You're simply blessed with indifferent ears and enough sense not to buy overpriced 'audiophile' stuff. (You know, even digital data must move over oxygen-free copper lines or be distorted.) -Wolfgang
You misread Martin on purpose, which tells us a lot about you. *sigh* The equivalent number of bits are the bits that encode the luminance for each pixel PLUS (at most) "an extra 2 bits of information beyond the luminance measurement depth at the pixel." With 11 bits of luminance any randomly picked bayer pixel contains at most 13 bits. A large group of bayer pixels (with 11 bits of luminance) actually contain well less than 13 bits (but more than 11 bits), since their colours follow a regular pattern which is very easy to compress to a description. Martin isn't always gentle to people who wrongly believe they are right and who don't grasp elementary details about the things they proclaim as truth. But then he isn't paid to be that way. They don't care to explain to the n'th person that a perpetuum mobile is impossible. Would you have some proof for your claims[1], or are you just proclaiming them to be true, speaking ex cathedra by the power invested in you by God? -Wolfgang [1] e.g. someone qualified saying the same or a strict mathematical proof
No, liar, you're too stupid to grasp the fact that lenses admit only a small cone of light and block the vast majority of the light that comes in at incident angles. But that's just another example of how you really understand nothing of the real world.
If I understand you correctly, you are postulating a microscopic capacitor device which when it is charged with one electron can be manipulated to have a potential difference of five volts. It might go from 0V to 2.7V or whatever. But then, presumably, you adjust the capacitor to get a final voltage of 5V. If the circuit has less than near-infinite impedance you will immediately discharge the capacitor. (What voltage do you need to make the electron jump on down the circuit? It depends on the materials.) If the circuit has more than this critical impedance there will be no current flow and whatever voltage there is will be unmeasurable. Any voltmeter will look like a short circuit and immediately discharge the capacitor (of its one electron). This is where Heisenberg comes in: you can either measure its voltage or know the location of the electron (i.e. in the capacitor) but you can't do both. We are now down at the quantum level and I think your question has become meaningless. Is voltage quantised? I suspect it may be. Regards, Eric Stevens
I seem to have not received this message to which you are replying. http://georgegraham.com/compress.html explains many of the problems with current digital recording. Well worth a read. Regards, Eric Stevens
So what does downtime consist off, what causes downtime. RAID is suitable for backups it might be less sutable for archives.
Oh OK then. But whisky dave is just a nickname I was given in a night club by a gril that needed to be able to tell one dave from another, she was already friends with a smelly dave and a violent dave, and as I was drinking whisky at the time.. and my entails are DW so WD didn;t see so bad. Also if someone recognises me in a drinking establishment and calls Hi Whisky Dave, I just say Yes thanks, but no ice and make it a single malt please ;-) And whisky is digital it's measrued in drams ;-) But I drink it in analogue measures....
Hardware failures, viruses, et cetera. No; besides downtime reduction, RAID's prime advantage can be speed (depending upon the level involved). Its main pitfall, regarding backups, is that the user may restore a system, without knowing it was already corrupted by malware and/or other culprits.
[/QUOTE] No need for it to be microscopic. Why should it be? I move one electron from one plate to the other, then adjust the distance. Yep. If a circuit has "near-infinite impedance" then it's not exactly a circuit. There are volt meters that do not need current flow, and in fact do not allow current to flow. Why should it look like a short circuit then? You do understand that Heisenberg isn't an either-or proposition? It's a statement of accuracy attainable. So of course I can get a voltage-to-some-degree-of-precision and a position-to-some-degree-of-precision. But I am not after the position (and "somewhere in the capacitor --- which can be easily as large as I want to make it" isn't exactly ... precise. Nor is 'somewhere along the conductors'.) To the contrary ... .... if it is quantised, then --- obviously --- not every voltage between 0V and 5V is reached even in normal conditions. And since charge is quantised and space is quantized and voltage depends on charge, area and distance in a plate capacitor ... -Wolfgang
No, *you* cannot. Voltage can. Of course, *you* cannot prove it cannot. *You* can not even explain how the voltage is supposed to rise gradually in the voltmeter. Go on stamping you foot. Dance, Floyd, dance. It's not my problem that you are unable or unwilling to understand. -Wolfgang
It's still not a viable >long term< backup solution, however. Too many things can go wrong, with such an on-the-fly approach.
Shifting goalposts? Use a disk for every backup, in whatever backup scheme you prefer. It's not THAT hard ... -Wolfgang
