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#1
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On Wed, 07 Apr 2004 19:27:08 -0500, Cecil Moore
wrote: I thought I was reading the frequency of visible red. I don't carry such things around in my head. Hardly any amateur radio operator does. And such is the point of my illustrating the shortfalls of your lack of experience. No one is challenging your amateur status. The recitation of any wavelength starting with a significant three is enough to set off alarms when there is a concurrent claim of its visibility. That is why I said it was impossible to be a decimal error. |
#2
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Richard Clark wrote:
The recitation of any wavelength starting with a significant three is enough to set off alarms when there is a concurrent claim of its visibility. That is why I said it was impossible to be a decimal error. Take a look at the frequency chart in the "Reference Data for Radio Engineers" and you will see why someone with poor eyesight might make that mistake. -- 73, Cecil, W5DXP -----= Posted via Newsfeeds.Com, Uncensored Usenet News =----- http://www.newsfeeds.com - The #1 Newsgroup Service in the World! -----== Over 100,000 Newsgroups - 19 Different Servers! =----- |
#3
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![]() "Cecil Moore" wrote in message ... Richard Clark wrote: The recitation of any wavelength starting with a significant three is enough to set off alarms when there is a concurrent claim of its visibility. That is why I said it was impossible to be a decimal error. The point of the thing, which you seem intent on missing, is that EM radiation is reflected by impedance discontinuities. Optical reflection is very similar to what happens at lower frequencies. I normally use this analogy when discussing the use of shielding and absorbent materials for EMI supression. |
#4
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On Thu, 8 Apr 2004 10:20:58 -0500, "Dave VanHorn"
wrote: "Cecil Moore" wrote in message ... Richard Clark wrote: The recitation of any wavelength starting with a significant three is enough to set off alarms when there is a concurrent claim of its visibility. That is why I said it was impossible to be a decimal error. The point of the thing, which you seem intent on missing, is that EM radiation is reflected by impedance discontinuities. Optical reflection is very similar to what happens at lower frequencies. I normally use this analogy when discussing the use of shielding and absorbent materials for EMI supression. Hi Dave, I certainly am missing something from this post. How do you get from my comment about visible wavelengths to one about shielding EMI? 73's Richard Clark, KB7QHC |
#5
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Richard Clark wrote:
I certainly am missing something from this post. How do you get from my comment about visible wavelengths to one about shielding EMI? Dave is probably missing the fact that you like to harp and pick the same nit for weeks before you get it out of your system. :-) -- 73, Cecil, W5DXP |
#6
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![]() "Richard Clark" wrote in message ... On Thu, 8 Apr 2004 10:20:58 -0500, "Dave VanHorn" wrote: "Cecil Moore" wrote in message ... Richard Clark wrote: The recitation of any wavelength starting with a significant three is enough to set off alarms when there is a concurrent claim of its visibility. That is why I said it was impossible to be a decimal error. The point of the thing, which you seem intent on missing, is that EM radiation is reflected by impedance discontinuities. Optical reflection is very similar to what happens at lower frequencies. I normally use this analogy when discussing the use of shielding and absorbent materials for EMI supression. Hi Dave, I certainly am missing something from this post. How do you get from my comment about visible wavelengths to one about shielding EMI? By the intervening sentence, about the similarity of optical reflection to reflection at lower wavelengths. |
#7
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Hi Dave,
On Tue, 06 Apr 2004 11:48:53 GMT, Dave Shrader wrote: My EM guys, Physics types, [from my working days] indicated that the three dielectric interfaces, adhesive to glass to adhesive, all with different dielectric coefficients create reflections at the boundaries. How much reflection is there from a device with sub wavelength dimension? Cecil can't answer this, can you? Let's cast this back to optics: You have a mirror with 60nm sides (a similar, proportional scale to the glass mount). The question becomes, how much light (percentage or dB) is reflected? For others following this: would you use this mirror to shave? ;-) 73's Richard Clark, KB7QHC |
#8
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Richard Clark wrote:
How much reflection is there from a device with sub wavelength dimension? Cecil can't answer this, can you? OK, genius, please tell us how much reflection is there from a Through-Glass Antenna/Ford Taurus. -- 73, Cecil, W5DXP |
#9
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On Thu, 08 Apr 2004 10:07:58 -0500, Cecil Moore
wrote: Take a look at the frequency chart in the "Reference Data for Radio Engineers" and you will see why someone with poor eyesight might make that mistake. I duly note at the top of the graphic the expanded "Visible Spectrum in Microns (Micrometers)" which clearly offers 4 cardinal points none of which has a leading three (3) as a wavelength specification. Now "clear" may be an unfortunate choice of wording, but the separation of topic which is part and parcel to the issue of glare with its corresponding distinct and uncluttered typeface offers far less ambiguity. [Others may wish to observe the 610nm specification offered as "red" in other correspondence is distinctly yellow here. Such is the plight of subjectivity and the illusion of human perception.] As glare is another illusion of human perception of visible light it follows that there is no specification of it, nor visibility for any wavelength starting with a 3. It matters little how many decimal points you slipped on the occasion of visiting the extreme lower edge of the infrared. If you had misquoted 6.35 MILLION Angstroms instead of 6350 Angstroms that would have passed with little comment. All of this is commonplace to a practitioner of the art of Optics and OptoElectronics. This, then, returns us to the topic of through-glass attachments, their loss, and the contribution of the layers to reflect (a la glare suppression) which you re-introduced to this thread, above. [I will suspend the absurdity of this logic for the moment.] What is the resonant frequency of this adhesive layer: in wavelengths, frequency, or color? I think we can all agree (barring the slipped decimal place) that it is not 2M nor 440 MHz. I will go one step beyond and ask, if this geometry of attachment is variable through curvatures (windshield are always curved) what are the prospects of Newton's Rings offering a variation in that determination? These are all garden variety questions that plague newbies to the art. |
#10
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Richard Clark wrote:
All of this is commonplace to a practitioner of the art of Optics and OptoElectronics. And completely irrelevant to the subject of matches in transmission lines. Thus, it is obviously only a nit-picking logical diversion on your part to avoid discussing match points on transmission lines. -- 73, Cecil, W5DXP |
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