Chip with simple program for Toy

[hrhofmann@att.net]

You have tio know the standards for each and every country you will be
marketing your product in.

The best thing to do is bite the bullet and use a contract test house
to do the testing for the countries in which you will be selling the
product.

You want to use a test house that has accreditation for the type of
product you manufacture. In the USA, that would be A2LA or NVLAP
accreditation.

H. R. (Bob) Hofmann

 

[Rich Grise]

On Thu, 14 Jul 2005 12:31:26 -0400, Spehro Pefhany wrote:

On Thu, 14 Jul 2005 11:29:57 -0400, the renowned Keith Williams

Actually, whatever points either side of the debate
make are moot when you consider that a receptacle
can be installed sideways. Then the debate will be
"does the ground prong go left or right?"

Neutral on top, of course. ;-)

Ok, now let's move onto the great switch direction debate. Up/down?
Left/right? ;-)

Rocker 'in' at the top or bottom for 'on'?

Top half of rocker pressed == on.
Bottom half of rocker pressed == off.

This dates back to the days of the wall switches with the two,
one-inch-long buttons sticking out of the wall, and the top
one had a pearl insert, and when it was "IN", the light was
on, and the bottom one was "OUT", and when you'd press the
bottom one (more like "shove,"), it would latch "IN", eject
the top button to the "OUT" position, and turn the light
off. These have been replaced, of course, by toggle and
rocker switches. Duh.

But I think on the up/down switch, I think that rocker switches
would use the same logic as the bat handle on ordinary light
switches - when it's up, it's on, when it's down, it's off.
I think the justification/rationale is that, when there's an
electrical "mishap", you can just take a blind lunge at the
switch, and if you manage to swat it on the way to the floor,
you might have a chance to de-energize the circuit. ;-)

Of course, if you've got your switches mounted sideways, all
bets are off. ;-)

Cheers!
Rich

 

[Rich Grise]

On Mon, 11 Jul 2005 11:24:35 +0200, OBones wrote:

Pooh Bear wrote:

With the exception of the USA. IEC norms apply. End of story. The USA is slowly
applying IEC norms in UL etc standards too.

Uho, the EEC has it's own standards with respect to safety. Sure, most
of them are based on IEC, but some of them are much more restrictive.
For instance, the plugs in the US or Australia could not even make it to
market in Europe, they are not safe enough. I mean, it's way too easy to
touch the metal part if the plug is not fully inserted.
And that's only one example.

In America our kids get taught at an early age to keep their thumb
out from between the prongs while plugging in a plug. I still remember
when I learned it! =:-Z

Thanks!
Rich

 

[redbelly]

Dan Mills wrote:

The real reason is that arc damage on opening is dependent on both current
and voltage. As the switch opens an arc forms between the contacts, and if
there is too much power being handled then this can destroy the switch.

Makes sense to me. Thanks Dan.

Mark

 

[art]

look here: http://artur9n9.com/hvl-32x.htm
-------------------------------------
kreb wrote:

Keith said:

I have purchased the Sony (hvl-f32x) 6 volt flash but found it, slow to

recharge the flash, the problem is old technology used by Sony. They
are
using 4 1.5 volt (AA) alkaline batteries for the 6 volts but I am
using
(AA) rechargeable cells at 1.2 volts. That gives me 4.8 volts so I am
only
able to get around 30 full shots "but" the worst problem is
the slow
recharge after 10 flashes.

I am going to build a power supply with 5 rechargeable giving me the 6
volts. But of course the problem is finding how to connect (hardwire)
to
the flash without too much disturbance of the body. I am not sure how
to
get into the unit and where to connect with the battery terminals as I
am
not in as yet. This unit has only 4 battery slots so I must improvise
if I
am going to be able to use the rechargeable.

Question:

1) Where can I get the power supply schematics for this unit.

2) Since I will be using 5 cells I am probably going to have to use a
six
volt battery charger as no 1.2 volt charger, that I know of, comes with
the
ability to charge such an odd number of individual 1.2 volt cells.

Can anyone tell me how to tear this unit apart without damage as I want
to
keep the warranty if possible, but that is not a key concern.

I would go with high mah AA's if the flash was able to hold the five
cells
needed but alas it doesn't.

Before I purchased this flash I had heard the concern about lack of
power
and recharge flash time and now I realize why. And, yes, I would use
another flash but this is the only one with TTL for the Sony f828, so,
I
need this flash.

If anyone could help I would appreciate it.

Keith

##-----------------------------------------------##
Article posted with Cabling-Design.com Newsgroup Archive
http://www.cabling-design.com/forums
no-spam read and post WWW interface to your favorite newsgroup -
sci.electronics.basics - 19963 messages and counting!
##-----------------------------------------------##

 

[Ewan Parsons]

-------------------------------------
Dean wrote:

Please help me if you can and send me documentation about
DECABIT ripple control communication protocol

used in some ripple control receivers.

##-----------------------------------------------##
Article posted with Cabling-Design.com Newsgroup Archive
http://www.cabling-design.com/forums
no-spam read and post WWW interface to your favorite newsgroup -
sci.electronics.basics - 19963 messages and counting!
##-----------------------------------------------##

 

[John Fields]

On Thu, 21 Jul 2005 17:56:31 GMT,
ewan_at_phase6_dot_com_dot_au@foo.com (Ewan Parsons) wrote:

-------------------------------------
Dean wrote:


Please help me if you can and send me documentation about
DECABIT ripple control communication protocol

used in some ripple control receivers.

---
Google is your friend.

--
John Fields
Professional Circuit Designer

 

[Rich Grise]

On Fri, 24 Jun 2005 15:26:35 -0700, obliquez wrote:

Lastly, what is an OP? Does it refer to me? If it does.. I'm a her not a
he. -smiles-

Original Poster. So, do you have a name other than "obliquez"?

Thanks,
Rich

 

[John Fields]

On Sat, 06 Aug 2005 05:58:56 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Fri, 5 Aug 2005 23:23:22 -0500, "scanner80"
scanner80@charter.net> wrote:

I just want to be able to measure the time it takes from when the resistance
leaves 1355 ohms and returns to 1355. I can do it with a meter and a
stopwatch, but I was looking for a more verifiable way to do it. I want to
remove human error. This is for calibration purposes.
Is there any type of simple timer or timer circuit that can will trigger as
the ohms changes and stop when it returns?

---
You could do it with something as simple as a voltage comparator
detecting the resistance change and using its (the voltage
comparator's) output to gate a counter, but you need to supply more
details about the resistance before we can help you much. For
example, what is the resistance, physically, and how is it being
used? Does it have a voltage across it?

---
BTW, I just noticed that you also posted your query to seb
separately. A more convenient way to post, if you have the same
message which you'd like to post to multiple groups is to
'crosspost'. All you have to do is type the names of all the
newsgroups you want the post to go to into the "newsgroups" (or
whatever it's called in your reader) line and it'll go to all of
them. It's a good way to do it because then the entire thread is
available to whoever wants to read it without having to hop back and
forth between groups. This post will go to alt.electronics and
sci.electronics.basics and, if you care to reply to it, so will your
reply, and so on.

--
John Fields
Professional Circuit Designer

 

[John Fields]

On Fri, 5 Aug 2005 21:12:21 -0500, "DBLEXPOSURE"
<celstuff@hotmail.com> wrote:

"scanner80" <scanner80@charter.net> wrote in message
newsEUIe.5352$Tt6.4270@fe04.lga...
Hello,
I'm looking for a circuit I can build or equipment I can buy
so I can time how long an ohms change is. I would like to build it if
possible . I would like to view it on a scope.
One example would be a starting ohms reading from a device of 1355 ohms.
It
will increase by approx. 20 ohms
and then return to 1355 ohms. The time it will take is approx. 19 seconds.
I
will need to be able to see a change as small as 1 ohm if possible , but
the
most important thing is to see the reading change from 1355 and return to
1355 ohms. I need to then measure the time with cursers on a scope.
I know a respiration monitor can see and display an ohms change , but I
need
a way to exactly measure the change.
I will be greatful for any help.
Thank you,
Jeff



Jeff:



To measure resistance you need to apply a voltage and then
measure the current. Another way is to use a "voltage divider", put your
changing resistance in series with a known resistance, apply a voltage to
the circuit and monitor the voltage dropped across the known resistor. As
your changing resistance decreases current in the circuit will increase and
the voltage drop across your known resistor will increase. This will all be
linear so you will be able to make notations on your scope graticule so that
you can directly readout the display in ohms.

---
That won't work, for several reasons.

The first is that the resistance change is too small to give a
detectable deflection of the scope trace with the scope vertical
input voltage range being whatever it needs to be to keep the trace
on the screen. For example:




E1
|
[R1]
|
+--->E2
|
[R2]
|
0V

If we let E1 equal 1 volt and R1 and R2 equal 1000 ohms and 1355
ohms, respectively, then with R2 at 1355 ohms we'll have:


E1R2 1V * 1355R
E2 = --------- = --------------- = 0.575 volts
R1 + R2 1100R + 1355R


and with R2 at the high end, (1375 ohms) W2 will be equal to 0.579V.

That's a change of only four millivolts, which would give you a
deflection of two boxes with an input sensitivity of 2mV per box.
That might be OK, but look at what that 4mV signal is riding on: a
voltage about 150 times higher, so no matter how you adjust the
vertical sensitivity and trace position controls, I don't believe
you're going to wind up with anything that works.


The second problem is going to be triggering the scope at the
instant the resistance starts to rise, unless an external trigger
can be rigged using a comparator, and the third problem is going to
be the accuracy of the timebase and marking the screen properly when
the trace crosses the reference. In other words, when was the last
time you had your scope calibrated and how good are you with that
grease pencil/sharpie?

BTW, that 4mV change _won't_ be linear, but it doesn't matter since
all he really wants to do is look for the crossing.
---

You will want to use a regulated voltage source. A look into Ohms law will
provide the math for you to make the calculations..

---
Woudn't have hurt for you to run the numbers, and it would have
saved me a post.


Just as an aside, there _is_ a way to do it using a couple of
voltage dividers to make a bridge, like this:


E1>--------+----------------+
| |
[R1] [R3]
| |
+--->E2 E3<---+
| |
[R2] [R4]
| |
0V>--------+----------------+


What happens here is that when the ratio of R1 to R2 equals the
ratio of R3 to R4, then E2 will equal E3 and you can connect a scope
between E2 and E3 and measure the voltage between them without
having to worry about the offset you get with a half-bridge.

For ease of use, R3R4 could be a pot which could be adjusted for
precisely zero volts across E2 and E3 when R2 was at 1355 ohms, or
for greater resolution,


E1>--------+------------------+
| |
[R1] [R3]
| |
+--->E2 E3<---[POT]
| |
[R2] [R4]
| |
0V>--------+------------------+

One caveat, the supply should be floating or, if the scope can do
it, it should be set to display E2 minus E3













--
John Fields
Professional Circuit Designer

 

[DBLEXPOSURE]

"John Fields" <jfields@austininstruments.com> wrote in message
newseb9f1t6ksl2drtjl6b7hdsplqjq3dr8s9@4ax.com...

On Fri, 5 Aug 2005 21:12:21 -0500, "DBLEXPOSURE"
celstuff@hotmail.com> wrote:


"scanner80" <scanner80@charter.net> wrote in message
newsEUIe.5352$Tt6.4270@fe04.lga...
Hello,
I'm looking for a circuit I can build or equipment I can buy
so I can time how long an ohms change is. I would like to build it if
possible . I would like to view it on a scope.
One example would be a starting ohms reading from a device of 1355 ohms.
It
will increase by approx. 20 ohms
and then return to 1355 ohms. The time it will take is approx. 19
seconds.
I
will need to be able to see a change as small as 1 ohm if possible , but
the
most important thing is to see the reading change from 1355 and return
to
1355 ohms. I need to then measure the time with cursers on a scope.
I know a respiration monitor can see and display an ohms change , but I
need
a way to exactly measure the change.
I will be greatful for any help.
Thank you,
Jeff



Jeff:



To measure resistance you need to apply a voltage and then
measure the current. Another way is to use a "voltage divider", put your
changing resistance in series with a known resistance, apply a voltage to
the circuit and monitor the voltage dropped across the known resistor. As
your changing resistance decreases current in the circuit will increase
and
the voltage drop across your known resistor will increase. This will all
be
linear so you will be able to make notations on your scope graticule so
that
you can directly readout the display in ohms.

---
That won't work, for several reasons.

The first is that the resistance change is too small to give a
detectable deflection of the scope trace with the scope vertical
input voltage range being whatever it needs to be to keep the trace
on the screen. For example:




E1
|
[R1]
|
+--->E2
|
[R2]
|
0V

If we let E1 equal 1 volt and R1 and R2 equal 1000 ohms and 1355
ohms, respectively, then with R2 at 1355 ohms we'll have:


E1R2 1V * 1355R
E2 = --------- = --------------- = 0.575 volts
R1 + R2 1100R + 1355R


and with R2 at the high end, (1375 ohms) W2 will be equal to 0.579V.

That's a change of only four millivolts, which would give you a
deflection of two boxes with an input sensitivity of 2mV per box.
That might be OK, but look at what that 4mV signal is riding on: a
voltage about 150 times higher, so no matter how you adjust the
vertical sensitivity and trace position controls, I don't believe
you're going to wind up with anything that works.


The second problem is going to be triggering the scope at the
instant the resistance starts to rise, unless an external trigger
can be rigged using a comparator, and the third problem is going to
be the accuracy of the timebase and marking the screen properly when
the trace crosses the reference. In other words, when was the last
time you had your scope calibrated and how good are you with that
grease pencil/sharpie?

BTW, that 4mV change _won't_ be linear, but it doesn't matter since
all he really wants to do is look for the crossing.
---

You will want to use a regulated voltage source. A look into Ohms law
will
provide the math for you to make the calculations..

---
Woudn't have hurt for you to run the numbers, and it would have
saved me a post.


Just as an aside, there _is_ a way to do it using a couple of
voltage dividers to make a bridge, like this:


E1>--------+----------------+
| |
[R1] [R3]
| |
+--->E2 E3<---+
| |
[R2] [R4]
| |
0V>--------+----------------+


What happens here is that when the ratio of R1 to R2 equals the
ratio of R3 to R4, then E2 will equal E3 and you can connect a scope
between E2 and E3 and measure the voltage between them without
having to worry about the offset you get with a half-bridge.

For ease of use, R3R4 could be a pot which could be adjusted for
precisely zero volts across E2 and E3 when R2 was at 1355 ohms, or
for greater resolution,


E1>--------+------------------+
| |
[R1] [R3]
| |
+--->E2 E3<---[POT]
| |
[R2] [R4]
| |
0V>--------+------------------+

One caveat, the supply should be floating or, if the scope can do
it, it should be set to display E2 minus E3













--
John Fields
Professional Circuit Designer

Ř The first is that the resistance change is too small to give a

detectable deflection of the scope trace with the scope vertical
input voltage range being whatever it needs to be to keep the trace
on the screen.

If a 30V source is used (My bench supply is 0-60Vdc regulated, BK 1623A) and
1K for the known resistance a difference in 20 Ohms of the resistor under
test will result in a rise of 100mV across the known resistance



30Vdc

|

|

R1 (1355-1375 Ohms)

|

|

R2 1K Ohm -ŕ

|

|

Gnd------------ŕ





1355+1000=2355 Ohms



30/2355=12.7mA



12.7mA X 1000 Ohms = 12.7V



And the high end



1375+1000=2375 Ohms



30/2375=12.6mA



12.6mA X 1000 Ohms = 12.6V



A difference of



12.7-12.6 = 100mV







Ř

The second problem is going to be triggering the scope at the
instant the resistance starts to rise, unless an external trigger
can be rigged using a comparator, and the third problem is going to
be the accuracy of the timebase and marking the screen properly when
the trace crosses the reference. In other words, when was the last
time you had your scope calibrated and how good are you with that
grease pencil/sharpie

Ř > will increase by approx. 20 ohms

and then return to 1355 ohms. The time it will take is approx. 19
seconds.

If this whole process takes 19 seconds, triggering shouldn't be an issue.
Remember, he was using a stopwatch to measure the time, so, just use a quick
time base and watch the green line rise and fall.



10mV/cm gives me full deflection and I have plenty of range on the Vert.
position to bring the trace to the bottom of the screen. My scope is now
measuring 2 ohms/cm. I don't know why this would not be linear. Scope is
calibrated annually.



As for not running the number last night, I was doing other things and
fiddled with this while on break. And, I didn't want to steal your thunder.
I knew you would come up with a more elegant way of doing this. I'm just a
hack anyway..



I do like your bridge idea...

 

[John Fields]

On Sat, 6 Aug 2005 11:09:05 -0500, "DBLEXPOSURE"
<celstuff@hotmail.com> wrote:

If a 30V source is used (My bench supply is 0-60Vdc regulated, BK 1623A) and
1K for the known resistance a difference in 20 Ohms of the resistor under
test will result in a rise of 100mV across the known resistance

snip

12.7-12.6 = 100mV

If this whole process takes 19 seconds, triggering shouldn't be an issue.
Remember, he was using a stopwatch to measure the time, so, just use a quick
time base and watch the green line rise and fall.

---
But, remember, he said he didn't want to do it that way any more.
---

10mV/cm gives me full deflection and I have plenty of range on the Vert.
position to bring the trace to the bottom of the screen. My scope is now
measuring 2 ohms/cm. I don't know why this would not be linear. Scope is
calibrated annually.

---
Did you actually try it?

I've got a Tektronix 2215, and with a 15V input, the best I can do
is 1V/box and still keep the trace on the scope with the position
control.

I didn't say the scope's response wasn't linear, I was talking about
the change in the output voltage from the divider as a function of
the change in resistance of the DUT.
---

I do like your bridge idea...

---
Thanks, but the credit should go to Samuel Hunter Christie and Sir
Charles Wheatstone.


--
John Fields
Professional Circuit Designer

 

[DBLEXPOSURE]

"John Fields" <jfields@austininstruments.com> wrote in message
news:hmq9f1p9auflc08j32183k8vmpgdnoc3kr@4ax.com...

On Sat, 6 Aug 2005 11:09:05 -0500, "DBLEXPOSURE"
celstuff@hotmail.com> wrote:



If a 30V source is used (My bench supply is 0-60Vdc regulated, BK 1623A)
and
1K for the known resistance a difference in 20 Ohms of the resistor under
test will result in a rise of 100mV across the known resistance

snip

12.7-12.6 = 100mV


If this whole process takes 19 seconds, triggering shouldn't be an issue.
Remember, he was using a stopwatch to measure the time, so, just use a
quick
time base and watch the green line rise and fall.

---
But, remember, he said he didn't want to do it that way any more.
---

10mV/cm gives me full deflection and I have plenty of range on the Vert.
position to bring the trace to the bottom of the screen. My scope is now
measuring 2 ohms/cm. I don't know why this would not be linear. Scope is
calibrated annually.

---
Did you actually try it?

I've got a Tektronix 2215, and with a 15V input, the best I can do
is 1V/box and still keep the trace on the scope with the position
control.

I didn't say the scope's response wasn't linear, I was talking about
the change in the output voltage from the divider as a function of
the change in resistance of the DUT.
---


I do like your bridge idea...


---
Thanks, but the credit should go to Samuel Hunter Christie and Sir
Charles Wheatstone.


--
John Fields
Professional Circuit Designer

Did you actually try it?

Yes, just now, and your correct, (as usual), I had to add another divider to
give me 12V to run into the B chnl. Then display (A-B).

I didn't say the scope's response wasn't linear, I was talking about
the change in the output voltage from the divider as a function of
the change in resistance of the DUT.

Wouldn't you consider an output change of 5mV for every 1 ohm change in the
DUT to be linear? To me, a nonlinear response would be something like, 5mV
for the first ohm of change, 7mV for the next ohm, 10 for the third, etc,
etc...

 

[JeffM]

Three of my 60-80 Gig Maxtor hard drives failed.
[their] covers were opened for a second in a...dust free room
Sam Nickaby

Does anyone have that diagram
DEC used to include with disk drive user/service info?
You know the one -
it shows the heads flying over the surface of the disk
next to a particle of dust and cigarette smoke,
which looked like boulders in comparison.
Sam Goldwasser

Yeah. That's the 1st thing that came to my mind.
I tried to Google it, but I'm not finding anything.
It looked like this:
: View in monospaced font (Courier).
: ________
: /
: /
: /
: /
: /
: /
: /
: __________ |
: / \ | human
: / \ | hair
: / \ |
: / \ |
:============== | dust | \
: |_____| ___ | particle | \
: head | | | \
: | \ / \
: air gap \ / \
: | \ / \
: platter | \ / \
:=================================================================
:=================================================================

 

[John Fields]

On Sat, 6 Aug 2005 13:00:57 -0500, "DBLEXPOSURE"
<celstuff@hotmail.com> wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:hmq9f1p9auflc08j32183k8vmpgdnoc3kr@4ax.com...

Did you actually try it?



Yes, just now, and your correct, (as usual), I had to add another divider to
give me 12V to run into the B chnl. Then display (A-B).


I didn't say the scope's response wasn't linear, I was talking about
the change in the output voltage from the divider as a function of
the change in resistance of the DUT.

Wouldn't you consider an output change of 5mV for every 1 ohm change in the
DUT to be linear?

---
Yes, but that's not what's happening.


For your 30 volt circuit:


30V
|
[1000R]
|
+----->E
|
[R]
|
0V


This is what is:


R E DELTA E
------+----------+--------
1355 17.26115 -------
1356 17.26655 0.00540
1357 17.27196 0.00541
1358 17.27735 0.00539
1359 17.28275 0.00540
1360 17.28814 0.00539
1361 17.29352 0.00538
1362 17.29890 0.00538
1363 17.30427 0.00537
1364 17.30964 0.00537
1365 17.31501 0.00537
1366 17.32037 0.00536
1367 17.32573 0.00536
1368 17.33108 0.00535
1369 17.33643 0.00535
1370 17.34177 0.00534
1371 17.34711 0.00534
1372 17.35245 0.00534
1373 17.35777 0.00532
1374 17.36310 0.00533
1375 17.36842 0.00532


So, we see that for equal increments of resistance we _don't_ get
equal increments of voltage, therefore the relationship between
resistance and voltage isn't linear

To me, a nonlinear response would be something like, 5mV
for the first ohm of change, 7mV for the next ohm, 10 for the third, etc,
etc...

Yup!

--
John Fields
Professional Circuit Designer

 

[John Doe]

et472@FreeNet.Carleton.CA (Michael Black) wrote:

John Doe (jdoe@usenet.love.invalid) writes:

....

Are you suggesting that small, old drives have more platters
than large, old drives?
Are you suggesting that the typical number of platters is
decreasing? I cannot imagine why.

Because designers get better at higher density.

So if everyone wanted the same capacity as before, the number of
platters would decrease. But users want higher capacity.

The reason for multiple platters is greater capacity and also
because the electronics is so much faster. With multiple platters,
the electronics do not have to wait as long for a single
platter/heads to retrieve data.

Look at floppy disks.

Haven't used them for years and I just bought a 256 MB USB flash
drive. The floppy disk drive is dead, at least here.

Current CDs/DVDs are the same size as a 5 1/4" inch floppy disk.
Yes, of course the capacity is greater.

Consumers pretty much dictate how large a device can be, according
to its usefulness/function. Consumers dictate how much space the
computer can occupy. When the electronics gets smaller, if the
consumer has the same amount of space, the consumer buys the same
size, more powerful computer.


Have fun.

 

[James Sweet]

"John Doe" <jdoe@usenet.love.invalid> wrote in message
news:Xns96ABD3AB231wisdomfolly@207.115.63.158...

I've opened a lot of dead drives to harvest the magnets and
it's true, older drives which are often smaller as well
typically have more platters than newer drives. I took apart
a 4 gig SCSI drive once that had 7 platters in it and it
seems like I once opened a 1 gig 5.25" SCSI drive that had
10-13 platters.

Are you suggesting that small, old drives have more platters
than large, old drives?

Are you suggesting that the typical number of platters is
decreasing? I cannot imagine why.

You're reading way too much into what I'm saying.

Nope.

All I'm suggesting is that
older drives typically have more platters than newer drives, and
it's because of much lower data density on the platters.

Apparently, consumer demand for greater capacity does not figure
into your equation. In fact, demand for greater capacity
increases.

I didn't see single
platter drives until just a few years ago,

The only one I have taken apart, over five years ago, had a single
platter.

Why do you have to see? All you have to do is look at the maker's
web site.

I don't have to see, I'm not the one who took apart working drives, I was
only chiming in that I've opened dozens of dead drives over the years to
harvest them for parts, and statistically the newer ones have fewer platters
than the older ones, I don't know why this is such a controversial statement
as it's simply what I've seen in a random sampling of drives that have
failed over the last 15 years or so.

 

[Michael A. Terrell]

John Doe wrote:

Apparently, consumer demand for greater capacity does not figure
into your equation. In fact, demand for greater capacity
increases.

Consumer demand has nothing to do with it. You can demand anything
you want but until its feasible to build at a reasonable price, it won't
exist. All any consumer can do is chose the an item somewhere in the
range of what is offered. I'd love to have a 100 million terabyte data
storage system with a 1024 bit wide buss and under a pico second access
time, but it isn't going to happen in my lifetime. I'll be lucky if I
live long enough to see really good OCR software.


<Snip a lot of nonsense.>

... the electronics is always much faster than the hardware, so
multiple platters means faster data transfer.

Really? It takes time to switch between heads, verify the head
location, synchronize the servo and collect the data. It would be like
reading more books at the same time, rather than finishing one before
starting another.

More heads slows it down unless you are reading each entire platter
at once, one side at a time. I have never seen this, even in the old 5
MB hard drives that I started with. If you want more speed you use RAID
arrays where you have multiple drives synched, and switch data access
between them.

--
Link to my "Computers for disabled Veterans" project website deleted
after threats were telephoned to my church.

Michael A. Terrell
Central Florida

 

[John Doe]

"Michael A. Terrell" <mike.terrell earthlink.net> wrote:

John Doe wrote:

Apparently, consumer demand for greater capacity does not
figure into your equation. In fact, demand for greater capacity
increases.

<snipped babbling about terabyte hard disk drives and OCR software>

... the electronics is always much faster than the hardware, so
multiple platters means faster data transfer.

Really? ...
More heads slows it down ...

http://www.wdc.com/en/products/productkit.asp?DriveID=40
http://www.wdc.com/en/products/productkit.asp?DriveID=65

Identical drives except the larger drive with two platters has
faster Read Seek Times and faster Track-To-Track Seek Times.

--
Link to my "Computers for disabled Veterans" project website
deleted after threats were telephoned to my church.

Michael A. Terrell
Central Florida

Path: newssvr33.news.prodigy.com!newsdbm06.news.prodigy.com!newsdst02.news.prodigy.com!newsmst01b.news.prodigy.com!prodigy.com!newscon06.news.prodigy.com!prodigy.net!border1.nntp.dca.giganews.com!border2.nntp.dca.giganews.com!nntp.giganews.com!elnk-atl-nf1!newsfeed.earthlink.net!stamper.news.atl.earthlink.net!newsread2.news.atl.earthlink.net.POSTED!b1a104da!not-for-mail
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[Michael A. Terrell]

John Doe wrote:

"Michael A. Terrell" <mike.terrell earthlink.net> wrote:

John Doe wrote:

Apparently, consumer demand for greater capacity does not figure
into your equation. In fact, demand for greater capacity
increases.


Consumer demand has nothing to do with it. You can demand
anything you want but until its feasible to build at a
reasonable price, it won't exist. All any consumer can do is
chose the an item somewhere in the range of what is offered.
I'd love to have a 100 million terabyte data storage system with
a 1024 bit wide buss and under a pico second access time, but it
isn't going to happen in my lifetime. I'll be lucky if I live
long enough to see really good OCR software.


Snip a lot of nonsense.


... the electronics is always much faster than the hardware, so
multiple platters means faster data transfer.


Really? It takes time to switch between heads, verify the head
location, synchronize the servo and collect the data. It would be like
reading more books at the same time, rather than finishing one before
starting another.

More heads slows it down unless you are reading each entire platter
at once, one side at a time. I have never seen this, even in the old 5
MB hard drives that I started with. If you want more speed you use RAID
arrays where you have multiple drives synched, and switch data access
between them.

--
Link to my "Computers for disabled Veterans" project website deleted
after threats were telephoned to my church.

Michael A. Terrell
Central Florida

Path: newssvr33.news.prodigy.com!newsdbm06.news.prodigy.com!newsdst02.news.prodigy.com!newsmst01b.news.prodigy.com!prodigy.com!newscon06.news.prodigy.com!prodigy.net!border1.nntp.dca.giganews.com!border2.nntp.dca.giganews.com!nntp.giganews.com!elnk-atl-nf1!newsfeed.earthlink.net!stamper.news.atl.earthlink.net!newsread2.news.atl.earthlink.net.POSTED!b1a104da!not-for-mail
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From: "Michael A. Terrell" <mike.terrell earthlink.net
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Nothing at all.

Plonk!
--
Link to my "Computers for disabled Veterans" project website deleted
after threats were telephoned to my church.

Michael A. Terrell
Central Florida