Should I have underclocked.

[kony]

Yes I would agree with that, posibbly or even better, however is not
power current squared X resistance in which case it wuld be
3/4 X 3/4 = 9/16 = about half the power?
If true thats quite a saving on heat and you might get away
without a heatsink even better to drop from 600 to 300
which would be only 25% of normal power.

No, that's incorrect. I've posted equations previously
about CPU power, Google should find them.
http://groups-beta.google.com/group/alt.comp.hardware/msg/4312ff0f9b3c4203?hl=en

 

[kony]

I would be better going to PC world and buying that lovely
Athlon64 methinks.

Yes, do treat yourself to a new system.

 

[Donald McTrevor]

No, that's incorrect.

No it isn't its correct.

because W=I x I x R

So...

X =I x I x R/ [(Freq)*(V)²]

I've posted equations previously

 

[Pen]

Power is E*I. If voltage is held constant then
power is not a function of current squared. The
impedance of the CPU is not a constant with
frequency making your theory invalid.

No, that's incorrect.

No it isn't its correct.

because W=I x I x R

So...

X =I x I x R/ [(Freq)*(V)2]

I've posted equations previously

about CPU power, Google should find them.
http://groups-beta.google.com/group/alt.comp.hardware/msg/4312ff0f9b3c4203?hl=en

 

[Donald McTrevor]

Power is E*I. If voltage is held constant then
power is not a function of current squared. The
impedance of the CPU is not a constant with
frequency making your theory invalid.

Yes I was not thinking of a constant voltage, however
maybe you can run the chip at a lower voltage at a lower
frequency.

Also with less current the chip will be cooler and hence
have a lower resistance? (not sure whether it would be higher
though!!!).

On Tue, 09 Aug 2005 23:03:38 GMT, "Donald McTrevor"


Yes I would agree with that, posibbly or even better,
however is not
power current squared X resistance in which case it wuld
be
3/4 X 3/4 = 9/16 = about half the power?
If true thats quite a saving on heat and you might get
away
without a heatsink even better to drop from 600 to 300
which would be only 25% of normal power.

No, that's incorrect.

No it isn't its correct.

because W=I x I x R

So...

X =I x I x R/ [(Freq)*(V)2]

I've posted equations previously

about CPU power, Google should find them.

http://groups-beta.google.com/group/alt.comp.hardware/msg/4312ff0f9b3c4203?hl=en

 

[Pen]

The chip doesn't have a resistance, but rather an Impedance
which will vary with frequency. Ohms law only really applies
to static conditions. You're leaping to conclusions to
support
what you think should be happening.

Power is E*I. If voltage is held constant then
power is not a function of current squared. The
impedance of the CPU is not a constant with
frequency making your theory invalid.

Yes I was not thinking of a constant voltage, however
maybe you can run the chip at a lower voltage at a lower
frequency.

Also with less current the chip will be cooler and hence
have a lower resistance? (not sure whether it would be
higher
though!!!).

On Tue, 09 Aug 2005 23:03:38 GMT, "Donald McTrevor"


Yes I would agree with that, posibbly or even better,
however is not
power current squared X resistance in which case it
wuld
be
3/4 X 3/4 = 9/16 = about half the power?
If true thats quite a saving on heat and you might
get
away
without a heatsink even better to drop from 600 to
300
which would be only 25% of normal power.

No, that's incorrect.

No it isn't its correct.

because W=I x I x R

So...

X =I x I x R/ [(Freq)*(V)2]

I've posted equations previously
about CPU power, Google should find them.

http://groups-beta.google.com/group/alt.comp.hardware/msg/4312ff0f9b3c4203?hl=en

 

[Donald McTrevor]

The chip doesn't have a resistance, but rather an Impedance
which will vary with frequency. Ohms law only really applies
to static conditions. You're leaping to conclusions to
support
what you think should be happening.

The chip does have a resistance, obviously.

Power is E*I. If voltage is held constant then
power is not a function of current squared. The
impedance of the CPU is not a constant with
frequency making your theory invalid.

Yes I was not thinking of a constant voltage, however
maybe you can run the chip at a lower voltage at a lower
frequency.

Also with less current the chip will be cooler and hence
have a lower resistance? (not sure whether it would be
higher
though!!!).

On Tue, 09 Aug 2005 23:03:38 GMT, "Donald McTrevor"


Yes I would agree with that, posibbly or even better,
however is not
power current squared X resistance in which case it
wuld
be
3/4 X 3/4 = 9/16 = about half the power?
If true thats quite a saving on heat and you might
get
away
without a heatsink even better to drop from 600 to
300
which would be only 25% of normal power.

No, that's incorrect.

No it isn't its correct.

because W=I x I x R

So...

X =I x I x R/ [(Freq)*(V)2]

I've posted equations previously
about CPU power, Google should find them.

http://groups-beta.google.com/group/alt.comp.hardware/msg/4312ff0f9b3c4203?hl=en

 

[CBFalconer]

Yes I would agree with that, posibbly or even better, however is not
power current squared X resistance in which case it wuld be
3/4 X 3/4 = 9/16 = about half the power?
If true thats quite a saving on heat and you might get away
without a heatsink even better to drop from 600 to 300
which would be only 25% of normal power.

No, power is current * voltage, and voltage is constant here. The
device is not ohmic, so it doesn't have a 'resistance'.

 

[Pen]

No I suggest you go back an learn some AC circuit theory.
There
may be a resistive component to the impedance, but the
impedance is
mainly reactive. If it were resistive then current drain
would remain
constant with frequency, which it clearly does not.

The chip doesn't have a resistance, but rather an
Impedance
which will vary with frequency. Ohms law only really
applies
to static conditions. You're leaping to conclusions to
support
what you think should be happening.

The chip does have a resistance, obviously.

Power is E*I. If voltage is held constant then
power is not a function of current squared. The
impedance of the CPU is not a constant with
frequency making your theory invalid.

Yes I was not thinking of a constant voltage, however
maybe you can run the chip at a lower voltage at a
lower
frequency.

Also with less current the chip will be cooler and
hence
have a lower resistance? (not sure whether it would be
higher
though!!!).





On Tue, 09 Aug 2005 23:03:38 GMT, "Donald McTrevor"


Yes I would agree with that, posibbly or even
better,
however is not
power current squared X resistance in which case
it
wuld
be
3/4 X 3/4 = 9/16 = about half the power?
If true thats quite a saving on heat and you might
get
away
without a heatsink even better to drop from 600 to
300
which would be only 25% of normal power.

No, that's incorrect.

No it isn't its correct.

because W=I x I x R

So...

X =I x I x R/ [(Freq)*(V)2]

I've posted equations previously
about CPU power, Google should find them.


http://groups-beta.google.com/group/alt.comp.hardware/msg/4312ff0f9b3c4203?hl=en

 

[Donald McTrevor]

No I suggest you go back an learn some AC circuit theory.
There
may be a resistive component to the impedance, but the
impedance is
mainly reactive. If it were resistive then current drain
would remain
constant with frequency, which it clearly does not.

Thats becayse V and I are out of phase, obviously silicon
is a material which has resistance. Do you deny that? (yes or no).

The chip doesn't have a resistance, but rather an
Impedance
which will vary with frequency. Ohms law only really
applies
to static conditions. You're leaping to conclusions to
support
what you think should be happening.

The chip does have a resistance, obviously.

Power is E*I. If voltage is held constant then
power is not a function of current squared. The
impedance of the CPU is not a constant with
frequency making your theory invalid.

Yes I was not thinking of a constant voltage, however
maybe you can run the chip at a lower voltage at a
lower
frequency.

Also with less current the chip will be cooler and
hence
have a lower resistance? (not sure whether it would be
higher
though!!!).





On Tue, 09 Aug 2005 23:03:38 GMT, "Donald McTrevor"


Yes I would agree with that, posibbly or even
better,
however is not
power current squared X resistance in which case
it
wuld
be
3/4 X 3/4 = 9/16 = about half the power?
If true thats quite a saving on heat and you might
get
away
without a heatsink even better to drop from 600 to
300
which would be only 25% of normal power.

No, that's incorrect.

No it isn't its correct.

because W=I x I x R

So...

X =I x I x R/ [(Freq)*(V)2]

I've posted equations previously
about CPU power, Google should find them.

http://groups-beta.google.com/group/alt.comp.hardware/msg/4312ff0f9b3c4203?hl=en

 

[Pen]

Get real. I've told you 3 times it's complex with a
resistive component.
It is NOT purely resistive, which is what made all your
conclusions erroneous. You are clearly in over your head
with this entire project.

No I suggest you go back an learn some AC circuit theory.
There
may be a resistive component to the impedance, but the
impedance is
mainly reactive. If it were resistive then current drain
would remain
constant with frequency, which it clearly does not.

Thats becayse V and I are out of phase, obviously silicon
is a material which has resistance. Do you deny that? (yes
or no).

The chip doesn't have a resistance, but rather an
Impedance
which will vary with frequency. Ohms law only really
applies
to static conditions. You're leaping to conclusions to
support
what you think should be happening.

The chip does have a resistance, obviously.



Power is E*I. If voltage is held constant then
power is not a function of current squared. The
impedance of the CPU is not a constant with
frequency making your theory invalid.

Yes I was not thinking of a constant voltage,
however
maybe you can run the chip at a lower voltage at a
lower
frequency.

Also with less current the chip will be cooler and
hence
have a lower resistance? (not sure whether it would
be
higher
though!!!).





On Tue, 09 Aug 2005 23:03:38 GMT, "Donald
McTrevor"


Yes I would agree with that, posibbly or even
better,
however is not
power current squared X resistance in which
case
it
wuld
be
3/4 X 3/4 = 9/16 = about half the power?
If true thats quite a saving on heat and you
might
get
away
without a heatsink even better to drop from 600
to
300
which would be only 25% of normal power.

No, that's incorrect.

No it isn't its correct.

because W=I x I x R

So...

X =I x I x R/ [(Freq)*(V)2]

I've posted equations previously
about CPU power, Google should find them.



http://groups-beta.google.com/group/alt.comp.hardware/msg/4312ff0f9b3c4203?hl=en

 

[Donald McTrevor]

Get real. I've told you 3 times it's complex with a
resistive component.
It is NOT purely resistive, which is what made all your
conclusions erroneous. You are clearly in over your head
with this entire project.

I barely have my toes, wet all materials have a resistance
a capacitance and an inductance. Or are you denying this too???

No I suggest you go back an learn some AC circuit theory.
There
may be a resistive component to the impedance, but the
impedance is
mainly reactive. If it were resistive then current drain
would remain
constant with frequency, which it clearly does not.

Thats becayse V and I are out of phase, obviously silicon
is a material which has resistance. Do you deny that? (yes
or no).

The chip doesn't have a resistance, but rather an
Impedance
which will vary with frequency. Ohms law only really
applies
to static conditions. You're leaping to conclusions to
support
what you think should be happening.

The chip does have a resistance, obviously.



Power is E*I. If voltage is held constant then
power is not a function of current squared. The
impedance of the CPU is not a constant with
frequency making your theory invalid.

Yes I was not thinking of a constant voltage,
however
maybe you can run the chip at a lower voltage at a
lower
frequency.

Also with less current the chip will be cooler and
hence
have a lower resistance? (not sure whether it would
be
higher
though!!!).





On Tue, 09 Aug 2005 23:03:38 GMT, "Donald
McTrevor"


Yes I would agree with that, posibbly or even
better,
however is not
power current squared X resistance in which
case
it
wuld
be
3/4 X 3/4 = 9/16 = about half the power?
If true thats quite a saving on heat and you
might
get
away
without a heatsink even better to drop from 600
to
300
which would be only 25% of normal power.

No, that's incorrect.

No it isn't its correct.

because W=I x I x R

So...

X =I x I x R/ [(Freq)*(V)2]

I've posted equations previously
about CPU power, Google should find them.

http://groups-beta.google.com/group/alt.comp.hardware/msg/4312ff0f9b3c4203?hl=en

 

[Donald McTrevor]

No, power is current * voltage, and voltage is constant here. The
device is not ohmic, so it doesn't have a 'resistance'.

Well I think it does, it will have a resistance a capacitance and an
inductance.

 

[kony]

Yes I was not thinking of a constant voltage, however
maybe you can run the chip at a lower voltage at a lower
frequency.

yes you often can reduce voltage at lower speed. There is
some variability in how much though.

Also with less current the chip will be cooler and hence
have a lower resistance? (not sure whether it would be higher
though!!!).

It's not very significant, the relative temp only changes
slightly. As with your K6-2 exploration and ideas about
performance, you again try to reinvent the wheel so to
speak. These are known facts and it would be better to
simply accept them until you understand what and why.

 

[kony]

The chip does have a resistance, obviously.

Regardless of what you do or don't want to accept, simply
plugging in the numbers and taking measurements, it's
obvious your calculations are incorrect.

 

[CBFalconer]

.... snip ...

Well I think it does, it will have a resistance a capacitance
and an inductance.

It has an impedance, which is frequency sensitive. Square waves
contain many harmonics. The phase relationship between current and
voltage marks things as capacitive, inductive, or resistive. After
you study some functions of complex variables and Laplace and
Fourier transforms you may be able to begin to appreciate the
complexity of so characterizing the device. I already told you the
primary current draining mechanism.

Since you obviously already know everything I suppose there is no
point in informing you.

 

[Donald McTrevor]

It has an impedance, which is frequency sensitive. Square waves
contain many harmonics. The phase relationship between current and
voltage marks things as capacitive, inductive, or resistive. After
you study some functions of complex variables and Laplace and
Fourier transforms you may be able to begin to appreciate the
complexity of so characterizing the device. I already told you the
primary current draining mechanism.

Since you obviously already know everything I suppose there is no
point in informing you.

Obviously I don't know everything I ws just stateing some basic
laws of physics. If in a physics class I said power was directly
proportional to currrent in an electrical circuit I would get my
knuckles rapped and deservidly so!!!
Probably be told to write out "Power equals I squared R" 1000 times too!!

 

[kony]

Obviously I don't know everything I ws just stateing some basic
laws of physics. If in a physics class I said power was directly
proportional to currrent in an electrical circuit I would get my
knuckles rapped and deservidly so!!!
Probably be told to write out "Power equals I squared R" 1000 times too!!

Rather than trying to comprehensively understand every tiny
mechanism involved in current, heat generation, it is easier
to simply take the opposite approach, to know that others
have measured current consumption to be very close (close
enough) to linear with frequency changes. If it did happen
to deviate a percent or two from that, what useful purpose
is there to implement this information into your K6-2
experiment?

 

[Donald McTrevor]

Rather than trying to comprehensively understand every tiny
mechanism involved in current, heat generation, it is easier
to simply take the opposite approach, to know that others
have measured current consumption to be very close (close
enough) to linear with frequency changes.

Well if you can run at a lower voltage at a lower frequency they you may
find is not so linear and more proportional to the square root of the
frequency, possibly.

If it did happen
to deviate a percent or two from that, what useful purpose
is there to implement this information into your K6-2
experiment?

Not much really since I have as good as given up on it.
It seemed to be running OK at 4.5 X 66 =300 but it
failed to boot second time so I am back on the trusty
cyrix.
I think that without an L2 cache the K6 is handicapped.
The AIDA32 prog has a benchmark thing in it.

Memory reads:-


Duron-850 Gigabyte GA-7IXE4 AMD750
PC100 SDRAM 460 MB/s
Celeron-466A IBM 628848U i810 Int.
PC66 SDRAM 390 MB/s
PII-266 Intel DK440LX i440LX
PC66 SDRAM 350 MB/s
K6-III-450 Asus P5A ALADDiN5
PC100 SDRAM 290 MB/s
K6-2-450 Gigabyte GA-5AX ALADDiN5
PC100 SDRAM 260 MB/s
C3-800 VIA EPIA PLE133
PC133 SDRAM 210 MB/s
PentiumMMX-166 Asus TX97-E i430TX
PC66 SDRAM 190 MB/s
K6-266 Asus SP97-V SiS5598 Int.
66 MHz EDO RAM 60 MB/s
This Computer K6-2 4.5X66=300
52 MB/s
This Computer Cyrix MII 300
49 MB/s

So not much in it and the Cyrix had ZoneALarm running which may have slowed
it a tad.
But it shows how slow my computer is compared to others.
Basically it needs a new motherboard, and I need to get that Athlon64.

 

[kony]

Well if you can run at a lower voltage at a lower frequency they you may
find is not so linear and more proportional to the square root of the
frequency, possibly.

No, it is not proportional to the square root.
You have zero evidence to suggest it is and this is getting
ridiculous. You are essentially trying to reinvent facts
already known. Your argument is not valid and it is a waste
of time to argue something already proven false.

Not much really since I have as good as given up on it.
It seemed to be running OK at 4.5 X 66 =300 but it
failed to boot second time so I am back on the trusty
cyrix.
I think that without an L2 cache the K6 is handicapped.
The AIDA32 prog has a benchmark thing in it.




Memory reads:-

That's not a cache benchmark.
I'd already suggested a benchmark for the cache but you are
intent on wasting time instead.

It could easily be that your motherboard or power supply are
simply aged to the point where they aren't stable with this
higher CPU load on them- something else I had also mentioned
briefly about the system age being a factor in upgrade
viability.

However, we can only assume what you have or haven't done to
set it up, you might recheck the jumpers to confirm they
were correct.

So not much in it and the Cyrix had ZoneALarm running which may have slowed
it a tad.
But it shows how slow my computer is compared to others.
Basically it needs a new motherboard, and I need to get that Athlon64.

It needs a 100% overhaul, not just a motherboard and CPU.
You might keep the floppy drive though, as one that old may
be out of alignment and any floppies made with it may not be
as readable with a new floppy drive.