We are the manufacturer of Circle which is made from stainless steel and others metals too. The materials we use to make these circles are of superior quality. It contains numbers of features such as reliability, durability, easy to wash, easy to install, corrosion resistance, high tolerance and many more. This is used in industries such as gas industry, automobile industry, oil industry as well as the construction of building or industries. It has the ability to resist low-temperature heat to high-temperature heat. We provide the product which doesn't need a conventional maintenance. It is specially made to use in underground water surfaces such as the sea, bore well and so on. Its size and diameter are from 15mm dia – 600 mm Dia.
If you want this product urgently, we can facilitate you delivery services in accordance with your time and schedule. We can accept any type of payment, such as net banking, cards etc. Using it you can complete whatever the project you have.
Nickel Alloy Circle, Hot Rolled Circle, Monel Circle Dealer, Hastelloy Circle, Titanium Circle, Tantalum Circle, Aluminium Circle, Cold Drawn Circle, Copper Circle, Copper Nickel Circle Exporter, Stainless Steel Circle, Duplex Steel Circle Stockist, Super Duplex Steel Circle, Wrought Austenitic Circle Supplier.
It is checked out in the laboratory by professionals and then given to our clients at best rates. While buying it, all the relevant duty information is given on the packaging box. Using big cartons it is packed and covered with plastic made tapes. It is also known as stainless steel discs because of its round size. It is available in different metals and provided by us, but stainless steel is better than that all. Buy Circle from "Sagar Prakash Alloys" and use repeatedly because it never gets corrode and damaged.
Dimensions AISI, ASME, BS, DIN, EN, JIS etc.
SS 304 Circle, ASTM A182 Steel Circle, ASME Stainless Steel Circle, DIN 1.4301 Circle, UNS S30403 Circle.
Duplex Circle, ASME Duplex 2205 Circle, Hot Rolled Duplex Circle, DIN 1.4462 Circle, UNS S31803 Circle.
Nickel 201 Circle, ASTM Nickel Alloy Circle, DIN 2.4066 Circle, Nickel Alloy 200 Circle, UNS N02200 Circle.
Titanium DIN 3.7035 Circle, Ti Grade 2 Circle, Titanium UNS R50400 Circle, ASME SB338 Titanium Gr-5 Circle.
Circles are manufactured and controlled to international standards such as ASTM, ASME, BS, MSS, EN, DIN or WERKSTOFFE. Our products undergo stringent quality tests before they are dispatched.
South Korea, Egypt, Puerto Rico, Kuwait, Russia, Angola, Norway, Yemen, Nepal, Finland, Gabon, Macau, Bulgaria, India, Algeria, Bahrain, Tibet, Slovakia, Ghana, Portugal, Bangladesh, Romania, Venezuela, Saudi Arabia, Gambia, Israel, Lithuania, Austria, New Zealand, Ecuador, Estonia, Qatar, Bhutan, Australia, Hungary, Sri Lanka, Turkey, Costa Rica, Greece, Azerbaijan, Ukraine, Colombia, Poland, Argentina, Brazil, Mongolia, Croatia, Belarus, United Kingdom, Poland, Iraq, Singapore, Netherlands, Malaysia, Nigeria, Germany, Jordan, Ireland, Mexico, Canada, Spain, Philippines, Czech Republic, Pakistan, Lebanon, Taiwan, Belgium, Namibia, Kenya, United Arab Emirates, Mexico, France, Serbia, Libya, Denmark, Vietnam, Japan, Hong Kong, South Africa, Trinidad & Tobago,Tunisia, Italy, Iran, Kazakhstan, Iran, Indonesia, Thailand, Chile, Oman, Bolivia, United States, Nigeria, Zimbabwe, Peru, China, Switzerland, Sweden, Chile, Afghanistan, Morocco.
Ernakulam(Kerala), Secunderabad(Hyderabad), Chennai(Tamil Nadu), Haryana(india), Ludhiana(Punjab), Faridabad(Haryana), Indore(Madhya Pradesh), Gurgaon(Haryana), Chandigarh(Punjab), Ahmedabad(Gujarat), Mumbai (maharashtra), Pune(Maharashtra), New Delhi, Navi Mumbai(Maharashtra), Thane(Maharashtra), Pimpri-Chinchwad(maharashtra), Coimbatore (Tamil Nadu), Surat(Gujarat), Bhopal(india), Howrah(West Bengal), Bengaluru(Karnataka), Thiruvananthapuram(Kerala), Visakhapatnam(Andhra Pradesh), Vadodara(Gujarat), Rajkot(Gujarat), Kolkata(West Bengal), Hyderabad(Andhra Pradesh), Noida(Uttar Pradesh).


19th December, 2012: An International Carnival for the First time in Hyderabad is here to appease your children's and family, Christmas and New year round the comer and the best option for your family to really freak out is FUNIVAL. Funival - A First of its kind Carnival featuring Hot Air Balloon Ride, Games, Games Desk, International Cuisines, Juke Box, and many many more- A Day of Fun, Frolic & Entertainment is all set to be here at Funival-An International Carnival featuring at HITEX Grounds from 20th to 23rd December, 2012.
Hitex Funival-An International Carnival now presents Duo-Effekt-An International Troupe Famous for Dance & Aerobics, dynamic duo from Eastern Europe have mesmerized audiences all over Europe and Asia with their unique and entertaining performances.
Their show includes: Duo Aerial Silk, Duo Aerial Ring, Balance Acrobatic Art and many more like fire shows, UV Lights show etc...DUO Effekt is all set to mesmerize the hyderabadi audience with their breath taking moves, dance and highly appreciated acrobatic skills.. .Funival is now all set to take performances of aerobics to the next level.
Funival is an exclusive international carnival featuring many games, cuisines, music, juke box etc. It is for the first time in Hyderabad Funival will be having Hot Air Balloon Ride which will be exclusively arranged from HITEX grounds, the carnival which here to provide a best of its kind experience for family and kids. The holiday time around Christmas is now all set to be more exciting & energized.
Speaking on the occasion of introducing team Duo-Effekts for Funival CEO HITEX Mr K.V Nagendra Prasad & Mr.Thomas VP Times informed that they look forward to provide the best of Entrainment and high Voltage performance from the DUO-EEFEKTS, he added that the visiting families and kids will sure be mesmerized with their breathtaking skills & acrobatic moves, he added that HITEX which are the best known convention centres of Hyderabad will ensure that this holiday season will be the best for kids & families.


Vromm has been producing music for more than 20 years and merges the realms of techno, jungle, abstract, sci-fi into his tracks in a unique style that is totally his own. His music career began when he started playing vinyls at Flexo Club in south Spain which then later lead to him producing music using on MS DOS with a PC486. He's now based in London and since moving has had releases on both 31 Recordings and Critical Music. He's now working on projects with Craig Richards for The Nothing Special along with an EP for Cosmic Bridge Records that will include a track with Om Unit.


Dubai to London flights will cost you a minimum of 785. This may extend up to 33320. Worry not! As you may enjoy a number of domestic flight booking offers exclusive for Cleartrip customers!
The price of a Dubai to London flight ticket varies between a minimum AED 33320 or up to AED 33320.
Is it true that Pegasus takes less time on a direct Dubai to London flight than other airlines?
What is the average range of Economy class tariffs on Dubai to London flight route?
The Economy class airfare ranges from AED 785 to AED 33320. American Airlines, British Airways, Aer Lingus, Iberia, Royal Brunei Air, Emirates, Sunstate Airlines, Virgin Atlantic, and Singapore Airlines provide tickets in this range.


A collection of comics depicting the everyday life of characters from the wuxia-themed MMORPG Jian Wang 3. Mainly SoL(slice of life) comic shenanigans set in the Tang Dynasty. =========================== Ep 1-42: standalone episodes that don't have continuity (aka one-shot/omake types) Ep 43 onwards: start of "story"
Chap Yi Chui Wu Yue Tiao Man Ji 85 Shall I Send You To Heaven?
Chap Yi Chui Wu Yue Tiao Man Ji 61 Rock, Paper, Scissors!
Chap Yi Chui Wu Yue Tiao Man Ji 57 Wanna Hug Me?


text:
EESM
Earth System Model Development
Regional & Global Model Analysis
MultiSector Dynamics
Publications & Highlights
Earth System Model Development ESMD
Regional & Global Model Analysis RGMA
MultiSector Dynamics MSD
Scoring Methods in the International Land Benchmarking (ILAMB) Package
Wednesday, December 13, 2017 at 8:00am
Office.com
iCal / MS Outlook
New Orleans Ernest N. Morial Convention Center - - Poster Hall D-F
Nathan Collier - Oak Ridge National Laboratory (ORNL)
Forrest M Hoffman - Oak Ridge National Laboratory (ORNL)
Gretchen Keppel-Aleks - University of California Irvine (UC Irvine)
David M Lawrence - National Center for Atmospheric Research (NCAR)
Mingquan Mu
William Riley - Lawrence Berkeley National Laboratory (LBNL)
James Tremper Randerson - University of California Irvine (UC Irvine)
The International Land Model Benchmarking (ILAMB) project is a model-data intercomparison and integration project designed to improve the performance of the land component of Earth system models. This effort is disseminated in the form of a python package which is openly developed (https://bitbucket.org/ncollier/ilamb). ILAMB is more than a workflow system that automates the generation of common scalars and plot comparisons to observational data. We aim to provide scientists and model developers with a tool to gain insight into model behavior. Thus, a salient feature of the ILAMB package is our synthesis methodology, which provides users with a high-level understanding of model performance.
Within ILAMB, we calculate a non-dimensional score of a model's performance in a given dimension of the physics, chemistry, or biology with respect to an observational dataset. For example, we compare the Fluxnet-MTE Gross Primary Productivity (GPP) product against model output in the corresponding historical period. We compute common statistics such as the bias, root mean squared error, phase shift, and spatial distribution. We take these measures and find relative errors by normalizing the values, and then use the exponential to map this relative error to the unit interval. This allows for the scores to be combined into an overall score representing multiple aspects of model performance. In this presentation we give details of this process as well as a proposal for tuning the exponential mapping to make scores more cross comparable.
However, as many models are calibrated using these scalar measures with respect to observational datasets, we also score the relationships among relevant variables in the model. For example, in the case of GPP, we also consider its relationship to precipitation, evapotranspiration, and temperature. We do this by creating a mean response curve and a two-dimensional distribution based on the observational data and model results. The response curves are then scored using a relative measure of the root mean squared error and the exponential as before. The distributions are scored using the so-called Hellinger distance, a statistical measure for how well one distribution is represented by another, and included in the model's overall score.
Link for More Information
https://agu.confex.com/agu/fm17/meetingapp.cgi/Paper/297887
Related Meeting/Event
AGU Fall Meeting 2017
Funding Program Area(s)
ESMD RGMA
Reducing Uncertainty in Biogeochemical Interactions through Synthesis and Computation (RUBISCO) Scientific Focus Area (SFA)
About EESM
Contact EESM
Energy.gov
Office of Science
Contact Office of Science

redpajama_set_name: RedPajamaCommonCrawl


Having difficulty with the keyhole section major time. For example in this questions, how is D the answer and yet B is wrong? They look EXACTLY the same.
Also the top and right/end views look correct as well! For the end view, C and E look exactly the same. All the views look correct in the answers/explanation and yet D is the correct view and answer.
What am I missing or not understanding??? There are a lot of correct answers in this section that are ambiguous with identical answers and yet one answer ends up being randomly correct than the other. There are also explanations where all the views are correct and yet one view and answer ends up being chosen as the correct answer.
Yikes..... I am very confused with this keyholes section and these GS keyhole questions are not helping either.
Its probably just a technical error while putting the test together. I had a similar issue with a question in the angle ranking section. Just live chat with them and they will get it checked out right away from their team and email you back with them either correcting it or thats how it is.


I received a copy of A Cookie to Celebrate by Jana Douglass for review. All opinions are my own. Grandma's Danish Sugar Cookies is a family recipe passed down the generations.
Family recipes. We cherish those meals passed down from generation to generation. The secret sauce, special ingredient, and particular ways to handle the food were written with love for a family that has continued to grow.
Unfortunately, I don't have very many of these recipes. My grandmothers were both fantastic cooks, but recipes were not written down and shared too often.
So a little over a year ago, my childhood BFF reached out and asked for my mom's sugar cookies recipe. She claimed them to be the best she's ever had, and wanted to make them with her children for the holidays. Not remembering the days of cookie baking myself, I didn't have the recipe.
A quick call (just kidding, calls with my mom are never quick) and I had the recipe written down along with it's history. The sugar cookie recipe my mom always used came from her grandmother, my great grandmother and was for Danish Sugar Cookies. I had no idea, but it's only made these cookies even sweeter.
My friend received the recipe, made them with her kids, and let me know they were just like she remembered. I was glad I could help bring those good memories back into her family.
It took me a year, but I finally decided it was high time I make these cookies. Even if sugar cookies aren't my favorite, I would still have fun making and then decorating them.
Thanks to A Cookie To Celebrate by Jana Douglass, I felt confident in my ability to decorate these cookies. They deserved to look as special as their history, and it helped me learn a new skill!
If you've wanted to learn cookie decorating, but feel overwhelmed with online tips and How-to's I would highly suggest Jana's book. It's straight forward, funny, and feels like your best friend is walking you through each step.
In a small bowl combine butter, sugar, eggs, and vanilla.
In a separate bowl combine flour, cream of tartar, baking soda, and salt.
Mix the wet and dry ingredients together. The dough gets stiff, so you'll have to use your hands to fully mix it.
Divide the dough in two. Wrap each section in saran wrap and chill for at least 1 hour, but up to 1 week.
Preheat the oven to 330°. Grease cookie sheets.
Roll the dough out on a lightly floured surface to 1/8-1/4 inch thick. Cut with cookies cutters.


Home Designer Comment : Virginia Mist Granite is a naturally occurring stone that formed thousand of years ago by the crystallization of molten rock from volcanic activity and pressure over time by times. This stunning stone is made up of interlocking mineral crystals, the most common being feldspar and quartz. Virginia Mist Granite can ensure good quality, strong, lightweight and very ideal for both interior and exterior use.
0 Response to "Virginia Mist Granite Countertops, Slab And Prices"


text:
/*
 * Driver for DORM1/REV7 direct motor drive.
 * TODO!!! Rename!
 */

#ifndef CONTENT_OTRADIOLINK_UTILITY_OTRADVALVE_TESTVALVE_H_
#define CONTENT_OTRADIOLINK_UTILITY_OTRADVALVE_TESTVALVE_H_

#include <stddef.h>
#include <stdint.h>
#include <OTV0p2Base.h>
#include "OTRadValve_AbstractRadValve.h"
#include "OTRadValve_ValveMotorDirectV1.h"

// Use namespaces to help avoid collisions.
namespace OTRadValve
    {

/**
 * @class   TestValveMotor
 * @brief   Generic high level motor driver with minimal logic. Intended for battery and motor drive testing.
 */
class TestValveMotor : public OTRadValve::HardwareMotorDriverInterfaceCallbackHandler
  {
  private:
    // Hardware interface instance, passed by reference.
    // Must have a lifetime exceeding that of this enclosing object.
    OTRadValve::HardwareMotorDriverInterface * const hw;

    // Direction the motor is running in.
    volatile bool state;

    // Flag set on signalHittingEndStop() callback from end-top / stall / high-current input.
    // Marked volatile for thread-safe lock-free access (with care).
    volatile bool endStopDetected;

    // Stores the number of times motor has hit an endstop.
    volatile uint32_t counter;

    // Run fast towards/to end stop as far as possible in this call.
    // Terminates significantly before the end of the sub-cycle.
    // Possibly allows partial recalibration, or at least re-homing.
    // Returns true if end-stop has apparently been hit,
    // else will require one or more further calls in new sub-cycles
    // to hit the end-stop.
    // May attempt to ride through stiff mechanics.
    // Prints counter to serial in hex each time an end stop has been reached.
    bool runFastTowardsEndStop(bool toOpen);

  public:
    // Create an instance, passing in a reference to the non-NULL hardware driver.
    // The hardware driver instance lifetime must be longer than this instance.
    TestValveMotor(OTRadValve::HardwareMotorDriverInterface * const hwDriver)
  	  : hw(hwDriver), state(false), endStopDetected(false), counter(0) {}

    /**
     * @brief   Updates the state of the motor logic.
     *          - Will run until end stop is detected and then reverse
     *          the direction of the motor and increment counter.
     */
    void poll();

    /**
     * @brief   Returns the end stop counter.
     */
    inline uint32_t getCounter() { return counter; }

    // Unused abstract functions.
    void signalHittingEndStop(bool) { endStopDetected = true; }
    void signalShaftEncoderMarkStart(bool) {}
    void signalRunSCTTick(bool) {}
  };

    }

#endif /* CONTENT_OTRADIOLINK_UTILITY_OTRADVALVE_TESTVALVE_H_ */

redpajama_set_name: RedPajamaGithub


Packaged with the Aeroccino, which provides a smooth, finely textured hot or cold fresh milk froth for your coffee recipes via an ultra simple and fast automatic system, in a few seconds, without noise or vibration. The Aeroccino also heats milk for your Caffè Latte or Cappuccino. Each machine includes a welcome set with 16 Nespresso capsules containing individual aromas.


text:
@interface TKPhotoSet()

@property (nonatomic, retain) NSArray *sortedPhotos;

@end

@implementation TKPhotoSet {
    NSMutableDictionary *_layoutInformation;
    NSMutableDictionary *_frameInformation;
}

@dynamic attribute;
@dynamic caption;
@dynamic layout;
@dynamic photos;
@dynamic post;

@synthesize sortedPhotos = _sortedPhotos;
@synthesize firstPhoto = _firstPhoto;

- (TKPhoto *)firstPhoto
{
    if (!_firstPhoto) {
        NSFetchRequest *fetchRequest = [[NSFetchRequest alloc] init];
        fetchRequest.entity = [TKPhoto entityWithContext:self.managedObjectContext];
        fetchRequest.predicate = [NSPredicate predicateWithFormat:@"photoSet = %@", self];
        fetchRequest.sortDescriptors = [NSArray arrayWithObject:[NSSortDescriptor sortDescriptorWithKey:@"sortID" ascending:YES]];
        fetchRequest.fetchLimit = 1;

        NSArray *results = [self.managedObjectContext executeFetchRequest:fetchRequest error:nil];
        if ([results count] > 0) {
            _firstPhoto = [results objectAtIndex:0];
        }
    }

    return _firstPhoto;
}

- (NSArray *)sortedPhotos
{
    if (!_sortedPhotos) {
        NSFetchRequest *fetchRequest = [[NSFetchRequest alloc] init];
        fetchRequest.entity = [TKPhoto entityWithContext:self.managedObjectContext];
        fetchRequest.predicate = [NSPredicate predicateWithFormat:@"photoSet = %@", self];
        fetchRequest.sortDescriptors = [NSArray arrayWithObject:[NSSortDescriptor sortDescriptorWithKey:@"sortID" ascending:YES]];

        _sortedPhotos = [self.managedObjectContext executeFetchRequest:fetchRequest error:nil];
    }

    return _sortedPhotos;
}

- (NSDictionary *)layoutInformation
{
    if (!_layoutInformation) {
        _layoutInformation = [NSMutableDictionary dictionary];
    } else {
        [_layoutInformation removeAllObjects];
    }

    NSArray *sortedPhotos = [self sortedPhotos];
    int sortedPhotosCount = [[self sortedPhotos] count];
    NSMutableArray *rows = [NSMutableArray array];
    NSString *layoutString = [NSString stringWithFormat:@"%@", self.layout];
    int totalCount = 0;
    int totalLayoutCount = 0;
    CGFloat y = 0;
    CGFloat width = 0;
    BOOL layoutPossible = YES;
    unsigned int len = [layoutString length];
    unichar buffer[len + 1];
    [layoutString getCharacters:buffer range:NSMakeRange(0, len)];

    // If the total number of photoset photos required for this layout is more than the actual photo count,
    // we can't lay it out properly
    // This exists for the following post:
    //   http://el-amor-de-tu-dia.tumblr.com/post/41262495464/diggydre-best-dad-everr
    // Layout is 27591, which is a total of 24 photos - but we only have 9 photos.
    // fuck knows why.
    for (int row = 0; row < len; ++row) {
        int rowCount = [[NSString stringWithFormat:@"%c", buffer[row]] intValue];
        totalLayoutCount = totalLayoutCount + rowCount;
    }

    if (totalLayoutCount > sortedPhotosCount) {
        layoutPossible = NO;
    }

    // If we can do the layout, do it.
    if (layoutPossible) {
        for (int row = 0; row < len; ++row) {
            int rowCount = [[NSString stringWithFormat:@"%c", buffer[row]] intValue];
            CGFloat rowX = 0;
            CGFloat rowWidth = 0;
            NSMutableArray *rowPhotos = [NSMutableArray array];
            NSNumber *rowHeight = [NSNumber numberWithInt:0];

            for (int i = 0; i < rowCount; i++) {
                TKPhoto *photo = [sortedPhotos objectAtIndex:totalCount];

                [rowPhotos addObject:@{
                    @"width" : [photo width],
                    @"height" : [photo height],
                    @"x" : [NSNumber numberWithFloat:rowX],
                    @"y" : [NSNumber numberWithFloat:y]
                }];

                rowHeight = [photo height];

                rowWidth = rowWidth + [[photo width] floatValue];

                totalCount = totalCount + 1;
                rowX = rowX + [[photo width] floatValue];
            }

            [rows addObject:@{
                @"photos" : rowPhotos,
                @"height" : rowHeight,
                @"width" : [NSNumber numberWithFloat:rowWidth]
            }];

            y = y + [rowHeight floatValue];
            width = MAX(width, rowWidth);
        }
    } else {
        // if not, just lay them out the full width
        for (int i = 0; i < sortedPhotosCount; i++) {
            NSMutableArray *rowPhotos = [NSMutableArray array];
            TKPhoto *photo = [sortedPhotos objectAtIndex:i];
            NSNumber *rowHeight = [photo height];
            NSNumber *rowWidth = [photo width];

            [rowPhotos addObject:@{
                @"width" : [photo width],
                @"height" : [photo height],
                @"x" : [NSNumber numberWithInt:0],
                @"y" : [NSNumber numberWithFloat:y]
            }];

            [rows addObject:@{
                @"photos" : rowPhotos,
                @"height" : rowHeight,
                @"width" : rowWidth
            }];

            y = y + [rowHeight floatValue];
        }
    }

    [_layoutInformation setObject:rows forKey:@"rows"];
    [_layoutInformation setObject:[NSNumber numberWithFloat:y] forKey:@"height"];
    [_layoutInformation setObject:[NSNumber numberWithFloat:width] forKey:@"width"];

    return (NSDictionary *)_layoutInformation;
}

- (NSDictionary *)frameInformationForWidth:(CGFloat)width spacing:(CGFloat)spacing
{
    if (!_frameInformation) {
        _frameInformation = [NSMutableDictionary dictionary];
    }

    NSString *key = [NSString stringWithFormat:@"%@-%@", [NSNumber numberWithFloat:width], [NSNumber numberWithFloat:spacing]];

    if (![_frameInformation objectForKey:key]) {
        NSMutableArray *frames = [NSMutableArray array];
        NSDictionary *layoutInformation = [self layoutInformation];

        __block CGFloat currentY = 0;

        NSArray *rows = [layoutInformation valueForKey:@"rows"];
        [rows enumerateObjectsUsingBlock:^(id obj, NSUInteger idx, BOOL *stop) {
            NSArray *photos = [(NSDictionary *)obj valueForKey:@"photos"];
            CGFloat totalWidth = [[(NSDictionary *)obj valueForKey:@"width"] floatValue];
            CGFloat photoWidth = totalWidth / [photos count];
            CGFloat ratio = (width - (([photos count] - 1) * spacing)) / totalWidth;
            __block CGFloat rowHeight = 0;

            [photos enumerateObjectsUsingBlock:^(id obj, NSUInteger photosIndex, BOOL *stop) {
                rowHeight = lroundf([[obj valueForKey:@"height"] floatValue] * ratio);

                NSDictionary *frame = @{
                    @"width" : [NSNumber numberWithFloat:lroundf(photoWidth * ratio)],
                    @"height" : [NSNumber numberWithFloat:rowHeight],
                    @"x" : [NSNumber numberWithFloat:lroundf(([[obj valueForKey:@"x"] floatValue] * ratio) + (spacing * photosIndex))],
                    @"y" : [NSNumber numberWithFloat:lroundf(currentY)]
                };

                [frames addObject:frame];
            }];

            currentY = currentY + rowHeight + spacing;
        }];

        [_frameInformation setObject:@{ @"frames" : (NSArray *)frames, @"height" : [NSNumber numberWithFloat:currentY - spacing] } forKey:key];
    }

    return [_frameInformation objectForKey:key];
}

- (CGFloat)heightForRow:(int)row
{
    return 0;
}

- (TKPhoto *)photoForRow:(int)row
{
    return nil;
}

@end


At the July 5th meeting of the Board of Education, the Board unanimously voted to extend the contract of Superintendent of Schools James Hunderfund, Ed.D. through June 30, 2020.
Beginning his tenth year as Malverne's School Superintendent, Dr. Hunderfund's vision and leadership have been a driving force for educational improvement. Recent notable academic achievements include: Malverne High School honored as the sole Schools of Opportunity Gold Award recipient on Long Island; MHS being cited as a "Reward School" for the first time and also being named by The College Board to the National Advanced Placement Honor Roll; being ranked – for four years in a row – as one of the top high schools in the nation on the Washington Post's "Most Challenging High Schools" list; a computer/STEM initiative in all four schools; 96% Regents Diploma graduation rate; and a record-breaking number of AP Scholars, All-State and All-County musicians, performing groups and artists.
During Dr. Hunderfund's tenure, the community experienced its first successful passage of a major bond in decades, as well as the reorganization of the educational structure so that all students would be integrated into a common learning experience, K-12.


text:
The Lord Governs the Hells (Continued)
On January 6, 2019 By chungsooIn Spiritual Learning
People in the world still believe that there is a devil who rules the hells and that he was created as an angel of light but was cast into hell with his gang after he led a rebellion. The reason for this belief is that the Word talks about the devil and Satan, and about Lucifer as well; and in these cases the Word is understood literally. However, in these passages the devil and Satan mean hell. The devil means the hell toward the rear where the worst people live, the people called evil demons; and Satan means the hell that is toward the front where the less malevolent people live, the people called evil spirits. Lucifer means the people who are from Babel or Babylon, the ones who extend their control all the way into heaven.
We can also see that there is no devil to whom the hells are subject from the fact that all the people who are in the hells, like all the people in the heavens, are from the human race (see Sections 311–317), that there are millions there from the beginning of creation to the present day, and that everyone there is the kind of devil he or she became by opposition to the Deity while in the world (see above, Section 311–312).
Sections 311-317: Published 5/11/2017-5/15/2017
DevilLuciferSatan
The Lord Does Not Cast Anyone into Hell: Spirits Cast Themselves In
3 thoughts on "The Lord Governs the Hells (Continued)"
babymix88
아주 재미있어요 감사합니다
이태성 입니다 고마워요
chungsoo
Thank you for your comments. Glad you liked it.


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text:
Gery History, Family Crest & Coats of Arms
Early Origins of the Gery family
The surname Gery was first found in Norfolk where one of the first records of the name was Guericus, Gueri who was listed in the Domesday Book of 1086. [1]
Almost one hundred years later Werri de Marinis was listed in Yorkshire in 1166 and a few years after that Werreis de Pilledona was listed in the Pipe Rolls of Dorset in 1179. [2]
Henry Werreys was Mayor of Sandwich in 1270. [3]
The Yorkshire Poll Tax Rolls of 1379 list Thomas Gery and Gerri de Planastre in Oxfordshire, and the following in Cambridgeshire: Warrin Gery; Werry de Cadamo; Herry Werri; and Peter Werri. [4]
Early History of the Gery family
This web page shows only a small excerpt of our Gery research. Another 230 words (16 lines of text) covering the years 1273, 1455, 1487, 1380, 1753, 1347, 1347, 1360, 1377, 1616, 1630, 1660, 1839, 1886, 1649, 1730, 1688 and 1713 are included under the topic Early Gery History in all our PDF Extended History products and printed products wherever possible.
Gery Spelling Variations
Spelling variations of this family name include: Gery, Warre, Werre, Gerry, Werry, Warry, Warriss and many more.
Early Notables of the Gery family (pre 1700)
Another 26 words (2 lines of text) are included under the topic Early Gery Notables in all our PDF Extended History products and printed products wherever possible.
Gery Ranking
In France, the name Gery is the 5,073rd most popular surname with an estimated 1,000 - 1,500 people with that name. [5]
Migration of the Gery family
Some of the first settlers of this family name or some of its variants were: settlers, who arrived along the eastern seaboard, from Newfoundland, to Maine, to Virginia, the Carolinas, and to the islands.
Contemporary Notables of the name Gery (post 1700) +
Sir Thomas Gery Cullum (1741-1831), 7th Baronet of Hastede, Suffolk, British medical doctor and politician, Bath King-at-Arms [6]
Sir Thomas Gery Cullum (1777-1855), 8th Baronet of Hastede, Suffolk, English peer
Yorkshire Poll Tax Rolls
Poll Tax
The Gery Motto +
The motto was originally a war cry or slogan. Mottoes first began to be shown with arms in the 14th and 15th centuries, but were not in general use until the 17th century. Thus the oldest coats of arms generally do not include a motto. Mottoes seldom form part of the grant of arms: Under most heraldic authorities, a motto is an optional component of the coat of arms, and can be added to or changed at will; many families have chosen not to display a motto.
Motto: Je trouve bien
Motto Translation: I find good.
^ Williams, Dr Ann. And G.H. Martin, Eds., Domesday Book A Complete Translation. London: Penguin, 1992. Print. (ISBN 0-141-00523-8)
^ Cleveland, Dutchess of The Battle Abbey Roll with some Account of the Norman Lineages. London: John Murray, Abermarle Street, 1889. Print. Volume 3 of 3
^ http://www.journaldesfemmes.com/nom-de-famille/nom/
^ Wikisource contributors. "Dictionary of National Biography, 1885-1900." Wikisource . Wikisource , 4 Jun. 2018. Web. 30 June 2020
Gery (English)


Protecting your skin from damaging UV rays begins with hats, sunglasses, and picking the right sunscreen.
Sunscreen can help protect your skin against harmful ultraviolet (UV) rays this summer, but only if it is applied correctly and often enough. The mistakes people make when they apply sunscreen can compromise the protective effects of sunscreen and can increase the risk of skin cancer.
"To get the best possible sun protection, it is important to wear protective clothing and apply sunscreen to all exposed skin, not just your face and arms," Ingrid Polcari, corresponding author of a study identifying the mistakes people make when using sunscreen, said in a statement.
It is a common misconception that clouds prevent sunburn. However, up to 80 percent of harmful UV rays can penetrate clouds and damage the skin.
Researchers set up free sunscreen dispensers at the Minnesota State Fair and observed over 2100 people using them over a 93-hour period. Only a third of those who used the dispensers applied sunscreen to all of their sun-exposed skin. The other two-thirds applied it to their arms (51 percent), face (42 percent), chest (six percent) and legs (3.6 percent). The average adult should generally apply about one ounce of sunscreen to all exposed skin, Polcari said.
About 38 percent of those who used the sunscreen dispensers were not wearing additional sun-protective clothing, such as hats, sunglasses and long sleeves.
Sunscreen use was related to the weather, with use of the free dispensers decreasing significantly on cloudy days. Polcari told TheDoctor in an email. It is a common misconception that clouds prevent sunburn. However, up to 80 percent of harmful UV rays can penetrate clouds and damage the skin. "We need to do a better job of reminding the public that sun protection is important in all kinds of weather," said Polcari, an assistant professor of pediatric dermatology at the University of Minnesota.
Women used the sunscreen dispensers more than men. Previous studies have found women are more likely than men to wear sunscreen, but it's important that men use sunscreen, too.
If you have sensitive skin, look for a sunscreen that contains the active ingredients zinc oxide or titanium dioxide.
So what should you look for in a sunscreen? "The best kind of sunscreen is one you'll use," said Rigel. Look for sunscreens that say they are "broad-spectrum" and "water-resistant" on the label. Find one you like and put it on before heading outside.
If you have sensitive skin, look for a sunscreen that contains the active ingredients zinc oxide or titanium dioxide. You should avoid ingredients like fragrances, oils or PABA — para-aminobenzoic acid.
Apply a broad-spectrum sunscreen with a Sun Protection Factor (SPF) of 30 or higher to exposed skin about 15 to 20 minutes before going outside. Sunscreen should be reapplied every two hours, or more often if you are swimming or sweating.
Skin cancers can also form on the lips, Polcari warns, so don't forget to apply a lip balm with SPF.
The study is published online recently in the Journal of the American Academy of Dermatology.


27b/6 – "yeah thats not what I was looking for at all." From: David Thorne Date: Monday 21 June 2010 10.28am To: Shannon Walkley Subject: Re: Re: Re: Re: Re: Poster Dear Shannon, It's a design thing. The cat is lost in the negative space. Regards, David. READ THE WHOLE THING.
Tomorrow's a big day. So I offer you just this: A dog playing soccer with a deer. Where is that dog's self respect. EAT THE DEER.


This site provides information about personal care homes currently operating in Saskatchewan. The Ministry of Health began using electronic technology for collecting inspection results on September 24, 2015. Information from inspections performed from that date onward is available at this site.
This information is not intended to guarantee the conditions of a home at all times and should not be relied upon for that purpose. Persons wishing to use the information provided within or through this website in selecting a personal care home, or with questions about the inspection results for a particular home, should contact the personal care home licensee directly.
Although efforts are made to ensure the accuracy of this information, the Saskatchewan Ministry of Health makes absolutely no warranty regarding the accuracy or completeness of the information. Neither the Ministry of Health nor the Government of Saskatchewan is liable for any errors or omissions in the information or for any losses or damages arising from the use of any information accessed within or through this website.


Public service announcement created for the WKU Centennial celebration. This video features agriculture, university farm and the Agriculture Exposition Center.
Western Kentucky University, "UA1B2/1 Agriculture at WKU Centennial PSA" (2006). WKU Archives Records. Paper 1808.


How Does Flexible Learning Work?
Get a rewarding career as a licensed practical nurse through blended delivery in Stettler/Drumheller and Rocky Mountain House! You will provide care in a variety of health care settings, such as hospitals, the community and other health care agencies.
RDC offers a variety of Flexible Learning courses to fit your schedule.
Excel in the Administrative Professional career with the Administrative Professional Certificate offered through Flexible Learning at RDC.


Awesome Buyer, thanks Solomon .
Thanks for your business. Look forward to your next campaign!
A very good buyer! Thanks for business Solomon!
Hey thanks for your business!
Thanks for letting me send you some traffic. I hope it was successful for you.


19/11/2013 · When you get a Toronto girl in the right situation be prepared for lots of LMR and usually some drama about an ex-bf yada yada. I would love for you to come here and see what it's like, being from LA will be the ultimate DHV and I am sure girls will be down.... 19/11/2013 · When you get a Toronto girl in the right situation be prepared for lots of LMR and usually some drama about an ex-bf yada yada. I would love for you to come here and see what it's like, being from LA will be the ultimate DHV and I am sure girls will be down.
Is anyone getting One Night Stands in Toronto?
Nataska Gunter, a 35-year-old trained nurse, had a one night stand with a man she'd met in a nightclub two years ago which resulted in the birth of her son, Tariq.... Nataska Gunter, a 35-year-old trained nurse, had a one night stand with a man she'd met in a nightclub two years ago which resulted in the birth of her son, Tariq.
19/11/2013 · When you get a Toronto girl in the right situation be prepared for lots of LMR and usually some drama about an ex-bf yada yada. I would love for you to come here and see what it's like, being from LA will be the ultimate DHV and I am sure girls will be down.
Nataska Gunter, a 35-year-old trained nurse, had a one night stand with a man she'd met in a nightclub two years ago which resulted in the birth of her son, Tariq.


You're welcome for the shot of Daisy eating. That one's a freebie.
In order to avoid eating my weight in salad every meal (I still have PLENTY in the garden), it was giveaway time for the lettuce, and blanch-and-freeze time for the spinach, which means it was large-batch produce cleaning time in my kitchen sink. I thought I'd share my method for doing this. It works great when you have a voluminous amount of greens like I did, but is also perfect for taking care of a week's worth of assorted produce, all at once. You can unload your grocery bag right into the sink with this method. I like it because it keeps the veggies crispy-crunchy until you use them – often a problem with organic produce that hasn't been treated to withstand a long, diesel truck drive to you.
"Victoria" variety butterhead lettuce leaves.
Step one: Fill a large bowl (or your very clean sink) with a 1:4 ratio of vinegar and water.
Step two: Mix up the vinegar and water, then dump in the goods. Swish them around and let them soak for a good five or ten minutes. This is where the magic happens.
This is the part of the process when you're really thankful you decided to do this.
My love. My salad spinner.
Give them a quick rinse. If you are using a colander, drain and pat the veggies dry. If you are using a salad spinner, spin away. Get the veggies completely dry, as you're likely to be saving them for future use.
The same process with giant spinach leaves.
Step four: Store as needed, and enjoy soon. I am going to blanch, drain, and freeze the spinach, for some lovely party appetizers at some point down the road. One bag of lettuce went into a Father's Day salad, and the other two went to obliging relatives.
I'd love to hear if any of you have had success with different ways to extend the life of your fresh produce. Chime in!
16 Jun	This entry was published on June 16, 2014 at 8:33 pm and is filed under Collection.	Bookmark the permalink.	Follow any comments here with the RSS feed for this post.


text:
Rock In a Hard Place: Top Ten Underrated Christian Rockers of the Past
By Andy_Argyrakis
Posted: Mon, 06/15/2009 - 16:23
Long before she married Jeremy Camp, Adrienne Liesching (now Adie Camp) fronted The Benjamin Gate.
By Andy Argyrakis, senior music editor, GospelMusicChannel.com
Sometimes critical praise doesn't add up to commercial success, and even if an act is lucky enough to score some singles, its legacy can often be confined to strictly those in the know. Looking back over the past 10 years, there are plenty of artists who deserved to make major waves, and while many did for a time, some have been relegated to the bargain bin or have broken up all together. Regardless of the reasons, here are 10 immensely underrated blasts from the past worth rediscovering or checking out for the very first time.
1. Burlap To Cashmere
Nowadays, Steven Delopoulos may have carved out a name for himself on the solo circuit, but about a decade ago, he was riding high as frontman for Burlap To Cashmere. The group was one of the first to ever spawn a reverse crossover, initially signing with A&M before inking a Christian market deal with the now defunct Squint Entertainment. The group's live shows were absolutely incendiary thanks to a robust blend of ethnic instrumentation and alternative rock, but considering the band thought so far outside of the box when it came to jamming, Christian radio would only play its tamer tunes.
2. Over the Rhine
Though never marketed specifically to a faith-based audience, Over the Rhine's music is loaded with soul-searching qualities. But the husband/wife duo Linford Detweiler and Karin Bergquist are amongst the most eloquent across the entire music industry (even scoring a coveted slot as Bob Dylan's opening act), while providing a plethora of earthy arrangements and organic experimentation.
3. Seven Day Jesus
In late 1997, hardly an hour went by when this band's ultra-catchy "Butterfly" wasn't played on the radio, but that simple song really didn't represent Seven Day Jesus' genuine sound. Instead, the group could be compared to Jeff Buckley with a gasoline fire lit under him, scoring with the socially conscious, and also forging the occasional controversial topic (including a commentary on rape recovery from a Christian perspective). After just two stellar albums, the group broke up and has since pursued a variety of less prominent solo projects.
4. The Benjamin Gate
Long before she married Jeremy Camp, Adrienne Liesching was fronting the ferocious alternative rock act The Benjamin Gate (complete with a gas mask as a band logo). In 2002, the group released its final (and best) effort called Contact, which featured a complex cover of Men At Work's mainstream smash "Overkill" and was promoted heavily on newsboys' Festival Con Dios tour. Though the group is on indefinite hiatus, Adie continues her individual career on BEC Recordings.
5. Aaron Sprinkle
Every true Christian rock fan is no doubt familiar with producer Aaron Sprinkle, who's been behind the boards on major albums including Anberlin, Kutless, Jeremy Camp and Hawk Nelson. But many younger listeners may not even realize he led several bands (Poor Old Lu, Rose Blossom Punch, Fair) and even had a streak of solo CDs from 1999–2004. Each of these efforts is worth exploring to see the multi-talent's singer/songwriter side, which certainly deserved radio airplay despite being confined to mostly indie appreciators.
~ (continued from page 1)
6. Chasing Furies
This year marks the 10th anniversary of Sparrow Records signing Chasing Furies for its sole CD With Abandon. These dreamy alternative darlings truly pushed the envelope of melody and dissonance, stirring the waters within the critics' pool, but finding a tougher time connecting with the masses. These days, frontwoman Sarah MacIntosh is on her own and recently dropped the slightly more accessible The Waiters, The Watchers, The Listeners, The Keepers and Me.
7. Luna Halo
Long before signing with Rick Rubin's American Recordings on the mainstream side of the fence, Luna Halo was also inked to Sparrow, dropping the one and only faith-based full-length, Shimmer. The hypnotic disc was instantly heralded as a triumph, though members soon sought to spread their wings beyond the Christian world, since their ambitious musical ideas tended to break the mold. After much delay, 2007 spawned a self-titled mainstream debut, but many early fans felt alienated by the switch from ethereal alternative to garage rock.
8. Bleach
Bleach is perhaps the best example of a modern rock band that consistently improved with each album released, starting off as a somewhat sloppy pop/rock outfit in 1995 that became immensely respected by its 2004 farewell. Radio dwellers will probably best recall the youth group-friendly 1998 single "Super Good Feeling," but latter projects found the focus leaning towards Weezer-esque power pop with a Jimmy Eat World undercurrent.
9. Stavesacre
This group's three Tooth & Nail recordings throughout the second half of the 1990s are often regarded as its best, with less ability to reclaim its brooding brand of hard rock when switching to secular labels like Nitro and Abacus. To this day, 1999's cover of The Cure's "Fascination Street" remains a concert staple that seamlessly bridges the gap between believers and the group's unchurched audience.
10. The W's
Remember when Brian Setzer Orchestra was bringing back swing just before the new millennium? Around the same time, Christian music had its own rhythm kings in the form of The W's, a flagship act on the now-stagnant Five Minute Walk Records. The smashing single "The Devil Is Bad" doesn't get much airplay these days, but it's a quintessential snapshot of the era and points to the players' underappreciated instrumental abilities.
Andy Argyrakis is a Chicago-based entertainment writer/photographer who appears in the Chicago Tribune, Illinois Entertainer, Daily Journal, Concert Livewire, Hear/Say Magazine and Image Chicago (to name few). His record label writing credits include Warner Brothers, Atlantic, Curb, EMI and Universal, with additional photo credits for Fuse TV, Live Nation, Nikon, Pollstar, Celebrity Access, Paste Magazine, MTV.com and Vibe.com. He's also the author/narrator of "Access Matthews" (an audio CD tracing the career of Dave Matthews Band) and spends considerable time on tour, including outings with Arlo Guthrie, The Guess Who, Madina Lake (on Linkin Park's Projekt Revolution) and Gospel Music Channel's very own "Gospel Dream" (where he served as season one judge).
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text:
For most of the last 15 years the route in paintball was to go to a rental day or two then if you wanted to play more regularly the only real option was to get onto a sup'air team and join into whatever series was running at the time. CPPS and it's ilk is a world away from the fun of playing in the woods as a rental customer and this jump is jarring for many people and puts a lot of people off.
Back in the early 2000s many sites were too scared of moving customers away from paying £120 a case to paying £35 at a walk on. But back then sites were flooded with birthday parties, stag dos and hundreds of new players who kept the easy money rolling in. But the novelty of paintball as new thing died out and the bottom fell out of the rental market with a tiny fraction of players of what there once was. Sites are now learning there is smarter ways of keeping the customers they have, which is why there's a heck of a lot more scenario events and walk ons in the UK than there used to be. Nowadays if a new rental baller wants to play more they see many more options, which are a lot more appealing to the sensibilities of someone who went to a rental day in the first place.
The resurgence of competitive woodsball is the natural progression for this new model. Of course there's the old farts coming back but I think this is an incredible opportunity to keep new players involved in our silly little game. I've gotta add Nick & Sally at Skirmish Sherwood Forest (Nottingham) to your list Pete. They've been doing great work in developing an end to end quality experience with rental, walk-on, sup'air (tournaments & training) and now woodland tournaments all available at a site that's very well run and looked after.
Reactions: toecutter, Care_Bears and Robbo
Agreed H, I agree that Nick should have been included, my bad !!
Reactions: H and Care_Bears
I think you are right re: getting more folk involved but I also think this may be a gradual transition into more woodland events rather than a big bang. This may come across as being self-interested but I honestly think that the fate of the UK Woodland Masters series in 2017 will be a sign-post as to whether other folk and the industry should support this further in 2018 and beyond. We've put together a multi-site, multi-date series centred around a two day international event at Paintfest - and that is a real leap forward from what we ran in 2016 as a single site, 4 date series.
Again, I think people may see this as self-interest but the two dayer at Paintfest has the potential to generate a lot of momentum for woodsball. Just as you and Marcus were talking about the size of our last event (possibly the biggest UK woodsball event for the last 17 years), if together with Ged we can get circa 40 teams out for a two day international event then I think that will grab the next level of attention from players and the industry! With some early interest from the States and Europe, I am excited to see what may happen!
Totally agree with all of your points re: a new growth area for the industry - as a product, competitive woodsball really offers a lot and the guys playing out there are by no means using starter equipment or low end paint - we see top end gear and paint getting used out in the woods!
2017 is a big step forward and fingers crossed will open the door to bigger steps forward for 2018 and beyond!
Edit - we've even managed to get Ian Miller out to play the odd event!
olliewidd
I recently spoke to a guy who plays with the American team Master Blasters (old school pro team apparently) and they're planning on coming over for a couple of events next year
Reactions: Dave S ECI
Dave S ECI said:
Dave, I'd like to think this resurgence follows the same trajectory as woodland paintball did during its genesis back in 1990 onwards .... I see no real reason why it shouldn't but at the back of my mind there's a fundamental difference between both growth curves.
Admittedly, this recent renaissance is just getting underway and so any parallel being drawn by myself is borne more out of wishful thinking than of natural market forces.
I know this much, we are entering a period that needs managing correctly if we are to optimise this opportunity - it's not as if we're under-resourced in this area, a lot of our industry have been here before and so logistically we're well placed.
The most important consideration is always going to be the people involved and to that end, our industry line-up looks good.
And with no disrespect intended Dave, I don't know you but you seem to be doing the right things, and saying what's needed.
The likes of Ged Green, Rob Hollington, Nick Sanders, Jim and Tim from Warped, Ainsley, yourself and Twizz, and it makes absolute sense to throw Ledzy into the mix if only because he's one of the very few people I know who can see things from a multiplicity of vantage points not least of which is from the new customer's requirements.
If anybody is able to make a success of what now confronts us then that list of industry faces are up to the job .....
It's very rare in any sport that turning the clock back when it comes to the fundamentals such as playing formats and/or locations proves successful but maybe paintball can carve out its own destiny in this respect, there's a lot riding on it, that's for sure !
Reactions: H and Dave S ECI


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The Principal's Desk
Greetings from the Principal of Hansraj College! As a premier institution of higher education, Hansraj College offers its students a unique blend of modern education anchored in hallowed Vedic values. The college is guided in its mission of combining the best of tradition and modernity by the vision of Mahatma Hans Raj. It is in the memory of Mahatma ji, a pioneer in the field of education, that the D.A.V. College Management Committee had founded the college on 26th July, 1948. The college has been contributing to the solemn task of nation-building since its very inception. The exemplary service the college has rendered society has received due recognition in several nationwide surveys that have consistently ranked Hansraj College among the top five out of about 5000 colleges in India.
Hansraj College is known for offering a state of art infrastructure to meet the pedagogic and research needs of its students and faculty members. These include a well-stocked Central Library and a Book Bank for economically disadvantaged students. The science departments have sprawling wings that house eighteen Science Labs and six Computer Labs.
Alongside its rigorous academic standards the college maintains a vibrant tradition of co-curricular activities. Our students complement their academic lives with active engagement in the fields of sports, theatre, debate, music, photography and the fine arts. The college auditorium, amphitheatre and the multiple seminar rooms equipped with latest sound and projection systems facilitate a plethora of academic and co-curricular exchanges. The college also boasts of an exemplary record of in the field of sports. We are the only college in the University to house an Electronic Shooting Range. Our sportspersons also have access to a sprawling play ground, an Indoor Sports Complex and several playrooms.
I am fortunate to be a part of the Hansraj family which untiringly assists me in my efforts to continually raise the benchmark of our success. A dedicated team of faculty members and non-teaching staff form the backbone of our institutional machinery. I view my role as an administrator and a friend who facilitates overall development of the students of the college.
I welcome the new batch of students and invite them to embark on an intellectually invigorating journey with us.
Dr. Rama


text:
Q: Calculating distances from latitude and longitude coordinates in R I am trying to calculate distance between one location and all other locations from their latitudes and longitudes.
I have been using distanceTrack from the argosfilter package to calculate the distance between consectuive locations which works very simply as this is a the code I need to make it work:
    lat<-lizard$Latitude
    lon<-lizard$Longitude
    distanceTrack(lat, lon)

But now I am wanting to calculate the distance between the first location and the second location, and then the distance between the first and third location, etc. for all locations.
I have been looking for a way to do this quickly as the only way I have been able to do it is manually copying and pasting the desired coordinates into a different csv file and then uploading it to R and running the above code, which is arduous. I am thinking an apply function or something similar might work. So would an apply function be the right way forward? And I don't know how to write the code to specify the rows to calculate the distance between, so any help with that would be great.

A:  earthDist <- function (lon1, lat1, lon2, lat2){
    rad <- pi/180
    a1 <- lat1 * rad
    a2 <- lon1 * rad
    b1 <- lat2 * rad
    b2 <- lon2 * rad
    dlon <- b2 - a2
    dlat <- b1 - a1
    a <- (sin(dlat/2))^2 + cos(a1) * cos(b1) * (sin(dlon/2))^2
    c <- 2 * atan2(sqrt(a), sqrt(1 - a))
    R <- 6378.145
    d <- R * c
    return(d)
}

earthDist(lon[1], lat[1], lon, lat)

A: Sample input data I'll use:
p <- data.frame(lat=runif(6,-90,90), lon=runif(6,-180,180) );
p;
##         lat        lon
## 1 -27.85808  160.23800
## 2  31.14363  -45.22589
## 3 -50.01119   48.84754
## 4  68.11402   71.46464
## 5  45.58087 -104.46365
## 6  87.49930  -88.12699

If you only want the distances between the first point and all subsequent points, you can do this:
d <- sapply(2:nrow(p), function(x) distance(p$lat[x],p$lat[1],p$lon[x],p$lon[1]) );
d;
## [1] 17517.58  9037.04 12806.45 12557.67 13196.02

If you want all combinations of points, it's more complicated. In the below, I've switched from a simple vector to a data.frame for the result, so the two point indexes can be stored alongside each distance value:
d <- setNames(do.call(rbind.data.frame,combn(1:nrow(p),2,simplify=F)),c('p1','p2'));
d$dist <- sapply(1:nrow(d), function(r) distance(p$lat[d$p1[r]],p$lat[d$p2[r]],p$lon[d$p1[r]],p$lon[d$p2[r]]) );
d;
##    p1 p2      dist
## 1   1  2 17517.583
## 2   1  3  9037.040
## 3   1  4 12806.448
## 4   1  5 12557.672
## 5   1  6 13196.020
## 6   2  3 12868.340
## 7   2  4  7815.027
## 8   2  5  5276.540
## 9   2  6  6338.329
## 10  3  4 13259.829
## 11  3  5 17961.347
## 12  3  6 15757.656
## 13  4  5  7363.244
## 14  4  6  2694.062
## 15  5  6  4669.714

redpajama_set_name: RedPajamaStackExchange


you have an advantage if you have important information with you, well-organized and written down. This step helps you record important information.
There are 2 sections to complete under Step # 1: My Contact Information (Include People I Live With and People Who Can Help Me ) and My Important Documents.
Who can help me during a disaster?
Have I shared my contact information and important documents with that person(s)?
Where should I store my important documents and contact information at home?
Fill in the tables below, and print them, to include as part of your personal Emergency Readiness Plan.


This year's survey continues to report on salary and bonus data, along with compensation and benefit information including insurance coverage, paid time off, retirement plans, turnover, work – life, etc.
Compensation Resources, Inc. (CRI) and Christian Leadership Alliance (CLA) have released the 2018 Compensation Survey Report for Christian Organizations. This year's survey continues to report on salary and bonus data, along with compensation and benefit information including insurance coverage, paid time off, retirement plans, turnover, work – life, etc. The data is reported in 8 Christian-based industrial classifications, and also by revenue, employee size, and region, to provide the user with various means by which to examine and evaluate results. Results are based on data provided by 84 participating organizations.
To learn more about the 2018 Compensation Survey Report for Christian Organizations, or to order the complete report, please contact Diana Neelman at (877) 934-0505 x105. You can also visit CRI's website at http://www.CompensationResources.com for more information and to download a free sample of this year's survey.
About Compensation Resources, Inc. (CRI): CRI provides compensation and human resource consulting services to mid- and small-cap public companies, private, family-owned, and closely held firms, as well as not-for-profit organizations. CRI specializes in executive compensation, sales compensation, pay-for-performance and incentive compensation, performance management programs, and expert witness services.
About Christian Leadership Alliance unites, trains, and equips Christian leaders to think higher for greater kingdom gain. CLA moves beyond good Christian management to empower Christian leadership that is catalytic in its transformation of people, their organizations, and ultimately the world. CLA is an alliance of mission focused Christians who lead in today's high-impact Christian nonprofit ministries, churches, educational institutes, and businesses. For more information about becoming a member, please contact CLA at (949) 487-0900 or visit us at http://www.ChristianLeadershipAlliance.org.


Water management officials of the Colorado Water Conservation Board (CWCB) and Reclamation wish to know the timing and volume of streamflows from snowmelt in Colorado rivers, for both water supply and flooding forecasts. A major limitation in these forecasts has been a lack of detailed knowledge of the spatial and temporal evolution of the snowpack's snow water equivalent (SWE) in the mountains. This R&D proposes to apply the most sophisticated modeling and data assimilation system available for SWE to answering this limitation. This system is the Snow Data Assimilation System or SNODAS.
SNODAS will be further enhanced by accurate, real-time, multisensor precipitation estimation in the form of a system known as QPE SUMS. This combined precipitation/snowpack measurement system will provide water managers with the most accurate and timely information about snowmelt water volumes, which represents crucial decision support for both CWCB and Reclamation water operations managers.
Snowfall monitoring and flood warning systems have traditionally relied on precipitation gauges to measure precipitation. A major problem with gauges is that they provide sparsely distributed point measurements; this is especially true in mountainous areas such as the Colorado Rockies. Another limitation for gauges is that they have more difficulties measuring snow vs. rain, and snow is a major part of the annual precipitation in the state, particularly in the mountains. The primary gauge system for snowfall measurements in the mountains is SNOTEL; these gauges are very few in number, particularly above 10,000 feet where much of the snowpack is found. Therefore it is difficult to accurately measure snowfall over wide areas of the mountains, and therefore difficult to gauge the snow water equivalent (SWE) in the mountain snowpack.
Even if snowfall is accurately measured, it is still a challenge to track ablation (melt and sublimation) of the snowpack and its SWE. This challenge owes to a situation analogous to that of snowfall measurement, i.e., the sparseness of measurement sites - current measurements are taken at SNOTEL sites and snow courses. This data sparseness leads to coarse and sometimes erroneous spring snowmelt and runoff forecasts. Since snowmelt contributes, on average, around 80% of water storage in Western reservoirs, such forecasts can be very problematic for Reclamation's water management. Not knowing how much snowmelt will enter the reservoirs can lead to inappropriate releases of water from dams, causing floods or reduced conservation pool storage and water delivery (volumes).


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Do you ask for the online price? I once went into Toys R Us and there was a huge discrepancy between store and online price. They said they could not price match themselves. So I ordered it online and picked it up.
In-store pickup can be a life saver! So many times I've been able to save time and multitask by ordering online and just running into the store quickly to pick up!
We've tried In-store pick up a few times and I have found the produce selection to be less than stellar. The bananas are overripe or very green, avocados are hard as a rock and mold on the raspberries. So for that reason we don't really do it. If it's dry good… then totally!
Oh, I agree about groceries and that I don't do. I'm talking about products, like electronics, handbags, toys.


text:
\section{Introduction and the main results}\label{introduction}

The Boussinesq system for a reactive flow is a  model describing
flame propagation in a gravitationally stratified medium
\cite{ZBLM85}. It consists of the reaction-advection-diffusion
equation for the reactant temperature $\theta$ (normalised  so that
$0\leq\theta\leq 1$) and the Navier-Stokes
equations for the evolution of the incompressible flow $u$. The
variables $u$ and $\theta$ are coupled through the advection velocity
in the reaction equation, and
through the force term in the fluid equation.
After passing to nondimensional variables
\cite{BCR06}, the simplified Stokes-Boussinesq system takes the form:
\begin{align}
&\theta_t +u\cdot\nabla\theta-\Delta\theta=f(\theta), \label{heat}\\
&u_t -\nu\Delta u +\nabla p=\theta\vec{\rho},  \label{fluid}\\
&\text{div}\, u =0. \label{solenoid}
\end{align}
Here, $\nu>0$ is the Prandtl number (inversely proportional to the
Reynolds number), while the reaction rate is given by a nonnegative,
nonlinear function $f(\theta)$  of ignition type. That is, $f$ is
Lipschitz continuous, and there is a threshold temperature $0<\vartheta_0<1$  such that:
 \begin{equation}\label{f}
 f(\theta)=0 \mbox{ for } \theta\le \vartheta_0 \mbox{ and } \theta\ge 1, \qquad
 f(\theta)>0 \mbox{ on } (\vartheta_0, 1), \qquad f'(1)<0.
 \end{equation}
The vector $\vec{\rho}=\rho \vec g$ corresponds
to the non-dimensional gravity $\vec g$ scaled by the Rayleigh number
$\rho$, and we assume that $\vec\rho$ is non-parallel to the unbounded
direction of the 3d channel $D$, which amounts to studying the system
(\ref{heat}) - (\ref{solenoid}) in:
$$D=(-\infty, \infty)\times \Omega = \{(x,\tilde x); ~ x\in
\mathbb{R}, ~ \tilde x\in\Omega\},$$
with:
$$\vec\rho \cdot e_3 \neq 0.$$
The crossection $\Omega$ is a sufficiently regular,
 connected and bounded domain in $\mathbb{R}^2$.
For the solutions to (\ref{heat}) - (\ref{solenoid}), we impose the Neumann  condition in
$\theta$, and the no-slip (Dirichlet) boundary condition in $u$:
\begin{equation}\label{bcond}
\frac{\partial \theta}{\partial\vec n} = 0 \quad \mbox{ and } \quad
u=0 \mbox{ on } \partial D.
\end{equation}

The classical result by Kanel \cite{K61, B03} states that there exists
the unique speed $c_0$ (necessarily positive) and a (necessarily
decreasing) traveling wave profile $\Phi:\mathbb{R}\longrightarrow
[0,1]$, satisfying:
\begin{equation*}\label{laminar}
 -c_0\Phi' - \Phi'' = f(\Phi), \quad \Phi(-\infty)=1, \quad \Phi(+\infty)=0.
\end{equation*}
We call $c_0$ and $\Phi$ the {\em laminar speed} and {\em laminar
  front}, and to fix
the ideas, we let $\Phi(0) = \vartheta_0$. Then $\theta(x,t) =
\Phi(x-c_0t)$ is the unique (up to translations) traveling-wave
solution to:
\begin{equation}\label{oned}
\theta_t - \theta_{xx} = f(\theta), \qquad \theta (x,0) = \theta_0(x)
\end{equation}
joining the two equilibria $1$ and $0$.
Moreover, if the initial data $\theta_0(x)$ differs from the Heaviside
function $H(x)$ by a compactly supported error, then there exists a
shift $x_0$ such that, for the solution $\theta$ to (\ref{oned}) there holds:
$$\|\theta(\cdot, t) - \Phi(\cdot + c_0t + x_0)\|_{L^\infty(\mathbb{R})}
\to 0  \quad \mbox{ as } t\to\infty.$$
Our purpose is to reproduce this result for the problem (\ref{heat}) -
(\ref{solenoid}), (\ref{bcond}).
Following \cite{CKR03} and \cite{BCR06}, we define the \emph{bulk
burning rate} $\bar B(t)$ and the \emph{Nusselt number}
$\bar N(t)$  by:
\begin{equation} \label{bn}
\begin{aligned}
&B(t) = \frac{1}{|\Omega|}\int_D f(\theta)~\mbox{d}x~\mbox{d}\tilde x,
\qquad &\bar B(t) = \frac1t \int_0^tB(s)~\mbox{d}s\\
&N(t) = \frac{1}{|\Omega|}\int_D |\nabla\theta|^2~\mbox{d}x~\mbox{d}\tilde x,
\qquad &\bar N(t) = \frac1t \int_0^tN(s)~\mbox{d}s,
\end{aligned}
\end{equation}
and the \emph{average  flow} $\bar U(t)$ by
\begin{equation} \label{ah}
 \bar U(t) = \frac1t \int_0^t\|u(\cdot,s)\|_\infty ~\mbox{d}s.
\end{equation}

\begin{theo}\label{main}
Assume that the initial temperature $\theta_0(x, \tilde x) \in [0,1]$ is
such that  $\theta_0(x,\tilde x) - H(x)$ is compactly supported in $D$.
Let $u_0\in W^{1,2}(D)$ and let $(\theta, u, p)$ be a global solution of
\eqref{heat} - \eqref{solenoid}, \eqref{bcond} with:
$\theta(\cdot, 0) = \theta_0$ and $u(\cdot, 0) = u_0$.
Then, there exists a constant $C_\Omega$, depending only on $\Omega$,
such that as $t\to+\infty$:
\begin{align}
&c_0 - C_\Omega\left(\frac{\rho}{\nu}+\frac{\rho^2}{\nu^2}\right) - o(1)\le
\bar B(t) \le c_0 +
C_\Omega\left(\frac{\rho}{\nu}+\frac{\rho^2}{\nu^2}\right) + o(1), \label{estb}\\
&\bar N(t)\le \left(C_\Omega\frac{\rho}{\nu}
  +\sqrt{\frac{c_0}{2}+C_\Omega\frac{\rho^2}{\nu^2}}\,\right)^2
+ o(1),\label{estn}\\
&\bar U(t)\le C_\Omega\left(\frac{\rho}{\nu}+\frac{\rho^2}{\nu^2}\right) + o(1).\label{estu}
\end{align}
\end{theo}
The above result shows that the solution of the
initial-boundary problem for \eqref{heat} - \eqref{solenoid}, with
small $\rho/\nu$ propagates
with finite speed close to the laminar front speed. Also, note that if
we replace $\theta$ with the laminar front $\Phi$, then a simple
integration of \eqref{oned}  gives:
$ \bar B(t)=c_0.$
This corresponds with the estimates in Theorem \ref{main}
when $\rho = 0$, i.e. the system
 \eqref{heat} - \eqref{solenoid} turns out to be
a regular perturbation of the reaction--diffusion equation.

Our next result states that with front-like initial data,
the solution to the studied system stays front-like:
\begin{theo}\label{secondtheo}
With the same assumptions as in Theorem \ref{main}, we have:
\begin{equation}
\begin{split}
&\Phi\left(x-c_0t+x_0+\bar U(t)t +C_{\Omega,0}\sqrt{t}\right)-\frac{C_{\Omega,0}}{\sqrt{t}}\le \theta(x,\tilde
x, t) \\ & \qquad\qquad\qquad
\le \Phi\left(x-c_0t-x_0+\bar U(t)t +C_{\Omega,
    0}\sqrt{t}\right)+\frac{C_{\Omega,0}}{\sqrt{t}}
\qquad \forall t>> 1,
\end{split}
\end{equation}
with appropriate $x_0>0$ and $C_{\Omega,0}$ depending on $\Omega$, $f$
and the initial data $\theta_0$.
\end{theo}

Finally, we have:

 \begin{theo}\label{remi}
The results in Theorem \ref{main} and Theorem \ref{secondtheo} remain
valid in either of the following cases:
\begin{enumerate}
\item[(i)] the channel $D=(-\infty, +\infty)\times [0,\lambda]$ is
  2-dimensional
\item[(ii)] the flow equation (\ref{fluid}) is replaced by the stationary
  Navier-Stokes:
$$-\nu\Delta u + u\cdot \nabla u + \nabla p = \theta\vec\rho,$$
and the crossection $\Omega$ is sufficiently ``thin'', i.e it
satisfies the same type of condition as in \cite{LM09}:
$$\frac{\sqrt{3}}{2}\frac{\rho}{\nu\sqrt{\pi \nu}} |\Omega|^{1/2} C_P \left(C_{PW} +
  \left(\fint|\vec g\cdot (0,\tilde x)|^2~\mbox{d}\tilde x\right)^{1/2}
\right) < 1.$$
\end{enumerate}
\end{theo}

Existence of traveling fronts for the Navier-Stokes-Boussinesq system
 in infinite channels has been the subject  of research during recent years
 \cite{CKR03, BCR06, TV01, CLR06, Lew07, LM09}.  Of interest has been also an
 understanding of the regularizing and mixing effect of convection
\cite{CNR08, CKRZ08}.  In \cite{CKR03} and \cite{BCR06}, the
 solutions of the system with front-like datum in a 2d strip have
been considered and uniform estimates for the full
Navier-Stokes-Boussinesq system
have been obtained for the \emph{stress-free}
boundary conditions on $u$.
In this paper we generalize these results to dimension 3 and the
more physically relevant \emph{no-slip} boundary conditions.
The analysis follows \cite{BCR06} closely; it first
seeks bounds for $\bar N$ and $\bar B$ using the
parabolic equation \eqref{heat}. Where our argument diverges
from that of  \cite{BCR06} is in finding an estimate for $\|u\|_{\infty}$.
In \cite{BCR06}, this has been done using
Poincar\'e's inequality for  vorticity,
based on the assumption that the vorticity vanishes at
the boundary, and using the rather simple form of the vorticity
equation in dimension two.
In the present case, we rely on the almost-uniform Xie bound
in Lemma \ref{supest}.

\bigskip

\noindent{\bf Acknowledgments.}
M.L. was partially supported by the NSF grants DMS-0707275 and DMS-0846996,
and by the Polish MN grant N N201 547438.
M.R. was partially supported by the IPM grant no 90350025.

\section{Auxiliary results}\label{estimates}

Our first result is a technical lemma, which extends Lemma
4.3 in \cite{BCR06}, proved there for 2d channels
(see also \cite{CKRZ08}). Here we prove a similar
result in dimension 3, showing that diffusion in a tube
behaves like 1d heat equation.
\begin{lem}\label{estymata}
Let $u\in W^{1,2}(D,\mathbb{R}^3)$ with $\mathrm{div } ~u = 0$ be a given
solenoidal flow satisfying $u=0$ on $\partial D$.
Let $\phi$ be the solution to the advection-diffusion equation:
\begin{equation} \label{phi}
\begin{aligned}
&\phi_t+u\cdot\nabla\phi-\Delta\phi = 0   \quad  \mbox{ for } (x,\tilde x) \in D \mbox{
  and } t>0\\
&\frac{\partial\phi}{\partial \vec n} (x,\tilde x, t)= 0 \quad \mbox{
  for } (x,\tilde x) \in \partial D\\
&\phi(x,\tilde x, 0)=\phi_0(x,\tilde x).
\end{aligned}
\end{equation}
Then there exists a constant $C_\Omega$ depending only on $\Omega$ (in
particular independent of $u$ and $\phi_0$) such that, for all
sufficiently large $t$, there holds:
\begin{equation*}\label{fee}
 \|\phi(\cdot,t)\|_\infty\le\frac{C_\Omega}{\sqrt{t}} \|\phi_0\|_{L^1(D)} \quad \forall
 t >> 1.
\end{equation*}
\end{lem}
\begin{proof}
{\bf 1.} We first prove a Nash-type inequality, valid for solutions of (\ref{phi}):
\begin{equation}\label{nash}
\|\nabla \phi(\cdot, t)\|_{L^2(D)}^2 \ge
C_\Omega\frac{\|\phi(\cdot, t)\|_{L^2}^6}{\|\phi(\cdot, t)\|^4_{L^1} +
  \|\phi(\cdot, t)\|_{L^1}\|\phi(\cdot, t)\|^3_{L^2}}  \qquad \forall t.
\end{equation}
To simplify the notation, in what follows we suppress the dependence
on $t$ and write $\phi$ instead of $\phi(\cdot, t)$, etc.

Define  $\psi(x)=\frac{1}{|\Omega|}\int_\Omega \phi(x, \tilde x) ~\mbox{d}\tilde x$ and
let $k(x,\tilde x)= \phi (x, \tilde x)-\psi (x)$, so that $\phi=\psi+k$
and $\int_\Omega k(x, \tilde x) ~\mbox{d}\tilde x=0$
for every $x\in\mathbb{R}$. Notice that, by Cauchy-Schwarz inequality:
\begin{equation}\label{2.6}
\|\psi\|^2_{L^2(D)}= |\Omega| \int_{\mathbb{R}} \left(\fint_\Omega \phi
~\mbox{d}\tilde x\right)^2 ~\mbox{d}x \leq \int_{\mathbb{R}} \int_\Omega
|\phi|^2 = \|\phi\|^2_{L^2(D)},
\end{equation}
Similarly:
\begin{equation}\label{2.6a}
\begin{split}
& \|\psi'\|_{L^2(D)}\le \|\phi_x\|_{L^2(D)}, \quad
\|\psi\|_{L^1(D)}\le \|\phi\|_{L^1(D)},\\
& \|k\|_{L^2}\le 2\|\phi\|_{L^2}, \quad \|\nabla k\|_{L^2}\le
2\|\nabla\phi\|_{L^2}, \quad  \|k\|_{L^1}\le 2\|\phi\|_{L^1}.
\end{split}
\end{equation}
Let $\hat\psi:\mathbb{R}\longrightarrow \mathbb{C}$ be the Fourier
transform of $\psi$, i.e.
$\hat\psi (\omega) = \int_\mathbb{R}\psi(s) e^{-2\pi i\omega s}
~\mbox{d}s.$
By Plancherel's identity, we have:
\begin{equation}\label{2.8}
\|\psi\|^2_{L^2(\mathbb{R})}= \|\hat\psi\|^2_{L^2(\mathbb{R})}, \qquad
\|\psi'\|^2_{L^2(\mathbb{R})}= \|2\pi i\omega\hat\psi\|^2_{L^2(\mathbb{R})}.
\end{equation}
For a given positive $m$, we now write:
\begin{equation*}
\begin{split}
\|\psi\|^2_{L^2(\mathbb{R})} &=
\int_{|\omega|\le m}|\hat\psi(\omega)|^2~\mbox{d}\omega
+ \int_{|\omega|> m}|\hat\psi(\omega)|^2~\mbox{d}\omega\\
&\leq 2 m \|\psi\|^2_{L^1(\mathbb{R})} + \frac{1}{m^2}
\int_\mathbb{R}\omega^2|\hat\psi(\omega)|^2 ~\mbox{d}\omega \\ &\leq
2 m \|\psi\|^2_{L^1(\mathbb{R})} +
 \frac{1}{4\pi^2m^2}\|\psi'\|^2_{L^2(\mathbb{R})}
\end{split}
\end{equation*}
where the estimate of the first term follows by: $|\hat\psi(\omega)
|\leq \|\psi\|_{L^1(D)}$, while to estimate the second term we used  (\ref{2.8}).

Setting $m= ||\psi'||^{2/3}_{L^2(\mathbb{R})}||\psi||^{-2/3}_{L^1(\mathbb{R})}$, we obtain:
\begin{equation*}
\|\psi\|^2_{L^2(\mathbb{R})}\le
(\frac{1}{4\pi^2}+2)\|\psi\|^{4/3}_{L^1(\mathbb{R})}\|\psi'\|^{2/3}_{L^2(\mathbb{R})},
\end{equation*}
which implies:
\begin{equation*}
\|\psi\|^2_{L^2(D)}\le
(\frac{1}{4\pi^2}+2) |\Omega|^{-2/3}\|\psi\|^{4/3}_{L^1(D)}\|\psi_x\|^{2/3}_{L^2(D)}.
\end{equation*}
Now, by Cauchy-Schwarz inequality, the Sobolev embedding
$W^{1,2}(D)\hookrightarrow L^4(D)$, and the Poincare-Wirtinger
inequality on $\Omega$, if follows that:
\begin{equation*}
\|k\|^2_{L^2(D)}\le \|k\|^{2/3}_{L^1(D)} \|k\|^{4/3}_{L^4(D)}\le C_\Omega \|k\|^{2/3}_{L^1(D)} \|\nabla k\|^{4/3}_{L^2(D)}.
\end{equation*}
Therefore, by (\ref{2.6}) and (\ref{2.6a}):
\begin{equation*}
\begin{split}
\|\phi\|^2_{L^2(D)} & \le 2(\|\psi\|^2_{L^2(D)}+ 2\|k\|^2_{L^2(D)})
\\ & \le C_\Omega
\left(\|\phi||^{4/3}_{L^1(D)}\|\nabla\phi\|^{2/3}_{L^2(D)} + \|\phi\|^{2/3}_{L^1(D)} \|\nabla \phi\|^{4/3}_{L^2(D)}\right).
\end{split}
\end{equation*}
We now argue as in \cite{BCR06}. Since
$y:=\|\nabla\phi\|^{2/3}_{L^2}$ satisfies: $ay^2 + by - c \geq 0$ with
appropriate $a,b,c\geq 0$, then: $y\geq \frac{-b + \sqrt{b^2+4ac}}{2a}
= \frac{2c}{b+\sqrt{b^2+4ac}} \geq \frac{c}{\sqrt{b^2+4ac}}$. Hence:
\begin{equation*}
 \|\nabla\phi\|_{L^2}^{2/3}\ge \|\phi\|^2_{L^2}\left(\|\phi||_{L^1}^{8/3}
   + \|\phi\|_{L^1}^{2/3}\|\phi\|^2_{L^2}\right)^{-1/2},
\end{equation*}
which gives:
\begin{equation*}
\|\nabla\phi\|_{L^2}^{2}\ge
C_\Omega\|\phi\|^6_{L^2}\left(\|\phi\|_{L^1}^{4} + \|\phi\|_{L^1}\|\phi\|^3_{L^2}\right)^{-1},
\end{equation*}
yielding exactly \eqref{nash}.

\smallskip

{\bf 2.} Recall now that:
\begin{equation}\label{2.3}
 \|\phi\|_{L^1(D)} \le \|\phi_0\|_{L^1(D)}.
\end{equation}
Indeed, the $L^1$ norm of a solution to \eqref{phi}  is conserved when
the initial data is positive.
In the general case one can write $\phi_0=\phi_0^+-\phi_0^-$ where $\phi_0^+$
and $\phi_0^-$ are positive,  with disjoint supports. Solving
\eqref{phi} for each, one obtains the inequality (\ref{2.3}).
Further, integrating (\ref{phi}) against $\phi$ and using
incompressibility of $u$ and the boundary condition, it follows that:
\begin{equation}\label{2.4}
\frac{d}{\mbox{d}t}\|\phi\|_{L^2}^2=-2 \|\nabla\phi\|_{L^2}^2.
\end{equation}
In view of (\ref{nash}), (\ref{2.3}), (\ref{2.4}) we now obtain:
\begin{equation*}
\frac{d\|\phi\|_{L^2}}{\mbox{d}t}
 \le C_\Omega\frac{\|\phi\|^5_{L^2}}{\|\phi_0\|^4_{L^1} + \|\phi_0\|_{L^1}\|\phi\|^3_{L^2}},
\end{equation*}
which, after integrating in time, gives:
\begin{equation}\label{2.18}
t\le C_\Omega\left(\frac{\|\phi_0\|^4_{L^1}}{\|\phi\|^4_{L^2}} + \frac{\|\phi_0\|_{L^1}}{\|\phi\|_{L^2}}\right).
\end{equation}
Call $\alpha = \|\phi\|_{L^2}/ \|\phi_0\|_{L^1}$. From (\ref{2.18}) it
follows that $\frac{\alpha^4}{\alpha^3 + 1}\leq \frac{C_\Omega}{t}$.
The function $\alpha\mapsto\frac{\alpha^4}{\alpha^3 +1}$ is increasing
and it converges to $0$ as $\alpha\to 0$. Let $\beta$ be the unique
solution to $\frac{\beta^4}{\beta^3 + 1} =  \frac{C_\Omega}{t}$, so
that $\alpha\leq \beta$. Now, for $t\to\infty$ clearly
$\frac{C_\Omega}{t}\to 0$, hence also $\beta\to 0$ and
$\frac{\beta^4}{2} \leq \frac{\beta^4}{\beta^3 + 1} = \frac{C_\Omega}{t}$.
Consequently, $\beta \leq \frac{C_\Omega}{t^{1/4}}$ and we arrive at:
$$ \|\phi\|_{L^2}\leq
\frac{C_\Omega}{t^{1/2}}\|\phi_0\|_{L^1} \qquad \forall t>>1.$$

\smallskip

{\bf 3.} We now argue as in \cite{BCR06}.
Let $\mathcal{P}_t$ be the solution operator for \eqref{phi}. Thus far, we have showed that:
\begin{equation*}\label{pt}
\|\mathcal{P}_t\|_{L^1\to L^2}\le \frac{C_\Omega}{t^{1/4}} \qquad \forall t>>1.
\end{equation*}
Now if $\mathcal{P}_t^*$ is the adjoint operator, then $\mathcal{P}_t^*$ is the
solution operator to (\ref{phi}),  with $u$ replaced by $-u$. Therefore the above argument works again:
\begin{equation*}\label{pt*}
\|\mathcal{P}^*_t\|_{L^1\to L^2}\le \frac{C_\Omega}{t^{1/4}} \qquad \forall t>>1.
\end{equation*}
Finally, we conclude the lemma:
\begin{equation*}\label{ptt}
\begin{split}
\|\mathcal{P}_{2t}\|_{L^1\to L^{\infty}} & \le \|\mathcal{P}_t\|_{L^1\to
  L^2}\|\mathcal{P}_t\|_{L^2\to L^{\infty}} \\ & =\|\mathcal{P}_t\|_{L^1\to
  L^2}\|\mathcal{P}^*_t\|_{L^1\to L^2} \leq \frac{C_\Omega}{t^{1/2}} \qquad
\forall t>>1.
\end{split}
\end{equation*}
\end{proof}

We now present a lemma taken from \cite{LM09}:
\begin{lem} \label{supest}
Let $g\in L^2(D)$.
There exists a constant $C_\Omega$, depending only on the crossection
$\Omega$, such that for any solenoidal flow $u\in W^{2,2}\cap
W^{1,2}_0(D)$ satisfying:
\begin{equation*}\label{upf}
-\nu \Delta u+\nabla p= g , \qquad \mbox{div } u = 0 \mbox{ in } D,
\end{equation*}
there holds:
\begin{equation}\label{ineq}
\|u\|_{\infty}\le \frac{\sqrt{2}}{\sqrt{\nu\pi}}\|\nabla u\|^{1/2}_{L^2(D)}
\|g\|^{1/2}_{L^2(D)}+ C_\Omega\|\nabla u\|_{L^2(D)}.
\end{equation}
\end{lem}

We remark that the proof of (\ref{ineq}) relies on Xie's estimate \cite{Xie92}:
\begin{equation*}\label{xie}
 \|u\|_\infty \le \frac{1}{\sqrt{2\pi}}\|\Delta u\|_{L^2(D)}^{1/2}\|\nabla u\|_{L^2(D)}^{1/2}
\end{equation*}
valid for $u\in W^{2,2}(D)\cap W^{1,2}_0(D),$ and on a commutator estimate
\cite{LLP07}:
\begin{equation*}\label{pll}
 \|(\mathcal{P}\Delta-\Delta\mathcal{P})u\|^2_{L^2(D)}\le \left(\frac12
   +\epsilon\right)\|\Delta u\|^2_{L^2(D)}  + C_{D,\epsilon}\|\nabla u\|^2_{L^2(D)}
\end{equation*}
where $\mathcal{P}$ is the Helmholz projection onto the space of solenoidal vector
fields.

The interest in the inequality (\ref{ineq}) lies in the independence of the
constant $\frac{\sqrt{2}}{\sqrt{\pi \nu}}$ at the term involving $g$.
Indeed, this was the key argument allowing to prove \cite{LM09} existence of traveling
waves for the full Navier-Stokes-Boussinesq system in 3d channels,
satisfying appropriate ``thinness'' condition on the crossection
$\Omega$.
The same argument is needed to obtain the uniform bounds for the stationary
Navier-Stokes-Boussinesq system in Theorem \ref{remi} (ii).

On the other hand, by elliptic estimates and the Sobolev imbedding, it follows directly that:
\begin{equation}\label{easy}
\|u\|_\infty \leq C_\Omega \|u\|_{W^{2,2}(D)} \leq C_\Omega \left(\|\nabla u\|_{L^2(D)}
+ \frac{1}{\nu}\|g\|_{L^2(D)}\right).
\end{equation}
In fact, already this inequality is sufficient for the
estimates in case of the Stokes-Boussinesq system.

\bigskip

We will also need the following result, in the line of Lemma 3.6 from \cite{LM09}:
\begin{lem}\label{h}
For each $t$, there exists a function $h\in W^{1,2}_{loc}(D)$, such that:
\begin{equation*}
\|\theta(\cdot, t) \vec{\rho}-\nabla h\|_{L^2(D)}\le C_\Omega \rho \|\nabla \theta(\cdot, t)\|_{L^2(D)},
\end{equation*}
with $C_\Omega$ depending only on  $\Omega$.
In fact:
\begin{equation}\label{const_exact}
C_\Omega = C_{PW}+ \left(\fint_\Omega|\vec g \cdot (0,\tilde
  x)|^2~\mbox{d}\tilde x\right)^{1/2},
\end{equation}
where $C_{PW}$ stands for the Poincare-Wirtinger constant of $\Omega$.
\end{lem}
\begin{proof}
Let $e_1, e_2, e_3$ be the standard basis for $\mathbb{R}^3$.
Suppressing the time variable $t$, we define:
\begin{equation*}
h(x,\tilde x)= \vec{\rho} \cdot e_1\int_0^x\fint_\Omega\theta(x,\tilde x)~\mbox{d}\tilde x ~\mbox{d}x +
\vec{\rho} \cdot (0,\tilde x)\fint_\Omega\theta(x,\tilde x) ~\mbox{d}\tilde x.
\end{equation*}
Therefore, the following identity concludes the proof of lemma:
\begin{equation*}
\theta\vec{\rho} -\nabla h= \vec{\rho}\left(\theta(x, \tilde x) - \fint_\Omega\theta(x,\tilde x) ~\mbox{d}\tilde x\right)-
\vec{\rho} \cdot (0,\tilde x) \fint_\Omega\theta_x(x,\tilde x)~\mbox{d}\tilde x~ e_1,
\end{equation*}
by the Poincare-Wirtinger inequality: $\|\theta -
\fint \theta \|_{L^2(\Omega)} \leq C_{PW} \|\nabla\theta\|_{L^2(\Omega)}$.
\end{proof}

\section{Proofs of the main result}
The following lemma has been proven in \cite{BCR06} for the case of 2d channels
and vorticity of the flow $u$ vanishing at $\partial D$. Exactly the
same proof, relying on the construction of super and sub-solutions to
(\ref{heat}) is valid also in the present 3d case. Since the argument
uses the estimate in Lemma \ref{estymata}, we partially reproduce it
below for the sake of completeness.

\begin{lem} \label{1stlem}
There exists a constant $C_{\Omega,0}$ depending on $\Omega$, $f$,
and on the initial data $\theta_0$, such that:
\begin{equation}\label{1st}
\bar N(t) \le \frac 12 \bar B(t) + \bar U(t) + C_{\Omega, 0}
\left(\frac1t+\frac{1}{\sqrt{t}}\right) \qquad \forall t>>1.
\end{equation}
Moreover, there exists $x_0>0$ and $q\in L^1(\mathbb{R})$, such that
for all sufficiently large $t>>1$:
\begin{equation}\label{uplowbound}
\begin{split}
& \Phi\left(x-c_0t+x_0+\bar U(t)t + C_{\Omega,0}\sqrt{t}\right) -
Q(x,\tilde x, t)
\leq \theta(x, \tilde x, t) \\
& \qquad\qquad \qquad \leq
\Phi\left(x-c_0t-x_0 - \bar U(t)t - C_{\Omega,0}\sqrt{t}\right) +
Q(x,\tilde x, t),
\end{split}
\end{equation}
where $Q$ is the solution to:
\begin{equation}\label{form}
\begin{split}
& Q_t + u\cdot\nabla Q - \Delta Q = 0 \quad \mbox{ in } D,\\
& \frac{\partial Q}{\partial \vec n} = 0 \mbox{ on } \partial D, \qquad
Q(x,\tilde x, 0) = q(x).
\end{split}
\end{equation}
\end{lem}
\begin{proof}
We only prove (\ref{uplowbound}), since it implies (\ref{1st}) as in
\cite{BCR06}, Lemma 4.2. Define:
$$\psi_l(x, \tilde x, t) = \Phi\left(x-c_0t+x_0+\bar U(t)t +
  C\sqrt{t}\right) - Q(x,\tilde x, t),$$
where $x_0, C>0$ are to be determined later, and $Q$ is as in (\ref{form})
with $q$ appropriately chosen.

To prove that
$\psi$ is a subsolution, we first need to show the non-positivity of
the following expression:
\begin{equation}\label{2.27}
\begin{split}
& (\psi_l)_t + u\cdot\nabla \psi_l - \Delta \psi_l - f(\psi_l)\\
& = \left( \|u(\cdot, t)\|_{L^\infty(D)} + \frac{C}{2\sqrt{t}} +
  u_1(x,\tilde x, t)\right)
\Phi'\left(x-c_0t+x_0+\bar U(t)t + C\sqrt{t}\right) \\
& \qquad\qquad\qquad\qquad \qquad\qquad\qquad\qquad
\qquad\qquad\qquad
+ f(\Phi) - f(\Phi - Q) \\
& \leq \frac{C}{2\sqrt{t}}\Phi' +  f(\Phi) - f(\Phi - Q).
\end{split}
\end{equation}
Take  $q\in L^1(\mathbb{R})$ such that $0\leq q(x)\leq \alpha = \min\left(\vartheta_0/2,
  (1-\vartheta_0)/4\right)$. By the maximum principle we have:
\begin{equation}\label{2.28}
0\leq Q(x,\tilde x, t) \leq \alpha.
\end{equation}
Let now $x_0>0$ be such that:
\begin{equation}\label{2.29}
\theta_0(x,\tilde x, 0) \geq \Phi(x+x_0) - q(x).
\end{equation}
Finally, let $C$ be large enough so that when $\Phi\in \left(\vartheta_0,
  1 - (1-\vartheta_0)/4\right)$ then:
\begin{equation*}
\begin{split}
\frac{C}{2\sqrt{t}}\Phi' & + f(\Phi)  - f(\Phi - Q)  \leq
\frac{C}{2\sqrt{t}}\Phi' + \|f'\|_{L^\infty} \|Q(\cdot, t)\|_{L^\infty(D)}
\\ & \leq - \frac{C}{2\sqrt{t}} ~ \min_{\Phi(s) \in (\vartheta_0,
  (3+\vartheta_0)/4)} |\Phi'(s)| ~ + \frac{C_\Omega |\Omega|}{\sqrt{t}}
\|f'\|_{L^\infty}\|q\|_{L^1(\mathbb{R})} \leq 0,
\end{split}
\end{equation*}
where we used Lemma \ref{estymata} to estimate $\|Q\|_\infty$.
On the other hand, for $\Phi\in(0,\vartheta_0)\cup (1-(1-\vartheta_0)/4,1)$  the
nonpositivity of the right hand side in (\ref{2.27}) follows directly,
via (\ref{2.28}). Concluding, (\ref{2.27}) and (\ref{2.29}) imply the lower bound in (\ref{uplowbound}).

\medskip

The upper bound follows by assuring that:
$$\psi_r(x, \tilde x, t) = \Phi\left(x-c_0t - x_0 - \bar U(t)t -
  C\sqrt{t}\right) + Q(x,\tilde x, t)$$
is a supersolution. Similarly as above, this follows by choosing
$x_0$ such that:
$$\theta_0(x,\tilde x, 0) \leq \Phi(x-x_0) + q(x)$$
in addition to (\ref{2.29}), and having $C$ large enough so that:
\begin{equation*}
\begin{split}
(\psi_r)_t + u\cdot&\nabla \psi_r - \Delta \psi_r - f(\psi_r)
\geq - \frac{C}{2\sqrt{t}}\Phi' -  \|f'\|_{L^\infty} \|Q(\cdot,
t)\|_{L^\infty(D)} \\ & \geq \frac{C}{2\sqrt{t}} ~ \min_{\Phi(s) \in (\vartheta_0/2,
  (1+\vartheta_0)/2)} |\Phi'(s)| ~ - \frac{C_\Omega |\Omega|}{\sqrt{t}}
\|f'\|_{L^\infty}\|q\|_{L^1(\mathbb{R})} \geq 0,
\end{split}
\end{equation*}
when $\Phi \in (\vartheta_0/2, 1-(1-\vartheta_0)/2)$.
\end{proof}

Clearly, Theorem \ref{secondtheo} follows from (\ref{uplowbound}) and
Lemma \ref{estymata}. We now have:

\begin{lem} \label{2ndlem}
There exists a constant $C_{\Omega,0}$ depending on $\Omega$, $f$, and
on the initial data $\theta_0$, such that for all sufficiently large $t>>1$:
\begin{equation}\label{2nd}
\bar B(t) \le c_0 + \bar U(t) +
C_{\Omega,0}\left(\frac1t+\frac{1}{\sqrt{t}}\right),
\end{equation}
and
\begin{equation}\label{3rd}
\bar B(t) \ge c_0 - \bar U(t) -
C_{\Omega,0}\left(\frac1t+\frac{1}{\sqrt{t}}\right).
\end{equation}
\end{lem}
\begin{proof}
Denoting $\xi(t) = \bar U(t) t + C_{\Omega,0}\sqrt{t}$and using the
bound  (\ref{uplowbound}), we obtain:
\begin{equation}\label{3.12}
\begin{split}
\bar B(t)&=\frac{1}{|\Omega|t}\int_0^t\int_D f(\theta)
=\frac{1}{|\Omega|t}\int_0^t\int_D \theta_t ~\mbox{d}x~\mbox{d}\tilde x
\\ &  =\frac{1}{|\Omega|t}\int_D \theta(x,\tilde x,t)-\theta(x,\tilde x,0) ~\mbox{d}x~\mbox{d}\tilde x\\
&\leq \frac{1}{|\Omega|t}\int_D \Phi(x-c_0t-x_0-\xi(t))+Q(x,\tilde x,
t)-\Phi(x+x_0)+Q(x,\tilde x, 0)~\mbox{d}x~\mbox{d}\tilde x.
\end{split}
\end{equation}
By (\ref{2.3}) and the construction of the corrector $Q$, it follows that:
\begin{equation}\label{ab}
\frac{1}{|\Omega|t} \int_D |Q(x,\tilde x, t) + Q(x, \tilde x, 0)|
~\mbox{d}x~\mbox{d}\tilde x \leq \frac{2}{t} \|q\|_{L^1(\mathbb{R})}
\leq \frac{C_{\Omega, 0}}{t}.
\end{equation}
Further:
\begin{equation}\label{abc}
\begin{split}
\frac{1}{|\Omega|t} \int_D & \Phi(x-c_0t-x_0-\xi(t)) -\Phi(x+x_0) \\ &
=
\frac{1}{t} \Bigg[\int_{-\infty}^{c_0t+x_0+\xi(t)} (\Phi(x-c_0t-x_0-\xi(t))-1)\\
& \qquad + \int_{-\infty}^{-x_0} (1-\Phi(x+x_0)) +
\int_{-x_0}^{c_0t+x_0+\xi(t)} (1-\Phi(x+x_0))\\
& \qquad + \int_{c_0t+x_0+\xi(t)}^{\infty} \Phi(x-c_0t-x_0-\xi(t)) \\
& \qquad - \int_{c_0t+x_0+\xi(t)}^{-x_0} \Phi(x + x_0)
- \int_{-x_0}^\infty\Phi(x+x_0)\Bigg]\\
& = \frac{1}{t} \int_{-x_0}^{c_0t+x_0+\xi(t)} 1 ~\mbox{d}s = \frac{|c_0t + 2x_0 +
  \xi(t)|}{t} \\ & \leq c_0 + \bar U(t) + C_{\Omega, 0} \left(\frac{1}{t} + \frac{1}{\sqrt{t}}\right),
\end{split}
\end{equation}
which together with (\ref{ab}) proves (\ref{2nd}).
To prove (\ref{3rd}), similarly as in (\ref{3.12}) we note that:
\begin{equation*}
\bar B(t) \geq \frac{1}{|\Omega|t}\int_D \Phi(x-c_0t + x_0 +\xi(t)) - Q(x,\tilde x,
t)-\Phi(x-x_0) - Q(x,\tilde x, 0)~\mbox{d}x~\mbox{d}\tilde x,
\end{equation*}
and as in (\ref{abc}) we obtain:
\begin{equation*}
\begin{split}
\frac{1}{|\Omega|t} \int_D &\Phi(x-c_0t+x_0+\xi(t)) -\Phi(x-x_0) =
\frac{|c_0t - 2x_0 - \xi(t)|}{t}\\ &
\geq c_0 - \bar U(t) - C_{\Omega, 0} \left(\frac{1}{t} + \frac{1}{\sqrt{t}}\right).
\end{split}
\end{equation*}
Together with (\ref{ab}) the above implies (\ref{3rd}).
\end{proof}

\bigskip

\noindent {\bf Proof of Theorem \ref{main}.}\\
Multiplying the fluid equation \eqref{fluid} by $u$ and integrating over $D$, one obtains:
\begin{equation}\label{help0}
\frac{\mbox{d}}{\mbox{d}t} \|u\|_{L^2}^2 + \nu\|\nabla u\|_{L^2}^2 \leq
C_\Omega \frac{\rho^2}{\nu}\|\nabla \theta\|^2_{L^2}.
\end{equation}
Integrating (\ref{fluid}) against $u_t$ gives, in turn:
\begin{equation*}
\|u_t\|_{L^2}^2  + \nu \frac{\mbox{d}}{\mbox{d}t} \|\nabla u\|_{L^2}^2 \leq
C_\Omega \rho^2\|\nabla \theta\|^2_{L^2},
\end{equation*}
where in both inequalities above we used Lemma \ref{h}
to ``replace'' the term $\theta\vec\rho$ by $\rho\nabla \theta$,
Taking averages in time, we get:
\begin{equation}\label{help}
\begin{split}
 & \frac{1}{t}\int_0^t \|\nabla u\|_{L^2}^2 ~\mbox{d}t\leq
 C_\Omega\frac{\rho^2}{\nu^2}N(t) + \frac{1}{\nu t} \|\nabla
 u_0\|^2_{L^2},\\
&  \frac{1}{t}\int_0^t \|u_t\|_{L^2}^2 ~\mbox{d}t\leq
 C_\Omega \rho^2 N(t) + \frac{\nu}{t} \|\nabla u_0\|^2_{L^2}.
\end{split}
\end{equation}
By Lemma \ref{supest} or (\ref{easy}), and Lemma \ref{h} it now follows
that:
\begin{equation*}
\begin{split}
\|u\|_\infty &\le C_\Omega \left(\|\nabla u\|_{L^2} + \frac{1}{\nu}\|u_t\|_{L^2}
  + \frac{\rho}{\nu} \|\nabla \theta\|_{L^2}\right).
\end{split}
\end{equation*}
Using (\ref{help}) we obtain:
\begin{equation}\label{help2}
 \begin{split}
\bar U(t)&\le C_\Omega\left(
\left(\frac{1}{t} \int_0^t
\|\nabla u\|_{L^2}^2\right)^{1/2} +
\frac{1}{\nu}\left(\frac{1}{t}\int_0^t\|u_t\|_{L^2}^2\right)^{1/2}
+ \frac{\rho}{\nu}\sqrt{N(t)}\right)\\ &
\leq C_\Omega\left( \frac{\rho}{\nu} \sqrt{N(t)} +
\frac{1}{\sqrt{\nu t}} \|\nabla u_0\|_{L^2}  +
\frac{1}{\sqrt{t}} \|\nabla u_0\|_{L^2}\right).
\end{split}
\end{equation}
By (\ref{1st}) and (\ref{2nd}) we hence get, for large $t>>1$:
\begin{equation}\label{help3}
 \begin{split}
\bar N(t) & \le \frac{1}{2} c_0 +
\frac{3}{2}\bar U(t) + C_{\Omega, 0}\left(\frac{1}{t} + \frac{1}{\sqrt{t}}\right)\\ &
\leq \frac{1}{2} c_0 + C_\Omega \frac{\rho}{\nu} \sqrt{N(t)} +
C_{all}\left( \frac{1}{t} + \frac{1}{\sqrt{t}} \right).
\end{split}
\end{equation}
where $C_{all}$ depends on $\Omega, f, \nu, \theta_0$ and $u_0$.
Consequently:
\begin{equation*}
 \begin{split}
\bar N(t) & \le \left( C_\Omega \frac{\rho}{\nu} + \sqrt{\frac{1}{2} c_0 +
 C_\Omega \frac{\rho^2}{\nu^2}} + C_{all} \sqrt{\frac{1}{t} +
\frac{1}{\sqrt{t}} }~\right)^2\\
& \leq \left(C_\Omega \frac{\rho}{\nu} +
\sqrt{\frac{1}{2} c_0 +  C_\Omega
    \frac{\rho}{\nu}}\right)^2
+ C_{all} \left(\frac{1}{t} + \frac{1}{\sqrt{t}} \right),
 \end{split}
\end{equation*}
and, returning to (\ref{help2}):
$$\bar U(t) \leq C_\Omega\left( \frac{\rho^2}{\nu^2} +
  \frac{\rho}{\nu} \sqrt{\frac{1}{2} c_0 +  C_\Omega
    \frac{\rho}{\nu}}\right) + C_{all} \left(\frac{1}{t} + \frac{1}{\sqrt{t}} \right). $$
In view of Lemma \ref{2ndlem}, Theorem \ref{main} is hence proven.
\hspace*{\fill}\mbox{\ \rule{.1in}{.1in}}\medskip

\bigskip

\noindent {\bf Proof of Theorem \ref{remi}.}\\
We only prove the assertion (ii), because the 2d case in (i) follows with exactly
the same calculations as in Theorem \ref{main} and Theorem \ref{secondtheo} .

For the stationary Navier-Stokes-Boussinesq system,
using the Poincar\'e inequality and
(\ref{const_exact}), we obtain the following counterpart of (\ref{help0}):
\begin{equation}\label{help11}
\|\nabla u\|_{L^2(D)} \leq \frac{\rho}{\nu} C_P\left(C_{PW} +
  \left(\fint|\vec g\cdot (0,\tilde x)|^2~\mbox{d}\tilde x\right)^{1/2}
\right) \|\nabla \theta\|_{L^2(D)},
\end{equation}
where $C_P$ and $C_{PW}$ are, respectively, the Poincar\'e and the
Poincar\'e-Wirtinger constants of $\Omega$.
By Lemma \ref{estymata} and argueing as (\ref{help2}), we arrive at:
\begin{equation*}
\bar U(t) \leq \frac{\rho^2}{\nu^3} \frac{|\Omega|}{2\pi} C_P^2 \left(C_{PW} +
  \left(\fint|\vec g\cdot (0,\tilde x)|^2~\mbox{d}\tilde x\right)^{1/2}
\right)^2 \bar N(t) + C_\Omega \frac{\rho}{\nu} \sqrt{\bar N(t)},
\end{equation*}
while the counterpart of (\ref{help3}) in the present case is:
\begin{equation*}
\begin{split}
\bar N(t) \leq \frac{1}{2}c_0 & + C_\Omega \frac{\rho}{\nu} \sqrt{\bar N(t)}
+ C_{\Omega, 0} \left(\frac{1}{t} + \frac{1}{\sqrt{t}}\right) \\
& + \frac{3|\Omega|}{4\pi}\frac{\rho^2}{\nu^3} C_P^2 \left(C_{PW} +
  \left(\fint|\vec g\cdot (0,\tilde x)|^2~\mbox{d}\tilde x\right)^{1/2}
\right)^2\bar N(t)
\qquad
\forall t>>1.
\end{split}
\end{equation*}
It is therefore clear that when the constant in front of $\bar N(t)$
in the last term of the right hand side above is smaller than $1$, the
results of Theorem \ref{main} and Theorem \ref{secondtheo}
follow as in the case of the Stokes-Boussinesq system.
\hspace*{\fill}\mbox{\ \rule{.1in}{.1in}}\medskip

redpajama_set_name: RedPajamaArXiv