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!
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 ?
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 ?
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 ?
"*.-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.....
"*.-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.....
"*.-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.....
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...
alors
comprendre les problèmatiques de rafaichissement de diffuseurs, c'est
même pas la peine...
> 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.....
> 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.....
> 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.....
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
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
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
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.
> 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!
un copier coller ?
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.
> 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!
un copier coller ?
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.
> 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!
un copier coller ?
On 8 fév, 18:08, *.-pipolin-.* 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
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
On 8 fév, 18:08, *.-pipolin-.* 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
kowalski a pensé très fort :
> On 8 fév, 18:08, *.-pipolin-.* 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 .
>>> 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!
>> un copier coller ?
> Houlà là c'est bien, tu te surpasses !!!! on a vachement avanc é... :-D
"on" ?
tu es tout seul, crétin...
kowalski a pensé très fort :
> 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 .
>>> 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!
>> un copier coller ?
> Houlà là c'est bien, tu te surpasses !!!! on a vachement avanc é... :-D
"on" ?
tu es tout seul, crétin...
kowalski a pensé très fort :
> On 8 fév, 18:08, *.-pipolin-.* 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 .
>>> 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!
>> un copier coller ?
> Houlà là c'est bien, tu te surpasses !!!! on a vachement avanc é... :-D
"on" ?
tu es tout seul, crétin...
kowalski avait soumis l'idée :
> On 8 fév, 18:45, *.-pipolin-.* wrote:
>> alors
>> comprendre les problèmatiques de rafaichissement de diffuseurs, c'es t
>> 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 délir es
de format et de pixel visibles...
kowalski avait soumis l'idée :
> On 8 fév, 18:45, *.-pipolin-.* <..pipoli...@DTC.com> wrote:
>> alors
>> comprendre les problèmatiques de rafaichissement de diffuseurs, c'es t
>> 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 délir es
de format et de pixel visibles...
kowalski avait soumis l'idée :
> On 8 fév, 18:45, *.-pipolin-.* wrote:
>> alors
>> comprendre les problèmatiques de rafaichissement de diffuseurs, c'es t
>> 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 délir es
de format et de pixel visibles...