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RAM running 2:3 want 1:1, help please! - Page 6

post #51 of 61
Quote:
Originally Posted by Nasgul
I only going to say it one last time, not to make anyone upset but I think it's not almost about benchmarks but real situations where the difference is seen.

http://www.overclock.net/842435-post16.html

http://www.overclock.net/836472-post16.html

http://www.overclock.net/836474-post17.html

http://www.overclock.net/841924-post26.html

I'll keep using the 3:5 ratio and 1:2, but overall I like the 1:2.

If a 1:1 I wanted, I should've gone for DDR2 400.

and QFT

We all have to agree to disagree.
LOL! All of those are benchmarks, not
Quote:
real situations
!!
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post #52 of 61
It's understandable. I remember how many people could not get past the numbers game in applied Physics. Many barely passed or outright failed the classes as they were not able to see past quantification and since Physics is a science of measurement there was clearly an inability to get past the boundaries with logic.

However, I will await and we will see if what I am saying is correct or not. I have time on my side as the proof will be revealed with higher output of processors.

See? If the 1066fsb shows less of a differential than the 800MHz fsb and the 1333 even less, then what I think makes more sense. If it is the opposite, then what I think makes less sense.

The truth will out as technology increases and the FSB reaches the external dram bandwidths on a 1:1 basis. If there is a disproportionate increase then compared to now then what I am stating is correct but if there is not a disproportionate (Percentage) then what I am stating is incorrect.

The reason why there is no hard and fast answer right now is because we only really have one side of the equation. Soon we will have the other side

R
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post #53 of 61
Quote:
Originally Posted by Ropey
It's understandable. I remember how many people could not get past the numbers game in applied Physics. Many barely passed or outright failed the classes as they were not able to see past quantification and since Physics is a science of measurement there was clearly an inability to get past the boundaries with logic.

However, I will await and we will see if what I am saying is correct or not. I have time on my side as the proof will be revealed with higher output of processors.

See? If the 1066fsb shows less of a differential than the 800MHz fsb and the 1333 even less, then what I think makes more sense. If it is the opposite, then what I think makes less sense.

The truth will out as technology increases and the FSB reaches the external dram bandwidths on a 1:1 basis. If there is a disproportionate increase then compared to now then what I am stating is correct but if there is not a disproportionate (Percentage) then what I am stating is incorrect.

The reason why there is no hard and fast answer right now is because we only really have one side of the equation. Soon we will have the other side

R
The reason that it is 64.99999km/h is because you have to add it relativistically.
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post #54 of 61
Quote:
Originally Posted by pauldovi
The reason that it is 64.99999km/h is because you have to add it relativistically.
Only in the Physics of Quanta would you be correct Paul. I am speaking linearly in the bounds of two dimensions.

R
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post #55 of 61
Quote:
Originally Posted by Ropey
Only in the Physics of Quanta would you be correct Paul. I am speaking linearly in the bounds of two dimensions.

R
Einstein's equations hold true in a dimensions as well as all inertial frames of reference. We use classical equations just because they are easier, and accurate enough at low velocities.
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post #56 of 61
Quote:
Originally Posted by pauldovi
Einstein's equations hold true in a dimensions as well as all inertial frames of reference. We use classical equations just because they are easier, and accurate enough at low velocities.
This is what I am saying. You are stating a transactional equation that does not really fit our frame of reference. There is no use to state Quantum transactions in a two dimensional reference.

Sort of like speaking English at a Spanish convention.

R
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post #57 of 61
im glad to see that were still right back where we started
so im going to try to settle this bluntly
ahem...

the benchmarks are higher, becuase they make use of the extra bandwidth for ECC, therfore giving a higherscore in whatever test

but under most circumstances, including gaming, the extra bandwidth is not utilised, and therefore unused,

THOU!! a few system tasks MIGHT use a little of the extra bandwidth, but it wont noticably impact preformnce


and......
im not even gonna get in on the quantum and equations argument
CYA LATER GUYS!!
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post #58 of 61
Quote:
Originally Posted by thehybridpyro
the benchmarks are higher, becuase they make use of the extra bandwidth for ECC, therfore giving a higherscore in whatever test

but under most circumstances, including gaming, the extra bandwidth is not utilised, and therefore unused,

THOU!! a few system tasks MIGHT use a little of the extra bandwidth, but it wont noticably impact preformnce
This is pretty much my view. There are those who say that they "feel" a difference but even with the benchmark numbers I find this hard to believe.

Take a 2 lb grapefruit in one hand and a 2.1 lb grapefruit in the other and say which is which. Yes, one can tell on a scale but by "feel". Hmmm . . .

R
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post #59 of 61
Quote:
Originally Posted by Ropey
This is what I am saying. You are stating a transactional equation that does not really fit our frame of reference. There is no use to state Quantum transactions in a two dimensional reference.

Sort of like speaking English at a Spanish convention.

R
I disagree. Einstein postulates sate that:

1. The laws of physics hold true in all interial frames of reference.

2. The speed of light in a vacuum is constant for all inertial frames of reference.

Now, the guy watching your person move on the train in is a different frame of reference, lets call it u. The train is also in a different frame of reference, v, and the person moving on the train is in u'. According to einsteins velocity vectors (it doesn't matter if they are in the same direction, velocity is still a vector), u = (v + u')/(1+(u'v)/c^2).

It doesn't matter wether we are talking about electrons or cars, Einstein's postulates hold true.
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post #60 of 61
that also puts it greatly
off topic: ropey, is your avatar what boots up with your windows?
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