What's the use of a 192 kHz sample rate?

On May 3, 9:16 am, rajesh <getrajes...@gmail.com> wrote:
>    Its also about how you store data.
>
>    here is an simplified analogy.

Yes, simplified to the point of being factually wrong.

> say you need 44.1k samples per second to hear properly.
> If the disk is corrupted with scrathes and 1 samples in his
> region are lost your sound is distorted or lost for that period
> of time.

Wrong. First, you have a pretty robust error correction
scheme built in to the disk. The encoding and decoding
is such that significant amounts of data can be lost
but can be EXACTLY reconstructed on playback with NO
loss. And if the disk is severely scratched to the point where
the error correction algorith fails, interpolation takes place.

One can see thousands of uncorrected errors in the raw
data coming of the disk, and once the error correction
has been applied, the result might be a SMALL handful
(like, oh, 4?)  uncorrectable but interpolated errors

> Now if there are 196k samples even if (196/44.1)
> samples are lost there is no difference to what you
> hear.

False. Since you're cramming more data into the same
area, and the physical faults take up the same area
regardless of the data density, more bits, according to
YOUR theory, will be lost on the higher density disk
than on the lower density disk.

That means MORE data is missing, that means the
error correction algorith is subject to higher rates of
non-correctable errors, and so on. Your theory is
bogus if for no other reason than it simply ignores the
facts.

But, in EITHER case, unless the disk is SERIOUSLY
damaged, the data loss in either case is repaired.

> DVD's come wih high density of data due to this
> they are highly vulnerable to scratches this can
> be avoided with better waveform matching achieved
> by high sampling rate.

Sorry, this is nothing but technobabble nonsense.

On May 3, 9:16 am, rajesh <getrajes...@gmail.com> wrote:
>    Its also about how you store data.
>
>    here is an simplified analogy.

Yes, simplified to the point of being factually wrong.

> say you need 44.1k samples per second to hear properly.
> If the disk is corrupted with scrathes and 1 samples in his
> region are lost your sound is distorted or lost for that period
> of time.

Wrong. First, you have a pretty robust error correction
scheme built in to the disk. The encoding and decoding
is such that significant amounts of data can be lost
but can be EXACTLY reconstructed on playback with NO
loss. And if the disk is severely scratched to the point where
the error correction algorith fails, interpolation takes place.

One can see thousands of uncorrected errors in the raw
data coming of the disk, and once the error correction
has been applied, the result might be a SMALL handful
(like, oh, 4?)  uncorrectable but interpolated errors

> Now if there are 196k samples even if (196/44.1)
> samples are lost there is no difference to what you
> hear.

False. Since you're cramming more data into the same
area, and the physical faults take up the same area
regardless of the data density, more bits, according to
YOUR theory, will be lost on the higher density disk
than on the lower density disk.

That means MORE data is missing, that means the
error correction algorith is subject to higher rates of
non-correctable errors, and so on. Your theory is
bogus if for no other reason than it simply ignores the
facts.

But, in EITHER case, unless the disk is SERIOUSLY
damaged, the data loss in either case is repaired.

> DVD's come wih high density of data due to this
> they are highly vulnerable to scratches this can
> be avoided with better waveform matching achieved
> by high sampling rate.

Sorry, this is nothing but technobabble nonsense.

On May 3, 9:16 am, rajesh <getrajes...@gmail.com> wrote:
>    Its also about how you store data.
>
>    here is an simplified analogy.

Yes, simplified to the point of being factually wrong.

> say you need 44.1k samples per second to hear properly.
> If the disk is corrupted with scrathes and 1 samples in his
> region are lost your sound is distorted or lost for that period
> of time.

Wrong. First, you have a pretty robust error correction
scheme built in to the disk. The encoding and decoding
is such that significant amounts of data can be lost
but can be EXACTLY reconstructed on playback with NO
loss. And if the disk is severely scratched to the point where
the error correction algorith fails, interpolation takes place.

One can see thousands of uncorrected errors in the raw
data coming of the disk, and once the error correction
has been applied, the result might be a SMALL handful
(like, oh, 4?)  uncorrectable but interpolated errors

> Now if there are 196k samples even if (196/44.1)
> samples are lost there is no difference to what you
> hear.

False. Since you're cramming more data into the same
area, and the physical faults take up the same area
regardless of the data density, more bits, according to
YOUR theory, will be lost on the higher density disk
than on the lower density disk.

That means MORE data is missing, that means the
error correction algorith is subject to higher rates of
non-correctable errors, and so on. Your theory is
bogus if for no other reason than it simply ignores the
facts.

But, in EITHER case, unless the disk is SERIOUSLY
damaged, the data loss in either case is repaired.

> DVD's come wih high density of data due to this
> they are highly vulnerable to scratches this can
> be avoided with better waveform matching achieved
> by high sampling rate.

Sorry, this is nothing but technobabble nonsense.

On May 3, 9:16 am, rajesh <getrajes...@gmail.com> wrote:
>    Its also about how you store data.
>
>    here is an simplified analogy.

Yes, simplified to the point of being factually wrong.

> say you need 44.1k samples per second to hear properly.
> If the disk is corrupted with scrathes and 1 samples in his
> region are lost your sound is distorted or lost for that period
> of time.

Wrong. First, you have a pretty robust error correction
scheme built in to the disk. The encoding and decoding
is such that significant amounts of data can be lost
but can be EXACTLY reconstructed on playback with NO
loss. And if the disk is severely scratched to the point where
the error correction algorith fails, interpolation takes place.

One can see thousands of uncorrected errors in the raw
data coming of the disk, and once the error correction
has been applied, the result might be a SMALL handful
(like, oh, 4?)  uncorrectable but interpolated errors

> Now if there are 196k samples even if (196/44.1)
> samples are lost there is no difference to what you
> hear.

False. Since you're cramming more data into the same
area, and the physical faults take up the same area
regardless of the data density, more bits, according to
YOUR theory, will be lost on the higher density disk
than on the lower density disk.

That means MORE data is missing, that means the
error correction algorith is subject to higher rates of
non-correctable errors, and so on. Your theory is
bogus if for no other reason than it simply ignores the
facts.

But, in EITHER case, unless the disk is SERIOUSLY
damaged, the data loss in either case is repaired.

> DVD's come wih high density of data due to this
> they are highly vulnerable to scratches this can
> be avoided with better waveform matching achieved
> by high sampling rate.

Sorry, this is nothing but technobabble nonsense.

On May 3, 9:16 am, rajesh <getrajes...@gmail.com> wrote:
>    Its also about how you store data.
>
>    here is an simplified analogy.

Yes, simplified to the point of being factually wrong.

> say you need 44.1k samples per second to hear properly.
> If the disk is corrupted with scrathes and 1 samples in his
> region are lost your sound is distorted or lost for that period
> of time.

Wrong. First, you have a pretty robust error correction
scheme built in to the disk. The encoding and decoding
is such that significant amounts of data can be lost
but can be EXACTLY reconstructed on playback with NO
loss. And if the disk is severely scratched to the point where
the error correction algorith fails, interpolation takes place.

One can see thousands of uncorrected errors in the raw
data coming of the disk, and once the error correction
has been applied, the result might be a SMALL handful
(like, oh, 4?)  uncorrectable but interpolated errors

> Now if there are 196k samples even if (196/44.1)
> samples are lost there is no difference to what you
> hear.

False. Since you're cramming more data into the same
area, and the physical faults take up the same area
regardless of the data density, more bits, according to
YOUR theory, will be lost on the higher density disk
than on the lower density disk.

That means MORE data is missing, that means the
error correction algorith is subject to higher rates of
non-correctable errors, and so on. Your theory is
bogus if for no other reason than it simply ignores the
facts.

But, in EITHER case, unless the disk is SERIOUSLY
damaged, the data loss in either case is repaired.

> DVD's come wih high density of data due to this
> they are highly vulnerable to scratches this can
> be avoided with better waveform matching achieved
> by high sampling rate.

Sorry, this is nothing but technobabble nonsense.

On May 3, 9:16 am, rajesh <getrajes...@gmail.com> wrote:
>    Its also about how you store data.
>
>    here is an simplified analogy.

Yes, simplified to the point of being factually wrong.

> say you need 44.1k samples per second to hear properly.
> If the disk is corrupted with scrathes and 1 samples in his
> region are lost your sound is distorted or lost for that period
> of time.

Wrong. First, you have a pretty robust error correction
scheme built in to the disk. The encoding and decoding
is such that significant amounts of data can be lost
but can be EXACTLY reconstructed on playback with NO
loss. And if the disk is severely scratched to the point where
the error correction algorith fails, interpolation takes place.

One can see thousands of uncorrected errors in the raw
data coming of the disk, and once the error correction
has been applied, the result might be a SMALL handful
(like, oh, 4?)  uncorrectable but interpolated errors

> Now if there are 196k samples even if (196/44.1)
> samples are lost there is no difference to what you
> hear.

False. Since you're cramming more data into the same
area, and the physical faults take up the same area
regardless of the data density, more bits, according to
YOUR theory, will be lost on the higher density disk
than on the lower density disk.

That means MORE data is missing, that means the
error correction algorith is subject to higher rates of
non-correctable errors, and so on. Your theory is
bogus if for no other reason than it simply ignores the
facts.

But, in EITHER case, unless the disk is SERIOUSLY
damaged, the data loss in either case is repaired.

> DVD's come wih high density of data due to this
> they are highly vulnerable to scratches this can
> be avoided with better waveform matching achieved
> by high sampling rate.

Sorry, this is nothing but technobabble nonsense.

On May 3, 9:16 am, rajesh <getrajes...@gmail.com> wrote:
>    Its also about how you store data.
>
>    here is an simplified analogy.

Yes, simplified to the point of being factually wrong.

> say you need 44.1k samples per second to hear properly.
> If the disk is corrupted with scrathes and 1 samples in his
> region are lost your sound is distorted or lost for that period
> of time.

Wrong. First, you have a pretty robust error correction
scheme built in to the disk. The encoding and decoding
is such that significant amounts of data can be lost
but can be EXACTLY reconstructed on playback with NO
loss. And if the disk is severely scratched to the point where
the error correction algorith fails, interpolation takes place.

One can see thousands of uncorrected errors in the raw
data coming of the disk, and once the error correction
has been applied, the result might be a SMALL handful
(like, oh, 4?)  uncorrectable but interpolated errors

> Now if there are 196k samples even if (196/44.1)
> samples are lost there is no difference to what you
> hear.

False. Since you're cramming more data into the same
area, and the physical faults take up the same area
regardless of the data density, more bits, according to
YOUR theory, will be lost on the higher density disk
than on the lower density disk.

That means MORE data is missing, that means the
error correction algorith is subject to higher rates of
non-correctable errors, and so on. Your theory is
bogus if for no other reason than it simply ignores the
facts.

But, in EITHER case, unless the disk is SERIOUSLY
damaged, the data loss in either case is repaired.

> DVD's come wih high density of data due to this
> they are highly vulnerable to scratches this can
> be avoided with better waveform matching achieved
> by high sampling rate.

Sorry, this is nothing but technobabble nonsense.

On May 3, 9:16 am, rajesh <getrajes...@gmail.com> wrote:
>    Its also about how you store data.
>
>    here is an simplified analogy.

Yes, simplified to the point of being factually wrong.

> say you need 44.1k samples per second to hear properly.
> If the disk is corrupted with scrathes and 1 samples in his
> region are lost your sound is distorted or lost for that period
> of time.

Wrong. First, you have a pretty robust error correction
scheme built in to the disk. The encoding and decoding
is such that significant amounts of data can be lost
but can be EXACTLY reconstructed on playback with NO
loss. And if the disk is severely scratched to the point where
the error correction algorith fails, interpolation takes place.

One can see thousands of uncorrected errors in the raw
data coming of the disk, and once the error correction
has been applied, the result might be a SMALL handful
(like, oh, 4?)  uncorrectable but interpolated errors

> Now if there are 196k samples even if (196/44.1)
> samples are lost there is no difference to what you
> hear.

False. Since you're cramming more data into the same
area, and the physical faults take up the same area
regardless of the data density, more bits, according to
YOUR theory, will be lost on the higher density disk
than on the lower density disk.

That means MORE data is missing, that means the
error correction algorith is subject to higher rates of
non-correctable errors, and so on. Your theory is
bogus if for no other reason than it simply ignores the
facts.

But, in EITHER case, unless the disk is SERIOUSLY
damaged, the data loss in either case is repaired.

> DVD's come wih high density of data due to this
> they are highly vulnerable to scratches this can
> be avoided with better waveform matching achieved
> by high sampling rate.

Sorry, this is nothing but technobabble nonsense.

On May 3, 9:16 am, rajesh <getrajes...@gmail.com> wrote:
>    Its also about how you store data.
>
>    here is an simplified analogy.

Yes, simplified to the point of being factually wrong.

> say you need 44.1k samples per second to hear properly.
> If the disk is corrupted with scrathes and 1 samples in his
> region are lost your sound is distorted or lost for that period
> of time.

Wrong. First, you have a pretty robust error correction
scheme built in to the disk. The encoding and decoding
is such that significant amounts of data can be lost
but can be EXACTLY reconstructed on playback with NO
loss. And if the disk is severely scratched to the point where
the error correction algorith fails, interpolation takes place.

One can see thousands of uncorrected errors in the raw
data coming of the disk, and once the error correction
has been applied, the result might be a SMALL handful
(like, oh, 4?)  uncorrectable but interpolated errors

> Now if there are 196k samples even if (196/44.1)
> samples are lost there is no difference to what you
> hear.

False. Since you're cramming more data into the same
area, and the physical faults take up the same area
regardless of the data density, more bits, according to
YOUR theory, will be lost on the higher density disk
than on the lower density disk.

That means MORE data is missing, that means the
error correction algorith is subject to higher rates of
non-correctable errors, and so on. Your theory is
bogus if for no other reason than it simply ignores the
facts.

But, in EITHER case, unless the disk is SERIOUSLY
damaged, the data loss in either case is repaired.

> DVD's come wih high density of data due to this
> they are highly vulnerable to scratches this can
> be avoided with better waveform matching achieved
> by high sampling rate.

Sorry, this is nothing but technobabble nonsense.

On May 3, 9:16 am, rajesh <getrajes...@gmail.com> wrote:
>    Its also about how you store data.
>
>    here is an simplified analogy.

Yes, simplified to the point of being factually wrong.

> say you need 44.1k samples per second to hear properly.
> If the disk is corrupted with scrathes and 1 samples in his
> region are lost your sound is distorted or lost for that period
> of time.

Wrong. First, you have a pretty robust error correction
scheme built in to the disk. The encoding and decoding
is such that significant amounts of data can be lost
but can be EXACTLY reconstructed on playback with NO
loss. And if the disk is severely scratched to the point where
the error correction algorith fails, interpolation takes place.

One can see thousands of uncorrected errors in the raw
data coming of the disk, and once the error correction
has been applied, the result might be a SMALL handful
(like, oh, 4?)  uncorrectable but interpolated errors

> Now if there are 196k samples even if (196/44.1)
> samples are lost there is no difference to what you
> hear.

False. Since you're cramming more data into the same
area, and the physical faults take up the same area
regardless of the data density, more bits, according to
YOUR theory, will be lost on the higher density disk
than on the lower density disk.

That means MORE data is missing, that means the
error correction algorith is subject to higher rates of
non-correctable errors, and so on. Your theory is
bogus if for no other reason than it simply ignores the
facts.

But, in EITHER case, unless the disk is SERIOUSLY
damaged, the data loss in either case is repaired.

> DVD's come wih high density of data due to this
> they are highly vulnerable to scratches this can
> be avoided with better waveform matching achieved
> by high sampling rate.

Sorry, this is nothing but technobabble nonsense.