Tuesday, October 31, 2006

Magic of the Internet

My friend Nancy and I decided to introduce her elderly mother to the magic of the Internet. Our first move was to access the popular Ask Jeeves website, and we told her it could answer any question she had.

Nancy's mother was very skeptical until Nancy said, "It's true, Mom. Think of something to ask it." As I sat with fingers poised over the keyboard, Nancy's mother thought a minute, then responded, "How is Aunt Helen feeling?"

Wednesday, October 25, 2006

Hacker unlocks Apple music download protection

A hacker who as a teen cracked the encryption on DVDs has found a way to unlock the code that prevents iPod users from playing songs from download music stores other than Apple Computer Inc.'s iTunes, his company said on Tuesday.

Jon Lech Johansen, a 22-year-old Norway native who lives in San Francisco, cracked Apple's FairPlay copy-protection technology, said Monique Farantzos, managing director at DoubleTwist, the company that plans to license the code to businesses.

"What he did was basically reverse-engineer FairPlay," she said. "This allows other companies to offer content for the iPod."

At the moment, Apple aims to keep music bought from its iTunes online music store only available for Apple products, while songs bought from other online stores typically do not work on iPods.

But Johansen's technology could help rivals sell competing products that play music from iTunes and offer songs for download that work on iPods as they seek to take a bite out of Apple's dominance of digital music.

ITunes commands an 88 percent share of legal song downloads in the United States, while the iPod dominates digital music player sales with more than 60 percent of the market.

Cupertino, California-based Apple, whose profits have soared in recent years on the strength of the iPod, declined to comment.

Johansen, known as DVD Jon, gained fame when at the age of 15 he wrote and distributed a program that cracked the encryption codes on DVDs. This allowed DVDs to be copied and played back on any device.

His latest feat could help companies such as Microsoft Corp., Sony Ericsson and Samsung Electronics Co. Ltd., which have all announced plans over the past few months for music download services combined with new devices to challenge Apple.

Source: http://news.yahoo.com

Sunday, October 22, 2006

Design of R.C.C members

 

 Design of R.C.C members                                    

      R.C.C members can be design by two approaches

F  Working stress method

F  Limit state method

 

Working stress method

Working stress method is based on classical elastic theory and developed for purely elastic material. The method ensures satisfactory behavior of the structure under working load but doesn't give actual margin of safety at collapse. Therefore design is generally done by limit state method.

 

Limit state of collapse in flexure

Assumption:

ü      A normal section, which is plane before bending, remains plane after bending.

ü      This implies that strain at any point on the cross-section is directly proportional to it

ü      The tensile strength of concrete is ignored.

ü      The maximum strain at the outermost compression fiber is taken as 0.0035 in bending.

ü      The maximum strain in the tension reinforcement in the section at failure shall not be less than    (f y /1.15Es)+0.002

Where,

   fy = characteristic strength of steel.

   Es = modulus of elasticity of steel. 

 

Design Of Beam

 

Design of rectangular section

 

F  Singly Reinforced Beam:  Design of singly and doubly reinforced rectangular section shall be made as below the compressive stress block for concrete is represented by the design stress strain curve. The triangular distribution of strain and non-linear stress diagram in compression concrete are as shown in the design aids.

 

Properties of the singly reinforced section:

 

i. Compressive force and the centroid of compressive force: It is seen from this stress block that the centroid of the compressive force in a rectangular section lies at the distance of 0.416*Xu from the extreme compression fiber and the total force of compression is 0.36*fck*b*Xu.

 

ii.Lever arm : The lever arm, the distance between the centroid of compressive force and centroid of tensile force.

                         Zu =  (d-0.416*Xu)

 

iii. Depth of the neutral axis (Xu): Depth of neutral axis below the highly compressed edge is obtained by equating compression in concrete with tension in steel to satisfy the condition of equilibrium.

                                   C = T

                Hence,

                           Xu = (fs*Ast)/(0.36*fck*b)

                                 

                                = Pt*b*d*(0.87fy)/100*0.36*fck*b

 

                  or, Xu/d =  (Pt/10)*(0.87fy/0.36fck)

                     

                          Where, fs = 0.87f

 

iv. Ultimate moment of resistance: The ultimate moment of resistance of a singly reinforced rectangular section is obtained by taking moment of total compression (C) or about CG of steel or moment of total tension (T) about CG of compression.

 

            Moment of resistance in terms of concrete strength is given by

 

Mu = C * Zu

      

        = 0.36* fck* b* Xu *(d-0.416Xu)

 

The ultimate moment of resistance, Mu in terms of steel and concrete strength can be found by

 

Mu  = T * Zu

       

         = fs *Ast *( d- 0.416Xu)

      

         = 0.87fy(Pt*b*d/ 100)* (1-0.416Xu)

 

          =0.87fy*(Pt/100)*(1-1.005(fy/fck)(Pt/100)*b*d2

 

Substituting for Xu/d, we get

 

Mu = 0.87fy*( Pt /100) * (1-1.005(fy/fck)( Pt /100)*b*d2

 

v. The area of steel is obtain as:

 

Ast  = Mu/(0.87*fy(d-0.416*Xu))

 

Design constants for balance or critical section: A balance or critical section is that section in which strain in concrete and steel reach their ultimate values simultaneously. As the design of a beam should be under – reinforced, the value Xu for the critical section is the maximum allowable value Xul and can be obtained from strain distribution diagram as,

 

             Xul=0.0035/(0.005+0.87fy/Es)*d

 

Where, Es = 2*105N/mm2

 

Once the value of Xul is known the other design constants can be obtained from

 

Mul = 0.87fy* Ast *(d-0.416 Xul) and, Mul  = 0.36fck/(.87fy)*b* Xul

 

FDoubly reinforced beam: Doubly reinforced sections are generally adopted when the dimensions of the beam predetermined from other considerations and the design moment exceeds the moment of the resistance of a singly reinforced section. The additional moment of resistance needed is obtained by providing compression reinforcement and additional tensile reinforcement.

 

Properties of doubly reinforced section:

 

i. Depth of neutral axis (Xu): The depth of neutral axis is obtained by equating       total compression in concrete and compression steel with total tension in tension steel.

                                                

                                                  C   =   T

                                  

                                 Or,  Cs  + Cc   =    T

 

Or,  0.36 * fck* b* Xu + (fsc-fcc)*Asc = 0.87fy*Ast

 

Where,  fsc = stress in compression steel (Asc) corresponding to strain

Where,  d  = effective cover to compression steel

             fcc= stress in concrete in compression at the level of compression steel

 

ii. Moment of resistance: The ultimate moment of resistance is obtained by taking moment of compressive force in concrete and compression steel about the centroid of tension steel as:

 

             M=   Mu1 + Mu2

Where, Mu1 = Mu(lim) = K*fck*b*d2

                Mu2  = the additional moment in compression failure

            Mu  = 0.36*fck*b*Xu*(d - 0.416Xu) + (fsc-fcc)*Asc*(d-d')

 

iii.Area of steel: In the design of doubly reinforced section, the section is kept balanced to make full utilization of the resistance of concrete. The area of tension steel and compression steel required for resisting Mu1 and Mu2 can be obtained as follows:

 

1. Tension steel: Area of tension steel (Ast1) corresponding to Mul is obtained as

                

               Ast1 = Mu1/(0.87fy*(d-0.416Xul))

 

And the area of tension steel (Ast2) is obtained by equating Mu2 with the moment about compression steel.

                 

                 Ast2=Mu2/(0.87fy(d-d'))

    

Thus total area tension steel = Ast1+Ast2

 

2. Compression steel:  The area of compression steel (Asc) is obtained by equating (Mu2) with the moment of compression C2 = (fsc-fcc)*Asc about tension steel.

         Mu2 = Asc*(fsc-fcc)(d-d')

Or,    Asc= Mu2/(fsc-fcc)*(d-d')

 

Design of Flanged section: When the beam and slab cast together in one operation and the main reinforcement of the slab extend into the beam and the integral connection between the slab and beam is formed due to this part of slab lying on the compression zone of slab act along with the beam in compression zone. Then the resulting section is the flanged section.

 

Singly reinforced flanged section:

 

Properties of flanged section

 

 i. Effective width of flange:

ü      For T-Beam: b= (Lo/6 + 6*Df)

ü      For L-Beam: b= (Lo/12 + bw + 3*Df)

ü      For isolated beams, effective width of beam is given by

¯     bf = Lo/(Lo/b + 4) + bw  or actual width for T-beam

¯     bf = 0.5 Lo(Lo/b + 4) + bw  or actual width for L-beam

 

ii.Neutral axis: The neutral axis lie either on the flange or in web, which may be ascertained by comparing the value Xu and Df. Where Xu may be determined by assuming that the neutral axis lies either in the flange or web.

 

Case-1: Neutral axis lies in the flange             

               C     =    T

 

             0.36fck*bf*Xu =.87fy*Ast

 

               Xu = (.87fy*Ast)/(.36fck*bf)

 

  If Xu ≤ Df, then the neutral axis is within the flange. Otherwise, neutral axis lies in the web and section is analyzed as flanged section then the position of neutral axis may be determined from equilibrium of compressive and tensile forces as,

   

         Case 2:Neutal axis below the flange and Df/d<0.2

                        C = T

 

                  C = C1 + C2

                   

       Or,      C = 0.36fck*bw*X + 0.446fck* (bf - bw)* Df

 

      Or,     0.87fy*Ast = 0.36*fck*bw*X + 0.446*fck*(bf - bw)*Df

 

     Case 3: Neutral axis below the flange and Df/d>0.2

 

In this case, the Df in the above formula is replaced by the term Yf .

       

    where,  Yf  = (0.15 * X + 0.65 * Df) or Df  whichever is smaller.

 

iii. Moment of resistance and area of tension steel

 

For case1: Moment of resistance is similar to that for singly reinforced rectangular beam.

 

              Mu = 0.36fck*bf (d - 0.416X)

 

Area of steel can be calculated by comparing force in compression to the force in tension. i.e,    

 

         0.36fck*bf*Xu =.87fy*Ast

 

For case 2: Moment of resistance is given by ,

         

                Mu = Mu1 + Mu2

               

                Mu = 0.36fck*bw*X*(d-0.416*X)+0.446fck*(bf-bw)*Df(d-Df/2)

 

Area of steel can be calculated by comparing force in compression to the force in tension. i.e,           

    

            0.87fy*Ast = 0.36fck* bw *X + 0.446fck*(bf - bw)*Df

 

Limit State Of Shear And Torsion

 

1. Limit state of collapse in shear    

 

i. Nominal Shear stress:  The nominal shear stress in the uniform depth of beam is given by the equation,

                                  tv = Vu/bd

Where,       Vu = Shear force due to design loads

                   B = Breadth of the member

                   D = Effective depth

 

i.Shear Reinforcement: The design  shear strength  of vertical stirrups is given by the following equation,

                       Vus =  0.87*fy*Asv*d/Sv

 

Where,  Asv = the total cross-section area of the vertical legs of stirrups

              Sv = the spacing (pitch) of the stirrups.

 

2. Design for torsion

 

i.Equivalent Bending  Moment (Me): The torsion moment acting on the member is converted into equivalent shear , which is added to the B.M and shear acting at the section . The member is designed for resulting B.M and shear.

 

Me1 = Mu (bending) ± Mt (Equivalent torsion)

 

Where , Mt =Tu(1+D/b)/1.7

              Tu    =  torsion moment at the section

              M=  bending moment at the section

When, Mt> Mu, steel will be provided on compression face to resist bending moment    

                   Me2  = Mt - Mu

 

ii.Equivelent Shear,

   

Ve=Vu+1.6*Tu/b

 

iii.Equivalent Shear stress

  

  tve= Ve/(b*d)

 

iv.Maximum Shear Stress

 

tcmax is from table 20 of IS:456:2000

tve should be less than tcmax

 

v.Design of Shear Strength of concrete: tc is from table 19 of Design Aids IS: 456:2000, for an assumed value of Pt . If the equivalent nominal shear stress, tve does not exceed  tgiven in table 19 of the code , minimum shear reinforcement shall be provided as per clause 26.5.1.6 of IS: 456 : 2000.

                          

If  tve exceeds tc given in table 19, both longitudinal and transverse shall be provided in accordance with clause 41.4.

 

vi.Longitudinal reinforcement: Longitudinal  reinforcement shall be provided to resist equivalent moment.

 

vii.Transverse reinforcement:The transverse reinforcement shall consist of vertical stirrups enclosing the longitudinal reinforcement.Its design requires determination of spacing for its chosen diameter is as according to 41.43 of I.S code.

 

Design of Column:

A column is a compression member subjected to compressive force in a direction parallel to its longitudinal axis .The column are further classified into various type in with the different criteria such as

Based on shape of cross-section

ü      Rectangular

ü      Circular

Based on type of footing

ü      Axially load

Based on slenderness ratio

ü      Short

ü      Slender

 

Design criteria

 

  i.  Minimum eccentricity: According to code, Emin=L/500+D/30,subjected to minimum of 20mm.

Where, L= unsupported length of column

            D= lateral dimension of column under consideration

 

ii.Clear cover to reinforcement: When the longitudinal reinforcement bars are not exceeding to code IS 456-1978 species that d' should not less than 40mm.

 

iii.Requirement Required:

 

a. Longitudinal reinforcement :The percentage of reinforcement shouldn't be less than 0.8% and maximum should  not exceed 4% up to 6% of the gross area of the section is allowed. The minimum number of bars should not be less than 4 in rectangular and 6 in circular column.

 

b.Transverse reinforcement: If the longitudinal bars are not spaced more than 75mm on either side the transverse reinforcement  needs only to go exceeding 48 times the diameter of the purpose of providing effective lateral support.

If the longitudinal bars spaced at a distance of not exceeding 48 times the diameter of tie are effectively tied in two directions by open ties.

        Where the longitudinal reinforcement bars in compression members are placed in more one row, effective lateral support to longitudinal bars in the row may be assumed to have been provided. If transverse reinforcement is provided for the outermost row and no of bar of inner row is closer to the nearest compression face three times the diameter of largest longitudinal bar in inner row.

 Where the longitudinal bars in the compression member are grouped and each adequately tied with transverse reinforcement for compression members as a whole may be provided on the assumption that each group is a single longitudinal for the purpose of determining the pitch and the diameter of the transverse reinforcement. The diameter of such traverse reinforcement need not however exceed 20mm.

 

FAxially loaded column: According to IS 456-1978 code following formula is used.

                

                    Pu= 0.4fck*Ac+0.6*fy*Asc

              

                Where,  Pu = Axial load

                             Ac =Area of concrete

                             Asc = Area of reinforcement

 

Column with unaxial bending : The formula used in this case is same as that of axially loaded column.

 

Column with biaxial bending: The theoretical limit strength of a section under bi-axial bending and compression is a function of three variable Pu, Mx,My which may be expressed in terms of axial load placing at eccentricity,

                           ex = Mx/Pu,                           ey =My/Pu

Moment Mx and My are acting along the x-axis and y-axis respectively.

The code IS 456-1978 utilize the load contour method and the general known dimensional method.

Equation for contour at constant P may be expressed as

(M x /M ux) æ+(Mu/Muy)æ≤1

where, Mx=Pu*ex

           My=Pu*ey

           Mux=maximum moment capacity for bending along x-axis at axial load Pa

           Muy=maximum moment capacity for bending along y-axis load Pa

æ= an exponent that depends upon the dimension of cross section, amount of reinforcement, concrete strength and yield stress of steel.

 

æ=.445+.155Pu/Puz

 

where,  Pu=.45fck.Ac+.155Pu/Puz

            Puz=.45fck.Ac+.75fy.Asc

 

DETAILING: Ductility is capacity to take additional displacement (without collapse) even after yield.

 

Ductility increases in:

ü      as concrete grade increases

ü      as steel grade increases

Structural ductility can be enhanced by

ü        Regular configuration

ü      More redundancy

ü      Avoiding yielding of column

ü      Avoiding failure of foundation

     So, ductility design and detailing is done in this project work which is based on IS:13920-1993.

 

i. Detailing of flexural member:

Requirements;

ü      The factor axial stress on the member under earthquake loading shall not exceed 0.1fck

ü       Members shall have a width to depth ratio of more than 3

ü      The width of member shall  not less than 200mm.

ü      The overall depth of member not more than ¼ of clear spans

 

Anchorage of beam bar: In external joint anchorage length beyond inner face of the column equal to development length in tension plus ten times bar diameters minus allowance for 90 degree bend.

 

Lap splice: Lap length should be greater than development length in tension

 

Lap splice shall not be provided

(a)    within a joints

(b)    within 2d from joint face

(c)    within ¼ of member lengthwhere flexure yielding may occur during earthquake.

 

Shear reinforcement

Hook length >10Ф or 75mm whichever is more

 Minimum diameter = 6mm

Minimum diameter = 8mm for clear span greater than 5m

Spacing as per ductility: At distance of 2d from face of column spacing should be more than 1/4d .at mid span spacing should not be more than 1/2d.

 

ii.Detailing of compression member:

-         minimum dimension should be greater than 300mm

-         B/D >0.4

-         Lap splice shall be provided only in the central half of the member length.

-         Spacing of hoop within lap splice<150m

-          

Requirement of longitudinal reinforcement

ü      The longitudinal reinforcement shall not be less than 0.8% or more than 4%of the gross cross-sectional area of column.

ü      The minimum number of longitudinal bars provided in a column shall be four in rectangular columns and six in a circular.

ü      Spacing of longitudinal bars measured along periphery of the column shall not exceed 300mm.

ü       

 Requirement ofTransverse reinforcement as per ductility design

ü      Up to distance as per ductility design 450mm, spacing should not be more than 100mm.

ü      At middle spacing should be less than ½ width of column

ü      Lap should be provided at distance more than ¼ clear height

 

Minimum and Maximum Sizes of Reinforcement: The minimum size of bars for column not less than 12mm in diameter.The diameter of the polygonal links  or ties shall  not be less than ¼ of the diameter of the largest longitudinal bar, and in no case less than 5mm.When a column terminates into a raft special confining reinforcement shall extend at least 300mm into raft.

http://i-geek.blogspot.com

Saturday, October 21, 2006

Scientists make water run uphill

This story is available at the BBC’s website.

Physicists have made water run uphill quite literally under its own steam.
The droplets propel themselves over metal sheets scored with a carefully designed array of grooves.

The US scientists did the experiment to demonstrate how the random motion of water molecules in hot steam could be channelled into a directed force.
But the team, writing in Physical Review Letters, believes the effect may be useful in driving coolants through overheating computer microchips.
The physics at work here has been witnessed by all of us in the kitchen.
Leave an empty pan on the stove for too long, and water, when you drip it over the scorching pan bottom, will hover over the surface on a bed of steam.
The effect was described in the 18th Century by a German scientist Johann Gottlob Leidenfrost.
What happens is that the heat is so intense, it boils the underside of the water droplet without any physical contact with the pan.
"We were interested in whether it would be possible to use this phenomenon to move liquids around," said Dr Heiner Linke, the intellectual power behind the self-propelled droplets.

An uphill struggle
The trick seems simple. Instead of using a smooth surface, the team scored it with a series of skewed triangular grooves. This gives it a kind of saw-tooth profile.
Now the water droplets appear to push themselves off the long-slope side of the grooves and rocket across the heated surface - instead of just dancing on the spot as they do in the kitchen pan.
The mechanism is a little more complicated and took a while to work out, Dr Linke told the BBC. "The vapour," he explained, "mostly flows in one direction, and the droplet sits on the flowing vapour, a bit like a boat carried along in a flowing river."
Droplets can also climb over steps, and up inclines of up to 12 degrees. Filmed with high-speed cameras, the droplets appear to take on a life of their own, sliding along like sloppy amoebae.
Although the original intention was to devise an arresting demonstration of how random energy can be rectified into directed motion - the focus of Dr Linke's main work is with molecular motors - the researchers now think there may be a use for the effect in cooling computer microchips.
The electrical currents now passing through microprocessors are so large the heat they generate can limit computing performance.
Many chips have cooling circuits nowadays, but these require pumps to drive the coolant, which in turn generate even more heat.
Suitably micro-patterned channels, argues Dr Linke, would make the coolant flow automatically.
"It would be very neat if we could use the heat from the chip to be the pump, because you would not need any additional power, but also because the pumping only happens when the thing is warm; it would also be a thermostat at the same time. So it would all be in one package."

Friday, October 20, 2006

Know these briefs!

  • The term "virtual reality" was coined by the composer, computer scientist and author Jaron Lanier.

  • The code name for the 12 engineers who designed the first IBM PC was "The Dirty Dozen".

  • It is possible to run a P III processor without a heat sink because Intel bundles along protection devices that clocks down the CPU to get back to a stable temperature.

  • Blue Gene is the nickname of the new supercomputer from IBM that would be 1000 times faster than Deep Blue.

  • 'Overburning' is the process of writing beyond the official capacity of a CD-R or CD-RW. This is possible by using the lead-out to store data- the lead-out is used to indicate the end of the disk and contains nothing but zeroes.

  • "To get amazonned" means that one has lost a significant portion of one's business to a dotcom enterprise.

How safe is your password?


The first step in protecting your online privacy is creating a safe password - i.e. one that
a computer program or persistent individual won't easily be able to guess in a short
period of time. To help you choose a secure password, we've created a feature that lets
you know visually how safe your password is as soon as you create it.
Tips for creating a secure password:
Include punctuation marks and/or numbers.
Mix capital and lowercase letters.
Include similar looking substitutions, such as the number zero for the letter 'O' or
'$' for the letter 'S'.
Create a unique acronym.
Include phonetic replacements, such as 'Luv 2 Laf' for 'Love to Laugh'.
Things to avoid:
Don't use a password that is listed as an example of how to pick a good
password.
Don't use a password that contains personal information (name, birth date, etc.)
Don't use words or acronyms that can be found in a dictionary.
Don't use keyboard patterns (asdf) or sequential numbers (1234).
Don't make your password all numbers, uppercase letters or lowercase letters.
Don't use repeating characters (aa11).
Tips for keeping your password secure:
Never tell your password to anyone (this includes significant others, roommates,
parrots, etc.).
Never write your password down.
Never send your password by email.
Periodically test your current password and change it to a new one.

Tuesday, October 17, 2006

Jokes that Blog

I was just, just amazed you will find hour from all fields,
 

What's on your lap?

What's on your lap? Is it a notebook?

Yesternight while I was reading digit, I was fascinated once again by the Rich Cinematic Quality Experience of the LG XNote Entertainer's 5.1 Channel Dolby Digital System. What I liked most about this laptop is Quality Graphics and hyper memory, maximum possible screen, Dual Hexa Band Antennae, sleek light design. And last not the least the Fingerprint Security.

The XNote is said to contain Battery Miser Technology- probably the best latest life for a battery at your lap. The mag goes saying- powered by the Intel Centrino Duo Technology- oops! I have trouble with these centrino processors and this must be my only disappointment with this laptop. LG XNote is  arange of new-age multimedia laptops that come loaded witht he most advanced entertainment features ever. The 5.1 Channel Dolby Digital Sound System, the exclusive LG XTS Pro Sound Technology and SRS Tru Surround XT give the LG XNote the ultimate edge in sound quality. The Next-Gen Multimedia features such as theatre-like Wide Screen and Cinematic Quality Graphics make hte LG XNote the best in picture quality. Details on this mobile entertainer can be found at http://www.lgindia.com

Wednesday, October 11, 2006

Google buys YouTube for $1.65bn

Google is buying video-sharing website YouTube for $1.65bn (£883m) in shares after a weekend of speculation that a deal was in the offing.

The two companies will continue to operate independently, Google said as it announced the news on Monday.

YouTube, launched in February 2005, has grown quickly into one of the most popular websites on the internet.

It has 100 million videos viewed every day and an estimated 72 million individual visitors each month.

'Natural partners'

"The YouTube team has built an exciting and powerful media platform that complements Google's mission to organize the world's information and make it universally accessible and useful," Google chief executive Eric Schmidt said in a statement.

He said the two companies were "natural partners" to offer a media entertainment service to users, content owners and advertisers.

Mr Schmidt also told investors that YouTube will be "one of many investments" Google plans to make in the video field.

However, the company will keep operating its own Google Video as a separate operation.

YouTube will retain its brand, and its 67 staff, including co-founders Chad Hurley and Steve Chen, will keep their jobs.

"Our community has played a vital role in changing the way that people consume media, creating a new clip culture," said Mr Hurley.

"By joining forces with Google, we can benefit from its global reach and technology leadership to deliver a more comprehensive entertainment experience for our users and to create new opportunities for our partners."

According to Comscore World Metrix, YouTube's audience has soared from 2.8 million unique users one year ago to 72 million users in August 2006.

Music tie-ups

The announcement came after a day of distribution deals drawn up by the pair.

Universal Music Group has signed a distribution deal with YouTube, which will protect the rights of the music firm's artists.

YouTube also says it has signed a deal with CBS, which will offer short-form video programming, including news, sport and entertainment on YouTube.

Google has also signed distribution deals of its own, with Sony BMG and Warner Music to offer music videos.

The Google deals should enable internet users in the US to view music videos, artist interviews, and other footage from the two firms on Google video for free from this month.

The content is sponsored through a Google advertising-supported revenue-sharing agreement.

Google also said that in addition to the advertising-supported video content, music videos from Warner would be available for purchase as downloads at $1.99 each.

As part of YouTube's deal with CBS, the companies will share revenue from advertising sponsorship of CBS Videos.

Source: http://news.bbc.co.uk/1/hi/business/6034577.stm

Google buys YouTube for US$1.65 billion.

COngratulations to YouTube owners! They have been bought over by Google. It's a good investment I guess but I think Google needs to work very hard to integrate YouTube into their search engine.

It's a good addition to the Google family and I love to see how Yahoo and Microsoft would react.

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