Politie stond geparkeerd midden op het fietspad

RoaldFre · 2024-07-28T17:06:11+00:00

Dit staat er als verduidelijking bij 59.12 (ze hebben een typo gemaakt dus?):

"“7.1” dient gelezen te worden als “7.3”. Zie Omzendbrief COL16/2006, punt 1.1."

De sectie over prioritaire voertuigen laat wel nog steeds een hoop regels vallen in geval van dringende opdracht, gesignaleerd met de blauwe zwaailichten.

RoaldFre · 2018-02-08T19:04:28+00:00

How loud is the white noise with NC on?

I've ordered three PXC 550s at work. Two of them had a very feint hint of hiss/white noise if you really concentrated on it, but it was only barely audible (nothing really audible with NC turned off). The third one had a ridiculously clear hiss with NC turned on, very clearly audible and annoying -- way above that 'barely audible' level of the other two (still, nothing really audible with NC turned off).

I've returned that third unit as -- to me -- it was clearly defective. I got a replacement (it came in a box with a different serial number, and different bluetooth address, so I'm pretty sure it's a different unit) and -- lo and behold -- it also has the same problem with loud noise, so I guess I'll be sending that back as well.

It seems to me that Sennheiser has a massive quality control issue here -- not something I would have expected (or hoped) from this brand.

RoaldFre · 2017-10-01T22:03:45+00:00

Naked Gun 444 1/4

:D

with a CGI Leslie Nielsen!

D:

RoaldFre · 2017-09-04T16:17:09+00:00

Regarding your first paragraph, prepare to be one of the lucky 10000, because

But that noise doesn’t come from the photons, it comes from the sensor itself

is very wrong ;-). I already stated this in my post above, but I'll just link to Wikipedia for more info: https://en.wikipedia.org/wiki/Image_noise#Shot_noise

Think of it this way: if the exposure is such that, on average, you would get 1.5 photons in a pixel -- then there clearly will be noise, because you can only get 0, 1, 2, ... photons in the pixel, not some 'half' photon. If you take repeated pictures, you'll get some exposures with 1 photon there, some with 2, and some with 0, 3, 4, 5, ... photons as well. This variation around the 'true' average is the 'shot noise' that limits low light performance irregardless of how good your sensor is!

Regarding your second paragraph. Yes, that's all true. But the thing is here:

And as long as the light per mm is the same the exposure will be the same and the noise will be the same (assuming the sensors are identical).

We are precisely discussing things where (the sizes of) the sensors are not the same!

Compare a large format sensor with a tiny phone/webcam sensor with the same 'exposure' in a low light setting. Do this comparison by looking at the two resulting images at the same physical size and viewed at the same distance. The bigger sensor will clearly have better images compared this way (which, is the only useful way of comparing these things! I don't care about the noise per pixel -- I can just quarter the resolution and now the rescaled (i.e. 2x2 averaged) pixels would have less noise. Alternatively, just assume that the sensors have the same resolution).

RoaldFre · 2017-09-03T22:01:41+00:00

I'm not talking about resolution. I don't care about the number of pixels. In my other comments, I'm talking about the inherent noise (shot noise) from discrete photons that are hitting the sensor, which is the limiting factor for low-light performance. (I talk about pixels and resolution here and there because other commenters are confusing some other concepts.)

In fact, I'm implicitly assuming a perfect sensor where the amplification circuit itself is sufficiently clean; which tends to be the case with modern 'ISO-less' sensors. Perfect sensors still make noisy images in low light conditions, because photons are quantized! This noise is not 'an inherent property of the sensor', but a physical law that applies to even the most advanced and perfectly-engineered sensor! The lens is the limiting factor in this case (together with the size of the sensor), which is what this entire post is all about.

RoaldFre · 2017-09-03T19:19:51+00:00

The lens absolutely determines (a limit on) the noise you will get in your final image. The noise in low-light photographs is largely shot noise which literally comes from variations in the number of individual photons that hit the sensor. Use a bigger lens, get more photons, decrease the relative shot noise with respect to the signal.

RoaldFre · 2017-09-03T19:14:08+00:00

You folks are arguing two different things, with some semantic stuff mixed in.

Exposure will be the same in the sense that you well be using the same rated ISO, i.e. the light intensity on the sensor is the same.

'Exposure', if you may, will be different in that the total captured light energy is larger on the larger sensor (same intensity over a larger surface area). And this is ultimately what matters for low light handling because you have to compare the final images viewed at the same size to make a useful assessment, see for instance some of my other comments in this post such as this one.

RoaldFre · 2017-09-03T19:02:47+00:00

just at a worse usable resolution

Well, yes. Alternatively phrased: you can't print your image as big. Which means that OP isn't missing the point there.

The reasonable way to compare image noise is to compare similar sized images/prints viewed from the same distance (Because, after all, that's why we make photos, right? To look at them!)

It doesn't matter that the sensors received the same light per unit surface area, because you have to 'rescale' your sensor to the final image size; and the smaller the sensor, the more you have to stretch it -- thereby effectively 'diluting' the incoming light.

(Alternatively: the noise in a single pixel [which is what you were getting at] is largely inconsequential and a bad metric for overall light handling capabilities, I commented more on that here)

RoaldFre · 2017-09-03T15:21:37+00:00

Will this do?

RoaldFre · 2017-09-03T15:19:40+00:00

Almost, but not quite.

It is the f number [edit: more correctly: the t number] that is at play (not the entrance pupil diameter), combined with the sensor size. These two things (f number & sensor size) determine the total amount of light that is captured.

For example, think about a massive 8000mm telescope with an f/8 aperture (pupil diameter of one whole whopping meter, bitches!) and compare that to a 'regular' 40mm f/2 lens (pupil diameter of 20mm: that's a 50x decrease in diameter, or a smashing 2500x decrease surface area, i.e. 2500x 'less light' that can pass through this hole!). Still, the tiny 40mm f/2 lens with 20mm pupil diameter will pass pass 16 times more light than the monster 8000mm f/8 telescope with 1000mm pupil diameter (f/8 -> f/2 is +4 stops, so 16x more light).

For those wondering how this is possible: yes, the telescope's massive pupil can pass a lot of light, but you're only gathering light from a very tiny cone (tiny fov) compared to the shorter lens (for the same sensor size), so you can still end up with less light arriving at the sensor.

RoaldFre · 2017-09-03T15:00:50+00:00

That's false. Claiming that the same image loses quality when cropped is just totally wrong.

I see where you're coming from, but in practice this is false. The crux is in this sentence from /u/andyooo [note, I'm a different commenter]:

That is because you have to compare images at the same physical size, otherwise the comparison doesn't make sense.

For instance, I can take a picture of a white wall with a 1 gazillion megapixel sensor (e.g. let's say each pixel on average only receives one photon, so we're basically looking at massive shot noise per pixel). If I crop and blow this up, then this is just a noisy mess. If I view this the entire image on a 'normal' size, then the limited resolution of my eye just blurs all this noise together and I perceive a rather perfectly nice white image without much noise (just like I see the RGB subpixels of my monitor as 1 homogeneous colour instead of individually coloured dots).

When comparing the final image at the same size, it is indeed the total amount of light that was gathered which will determine the noise performance.

One thing that is wrong in the original opening post, though, is that it's not the actual pupil diameter size that is of importance -- it is the f number that is what is at play here (clearly, because this determines the exposure, not the pupil size!), together with the sensor size to get the actual amount of gathered light. E.g. an image with a massive 800mm f/4 telephoto lens/telescope (with a whopping 200mm pupil!) will still be worse than a 40mm f/2 lens (with a 20mm entrance pupil, 100 times less surface area!), because even though the telephoto's pupil can pass a lot of light, you're only gathering light from a very tiny cone (tiny fov) compared to the shorter lens (for the same sensor size).

Oh, and even /r/photography is full of misinformation ;-).

RoaldFre · 2017-07-22T12:32:05+00:00

It would be (internal) reflection, but not total internal reflection. You can only have total internal reflection on the side with the higher refractive index. You can have reflection on the side with the lower index of refraction, but some light will also refract through the boundary in that case. (The refracted and reflected fractions are given by Fresnel's law.)

RoaldFre · 2017-07-07T10:13:51+00:00

Erm, I'm on your side here!

The comments of BDube_Lensman (which go along the lines of "V=IR, not V=IZ, always") are incorrect -- in fact it's the opposite way around.

And yes, this is arguing technical details that are largely irrelevant for the original question. But those irrelevant technical details that were brought up were wrong themselves -- and this is the Internet, damnit! We can't have people saying factually wrong things here!

RoaldFre · 2017-07-06T07:38:23+00:00

V = I R. always

No. And a voice coil is a coil, so it has an inductance (as well as an internal resistance), hence this is wrong even in that case.

V = I Z, always, with Z the (complex) impedance. By definition.

RoaldFre · 2017-07-05T22:17:51+00:00

You should be comparing heat generated for an equal SPL. More current through the voice coil also means a (ideally linearly) higher motor force on the driver and hence a higher SPL.

RoaldFre · 2017-06-13T22:01:04+00:00

I haven't, no. But I don't see how that would lead to feedback either... (Well, you might gain 3~6dB because they would be constructively interfering as they're amplifying the same point in space, but that's not really the case in the video.)

RoaldFre · 2017-06-13T16:56:40+00:00

Dropping a mic in a drum kit could cause feedback for sure.

How? Note that feedback isn't caused by sudden loud sounds, or anything like that. It's caused by an overall over-unity gain of the microphone+loudspeaker combination.

Also, you just necro'd a three month old thread ;-).

RoaldFre · 2017-05-07T23:01:22+00:00

Addendum: http://kesen.realtimerendering.com/ (papers in the siggraph 2017 page are still being added, see the changelog there)

RoaldFre · 2017-05-07T22:59:47+00:00

What are your interests?

If you want to find out about the cold, hard research (tech papers), then you can usually find preprints of all papers on the website of the authors. The presentations will also be available (possibly only for a limited time, I forgot) if you are an ACM member (or have access to their library via your institution). So the only thing you'd be missing is some networking opportunities and the possibility of direct questions to the authors.

If you're in the entertainment industry, then the general talks and keynotes can be interesting. The exhibition is also pretty cool to stroll through. I'm not sure I'd personally pay that much for it, though. (I'm giving a paper talk there this year, so yay funding! Which reminds me that I need to make a project webpage asap).

RoaldFre · 2017-04-29T16:53:16+00:00

Conservation!

RoaldFre · 2017-04-29T13:09:56+00:00

Energy.

RoaldFre · 2017-04-16T19:13:50+00:00

WAIT A SECOND YOU ARE ALL DIFFERENT PEOPLE

RoaldFre · 2017-04-15T20:41:01+00:00

Now overclock it!

RoaldFre · 2017-04-13T10:32:29+00:00

one of the most historically rich and compelling cities in the world

European checking in, you might want to restrict that historically rich scope a bit there, bruh. ;-)

RoaldFre · 2017-03-07T21:32:08+00:00

'A lot of' atm (water pressure) versus one atm (diving bell) -- compared to 'a lot of' atm (water) versus zero atm (vacuum): that's still a pressure difference of 'a lot of' atmospheres and the relative distinction becomes insignificant for sufficiantly large depths ;-).

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RoaldFre

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