1

New Theories / Re: Global worming theory

« on: 24/03/2023 16:32:45 »

Hi.

There may be a small spelling mistake in the title of this thread.   @vdblnkr34  , you might be able to edit that or ask a moderator to do it.   It's not vital but people may find the thread more easily.

Best Wishes.

2

Technology / Re: Can you explain this demonstration of a capacitor?

« on: 24/03/2023 16:20:43 »

Hi.

Welcome back @paul cotter .

About stress mentioned by @alancalverd :   I can see there would be some extension ("stretch") along the axis from plate to plate but are there any other forces?  Shear forces, torque around some axis, a net force acting radially?    For example, does the dielectric need to be glued in to prevent the capacitor spitting out bits of dielectric?

Best Wishes.

3

Technology / Re: Can you explain this demonstration of a capacitor?

« on: 23/03/2023 20:16:23 »

Hi again.

   Just realised we haven't actually discussed part (c) of the original task.
So, the basic capacitor equations are, as before:    C =  εA/d      [Eqn 1]         and          C =  Q/V    [Eqn 2]
Where this time, the insertion of the dielectric is going to change  ε.
We have:      insert dielectric  =>  ε increases  =>  [Eqn 1] C increases =>  [Eqn 2]  Q increases   OR   V decreases.
We see from the experiment that both happens,  Q increases (we saw the ammeter deflect one way in the video)  AND ALSO   V decreases (the electroscope leaf fell down).
The basic electronics equations accepts either or both of these but doesn't explain why it would be both and not just one or the other.

Some simple electrostatics arguments can be used to explain why you get both effects.  Basically, the E field can and will change instantly when the dielectric is pushed further into the capacitor, since the dipole sources in the dielectric are changing their location in space.   The Voltage is just an integral of the E field, so some change in V can and will happen instantly.   Local changes force a current to flow from plate to plate, so you will also see a current flow and that would be immediate.  However, a current does not mean the new charges arrive on the plates instantly, the current will bring them but over some time.   So there was never any chance a new charge could arrive on the plates fast enough to maintain a constant V and satisfy the constraint of [Eqn 2] where C was increasing.  We will see the voltage V change slightly.    However the current was there, so Q will also change over time.   Overall both Q and V will be seen to change as the dielectric is inserted.

   We reverse the reasoning and expectations when the dielectric is removed.   That's all fine and the video seems to show that.

   Now the main question that remains is:    What happens to the energy?
The Energy stored in a capacitor is given by      
We had Q went up and  V went down when the dielectric was inserted.   How did the overall stored energy change?
Were there forces resisting the experimenter inserting the dielectric or did the apparatus try to pull the thing out of his hand?

Best Wishes.

4

Technology / Re: Can you explain this demonstration of a capacitor?

« on: 23/03/2023 15:02:26 »

Hi.

    Thanks for all the replies.
    I've recently found an MIT website where they describe the way the Braun electroscope was connected:
https://tsgphysics.mit.edu/demo.html?E_4

    It looks like @vhfpmr got the diagram right in post #4 and deserves a star or medal of honour rather than the suspicion I displayed in post #8. I will edit my post to show that.

     Overall, I think we have the right general idea of how the circuit was connected and where some of the problems have arisen mainly as a consequence of the capacitance and static charge on the experimenter together with the placement of the ammeter in the circuit.

   Thank you to everyone.

5

New Theories / Re: What makes Riemann's Hypothesis Hard to Prove?

« on: 22/03/2023 23:28:14 »

Hi.

Post #43  About recommending video(s).

    They have no significant risk of needing to make the payment  - but there is a small question, let's call it a theory, as to whether they even had the intention of making the payment.   That's a question we may never know the answer to.  It is hard to prove and no more likely than the Riemann Hypothesis.

Best Wishes.

6

Technology / Re: Can you explain this demonstration of a capacitor?

« on: 22/03/2023 20:11:02 »

LATE EDITING: See post #12,  @vhfpmr  gets a gold star.

Hi.

Thanks for all replies.

No, you can see that every time his finger makes contact with the knob, the meter gives a kick to the right.

   Allowing for noise, taking an average etc. , you could be right.   It's not crystal clear to me but I'm old and can't really watch two places (hands and ammeter) at the same time.   The static charge and capacitance of the experimenter together with a wiring diagram something like yours is beginning to look like a clear favourite to explain the right deflections at the beginning when the plates were moved apart.

Yes you do, follow the wiring:   (with a wiring diagram shown)

    I'm not sure that is the wiring.
    If the electroscope (the leaf style voltmeter) works the way I think it does then it doesn't even need to be connected more than once, it may not have a positive and negative lead (or an earth side and live side using your terminology).   There can be one lead from the outer part (or top part) of the stem of the device to one place in the circuit where you want to determine the voltage.   The device is really responding to charge that is driven into the deeper part of the stem and thus also the leaf which was connected to that stem at a pivot point.   It's like a capacitor with just one plate.  It is an indicator of surplus (or net) charge being present on the stem (and leaf) sections that are deep inside the device.   
    It is indirectly useable as a voltage indicator since (negative or positive) charge will tend to be driven into it if the place it's connected to in the circuit evolves over time to have a different (lower or higher, respectively) voltage.

   However, I just don't know how the thing was wired up,  the video never bothered to show any of that.  We seem to be left guessing.   Forced to accept that somehow every piece of kit does precisely and only what they claim, charge Q stays on the plate  etc.   I'm grateful for any attempt to show what the wiring may have been like, thank you.

ES said:     So there is at least one path from plate to plate through the ammeter.
vhfpmr replied:   No there isn't, the power supply has been disconnected, so the only connection to the live plate is the insulated support on the stand, and the electroscope.

   If that's correct then the ammeter is virtually useless, unable to tell us anything from the moment the power supply was disconnected.   At least that would solve one problem, the deflections were random during this part of the experiment.   However, they did seem to make sense for the later part where the dielectric was inserted and removed.

Best Wishes.

7

Technology / Re: Can you explain this demonstration of a capacitor?

« on: 22/03/2023 13:13:59 »

Hi.

Static charge on the experimenter might be as good as any other explanation and I hadn't thought of that.
In favour:   They did walk fast.  The lab coat looks shiny, could be polyester.
Concerns:  I don't know where the ammeter is connected.   Current must have passed through that when they touched the handle.   This suggests the ammeter does allow a good path between the two plates - which will be a problem for later.

I was going to move on to part (b) of the exercise:
   The capacitor is disconnected from the Power supply and a fixed charge Q is deposited on it.
Taken at face value,  we have the same equations as before    Q = VC      [Eqn 2' ]
    C =          [Eqn 1]

When moving the plates apart,  d increases => C decreases from [Eqn 1]  =>  V must increase from [Eqn 2'] since Q was constant.
   Overall,  V ~ d   but the electroscope has no scale and it's not obvious.
    I think the original exercise wanted E fields and electrostatics discussed directly instead of just electronics.  So, we have some large plates.   Near the centre of a plate draw a Gaussian surface, usual textbook stuff,  we have  E = σ / ε in the direction normal to the surface (with σ = charge density = Q / Area ).  The E field is completely independent of distance from the plane (for a large plane and staying near the centre).   σ never changes, so over most of the central region of the plates at least we obtain
  E(at any location between the plates) = a constant independent of d (the distance between plates).
Then the potential difference, V, from one plate to the other is just  E . d .    So  we have V ~ d exactly as before.
    We see the electroscope lift up and down in the right places, so taking everything at face value, that's all fine.

Now the main question has to be:    How did they tell the charge Q to stay on the plate?
   There is an ammeter in the circuit and they claim it measures current flowing from one plate to the other.  So there is at least one path from plate to plate through the ammeter.   I would have expected the capacitor to discharge in the usual pattern, an exponential decay with time.   I'm not seeing much evidence of that in the video.   The ammeter has some huge resistance? - but then it's unlike the properties you would want or expect in an ammeter.   It shows a flow of current from one plate to the other during many parts of the video, it clearly doesn't put a stop to all of them.

The second question would be:   Is there any discernible pattern or explanation for the deflections shown on the ammeter during this part of the demonstration?  It seems to move left or right a bit almost at random.   Just noise  or just there to deliberately make you ask why the system can't satisfy the restrictions from [Eqn 1] and [Eqn 2] by simply discharging some Q from the first plate to the other when the distance between plates is increased.      i.d.k.
   Let's take an argument based on E fields.   The charge on the first plate is essentially on the surface of the plate only (because of the usual arguments about conductors and electrostatics).   So those surplus charges are naturally tending to repel each other and would ordinarily flow through the conductor,  ultimately reaching the other plate through some path like that which the ammeter should provide.   However, there's a non-zero E field between the plates and it's precisely enough to match the force that would propel those surface charges away, they are held on the surface.   If this was a stable equilibrium we'd all be happy - no current will appear.   However it's not, it's an unstable equilibrium.  If some small charge δQ is able to leave the first plate then it can reach the other plate and that will reduce the E field between the plates rather than increasing it.   Basically, capacitors do discharge when there is a path from one plate to the other and, as far I can see, they would discharge even faster when you move the plates further apart.
   Now, despite giving the capacitor every encouragement to discharge while moving the plates apart, at the end of the entire experiment it is still claimed that the charge Q is still there.
However, when the plexi is removed, the voltage rises back up again, showing that the charge is still there.
  - taken from the info. box on the YouTube video.

   I must be missing some physics here.... some clever way in which they have connected this whole circuit together,  used the electroscope and/or galvanometer.    i.d.k.

Best Wishes.

8

Technology / Can you explain this demonstration of a capacitor?

« on: 22/03/2023 05:02:21 »

Hi.

The Tech section doesn't get used as much as the Physics section.  So I've put this in here rather than in the usual Physics section.

Declaration:   This has been set as an exercise by a UK University.

Background:
1.    It's not my homework, just something I saw a month ago, became interested in and still can't explain to my satisfaction.
2.   It was never assessed but just intended as something a student should do in their own time, for their own benefit.   
3.   The general spirit of the exercise is that a student should be able to explain the video to a friend.
   Overall, I would think that a discussion here causes no harm and is perfectly consistent with the spirit of the exercise. Even if next year's students find this thread, it should help not hinder.

Here's the original phrasing of the task:

      This is the video we watched at the start of the live session. You should be able to explain what happens to the current through the circuit and the voltage between the plates at the following times:

a) 0 -> 1m, the plates are connected to a power supply and moved apart from / towards each other;

b) 1m5s -> 1m45s, The capacitor is detached from the power supply, a fixed amount of charge is deposited, and the plates are moved apart from each other;

c) 1m45s -> 2m05s, with the capacitor plates still isolated, a sheet of plexiglass is placed between the plates.

Here's the video:     "MIT Physics Demo -- Adjustable Capacitor with Dielectric",    available on YouTube,  duration ~2 minutes.

- - - - - - - - - - - - - - -

Brief response:

a)    We have      C = εA/d                  [1]
                        V =  Q/C                     [2]
 d increases,  Eqn [1]  =>  C decreases  =>  Eqn [2]  Q must decrease  (because V is held constant) =>  we have a current flow.

    OK... so the Ammeter should show a deflection in the same direction from the central position every time the experimenter turns the handle to move the plates apart.
   BUT....if you look carefully when the plates move apart (at about time 0:27),  the first deflection of the Ammeter during the first twist is to the right.   All the other twists and you have a deflection to the left.    WHY?    i.d.k.   The demonstrator knocked the handle?  I am missing some physics?

   I would blame it on a faulty Ammeter but the Ammeter is actually pretty good in that it manages to show something else that demanded precision from the instrument:   Combining [Eqn 1] and [Eqn 2]  we obtain   Q ~  1/d   meaning that the biggest changes in Q (associated with the bigger currents and deflection) should be seen when the gap was smallest.   (Assuming they turn the handle roughly the same number of rotations and roughly the same speed each time they twist it).   You DO see that in the video clearly enough...... the deflections on the Ammeter are bigger when the plates are close.

   Reverse the behaviour when the plates are moved back together.   That bit seems to give the expected results.   We get deflections to the right only and they become massive, almost to the limit of deflection, when the plates are closest.

----The post is already getting long, so I'll leave discussion of parts b and c for now  ---
Main question from the first part is   Why does the ammeter deflect right on the first twist of the handle when the plates were moved apart?

Best Wishes.

9

Physics, Astronomy & Cosmology / Re: Why can't i understand the andromeda paradox?

« on: 20/03/2023 00:40:08 »

Hi.

I'm sorry but maybe you do have a subscription to PBS or NOVA or whatever it is.   I get asked to type in my log-in details, skipping or closing that and you get this:


pbs-service.jpg (43.98 kB . 1011x568 - viewed 121 times)

Best Wishes.

10

Physics, Astronomy & Cosmology / Re: Do you change the spectrum of radiation inside an oven if you change its shape?

« on: 18/03/2023 16:46:39 »

Hi.

What's the material of the cavity wall?

    The articles are described as concerning blackbody radiation and could be theoretical    BUT   they could also involve some experimentation.   I can't answer further questions, as stated earlier, those articles are paywalled and I'm not going to a suitable library for a while.
   
- - - - - - - - - - - - -

Fundamentally, how does one wall know where the other wall is in order to change its emission spectrum?

    1.   Why does it need to?  The amount of radiation of frequency f can grow if that frequency can be supported as a standing wave inside the cavity.   
     Conversely if it can't be supported as a standing wave then it can't grow.   Only an amount of radiation equal to (proportional to) the rate of production times the flight time before it hits a wall can exist inside the cavity.  Making the cavity bigger (moving the walls further away) shouldn't matter because it's the energy density  (energy per unit volume of the cavity) that is considered for the blackbody spectrum calculations and you'd only be making the volume bigger in proportion to the increased flight time.   So a classical interpretation may argue that frequencies not capable of forming standing waves never become significant.


    2.    For all I know, you were starting a discussion about more general aspects of Quantum Mechanics.    A photon is a QM particle and as such its properties may not be local.  To say this a different way, if we use something like QFT then a photon should be an excitation in the underlying photon field.  If the space available or boundary conditions are such that no solution exists that looks like a photon of frequency f located at a given place then nothing that looks like a photon of frequency f can be found in that place.  It wouldn't matter if the restrictive boundary conditions were located on the other side of the galaxy.
    The issue you were describing is only an issue if you consider a photon as being something that an atom can create inherently or independantly of the rest of the universe.

Best Wishes.

11

Physics, Astronomy & Cosmology / Re: Do you change the spectrum of radiation inside an oven if you change its shape?

« on: 18/03/2023 04:28:23 »

Hi.

I've not had a lot of time to read the new posts.   At a glance quite a lot is tangential to the original question - which is fine,  I'm not troubled at all  -  it's just that I can post this bit without needing to address those other comments because there isn't any conflict etc.

   I'm starting to find a few papers which suggest the shape of cavity does matter and can influence the spectrum of radiation within it.   Sadly, a lot of them are behind paywalls.   Best I can do is pull some details out of the abstracts because I'm not going pay for access or get to a good academic library for at least a month.

   

....Under the assumption that the cavity is a spherical one, the intensity of the blackbody radiation at some frequency is obtained and found to be uniform only in a small region around the center of the cavity. With the help of the theorem of equipartition, the intensity, or the spectrum of the blackbody radiation, is then expressed as a function of the temperature of the cavity and shown to satisfy the familiar Rayleigh–Jeans’ law. Some other properties of the blackbody radiation are also discussed.

    Taken from:    "Non-uniform distribution of low-frequency blackbody radiation inside a spherical cavity",  Journal of the Optical Society of America A Vol. 37, Issue 9, pp. 1428-1434 (2020)

      Other than the obvious issue they mention with a spherical cavity,  it would also be interesting to see how they derived the Rayleigh-Jeans law.    That is usually derived classically and it is only an approximation to the Planck's law (the usual description of Blackbody radiation).

- - - - - - -

By taking into account all of the standing electromagnetic wave frequencies inside cubical and spherical cavities, generalized expressions for the spectral and total radiation from cubical and spherical blackbodies are derived. It is found that the Stefan–Boltzmann law becomes valid only when χT≫hc/k, where χ denotes the length of a cube edge or the diameter of the sphere, T is the blackbody temperature, h is Planck’s constant, c is the speed of light, and k is Boltzmann’s constant. When χT≲hc/k, the radiated power per unit area is less than that predicted by the SB law.

   Taken from:    "Blackbody radiation from cubes and spheres with application to rapid solidification of microspheres", Journal of Applied Physics 56, 1347 (1984)
 
   Basically, the size of the cavity does seem to matter.   If they were radiating like Blackbodies following Planck's law then the Stefan-Boltzmann law would follow.

- - - - - - - - - - - -
    There's a few other examples but overall,  it seems that -
(i)  Very small cavities will deviate from the usual spectrum and,
(ii)  the low frequency radiation in some larger but non-cubical shaped cavities seems to show some deviation.

Best Wishes.

12

Just Chat! / Re: Generally speaking, how would Brits respond if...

« on: 16/03/2023 21:44:16 »

Hi.

Currently, about £2 per capita per annum

   As I mentioned I'm sufficiently in support of the Royal Family that I don't mind paying that much.
   Just to be fair, you could have phrased it differently,  it's over £ 100 million per year   AND some property.  The land and property assests (like Buckingham palace) used to keep the Queen right at the top of the list of wealthy people in Britain a few years ago (I don't read all the latest magazines these days).  Presumably, the anti-Royalists would intend to sell or re-purpose the property.

Best Wishes.

13

Physics, Astronomy & Cosmology / Re: Do you change the spectrum of radiation inside an oven if you change its shape?

« on: 16/03/2023 20:44:45 »

Hi.

Ummm.. yes some will be some light transmitted out that side.   An earlier post mentioned that there was no return path to the laser,  but there could be paths coming back to the left just not aimed at the original source, much as shown in your digram.   So there is no disagreement that there could be some light coming out of the left side of the etalon.

   Overall, the relevance or connection between etalons and the interior space of an oven is somewhat oblique or marginal.   I'm not sure worrying about etalons is vital to the original issue.  I was only considering that the interior of the oven has some comparisons with the interior of an etalon.  In particular, changing the gap between mirrors of an etalon does seriously influence the amplitude of a given frequency of light that would be found inside the etalon.   As you stated, it does not change that frequency of light to a new frequency - but simply reducing the amplitude can be important  (as outlined in a previous post with a gamma-emitter in the oven).

Best Wishes.

14

Physics, Astronomy & Cosmology / Re: Have you ever seen a FIREBALL?

« on: 16/03/2023 19:40:36 »

Hi.

Welcome to the site, I'm not sure I've spoken to you before.

Have you ever seen a FIREBALL?
   No, sorry.
....But your enthusiasm is apparent from your description and it was a pleasure to read.   Thank you for sharing it. 
(I'm not entirely sure it needed to be in this section rather than "Just Chat" or somewhere else - but that's a separate issue and I'm not a moderator). 

Best Wishes.

15

Physics, Astronomy & Cosmology / Re: Do you change the spectrum of radiation inside an oven if you change its shape?

« on: 15/03/2023 17:32:46 »

Hi.

They don't absorb it; where has it gone?

   Heat.   The Etalons get hot when light is shone in one side but none is transmitted out of the other side.   They can run much cooler if the gap is adjusted so that all the light is transmitted out through the other side.

----
   I haven't done the experiment, or found any theory, for the oven / blackbody cavity situation but you'd guess for something similar.

Best Wishes.

16

Physics, Astronomy & Cosmology / Re: Do you change the spectrum of radiation inside an oven if you change its shape?

« on: 15/03/2023 12:46:43 »

Hi.

@Bored chemist  said something like this:
  Etalons / F-P interferometers  don't change the frequency of light.   
--->  Agreed  but it can be enough that they will reduce the amplitude of light found between the mirrors  if  frequency ≠ a resonance    just because there will be a lot of destructive interference.

   Recall that a blackbody spectrum can have some blue light, it just can't be too much.

   Moving to a general example instead of worrying about the F-P interferometer too much.....
Suspend a strong γ emitting radioisotope inside an oven and you would have thought the spectrum of radiation inside the oven would be seriously perturbed.   I mean you've got plenty of high energy γ frequency radiation in there and it doesn't seem to ask or be concerned about whether you can fit a standing wave of that frequency inside the oven cavity.
However, the usual blackbody spectrum calculations (where only standing waves are considered as modes) only claims to predict the spectrum inside the oven AT EQUILIBRIUM.   You're not at an equilibrium situation at the moment.   Assume the walls of the oven are capable of fully absorbing the γ rays and wait a while.   Two things happen:
    (i)   The walls get warm because they are absorbing γ rays.   This begins to shift the predicted blackbody spectrum to somewhere where some measurable intensity of γ frequencies is expected.
    (ii)  Now that the walls are producing some γ rays, you can get some destructive interference with those produced by the radioisotope,  lowering the overall intensity of γ frequencies.

    Those two things together can get the final spectrum, at equilibrium, back in reasonable agreement with a blackbody spectrum.

    Now to my mind, it still doesn't really explain WHY we do the calculations for a blackbody spectrum considering "modes" as being ONLY that which can be supported as a standing wave   but it does seem to suggest that you CAN make those assumptions and you will still get the right spectrum in the end.

   That was rushed and may not have linked smoothly with previous comments, sorry.   I seem to have some other things to do and may not respond or put "likes" on future comments for a while, sorry.

Best Wishes.

17

Just Chat! / Re: Generally speaking, how would Brits respond if...

« on: 14/03/2023 00:29:33 »

Hi.

I really don't know but overall the question could have been stopped after the first half...

How would most Brits react if King Charles summoned a press conference?

     Unless it was Christmas, they may not even switch on the TV set to listen.   Are you from the U.S. by any chance, you do seem to use some US spellings.   It's been a few years since the King really acted like a head of state for the United Kingdom.   Personally I'm reasonably happy with the Royal Family but there's a good number of Brits who are not and are described as "anti-Royalists".  Don't get me wrong, they aren't going to put the Royal Family on a Guillotine but they believe there's no need to support the Royal Family in the style that we currently do when tax-payers money can be spent on other things.

....to announce that he has begun taking GnRH modulators, estrogens, antiandrogens, and progestogens so he can become a woman and become Queen Charles?

    To be honest, the first thing to cross my mind would be that this was a bit of an over-inflated PR exercise.   I mean it would do no harm to have the Royal family seen to be involved in current issues being explored in society.   There would no doubt be books written and films made about their struggle with gender identity but they were never allowed to express themselves because of their duties and role in the Royal Family.   I can see the end scene - a real tearjerker.... they heroically stand up in front of the whole nation and say "I am what I am"   (...so you just know that the later stage versions will use the Gloria Gaynor song of that name but you're going to have wait for that).
   Would it cause the majority of U.K. citizens to recoil in horror?  I wouldn't have thought so.  Get a good ghostwriter for the book and some good actors and screenwriters for the film version and it will probably put Britain back on the maps again, an international "tour de force".

Best Wishes.

18

Physics, Astronomy & Cosmology / Re: Do you change the spectrum of radiation inside an oven if you change its shape?

« on: 13/03/2023 18:47:34 »

Hi and thanks for everyone's time.

There's no population inversion (as you would find in a laser)...

     Sorry, I'm fairly sure that is not really required.   Lasers are often used just to provide a good clean monochromatic source of light but there's nothing about the laser or population inversion that needs to be involved other than that.   This is most easily seen just by having the laser outside the interferometer and tilted at some small angle to the mirrored surfaces so that there is no path along which any kind of return ray of light could be passed back into the laser and influence it.


Fabry-Perot.jpg (74.49 kB . 783x480 - viewed 508 times)

Image based on:   http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/fabry.html

    Shine a monochromatic source of light into one side of a Fabry- Perot interferometer and then set the gap correctly and you get interesting results - exactly what the source of that monochromatic light was doesn't matter.   Depending on the gap and angle of incidence of the incoming ray, you can have all, some or none of the light being emitted out of the other side and brought to focus on the screen,  along with there being  no, some, or lots of light to be found inside the gap between the mirrors.    In situations where the light is effectively eliminated, the interferometer itself can get hot  (the air inside it and/or the mirrored walls of it will get hot) the laser (if that was the original source) should not be affected by what is happening in the interferometer.

Best Wishes.

19

Just Chat! / Re: Advertisements on the forum

« on: 13/03/2023 18:07:01 »

Hi.

Yes I do get the humour.   
   I still have to mention that the decision about "relevance" has got more to do with what THEY think they can sell to a user of this forum and not what YOU would choose to see as an advert.   
   Actually, Google (if that's what they are using) is amazingly efficient at targetting adverts, they know what you might be thinking about buying better than you do yourself - just from analysing your cookies (browsing history), some words or phrases you seem to use often and any personal information you may have deliberately or accidentally given over.  Their psychological profiling is second to none.   I think I'd better stop talking.

Best Wishes.

20

Just Chat! / Re: Advertisements on the forum

« on: 13/03/2023 17:18:23 »

Hi again.

    These adverts seem to have seriously slowed down performance.   Move the mouse over something and you really wait for the drop-down menu to appear, click on a thing and you start to wonder if the computer has actually crashed.    Presumably some (Google based?) server somewhere in the world gets accessed, the web content from the Nakedscientist end of things is analysed, a choice of relevant advert is finally made and then all the data for some high resolution and sometimes animated advert is eventually fed back the other way and re-assembled by some server to display the final web page.   It's a noticeable delay.   
     Are the administrative staff unconcerned about this?   It's a mild inconvenience for me but for some people it will be a physical barrier to using this site.    Nakedscientist's frequently go out of their way to involve schools or organisations from countries that might be described as developing economies.   That's great but what are you doing to address the problem that is known as "Digital Inclusion"?   Some people do not have a fast internet connection and may be charged by the kilobyte to download the irrelevant data that is used to make the advertisements.   Digital inclusion is comparable to inclusion of all other kinds.   We would not tolerate it if you forced a disabled person to crawl up a flight of stairs just because someone had a cunning plan to display an advert on every step - but you have done almost exactly that.

Rant over.  Sorry.