A revolution in thought - Page 615

Spacemonkey · #**15351** 03-12-2012, 12:58 AM

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Originally Posted by thedoc

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Originally Posted by Spacemonkey

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Originally Posted by thedoc

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Originally Posted by peacegirl

If you can't control yourself, I'm not going to talk to you.

Yes Spacemonkey, don't you know that you shouldn't ask questions that Peacegirl can't answer or that expose her's and Lessans ignorance?

Indeed. Otherwise she'll ignore my questions instead of just not answering them.

What's the difference, you get no response either way? But in reality she has no coherent response, just nonsense.

Exactly. There is no difference, yet somehow her ignoring my questions instead of dodging them is supposed to constitute some kind of punishment for speaking the harsh truth about her father's abilities.

thedoc · #**15352** 03-12-2012, 01:05 AM

Ahhh, much better.

thedoc · #**15353** 03-12-2012, 01:09 AM

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Originally Posted by Spacemonkey

Exactly. There is no difference, yet somehow her ignoring my questions instead of dodging them is supposed to constitute some kind of punishment for speaking the harsh truth about her father's abilities.

Aren't you glad she wasn't your grade school teacher? Can you imagine how fucked-up your head would be right now?

Vivisectus · #**15354** 03-12-2012, 01:26 AM

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Originally Posted by peacegirl

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Originally Posted by Vivisectus

Another fun fact I just stumbled across: most monitors and TV work at a frequency of about 60 Hz. To us, that makes the image look stable, and movement fluid.

For dogs, the frequency needs to be much higher as they process images faster than we do. What looks like a stable image to us, is flickery and unreal to them, and movement very jerky.

Good try Vivisectus. So let's go back to pictures.

Was that where you were explaining to me how your 5 minutes with skype and the family dog was WAY more convincing than a carefully controlled study, because of levers? I am still waiting for your explanation about how these pesky levers are such a bane to scientific testing, by the way.

5 minutes with skype in a situation where the dog's reaction - by your own admission! - could have been interpreted in any way you like?

Done by a person with a very clear interest in seeing a particular result?

With exactly 1 test subject? It does make the "statistics" easy to calculate, I have to say!

And for extra convenience, without finding out how your test group reacts to strangers? It DOES make things so much simpler, doesn't it?

Sure! Why don't you start by telling me all about those darn levers?

thedoc · #**15355** 03-12-2012, 02:29 AM

Hmmm, 'Fur Elise' went much better but a little less Fleischmann's and Ginger Ale would help.

LadyShea · #**15356** 03-12-2012, 04:14 AM

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Does that mean light from all different sources is striking the tree?

Yes, of course light from all different sources is striking the tree, what a strange question to ask

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Occam's razor would say the simplest explanation is probably the right explanation, and this explanation is definitely simpler.

Actually, Occams razor states that the explanation with the fewest assumptions is probably correct.

So, in the standard model, the assumption is that all light behaves the same way all the time, and those properties are observable, measurable, and make predictions and testing shows that those predictions are accurate

In your model the assumption is that some light does one thing, while other light does another thing, and nothing is predictable or measurable and there is a lot of "something else, unknown, going on", and light seems to have been created for the sole purpose of revealing matter to human eyes which can utilize it in ways that contradict the observed properties of light.

Nope, your idea makes more assumptions and is way more complex.

thedoc · #**15357** 03-12-2012, 05:02 AM

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Originally Posted by LadyShea

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Does that mean light from all different sources is striking the tree?

Yes, of course light from all different sources is striking the tree, what a strange question to ask

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Occam's razor would say the simplest explanation is probably the right explanation, and this explanation is definitely simpler.

Actually, Occams razor states that the explanation with the fewest assumptions is probably correct.

So, in the standard model, the assumption is that all light behaves the same way all the time, and those properties are observable, measurable, and make predictions and testing shows that those predictions are accurate

In your model the assumption is that some light does one thing, while other light does another thing, and nothing is predictable or measurable and there is a lot of "something else, unknown, going on", and light seems to have been created for the sole purpose of revealing matter to human eyes which can utilize it in ways that contradict the observed properties of light.

Nope, your idea makes more assumptions and is way more complex.

And therefore fails the test.

thedoc · #**15358** 03-12-2012, 05:03 AM

As does everything else in Lessans book.

thedoc · #**15359** 03-12-2012, 05:04 AM

And all of Peacegirls posts.

thedoc · #**15360** 03-12-2012, 05:04 AM

If we take a vote, does that make it so?

LadyShea · #**15361** 03-12-2012, 05:04 AM

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Originally Posted by peacegirl

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Originally Posted by LadyShea

Bump

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Originally Posted by LadyShea

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Originally Posted by peacegirl

The answer given was not adequate and nobody countered it with a better

I did not make that up. So if that wasn't the answer, then give me the answer as to why we cannot get an image of a person (a real human being with substance) when he steps out of visual range but he is in a straight line with the observer. According to the current belief, the person should be reflecting said light toward the film/retina, and we should get an image. Tell me why we don't. The inverse square law isn't the reason because the person isn't that far from the observer to be the cause.

I just posted it, plus all the many links to it, a few posts up. Here it is again, in red and broken down in case a paragraph is too much to read. "Too far" depends on the sensor/detector. What may be too far for human eyes, isn't too far for an Eagle's eyes, or what is too far for a telephoto lens isn't too far for a super telephoto lens, etc.

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Originally Posted by LadyShea

If clicking links is beyond your ability, here's the answer yet again. What is inadequate?

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Originally Posted by Spacemonkey

imagine the camera as having thousands of individual detectors each corresponding to an individual pixel of its resolution. The resulting image is a collection of dots, each of which has to be one specific color.

Stick a red ball right in front of the camera, and every detector will see red, and the photo will be all red.

Progressively move the ball away from the camera, and the outer detectors cease to see red, with only a progressively smaller group of central detectors seeing red, such that we get a smaller and smaller red circle in the center of the photo.

Eventually the red circle gets smaller than the size of the single central detector, such that all the other detectors are not detecting red, and this one central detector is receiving more non-red light (from the areas around the ball) than red light (from the ball itself).

At that point the ball will represent a smaller part of the image than the smallest detector, and the ball will cease to show up on the image. That central detector will have a decision to make as to whether or not to create a red dot in the image, and as it is receiving more non-red than red light, it will not indicate red. So even though the camera is still receiving (a small amount) of red light from the ball, the resulting image will not show the ball at all.

Too far does have to do with the sensor LadyShea, but it also has to do with the object. I can't believe how little you have understood.

It has nothing to do with the object, only the detector...configuration, sensitivity, size, etc..

thedoc · #**15362** 03-12-2012, 05:10 AM

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Originally Posted by LadyShea

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Originally Posted by peacegirl

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Originally Posted by LadyShea

Bump

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Originally Posted by LadyShea

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Originally Posted by peacegirl

The answer given was not adequate and nobody countered it with a better

I did not make that up. So if that wasn't the answer, then give me the answer as to why we cannot get an image of a person (a real human being with substance) when he steps out of visual range but he is in a straight line with the observer. According to the current belief, the person should be reflecting said light toward the film/retina, and we should get an image. Tell me why we don't. The inverse square law isn't the reason because the person isn't that far from the observer to be the cause.

I just posted it, plus all the many links to it, a few posts up. Here it is again, in red and broken down in case a paragraph is too much to read. "Too far" depends on the sensor/detector. What may be too far for human eyes, isn't too far for an Eagle's eyes, or what is too far for a telephoto lens isn't too far for a super telephoto lens, etc.

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Originally Posted by LadyShea

If clicking links is beyond your ability, here's the answer yet again. What is inadequate?

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Originally Posted by Spacemonkey

imagine the camera as having thousands of individual detectors each corresponding to an individual pixel of its resolution. The resulting image is a collection of dots, each of which has to be one specific color.

Stick a red ball right in front of the camera, and every detector will see red, and the photo will be all red.

Progressively move the ball away from the camera, and the outer detectors cease to see red, with only a progressively smaller group of central detectors seeing red, such that we get a smaller and smaller red circle in the center of the photo.

Eventually the red circle gets smaller than the size of the single central detector, such that all the other detectors are not detecting red, and this one central detector is receiving more non-red light (from the areas around the ball) than red light (from the ball itself).

At that point the ball will represent a smaller part of the image than the smallest detector, and the ball will cease to show up on the image. That central detector will have a decision to make as to whether or not to create a red dot in the image, and as it is receiving more non-red than red light, it will not indicate red. So even though the camera is still receiving (a small amount) of red light from the ball, the resulting image will not show the ball at all.

Too far does have to do with the sensor LadyShea, but it also has to do with the object. I can't believe how little you have understood.

It has nothing to do with the object, only the detector...configuration, sensitivity, size, etc..

Peacegirl's understanding has nothing to do with anything that is not written in Lessans book, that is the sum total of truth and knowledge. Anything beyond that is questionable if not rejected on principle.

Angakuk · #**15363** 03-12-2012, 06:39 AM

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Originally Posted by peacegirl

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Originally Posted by peacegirl

I did not make that up. So if that wasn't the answer, then give me the answer as to why we cannot get an image of a person (a real human being with substance) when he steps out of visual range but he is in a straight line with the observer. According to the current belief, the person should be reflecting said light toward the film/retina, and we should get an image. Tell me why we don't. The inverse square law isn't the reason because the person isn't that far from the observer to be the cause.

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Originally Posted by Angagkuk

How far is "that far"? In other words, how do you, peacegirl, go about calculating, for any particular object, how far it needs to be from the observer before it is "too far" away for its image to be resolved by the film/retina? Additionally, if the non-absorbed photons do not bounce/reflect off the object then why and how is their subsequent activity governed by the inverse square law, a law that applies specifically to movement of photons?

I already went over this. The field of view is what is capable of being seen, even if it's magnified. Too far means that the object is too small or too dim to see by the lens. If a telescope is used, it will still meet the definition of "in range", therefore the inverse square law works regardless of how far away an object is. In efferent vision, light energy continues to flow, but the non-absorbed light does not travel and get carried along through time and space. It fades the farther away it gets from the object.

I think you may have missed the point of my question. You state that the person in your example "isn't that far from the observer". That appears to suggest that you have some specific distance in mind, otherwise you have no way of determining whether or not the person is "that far from the observer". If all you meant to say is that at point A the person is not too far away to be seen by the observer, but at point B the person is to far away to be seen by the observer then all you are saying is that when something is close enough to be seen we can see it and when it is not close enough to be seen we can't see it. That is not even an interesting observation.

Angakuk · #**15364** 03-12-2012, 06:43 AM

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Originally Posted by thedoc

Well, enough of this frivolity and mirth, it's back to the Piano. I was playing the first movement of 'Moonlight Sonata' and totally blanked out at the last 2 measures, got to work on that.

What is this "Moonlight" of which you speak? Are you introducing yet another kind of light? Is it not enough that we have to deal with (N) light and (P) light, now we have to contend with (M) light as well? Is there no end to this profligate proliferation of different categories of light?

davidm · #**15365** 03-12-2012, 06:45 AM

LOL, peacegirl can't believe how little LadyShe has understood.

Of course peacegirl has an idiosyncratic understanding of understanding, viz: Uncritically accepting what the buffoon Lessans wrote, even though everything he wrote was demonstrably wrong.

LadyShea · #**15366** 03-12-2012, 01:19 PM

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the non-absorbed light does not travel and get carried along through time and space. It fades the farther away it gets from the object.

All light always travels.

The light doesn't "get carried along" by what? What does that mean?

The inverse square law has to do with traveling. Without traveling, there is no "fading"

LadyShea · #**15367** 03-12-2012, 01:37 PM

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Originally Posted by peacegirl

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Originally Posted by LadyShea

Where are the actual photons with the specific wavelengths that weren't absorbed now?

When you say they spread out "until there is no more (P) light", do you mean they cease existing at that time? They cease traveling at that time? They cease to have their specific wavelength? They rejoin the other wavelengths that do travel? Where are they and what are they doing now?

You are still confused. These photons are not traveling. White light is traveling. The object is revealing, therefore these photons do not continue. The light fades out when the object cannot be seen. Do you see what you're doing LadyShea? You are saying that this light cannot fade out because light continues on. That is true, but it's white light that continues on, not P light.

Photons cannot stop existing, so the specific photons must be somewhere. Photons must have a wavelength. Do you understand what light is and what it's properties are? Do you understand frequencies and wavelength at all?

All I am "doing" is trying to get you to explain your model in a way that doesn't break the laws of physics and doesn't change the known properties of light.

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They are somewhere, they get spread out until they no longer reveal the object at that distance.

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Originally Posted by LadyShea

Where?

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They don't spread out beyond the visual range because this (P) light is revealing the object, it is not bringing the pattern of the light to the eye over long distances. That's where science got it wrong.

Where are they? I am not asking about eyes and visual ranges, I am asking about light. Where are the photons that were not absorbed?

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These photons do not bounce and travel, therefore when they fade out (which means they are not resolved on the film/retina), white light continues on.

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Originally Posted by LadyShea

That makes no sense.

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To you it doesn't, because you believe these photons bounce and travel. But they don't. You're never going to understand this model because you refuse to budge off of your position.

"Fading out" (becoming less intense due to divergence) only happens when something is traveling. If they aren't traveling, they aren't fading out. They have a location whether they are or are not traveling, unless they've ceased to exist. WHAT IS THE LOCATION? Where are they?

If you start making sense, and stop contradicting yourself, maybe I'll understand what you are positing here.

LadyShea · #**15368** 03-12-2012, 02:00 PM

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Originally Posted by LadyShea

Why would non-absorbed wavelengths have different properties than the wavelengths traveling together as white light? Why are non absorbed photons different from other photons?

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They are replaced by Sunlight in a constant procession, but they do not travel far and wide because they do not get reflected. You're having a very hard time with this concept because you are wondering where the photons go. They go back to white light when the object is too far away to be resolved.

Yes, I am wondering where the photons go, because they have to have a location unless they ceased to exist. Light always travels unless it isn't light, or has been absorbed and transformed, or unless you are changing the known laws of physics.

What do you mean they "go back to white light"? White is not a wavelength, white light is all wavelengths being present in equal intensities.

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Originally Posted by LadyShea

Where are the photons that were not absorbed? If they don't travel, and they weren't absorbed, what is their location?

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The location is exactly where they are WHEN THEY ARE AT THE FILM/RETINA. If they are no longer at the film/retina, they do not continue traveling.

How did they get to the film/retina? What if there is no film or retina anywhere around? What happens to them then?

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This is what separates efferent vision from afferent. This does not violate physics because white light does travel on, just not light that is reflected off of objects consisting of non-absorbed wavelength light.

White light is just light of all wavelengths. Light with a blue wavelength is still light. All light travels until/unless it is absorbed, that's a law of physics.

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The new replacement of photons that get absorbed and (P) reflected continues as long as the object is in existence.

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Originally Posted by LadyShea

So traveling photons cease to exist of some object they once interacted with ceases to exist? That breaks the laws of physics, energy cannot be destroyed, light doesn't cease to exist unless it is transformed into some other kind of energy.

So, you are positing brand new laws of physics. Just admit it. This is why you sound like a lunatic, you say one minute that the laws of physics remain unchanged, then you happily break them into pieces.

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I'm sorry about that; but it's YOU that is not getting it.

My point stands. Your model requires a brand new set of physical laws. One where some light has one set of properties and other light has a different set of properties. I don't "get it" because you keep saying you are not changing physics, then you go on to posit new physics.

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The new photons are constantly being replaced as the Sun's light is being emitted. Nothing is stationary. But the farther away these photons become, the more dispersed they get until they are no longer a condition of sight. At that point, white light is all there is.

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Originally Posted by LadyShea

You are avoiding the question, weasel. Where are the non absorbed photons?

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I'm trying to explain where physics still works, even though the light from objects do not travel away from the object beyond the inverse square law. White light continues where the last photons on the sensor can not be resolved.

Non-traveling light is not physics at work. That's new physics.

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Some of these laws will need to be rethought if it turns out that objects do not (N) reflect their image indefinitely. Light does not fade away, but the (P) reflection does. I never said light cannot be destroyed, therefore thermodynamics doesn't enter into it. Snell's law is not negated either since we would be seeing the effects of refraction, but just in real time.

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Originally Posted by LadyShea

Then quit saying efferent vision is consistent with the laws of physics because it clearly is not, at all.

It is very much consistent except for the part that is inaccurate, which is being contested.

It doesn't break the laws of physics except when it does. LOL

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That is what is being disputed LadyShea, so you can't just repeat the very thing that is being contested.

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Originally Posted by LadyShea

Then quit saying efferent vision is consistent with the laws of physics because it clearly is not, at all.

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Okay. Efferent vision does not change how light works, and it doesn't change the laws of physics, although it does change what science believes is occurring.

It clearly does require a change to physics. It clearly does require a big change in how light works.

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Originally Posted by LadyShea

Why would the object or matter stop absorbing light? If the light is not traveling, is it stationary? If it's neither traveling nor stationary WHERE IS IT AND HOW DID IT GET THERE?

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I told you that nothing is stationary. Those photons are present as long as the object is present and absorbing light, but those non-absorbed photons do not get reflected.

Then where are they and how did they get there?

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You're making it appear as if I'm giving multiple explanations and being inconsistent, but I'm really not.

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Originally Posted by LadyShea

What you are trying to get across makes no sense, whatsoever and is very inconsistent both internally (brain's ability or the ability of lenses?) and within reality (light isn't reflected). You can't even answer questions about it. Can you draw a diagram or something?

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I don't have the software to do this.

You don't need special software. Draw it on a piece of paper and scan it into your computer then upload it...do you have a scanning printer?

peacegirl · #**15369** 03-12-2012, 02:08 PM

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Originally Posted by Spacemonkey

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Originally Posted by thedoc

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Originally Posted by Spacemonkey

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Originally Posted by thedoc

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Originally Posted by peacegirl

If you can't control yourself, I'm not going to talk to you.

Yes Spacemonkey, don't you know that you shouldn't ask questions that Peacegirl can't answer or that expose her's and Lessans ignorance?

Indeed. Otherwise she'll ignore my questions instead of just not answering them.

What's the difference, you get no response either way? But in reality she has no coherent response, just nonsense.

Exactly. There is no difference, yet somehow her ignoring my questions instead of dodging them is supposed to constitute some kind of punishment for speaking the harsh truth about her father's abilities.

It has nothing to do with punishment, but I will not tolerate you calling him names like idiot. You'll regret it one day when he turns out to be right.

LadyShea · #**15370** 03-12-2012, 02:10 PM

You will rue the day! :melo:

peacegirl · #**15371** 03-12-2012, 02:28 PM

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Originally Posted by LadyShea

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Originally Posted by peacegirl

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Originally Posted by LadyShea

Where are the actual photons with the specific wavelengths that weren't absorbed now?

When you say they spread out "until there is no more (P) light", do you mean they cease existing at that time? They cease traveling at that time? They cease to have their specific wavelength? They rejoin the other wavelengths that do travel? Where are they and what are they doing now?

You are still confused. These photons are not traveling. White light is traveling. The object is revealing, therefore these photons do not continue. The light fades out when the object cannot be seen. Do you see what you're doing LadyShea? You are saying that this light cannot fade out because light continues on. That is true, but it's white light that continues on, not P light.

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Originally Posted by LadyShea

Photons cannot stop existing, so the specific photons must be somewhere. Photons must have a wavelength. Do you understand what light is and what it's properties are? Do you understand frequencies and wavelength at all?

You're still not getting it, and the reason you're not getting it is because you are thinking in terms of the blue light traveling. It is not. It is revealing. We cannot resolve the image if the object is not present. As white light passes over substance, it absorbs some of it and the remaining light allows us to see that substance. It does not travel although the photons are being replaced as white light continues to strike the object. When we can no longer see the object, or material substance, because it cannot be resolved on the film/retina, since it's too far away, white light appears. The blue light now becomes part of white light which does travel through space and time.

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Originally Posted by LadyShea

All I am "doing" is trying to get you to explain your model in a way that doesn't break the laws of physics and doesn't change the known properties of light.

That's what I'm trying to do as best I can.

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They are somewhere, they get spread out until they no longer reveal the object at that distance.

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Originally Posted by LadyShea

Where?

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They don't spread out beyond the visual range because this (P) light is revealing the object, it is not bringing the pattern of the light to the eye over long distances. That's where science got it wrong.

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Originally Posted by LadyShea

Where are they? I am not asking about eyes and visual ranges, I am asking about light. Where are the photons that were not absorbed?

They are present LadyShea, but when they get beyond visual range, they become white light because the blue light is not bouncing off of the object and traveling. To understand this, you have to get that picture out of your mind.

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These photons do not bounce and travel, therefore when they fade out (which means they are not resolved on the film/retina), white light continues on.

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Originally Posted by LadyShea

That makes no sense.

It actually does make sense if you understand that the blue light doesn't travel. It becomes a condition of sight. Because of the blue light we are able to see the blue ball in real time.

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To you it doesn't, because you believe these photons bounce and travel. But they don't. You're never going to understand this model because you refuse to budge off of your position.

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Originally Posted by LadyShea

"Fading out" (becoming less intense due to divergence) only happens when something is traveling. If they aren't traveling, they aren't fading out. They have a location whether they are or are not traveling, unless they've ceased to exist. WHAT IS THE LOCATION? Where are they?

I told you that this does not violate physics because light energy is continually in motion even when it's passing over the material world. The fact that photons are dispersing, which cause less to show up on the film/retina, does not mean that these blue photons are traveling. It is still white light that is traveling, but it gets interrupted, so to speak, by the object that absorbs part of that light, which allows the other part to become a condition that will reveal the object if it is in range. This in no way means that white light isn't intact and stays intact as it travels.

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Originally Posted by LadyShea

If you start making sense, and stop contradicting yourself, maybe I'll understand what you are positing here.

I'm not contradicting myself, so please don't make me the bad guy here.

peacegirl · #**15372** 03-12-2012, 02:38 PM

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Originally Posted by LadyShea

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Originally Posted by peacegirl

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Originally Posted by LadyShea

Bump

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Originally Posted by LadyShea

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Originally Posted by peacegirl

The answer given was not adequate and nobody countered it with a better

I did not make that up. So if that wasn't the answer, then give me the answer as to why we cannot get an image of a person (a real human being with substance) when he steps out of visual range but he is in a straight line with the observer. According to the current belief, the person should be reflecting said light toward the film/retina, and we should get an image. Tell me why we don't. The inverse square law isn't the reason because the person isn't that far from the observer to be the cause.

I just posted it, plus all the many links to it, a few posts up. Here it is again, in red and broken down in case a paragraph is too much to read. "Too far" depends on the sensor/detector. What may be too far for human eyes, isn't too far for an Eagle's eyes, or what is too far for a telephoto lens isn't too far for a super telephoto lens, etc.

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Originally Posted by LadyShea

If clicking links is beyond your ability, here's the answer yet again. What is inadequate?

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Originally Posted by Spacemonkey

imagine the camera as having thousands of individual detectors each corresponding to an individual pixel of its resolution. The resulting image is a collection of dots, each of which has to be one specific color.

Stick a red ball right in front of the camera, and every detector will see red, and the photo will be all red.

Progressively move the ball away from the camera, and the outer detectors cease to see red, with only a progressively smaller group of central detectors seeing red, such that we get a smaller and smaller red circle in the center of the photo.

Eventually the red circle gets smaller than the size of the single central detector, such that all the other detectors are not detecting red, and this one central detector is receiving more non-red light (from the areas around the ball) than red light (from the ball itself).

At that point the ball will represent a smaller part of the image than the smallest detector, and the ball will cease to show up on the image. That central detector will have a decision to make as to whether or not to create a red dot in the image, and as it is receiving more non-red than red light, it will not indicate red. So even though the camera is still receiving (a small amount) of red light from the ball, the resulting image will not show the ball at all.

Too far does have to do with the sensor LadyShea, but it also has to do with the object. I can't believe how little you have understood.

It has nothing to do with the object, only the detector...configuration, sensitivity, size, etc..

But in efferent vision, you cannot just detect light without the object, because it would show up as white light, no image. And no one has answered my question, which is why does an image not show up on film if a person steps out of visual range slightly, but is in a straight line with the lens of a camera, but when he steps within range, his image is resolved on film. This proves that it's not only light that is necessary here. It is the substance that must be within the camera's field of view.

Vivisectus · #**15373** 03-12-2012, 02:46 PM

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The fact that photons are dispersing, which cause less to show up on the film/retina, does not mean that these blue photons are traveling.

That is a contradiction right there. For particles to disperse they have to travel.

thedoc · #**15374** 03-12-2012, 02:50 PM

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Originally Posted by Angakuk

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Originally Posted by thedoc

Well, enough of this frivolity and mirth, it's back to the Piano. I was playing the first movement of 'Moonlight Sonata' and totally blanked out at the last 2 measures, got to work on that.

What is this "Moonlight" of which you speak? Are you introducing yet another kind of light? Is it not enough that we have to deal with (N) light and (P) light, now we have to contend with (M) light as well? Is there no end to this profligate proliferation of different categories of light?

Ahh, I'm glad you asked. 'M' light is the kind of light that is absorbed by piano strings, and allows me to know where the keys are so that I can play the right notes. Since my piano is about 100 years old the light that reaches my ears to be intrepreted by my brain takes a long time to get there, (must be very tired light) and so my playing is a bit slow sometimes. The really interesting thing is that now, because each string 'M' reflects light at a different frequency, I am hearing many different colors when I play, but for some reason when I hit a wrong note everything goes white?

peacegirl · #**15375** 03-12-2012, 03:18 PM

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Originally Posted by peacegirl

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Originally Posted by Vivisectus

Another fun fact I just stumbled across: most monitors and TV work at a frequency of about 60 Hz. To us, that makes the image look stable, and movement fluid.

For dogs, the frequency needs to be much higher as they process images faster than we do. What looks like a stable image to us, is flickery and unreal to them, and movement very jerky.

Good try Vivisectus. So let's go back to pictures.

Quote:

Originally Posted by Vivisectus

Was that where you were explaining to me how your 5 minutes with skype and the family dog was WAY more convincing than a carefully controlled study, because of levers? I am still waiting for your explanation about how these pesky levers are such a bane to scientific testing, by the way.

It would not be a bane to scientific testing if dogs had the ability to connect the dots of recognition with the concept of pushing a lever, but that is a presupposition that hasn't been fully supported.

Quote:

Originally Posted by Vivisectus

5 minutes with skype in a situation where the dog's reaction - by your own admission! - could have been interpreted in any way you like?

Done by a person with a very clear interest in seeing a particular result?

With exactly 1 test subject? It does make the "statistics" easy to calculate, I have to say!

And for extra convenience, without finding out how your test group reacts to strangers? It DOES make things so much simpler, doesn't it?

Sure! Why don't you start by telling me all about those darn levers?

I just did. You are the one trying to see a particular result, because observation [yes, everyday observation counts; this was one example but I said nobody I know ever saw a dog recognize his master on a monitor or a photograph] is being disregarded in favor of an empirical test that was flawed in the way it was constructed.