Cost of DVD as data storage versus HDD (UK)

"Rob Morley" wrote in message
t...
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?


Yes!! Now imagine a file which occupies 1 degree of the track
(on *my* drive) and it is 180%s away (%=degree here)
the disk has to spin 181%s to get the data, however on your
'faster' drive it only has to spin 180.5 degrees, wow!!!!!
thats a great improvement!!!!! your drive is 181/180.5 faster
than mine, that is 1.00277 or 0.277% (back to real percents now).
So you are a quarter of a percent faster than me!!
Big deal!!!!!!! you would never notice it!!!!

Are you still with me?

Maybe I should rename myself 'quarter_pint'?

"Neil Maxwell" wrote in message
...
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...

I think it's about time you quit taking hallucination inducing
drugs as it appears to have become a permenant condition.


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

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


Yes!! Now imagine a file which occupies 1 degree of the track
(on *my* drive) and it is 180%s away (%=degree here)
the disk has to spin 181%s to get the data, however on your
'faster' drive it only has to spin 180.5 degrees, wow!!!!!
thats a great improvement!!!!! your drive is 181/180.5 faster
than mine, that is 1.00277 or 0.277% (back to real percents now).
So you are a quarter of a percent faster than me!!
Big deal!!!!!!! you would never notice it!!!!

Are you still with me?

Indeed. Now take into account that the head of the 1GB drive in my
example has to move from one track to the next before it can read the
second track - by the time it gets there it might have missed the start
of the data, in which case it would need another revolution before it
was read. This makes the 1GB drive a third of the speed of the 4GB
drive.
Now lets look at the combined effect of seek time and access time.
Assume an average seek time of 11 milliseconds for both disks.
For a 5400RPM drive one revolution takes 11 milliseconds. So for the
scenario in my example you have
for the 1GB disk: 11 + 11 + 11 + 11 = 44mS
for the 4GB disk: 11 + 11 = 22 mS
That's the worst-case scenario on an unfragmented disk. As you said,
the best-case scenario sees very little difference. So on average we
might expect the 4GB drive to have completed its read in 66% of the time
that it takes the 1GB drive. Remember we're talking a 4x difference in
arial density - the drives you were originally comparing were ISTR 5GB
per platter versus 60 GB per platter, which gives a difference in linear
density of around 330%, while my example used 200%.

Now tell me that the small drive is as fast as the big one - show your
workings.

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?


Yes!! Now imagine a file which occupies 1 degree of the track
(on *my* drive) and it is 180%s away (%=degree here)
the disk has to spin 181%s to get the data, however on your
'faster' drive it only has to spin 180.5 degrees, wow!!!!!
thats a great improvement!!!!! your drive is 181/180.5 faster
than mine, that is 1.00277 or 0.277% (back to real percents now).
So you are a quarter of a percent faster than me!!
Big deal!!!!!!! you would never notice it!!!!


Actually nobody will notice that because the likelyhood of above situation
is extremely small. By the way, what you describe is more likely a random,
not sequential access pattern. As we all know random access depends mostly
on seek time and latency. For a single track that would mean latency only
(assuming the same settle time, command overhead and others). So reading
very small chunks of data randomly will be done with the same speed. I guess
your hard drive contains only very small files. And all of them are placed
on the single track. The rest of the disk is empty.

"Peter" wrote in message
...


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?


Yes!! Now imagine a file which occupies 1 degree of the track
(on *my* drive) and it is 180%s away (%=degree here)
the disk has to spin 181%s to get the data, however on your
'faster' drive it only has to spin 180.5 degrees, wow!!!!!
thats a great improvement!!!!! your drive is 181/180.5 faster
than mine, that is 1.00277 or 0.277% (back to real percents now).
So you are a quarter of a percent faster than me!!
Big deal!!!!!!! you would never notice it!!!!


Actually nobody will notice that because the likelyhood of above situation
is extremely small. By the way, what you describe is more likely a random,
not sequential access pattern. As we all know random access depends mostly
on seek time and latency. For a single track that would mean latency only
(assuming the same settle time, command overhead and others). So reading
very small chunks of data randomly will be done with the same speed. I
guess
your hard drive contains only very small files. And all of them are placed
on the single track. The rest of the disk is empty.

I am not 100% sure of the term seek time and latency in this context.
One is changing track anf the other waiting for the disk to spin?

Reading large files is not generally a problem as their processing is
generally
slower than the read.
It is the constant switching between files etc... (seeking?) which slows
things
down considerabley.
Web browsing, for istance involves a lot of very small files, scattered
"all over the shop", reading the data is not a problem, it is the finding
the data to read which is the big problem.

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


Yes!! Now imagine a file which occupies 1 degree of the track
(on *my* drive) and it is 180%s away (%=degree here)
the disk has to spin 181%s to get the data, however on your
'faster' drive it only has to spin 180.5 degrees, wow!!!!!
thats a great improvement!!!!! your drive is 181/180.5 faster
than mine, that is 1.00277 or 0.277% (back to real percents now).
So you are a quarter of a percent faster than me!!
Big deal!!!!!!! you would never notice it!!!!

Are you still with me?

Indeed. Now take into account that the head of the 1GB drive in my
example has to move from one track to the next before it can read the
second track - by the time it gets there it might have missed the start
of the data, in which case it would need another revolution before it
was read. This makes the 1GB drive a third of the speed of the 4GB
drive.
Now lets look at the combined effect of seek time and access time.
Assume an average seek time of 11 milliseconds for both disks.
For a 5400RPM drive one revolution takes 11 milliseconds. So for the
scenario in my example you have
for the 1GB disk: 11 + 11 + 11 + 11 = 44mS
for the 4GB disk: 11 + 11 = 22 mS
That's the worst-case scenario on an unfragmented disk. As you said,
the best-case scenario sees very little difference. So on average we
might expect the 4GB drive to have completed its read in 66% of the time
that it takes the 1GB drive. Remember we're talking a 4x difference in
arial density - the drives you were originally comparing were ISTR 5GB
per platter versus 60 GB per platter, which gives a difference in linear
density of around 330%, while my example used 200%.

Now tell me that the small drive is as fast as the big one - show your
workings.

I will do that tomorrow when I have more time :O)

half_pint wrote:


"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).

You don't even know the correct terminology and yet you blithely assert that
"data transfer time is pretty negligible compared to seek time for most
applications" as if by asserting this you make it true.


Isaac

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

Yes!! Now imagine a file which occupies 1 degree of the track
(on *my* drive) and it is 180%s away (%=degree here)
the disk has to spin 181%s to get the data, however on your
'faster' drive it only has to spin 180.5 degrees, wow!!!!!
thats a great improvement!!!!! your drive is 181/180.5 faster
than mine, that is 1.00277 or 0.277% (back to real percents now).
So you are a quarter of a percent faster than me!!
Big deal!!!!!!! you would never notice it!!!!


Actually nobody will notice that because the likelyhood of above
situation
is extremely small. By the way, what you describe is more likely a
random,
not sequential access pattern. As we all know random access depends
mostly
on seek time and latency. For a single track that would mean latency
only
(assuming the same settle time, command overhead and others). So reading
very small chunks of data randomly will be done with the same speed. I
guess
your hard drive contains only very small files. And all of them are
placed
on the single track. The rest of the disk is empty.

I am not 100% sure of the term seek time and latency in this context.
One is changing track anf the other waiting for the disk to spin?

Reading large files is not generally a problem as their processing is
generally
slower than the read.
It is the constant switching between files etc... (seeking?) which slows
things
down considerabley.
Web browsing, for istance involves a lot of very small files, scattered
"all over the shop", reading the data is not a problem, it is the finding
the data to read which is the big problem.

Latency [ms] = 30000 / SpindleSpeed [RPM]

In article ,
says...
Latency [ms] = 30000 / SpindleSpeed [RPM]

Where does the 30000 figure come from?

"Toshi1873" wrote in message
.. .
In article ,
says...
Latency [ms] = 30000 / SpindleSpeed [RPM]

Where does the 30000 figure come from?

If SpindleSpeed is in rpm, we get:
Time for one revolution in minutes = 1 / SpindleSpeed

There are 60000 ms in a minute, so this is equivalent to:
Time for one revolution in ms = 60000 / SpindleSpeed

On average, rotational latency is the time for half a revolution, hence:
Latency in ms = (1/2) * 60000 / SpindleSpeed
= 30000 / SpindleSpeed

HTH,
Alex