Eric Holder's corruption of the American electoral system goes well beyond one precinct in Philadelphia and the racial intimidation he allowed a pair of armed Black Panthers to engage in. This corruption extends to the very concept of clean elections. Just more reason to impeach Holder is that during his time as attorney general, the Justice Department has provided cover for an attack on the basic integrity of our voting rolls.
Some background: the so-called motor-voter law passed in 1993 was designed to make it easier to register to vote. When someone goes into a state motor vehicle department, he or she is asked if they want to register. (The same question is asked in welfare agencies, food stamp offices and even methadone clinics under an addition to the law championed, incidentally, by the infamous 1960s radical academic Frances Fox Piven.) Another part of motor-voter, however, required states to keep their voter rolls free of dead and ineligible voters.
During the Bush administration, both provisions of the law - the voter registration requirement and the voter roll cleanup provisions - were enforced by the Justice Department. But once Holder, became attorney general, that changed.
In 2009, the employees of the voting section were assembled in a conference room and informed by the political appointees overseeing their work that the Justice Department would no longer be enforcing the voter roll cleanup provisions. This was surprising news, because the chief of the voting section, Christopher Coates, had just recommended opening up eight investigations into states that had more people on the voter rolls than people actually alive. The Holder Justice Department spiked those investigations, and the matter has remained in a black hole since then.
Today, more than 200 counties in America have more people on the voter rolls than people age 18 or older who are alive. Nothing could more deeply corrupt our electoral process than allowing voter rolls to be filled with ineligible dead voters, felons, fictitiously named voters and people who now live in other states; it's an invitation to those with an agenda to vote in the names of other people, or pay or coerce others to do so.
Voter rolls are allowed to be polluted by DOJ inaction, while DOJ action fills the rolls with welfare agency registrations. Rolls corrupted by the presence of felons and dead voters tend to benefit only one political party. Under Holder, the Justice Department walks a one-way political street.
Holder has made it plain that the DOJ Voting Section's agenda - to block voter ID requirements, to flood the rolls with welfare agency registrations and to aggressively enforce the Voting Rights Act against state measures to promote election integrity - is also his agenda. These acts are his top priority at Justice, where he has allowed a biased lawlessness to flourish and threaten the integrity of the presidential election, and also make it more likely that his friend Barack Obama will be re-elected.
Robert Yost


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Kingston MobileLite Wireless G3 / Pro 05-17-2016
With the popularity of mobile devices like smartphones, tablets and other handhelds (most of which have a fixed amount of storage built in), Kingston realized there was a need for an easy-to-access-external storage device. While cloud-based storage is often the most simple way to back up or transfer your digital media, there are occasions where access to the Internet or cloud service may not be available. The Kingston MobileLite Wireless devices address these concerns.
Gigabyte Z170X-UD5 Socket 1151 Motherboard 01-12-2016
With the introduction of the Intel Z170 chipset, we see a number of changes from the previous Z97 and several improvements. First and foremost is a new socket specification for the CPU, the 1151. Although not compatible with previous socket interfaces as far as the CPU is concerned, the overall dimensions are nearly identical, allowing for interchangeability with other 1150/1155/1156 compatible coolers. Another change is the switch to DDR4 memory. The DMI version 3.0 grants a greater number of PCI-E lanes between CPU and chipset, allowing manufacturers to implement more M.2 or SATA Express, as well as more USB 3.0 ports. For our Skylake build, Gigabyte has sent over their Z170X-UD5 motherboard.
Kingston HyperX Fury DDR4-2666 16GB Memory 11-10-2015
With the release of its Skylake Core processors, Intel introduced a new 1151 socket and Z170 chipset design. Rather than previous generation mainstream desktop configurations using DDR3, Intel opted to adopt the latest DDR4 also used in their high-end/enthusiast X99 platform. The DDR4 specification starts at 2133 and currently runs up to 3400 and will certainly continue higher as frequencies improve. For our Skylake build, Kingston sent over a 2x8GB kit of their latest HyperX Fury modules in DDR4-2666 speed. Join us as we take a closer look at the Kingston HyperX Fury DDR4-2666 memory.
Gigabyte Z97X-SLI Motherboard 04-13-2015
The Gigabyte Z97X-SLI is an SLI-capable motherboard priced very attractively for budget gamers. Due to licensing models with Nvidia, not all chipsets and boards are available to support SLI. Additionally, the board will support CrossFire natively as all Z97 boards do. This model also features high-end ALC1150 115dB SNR audio, a high quality headphone amplifier and a dedicated audio hardware zone to protect from interference. Intel Gigabit LAN, all solid caps and a new heatsink design round out the feature list.
Swiftech Komodo R9 LE GPU Water Block 11-30-2014
Swiftech recently released their latest GPU block design, the Komodo R9 LE and Komodo NV GTX9, which match the aesthetics of their Apogee XL. The Komodo blocks are a full coverage design and offer similar illumination effects to the Apogee XL, with changeable color scheme between red, blue, green and white. The base is manufactured of copper that has been chrome plated instead of the more commonly used nickel and the housing is acetal with transparent window.


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Dead Doctors Dont Lie 21 Jan 2015
Pharmacist Ben Fuchs fills in for Dr. Joel Wallach beginning the show discussing connective tissue. Outlining where connective tissue is found in the body. Citing various disease states that become possible when connective tissue breaks down. Stating what nutrients that are necessary for building strong connective tissue.Pearls of Wisdom
Doug Winfrey and Ben Fuchs discuss a news article concerning a study of sedentary lifestyles. Researchers in Canada analyzed the data from 47 previous studies of sedentary behavior. Finding that sitting several hours a day increases the risk of death from cardiovascular disease, cancer or type 2 diabetes. In fact sitting for 8 to 12 hours daily increases the risk of type 2 diabetes by 90%.Callers
Stephanie has a friend diagnosed with kidney failure and is on dialysis.
Donna's father has torn rotator cuffs in both shoulders.
Gary has a friend whose child has been diagnosed with arthritis.
Amir has a friend that wants to prevent getting breast cancer.
Trina's sister has had fibromyalgia for several years.
Call Dr. Wallach's live radio program weekdays from noon until 1pm pacific time at 831-685-1080 or toll free at 888-379-2552.


text:
I want to make a formula for the population size $N(t)$,
with the following ingredients: Let $N_t$ be the random variable denoting number of individuals at time $t$. I then call $$ N(t) =\mathbb{E}[N_t]=\int_{0}^{\infty}\mathbb{P}(N_t>z)dz$$
Let $N_0$ the the population-size at time $0$. Furthermore let $B(t)$ be the birthrate at time $t$, and $\pi(a',a)$ the probability that an individual of age $a$ survives upto $a'\geq a$. Assume that $$\lim_{a'\to\infty}\pi(a',a)=0 $$ Also let $n_0(a)$ be the initial age distribution.
So far all my attempts to write down a formula failed...I think im missing some simple logical insight.
So actually I want to have a formula for $\mathbb{P}(N_t>z)$ for all $z>0$.


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text:
I was reading a novel (The Stranger) that included the usage, "...speaking the French."; another (Radio Free Albemuth) included, "We were listening to an LP of the Jefferson Airplane at that moment."
I would have omitted "the" in both cases, or extended them to their full expression: "the French tongue/language", "the rock band Jefferson Airplane".
I can't see their decisions as purposefully colloquial, or as an intended stylistic tool. I suspect something larger is afoot - some dastardly rule.
(I'm also reminded of The Hague, which may or may not be related.)
Clearly I'm missing something. What?


text:
you are 100% right.... but this is something that will continue to haunt every one who fits bubble levels.
If you went to the trouble of offering an adjustable bubble, people would not want to pay the price, and even fewer would use it.
If they actually understood how their software functioned they would not worry at all about bubbles.


text:
This Is Your Cornfield On Coke
- What happens when you're high on coke and eluding cops in a cornfield? Some maize-y shit! [Daily Mail]
- Proving himself to be the poster child for special-needs reptiles everywhere, a one-eyed alligator attacked a golfer in Florida today. Both man and gator are doing fine. [CNN]
- Penguins used to be almost 5 ft tall!…


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TROY (October 1, 2012)—Work has resumed along Interstate 35 in Troy after the remains of a child and three adults were recovered from unmarked graves discovered at the end of August by a contractor doing construction on the southbound frontage road of the highway near Roberts Cemetery.
Work in that area was suspended while a team of Texas Department of Transportation archaeologists excavated the graves and removed the remains.
TxDOT was working with the Roberts Cemetery Board of Trustees to review cemetery records to see if they hold a clue to the identity of those whose graves were uncovered.
The four were evidently buried sometime in the late 1800s or early 1900s, the Texas Department of Transportation said Monday.
The remains will be stored temporarily in TxDOT's Cultural Resources Consultant firm's archaeological laboratory, in Austin for additional forensic study and then will be re-interred, the state said.
"In the interim we will continue with identification efforts, and any descendants identified would be free to petition the court for permission to remove them for burial elsewhere," TxDOT said in a press release Monday.
TxDOT said five people have expressed an interest in the possible identification of previously lost ancestors.


text:
At work, it is wrong to simply say 分かる to say that you understand something. In what situation should I opt to use one of the previously mentioned forms?
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かしこまりました is by far the most formal, and is a humble form (謙遜語). It says that you are inferior to the listener. Most specifically this should be used to interface with customers (hence why wait staff at a restaurant may say it).
承知しました is polite (〜します), but not humble. It is also appropriate to use with customers or superiors.
了解です is also polite in form, and is not humble, but has a certain curt feeling to it (this could be from its usage by the armed services as "Roger"). Informally friends and family often say/text 了解 (without the です) in a context where they want to say "Yep, got it".
The difference between the latter 2 is nuance, I suppose. Both seem to translate as "I acknowledge and understand". My gut feeling having worked in a Japanese office is that 了解です, while polite, isn't formal enough for customers. 承知 is better for customers or superiors.
I just want to add one point to other nice answers: 了解です is not a proper polite form for this meaning. The proper polite form is 了解しました. Saying 了解です instead of 了解しました is acceptable and many young people use it but decreases the formality level. It shows an attempt to be polite, but at least if it is used by an educated native speaker of Japanese, it may imply that the speaker is not bothering to use a proper form, which may decrease the politeness. If you go this informality one step further, you get 了解っす which is even less formal (see another question).
かしこまる/承知する are used mostly for answering requests from superiours (bosses, clients, etc.). Like, "Yes, I understand what you're asking me to do (and I'll do it)." 分かる just implies you mentally understand.
了解 is not formal as far as I know, and I hear it quite often. I learned it to mean almost like "Roger!" or "copy that".
Mostly the usage. Remember, first, that Sino-Japanese is always more formal than native Japanese.


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5 (Thankfully) Extinct Giant Versions of Modern Animals
As we've pointed out before, a lot of the animals we see today used to be much larger in olden times. Like, a LOT larger. Also, a lot meaner, hungrier, and ripped-straight-from-our-nightmares-ier. So, in case you've been sleeping well these past few months, here's a brand-new set of prehistoric colossi to help you stare at the ceiling all night, being thankful that mankind showed up so late in the evolutionary parade. Otherwise you could unexpectedly run into ...
#5. Megapiranha paranensis, the Giant Piranha
Jupiterimages/Photos.com/Getty Images
Modern piranhas are scary enough in their own right. If you have no sense of self-preservation and decide to swim in one of their rivers, these bastards will teach you never to do so again, at the not-unreasonable cost of toes, genitals, or whatever other chunks of flesh look appetizing. But not all that many people have actually died at the hands of the tiny, modern version of these ravenous machines.
That's why the idea of taking these little balls of hate and turning them into humongous slimy chainsaws with bites powerful enough to crush bone is a popular trope among direct-to-video horror movies. Which is funny, because nature beat us to the punch a good long time ago. Behold:
Via Silk666
We can't be the only ones who insert the sound of angels singing when we see a big scary mouth open like that.
That is Megapiranha, which lived 8 to 10 million years ago in the rivers of South America and grew to a meter in length, four times that of the cow skeletonizers of today. And if it ever finds out we gave it a lazy-ass name like "megapiranha," it'll find a way to revive itself and tear us to shreds in one or two bites.
Via Prehistoric-wildlife.com
Pretty sure this diagram is mistaken on the whole "swimming with man-eaters" thing.
See, not only was Megapiranha huge, but its bite was among the strongest in history, with a force-to-body-weight ratio unmatched by even the mightiest dinosaur. Whereas modern-day piranhas peak at a bite force of 70 pounds, a Megapiranha is estimated to have bitten with a force of 1,000 pounds.
To further illustrate how insane that is, a T. rex could deliver a bite force of just over 3,000 pounds, roughly three times that of a Megapiranha. Only thing is, the T. rex weighed seven tons. The Megapiranha? Twenty pounds, maybe 25 if it started slacking off at the gym. If it were as big as T. rex, it probably could've chewed up an Apatosaurus and swallowed it whole, while barely breaking a sweat.
And here's our friend's bite when compared to other feared creatures of the sea:
Via Oceanminds.com
There were 47 more entries, but the Megapiranha ate them all.
See how no other fish comes close? Precisely. It's a damn good thing humans hadn't been invented yet, because we probably wouldn't have lasted two years before evolving into steaming chunks of piranha shit.
#4. Giant Birds of Prey
Tom Brakefield/Stockbyte/Getty Images
One thing we rarely have to worry about today is giant birds of prey swooping down and flying away with our children. That's mainly because there are no Haast's eagles around. But that is a very recent development.
Via Badassoftheweek.com
This guy is building one from memory.
The average Haast's eagle had a 10-foot wingspan. Its gigantic claws would impale, rip, and tear anything it deemed delicious; that something was usually quite big and meaty, like the moa bird. If the below picture doesn't look impressive to you, try to keep in mind that the moa it's about to munch on are 10 to 12 feet tall:
Via John Megahan
My God, it's already eaten their arms off!
What else did it eat? Apparently, us. The Haast's eagle definitely lived among humans, up until 1400 or so. So yes, a mere 600 years ago, these things roamed the skies, snatching our babies, and maybe even our mommies and daddies. Here's a comparison between the Haast's talons and a modern eagle, if you want to imagine them clamping down on your neck:
PLOS via Wikipedia
Damn, it's more than 10 cinnamon long!
New Zealand folk tales speak in hushed whispers of giant man-eating birds ruling the land, and these stories have all but been proven to be about Haast's eagle. It was only after the Maori tribe hunted its primary food source -- the moa -- to extinction that Haast's eagle met its own demise.
Of course, if you keep rewinding the clock by millions of years, you find murderous birds getting bigger, and stranger. You could, for instance, step out of your time machine and find yourself running from Gastornis:
Via BBC
Or finding a way to kill it, because just one hot wing from that bastard would feed a whole Super Bowl party.
That ridiculous dinosaur/dodo hybrid stood over 6 feet tall, could in no way fly with those stupid little stump wings, and had a keen taste for meat.
Giant birds like these were among the Earth's top predators after the extinction of the dinosaurs, gulping down our poor mammalian ancestors in the forests of Europe some 50 million years ago. If they'd tried bringing one of these bastards back to life in Jurassic Park, the audience would have been laughing their asses off, right up until this Gastornis severed Jeff Goldblum's head with one effortless snap of its beak.
Via Everythingdinosaur
OK, we'd still be laughing, but in a horrified sort of way.
#3. Megalania prisca, the Giant Lizard
Jupiterimages/Photos.com/Getty Images
Komodo dragons of today are not to be fucked with, reaching 10 feet in length and weighing up to 150 pounds. Also, they're quite poisonous and can very much kill you. Now take one of those things and double its size. Then make it anywhere from five to 30 times as heavy. That's the Megalania, aka the giant ripper lizard. And us modern-day humans just missed them.
Via Thecryptozoologist
What a shame; they looked playful.
As recently as 40,000 years ago, Megalania patrolled the Australian Outback, living up to its "ripper" name with little difficulty. And while anyone who knows their history laughs at the idea of humans and dinosaurs walking the Earth at the same time, your Australian ancestors experienced something pretty damned close -- it's likely the earliest human settlers in the area ran into these monsters, and then promptly ran away. As well they should have.
And unlike today's Komodo dragons, whose saliva may or may not be somewhat poisonous, Megalania were absolutely venomous, and it was potent as hell. If the soft marsupials of the time ever questioned who ruled the roost, Megalania and its toxic spit was more than happy to remind them. Again and again, if need be.
Via Wikimedia Commons
And, as shown here, they could live for tens of thousands of years without skin or internal organs.
Now remember: All of this occurred 40,000 measly years ago. As far as history is concerned, we crashed the party a mere millisecond after these gigantic fuck-yous from Mother Nature stepped out for a smoke. They were here so recently, in fact, that some people think they still exist. Those people better hope to God that they're wrong. With the seemingly infinite number of murderous creatures that Australia currently sports, do we really need a real-life poisonous Smaug making some grand return to ensure our final days on Earth are completely unpleasant?


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I'm having an issue with the internal speaker not muting when speakers are plugged into the headphone jack. if i unplug from the headphone jack and then plug in again it mutes. but once i reboot it unmutes.
I've tried gnome-alsa-mixer. it works until i reboot
I've tried hda-jack-retask. it works even after reboot but not after a complete shutdown.
i'm on 12.10 with kernel 3.5.0-17-generic. and my card is a Realtek ALC275
any ideas? maybe i need to put a new line in alsa-base.conf??


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Every real symmetric matrix has at least one real eigenvalue. Does anyone know how to prove this elementary, that is without the notion of complex numbers?
If "elementary" means not using complex numbers, consider this.
But of course any proof requires some topology. The standard proof requires Fundamental theorem of Algebra, this proof requires existence of a minimum.
How about Jacobi's proof?
See, e.g., Folkmar Bornemann, ``Teacher's Corner - kurze Beweise mit langer Wirkung,'' DMV-Mitteilungen 3-2002, Seite 55 (in German, sorry). Sorry, I don't have the original reference.
The idea is simple, define $\Sigma(A)=\sum_{i=1}^n\sum_{j=i+1}^n a_{ij}^2$ for $A=(a_{ij})$ a symmetric real matrix. Then minimize the function $O(n)\ni J \mapsto \Sigma(J^TAJ)$ over the orthogonal group $O(n)$. The function is continuous and bounded below by zero, and $O(n)$ is compact, so the minimum is attained. But it can not be strictly positive, because if there is an $a_{ij}\not=0$, $i\not=j$, then you can make it zero by a rotation that acts only on the $i$-th and $j$-th row and column, so that it decreases $\Sigma$ (this is a simple little calculation with $2\times 2$ matrices). Therefore the minimum is zero and it is attained in a matrix $J$ for which $J^TAJ$ is diagonal.
The eigenvalues of $A$ are now the (diagonal) entries of $J^TAJ$. No complex numbers are used, but you have to know that the minimum exists. We get the existence of an orthonormal basis consisting of eigenvectors with real eigenvalues.
Let me give it a try. This one only uses the existence of a maximum in a compact set, and the Cauchy-Schwarz inequality.
Let $T$ be a selfadjoint operator in a finite dimensional inner product space.
Claim: $T$ has an eigenvalue $\pm\|T\|$.
Proof: Let $v$ in the unit sphere be such that $\|Tv\|$ attains its maximum value $M=\|T\|$. Let $w$ also in the unit sphere be such that $Mw=Tv$ (which is like saying that $w=\frac{Tv}{\|Tv\|}$, except in the trivial case $T=0$).
This implies that $\langle w,Tv\rangle=M$. In fact, the only way that two unit vectors $v$ and $w$ can satisfy this equation is to have $Tv=Mw$. (Since we know that $\|w\|=1$ and $\|Tv\|\leq M$, the Cauchy-Schwarz inequality tells us that $|\langle w,Tv\rangle|\leq M$, and the equality case is only attainable when $Tv$ is a scalar multiple of $w$, with the scalar $\lambda$ determined by the computation $M=\langle w,Tv\rangle=\langle w,\lambda w\rangle=\lambda\langle w,w\rangle=\lambda$.)
But by selfadjointness of $T$, we also know that $\langle v,Tw\rangle=M$, and this implies, by the same Cauchy-Schwartz-equality reasoning, that $Tw=Mv$.
Now, one of the two vectors $v\pm w$ is nonzero, and we can compute
$T(v\pm w)=Tv\pm Tw=Mw\pm Mv=M(w\pm v)=\pm M(v\pm w)$.
This concludes the proof that $\pm\|T\|$ is eigenvalue with eigenvector $v\pm w$. The reality of the other eigenvalues can be proved by induction, restricting to $(v\pm w)^\bot$ as in the usual proof of the spectral theorem.
Remark: The proof above works with real or complex spaces, and also for compact operators in Hilbert spaces.
Comment: I would like to know if this proof can be found in the literature. I obtained it while trying to simplify a proof of the fact that if $T$ is a bounded selfadjoint operator, then $\|T\|=\sup_{\|v\|\leq 1} \langle Tv,v\rangle$ (as found, for example, on p.32 of Conway J.B., "An Introduction to Functional Analysis"). In the case of non-compact operators, one can only prove that $T$ has as an approximate eigenvalue one of the numbers $\pm\|T\|$. The argument is similar to the one above, but knowledge of the equality case of Cauchy-Schwarz is not enough. One has to know that near-equality implies near-dependence. More precisely, let $v$ be a fixed unit vector, $M\geq 0$ and $\varepsilon\in[0,M]$. If $z$ is a vector with $\|z\|\leq M$ such that $|\langle v,z\rangle|\geq \sqrt{M^2-\epsilon^2}$, then it can be proved that $z$ is within distance $\varepsilon$ of $\langle v,z\rangle v$.
Exercise: Follow the proof (find the possible vectors $v$ and $w$) for the cases in which $T:\mathbb R^2\to\mathbb R^2$ is given by any of the matrices $\begin{pmatrix}2&0\\0&1\end{pmatrix}$, $\begin{pmatrix}-2&0\\0&1\end{pmatrix}$, $\begin{pmatrix}2&0\\0&-2\end{pmatrix}$. This may make clear how the proof was made. Notice that $v$ and $w$ are already eigenvectors in some ("most") cases.
This is just the details of the first step of Alexander Eremenko's answer (so upvote his answer if you like mine), which I think is by far the most elementary. You only need two facts: A continuous function on a compact set in $R^n$ achieves its maximum (or minimum), and the derivative of a smooth function vanishes at a local maximum. And there's no need for Lagrange multipliers at all.
Let $C$ be any closed annulus centered at $0$. The function $$ R(x) = \frac{x\cdot Ax}{x\cdot x}, $$ is continuous on $R^n\backslash\{0\}$ and therefore achieves a maximum on $C$. Since $R$ is homogeneous of degree $0$, any maximum point $x \in C$ is a maximum point on all of $R^n\backslash\{0\}$. Therefore, for any $v \in R^n$, $t = 0$ is a local maximum for the function $$ f(t) = R(x + tv). $$ Differentiating this, we get $$ 0 = f'(0) = \frac{2}{x\cdot x}[Ax - R(x) x]\cdot v $$ This holds for any $v$ and therefore $x$ is an eigenvector of $A$ with eigenvalue $R(x)$.
Another elementary proof, based on the order structure of symmetric matrices. Let me first recall the basic definitions and facts to avoid misunderstandings: we define $A\ge B$ iff $(A-B)x\cdot x\ge0$ for all $x\in\mathbb{R}^n$). Also, a lemma:
A quick proof passes through the square root of $A$: $(Ax\cdot x)=\|A^{1/2} x\|^2 \ge \|A^{-1/2}\|^{-2} \| x\|^2$; one has to construct $A^{1/2}$ before, without diagonalization, of course.
We may rephrase the lemma saying equivalently: if $A$ is positive but, for any $\epsilon >0$, the matrix $A-\epsilon I$ is not, then $A$ is not invertible.
As a consequence, $\alpha_*:=\inf_{|x|=1}(Ax \cdot x)$ is an eigenvalue of $A$, because $A-\alpha_*I$ is positive and $(A-\alpha_*I)-\epsilon I$ is not (and $\alpha ^ *:=\sup _ {|x|=1}(Ax \cdot x)$ too, for analogous reasons).
The complete diagonalization is then performed inductively, as in other proofs.
This is quite an interesting question, perhaps a research problem. I think an elementary answer should be a high school algebra answer in the sense of Abhyankar and it would have to be in the spirit of what follows. But first a little story.
I was teaching linear algebra and had just covered eigenvalues and characteristic polynomials but was not yet at the chapter on the spectral theorem for real symmetric matrices. I was looking for problems to assign for my students as homework in the textbook we were using. One of the exercises was to show that a real matrix $$ A=\left[ \begin{array}{cc} \alpha & \beta \\\ \beta & \gamma \end{array} \right] $$ only had real eigenvalues. Not too hard. Write the characteristic polynomial $$ \chi(\lambda)=det(\lambda I-A)=\lambda^2-(\alpha+\gamma)\lambda+\alpha\gamma-\beta^2 $$ then its discriminant is $$ \Delta=(\alpha+\gamma)^2-4(\alpha\gamma-\beta^2)=(\alpha-\gamma)^2+4\beta^2\ge 0\ . $$ Hence two real roots.
The next problem in the book was to do the same for $$ A=\left[ \begin{array}{ccc} \alpha & \beta & \gamma\\\ \beta & \delta & \varepsilon \\\ \gamma & \varepsilon & \zeta \end{array} \right] $$ and (silly me) I also assigned it...
Here is the solution in the 3X3 case. All roots are real if the discriminant (for a binary cubic) is nonnegative. The discriminant of the characteristic polynomial is $$ \Delta = (\delta \varepsilon ^{2} + \delta \zeta ^{2} - \zeta \delta ^{2} - \zeta \varepsilon ^{2} + \zeta \alpha ^{2} + \zeta \gamma ^{2} - \alpha \gamma ^{2} - \alpha \zeta ^{2} + \alpha \beta ^{2} + \alpha \delta ^{2} - \delta \alpha ^{2} - \delta \beta ^{2})^{2} \\\ \mbox{} + 14(\delta \gamma \varepsilon - \beta \varepsilon ^{2} + \beta \gamma ^{2} - \alpha \gamma \varepsilon )^{2} \\\ \mbox{} + 2(\delta \alpha \gamma + \delta \beta \varepsilon + \delta \gamma \zeta - \gamma \delta ^{2} - \gamma \varepsilon ^{2} + \gamma ^{3} - \alpha \beta \varepsilon - \alpha \gamma \zeta )^{2} \\\ \mbox{} + 2(\delta \beta \gamma + \delta \varepsilon \zeta - \varepsilon ^{3} + \varepsilon \alpha ^{2} + \varepsilon \gamma ^{2} - \alpha \beta \gamma - \alpha \delta \varepsilon - \alpha \varepsilon \zeta )^{2} \\\ \mbox{} + 2(\zeta \alpha \beta + \zeta \beta \delta + \zeta \gamma \varepsilon - \beta \varepsilon ^{2} - \beta \zeta ^{2} + \beta ^{3} - \delta \alpha \beta - \alpha \gamma \varepsilon )^{2} \\\ \mbox{} + 14(\zeta \beta \varepsilon - \gamma \varepsilon ^{2} + \gamma \beta ^{2} - \alpha \beta \varepsilon )^{2} \\\ \mbox{} + 2(\zeta \beta \gamma + \delta \varepsilon \zeta - \varepsilon ^{3} + \varepsilon \alpha ^{2} + \varepsilon \beta ^{2} - \alpha \beta \gamma - \alpha \delta \varepsilon - \alpha \varepsilon \zeta )^{2} \\\ \mbox{} + 14(\varepsilon \beta ^{2} + \zeta \beta \gamma - \delta \beta \gamma - \varepsilon \gamma ^{2})^{2} \\\ \mbox{} + 2(\zeta \alpha \beta + \zeta \beta \delta + \zeta \gamma \varepsilon - \beta \gamma ^{2} - \beta \zeta ^{2} + \beta ^{3} - \delta \alpha \beta - \delta \gamma \varepsilon )^{2} \\\ \mbox{} + 2(\alpha \gamma \zeta + \zeta \beta \varepsilon - \gamma ^{3} + \gamma \beta ^{2} + \gamma \delta ^{2} - \delta \alpha \gamma - \delta \beta \varepsilon - \delta \gamma \zeta )^{2}\ . $$
This formula comes from a paper by Ilyushechkin in Mat. Zametki, 51, 16-23, 1992.
I suspect the elementary answer should be as follows. First find a list of invariants or covariants of binary forms $C_1,C_2,\ldots$ such that a form with real coefficients has only real roots iff these covariants are nonnegative. Apply this to the characteristic polynomial of a general real symmetric matrix and show that you get sums of squares. I suppose these covariants, via Sturm's sequence type arguments, should correspond to subresultants or rather subdiscriminants. This seems also related to Part 2) of Godsil's answer.
Edit: Another recent research reference which relates to the above sum-of-squares formula is the article The entropic discriminant by Sanyal, Sturmfels and Vinzant.
Edit 2: I just found out that the problem I mentioned above has been completely solved! See Proposition 4.50 page 127 in the book by Basu, Pollack and Roy on real algebraic geometry. The connection with classical invariants/covariants of binary forms is not apparent but it is there: their proof is based on subresultants and subdiscriminants which are leading terms of $SL_2$ covariants.
Here's one inspired by the Schur-Horn theorem and by Jacobi's proof as described by Uwe Franz: Fix real numbers $a_1>a_2>\cdots>a_n$. For $X$ an $n \times n$ symmetric matrix, define $\psi(X) = \sum_i a_i X_{ii}$.
Let $M$ be the matrix we're trying to diagonalize. Maximize $\psi(J M J^T)$ over $J$ in $SO(n)$. Since $SO(n)$ is compact, $\psi$ has a maximum value; let $X = JMJ^T$ achieve this maximum. For any skew symmetric matrix $Y$, we compute: $$\psi \left( \exp(Y) X \exp(-Y) \right) =\psi \left( X + (YX-XY) + O(|Y|^2) \right) = $$ $$\psi(X) + \sum_{i,j} \left(a_{i} Y_{ij} X_{ji} - a_i X_{ij} Y_{ji} \right) +O(|Y|^2) = \psi(X) + 2 \sum_{i<j} (a_i-a_j) Y_{ij} X_{ij} +O(|Y|^2).$$ (Recall that $X$ is symmetric and $Y$ is skew-symmetric.) So $$\left. \frac{\partial \psi}{\partial Y_{ij}} \right|_{Y=0} = 2 (a_i - a_j) X_{ij}.$$ We see that, at a critical point, all the off diagonal $X_{ij}$ are zero. One can also compute that the Hessian is positive definite only when $X_{11} > X_{22} > \cdots > X_{nn}$. So the maximum occurs at the unique diagonalization for which the eigenvalues appear in order. (If there are repeated eigenvalues, then there is still a unique maximum on the orbit $J M J^T$, but it is achieved by multliple values of $J$, so the Hessian is only positive semi-definite.)
We can do it in two steps.
Step 1: show that if $A$ is a real symmetric matrix, there is an orthogonal matrix $L$ such that $A=LHL^T$, where $H$ is tridiagonal and its off-diagonal entries are non-negative. (Apply Gram-Schmidt to sets of vectors of the form $\{x,Ax,\ldots,A^mx\}$, or use Householder transformations, which is the same thing.)
Step 2. We need to show that the eigenvalues of tridiagonal matrices with non-negative off-diagonal entries are real. We can reduce to the case where $H$ is indecomposable. Assume it is $n\times n$ and let $\phi_{n-r}$ the the characteristic polynomial of the matrix we get by deleting the first $r$ rows and columns of $H$. Then $$ \phi_{n-r+1} = (t-a_r)\phi_{n-r} -b_r \phi_{n-r-1}, $$ where $b>0$. Now prove by induction on $n$ that the zeros of $\phi_{n-r}$ are real and are interlaced by the zeros of $\phi_{n-r-1}$. The key here is to observe that this induction hypothesis is equivalent to the claim that all poles and zeroes of $\phi_{n-r-1}/\phi_{n-r}$ are real, and in its partial fraction expansion all numerators are positive. From this it follows that the derivative of this rational function is negative everywhere it is defined and hence, between each consecutive pair of zeros of $\phi_{n-r-1}$ there must be a real zero of $\phi_{n-r}$.
Just found in Godsil-Royle's Algebraic graph theory: One first proves that two eigenvectors associated with two different eigenvalues are necessarily orthogonal to each other (pretty standard), then observes that if $u$ is eigenvector associated with eigenvalue $\lambda$, then $\bar u$ is eigenvector associated with eigenvalue $\bar\lambda$. Now the eigenvalues $\lambda,\bar\lambda$ cannot be different, for otherwise by the above observation $0=u^T u=\|u\|^2$ although $u\not=0$.
(It does contain complex numbers, but is still amazingly straightforward).
The fact that real symmetric matrix is ortogonally diagonalizable can be proved by induction. The crucial part is the start. Namely, the observation that such a matrix has at least one (real) eigenvalue. But this can be done in three steps.
(1) An easy observation (using direct matrix multiplication) shows that all columns of a matrix $\mathbf{A}\in\mathbb{R}_{m\times n}$ are orthogonal to any vector $z\in\mathbb{R}_{m\times 1}$ iff $z$ belongs to the null space of the transpose $\mathbf{A}^{\sf T}$, i.e. $\mathcal{N}(\mathbf{A}^{\sf T})=\mathcal{R}(\mathbf{A})^{\perp}$.
(2) If $\mathbf{S}^{\sf T}=\mathbf{S}$ and $\mathbf{S}x \neq 0$ for every $x\neq 0$, then the dot product $\langle\mathbf{S}x,x\rangle\neq 0$ for any $x\neq 0$ as well. Otherwise, if $\langle \mathbf{S}z,z\rangle=0$ for some $z\neq 0$, then we have, using (1), $z\in\mathcal{R}(\mathbf{S})^{\perp}=\mathcal{N}(\mathbf{S}^{\sf T})= \mathcal{N}(\mathbf{S})$, i.e. a contradiction $z\ne0$ and $\mathbf{S}z=0$.
(3) If matrix $\mathbf{A}=\mathbf{A}^{\sf T}\in\mathbb{R}_{n\times n}$ has no (real) eigenvalue, then $(t\mathbf{I}-\mathbf{A})x\neq 0$ for any $x\neq 0$ and every $t\in\mathbb{R}$. Consequently, according to (2), we have $\langle(t\mathbf{I}-\mathbf{A})y,y\rangle\neq 0$ for fixed $y\neq 0$ and $t\in\mathbb{R}$. Therefore $t\|y\|^2 -\langle\mathbf{A}y,y\rangle \neq 0$ for every $t\in\mathbb{R}$, which is impossible.


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Committee Site Visits
Members of the Planning Committee may sometimes defer a planning application until its next meeting so that they can visit the site.
Why might it be deferred for a site visit?
Members may wish to visit the site before making a decision on an application so that they can have a better understanding of the effects an application may have on an area or its immediate neighbours.
When will the site visit be?
Site visits usually take place on a Thursday afternoon as soon as practical after the application is deferred. If you have commented on an application, you will be notified of the date of the site visit.
Can I attend?
Anyone with an interest in the application is welcome to attend the site visit.
What happens at the site visit?
A planning officer will explain the application and reasons for the visit. They will also make a note of all the issues discussed during the visit.
When on site, Members of the Committee will:-
- Allow the applicant to explain their proposals
- Visit all areas of concern to members of the public (this may entail entering adjoining properties)
- Hear the concerns of local residents
What happens then?
The application will be discussed at the next Committee meeting. Members will discuss what they saw at the site visit and any comments made by interested parties.
Full details of the site visit procedures can be found in the Code of Practice below.
B11 - Planning Code (Size: 127 Kb Type: PDF )
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Last Updated: 18th November 2013


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Trail Map of California Coastal Trail - San Francisco
Hike along the Coastal Trail from the Cliff House at Lands End to the Golden Gate Bridge
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8.9 miles, Half day
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Dog Friendly
This section of the California Coastal Trail runs from the Cliff House near the Sutro Baths all the way to the Golden Gate Bridge and passes many landmarks including Baker Beach, the Seacliff area, Marshall Beach, and the Batteries to Bluffs Trail.
This area of San Francisco is unknown to so many residents, even though it is one of the most beautiful areas in all of the bay area. You will walk along trails on the cliffs above the water where you can watch crashing waves, sailboats, and the occasional surfers, all while having perfect views of the Golden Gate Bridge and the Marin Headlands across the bay--it's hard to beat.
There are several small detours to small beaches along the trail as well, if you care to take a relaxing break.
There are several points along the trail where you can leave the trail or turn back to hike a shorter loop. It is 9 miles from start to finish, so it would be wise to arrange a ride back (there are busses at the end that can take you back to Golden Gate Park).
Park in the parking lot just off of Point Lobos Avenue and walk north toward the trailhead. If this lot is full, you can turn north onto El Camino Del Mar where there is a larger parking lot.
Seal Rocks
In 1887, at Adolph Sutro's urging, Congress designated Seal Rocks a marine preserve--one of the nation's first--to protect the colony of sea lions that resided there. The colony has since moved around the Golden Gate to Pier 39, but you can still spot sea lions and seals off Lands End.
Sutro Baths
The Sutro Baths were a large public bath house constructed in 1896 and served the San Francisco community until 1966 when the building burned down (how ironic).
Take the short path down to the shore to get a closer look. There is a small tunnel you can walk through under the main platform to see the waves crashing on the nearby rocks.
When you are done, continue back up the path to meet up with the main coastal trail.
Vista Points
Sweeping views of the entire Golden Gate are offered along the trails at Lands End. Vistas include the Marin Headlands, Golden Gate Bridge, Pacific Ocean and the Mile Rock lighthouse, all of which can be seen from the California Coastal Trail.
Three overlooks along the trail offer seating and photo opportunities. At Eagles Point, at the east end of Lands End, a wooden viewing platform and staircase on the cliff face provides a sweeping view of the entire Golden Gate--the entrance to San Francisco Harbor.
Drive-in vista points are located on El Camino del Mar and the Lands End parking lot and offer similar views. Continue north along the Lands End overlook with its view of the Sutro Baths, Cliff House, Seal Rocks and Pacific Ocean to the trailhead where more information is available.
Behind you, you will find a staircase leading to the USS San Francisco Memorial, which you will visit later in the tour.
About halfway along the trail you will reach an optional detour that allows you to walk to the Labyrinth. The Labyrinth was created by a local artist and also offers great views of the Marin Headlands across the bay and the Golden Gate Bridge in the distance.
Take your time to walk the maze, but don't cheat!
If you want more time at the beach, there is a trail leading to the shore where you can get closer to the waves crashing on the rocks.
Lands End meets Sea Cliff
Lands end stops here and the wealthy Sea Cliff area begins. Follow the sidewalk and stay left and you will soon arrive at an entrance to Baker Beach.
Steps to Baker Beach
Here you will find some steps leading down to the start of Baker Beach. The next section can be walked along the beach or on harder ground near the parking lots.
Take these stairs up to the main road where you will turn to your left and find the beginning of the Batteries to Bluffs trail.
Batteries and Views
Soon you will reach the batteries and you can take some time to explore and walk around them.
You will also have great views of the beach below, as well as of the Golden Gate Bridge and Marin Headlands in the distance to the right.
Marshall Beach
You can take a short sidetrip down to Marshall Bridge where you can watch the waves crash on the sand or just relax and take a break.
View Point
Here is another great view of the Golden Gate Bridge. The trail turns to the right here, where it will end at a parking lot. You can choose to continue along the Coastal Trail to the Golden Gate Bridge, or turn back the way you came.
Golden Gate Bridge
This section of the Coastal Trail ends at the Golden Gate Bridge (but the full trail crosses the bridge and continues into Marin Headlands, but that hike is for another day.
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Discover useful tools for those who submit data for students with disabilities to the Student Accountability Information System (SAIS). Find relevant SAIS codes, tables that illustrate which needs may be reported with which service codes, and more.
ASDB Student Reporting
FAQs: Least Restrictive Environment (LRE) for Students with Disabilities Ages 6-21
Flowchart: Determining Preschool Service Codes for SAIS Reporting
Information Regarding Replacement of "Mental Retardation" with "Intellectual Disability"
Instructions for updating contact information in the LEA Profile / Enterprise Database
Need and Service Code Eligibility
SAIS Online for Secure Care Facilities
SPED 09-10 Memo: Developmental Delay Need Category
SPED School Concurrency and Federal Primary Need Indicator
Student Accountability Information System (SAIS)
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A Roaring Affair
A vintage Lincoln Zephyr convertible parked in front of the Big Canyon Country Club was just one of many details that transported more than 440 ladies and gents back to the era of hot jazz and flappers at CHOC Children's Glass Slipper Guild Gala, which raised more than $330,000 for the nonprofit pediatric medical institution. Take the Ole Smokey Moonshine Bar. Here, guests like Blaine and Renetta Caya, and Janet and Jim Rome were served sparkling flutes of bubbly before sitting down to dinner in a beautifully draped tent outfitted with speakeasy-inspired decor. (What else?) A live auction, meanwhile, wowed with winnings such as a ski trip to Aspen, and guests later put their hands together for group member Kara Kipp, who shared her son's inspiring story of surviving cancer with the help of CHOC. And that's the cat's meow! –Photos by Juls Megill Photography


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BELTON (October 7, 2012)--Those who know 57-year-old Curtis "Bubba" Knight say he has made an impact on everyone he's met, especially his family.
"Bubba" is funny, charming, inspirational and so much more.
To his family, he's anything but someone with Down syndrome.
"He's a tremendous gift from God," said Marjorie Knight, Bubba's mother.
"God gave him to us to take care of, and we tried our best to do that."
Bubba's parents, Bill and Marjorie Knight have been taking care of their son for the past 57 years.
"They've taken care of Bubba since he was born when people said they shouldn't," said Sandra Rourke, Bubba's sister.
Advocates for Down syndrome awareness say too many people do not understand Down syndrome and the special needs for those who have it.
Down syndrome usually occurs when a child has an extra copy of chromosome 21.
The extra chromosome causes problems with the way the body and brain develop, often causing mental and social issues.
Through it all, Bubba has made an incredible impact on the entire Knight Family.
"He's the oldest of five, and we're a very close knit family, so he has been an inspiration to us," said Gary Knight, the youngest sibling.
Advocates for children with special needs say reminding the children about their uniqueness and talents is very important to their lives and well-being.
"That's one of the things our children need to know: to celebrate them as people and not their disabilities," said Gwen Tanner, founder of Downs But Not Out, a support group for parents of children with special needs.
Downs But Not Out sponsors an awareness walk every year to celebrate the special needs children.
More than 200 people gathered on a chilly Sunday afternoon to walk around the Heritage Park Trail, play carnival games, eat barbeque and share information about special needs children.


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John Kaase's 2008 Ford Mustang - Blue Crescent BeheaderWritten by on April 1, 2009 Contributors: Robert McGaffin
"I have always wanted a Boss 429 Mustang because I consider it to be Ford's masterpiece of engineering. The problem is, considering their present value [approximately $500K], I was as far away from being able to afford one in 1969 as I am today. If I could buy an original, it would be far too valuable to drive. With the release of our Boss Hemi engine parts, the next logical step was to build our own version, rationalizing that a complete running car would showcase the new parts and create a new market for our parts sales," said the gleeful Jon Kaase.
"Let's be honest here. I wanted this car! I want to drive it home and give my family and neighbors a ride. Most of them think I have a garage that does quick oil changes. They have no idea. I hadn't driven anything with over 360hp since 1974. With 800 ft-lbs of torque and 900 hp, this car is a real handful."
We were yakking with Jon at the recent PRI show about the black beast in his booth. Then the thread wound into a trip we'd made to his old shop a dozen years past. We had gone there to document the building of an 812-inch IHRA Pro Stock motor. Forgot what it made, something like 1,400 on the motor. Kaase said that times have changed, drastically, and that same configuration outfitted with the latest parts would put out at least 1,900. We'll come back to this in a minute.
Look at the Boss Nine Mustang. Kaase's vision was based on a stock-appearing car, one that you would buy from a dealer with the biggest motor available. It was rated at a paltry 375 hp. Kaase's version makes 900. In the day, the 1969-70 Boss 429 came with 7-inch Magnum 500 rims and F60 Goodyear Polyglas GT rubber, and perhaps the most arresting thing about it was the distinct lack of stripes, black-out trim, wheel or rocker panel moldings, or chrome exhaust tips. All it said was "Boss 429" on the fenders, and it carried a functional hood scoop that practically disappeared when the body wore dark livery. Kaase's Boss Nine fairly reeks that aura and looks for all the world like it just rolled off the Kar Kraft (Brighton, Michigan) "assembly" line, modern Magnum rims and all.
Part of that image comes from a new Mustang that Kaase bought from B.F. Evans Ford in Livermore, Kentucky. It had but 180 miles on the odometer, and only because Evans' guys drove it to Nashville, where Kaase's guys loaded it on a trailer for the final miles home. All the build tags and stickers are still in place, adding to the dj vu mystique. "It was worth it because it's new and it's a pleasure to work on, no dirt, oil, or road gunk," quipped the low-key Kaase.
The biggest obstacle for Kar Kraft was that skinny space between the shock towers, a place never meant for an engine with the girth of the Boss (see "Total Performance," p. 44). Their techs had to cut out, rebuild, and relocate the towers to make room for the big Blue Crescent (semi-hemispherical combustion chambers) cylinder heads, and with little space to spare even at that. Kar Kraft also lowered the suspension by about an inch, using control arms and spindles specific to the Boss 429. We asked Jon about how the Boss Nine fit in that modern engine compartment. He smiled, Cheshire cat-like. "You could say it just sort of fell in there. We didn't have to trim, cut, or rebuild anything for it to take the motor. We didn't even have to take the hood off."
Then: "I should have measured the stock engine crank centerline at the front, with respect to the sides. The tailshaft and rear crossmember measured to be in the centerline of the car, and the pinion is also centered between the wheels. The thing is, I remember the valve covers to be unequal distances from the shock towers, offset to the right. Is the engine in these [new] cars at an angle? With the Boss engine in place, the tailshaft is centered and the valve covers are closer to the right, just like stock.
"This car is a natural for the Boss Hemi swap. There is plenty of room between the shock towers, enough room for headers [you can reach past them with your hand], and enough room between the accessory drives and the radiator. The hood needs no modification for the air filter. It's close, but the right filter assembly will fit. A stock rear sump Moroso pan fits over the steering and crossmember perfectly. The cooling system is completely stock. All we did was space the bottom of the core outward an inch or so, but the top of the radiator is in the stock location."
Kaase stripped it down and installed a mock-up engine, then sent it to fabricator pal Chuck Lawrence in Hiram, Georgia, for motor mounts and headers. Back at Jon's shop, they built and dyno tested the 589 and installed it with the clutch, transmission, and driveshaft. It went over to Lawrence's for the close-up work--hoses, wiring, gauges, and the rest of the finish work.
Kaase: "The front drive is from Billet Specialties. It comes with a billet front cover, which has extra bosses for attaching the accessories and brackets. It comes with the front cover, water pump, power steering pump, an alternator, A/C compressor, and all the pulleys and brackets. Nothing bolts to the heads. As complicated as it looks, it only took an hour to completely assemble it. Like any good racers, we threw the instructions out and just winged it. I was really impressed with the quality and fit of this assembly."
As the Boss 302 was created to champion small-block performance and to rub fenders with the Chevrolet Z28 in the Trans Am series, in the big-block world, the Boss 429 was really a homologation special required for legality in NASCAR. Production for '69 was 859 units (including two Boss 429 Cougars). In '70, the total was but 499. Automatic transmission and air conditioning were not offered, but at least it had an engine oil cooler and the battery was moved to the trunk, largely to free up space in the motor room. Well, they still don't race Mustangs in NASCAR, so how come it wound up in one instead of the obvious Torino? Sorry, we can only guess at the back story on this one, but Ford marketers figured heavily on boosting the Mustang's image, so that's the way it was.
Kaase's remarks about the efficacy of modern speed parts made us think. The Boss 429 had huge ports and a marginal camshaft that really didn't take effect until the engine was really winding up, so low-speed torque pretty much sucked. His Boss Nine heads (which will fit on any 429-460 block) have holes just as large, but the advance in camshaft phasing and ignition technology "adjust" for these ancient and potential shortcomings. The same goes for the completely stock suspension. When the weak stuff breaks, they will amend the woe with stronger stuff.
Since the Boss Nine was finished a couple of weeks before it debuted at the PRI show, it has some work-in-progress teething issues. "It doesn't want to run very well below 2,500 rpm," said Kaase. "It farts and bucks. It's like driving something between a Pro Stocker and a pulp truck. No matter how careful I am with the throttle, and I'm talking Third and Fourth gear now, it literally wants to rip the tires off (laughs). Now that I've felt this thing, realistically, a 466-inch Boss Hemi would be about right for this car, but, as they say, anything worth doing is worth doing in excess."
Can we get an AMEN!
By The Numbers
|'08 Ford Mustang
|John Kaase • Winder, GA
|Vehicle weight w/driver: 3,800 pounds
|Ford displacing 589 cubic inches (4.627 bore x 4.375 stroke)
|C &C MotorSports aluminum, siamesed bores, iron liners, stock 10.32-inch deck height
|Compression ratio:
|Moroso catalog pan, 7-quart, Kaase oil pump
|Rotating assembly:
|Sonny Bryant 4.375-inch forged crank, 6.800-inch Oliver steel I-beam rods,dished Diamond pistons, Ackerly & Childs ring packs
|Cylinder heads:
|Kaase Boss Nine, as-cast aluminum, 88cc combustion chambers, Manley 2.30/1.90 valves, Kaase/W.W. Engineering 1.75:1 rocker arms, Boss Nine rocker covers
|Comp Cams mechanical roller, .699-inch lift, 251/251 degrees duration at .050, 110-degree C/L, Trend 3/8-inch pushrods, Ford Motorsport timing gear
|Kaase Boss Nine intake manifold, Holley 1150 HP carburetor, BSR Racing Products 16-inch air cleaner (to clear Dominator and hood)
|Power adder:
|are you kidding?
|Mallory distributor, MSD 6AL-2 box, 34 degrees total timing
|Chuck Lawrence–built 2 1/8-inch primaries, 3.5-inch collectors, Granatelli stainless 3-inch exhaust system and mufflers
|Output at flywheel:
|902 hp at 6,800 rpm, 800 lb-ft at 5,100rpm
|Built by:
|Jon Kaase Racing Engines, Inc.
|D &D Performance (2.66, 1.78, 1.30, 1.00, 0.80, 0.62:1) T56 Viper-spec transmission, Quick Times steel bellhousing, McLeod 11-inch twin-disc clutch, pressure plate, flywheel, hydraulic throwout bearing
|Dynomax custom aluminum one-piece, 1350-series yokes
|Rear axle:
|stock Ford 8.8-inch housing, clutch-type differential, 3.31:1 ring-and-pinion
|Front suspension:
|stock, MacPherson struts, lower control arms, antisway bar
|Rear suspension:
|stock, coil springs, three-link, Panhard rod
|stock 12-inch disc, front; 12-inch disc, rear
|Wheels & Tires
|American Muscle Black Magnum 500, 18x9, front; 18x10, rear
|Pirelli PZero Rosso Asimmetrico, 255/45R18, front; 285/40R18, rear, Y-rated
More Photos From This Article View Slideshow
As a work in progress, the Mustang chassis/suspension is bone stock. The Boss Nine is equipped with latter-day 18-inch American Muscle Magnum 500 wheels and thick Pirelli PZero Rosso rubber, but the motor has enough moxie to break 'em in Fourth gear.About as slick as a moleskin biscuit, the Boss Nine has every bit of its assembly line interior. Colorful it ain't; functional it is. Note the just-the-right-height Hurst gear changer.D&D Performance supplied the blast-shield/bellhousing and the Viperized T56. Kaase made the crossmember. The McLeod clutch assembly is twin-disc on an 11-inch steel flywheel.Original cars had a working hood scoop. Kaase wanted to stay away from that, so he used a NASCAR-flavored BSR air cleaner and bottom plate that drops the whole assembly cleanly away from the underside of the hood. Big Boss is nearly 600 ci large. Kaase's pal Chuck Lawrence did all the wiring and considerable finessing needed to make it all work.


text:
From award-winning poet John Koethe, a rich and resonant new collection that moves easily between autobiographical anecdote and philosophical reflection.
About the Author
John Koethe is distinguished professor of philosophy emeritus at the University of Wisconsin-Milwaukee and the first Poet Laureate of Milwaukee. His collections include Falling Water, which won the Kingsley-Tufts Award, North Point North, a finalist for the Los Angeles Times Book Prize, and Ninety-fifth Street, winner of the Lenore Marshall Prize. In 2011, he received a Literature Award from the American Academy of Arts and Letters. He lives in Milwaukee, Wisconsin.
Praise For…
"In this ninth collection of poems, very much obsessed with time's passage and its ending, with aging, remembering and preparing to die, poet-philosopher Koethe is in full, lucid command of his voice, an amiable hybrid of late Wallace Stevens, late John Ashbery, and William Bronk."
-Publishers Weekly (starred review)
"Koethe's winding syntax with its sudden, animating insights is like the clear, radiant path of a sunbeam; he illuminates another life inside our lives--the generous and capacious life of hte imagination that is his true province and home."
-Susan Stewart, author of COLUMBARIUM and RED ROVER
"John Koethe's poems in which he describes his past are more than simple memorials or langurous retrievals; rather, they are profound meditations on time and the curious hold it has on the human psyche."
-American Academy of Arts and Letters
"Koethe is a beautiful writer, one whose subtle inventiveness can give new life to persistent images, nail a complex feeling in just a few words, or make the basic tools of the poetic trade into sources of pleasure and persuasion."
"Koethe . . . might be unmatched in the manner in which he thinks in his poems. Or, more than thinks, he meditates. He dramatizes the emotional vitality of an idea which nowadays is strangely and sadly rare."
-The Rumpus
"He handles big themes deftly, like the professional philosopher he is. . . . The poet seems comfortable with uncomfortable insights; readers may take comfort in Koethe's lyric musings."
"I can think of no one whose company I'd rather keep-for Koethe writes in the water of his time, knowing all is temporary."
-Washington Independent Review of Books