I Miss my Viewfinder !

Yes what atre you talking about ?

 

How can AC act like a very small battery ?

Why are you saying inductance reactance and capacitance is mor
eimportant than resistance ?

Easy claim to make.
Of coursse if your job is testing old car batteries you might be
correct.

Wh,y what are you saying magnitude is ?
Show evidence of what you say.
Frequency is defined as 1/periodic time we used to call it cycles per
second
then it got renamed Hertz.

The design engineers in your head.

NO it does not.
You really are talking rubbish.

That's hardly suprises me, what job do you do or what sort of industry
do you work in.

A collegue of mine worked for a company they built switch mode PSUs

 

An "extra two bits of information beyond the luminance
measurement" is clearly not enough to provide full color
information at an adaquate level.

Didn't you notice that?

 

Classic!

 

I've also heard that such channels actual cpompress ads less than the
TV show,
this makes the Ads volume appear to be higher but I was told they
don't transmit ads
at a higher volume they just compress them less which makes them sound
louder.

 

I noticed that it seems wrong and ridiculous.

Bottom line: you get 11 bits of information from (each pixle in) the sensor
and a very small amount of extra information from knowing about where the
color filters are.

If you think that a Bayer sensor coughs up significantly more bits per pixel
than 11 + a small amount, I have a bridge I'd like to sell you.

 

AC is a current, and does not "act like a battery"; and
nobody has suggested that except you.

A a reactance, either a capacitor or an inductor, stores
energy. That is much the same as a battery.

Please read what you have quoted. Here is what I said:
"a resistor is no more, or less, important than reactance"

Nope. I have very little experience with batteries that small.

You are arguing that the dictionary is wrong. That's a bit silly.

Are you ever able to carry on an honest, logical,
non-nonsense, conversation?

It's pretty basic electric theory.

Wow. That is impressive. I'm glad you at least have
*some* association with electronic/electrical practice
and theory.

He swept floors?

 

You probably have quite a collection of bridges too.

 

Which light is that?

 

Now you're just playing stupid because you can't defend your idiocy.

 

Things don't work that way. Really not. Otherwise highly
compressed MP3s would be unbearable loud at the same volume
setting than low compressed MP3s.

(And JPEGs aren't brighter if higher compressed either.)

You can, however, raise the audio amplification, oversaturating the
audio channels, causing lots of clipping. That sounds bad[1], but
louder at the same volume knob setting. Pop groups have been doing
that a lot to CDs, and well ... it's a bad idea, but loudness beats
quality there. (compare an old CD versus a new CD for volume.)

-Wolfgang

[1] and tires the listener.

 

Things don't work that way. Really not. Otherwise highly

There may be a confusion here:

- compression as in bit-rate reduction where lower bit-rate MP3s or JPEGs
have a lower quality through carrying less information.

- compression as in reducing the difference between loud and soft sounds,
a reduction of the dynamic range. Makes some "music" more acceptable for
in-car lsitening.

Cheers,
David

 

Now you're resorting to a personal attack because you cannot answer the
question.

 

I believe it's quite possible in the lab. We can "create" and
transport single electrons.

Actually, no. While "charging up" the voltage will of course
depend on the distance of the 2 plates --- and hence not jump
from 0V to 5V.

What I am asking, is: Connect this plate capacitor to a circuit.
How does the voltage, measured on some place in the circuit behave?
Does it "gradually" rise? If so, how --- we have only one
electron. Does it suddenly jump? If so, then all the fuzz about
rise time and "every voltage between 0V and 5V" is rendered false.

-Wolfgang

 

[Floyd proves he's out of arguments. Again.]

Hey Floydy, how expensive is a couple of MBs storage space?
$100 or $10,000?

-Wolfgang

 

So you are talking about the message in which you wrote "You
won't understand the answer, which has already been explained to
you without effect", right? And that then is your explanation?
Really weak, really stupid.

Of course, I was right, you are too stupid to find a message ID.

I call you a liar: it was not explained, contrary to what you said.
A liar because you claim things that aren't true and which you
cannot support with facts like a message ID.

So you agree that maybe electrons transport the potential
difference. Now the question (which you cannot answer and thus
will call irrelevant or nonsense): When there is only one electron
put into the conductor, how is the potential difference raised
through all values between 0V and the voltage the electron was
sent with?

As I predicted it, you're too stupid to find a message ID and are
wildly flailing. You're good at copy-and-pasting half-understood
snippets from what you claim are industry standards. But your
half-knowledge trips you up every time.

Next up: let's hear how you produce out of a total of 1200 bits
(10x10 pixel, bayer pattern, 10 bits of data each, let's allow 2
bits for color coding each) of information 9000 bits (sum up a 3x3
for each pixel) or even 12500 bits (sum up a 5x5 for each pixel)
of information, without inventing anything, and without adding
redundant information.

-Wolfgang

 

That's Floyd L. Davidson, admitting defeat.
I just used his own argumentation ... and he has no answer to it.

-Wolfgang

 

Nope.
It's the luminance data of the integral over the wavelength,
multiplied with the filter's pass factor for each given wavelength.

Otherwise you'd never get colour out of it.

You have a camera that produces 16 bit TIFFs, but not the more
valuable (non-demosaiced yet) and much much smaller RAW, which can
(in almost all cameras) be loslessly compressed --- and contains
*more* information than the TIFF?

$50 for a 32GB card
http://preview.tinyurl.com/6cbnjsu

$145 for a 64 GB card
http://preview.tinyurl.com/6jvfbhu

A 21 MPix camera, like the 5D Mark II uses 25 MB (JPEG and RAW
normal) and ca. whopping 40MB for JPEG + RAW at high ISO settings
(> ISO 3200).

So you'll get 800-1300 shots on a $50 card --- that's very
expensive. Costs almost as much as about 10 high end DSLRs.
And like film, the memory is not reusable.

in DX. A whopping 5 MPix. That's just 55 MPix/s.

The 5D Mark II does just 3.9 fps ... @ 21 MPix. That's 81.9
MPix/s.
The 1D Mark IV: 16.1 MPix @ 10 fps: 161 MPix/s, almost three
times more than the 3D.

Now, what's impressive?

Anyway, the D3 can write almost 30MB/s (aka CF 200x).
And it's got a buffer.

Here's a 200x CF card, 32 GB for USD 49.99:
http://preview.tinyurl.com/4ywxvp4

Expensive?

I've paid more for a 40 MB (yep Megabyte) harddrive, which was
way slower. And needed an external controller.

I eagerly await your explanation why $50 for 800+ shots is
expensive.

So you're fresh out of arguments.

One operation is one operation, especially in RISC.
Of course, you wouldn't understand and of course you have no
proof except stamping your foot.

Nope. They just have a smaller range of values they can represent.
They still calculate 1+1 as exactly 2.

I'm saying that using less bits than you need to represent the
numbers you want is causing miscalculation or forcing you to a
mantissa/exponent recording, which is less accurate when rounding
is needed --- or you use routines that work accurately with values
that need more bits to represent than the machine supports in
hardware (which is slower).
Note that this also applies to 64, 128 and ten million bits.

Not everything you do not grasp is nonsense.

Nor is JPEG. Or TIFF. Or BMP. It's all just data, which can be
interpreted to form an image on the screen or on the printer.

And a zipped .exe is also not an executable, even though it
contains all the info of the .exe and is (usually) smaller.

In other words, completely irrelevant, and you are avoiding again.

I see ... your RAW converter travels back in time, sneaks into
the camera as the shutter is opened, catches the photons blocked
by the Bayer filter, sorts them to colours, and then inserts that
information *missing from the RAW* into the RGB.

I see, you have not yet progressed to the marker notes, where the
camera settings are stored, that will tell the initiated exactly
how to turn the RAW into the one image.

So where does it get the information from? Not in any form
from from the RAW --- that would add no information, because the
information is already in the RAW.

You are caught by your own dogmata and ignorance --- you are too
ignorant to grasp you don't know or to grasp you might be wrong.

Maybe you should educate yourself a bit about the things you talk,
that would help you.

So you agree you don't understand what he was doing.

The Brooklyn Bridge, for example. I sell it to stupid people.
The stupider, the more money I can extract.
I'll make you a special deal: only $100.000.000.

Hmm. You can't answer and turn to lies and abuse instead of
admitting you don't even know what you talk about. You must feel
really insecure, Floyd.

BTW, adding (binary) 111 and 111 doesn't give 111111, but 1110. i.e. at
most 1 bit extra, and that is true for any length of bits --- and yes,
addition us a type of arithmetic.

In other words, you don't know your ellbow from your ass. Again.

No URL, and only Floyd (i.e. completely unqualified).

Yes, you wrote a load of stuff that every participant of computer
science 101 can debunk, only you hold your ears and go la la la.

You're like the clever guy that proves 1 == 2 by
1 == 2 | * 0
<=> 1*0 == 2*0
<=> 0 == 0 QED.
and don't even grasp that you're using an invalid step, though
everyone is telling you.

Yes, you have explained why *you* think so.

And tellingly, you cannot find any other source ("URL") agreeing
with you, especially no source with any qualifications.

And I (and others) have been explainig to you over and over
again why you are wrong. Data B gathered from other data A
using a method (like 3x3 pixels or whatever) does not have more
information than data A + a representation of the method, because
it can be compressed to them.

A compressed file (RAW file) + the decompressor (demosaicer) does
not have less information than the resulting uncompressed file
(16 bit (or even 1024 bit) per channel, 3 (RGB) (or even 10)
channel linear RAW, and yes, you can amortise the decompressor
over all files sharing the decompressor. Wanna URL for that?
Just say so.

And you ask us to believe that your single, dissenting opinion
and new information theory is valid, and everyone else is wrong?
Just on your word?

Let's not kid ourselves, the likelihood of you contributing a
major thing like that is, ah, rather low (to paraphrase you).

Still, it could be ... stranger things have happened.


But I use exactly your method of creating extra data, just a
bit larger, you dare to tell me I'm talking nonsense? Come on!
Either your method is valid, then I'm not talking nonsense (or
you could point out exactly where and how I strayed from the true
course), or your method is nonsense, then why are you spouting
that stuff?


-Wolfgang

 

It is valid, you just don't understand it.

Why, with 11 bit of filtered luminance data and enough pixels
to catch it, about 8.6 billion colours (American billion, old
Englisch milliard).

That enough for you?

-Wolfgang

 

It's called morse code. That's a digital system.

If you however smoothly vary the strength of the signal or the
frequency of the signal to mimic an analog signal, like, say,
one carrying the image in lines, then you're looking at analog.

Sorry, the bits and bytes carrying my alledged 'misinformation' are
digital. Even Mxsmanic agrees the written word is digital ...

-Wolfgang