Bourdeau, pipolin, pourriez vous m'expliquer que quoi s'agit ? :-)
31 réponses
kowalski
Back to ITU
We have now have the 576 active lines at 52 =CE=BCs per line length, in
which the picture
information is contained, which we need for our digital final product.
=E2=97=8F 576 lines a 52=CE=BCs
The International Telecommunication Union says that of the 864
horizontal samples, 720
are (line-) active. Those go into the final product, remaining 144 are
"digital blanking".
That is obviously a contradiction!
We have a line of 64 =CE=BCs "length", which after (ITU-conform) sampling
consists of 864 digital
values.
So again we calculate:
64=CE=BCs / 864 =3D 74.074ns
So, in a line of 64 =CE=BCs there are 864 samples with a =E2=80=9Clength=E2=
=80=9D of 74.074
ns each.
=E2=97=8F Period duration for a single pixel =3D 74.074 ns
we multiply this value by the 720 pixels, which ITU recommends.
74.074ns * 720 =3D 53.3333=CE=BCs
The "window" suggested by the ITU is thus larger, than the 52 =CE=BCs
existing video
information.
=E2=97=8F - 720 pixel =3D 53.3333 =CE=BCs
We now divide the "genuine" active 52 =CE=BCs by the calculated period
duration/pixel:
52=CE=BCs / 74.074ns =3D 702.0007 =3D ~702 pixel
There are only 702 active pixels!
Back to ITU We have now have the 576 active lines at 52 μs per line length, in which the picture information is contained, which we need for our digital final product. ● 576 lines a 52μs The International Telecommunication Union says that of the 864 horizontal samples, 720 are (line-) active. Those go into the final product, remaining 144 are "digital blanking". That is obviously a contradiction! We have a line of 64 μs "length", which after (ITU-conform) sampling consists of 864 digital values. So again we calculate: 64μs / 864 = 74.074ns So, in a line of 64 μs there are 864 samples with a “length” of 74.074 ns each. ● Period duration for a single pixel = 74.074 ns we multiply this value by the 720 pixels, which ITU recommends. 74.074ns * 720 = 53.3333μs The "window" suggested by the ITU is thus larger, than the 52 μs existing video information. ● - 720 pixel = 53.3333 μs We now divide the "genuine" active 52 μs by the calculated period duration/pixel: 52μs / 74.074ns = 702.0007 = ~702 pixel There are only 702 active pixels!
Back to ITU
We have now have the 576 active lines at 52 μs per line length, in
which the picture
information is contained, which we need for our digital final product.
● 576 lines a 52μs
The International Telecommunication Union says that of the 864
horizontal samples, 720
are (line-) active. Those go into the final product, remaining 144 are
"digital blanking".
That is obviously a contradiction!
We have a line of 64 μs "length", which after (ITU-conform) sampling
consists of 864 digital
values.
So again we calculate:
64μs / 864 = 74.074ns
So, in a line of 64 μs there are 864 samples with a “length” of 74.074
ns each.
● Period duration for a single pixel = 74.074 ns
we multiply this value by the 720 pixels, which ITU recommends.
74.074ns * 720 = 53.3333μs
The "window" suggested by the ITU is thus larger, than the 52 μs
existing video
information.
● - 720 pixel = 53.3333 μs
We now divide the "genuine" active 52 μs by the calculated period
duration/pixel:
52μs / 74.074ns = 702.0007 = ~702 pixel
There are only 702 active pixels!
Back to ITU We have now have the 576 active lines at 52 μs per line length, in which the picture information is contained, which we need for our digital final product. ● 576 lines a 52μs The International Telecommunication Union says that of the 864 horizontal samples, 720 are (line-) active. Those go into the final product, remaining 144 are "digital blanking". That is obviously a contradiction! We have a line of 64 μs "length", which after (ITU-conform) sampling consists of 864 digital values. So again we calculate: 64μs / 864 = 74.074ns So, in a line of 64 μs there are 864 samples with a “length” of 74.074 ns each. ● Period duration for a single pixel = 74.074 ns we multiply this value by the 720 pixels, which ITU recommends. 74.074ns * 720 = 53.3333μs The "window" suggested by the ITU is thus larger, than the 52 μs existing video information. ● - 720 pixel = 53.3333 μs We now divide the "genuine" active 52 μs by the calculated period duration/pixel: 52μs / 74.074ns = 702.0007 = ~702 pixel There are only 702 active pixels!
Back to ITU We have now have the 576 active lines at 52 ?s per line length, in which the picture information is contained, which we need for our digital final product. ? 576 lines a 52?s The International Telecommunication Union says that of the 864 horizontal samples, 720 are (line-) active. Those go into the final product, remaining 144 are "digital blanking". That is obviously a contradiction! We have a line of 64 ?s "length", which after (ITU-conform) sampling consists of 864 digital values. So again we calculate: 64?s / 864 = 74.074ns So, in a line of 64 ?s there are 864 samples with a "length" of 74.074 ns each. ? Period duration for a single pixel = 74.074 ns we multiply this value by the 720 pixels, which ITU recommends. 74.074ns * 720 = 53.3333?s The "window" suggested by the ITU is thus larger, than the 52 ?s existing video information. ? - 720 pixel = 53.3333 ?s We now divide the "genuine" active 52 ?s by the calculated period duration/pixel: 52?s / 74.074ns = 702.0007 = ~702 pixel There are only 702 active pixels!
un copier coller ?
ça vient de là http://www.arachnotron.nl/videocap/doc/Karl_cap_v1_en.pdf
Mais il n'a pas tout mis car il ne comprend pas la suite.....
"*.-pipolin-.*" <..pipolin..@DTC.com> a écrit dans le message de news:
mn.44407da262975161.73628@DTC.com...
Dans son message précédent, kowalski a écrit :
Back to ITU
We have now have the 576 active lines at 52 ?s per line length, in
which the picture
information is contained, which we need for our digital final product.
? 576 lines a 52?s
The International Telecommunication Union says that of the 864
horizontal samples, 720
are (line-) active. Those go into the final product, remaining 144 are
"digital blanking".
That is obviously a contradiction!
We have a line of 64 ?s "length", which after (ITU-conform) sampling
consists of 864 digital
values.
So again we calculate:
64?s / 864 = 74.074ns
So, in a line of 64 ?s there are 864 samples with a "length" of 74.074
ns each.
? Period duration for a single pixel = 74.074 ns
we multiply this value by the 720 pixels, which ITU recommends.
74.074ns * 720 = 53.3333?s
The "window" suggested by the ITU is thus larger, than the 52 ?s
existing video
information.
? - 720 pixel = 53.3333 ?s
We now divide the "genuine" active 52 ?s by the calculated period
duration/pixel:
52?s / 74.074ns = 702.0007 = ~702 pixel
There are only 702 active pixels!
un copier coller ?
ça vient de là http://www.arachnotron.nl/videocap/doc/Karl_cap_v1_en.pdf
Mais il n'a pas tout mis car il ne comprend pas la suite.....
Back to ITU We have now have the 576 active lines at 52 ?s per line length, in which the picture information is contained, which we need for our digital final product. ? 576 lines a 52?s The International Telecommunication Union says that of the 864 horizontal samples, 720 are (line-) active. Those go into the final product, remaining 144 are "digital blanking". That is obviously a contradiction! We have a line of 64 ?s "length", which after (ITU-conform) sampling consists of 864 digital values. So again we calculate: 64?s / 864 = 74.074ns So, in a line of 64 ?s there are 864 samples with a "length" of 74.074 ns each. ? Period duration for a single pixel = 74.074 ns we multiply this value by the 720 pixels, which ITU recommends. 74.074ns * 720 = 53.3333?s The "window" suggested by the ITU is thus larger, than the 52 ?s existing video information. ? - 720 pixel = 53.3333 ?s We now divide the "genuine" active 52 ?s by the calculated period duration/pixel: 52?s / 74.074ns = 702.0007 = ~702 pixel There are only 702 active pixels!
un copier coller ?
ça vient de là http://www.arachnotron.nl/videocap/doc/Karl_cap_v1_en.pdf
Mais il n'a pas tout mis car il ne comprend pas la suite.....
*.-pipolin-.*
Dominique Bourdeau a écrit :
"*.-pipolin-.*" a écrit dans le message de news:
Dans son message précédent, kowalski a écrit :
Back to ITU We have now have the 576 active lines at 52 ?s per line length, in which the picture information is contained, which we need for our digital final product. ? 576 lines a 52?s The International Telecommunication Union says that of the 864 horizontal samples, 720 are (line-) active. Those go into the final product, remaining 144 are "digital blanking". That is obviously a contradiction! We have a line of 64 ?s "length", which after (ITU-conform) sampling consists of 864 digital values. So again we calculate: 64?s / 864 = 74.074ns So, in a line of 64 ?s there are 864 samples with a "length" of 74.074 ns each. ? Period duration for a single pixel = 74.074 ns we multiply this value by the 720 pixels, which ITU recommends. 74.074ns * 720 = 53.3333?s The "window" suggested by the ITU is thus larger, than the 52 ?s existing video information. ? - 720 pixel = 53.3333 ?s We now divide the "genuine" active 52 ?s by the calculated period duration/pixel: 52?s / 74.074ns = 702.0007 = ~702 pixel There are only 702 active pixels!
un copier coller ?
ça vient de là http://www.arachnotron.nl/videocap/doc/Karl_cap_v1_en.pdf
Mais il n'a pas tout mis car il ne comprend pas la suite.....
il n'a toujours pas compris que l'on ne mets pas de nom ni de pseudo dans les sujets, et pourtant, c'est simple a comprendre, alors comprendre les problèmatiques de rafaichissement de diffuseurs, c'est même pas la peine...
"*.-pipolin-.*" <..pipolin..@DTC.com> a écrit dans le message de news:
mn.44407da262975161.73628@DTC.com...
Dans son message précédent, kowalski a écrit :
Back to ITU
We have now have the 576 active lines at 52 ?s per line length, in
which the picture
information is contained, which we need for our digital final product.
? 576 lines a 52?s
The International Telecommunication Union says that of the 864
horizontal samples, 720
are (line-) active. Those go into the final product, remaining 144 are
"digital blanking".
That is obviously a contradiction!
We have a line of 64 ?s "length", which after (ITU-conform) sampling
consists of 864 digital
values.
So again we calculate:
64?s / 864 = 74.074ns
So, in a line of 64 ?s there are 864 samples with a "length" of 74.074
ns each.
? Period duration for a single pixel = 74.074 ns
we multiply this value by the 720 pixels, which ITU recommends.
74.074ns * 720 = 53.3333?s
The "window" suggested by the ITU is thus larger, than the 52 ?s
existing video
information.
? - 720 pixel = 53.3333 ?s
We now divide the "genuine" active 52 ?s by the calculated period
duration/pixel:
52?s / 74.074ns = 702.0007 = ~702 pixel
There are only 702 active pixels!
un copier coller ?
ça vient de là http://www.arachnotron.nl/videocap/doc/Karl_cap_v1_en.pdf
Mais il n'a pas tout mis car il ne comprend pas la suite.....
il n'a toujours pas compris que l'on ne mets pas de nom ni de pseudo
dans les sujets, et pourtant, c'est simple a comprendre, alors
comprendre les problèmatiques de rafaichissement de diffuseurs, c'est
même pas la peine...
Back to ITU We have now have the 576 active lines at 52 ?s per line length, in which the picture information is contained, which we need for our digital final product. ? 576 lines a 52?s The International Telecommunication Union says that of the 864 horizontal samples, 720 are (line-) active. Those go into the final product, remaining 144 are "digital blanking". That is obviously a contradiction! We have a line of 64 ?s "length", which after (ITU-conform) sampling consists of 864 digital values. So again we calculate: 64?s / 864 = 74.074ns So, in a line of 64 ?s there are 864 samples with a "length" of 74.074 ns each. ? Period duration for a single pixel = 74.074 ns we multiply this value by the 720 pixels, which ITU recommends. 74.074ns * 720 = 53.3333?s The "window" suggested by the ITU is thus larger, than the 52 ?s existing video information. ? - 720 pixel = 53.3333 ?s We now divide the "genuine" active 52 ?s by the calculated period duration/pixel: 52?s / 74.074ns = 702.0007 = ~702 pixel There are only 702 active pixels!
un copier coller ?
ça vient de là http://www.arachnotron.nl/videocap/doc/Karl_cap_v1_en.pdf
Mais il n'a pas tout mis car il ne comprend pas la suite.....
il n'a toujours pas compris que l'on ne mets pas de nom ni de pseudo dans les sujets, et pourtant, c'est simple a comprendre, alors comprendre les problèmatiques de rafaichissement de diffuseurs, c'est même pas la peine...
alors comprendre les problèmatiques de rafaichissement de diffuseurs, c'est même pas la peine...
arf c'est doooommage ! ça ne parle pas de rafraichissement d'écran.... Tu as toujours des problèmes de lecture, visiblement... :-D
kowalski
On 8 fév, 18:33, "Dominique Bourdeau" wrote:
> Dans son message précédent, kowalski a écrit : >> Back to ITU >> We have now have the 576 active lines at 52 ?s per line length, in >> which the picture >> information is contained, which we need for our digital final product. >> ? 576 lines a 52?s >> The International Telecommunication Union says that of the 864 >> horizontal samples, 720 >> are (line-) active. Those go into the final product, remaining 144 are >> "digital blanking". >> That is obviously a contradiction! >> We have a line of 64 ?s "length", which after (ITU-conform) sampling >> consists of 864 digital >> values. >> So again we calculate: >> 64?s / 864 = 74.074ns >> So, in a line of 64 ?s there are 864 samples with a "length" of 74.074 >> ns each. >> ? Period duration for a single pixel = 74.074 ns >> we multiply this value by the 720 pixels, which ITU recommends. >> 74.074ns * 720 = 53.3333?s >> The "window" suggested by the ITU is thus larger, than the 52 ?s >> existing video >> information. >> ? - 720 pixel = 53.3333 ?s >> We now divide the "genuine" active 52 ?s by the calculated period >> duration/pixel: >> 52?s / 74.074ns = 702.0007 = ~702 pixel >> There are only 702 active pixels!
ça vient de làhttp://www.arachnotron.nl/videocap/doc/Karl_cap_v1_en.p df
Mais il n'a pas tout mis car il ne comprend pas la suite.....
...La suite, tu nous l'expliqueras après... pour le moment ce qui m'intéresse c'est ce que j'ai posté. Tu as une idée de quoi ça parle pour me faire un topo ?
On 8 fév, 18:33, "Dominique Bourdeau" <dominique.bourd...@orange.fr>
wrote:
> Dans son message précédent, kowalski a écrit :
>> Back to ITU
>> We have now have the 576 active lines at 52 ?s per line length, in
>> which the picture
>> information is contained, which we need for our digital final product.
>> ? 576 lines a 52?s
>> The International Telecommunication Union says that of the 864
>> horizontal samples, 720
>> are (line-) active. Those go into the final product, remaining 144 are
>> "digital blanking".
>> That is obviously a contradiction!
>> We have a line of 64 ?s "length", which after (ITU-conform) sampling
>> consists of 864 digital
>> values.
>> So again we calculate:
>> 64?s / 864 = 74.074ns
>> So, in a line of 64 ?s there are 864 samples with a "length" of 74.074
>> ns each.
>> ? Period duration for a single pixel = 74.074 ns
>> we multiply this value by the 720 pixels, which ITU recommends.
>> 74.074ns * 720 = 53.3333?s
>> The "window" suggested by the ITU is thus larger, than the 52 ?s
>> existing video
>> information.
>> ? - 720 pixel = 53.3333 ?s
>> We now divide the "genuine" active 52 ?s by the calculated period
>> duration/pixel:
>> 52?s / 74.074ns = 702.0007 = ~702 pixel
>> There are only 702 active pixels!
ça vient de làhttp://www.arachnotron.nl/videocap/doc/Karl_cap_v1_en.p df
Mais il n'a pas tout mis car il ne comprend pas la suite.....
...La suite, tu nous l'expliqueras après... pour le moment ce qui
m'intéresse c'est ce que j'ai posté.
Tu as une idée de quoi ça parle pour me faire un topo ?
> Dans son message précédent, kowalski a écrit : >> Back to ITU >> We have now have the 576 active lines at 52 ?s per line length, in >> which the picture >> information is contained, which we need for our digital final product. >> ? 576 lines a 52?s >> The International Telecommunication Union says that of the 864 >> horizontal samples, 720 >> are (line-) active. Those go into the final product, remaining 144 are >> "digital blanking". >> That is obviously a contradiction! >> We have a line of 64 ?s "length", which after (ITU-conform) sampling >> consists of 864 digital >> values. >> So again we calculate: >> 64?s / 864 = 74.074ns >> So, in a line of 64 ?s there are 864 samples with a "length" of 74.074 >> ns each. >> ? Period duration for a single pixel = 74.074 ns >> we multiply this value by the 720 pixels, which ITU recommends. >> 74.074ns * 720 = 53.3333?s >> The "window" suggested by the ITU is thus larger, than the 52 ?s >> existing video >> information. >> ? - 720 pixel = 53.3333 ?s >> We now divide the "genuine" active 52 ?s by the calculated period >> duration/pixel: >> 52?s / 74.074ns = 702.0007 = ~702 pixel >> There are only 702 active pixels!
ça vient de làhttp://www.arachnotron.nl/videocap/doc/Karl_cap_v1_en.p df
Mais il n'a pas tout mis car il ne comprend pas la suite.....
...La suite, tu nous l'expliqueras après... pour le moment ce qui m'intéresse c'est ce que j'ai posté. Tu as une idée de quoi ça parle pour me faire un topo ?
*.-pipolin-.*
kowalski avait soumis l'idée :
On 8 fév, 18:45, *.-pipolin-.* wrote:
alors comprendre les problèmatiques de rafaichissement de diffuseurs, c'est même pas la peine...
arf c'est doooommage ! ça ne parle pas de rafraichissement d'écran.... Tu as toujours des problèmes de lecture, visiblement... :-D
sans blague, c'est bien que tu en parles, parce que ca illustre très bien le fait que lorsque tu es coincé sur un sujet, tu en changes, les rafraichissement te poses un problème, alors tu reviens sur tes délires de format et de pixel visibles...
alors, celon toi, comment sa se passe pour afficher une source en deux canaux a 24 images seconde sur un diffuseur avec un rafraichissement a 120 hz ?
On 8 fév, 18:45, *.-pipolin-.* <..pipoli...@DTC.com> wrote:
alors
comprendre les problèmatiques de rafaichissement de diffuseurs, c'est
même pas la peine...
arf c'est doooommage ! ça ne parle pas de rafraichissement d'écran....
Tu as toujours des problèmes de lecture, visiblement... :-D
sans blague, c'est bien que tu en parles, parce que ca illustre très
bien le fait que lorsque tu es coincé sur un sujet, tu en changes, les
rafraichissement te poses un problème, alors tu reviens sur tes délires
de format et de pixel visibles...
alors, celon toi, comment sa se passe pour afficher une source en deux
canaux a 24 images seconde sur un diffuseur avec un rafraichissement a
120 hz ?
alors comprendre les problèmatiques de rafaichissement de diffuseurs, c'est même pas la peine...
arf c'est doooommage ! ça ne parle pas de rafraichissement d'écran.... Tu as toujours des problèmes de lecture, visiblement... :-D
sans blague, c'est bien que tu en parles, parce que ca illustre très bien le fait que lorsque tu es coincé sur un sujet, tu en changes, les rafraichissement te poses un problème, alors tu reviens sur tes délires de format et de pixel visibles...
alors, celon toi, comment sa se passe pour afficher une source en deux canaux a 24 images seconde sur un diffuseur avec un rafraichissement a 120 hz ?
> Back to ITU > We have now have the 576 active lines at 52 μs per line length, in > which the picture > information is contained, which we need for our digital final product. > â 576 lines a 52μs > The International Telecommunication Union says that of the 864 > horizontal samples, 720 > are (line-) active. Those go into the final product, remaining 144 are > "digital blanking". > That is obviously a contradiction! > We have a line of 64 μs "length", which after (ITU-conform) sampli ng > consists of 864 digital > values. > So again we calculate: > 64μs / 864 = 74.074ns > So, in a line of 64 μs there are 864 samples with a âlengt hâ of 74.074 > ns each. > â Period duration for a single pixel = 74.074 ns > we multiply this value by the 720 pixels, which ITU recommends. > 74.074ns * 720 = 53.3333μs > The "window" suggested by the ITU is thus larger, than the 52 μs > existing video > information. > â - 720 pixel = 53.3333 μs > We now divide the "genuine" active 52 μs by the calculated period > duration/pixel: > 52μs / 74.074ns = 702.0007 = ~702 pixel > There are only 702 active pixels!
> Back to ITU
> We have now have the 576 active lines at 52 μs per line length, in
> which the picture
> information is contained, which we need for our digital final product.
> â 576 lines a 52μs
> The International Telecommunication Union says that of the 864
> horizontal samples, 720
> are (line-) active. Those go into the final product, remaining 144 are
> "digital blanking".
> That is obviously a contradiction!
> We have a line of 64 μs "length", which after (ITU-conform) sampli ng
> consists of 864 digital
> values.
> So again we calculate:
> 64μs / 864 = 74.074ns
> So, in a line of 64 μs there are 864 samples with a âlengt hâ of 74.074
> ns each.
> â Period duration for a single pixel = 74.074 ns
> we multiply this value by the 720 pixels, which ITU recommends.
> 74.074ns * 720 = 53.3333μs
> The "window" suggested by the ITU is thus larger, than the 52 μs
> existing video
> information.
> â - 720 pixel = 53.3333 μs
> We now divide the "genuine" active 52 μs by the calculated period
> duration/pixel:
> 52μs / 74.074ns = 702.0007 = ~702 pixel
> There are only 702 active pixels!
> Back to ITU > We have now have the 576 active lines at 52 μs per line length, in > which the picture > information is contained, which we need for our digital final product. > â 576 lines a 52μs > The International Telecommunication Union says that of the 864 > horizontal samples, 720 > are (line-) active. Those go into the final product, remaining 144 are > "digital blanking". > That is obviously a contradiction! > We have a line of 64 μs "length", which after (ITU-conform) sampli ng > consists of 864 digital > values. > So again we calculate: > 64μs / 864 = 74.074ns > So, in a line of 64 μs there are 864 samples with a âlengt hâ of 74.074 > ns each. > â Period duration for a single pixel = 74.074 ns > we multiply this value by the 720 pixels, which ITU recommends. > 74.074ns * 720 = 53.3333μs > The "window" suggested by the ITU is thus larger, than the 52 μs > existing video > information. > â - 720 pixel = 53.3333 μs > We now divide the "genuine" active 52 μs by the calculated period > duration/pixel: > 52μs / 74.074ns = 702.0007 = ~702 pixel > There are only 702 active pixels!
Back to ITU We have now have the 576 active lines at 52 μs per line length, in which the picture information is contained, which we need for our digital final product. ● 576 lines a 52μs The International Telecommunication Union says that of the 864 horizontal samples, 720 are (line-) active. Those go into the final product, remaining 144 are "digital blanking". That is obviously a contradiction! We have a line of 64 μs "length", which after (ITU-conform) sampling consists of 864 digital values. So again we calculate: 64μs / 864 = 74.074ns So, in a line of 64 μs there are 864 samples with a “length” of 74.074 ns each. ● Period duration for a single pixel = 74.074 ns we multiply this value by the 720 pixels, which ITU recommends. 74.074ns * 720 = 53.3333μs The "window" suggested by the ITU is thus larger, than the 52 μs existing video information. ● - 720 pixel = 53.3333 μs We now divide the "genuine" active 52 μs by the calculated period duration/pixel: 52μs / 74.074ns = 702.0007 = ~702 pixel There are only 702 active pixels!
un copier coller ?
Houlàlà c'est bien, tu te surpasses !!!! on a vachement avancé... :-D
On 8 fév, 18:08, *.-pipolin-.* <..pipoli...@DTC.com> wrote:
Dans son message précédent, kowalski a écrit :
Back to ITU
We have now have the 576 active lines at 52 μs per line length, in
which the picture
information is contained, which we need for our digital final product.
● 576 lines a 52μs
The International Telecommunication Union says that of the 864
horizontal samples, 720
are (line-) active. Those go into the final product, remaining 144 are
"digital blanking".
That is obviously a contradiction!
We have a line of 64 μs "length", which after (ITU-conform) sampling
consists of 864 digital
values.
So again we calculate:
64μs / 864 = 74.074ns
So, in a line of 64 μs there are 864 samples with a “length” of 74.074
ns each.
● Period duration for a single pixel = 74.074 ns
we multiply this value by the 720 pixels, which ITU recommends.
74.074ns * 720 = 53.3333μs
The "window" suggested by the ITU is thus larger, than the 52 μs
existing video
information.
● - 720 pixel = 53.3333 μs
We now divide the "genuine" active 52 μs by the calculated period
duration/pixel:
52μs / 74.074ns = 702.0007 = ~702 pixel
There are only 702 active pixels!
un copier coller ?
Houlàlà c'est bien, tu te surpasses !!!! on a vachement avancé... :-D
Back to ITU We have now have the 576 active lines at 52 μs per line length, in which the picture information is contained, which we need for our digital final product. ● 576 lines a 52μs The International Telecommunication Union says that of the 864 horizontal samples, 720 are (line-) active. Those go into the final product, remaining 144 are "digital blanking". That is obviously a contradiction! We have a line of 64 μs "length", which after (ITU-conform) sampling consists of 864 digital values. So again we calculate: 64μs / 864 = 74.074ns So, in a line of 64 μs there are 864 samples with a “length” of 74.074 ns each. ● Period duration for a single pixel = 74.074 ns we multiply this value by the 720 pixels, which ITU recommends. 74.074ns * 720 = 53.3333μs The "window" suggested by the ITU is thus larger, than the 52 μs existing video information. ● - 720 pixel = 53.3333 μs We now divide the "genuine" active 52 μs by the calculated period duration/pixel: 52μs / 74.074ns = 702.0007 = ~702 pixel There are only 702 active pixels!
un copier coller ?
Houlàlà c'est bien, tu te surpasses !!!! on a vachement avancé... :-D
>>> Back to ITU >>> We have now have the 576 active lines at 52 μs per line length, in >>> which the picture >>> information is contained, which we need for our digital final product . >>> â 576 lines a 52μs >>> The International Telecommunication Union says that of the 864 >>> horizontal samples, 720 >>> are (line-) active. Those go into the final product, remaining 144 ar e >>> "digital blanking". >>> That is obviously a contradiction! >>> We have a line of 64 μs "length", which after (ITU-conform) samp ling >>> consists of 864 digital >>> values. >>> So again we calculate: >>> 64μs / 864 = 74.074ns >>> So, in a line of 64 μs there are 864 samples with a âlen gthâ of 74.074 >>> ns each. >>> â Period duration for a single pixel = 74.074 ns >>> we multiply this value by the 720 pixels, which ITU recommends. >>> 74.074ns * 720 = 53.3333μs >>> The "window" suggested by the ITU is thus larger, than the 52 μs >>> existing video >>> information. >>> â - 720 pixel = 53.3333 μs >>> We now divide the "genuine" active 52 μs by the calculated perio d >>> duration/pixel: >>> 52μs / 74.074ns = 702.0007 = ~702 pixel >>> There are only 702 active pixels!
>>> Back to ITU
>>> We have now have the 576 active lines at 52 μs per line length, in
>>> which the picture
>>> information is contained, which we need for our digital final product .
>>> â 576 lines a 52μs
>>> The International Telecommunication Union says that of the 864
>>> horizontal samples, 720
>>> are (line-) active. Those go into the final product, remaining 144 ar e
>>> "digital blanking".
>>> That is obviously a contradiction!
>>> We have a line of 64 μs "length", which after (ITU-conform) samp ling
>>> consists of 864 digital
>>> values.
>>> So again we calculate:
>>> 64μs / 864 = 74.074ns
>>> So, in a line of 64 μs there are 864 samples with a âlen gthâ of 74.074
>>> ns each.
>>> â Period duration for a single pixel = 74.074 ns
>>> we multiply this value by the 720 pixels, which ITU recommends.
>>> 74.074ns * 720 = 53.3333μs
>>> The "window" suggested by the ITU is thus larger, than the 52 μs
>>> existing video
>>> information.
>>> â - 720 pixel = 53.3333 μs
>>> We now divide the "genuine" active 52 μs by the calculated perio d
>>> duration/pixel:
>>> 52μs / 74.074ns = 702.0007 = ~702 pixel
>>> There are only 702 active pixels!
>>> Back to ITU >>> We have now have the 576 active lines at 52 μs per line length, in >>> which the picture >>> information is contained, which we need for our digital final product . >>> â 576 lines a 52μs >>> The International Telecommunication Union says that of the 864 >>> horizontal samples, 720 >>> are (line-) active. Those go into the final product, remaining 144 ar e >>> "digital blanking". >>> That is obviously a contradiction! >>> We have a line of 64 μs "length", which after (ITU-conform) samp ling >>> consists of 864 digital >>> values. >>> So again we calculate: >>> 64μs / 864 = 74.074ns >>> So, in a line of 64 μs there are 864 samples with a âlen gthâ of 74.074 >>> ns each. >>> â Period duration for a single pixel = 74.074 ns >>> we multiply this value by the 720 pixels, which ITU recommends. >>> 74.074ns * 720 = 53.3333μs >>> The "window" suggested by the ITU is thus larger, than the 52 μs >>> existing video >>> information. >>> â - 720 pixel = 53.3333 μs >>> We now divide the "genuine" active 52 μs by the calculated perio d >>> duration/pixel: >>> 52μs / 74.074ns = 702.0007 = ~702 pixel >>> There are only 702 active pixels!
