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wave polarisation
I start reading about acoustic analogy.
I found that: "Over long distances, the atmosphere can cause the polarization of a radio wave to fluctuate, so the distinction between horizontal and vertical becomes less significant." From: http://whatis.techtarget.com/definit...843762,00.html The my question a 1. What means "long distances" in km (or miles), 2. What is the best orientation of the antenna for long distances. S* |
wave polarisation
On 8 mayo, 10:35, Szczepan Białek wrote:
I start reading about acoustic analogy. I found that: "Over long distances, the atmosphere can cause the polarization of a radio wave to fluctuate, so the distinction between horizontal and vertical becomes less significant." From:http://whatis.techtarget.com/definit...843762,00.html The my question a 1. What means "long distances" in km (or miles), 2. What is the best orientation of the antenna for long distances. S* Hello, Under normal circumstances, polarization change in line-off-site conditions (think of max 40 mile) is not that much, so antenna polarization does matter (unless you use at least circular polarization on one side). In a propagation path that is dominated by multi-path effects (reflection at buildings, hills, foliage, etc), you get almost random polarization and then the polarization is not that important. Your cell phone and indoor WIFI are examples. Extreme weather conditions can also lead to polarization changes or a random polarization component (ducting superrefraction). For sea water up to VHF, reflection depends on polarization. For ground-ground links (for example ship shore) mostly vertical polarization is used (as the sea water helps in this case). So if you want to receive these communication, you use a vertical polarized antenna. The largest change in polarization you will get when the waves have to travel through the ionosphere. At HF (ground-ground link via ionosphere), the polarization vector rotates many times. This is due to Faraday rotation. Also ground-satellite links suffer from this effect. The higher the frequency, the less the change in polarization. For example at 100 MHz you should think about 30 full rotations (that is more then 10k degrees), while at 10 GHz the change in polarization will be about 1 degree. Circular polarization may help to mitigate the influence of Faraday rotation. At HF sky wave (100....1000 mile via ionosphere) polarization of the antenna matters. This is not because of the polarization change of the waves due to Faraday rotation, but because of the reflection characteristics of mother earth. In HF antennas, reflection on mother earth is used (in combination with antenna height) to get the required elevation radiation pattern of the antenna. Reflection on earth depends on polarization. Hopefully this helps you a bit. Best regards, Wim PA3DJS www.tetech.nl don't forget to remove a, b and c from the mail address |
wave polarisation
Użytkownik napisał w wiadomo¶ci ... On 8 mayo, 10:35, Szczepan Białek wrote: I start reading about acoustic analogy. I found that: "Over long distances, the atmosphere can cause the polarization of a radio wave to fluctuate, so the distinction between horizontal and vertical becomes less significant." From:http://whatis.techtarget.com/definit...843762,00.html The my question a 1. What means "long distances" in km (or miles), 2. What is the best orientation of the antenna for long distances. S* Hello, Under normal circumstances, polarization change in line-off-site conditions (think of max 40 mile) is not that much, so antenna polarization does matter (unless you use at least circular polarization on one side). In a propagation path that is dominated by multi-path effects (reflection at buildings, hills, foliage, etc), you get almost random polarization and then the polarization is not that important. Your cell phone and indoor WIFI are examples. Extreme weather conditions can also lead to polarization changes or a random polarization component (ducting superrefraction). For sea water up to VHF, reflection depends on polarization. For ground-ground links (for example ship shore) mostly vertical polarization is used (as the sea water helps in this case). So if you want to receive these communication, you use a vertical polarized antenna. The largest change in polarization you will get when the waves have to travel through the ionosphere. At HF (ground-ground link via ionosphere), the polarization vector rotates many times. This is due to Faraday rotation. Also ground-satellite links suffer from this effect. The higher the frequency, the less the change in polarization. For example at 100 MHz you should think about 30 full rotations (that is more then 10k degrees), while at 10 GHz the change in polarization will be about 1 degree. Circular polarization may help to mitigate the influence of Faraday rotation. At HF sky wave (100....1000 mile via ionosphere) polarization of the antenna matters. This is not because of the polarization change of the waves due to Faraday rotation, but because of the reflection characteristics of mother earth. In HF antennas, reflection on mother earth is used (in combination with antenna height) to get the required elevation radiation pattern of the antenna. Reflection on earth depends on polarization. Hopefully this helps you a bit. You do not use the words "transversal" and "EM". The only evidence of polarization is antenna directional sensitivity. In the acoustic analogy a radio waves are normal spherical electric waves emitted from the two sources (ends of the dipole). So the sources are polarised, not the waves. Waves interfere. Do you agree? See my topic "frequency doubling" . I am only a science hobyist. The second question was: " What is the best orientation of the antenna for long distances? For old radio antennas. Very long horizontal wire. Best regards, S* |
wave polarisation
On 9 mayo, 10:02, Szczepan Bia©©ek wrote:
U˘Żytkownik napisa©ř w ... On 8 mayo, 10:35, Szczepan Bia©řek wrote: I start reading about acoustic analogy. I found that: "Over long distances, the atmosphere can cause the polarization of a radio wave to fluctuate, so the distinction between horizontal and vertical becomes less significant." From:http://whatis.techtarget.com/definit...843762,00.html The my question a 1. What means "long distances" in km (or miles), 2. What is the best orientation of the antenna for long distances. S* Hello, Under normal circumstances, polarization change in line-off-site conditions (think of max 40 mile) is not that much, so antenna polarization does matter (unless you use at least circular polarization on one side). In a propagation path that is dominated by multi-path effects (reflection at buildings, hills, foliage, etc), you get almost random polarization and then the polarization is not that important. Your cell phone and indoor WIFI are examples. Extreme weather conditions can also lead to polarization changes or a random polarization component (ducting superrefraction). For sea water up to VHF, reflection depends on polarization. For ground-ground links (for example ship shore) mostly vertical polarization is used (as the sea water helps in this case). So if you want to receive these communication, you use a vertical polarized antenna. The largest change in polarization you will get when the waves have to travel through the ionosphere. At HF (ground-ground link via ionosphere), the polarization vector rotates many times. This is due to Faraday rotation. Also ground-satellite links suffer from this effect. The higher the frequency, the less the change in polarization. For example at 100 MHz you should think about 30 full rotations (that is more then 10k degrees), while at 10 GHz the change in polarization will be about 1 degree. Circular polarization may help to mitigate the influence of Faraday rotation. At HF sky wave (100....1000 mile via ionosphere) polarization of the antenna matters. This is not because of the polarization change of the waves due to Faraday rotation, but because of the reflection characteristics of mother earth. In HF antennas, reflection on mother earth is used (in combination with antenna height) to get the required elevation radiation pattern of the antenna. Reflection on earth depends on polarization. Hopefully this helps you a bit. You do not use the words "transversal" and "EM". The only evidence of polarization is antenna directional sensitivity. You talked about radiowaves, that are EM waves. In free space, only progapation mode is transversal (that means both E- and H-field are perpendicular to the direction of energy propagation. With regards to audio, in gas, only lossless propagation mode is longitudinal (molecule movement and pressure vectors are parallel to the direction of energy propagation) . In the acoustic analogy a radio waves are normal spherical electric waves emitted from the two sources (ends of the dipole). So the sources are polarised, not the waves. Waves interfere. Do you agree? Not agree, the waves are also polarized, that can be physically measured. Polarization is determined by the E-field vector. See my topic "frequency doubling" . I am only a science hobyist. The second question was: " What is the best orientation of the antenna for long distances? For old radio antennas. Very long horizontal wire. On UHF (for example 2450 MHz), long distance can be 20 km, but on HF 500 km is not called long distance. So the meaning of long distance depends on the frequency band. You should distinguish between the actual polarization of the antenna and the physical appearance. depending on how you feed it, a very long horizontal wire can be sensitive to vertical or horizontal polarized waves. Though the equations for acoustical waves look similar to those of EM waves, the orientation of the field components is completely different. When you require a more specific answer, you should make your question more specific. I tried to give you a general answer for the various forms of radio wave propagation. Best regards, S* Best regards, Wim PA3DJS www.tetech.nl |
wave polarisation
wrote ... On 9 mayo, 10:02, Szczepan Bia©©ek wrote: U˘Żytkownik napisa©ř w ... On 8 mayo, 10:35, Szczepan Bia©řek wrote: I start reading about acoustic analogy. I found that: "Over long distances, the atmosphere can cause the polarization of a radio wave to fluctuate, so the distinction between horizontal and vertical becomes less significant." From:http://whatis.techtarget.com/definit...843762,00.html The my question a 1. What means "long distances" in km (or miles), 2. What is the best orientation of the antenna for long distances. S* Hello, Under normal circumstances, polarization change in line-off-site conditions (think of max 40 mile) is not that much, so antenna polarization does matter (unless you use at least circular polarization on one side). In a propagation path that is dominated by multi-path effects (reflection at buildings, hills, foliage, etc), you get almost random polarization and then the polarization is not that important. Your cell phone and indoor WIFI are examples. Extreme weather conditions can also lead to polarization changes or a random polarization component (ducting superrefraction). For sea water up to VHF, reflection depends on polarization. For ground-ground links (for example ship shore) mostly vertical polarization is used (as the sea water helps in this case). So if you want to receive these communication, you use a vertical polarized antenna. The largest change in polarization you will get when the waves have to travel through the ionosphere. At HF (ground-ground link via ionosphere), the polarization vector rotates many times. This is due to Faraday rotation. Also ground-satellite links suffer from this effect. The higher the frequency, the less the change in polarization. For example at 100 MHz you should think about 30 full rotations (that is more then 10k degrees), while at 10 GHz the change in polarization will be about 1 degree. Circular polarization may help to mitigate the influence of Faraday rotation. At HF sky wave (100....1000 mile via ionosphere) polarization of the antenna matters. This is not because of the polarization change of the waves due to Faraday rotation, but because of the reflection characteristics of mother earth. In HF antennas, reflection on mother earth is used (in combination with antenna height) to get the required elevation radiation pattern of the antenna. Reflection on earth depends on polarization. Hopefully this helps you a bit. You do not use the words "transversal" and "EM". The only evidence of polarization is antenna directional sensitivity. You talked about radiowaves, that are EM waves. In free space, only progapation mode is transversal (that means both E- and H-field are perpendicular to the direction of energy propagation. With regards to audio, in gas, only lossless propagation mode is longitudinal (molecule movement and pressure vectors are parallel to the direction of energy propagation) . EM is the hydraulic analogy (by Heaviside). It is a "piece to teach" a field method. In that time the electricity was incompressble and massles. Now the electrons are compressible and have mass. We need a new analogy. It can be call the Gas analogy or Acoustic analogy. In www.tetech.nl is wrote that are many analogies for EM. In the acoustic analogy a radio waves are normal spherical electric waves emitted from the two sources (ends of the dipole). So the sources are polarised, not the waves. Waves interfere. Do you agree? Not agree, the waves are also polarized, that can be physically measured. Polarization is determined by the E-field vector. A dipole has the E-field (in electrostatics). The equations are by Gauss. The same equations we can use for the Hertz dipole. The E-field will be alternate. At long distances the frequency in receiving antennas will be twice more. See my topic "frequency doubling" . I am only a science hobyist. The second question was: " What is the best orientation of the antenna for long distances? For old radio antennas. Very long horizontal wire. On UHF (for example 2450 MHz), long distance can be 20 km, but on HF 500 km is not called long distance. So the meaning of long distance depends on the frequency band. You should distinguish between the actual polarization of the antenna and the physical appearance. depending on how you feed it, a very long horizontal wire can be sensitive to vertical or horizontal polarized waves. Though the equations for acoustical waves look similar to those of EM waves, the orientation of the field components is completely different. When you require a more specific answer, you should make your question more specific. I tried to give you a general answer for the various forms of radio wave propagation. Now is XXI century. EM is a beautiful theory from XIX century. In Tetech products no incompressible massless fluid. So the most specific and important question is: How is frequency in receiving antenna. Is it doubled? Best regards, and sorry for my style S* |
wave polarisation
"Szczepan Bialek" wrote in message ... So the most specific and important question is: How is frequency in receiving antenna. Is it doubled? no. i transmit a given frequency and that is what is received. easily measured even with simple instruments. |
wave polarisation
On 9 mayo, 20:34, Szczepan Białek wrote:
... On 9 mayo, 10:02, Szczepan Bia©©ek wrote: U¢¯ytkownik napisa©ø w ... On 8 mayo, 10:35, Szczepan Bia©øek wrote: I start reading about acoustic analogy. I found that: "Over long distances, the atmosphere can cause the polarization of a radio wave to fluctuate, so the distinction between horizontal and vertical becomes less significant." From:http://whatis.techtarget.com/definit...843762,00.html The my question a 1. What means "long distances" in km (or miles), 2. What is the best orientation of the antenna for long distances. S* Hello, Under normal circumstances, polarization change in line-off-site conditions (think of max 40 mile) is not that much, so antenna polarization does matter (unless you use at least circular polarization on one side). In a propagation path that is dominated by multi-path effects (reflection at buildings, hills, foliage, etc), you get almost random polarization and then the polarization is not that important. Your cell phone and indoor WIFI are examples. Extreme weather conditions can also lead to polarization changes or a random polarization component (ducting superrefraction). For sea water up to VHF, reflection depends on polarization. For ground-ground links (for example ship shore) mostly vertical polarization is used (as the sea water helps in this case). So if you want to receive these communication, you use a vertical polarized antenna. The largest change in polarization you will get when the waves have to travel through the ionosphere. At HF (ground-ground link via ionosphere), the polarization vector rotates many times. This is due to Faraday rotation. Â*Also ground-satellite links suffer from this effect. The higher the frequency, the less the change in polarization. For example at 100 MHz you should think about 30 full rotations (that is more then 10k degrees), while at 10 GHz the change in polarization will be about 1 degree. Circular polarization may help to mitigate the influence of Faraday rotation. At HF sky wave (100....1000 mile via ionosphere) polarization of the antenna matters. This is not because of the polarization change of the waves due to Faraday rotation, but because of the reflection characteristics of mother earth. In HF antennas, reflection on mother earth is used (in combination with antenna height) to get the required elevation radiation pattern of the antenna. Reflection on earth depends on polarization. Hopefully this helps you a bit. You do not use the words "transversal" and "EM". The only evidence of polarization is antenna directional sensitivity. You talked about radiowaves, that are EM waves. In free space, only progapation mode is transversal (that means both E- and H-field are perpendicular to the direction of energy propagation. With regards to audio, in gas, only lossless propagation mode is longitudinal (molecule movement and pressure vectors are parallel to the direction of energy propagation) . EM is the hydraulic analogy (by Heaviside). It is a "piece to teach" a field method. In that time the electricity was incompressble and massles. Now the electrons are compressible and have mass. We need a new analogy. It can be call the Gas analogy or Acoustic analogy.. Inwww.tetech.nlis wrote that are many analogies for EM. In the acoustic analogy a radio waves are normal spherical electric waves emitted from the two sources (ends of the dipole). So the sources are polarised, not the waves. Waves interfere. Do you agree? Not agree, the waves are also polarized, that can be physically measured. Polarization is determined by the E-field vector. A dipole has the E-field (in electrostatics). The equations are by Gauss. The same equations we can use for the Hertz dipole. The E-field will be alternate. At long distances the frequency in receiving antennas will be twice more. See my topic "frequency doubling" . I am only a science hobyist. The second question was: " What is the best orientation of the antenna for long distances? For old radio antennas. Very long horizontal wire. On UHF (for example 2450 MHz), long distance can be 20 km, but on HF 500 km is not called long distance. So the meaning of long distance depends on the frequency band. You should distinguish between the actual polarization of the antenna and the physical appearance. depending on how you feed it, a very long horizontal wire can be sensitive to vertical or horizontal polarized waves. Though the equations for acoustical waves look similar to those of EM waves, the orientation of the field components is completely different. Â*When you require a more specific answer, you should make your question more specific. I tried to give you a general answer for the various forms of radio wave propagation. Now is XXI century. EM is a beautiful theory from XIX century. In Tetech products no incompressible massless fluid. So the most specific and important question is: How is frequency in receiving antenna. Is it doubled? Â*Best regards, and sorry for my style Â*S* Hello Szczepan, You are right, charge is compressible. The charge that is required to charge (for example) a sphere seems to break the coninuity equition as is used for incompressible fluid in hydraulics. Continuity in electromagnetism is regained by introducing the D-field (dielectric displacement). The D-field is responsible for the capacitive current in case of varying E-field. Regarding frequency doubling. We can be lucky. Antennas and propagation behaves in virtually all cases linearly. From linear systems you might know that input and output frequency are the same, so no doubling in frequency. In case of non-linear parts in a system (for example a corroded connector in an antenna cable that is used by two or more transmitters, that may behave as a semiconductor), you might get so called mixer products (sum frequencies, harmonics, difference frequencies, etc). If you would like to know more about EM-fields related to antennas and electronics, just start with classical EM theory. This is a solid tool, existing over 100 years and is used by many people with succes to predict behaviour of circuits and antennas. If this will change of today, I will close my business activities next monday. Best regards, Wim PA3DJS www.tetech.nl don't forget to remove the first three letters of the alphabet in case of PM. |
wave polarisation
Użytkownik "Dave" napisał w wiadomo¶ci ... "Szczepan Bialek" wrote in message ... So the most specific and important question is: How is frequency in receiving antenna. Is it doubled? no. i transmit a given frequency and that is what is received. easily measured even with simple instruments. Yes. But an antenna receive the doubled impulses (from the ends of the Hertz dipole). The simple instruments must distinguish the doubled frequency from the second harmonics. S* |
wave polarisation
wrote ... On 9 mayo, 20:34, Szczepan Białek wrote:  ... On 9 mayo, 10:02, Szczepan Bia©©ek wrote: U¢¯ytkownik napisa©ø w ... On 8 mayo, 10:35, Szczepan Bia©øek wrote: I start reading about acoustic analogy. I found that: "Over long distances, the atmosphere can cause the polarization of a radio wave to fluctuate, so the distinction between horizontal and vertical becomes less significant." From:http://whatis.techtarget.com/definit...843762,00.html The my question a 1. What means "long distances" in km (or miles), 2. What is the best orientation of the antenna for long distances. S* Hello, Under normal circumstances, polarization change in line-off-site conditions (think of max 40 mile) is not that much, so antenna polarization does matter (unless you use at least circular polarization on one side). In a propagation path that is dominated by multi-path effects (reflection at buildings, hills, foliage, etc), you get almost random polarization and then the polarization is not that important. Your cell phone and indoor WIFI are examples. Extreme weather conditions can also lead to polarization changes or a random polarization component (ducting superrefraction). For sea water up to VHF, reflection depends on polarization. For ground-ground links (for example ship shore) mostly vertical polarization is used (as the sea water helps in this case). So if you want to receive these communication, you use a vertical polarized antenna. The largest change in polarization you will get when the waves have to travel through the ionosphere. At HF (ground-ground link via ionosphere), the polarization vector rotates many times. This is due to Faraday rotation.  Also ground-satellite links suffer from this effect. The higher the frequency, the less the change in polarization. For example at 100 MHz you should think about 30 full rotations (that is more then 10k degrees), while at 10 GHz the change in polarization will be about 1 degree. Circular polarization may help to mitigate the influence of Faraday rotation. At HF sky wave (100....1000 mile via ionosphere) polarization of the antenna matters. This is not because of the polarization change of the waves due to Faraday rotation, but because of the reflection characteristics of mother earth. In HF antennas, reflection on mother earth is used (in combination with antenna height) to get the required elevation radiation pattern of the antenna. Reflection on earth depends on polarization. Hopefully this helps you a bit. You do not use the words "transversal" and "EM". The only evidence of polarization is antenna directional sensitivity. You talked about radiowaves, that are EM waves. In free space, only progapation mode is transversal (that means both E- and H-field are perpendicular to the direction of energy propagation. With regards to audio, in gas, only lossless propagation mode is longitudinal (molecule movement and pressure vectors are parallel to the direction of energy propagation) . EM is the hydraulic analogy (by Heaviside). It is a "piece to teach" a field method. In that time the electricity was incompressble and massles. Now the electrons are compressible and have mass. We need a new analogy. It can be call the Gas analogy or Acoustic analogy. Inwww.tetech.nlis wrote that are many analogies for EM. In the acoustic analogy a radio waves are normal spherical electric waves emitted from the two sources (ends of the dipole). So the sources are polarised, not the waves. Waves interfere. Do you agree? Not agree, the waves are also polarized, that can be physically measured. Polarization is determined by the E-field vector. A dipole has the E-field (in electrostatics). The equations are by Gauss. The same equations we can use for the Hertz dipole. The E-field will be alternate. At long distances the frequency in receiving antennas will be twice more. See my topic "frequency doubling" . I am only a science hobyist. The second question was: " What is the best orientation of the antenna for long distances? For old radio antennas. Very long horizontal wire. On UHF (for example 2450 MHz), long distance can be 20 km, but on HF 500 km is not called long distance. So the meaning of long distance depends on the frequency band. You should distinguish between the actual polarization of the antenna and the physical appearance. depending on how you feed it, a very long horizontal wire can be sensitive to vertical or horizontal polarized waves. Though the equations for acoustical waves look similar to those of EM waves, the orientation of the field components is completely different.  When you require a more specific answer, you should make your question more specific. I tried to give you a general answer for the various forms of radio wave propagation. Now is XXI century. EM is a beautiful theory from XIX century. In Tetech products no incompressible massless fluid. So the most specific and important question is: How is frequency in receiving antenna. Is it doubled?  Best regards, and sorry for my style  S* Hello Szczepan, You are right, charge is compressible. The charge that is required to charge (for example) a sphere seems Seems or unquestionable? to break the coninuity equition as is used for incompressible fluid in hydraulics. Continuity in electromagnetism is regained by introducing the D-field (dielectric displacement). The D-field is responsible for the capacitive current in case of varying E-field. So in your products is the dielectric displacement or compressed electrons? Regarding frequency doubling. We can be lucky. Antennas and propagation behaves in virtually all cases linearly. From linear systems you might know that input and output frequency are the same, so no doubling in frequency. Try understand me. Your Hertz dipole emits electrc waves from the TWO ends (opposite phases). So the electrons in a receiving antenna are kicked twice more frequent. In case of non-linear parts in a system (for example a corroded connector in an antenna cable that is used by two or more transmitters, that may behave as a semiconductor), you might get so called mixer products (sum frequencies, harmonics, difference frequencies, etc). Harmonics may be the reason that nobody have seen the Phenomenon. If you would like to know more about EM-fields related to antennas and electronics, just start with classical EM theory. This is a solid tool, existing over 100 years and is used by many people with succes to predict behaviour of circuits and antennas. If this will change of today, I will close my business activities next monday. EM existing over 100 years and will be used the next as the "piece to teach". Your business base on experiments. Now You have the opportunity to make the most famous experiment in the history. If the result will be null I will change my hobby. Best regards, S* |
wave polarisation
"Szczepan Bialek" wrote in message ... Użytkownik "Dave" napisał w wiadomo¶ci ... "Szczepan Bialek" wrote in message ... So the most specific and important question is: How is frequency in receiving antenna. Is it doubled? no. i transmit a given frequency and that is what is received. easily measured even with simple instruments. Yes. But an antenna receive the doubled impulses (from the ends of the Hertz dipole). The simple instruments must distinguish the doubled frequency from the second harmonics. S* sure, they are double kicked... first one direction then the other within one cycle. you and art should get together, maybe he could straighten you out. |
wave polarisation
Szczepan Białek wrote:
EM existing over 100 years ... EM waves have existed ever since space became transparent more than 10 billion years ago. :-) -- 73, Cecil, IEEE, OOTC, http://www.w5dxp.com |
wave polarisation
Użytkownik "Cecil Moore" napisał w wiadomo¶ci ... Szczepan Białek wrote: EM existing over 100 years ... EM waves have existed ever since space became transparent more than 10 billion years ago. :-) It is proposition by Maxwell. Do not verified to now. Electric waves have more chance. For some it is obvious that in space may be the only one mechanism. Aepinus proved that gravity and electrostatics are the same. (the coefficient in the Coulomb equation has the tree different values) Ampere proved that the magnetism is an ilusion. So no place for the proposition. S* |
wave polarisation
Użytkownik "Dave" napisał w wiadomo¶ci ... "Szczepan Bialek" wrote in message ... Użytkownik "Dave" napisał w wiadomo¶ci ... "Szczepan Bialek" wrote in message ... So the most specific and important question is: How is frequency in receiving antenna. Is it doubled? no. i transmit a given frequency and that is what is received. easily measured even with simple instruments. Yes. But an antenna receive the doubled impulses (from the ends of the Hertz dipole). The simple instruments must distinguish the doubled frequency from the second harmonics. S* sure, they are double kicked... first one direction then the other within one cycle. you and art should get together, maybe he could straighten you out. I am not fluent in English. So try to understand. The electrons (they have mass and inertia) colect in one end and next disappear. It is the one cycle. Next they travel to the other end. It takes time. So in space are send the two signals.(in one your cycle). In the same time 9Your cycle) the current flow to and fro. In unverified EM the current cause the spherical wave. In reality the ends. Only Your simple measurements can verify it. S* |
wave polarisation
"Szczepan Bialek" wrote in message ... Użytkownik "Dave" napisał w wiadomo¶ci ... "Szczepan Bialek" wrote in message ... Użytkownik "Dave" napisał w wiadomo¶ci ... "Szczepan Bialek" wrote in message ... So the most specific and important question is: How is frequency in receiving antenna. Is it doubled? no. i transmit a given frequency and that is what is received. easily measured even with simple instruments. Yes. But an antenna receive the doubled impulses (from the ends of the Hertz dipole). The simple instruments must distinguish the doubled frequency from the second harmonics. S* sure, they are double kicked... first one direction then the other within one cycle. you and art should get together, maybe he could straighten you out. I am not fluent in English. So try to understand. The electrons (they have mass and inertia) colect in one end and next disappear. It is the one cycle. Next they travel to the other end. It takes time. So in space are send the two signals.(in one your cycle). In the same time 9Your cycle) the current flow to and fro. In unverified EM the current cause the spherical wave. In reality the ends. Only Your simple measurements can verify it. S* i have made my simple measurements, the frequency is not doubled by the antenna. |
wave polarisation
"Dave" wrote ... "Szczepan Bialek" wrote in message ... I am not fluent in English. So try to understand. The electrons (they have mass and inertia) colect in one end and next disappear. It is the one cycle. Next they travel to the other end. It takes time. So in space are send the two signals.(in one your cycle). In the same time 9Your cycle) the current flow to and fro. In unverified EM the current cause the spherical wave. In reality the ends. Only Your simple measurements can verify it. S* i have made my simple measurements, the frequency is not doubled by the antenna. You are to speedy. Look at the original Hertz experiment: http://people.seas.harvard.edu/~jone...Hertz_exp.html There is: "According to theory, if electromagnetic waves were spreading from the oscillator sparks, they would induce a current in the loop that would send sparks across the gap" In EM waves are produced by the current (oscllator sparks in the Hertz apparatus). One cycle is completed when the current flow to and fro. But there is possible the other theory. Electric waves are spreading from the ends (plates or big balls in Hertz apparatus). Now we know that electrons have mass and are compressible. So at the ends appear and disappear the huge charges. In that case an electric impulse is send when the current flows to (from one end) , and the next when the current flows fro (from the other end). So in one EM cycle are the two electric cycles. So the frequency is not doubled. The electric is twice more. Does your measurements distinguish radiation from the spakrks from that from the plates? S* |
wave polarisation
On Mon, 11 May 2009 19:21:32 +0200, Szczepan Bia?ek
wrote: You are to speedy. Look at the original Hertz experiment: http://people.seas.harvard.edu/~jone...Hertz_exp.html Poor citation to include broken links and missing material. There is: "According to theory, if electromagnetic waves were spreading from the oscillator sparks, they would induce a current in the loop that would send sparks across the gap" If? If is wrong. In EM waves are produced by the current (oscllator sparks in the Hertz apparatus). One cycle is completed when the current flow to and fro. You say this often, but it adds nothing important. But there is possible the other theory. Electric waves are spreading from the ends (plates or big balls in Hertz apparatus). This is not ANOTHER theory. Your first "If" is not a theory at all. It is an incorrect statement surrounded by poor writing. It is like this next statement: Now we know that electrons have mass and are compressible. Nonsense. So at the ends appear and disappear the huge charges. In that case an electric impulse is send when the current flows to (from one end) , and the next when the current flows fro (from the other end). Nonsense. So in one EM cycle are the two electric cycles. Nonsense. So the frequency is not doubled. The electric is twice more. This is either very poor English, or more nonsense. Does your measurements distinguish radiation from the spakrks from that from the plates? S* Of course they do. More the question: can you measure them too? If you cannot, then this explains the nonsense. 73's Richard Clark, KB7QHC |
wave polarisation
"Szczepan Bialek" wrote in message ... "Dave" wrote ... "Szczepan Bialek" wrote in message ... I am not fluent in English. So try to understand. The electrons (they have mass and inertia) colect in one end and next disappear. It is the one cycle. Next they travel to the other end. It takes time. So in space are send the two signals.(in one your cycle). In the same time 9Your cycle) the current flow to and fro. In unverified EM the current cause the spherical wave. In reality the ends. Only Your simple measurements can verify it. S* i have made my simple measurements, the frequency is not doubled by the antenna. You are to speedy. Look at the original Hertz experiment: http://people.seas.harvard.edu/~jone...Hertz_exp.html There is: "According to theory, if electromagnetic waves were spreading from the oscillator sparks, they would induce a current in the loop that would send sparks across the gap" In EM waves are produced by the current (oscllator sparks in the Hertz apparatus). One cycle is completed when the current flow to and fro. But there is possible the other theory. Electric waves are spreading from the ends (plates or big balls in Hertz apparatus). Now we know that electrons have mass and are compressible. So at the ends appear and disappear the huge charges. In that case an electric impulse is send when the current flows to (from one end) , and the next when the current flows fro (from the other end). So in one EM cycle are the two electric cycles. So the frequency is not doubled. The electric is twice more. Does your measurements distinguish radiation from the spakrks from that from the plates? S* I'm sorry, but i don't have a spark transmitter, those have been rather illegal for many years because of the extreme bandwidth and noise they generate. I use a nice clean sine wave so there is no harmonic generated. if you use a spark generator there will be many harmonics, but they come from the waveform being generated. |
wave polarisation
"Richard Clark" wrote ... On Mon, 11 May 2009 19:21:32 +0200, Szczepan Bia?ek wrote: Now we know that electrons have mass and are compressible. Nonsense. Millikan measured it in XIX centuary. Latelly Wim wrote: "Hello Szczepan, You are right, charge is compressible. The charge that is required to charge (for example) a sphere seems to break the coninuity equition as is used for incompressible fluid in hydraulics". So at the ends appear and disappear the huge charges. In that case an electric impulse is send when the current flows to (from one end) , and the next when the current flows fro (from the other end). Nonsense. I was told that the electrons want to escape from the ends of a dipole ( balls are a remedy). So in one EM cycle are the two electric cycles. Nonsense. So the frequency is not doubled. The electric is twice more. This is either very poor English, or more nonsense. Let assume that something is radiated from the end when the charge (compressed electrons) appears. There are the two ends. How many pulses will be send in space ? Does your measurements distinguish radiation from the spakrks from that from the plates? S* Of course they do. More the question: can you measure them too? No. The only instrument I have is the comb. But it was enough to verify the second meaning of word "polarisation". Everywhere is written that a charged comb attract a stream of water. And that it is caused by polarisation. It is caused by electrostriction. Decreasing E-field deform water stream. Of course the electrostriction is caused by the polarisation of H2O particles. Charged comb do not attract water drops. If you cannot, then this explains the nonsense. Something is radiated from the ends of the dipole. Can you detect it? S* |
wave polarisation
On Tue, 12 May 2009 09:56:18 +0200, Szczepan Bia?ek
wrote: "Richard Clark" wrote .. . On Mon, 11 May 2009 19:21:32 +0200, Szczepan Bia?ek wrote: Now we know that electrons have mass and are compressible. Nonsense. Millikan measured it in XIX centuary. Latelly Wim wrote: "Hello Szczepan, You are right, charge is compressible. The charge that is required to charge (for example) a sphere seems to break the coninuity equition as is used for incompressible fluid in hydraulics". Now we are into utter nonsense. So at the ends appear and disappear the huge charges. In that case an electric impulse is send when the current flows to (from one end) , and the next when the current flows fro (from the other end). Nonsense. I was told that the electrons want to escape from the ends of a dipole ( balls are a remedy). Then you were told nonsense. At least study something with accurate fundamentals. Let assume that something is radiated from the end This alone explains why you have a very poor understanding of the dynamics. An antenna radiates in ALL directions from EVERYPOINT of the antenna. No one has to make special "assumptions" that are limiting and with the appearance of being special knowledge. when the charge (compressed electrons) appears. There are the two ends. How many pulses will be send in space ? Radiation is not pulses. Does your measurements distinguish radiation from the spakrks from that from the plates? S* Of course they do. More the question: can you measure them too? No. Then you have very limited resources and absolutely no basis for discussion of the topic. The only instrument I have is the comb. But it was enough to verify the second meaning of word "polarisation". That adds nothing to the topic except you are working at very crude levels with very limited knowledge about a vastly more complex issue. If you cannot, then this explains the nonsense. Something is radiated from the ends of the dipole. Can you detect it? I already said I could, and you said you couldn't. Why do you ask again? 73's Richard Clark, KB7QHC |
wave polarisation
"Richard Clark" ... On Tue, 12 May 2009 09:56:18 +0200, Szczepan Bia?ek wrote: The only instrument I have is the comb. But it was enough to verify the second meaning of word "polarisation". That adds nothing to the topic except you are working at very crude levels with very limited knowledge about a vastly more complex issue. The topics is polarisation. This word has the two meaning. The both have wrong explanations in texbooks. The one I have verified with the comb. The second "wave polarisation" is explained with transverse waves. No transverse waves. If receiver (resonator) must be parallel to emmiter you can explain it in many ways. But to verify it the comb is not enough. So I need help. If you cannot, then this explains the nonsense. Something is radiated from the ends of the dipole. Can you detect it? I already said I could, and you said you couldn't. Why do you ask again? I hope that somebody else reads us. S* |
wave polarisation
On Tue, 12 May 2009 18:00:09 +0200, Szczepan Bia?ek
wrote: The topics is polarisation. This word has the two meaning. The both have wrong explanations in texbooks. The one I have verified with the comb. The second "wave polarisation" is explained with transverse waves. No transverse waves. If receiver (resonator) must be parallel to emmiter you can explain it in many ways. But to verify it the comb is not enough. So I need help. That is fair. Let's start with some serious misunderstandings with a few questions to test them. First, let us return to that link you offered with the Hertzian Loop with its spark gap. Let us say that this loop is 1 meter of wire (about the actual size anyway). Let us say there is a current detector at each end of this loop. Let us say we have closed a switch that applies voltage to the loop, and the first meter has indicated current flow. This is our time reference point. Now the questions: 1. For the electron that went through the first current detector, how long does it take for that SAME electron to get to the second detector? 2. How long does it take for the second detector to indicate there is current flow? Hint: the answer for 1. is very, very different for the answer for 2. Now, let us say that before that SAME electron gets to the second current detector, that path is broken open (maybe 1 pico second before the SAME electron arrival). The SAME electron sees an open circuit. What is the amount of energy required for the electron to break out of the metal conductor, and into the air? 73's Richard Clark, KB7QHC |
wave polarisation
"Richard Clark" wrote ... On Tue, 12 May 2009 18:00:09 +0200, Szczepan Bia?ek wrote: The topics is polarisation. This word has the two meaning. The both have wrong explanations in texbooks. The one I have verified with the comb. The second "wave polarisation" is explained with transverse waves. No transverse waves. If receiver (resonator) must be parallel to emmiter you can explain it in many ways. But to verify it the comb is not enough. So I need help. That is fair. Let's start with some serious misunderstandings with a few questions to test them. First, let us return to that link you offered with the Hertzian Loop with its spark gap. Let us say that this loop is 1 meter of wire (about the actual size anyway). Let us say there is a current detector at each end of this loop. Let us say we have closed a switch that applies voltage to the loop, and the first meter has indicated current flow. This is our time reference point. Now the questions: 1. For the electron that went through the first current detector, how long does it take for that SAME electron to get to the second detector? 2. How long does it take for the second detector to indicate there is current flow? Hint: the answer for 1. is very, very different for the answer for 2. Now, let us say that before that SAME electron gets to the second current detector, that path is broken open (maybe 1 pico second before the SAME electron arrival). The SAME electron sees an open circuit. What is the amount of energy required for the electron to break out of the metal conductor, and into the air? You went to details. Early you wrote: "An antenna radiates in ALL directions from EVERYPOINT of the antenna. " Textbooks say that EM transversial waves are emitted by current (the sparks in Hertz apparatus - not from the ends). I say that from the ends (as electric waves similar to acoustics). The directional pattern must be different. The directional patterns of loudspeakers and Herts dipoles are very similar. So I try to find evidences. Now I do not know if you prefer EM or electric waves. S* |
wave polarisation
On Tue, 12 May 2009 21:29:31 +0200, Szczepan Bia?ek
wrote: You went to details. Early you wrote: "An antenna radiates in ALL directions from EVERYPOINT of the antenna. " They do. Textbooks say that EM transversial waves are emitted by current (the sparks in Hertz apparatus - not from the ends). That is not the same thing as radiation. Repeating poor quotations does not make it better. I say that from the ends (as electric waves similar to acoustics). Compressional waves or longitudinal waves? In solid or air or in liquid? The answers to these questions lead to very, very different behavior. As I say, these simplicities you use are nonsense. The directional pattern must be different. The directional pattern is a combination of EVERYPOINT radiating in ALL directions. The differences in their position contribute to an unique pattern. This is the whole basis of the "method of moments" application of modeling radiation emitters. The directional patterns of loudspeakers and Herts dipoles are very similar. The are more differences than similarities. So I try to find evidences. Now I do not know if you prefer EM or electric waves. That shouldn't keep you from answering the simple physics of: Let's start with some serious misunderstandings with a few questions to test them. First, let us return to that link you offered with the Hertzian Loop with its spark gap. Let us say that this loop is 1 meter of wire (about the actual size anyway). Let us say there is a current detector at each end of this loop. Let us say we have closed a switch that applies voltage to the loop, and the first meter has indicated current flow. This is our time reference point. Now the questions: 1. For the electron that went through the first current detector, how long does it take for that SAME electron to get to the second detector? 2. How long does it take for the second detector to indicate there is current flow? Hint: the answer for 1. is very, very different for the answer for 2. Now, let us say that before that SAME electron gets to the second current detector, that path is broken open (maybe 1 pico second before the SAME electron arrival). The SAME electron sees an open circuit. What is the amount of energy required for the electron to break out of the metal conductor, and into the air? 73's Richard Clark, KB7QHC |
wave polarisation
We will take this by parts:
On Tue, 12 May 2009 18:00:09 +0200, Szczepan Bia?ek wrote: The topics is polarisation. This word has the two meaning. The both have wrong explanations in texbooks. Blaming references is a very strong indicator of you having the problem, not the textbooks. The one I have verified with the comb. I've read that, and it proves nothing that is part of the topic (I shall explain below). The second "wave polarisation" is explained with transverse waves. No transverse waves. If "no," then what "yes?" Actually you have mixed up two different characteristics. Polarity and polarization are NOT the same thing. With RF radiation, the wave is constantly changing polarity (that is why the source of RF is called alternating current), but within the "line of sight" of the antenna, the polarization for a dipole is defined by its angle to the earth as viewed by the observer. If you see an horizontal dipole, it produces alternating polarities of waves with horizontal polarization. If you see a vertical dipole, it produces alternating polarities of waves with vertical polarization. RF energy is ALWAYS changing polarity. If receiver (resonator) must be parallel to emmiter you can explain it in many ways. But to verify it the comb is not enough. So I need help. The comb was useless in proving anything about RF, antennas, or polarization. What it demonstrates is induced polarity in a dielectric. Interesting, but entirely unrelated to the topic. 73's Richard Clark, KB7QHC |
wave polarisation
"Richard Clark" wrote ... We will take this by parts: On Tue, 12 May 2009 18:00:09 +0200, Szczepan Bia?ek wrote: The topics is polarisation. This word has the two meaning. The both have wrong explanations in texbooks. Blaming references is a very strong indicator of you having the problem, not the textbooks. Is in old books the same as in todays? Sometimes they are changed. The one I have verified with the comb. I've read that, and it proves nothing that is part of the topic (I shall explain below). The second "wave polarisation" is explained with transverse waves. No transverse waves. If "no," then what "yes?" You wrote: "Compressional waves or longitudinal waves? In solid or air or in liquid? The answers to these questions lead to very, very different behavior. As I say, these simplicities you use are nonsense" In EM are many simplifications. Electric waves are like the acoustic. Of course they kick the electrons in antennas not a membranes. Actually you have mixed up two different characteristics. Polarity and polarization are NOT the same thing. With RF radiation, the wave is constantly changing polarity In transvers waves something changes the direction of rotation. The source makes rotating oscillations. (that is why the source of RF is called alternating current), but within the "line of sight" of the antenna, the polarization for a dipole is defined by its angle to the earth as viewed by the observer.) You decribe the electric waves. If you see an horizontal dipole, it produces alternating polarities of waves with horizontal polarization. If you see a vertical dipole, it produces alternating polarities of waves with vertical polarization. !00% agreement with me. RF waves are electrical. RF energy is ALWAYS changing polarity. If receiver (resonator) must be parallel to emmiter you can explain it in many ways. But to verify it the comb is not enough. So I need help. The comb was useless in proving anything about RF, antennas, or polarization. What it demonstrates is induced polarity in a dielectric. Interesting, but entirely unrelated to the topic. The Acoustic analogy will be also interesting. S* |
wave polarisation
Richard Clark wrote:
If you see an horizontal dipole, it produces alternating polarities of waves with horizontal polarization. That is broadside to the antenna. Directly off the ends of the dipole, the polarization is vertical. -- 73, Cecil, IEEE, OOTC, http://www.w5dxp.com |
wave polarisation
On Wed, 13 May 2009 09:01:35 +0200, Szczepan Bia?ek
wrote: Blaming references is a very strong indicator of you having the problem, not the textbooks. Is in old books the same as in todays? Sometimes they are changed. Engineering? Never. You wrote: "Compressional waves or longitudinal waves? In solid or air or in liquid? The answers to these questions lead to very, very different behavior. As I say, these simplicities you use are nonsense" In EM are many simplifications. Electric waves are like the acoustic. They are not. You are repeating your nonsense again. Of course they kick the electrons in antennas not a membranes. This is more nonsense. Actually you have mixed up two different characteristics. Polarity and polarization are NOT the same thing. With RF radiation, the wave is constantly changing polarity In transvers waves something changes the direction of rotation. The source makes rotating oscillations. You are showing very little capacity to understand alternating current, RF, and radiation. (that is why the source of RF is called alternating current), but within the "line of sight" of the antenna, the polarization for a dipole is defined by its angle to the earth as viewed by the observer.) You decribe the electric waves. I am describing radiation. The Acoustic analogy will be also interesting. S* No, it would not. Radiation is wholly unlike acoustics and very few people actually understand acoustics. Having the experience of "hearing" does not qualify you as being proficient in its discussion. Human sensation is vastly more illusion than science. This is quite evident in your postings where you rely on crude simplifications and incorrect metaphors. The proof of their failure in your being to understand the topic of wave polarization is you cannot answer: First, let us return to that link you offered with the Hertzian Loop with its spark gap. Let us say that this loop is 1 meter of wire (about the actual size anyway). Let us say there is a current detector at each end of this loop. Let us say we have closed a switch that applies voltage to the loop, and the first meter has indicated current flow. This is our time reference point. Now the questions: 1. For the electron that went through the first current detector, how long does it take for that SAME electron to get to the second detector? 2. How long does it take for the second detector to indicate there is current flow? Hint: the answer for 1. is very, very different for the answer for 2. Now, let us say that before that SAME electron gets to the second current detector, that path is broken open (maybe 1 pico second before the SAME electron arrival). The SAME electron sees an open circuit. What is the amount of energy required for the electron to break out of the metal conductor, and into the air? These are very fundamental concepts that, if you cannot respond to them, reveal your range of involvement is extremely thin and you probably have no real interest in the topic. Why are you posting here? 73's Richard Clark, KB7QHC |
wave polarisation
"Richard Clark" wrote ... On Wed, 13 May 2009 09:01:35 +0200, Szczepan Bia?ek wrote: Blaming references is a very strong indicator of you having the problem, not the textbooks. Is in old books the same as in todays? Sometimes they are changed. Engineering? Never. Now we have the two different physics. One for students and the second in engineering. Textbooks are for students. In engineering are Handbooks and specifications. You wrote: "Compressional waves or longitudinal waves? In solid or air or in liquid? The answers to these questions lead to very, very different behavior. As I say, these simplicities you use are nonsense" In EM are many simplifications. Electric waves are like the acoustic. They are not. You are repeating your nonsense again. Of course they kick the electrons in antennas not a membranes. This is more nonsense. Actually you have mixed up two different characteristics. Polarity and polarization are NOT the same thing. With RF radiation, the wave is constantly changing polarity In transvers waves something changes the direction of rotation. The source makes rotating oscillations. You are showing very little capacity to understand alternating current, RF, and radiation. (that is why the source of RF is called alternating current), but within the "line of sight" of the antenna, the polarization for a dipole is defined by its angle to the earth as viewed by the observer.) You decribe the electric waves. I am describing radiation. In engineering sense. I agree with you. The Acoustic analogy will be also interesting. S* No, it would not. Radiation is wholly unlike acoustics and very few people actually understand acoustics. Having the experience of "hearing" does not qualify you as being proficient in its discussion. Human sensation is vastly more illusion than science. This is quite evident in your postings where you rely on crude simplifications and incorrect metaphors. The proof of their failure in your being to understand the topic of wave polarization is you cannot answer: First, let us return to that link you offered with the Hertzian Loop with its spark gap. Let us say that this loop is 1 meter of wire (about the actual size anyway). Let us say there is a current detector at each end of this loop. Let us say we have closed a switch that applies voltage to the loop, and the first meter has indicated current flow. This is our time reference point. Now the questions: 1. For the electron that went through the first current detector, how long does it take for that SAME electron to get to the second detector? 2. How long does it take for the second detector to indicate there is current flow? Hint: the answer for 1. is very, very different for the answer for 2. Now, let us say that before that SAME electron gets to the second current detector, that path is broken open (maybe 1 pico second before the SAME electron arrival). The SAME electron sees an open circuit. What is the amount of energy required for the electron to break out of the metal conductor, and into the air? These are very fundamental concepts that, if you cannot respond to them, reveal your range of involvement is extremely thin and you probably have no real interest in the topic. Why are you posting here? To lern something from engineering people. I was sure that people in this group must know such phenomenon. It was Brian Howie. He wrote: "You can get ionospheric mixing of radio waves. e.g Luxembourg Effect; so doubling is possible." Today I found this: http://durenberger.com/resources/doc...EFFECT0235.pdf So I have known what I want to know. Of course my explanation of the phenomenon is as I described here. It is not easy to understand me. I am sure that Wim does. He wrote "If you would like to know more about EM-fields related to antennas and electronics, just start with classical EM theory. This is a solid tool, existing over 100 years and is used by many people with succes to predict behaviour of circuits and antennas. If this will change of today, I will close my business activities next monday" Most engineering people very quickly forget the "classical EM theory" as was tought in schools. In the "classical EM theory" no electrons. Engineering use his own physics with electrons. S* |
wave polarisation
On Wed, 13 May 2009 21:41:55 +0200, Szczepan Bia?ek
wrote: Blaming references is a very strong indicator of you having the problem, not the textbooks. Is in old books the same as in todays? Sometimes they are changed. Engineering? Never. Now we have the two different physics. One for students and the second in engineering. The students are different, not the Physics. There is also college physics and University physics. The sense here is that you are adept at calculus (University) or you are not (college). Your explanations of physics are not even High School (Gymnasium) level. Why are you posting here? To lern something from engineering people. I was sure that people in this group must know such phenomenon. It was Brian Howie. He wrote: "You can get ionospheric mixing of radio waves. e.g Luxembourg Effect; so doubling is possible." This is useless. It is like you asking "Will it rain?", and someone telling you the story of Noah and the flood. Would you build an Ark based on this answer to your question? Today I found this: http://durenberger.com/resources/doc...EFFECT0235.pdf Again, a very poor source. You are merely wandering along the shelves and picking up random information. So I have known what I want to know. Of course my explanation of the phenomenon is as I described here. It is not easy to understand me. I am sure that Wim does. He wrote "If you would like to know more about EM-fields related to antennas and electronics, just start with classical EM theory. This is a solid tool, existing over 100 years and is used by many people with succes to predict behaviour of circuits and antennas. If this will change of today, I will close my business activities next monday" What was the point of quoting this when you have done nothing from his advice? Most engineering people very quickly forget the "classical EM theory" as was tought in schools. In the "classical EM theory" no electrons. Engineering use his own physics with electrons. S* This worn out phrase is repeated so much that I don't think you are really interested in anything more than trolling. If you actually found a solution in the Luxembourg effect you could answer: First, let us return to that link you offered with the Hertzian Loop with its spark gap. Let us say that this loop is 1 meter of wire (about the actual size anyway). Let us say there is a current detector at each end of this loop. Let us say we have closed a switch that applies voltage to the loop, and the first meter has indicated current flow. This is our time reference point. Now the questions: 1. For the electron that went through the first current detector, how long does it take for that SAME electron to get to the second detector? 2. How long does it take for the second detector to indicate there is current flow? Hint: the answer for 1. is very, very different for the answer for 2. Now, let us say that before that SAME electron gets to the second current detector, that path is broken open (maybe 1 pico second before the SAME electron arrival). The SAME electron sees an open circuit. What is the amount of energy required for the electron to break out of the metal conductor, and into the air? 73's Richard Clark, KB7QHC |
wave polarisation
"Richard Clark" wrote ... On Wed, 13 May 2009 21:41:55 +0200, Szczepan Bia?ek wrote: Blaming references is a very strong indicator of you having the problem, not the textbooks. Is in old books the same as in todays? Sometimes they are changed. Engineering? Never. Now we have the two different physics. One for students and the second in engineering. The students are different, not the Physics. There is also college physics and University physics. The sense here is that you are adept at calculus (University) or you are not (college). Your explanations of physics are not even High School (Gymnasium) level. On all levels is: "James Clerk Maxwell's mathematical theory of 1873 had predicted that electromagnetic disturbances should propagate through space at the speed of light and should exhibit the wave-like characteristics of light propagation." Next on all levels is that the light waves are transversal because they can be polarised. Maxwell described it in his Treatise in words and in equations (are on-line) Why are you posting here? To lern something from engineering people. I was sure that people in this group must know such phenomenon. It was Brian Howie. He wrote: "You can get ionospheric mixing of radio waves. e.g Luxembourg Effect; so doubling is possible." This is useless. It is like you asking "Will it rain?", and someone telling you the story of Noah and the flood. Would you build an Ark based on this answer to your question? I have never heard about the Luxembourg Effect. I predict it and ask proper people (if is "frequency doubling). Today I found this: http://durenberger.com/resources/doc...EFFECT0235.pdf Again, a very poor source. You are merely wandering along the shelves and picking up random information. The only I need. So I have known what I want to know. Of course my explanation of the phenomenon is as I described here. It is not easy to understand me. I am sure that Wim does. He wrote "If you would like to know more about EM-fields related to antennas and electronics, just start with classical EM theory. This is a solid tool, existing over 100 years and is used by many people with succes to predict behaviour of circuits and antennas. If this will change of today, I will close my business activities next monday" What was the point of quoting this when you have done nothing from his advice? I know the classical EM theory. Most engineering people very quickly forget the "classical EM theory" as was tought in schools. In the "classical EM theory" no electrons. Engineering use his own physics with electrons. S* This worn out phrase is repeated so much that I don't think you are really interested in anything more than trolling. If you actually found a solution in the Luxembourg effect you could answer: First, let us return to that link you offered with the Hertzian Loop with its spark gap. Let us say that this loop is 1 meter of wire (about the actual size anyway). Let us say there is a current detector at each end of this loop. Let us say we have closed a switch that applies voltage to the loop, ....voltage by additionl wire ( in what point of the ring?) or by space? S* |
wave polarisation
On May 14, 3:12*am, Szczepan Białek wrote:
*"Richard Clark" *wrotenews:km8m05tns3cssc5r6sf3kf517onh05no3e@4ax. com... On Wed, 13 May 2009 21:41:55 +0200, Szczepan Bia?ek wrote: Blaming references is a very strong indicator of you having the problem, not the textbooks. Is in old books the same as in todays? Sometimes they are changed. Engineering? *Never. Now we have the two different physics. One for students and the second in engineering. The students are different, not the Physics. *There is also college physics and University physics. *The sense here is that you are adept at calculus (University) or you are not (college). *Your explanations of physics are not even High School (Gymnasium) level. On all levels is: "James Clerk Maxwell's mathematical theory of 1873 had predicted that electromagnetic disturbances should propagate through space at the speed of light and should exhibit the wave-like characteristics of light propagation." Next on all levels is that the light waves are transversal because they can be polarised. Maxwell described it in his Treatise in words and in equations (are on-line) Why are you posting here? To lern something from engineering people. I was sure that people in this group must know such phenomenon. It was Brian Howie. He wrote: "You can get ionospheric mixing of radio waves. e.g Luxembourg Effect; so doubling is possible." This is useless. *It is like you asking "Will it rain?", and someone telling you the story of Noah and the flood. *Would you build an Ark based on this answer to your question? I have never heard about the Luxembourg Effect. I predict it and ask proper people (if is "frequency doubling). Today I found this: http://durenberger.com/resources/doc...EFFECT0235.pdf Again, a very poor source. *You are merely wandering along the shelves and picking up random information. The only I need. So I have known what I want to know. Of course my explanation of the phenomenon is as I described here. It is not easy to understand me. I am sure that Wim does. He wrote "If you would like to know more about EM-fields related to antennas and electronics, just start with classical EM theory. This is a solid tool, existing over 100 years and is used by many people with succes to predict behaviour of circuits and antennas. If this will change of today, I will close my business activities next monday" What was the point of quoting this when you have done nothing from his advice? I know the classical EM theory. Most engineering people very quickly forget the "classical EM theory" as was tought in schools. In the "classical EM theory" no electrons. Engineering use his own physics with electrons. S* This worn out phrase is repeated so much that I don't think you are really interested in anything more than trolling. If you actually found a solution in the Luxembourg effect you could answer: First, let us return to that link you offered with the Hertzian Loop with its spark gap. *Let us say that this loop is 1 meter of wire (about the actual size anyway). *Let us say there is a current detector at each end of this loop. *Let us say we have closed a switch that applies voltage to the loop, ...voltage by additionl wire ( in what point of the ring?) or by space? S* - Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text -- Hide quoted text - - Show quoted text - S" needs to go back and read in totality the paper on Luxenbourg Effect. It explains that you DO in fact have a non-linear medium in the form of a charged ionosphere during periods of the effect. It is also difficult to tell whether or not what seems to be a case of the Luxenbourg effect is in fact just intermodulation distortion occurng in the receiver. Jimmie |
wave polarisation
On Thu, 14 May 2009 09:12:19 +0200, Szczepan Bia?ek
wrote: On all levels is: "James Clerk Maxwell's mathematical theory of 1873 had More dumpster diving. First, let us return to that link you offered with the Hertzian Loop with its spark gap. Let us say that this loop is 1 meter of wire (about the actual size anyway). Let us say there is a current detector at each end of this loop. Let us say we have closed a switch that applies voltage to the loop, ...voltage by additionl wire ( in what point of the ring?) or by space? Let's just skip the extremely difficult matters for you. It was your reference source but you don't seem to have read it yet (apparently the Luxembourg paper suffers your attention as well). If this were an audio based group instead of a text based one, maybe you could find a lectrice. 73's Richard Clark, KB7QHC |
wave polarisation
"JIMMIE" wrote ... S" needs to go back and read in totality the paper on Luxenbourg Effect. It explains that you DO in fact have a non-linear medium in the form of a charged ionosphere during periods of the effect. It is also difficult to tell whether or not what seems to be a case of the Luxenbourg effect is in fact just intermodulation distortion occurng in the receiver. The paper is from 1935. The year and half after the first observation. Now we have 2009. Probably were the next observations. Some of you may have yours own. What we need to analyse: 1. Type of mast (half-wave, or 1/4) and the frequency, 2. The frequency at which on the receiver a faint background appears. Of course only the cases where the frequency doubling take place. S* |
wave polarisation
"Szczepan Białek" wrote ... "JIMMIE" wrote ... S" needs to go back and read in totality the paper on Luxenbourg Effect. It explains that you DO in fact have a non-linear medium in the form of a charged ionosphere during periods of the effect. It is also difficult to tell whether or not what seems to be a case of the Luxenbourg effect is in fact just intermodulation distortion occurng in the receiver. The paper is from 1935. The year and half after the first observation. Now we have 2009. Probably were the next observations. Some of you may have yours own. What we need to analyse: 1. Type of mast (half-wave, or 1/4) and the frequency, 2. The frequency at which on the receiver a faint background appears. Of course only the cases where the frequency doubling take place. S* I have found the first: http://books.google.pl/books?id=QSke...P RA1-PA53,M1 There on the page 53 you can find that the medium-wave were disturbed by the long-vave (halve-way between). S* |
wave polarisation
"Richard Clark" wrote in message ... On Thu, 14 May 2009 09:12:19 +0200, Szczepan Bia?ek wrote: On all levels is: "James Clerk Maxwell's mathematical theory of 1873 had More dumpster diving. First, let us return to that link you offered with the Hertzian Loop with its spark gap. Let us say that this loop is 1 meter of wire (about the actual size anyway). Let us say there is a current detector at each end of this loop. Let us say we have closed a switch that applies voltage to the loop, ...voltage by additionl wire ( in what point of the ring?) or by space? Let's just skip the extremely difficult matters for you. It was your reference source but you don't seem to have read it yet (apparently the Luxembourg paper suffers your attention as well). If this were an audio based group instead of a text based one, maybe you could find a lectrice. 73's Richard Clark, KB7QHC Hi Richard I did some web searching on frequendy doubling and found that Ireland's capitol *is* Doublin.and want to point out that anyone with extra money could consider investing there in Ireland. Jerry |
wave polarisation
On Thu, 14 May 2009 17:36:38 GMT, "Jerry"
wrote: I did some web searching on frequendy doubling and found that Ireland's capitol *is* Doublin.and want to point out that anyone with extra money could consider investing there in Ireland. Hi Jerry, That could prove it! I think I had my fill of this. 73's Richard Clark, KB7QHC |
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