Cost of DVD as data storage versus HDD (UK)

half_pint wrote:

.

Ok, so my drive spins at 7200 rpm, the same as yours. How big are your
platters? Lets be VERY generous, and say the full 5 gig capacity of your
drive is on a single platter. My smallest drive is 180 gigs - lets say
there are 3 platters there. My platters therefore hold 60 gigs each,
despite being the same physical size as your platters. Therefore the
data density on my platters is 12 times greater than on yours.

Therefore, for each revolution of the platter, my drive can read 12 times
more data. That`s 12 times the amount of data in the same amount of
time, making the data transfer rate 12 times greater.

Is that simple enough for you, or is it still too complicated for you to
understand?

You have demonstrated how stupid you are, you have no idea how a computer
works, statistacially the data will be on the other side of the drive and
it will take
your drive just as long to assess it as mine. (aprox bearing in mind your
marginally
faster spin speed).

Actually, _statistically_ it will be halfway to the "other side of the
drive", if in fact it is truly being sought at random. That model is
sometimes valid but not always--sometimes sequential access going from one
track to the next with interleaving to reduce latency to near zero is the
correct model.

Then there's the length of the read. If it's a single sector accessed at
random then things would be more or less as you say. However if there are
multiple sectors to be read then the higher transfer rate of the newer
drive becomes a factor as well.

End of story.

No, not end of story. You're grossly oversimplifying.

Why can you not admit you are wrong?

Because he's not wrong?

--
--John
Reply to jclarke at ae tee tee global dot net
(was jclarke at eye bee em dot net)

half_pint wrote:


"Rob Morley" wrote in message
...
In article , "half_pint"
says...

"Rob Morley" wrote in message
t...
In article , "half_pint"
says...
snip
However what you fail to realise is that data just behind the read
head
requires one revolution for it to be read (unless it has multipule
read
heads).
So my 5400 is only about 33% slower than a 'modern' 7200 drive.

A fact which even the most persistant of trolls cannot deny.

That's the maximum time that the head will take to start reading the
data after it has been positioned over the track. You have not taken
account of the time for the head to position over the track or the
speed
of data transfer once it has started reading.

WEll they are not relevant to my point so obviously not.
But thanks for verifying that i am correct anyway.

You originally wrote "I dont think new harddrives will be any faster
than mine ( speeds are basically the same 5400 or 7200 ) so I cant see
them writing any faster". So you are either stupid, trolling or
deranged.

No you obviously have no idea of how a computer works.

"A little knowledge is a dangerous thing". Perhaps rather than having no
idea, he has a much better idea of it than you do?

--
--John
Reply to jclarke at ae tee tee global dot net
(was jclarke at eye bee em dot net)

half_pint wrote:


"chrisv" wrote in message
...
"half_pint" wrote:

Seeing as the density of data is far more in a modern multi-platter
drive, the amount of data read in one revolution will be so much
more,
the spin speed of your ancient drive becomes irelevant when trying
to
suggest your drive reads and writes the same volume and speed as a
modern drive.

SO you agree that "my 5400 is only about 33% slower than a 'modern'
7200
drive."

Your drive might spin at 33 less speed, but that has no relevance in
your claim your ancient 3 gig drives performance is equal to modern
drives. Are you trying to move the goalposts? Every modern 5400rpm
drive will outperform your ancient drive, even though the spin speed
is the same.

Spin speed is a critical factor.

Clueless idiot. You're evading the point. Learn how to read and
think. Or maybe you just enjoy making a fool of yourself in public.

you are talking ********, there has been no significant improvement in
drive speeds, spin speed is the most important factor

Please provide the test results to support that argument.

and new drives don't
spin appreciateable
faster than old drives (not more than twice the speed) whilst other
components have
improved by several factors (about 10 times better).

I can wander into any PCfashionvictim store and click on a file, it will
appear
not faster than on my ancient PC.

ROF,L. You've timed this with a stopwatch for large files of course.

You obviously don't really understand computers, like many other wannabes
in this thread.

Actually, you only _think_ that you do.

My knowledge stems from intelligence, not listening to to a PC salesman
or reading expensive glossy PC magazines (you would be probably find
something
more suitable to your abilities on the top shelf).

I see. The rest of us seem to be getting our knowledge from experience.

--
--John
Reply to jclarke at ae tee tee global dot net
(was jclarke at eye bee em dot net)

"Eric Gisin" wrote in message
...
"half_pint" wrote in message
...

Do you know what a hard drive is for and what the definition
"performance" means?

Yes ,but you will to find that out for yourself as I don't have time to
explain, you are asking in the wrong forum anyway. (alt.hardrives
maybe?).

Right forum. You must be posting from alt.kooks.

Your drive might spin at 33 less speed, but that has no relevance in
your claim your ancient 3 gig drives performance is equal to modern
drives. Are you trying to move the goalposts? Every modern 5400rpm
drive will outperform your ancient drive, even though the spin speed
is the same.

Spin speed is a critical factor.

Nope, access time and STR for IDE drives.

Yes one is writing to ram, you need to find the speed at which
a head writes a track not to a data buffer.

Incomprehensible.

It would appear your argument is based on a flawed premise that
technology has not advanced. If you want to believe that, then be my
guest.

I am sorry to tell you that hardrives spin at aproximately the same
speed
they did ten years ago. Fact.

There is medication for delusions.

I think you would find councilling a better solution in the long term.

"half_pint" wrote in message
...
There is medication for delusions.


Maybe you should try councilling then.


Opps I read "is" as "is no", hence my reformed response :O)
Just as good I think you will agree.

On Thu, 4 Nov 2004 01:21:39 -0000, "half_pint" wrote:

I can wander into any PCfashionvictim store and click on a file, it will

half_pint stumbles and wobbles, swinging wildly, clearly ready to go
down for the count. The crowd would go wild, but it got bored and
wandered off in search of a fair fight some time ago. The locals
circle lazily, wondering if it's worth the effort of a few more
punches...


--
Neil Maxwell - I don't speak for my employer

In article , "half_pint"
says...

"Rob Morley" wrote in message
...
In article , "half_pint"
says...

"Rob Morley" wrote in message
t...
In article , "half_pint"
says...
snip
However what you fail to realise is that data just behind the read
head
requires one revolution for it to be read (unless it has multipule
read
heads).
So my 5400 is only about 33% slower than a 'modern' 7200 drive.

A fact which even the most persistant of trolls cannot deny.

That's the maximum time that the head will take to start reading the
data after it has been positioned over the track. You have not taken
account of the time for the head to position over the track or the
speed
of data transfer once it has started reading.

WEll they are not relevant to my point so obviously not.
But thanks for verifying that i am correct anyway.

You originally wrote "I dont think new harddrives will be any faster
than mine ( speeds are basically the same 5400 or 7200 ) so I cant see
them writing any faster". So you are either stupid, trolling or
deranged.

No you obviously have no idea of how a computer works.

Let's try this using really basic concepts and small words, shall we?

Imagine two hard drives - both single platter single-sided, same
rotational speed. One is 1GB, the other is 4GB. In order to get 4GB in
the same space as 1GB, the larger drive has twice as many tracks, and
each track holds twice as much data as a track on the 1GB drive. Now
imagine that you want to read a file that occupies 3/4 track on the 4GB
drive - this will occupy 1 1/2 tracks on the 1GB drive, so while the 4GB
drive can read it in a single revolution, the 1GB drive will need to
make up to two revolutions to read the same amount of data. So, all
other things being equal (which they are not) the 4GB drive is up to
twice as fast as the 1GB drive.

Are you with me so far?

On Wed, 03 Nov 2004 23:41:43 -0500, J. Clarke wrote:
half_pint wrote:

.

Ok, so my drive spins at 7200 rpm, the same as yours. How big are your
platters? Lets be VERY generous, and say the full 5 gig capacity of your
drive is on a single platter. My smallest drive is 180 gigs - lets say
there are 3 platters there. My platters therefore hold 60 gigs each,
despite being the same physical size as your platters. Therefore the
data density on my platters is 12 times greater than on yours.

Therefore, for each revolution of the platter, my drive can read 12 times
more data. That`s 12 times the amount of data in the same amount of
time, making the data transfer rate 12 times greater.

Is that simple enough for you, or is it still too complicated for you to
understand?

You have demonstrated how stupid you are, you have no idea how a computer
works, statistacially the data will be on the other side of the drive and
it will take
your drive just as long to assess it as mine. (aprox bearing in mind your
marginally
faster spin speed).

Actually, _statistically_ it will be halfway to the "other side of the
drive", if in fact it is truly being sought at random. That model is
sometimes valid but not always--sometimes sequential access going from one
track to the next with interleaving to reduce latency to near zero is the
correct model.

In fact for situations where you are burning data to a CD or DVD, most likely
the next needed chunk of data is right next to the last accessed piece unless
your drive is extremely fragmented. On average the transfer rate is not
determined by the average time to get to a randomly located piece of data.
The average transfer rate is going to predominately related to how fast
the drive can reach adjacent chunks.

At this point you have to assume that Half's obtuseness is deliberate.

Isaac

half_pint wrote:


"Eric Gisin" wrote in message
...
"half_pint" wrote in message
...

Do you know what a hard drive is for and what the definition
"performance" means?

Yes ,but you will to find that out for yourself as I don't have time to
explain, you are asking in the wrong forum anyway. (alt.hardrives
maybe?).

Right forum. You must be posting from alt.kooks.

Your drive might spin at 33 less speed, but that has no relevance in
your claim your ancient 3 gig drives performance is equal to modern
drives. Are you trying to move the goalposts? Every modern 5400rpm
drive will outperform your ancient drive, even though the spin speed
is the same.

Spin speed is a critical factor.

Nope, access time and STR for IDE drives.

Yes one is writing to ram, you need to find the speed at which
a head writes a track not to a data buffer.

Incomprehensible.

It would appear your argument is based on a flawed premise that
technology has not advanced. If you want to believe that, then be my
guest.

I am sorry to tell you that hardrives spin at aproximately the same
speed
they did ten years ago. Fact.

There is medication for delusions.

I think you would find councilling a better solution in the long term.

Maybe the Storage Performance Council? For which you are clearly not
qualified in any regard?

--
--John
Reply to jclarke at ae tee tee global dot net
(was jclarke at eye bee em dot net)

"Isaac" wrote in message
.. .
On Wed, 03 Nov 2004 23:41:43 -0500, J. Clarke
wrote:
half_pint wrote:

.

Ok, so my drive spins at 7200 rpm, the same as yours. How big are
your
platters? Lets be VERY generous, and say the full 5 gig capacity of
your
drive is on a single platter. My smallest drive is 180 gigs - lets
say
there are 3 platters there. My platters therefore hold 60 gigs each,
despite being the same physical size as your platters. Therefore the
data density on my platters is 12 times greater than on yours.

Therefore, for each revolution of the platter, my drive can read 12
times
more data. That`s 12 times the amount of data in the same amount of
time, making the data transfer rate 12 times greater.

Is that simple enough for you, or is it still too complicated for you
to
understand?

You have demonstrated how stupid you are, you have no idea how a
computer
works, statistacially the data will be on the other side of the drive
and
it will take
your drive just as long to assess it as mine. (aprox bearing in mind
your
marginally
faster spin speed).

Actually, _statistically_ it will be halfway to the "other side of the
drive", if in fact it is truly being sought at random. That model is
sometimes valid but not always--sometimes sequential access going from
one
track to the next with interleaving to reduce latency to near zero is
the
correct model.

In fact for situations where you are burning data to a CD or DVD, most
likely
the next needed chunk of data is right next to the last accessed piece
unless
your drive is extremely fragmented. On average the transfer rate is not
determined by the average time to get to a randomly located piece of data.
The average transfer rate is going to predominately related to how fast
the drive can reach adjacent chunks.

At this point you have to assume that Half's obtuseness is deliberate.


Well not really, I do actually have a point but I should mention we
are talking about hard drives rather than CDs or DVD.

My point is that that there are two aspects to drive speed, data transfer
rate and 'seek time' (is that the correct expression?).
Seek times have not improved much at all as they depend on spin
speed (and other *mechanical* things which have not improved
by much at all (I doubt changing track has improved much either)).
Data transfer time is pretty negligle compared to seek time for
most applications so I am basically correct (as usual).

Isaac