Cost of DVD as data storage versus HDD (UK)

[half_pint]

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'?

 

[half_pint]

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.

 

[Rob Morley]

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.

 

[Peter]

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.

 

[half_pint]

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.

 

[half_pint]

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)

 

[J. Clarke]

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.

 

[Peter]

Yes!! Now imagine a file which occupies 1 degree of the track

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]

 

[Toshi1873]

Latency [ms] = 30000 / SpindleSpeed [RPM]

Where does the 30000 figure come from?

 

[Alex Fraser]

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

 

[half_pint]

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

I wondered where he plucked that figure from, I might have
guessed if it was 60000, mind you I did say somewhere in this
thread, that on average the data you want to read will be at
the other side of the disk, so it is nice to see my 'observation'
is part of an industry standard :O)

 

[Peter]

I wondered where he plucked that figure from, I might have

guessed if it was 60000, mind you I did say somewhere in this
thread, that on average the data you want to read will be at
the other side of the disk, so it is nice to see my 'observation'
is part of an industry standard :O)

Oh, it was here:
http://www.pcguide.com/ref/hdd/perf/perf/spec/posLatency-c.html
http://en.tldp.org/HOWTO/Unix-Hardware-Buyer-HOWTO/optimize.html

 

[Simon Finnigan]

.

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).

As someone else has stated, on average the drive will have to seek across
half of the platter. Depending what you`re doing, this could be reduced a
lot (if you`re accessing a load of data in sequence on a fairly well
de-fragged drive), but I can`t see many real-world situations that would
increase that figure by any significant margin. Why do you think that the
only issue here is the access time? Incidentally, if my drive spins at
7200rpm, and you`re spins at 5400rpm, that`s not a marginal increase in spin
speed, it`s almost a 50% difference.

What is the access time on your drive, can you find out?

End of story.

Why can you not admit you are wrong?

Because I`m not wrong.

Out of interest, why do you spend so much of your life trolling? You REALLY
should try going out and getting laid sometime, I`m sure it`d help your
mental issues.

 

[half_pint]

As someone else has stated, on average the drive will have to seek across
half of the platter. Depending what you`re doing, this could be reduced a
lot (if you`re accessing a load of data in sequence on a fairly well
de-fragged drive), but I can`t see many real-world situations that would
increase that figure by any significant margin. Why do you think that the
only issue here is the access time? Incidentally, if my drive spins at
7200rpm, and you`re spins at 5400rpm, that`s not a marginal increase in spin
speed, it`s almost a 50% difference.

What is the access time on your drive, can you find out?


Because I`m not wrong.

Out of interest, why do you spend so much of your life trolling? You REALLY
should try going out and getting laid sometime, I`m sure it`d help your
mental issues.

Bit rich coming from you sunshine ;O)

 

[Rob Morley]

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

Still waiting ...

 

[half_pint]

Still waiting ...

It will be ready to read by Thursday.

 

[J. Clarke]

I can play from mp3 on my pc, I won't buy a cd player (.wav) ever again
only mp3 playing devices, so I will never need to convert.



Whatever I have tried alll that, its too much hassel and it
doesn't get done when you are in rush and it is a job in it self.
How many hours (days more like) will it take to catalogue 300 DVD?

And then you cannot reorganise your files as you can on a HDD



Ah so I heed to add the cost of a new PC into the equation now?
Well thats another £500 minimum, mind you as it will have a HDD
inside I won't need to burn!!

If yours is so old that is doesn't have underrun protection then maybe it's
time to upgrade on general principle.

Well I find HHD's infinitely more reliable so I will invest my money
in that direction, incidently that is not the original CD drive, the
original failed within the first year and I got a free replacent. So in
My experience they are not very reliable (also have a failed
portable CD radio thing).
And whilst it *may* be my fault they failed, none of my
HDD's have ever had a single problem!!!

So the drives break. So what? When they break they don't lose your data.

Seconds? you have to read 300 lists, you must be a quick and flawless
reader, like my PC find files function.
It is a major hassle manintaining allyou lists and puting the corect
CD in the correct box.

Why would he have to read 300 lists? Just find the file that he wants in
the master directory, pull the numbered case he wants off the shelf, he's
done.

Or do you assume that the only record he has of what's on each CD is a paper
list? If so, you have no idea how to use the computer that is sitting in
front of you. Keeping that kind of information and making it accessible is
what computers _do_.

Yes but you don't need a set top DVD player as you can play
direct from your PC.

But now you need two, one in your workroom and one in the living room next
to the TV. Or you have to lug it back and forth. And how do you get the
data from one to the other? Now you have a network to worry about. And
given that you are too cheap to get yourself a five year old computer for
20 bucks that has enough gumption to support underrun protection I doubt
that you're going to pay for a second computer just to play videos on the
TV.

Oh, you don't have two rooms of your own? That would explain much.

Dont think I would trust a CD or a DVD, wont take up
much space on a HDD though.

Until that disk crashes and you lose _all_ of it, not just one DVD worth.

You do have your drives mirrored or RAIDEd I presume? No? Well, I predict
that in a few weeks or months or years you're going to be one of those
pitiful twits who comes on here with "I lost all my data, how can I get it
back without it costing me anything". To which the answer is "You're
screwed buddy".