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The dish problem
Hi All,
Modern engineering text indicates that for the same frequency, a large fully illuminated dish will provide more gain than it’s smaller equivalent. Why is this ? Consider the following scenario: ====== Take a 10GHz RF source, and attach an appropriate feed horn to fully illuminate a 2m dia dish. Adjust focus so that the wave front out of the dish is parallel (2m diameter ?) At a distance, arrange the same setup, except the RF source is replaced by a detector. Make a note of received signal strength. ====== Repeat the same experiment at the same frequency, but with a 3mdiameter dish at each end, and appropriate feed horns . Make a note of received signal strength. ====== Also, see http://www.idesignz.org/misc/dishproblem.pdf In both cases all of the TX energy is transmitted in a parallel beam, whose diameter is the same as the respective dish. It would seem that if you can fully convert all of the TX RF into parallel waves, at some point the diameter of the dishes should not make any difference to the transfer efficiency of the system. So, why would the received signal for the 3m + 3m dish be greater than that of the 2m + 2m dish experiment ? Thoughts please. Mark http://www.idesignz.org/AMPS/AMPS_BS.html – Experimental AMPS micro Base Station project http://www.idesignz.org/UAV/index.html – Early work on the four rotor FlyingThingy |
The dish problem
On Mon, 12 Oct 2009 21:45:39 -0700 (PDT), MarkAren
wrote: Hi All, Modern engineering text indicates that for the same frequency, a large fully illuminated dish will provide more gain than it’s smaller equivalent. Why is this ? Consider the following scenario: ====== Take a 10GHz RF source, and attach an appropriate feed horn to fully illuminate a 2m dia dish. Adjust focus so that the wave front out of the dish is parallel (2m diameter ?) At a distance, arrange the same setup, except the RF source is replaced by a detector. Make a note of received signal strength. ====== Repeat the same experiment at the same frequency, but with a 3mdiameter dish at each end, and appropriate feed horns . Make a note of received signal strength. ====== Also, see http://www.idesignz.org/misc/dishproblem.pdf In both cases all of the TX energy is transmitted in a parallel beam, whose diameter is the same as the respective dish. A parallel beam is not formed, rather a slightly expanding beam due to diffraction. Diffraction also limits the resolving power of an astronomical telescope, so when looking at some binary stars, a small telescope will show only a single blob, a slightly larger telescope will show an elongated figure (e.g. figure of 8) and an even larger telescope will show two separate stars. The diffraction limit is defined as 1.22*wavelength/diemeter radians or about 70*wavelength/diameter degrees. Diffraction also controls the beam spreading from a parabolic disk or laser. For this reason, a laser (with an aperture less than 1 cm) can not be used to illuminate the reflectors on the moon, but typically the laser beam is transmitted through a telescope with typically 1 m diameter. The beam is 100-1000 times narrower than the beam from the laser alone. The area illuminated on the moon is 10,000-1,000,000 times smaller and hence reflected power that much stronger than with a bare laser. Paul OH3LWR |
The dish problem
Hey OM:
Yes how is it? Bigger is better. I mean you are using the same source for both antennas but how does the larger one have all that extra power, being emitted and where does all that extra power come from? So instead of using that 400 watt microwave oven I just get me a 30 foot dish and a 20 milliwatt emitter and save all that money I would have wasted using the microwave oven, cooking my turkey. 73 OM de n8zu On Oct 13, 12:45 am, MarkAren wrote: Hi All, Modern engineering text indicates that for the same frequency, a large fully illuminated dish will provide more gain than it’s smaller equivalent. Why is this ? Consider the following scenario: ====== Take a 10GHz RF source, and attach an appropriate feed horn to fully illuminate a 2m dia dish. Adjust focus so that the wave front out of the dish is parallel (2m diameter ?) At a distance, arrange the same setup, except the RF source is replaced by a detector. Make a note of received signal strength. ====== Repeat the same experiment at the same frequency, but with a 3mdiameter dish at each end, and appropriate feed horns . Make a note of received signal strength. ====== Also, seehttp://www.idesignz.org/misc/dishproblem.pdf In both cases all of the TX energy is transmitted in a parallel beam, whose diameter is the same as the respective dish. It would seem that if you can fully convert all of the TX RF into parallel waves, at some point the diameter of the dishes should not make any difference to the transfer efficiency of the system. So, why would the received signal for the 3m + 3m dish be greater than that of the 2m + 2m dish experiment ? Thoughts please. Mark http://www.idesignz.org/AMPS/AMPS_BS.html– Experimental AMPS micro Base Station project http://www.idesignz.org/UAV/index.html– Early work on the four rotor FlyingThingy |
The dish problem
Paul Keinanen wrote:
On Mon, 12 Oct 2009 21:45:39 -0700 (PDT), MarkAren wrote: Hi All, Modern engineering text indicates that for the same frequency, a large fully illuminated dish will provide more gain than it’s smaller equivalent. Why is this ? ... In both cases all of the TX energy is transmitted in a parallel beam, whose diameter is the same as the respective dish. **** A parallel beam is not formed, rather a slightly expanding beam due to diffraction. Diffraction also limits the resolving power of an astronomical telescope, so when looking at some binary stars, a small telescope will show only a single blob, a slightly larger telescope will show an elongated figure (e.g. figure of 8) and an even larger telescope will show two separate stars. The diffraction limit is defined as 1.22*wavelength/diemeter radians or about 70*wavelength/diameter degrees. Diffraction also controls the beam spreading from a parabolic disk or laser. For this reason, a laser (with an aperture less than 1 cm) can not be used to illuminate the reflectors on the moon, but typically the laser beam is transmitted through a telescope with typically 1 m diameter. The beam is 100-1000 times narrower than the beam from the laser alone. The area illuminated on the moon is 10,000-1,000,000 times smaller and hence reflected power that much stronger than with a bare laser. Paul OH3LWR Great response! Brian W |
The dish problem
raypsi wrote:
Hey OM: Yes how is it? Bigger is better. I mean you are using the same source for both antennas but how does the larger one have all that extra power, being emitted and where does all that extra power come from? So instead of using that 400 watt microwave oven I just get me a 30 foot dish and a 20 milliwatt emitter and save all that money I would have wasted using the microwave oven, cooking my turkey. 73 OM de n8zu It comes from "focusing" the existing energy. No new energy is created. Yes, you *could* do that with a turkey. Just like you *could* cook it with a magnifying glass out in the sun. :) -Bill |
The dish problem
Hey OM:
Take a 60db gain dish illuminate that with 20 milliwatts of power. Your ERP is now 20,000 watts. My question raises the bar, how can you focus 20 milliwatts of EM particles and waves, to end up with 20,000 watts of EM particles and waves? And I can bet that it's all those extra particles that are going to be cooking my turkey. There's only one way that can happen: the dish is a storage device: the size and frequency of which determine the storage capacity in this case the storage capacity gives me a 60db gain. And will always give you 60db more than you put in. 73 OM de n8zu It comes from "focusing" the existing energy. No new energy is created. Yes, you *could* do that with a turkey. Just like you *could* cook it with a magnifying glass out in the sun. :) -Bill |
The dish problem
On Oct 15, 3:02*pm, raypsi wrote:
Hey OM: Take a 60db gain dish illuminate that with *20 milliwatts of power. Your ERP is now 20,000 watts. My question raises the bar, how can you focus 20 milliwatts of EM particles and waves, to end up with 20,000 watts of EM particles and waves? And I can bet that it's all those extra particles that are going to be cooking my turkey. There's only one way that can happen: the dish is a storage device: the size and frequency of which determine the storage capacity in this case the storage capacity gives me a 60db gain. And will always give you 60db more than you put in. 73 OM de n8zu 60db in one specific direction more than if same amount of RF was dissipated equally in all directions. Correct? Would not matter if starting with one milliwatt, one watt, one megawatt. The beaming effect will concentrate the available energy from all directions in a specific one. |
The dish problem
raypsi writes:
Hey OM: Take a 60db gain dish illuminate that with 20 milliwatts of power. Your ERP is now 20,000 watts. EIRP, actually ... Radiated power, relative to a (abstract) isotropic radiator. My question raises the bar, how can you focus 20 milliwatts of EM particles and waves, to end up with 20,000 watts of EM particles and waves? You can't. You're just focusing that 20 mW of power in the "useful direction" Imagine an incandescent light bulb hanging from a wire -- this is a reasonable metaphor for an isotropic radiator - it puts out (nearly) the same amount of light in all directions ... If the light isn't bright enough to read a book, you might put a reflector behind it - there is no more light being generated, but the page just got brighter, because some of the energy that was illuminating the garage roof, your head, the walls and the junk on the shelves is now illuminating the page. And I can bet that it's all those extra particles that are going to be cooking my turkey. There are no particles, and there are no "extra" particles. de n8zu It comes from "focusing" the existing energy. No new energy is created. Yes, you *could* do that with a turkey. Just like you *could* cook it with a magnifying glass out in the sun. :) Indeed - first you heat one molecule of water, then refocus the beam, heat the next, repeat 6 x 10**23 times for each mole of turkey. -Bill |
The dish problem
Hey OM
What you are saying is that I can take a 20 mW emitter and with an infinite size dish I can get infinite power from a 20 mW emitter? No way you can get infinite power from a 20 mW source unless the dish is supplying the extra power. In other words it's not possible to get infinite power focused from a 20 mW source, and that all that power is coming from the 20 mW emitter. Take that 20 mW emitters' isotropic pattern, there is a potential of an infinite amount of power in that pattern because it can be focused? Frying that turkey only takes 20 mW of total power period. And it may or may not be from focusing the 20 mW. The turkey gets cooked with 20 mW is still a fact that can't be refutted. 73 OM de n8zu You can't. *You're just focusing that 20 mW of power in the "useful direction" * -Bill- Hide quoted text - - Show quoted text - |
The dish problem
raypsi wrote:
Hey OM What you are saying is that I can take a 20 mW emitter and with an infinite size dish I can get infinite power from a 20 mW emitter? You can get infinite *Effective Isotropic Radiated Power* from 20 mW with an infinite dish. Not practical, but theoretically correct. You're misunderstanding EIRP, Effective Isotropic Radiated Power. If you feed a 60 dB parabolic antenna with 20 mW you get the equivalent of feeding a theoretical *isotropic* radiator with 20 kW, the key word being 'isotropic,' meaning that 20 kW is being radiated in all directions. Your 60 dB dish directs your 20 mW *in a specific direction* with a *specific* beam width. *In that beam* the field strength is the same as it would be from 20 kW *going in every direction*. The ratio of every direction to your dish's beam width is 60 dB. No way you can get infinite power from a 20 mW source unless the dish is supplying the extra power. In other words it's not possible to get infinite power focused from a 20 mW source, and that all that power is coming from the 20 mW emitter. You're not getting infinite power, you're getting 20 mW focused into a finite beam width and in that beam width it has the same field strength that 20 kW going in all directions would give you. Take that 20 mW emitters' isotropic pattern, there is a potential of an infinite amount of power in that pattern because it can be focused? Only if it can be focused into an infinitely small point which you won't get from any finite gain antenna. Frying that turkey only takes 20 mW of total power period. And it may or may not be from focusing the 20 mW. The turkey gets cooked with 20 mW is still a fact that can't be refutted. Part of that fact is it will take a long, long time to cook it with only 20 mW because you can only cook (defined as raising the temperature to about 160 F) a very, very small portion of the turkey at a time with only 20 mW. - W8LNA 73 OM de n8zu You can't. You're just focusing that 20 mW of power in the "useful direction" -Bill- Hide quoted text - - Show quoted text - |
The dish problem
Hi Mark,
I looked over the thread that your note kicked off. Unusually, this was not very informative. People were sticking with antenna measures, which weren't helping much in this case. I'll give it a try, introducing the missing concept: energy (and power) DENSITY. If you sit in the bright sun for 2 hours, you are likely to get a sunburn. If you sit in a dark tent, with an opening for a magnifying glass of 2 inch diameter ( = 5 cm), you will get a small burn within 2 minutes. Looking now at the numbers, you know that the effective power density of the Sun at the surface is 700 watts per square meter. Your skin represents say 1.5 sq meters, so you are getting a total power of say 1 kilowatt. So we are suggesting, at a particular place it takes 1 kilowatt of solar power for 2 hours to burn you all over. Now lets look at that dark tent with a hole for a magnifying glass: the power density at the glass is pi times 1 times 1 inches or pi X .025 X 0.025 square meters = 0.002 sq meters. The Solar power at the glass is 700w X 0.002 m^2 = 1.4 watts How the hell can 1.4 watts burn us in 2 minutes, but 1000 watts cannot burn us in less than 2 hours? Answer power density: That 1.4 watts is focussed on a spot that is 1 mm diameter ( 40 / 1000 inch diameter) so its power density is 1.4 / pi X 0.001 X 0.001 watts per sq meter ....and THAT is 446 kilowatts /sq meter. Oh look: the power density of the spot in the dark tent is about SIXTY times as great as plain sunlight, so waddaya know, it burns us 60 times as fast (but only over a LITTLE spot.) And that's what this thread was all about - the effects of focussing energy over a particular area. I hope this helps. Brian W MarkAren wrote: Hi All, Modern engineering text indicates that for the same frequency, a large fully illuminated dish will provide more gain than it’s smaller equivalent. Why is this ? Consider the following scenario: ====== Take a 10GHz RF source, and attach an appropriate feed horn to fully illuminate a 2m dia dish. Adjust focus so that the wave front out of the dish is parallel (2m diameter ?) At a distance, arrange the same setup, except the RF source is replaced by a detector. Make a note of received signal strength. ====== Repeat the same experiment at the same frequency, but with a 3mdiameter dish at each end, and appropriate feed horns . Make a note of received signal strength. ====== Also, see http://www.idesignz.org/misc/dishproblem.pdf In both cases all of the TX energy is transmitted in a parallel beam, whose diameter is the same as the respective dish. It would seem that if you can fully convert all of the TX RF into parallel waves, at some point the diameter of the dishes should not make any difference to the transfer efficiency of the system. So, why would the received signal for the 3m + 3m dish be greater than that of the 2m + 2m dish experiment ? Thoughts please. Mark http://www.idesignz.org/AMPS/AMPS_BS.html – Experimental AMPS micro Base Station project http://www.idesignz.org/UAV/index.html – Early work on the four rotor FlyingThingy |
The dish problem
On Oct 16, 5:08*pm, brian whatcott wrote:
Hi Mark, I looked over the thread that your note kicked off. Unusually, this was not very informative. People were sticking with antenna measures, which weren't helping much in this case. I'll give it a try, introducing the missing concept: * energy (and power) DENSITY. If you sit in the bright sun for 2 hours, you are likely to get a sunburn.. If you sit in a dark tent, with an opening for a magnifying glass of 2 inch diameter ( = 5 cm), you will get a small burn within 2 minutes. Looking now at the numbers, you *know that the effective power density of the Sun at the surface is 700 watts per square meter. Your skin represents say 1.5 sq meters, so you are getting a total power of say 1 kilowatt. So we are suggesting, at a particular place it takes 1 kilowatt of solar power for 2 hours to burn you all over. Now lets look at that dark tent with a hole for a magnifying glass: the power density at the glass is pi times 1 times 1 inches or pi X .025 X 0.025 *square meters = 0.002 sq meters. The Solar power at the glass is 700w X *0.002 m^2 = 1.4 watts How the hell can 1.4 watts burn us in 2 minutes, but 1000 watts cannot burn us in less than 2 hours? Answer power density: That 1.4 watts is focused on a spot that is 1 mm diameter ( 40 / 1000 inch diameter) so its power density is * 1.4 */ pi X 0.001 X 0.001 watts per sq meter ...and THAT is 446 kilowatts /sq meter. Oh look: the power density of the spot in the dark tent is about SIXTY times as great as plain sunlight, so whaddaya know, it burns us 60 times as fast (but only over a LITTLE spot.) And that's what this thread was all about - the effects of focusing energy over a particular area. I hope this helps. Brian W MarkAren wrote: Hi All, Modern engineering text indicates that for the same frequency, a large fully illuminated dish will provide more gain than it’s smaller equivalent. Why is this ? Consider the following scenario: ====== Take a 10GHz RF source, and attach an appropriate feed horn to fully illuminate a 2m dia dish. Adjust focus so that the wave front out of the dish is parallel (2m diameter ?) At a distance, arrange the same setup, except the RF source is replaced by a detector. Make a note of received signal strength. ====== Repeat the same experiment at the same frequency, but with a 3m diameter dish at each end, and appropriate feed horns . Make a note of received signal strength. ====== Also, seehttp://www.idesignz.org/misc/dishproblem.pdf In both cases all of the TX energy is transmitted in a parallel beam, whose diameter is the same as the respective dish. It would seem that if you can fully convert all of the TX RF into parallel waves, at some point the diameter of the dishes should not make any difference to the transfer efficiency of the system. So, why would the received signal for the 3m + 3m dish be greater than that of the 2m + 2m dish experiment ? Thoughts please. Mark http://www.idesignz.org/AMPS/AMPS_BS.html– Experimental AMPS micro Base Station project http://www.idesignz.org/UAV/index.html– Early work on the four rotor FlyingThingy- Hide quoted text - - Show quoted text - A good read thank you. A bit lengthy; but well done. While illuminating the subject it does focus attention on the important point; the 'beaming' or directionalizing of the available energy, whether it be sunlight or any other frequency. However there will always be some who 'can't' or 'won't' understand. Witness some of the nonsense that is posted on various news groups all the time. PS. That 700 watts per sq. metre is an interesting number? Bright sun at midday at the equator perhaps? But I doubt that here at around 48 degrees north and with our climate we get anything close to that? Also now wondering how efficient a solar panel is at converting that 700 watts into electric power (for say home use). Which may be why some 'solar' collectors are both photo-voltaic and heat collecting. |
The dish problem
terry wrote:
.... Looking now at the numbers, you know that the effective power density of the Sun at the surface is 700 watts per square meter. Your skin represents say 1.5 sq meters, so you are getting a total power of say 1 kilowatt. ..... That 700 watts per sq. metre is an interesting number? Bright sun at midday at the equator perhaps? But I doubt that here at around 48 degrees north and with our climate we get anything close to that? Also now wondering how efficient a solar panel is at converting that 700 watts into electric power (for say home use). Which may be why some 'solar' collectors are both photo-voltaic and heat collecting. The scientific name for this measure is Solar Irradiance. If you hold up a square frame 1 meter on a side, perpendicular to the Sun's direction, in space near Earth's orbital distance, about 1400 watts passes the frame - and twice as far from the Sun, that frame only frames a quarter of that amount (of course!) That number 700 W/m^2 takes into account the absorption of power by stuff in the atmosphere, more at slant angles of course. This URL can show the kind of Solar power you can expect at various geographical coordinates. http://re.jrc.ec.europa.eu/pvgis/apps/radday.php You'll see a peak value around noon - and this can be more than 700 W/m^2 at that 48N latitude. Brian W |
The dish problem
Hey OM
If I look at the node of a source of 20mW and connect 10 wires to that node with equal loads, I will have 2mW going down each wire. No matter how I try and add them up I can't get more than 20 mW. Now I take a 20 mW isotropic emitter, now your'e saying I feed a dish with that 20mW emitter through a feed horn and now all of a sudden I have gain, I have more than 20 mW? Because the dish and feed horn are able to add the particles and waves, where as with 10 wires I cannot add the waves and particles? The only difference in both these are the shape of the dish and the medium used to convey the wave, particles. And you can add the waves particles until you run out of dish. 73 OM de n8zu On Oct 18, 8:40 pm, brian whatcott wrote: terry wrote: ... Looking now at the numbers, you know that the effective power density of the Sun at the surface is 700 watts per square meter. Your skin represents say 1.5 sq meters, so you are getting a total power of say 1 kilowatt. .... That 700 watts per sq. metre is an interesting number? Bright sun at midday at the equator perhaps? But I doubt that here at around 48 degrees north and with our climate we get anything close to that? Also now wondering how efficient a solar panel is at converting that 700 watts into electric power (for say home use). Which may be why some 'solar' collectors are both photo-voltaic and heat collecting. The scientific name for this measure is Solar Irradiance. If you hold up a square frame 1 meter on a side, perpendicular to the Sun's direction, in space near Earth's orbital distance, about 1400 watts passes the frame - and twice as far from the Sun, that frame only frames a quarter of that amount (of course!) That number 700 W/m^2 takes into account the absorption of power by stuff in the atmosphere, more at slant angles of course. This URL can show the kind of Solar power you can expect at various geographical coordinates.http://re.jrc.ec.europa.eu/pvgis/apps/radday.php You'll see a peak value around noon - and this can be more than 700 W/m^2 at that 48N latitude. Brian W |
The dish problem
raypsi wrote:
Hey OM If I look at the node of a source of 20mW and connect 10 wires to that node with equal loads, I will have 2mW going down each wire. No matter how I try and add them up I can't get more than 20 mW. Now I take a 20 mW isotropic emitter, now your'e saying I feed a dish with that 20mW emitter through a feed horn and now all of a sudden I have gain, I have more than 20 mW? Because the dish and feed horn are able to add the particles and waves, where as with 10 wires I cannot add the waves and particles? The only difference in both these are the shape of the dish and the medium used to convey the wave, particles. And you can add the waves particles until you run out of dish. 73 OM de n8zu Actually your example tends to prove the point that you're questioning. 10 points of 2mw DO add up to 20 mw. The difference with an isotropic radiator is that there are an infinite number of points each at say 20mw. Add 'em up by focusing those points into a single path. If you can rationalize a 20mw transmitter feeding one direction down a feedline or waveguide and still getting 20mw EIRP in ALL directions with an isotropic radiator then it should make sense. -Bill |
The dish problem
raypsi wrote:
Hey OM If I look at the node of a source of 20mW and connect 10 wires to that node with equal loads, I will have 2mW going down each wire. No matter how I try and add them up I can't get more than 20 mW. Correct. Now I take a 20 mW isotropic emitter, now you're saying I feed a dish with that 20mW emitter through a feed horn and now all of a sudden I have gain, I have more than 20 mW? Not exactly, I am saying you have a gain in power DENSITY i.e watts per sq meter. In a desired direction only, naturally. Like when you use a magnifying glass to collect the Sun's power into a small spot. 73 OM de n8zu |
The dish problem
Hey OM:
I tell you what. I took just 1 particle wave and shoot the particle wave into a 60db dish. At the focus I got 1,000,000 partivle waves. I mean that is 60 db right? So where did the other 999,999 particles come from? Energy density from focusing just one particle? The two slot experiment buttresses the above. Take 2 parallel slots shoot just one particle wave into just one slot and 2 particle waves come out the other side. 73 OM de n8zu |
The dish problem
Take a small light bulb, it shines light in many directions.
Place this at the focal point of a mirrored parabolic dish and the light is predominantly directed in one direction. In the direction that the dish is pointing, and at a distance, measure the received light using a light meter. Remove the dish, and at the same distance, measure the received light using a light meter from the naked bulb. One measurement will be higher than the other. Why do you think this might be ? On Oct 24, 5:24*am, raypsi wrote: Hey OM: I tell you what. I took just 1 particle wave and shoot the particle wave into a 60db dish. At the focus I got 1,000,000 partivle waves. I mean that is 60 db right? So where did the other 999,999 particles come from? Energy density from focusing just one particle? The two slot experiment buttresses the above. Take 2 parallel slots shoot just one particle wave into just one slot and 2 particle waves come out the other side. 73 OM de n8zu |
The dish problem
raypsi wrote:
Hey OM: I tell you what. I took just 1 particle wave [ per sq meter] and shoot the particle wave into a 60db dish. At the focus I got 1,000,000 particle waves [ per sq meter]. I mean that is 60 db right? Yes. So where did the other 999,999 particles come from? Energy density from focusing just one particle? NOW, you're getting it! No more particles, but more particle density, when the flux is directed tightly... The two slot experiment buttresses the above. Take 2 parallel slots shoot just one particle wave into just one slot and 2 particle waves come out the other side. 73 OM de n8zu Actually no. But you probably realized that. Or did you? I am not pursuing this much further, because you are starting to respond like a person who wants to drag out an argument. Listen to me when I say: "I can be wrong. I often am." Can you say that too, with some sincerity? :-) Brian W :-) |
The dish problem
Nice concrete example, Mark.
The same conclusion holds if the bulb is so dim, that it only emits one photon at intervals. Brian W MarkAren wrote: Take a small light bulb, it shines light in many directions. Place this at the focal point of a mirrored parabolic dish and the light is predominantly directed in one direction. In the direction that the dish is pointing, and at a distance, measure the received light using a light meter. Remove the dish, and at the same distance, measure the received light using a light meter from the naked bulb. One measurement will be higher than the other. Why do you think this might be ? On Oct 24, 5:24 am, raypsi wrote: Hey OM: I tell you what. I took just 1 particle wave and shoot the particle wave into a 60db dish. At the focus I got 1,000,000 partivle waves. I mean that is 60 db right? So where did the other 999,999 particles come from? Energy density from focusing just one particle? The two slot experiment buttresses the above. Take 2 parallel slots shoot just one particle wave into just one slot and 2 particle waves come out the other side. 73 OM de n8zu |
The dish problem
Hey OM By golly ole timer you hit the nail on the head. I don't pay no stinking taxes, in the W-4 form I put "EXEMPT" where it asks for number of deductions, and when I file my 1040 it says at the end of the form REFUND DUE, in four figures, for the last several years. Now that is legit, but it sure is WRONG, at least that is what everyone tells me. Your trouble is you pay taxes, where as I don't. Now that is energy density. 73 OM de n8zu On Oct 23, 8:25 pm, brian whatcott wrote: raypsi wrote: Hey OM: I tell you what. I took just 1 particle wave [ per sq meter] and shoot the particle wave into a 60db dish. At the focus I got 1,000,000 particle waves [ per sq meter]. I mean that is 60 db right? Yes. So where did the other 999,999 particles come from? Energy density from focusing just one particle? NOW, you're getting it! No more particles, but more particle density, when the flux is directed tightly... The two slot experiment buttresses the above. Take 2 parallel slots shoot just one particle wave into just one slot and 2 particle waves come out the other side. 73 OM de n8zu Actually no. But you probably realized that. Or did you? I am not pursuing this much further, because you are starting to respond like a person who wants to drag out an argument. Listen to me when I say: "I can be wrong. I often am." Can you say that too, with some sincerity? :-) Brian W :-) |
The dish problem
raypsi wrote:
Hey OM By golly ole timer you hit the nail on the head. I don't pay no stinking taxes, in the W-4 form I put "EXEMPT" where it asks for number of deductions, and when I file my 1040 it says at the end of the form REFUND DUE, in four figures, for the last several years. Now that is legit, but it sure is WRONG, at least that is what everyone tells me. Your trouble is you pay taxes, where as I don't. Now that is energy density. 73 OM de n8zu Ah, now I see the real 'problem' :) -Bill |
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