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#2
October 14th 14, 05:47 AM
posted to rec.radio.amateur.antenna
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Short antenna = reduced power
On 10/13/2014 11:45 PM, wrote:
rickman wrote: On 10/13/2014 1:36 PM, wrote: gareth wrote: Quoting from Electromagnetism By F.N.H.Robinson in the Oxford Physics Series 1973 edition ISBN 0 19 8518913 Chapter 11, Radiation, page 102 Formula 11.11 Has in the equation for radiated power the term (2*PI*L/LAMBDA)**2 where L is the antenna length and LAMBDA the wavelength, thereby showing that the radiated power decreases when the antenna length decreases. I will read up further and report further... You do that and while you are at it take note of the fact that the expression you give is unitless and can not be power. You will also find that the total power radiated by an antenna is the surface integral of the average Poynting vector over a surface enclosing the antenna. The surface usually chosen is a sphere in the far field to keep the equations "simple". He is taking a portion of the equation and presenting it out of context assuming that this is a valid way to consider what he wishes to show. I would like to see the full equation. The devil is in the details. Actually, there is no "the" equation for the power radiated by an antenna other than the surface integral of the average Poynting vector over a surface enclosing the antenna. There are some approximate rules for specific cases and limiting conditions, but this isn't one of them. What he presented is for a 1/2 wavelegth antenna 9.87 and a full wave antenna 39.48. WTF is that?? I have no idea what you are talking about. Where did you get these numbers? 9.87 what? -- Rick |
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#3
October 14th 14, 06:21 AM
posted to rec.radio.amateur.antenna
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Short antenna = reduced power
rickman wrote:
On 10/13/2014 11:45 PM, wrote: rickman wrote: On 10/13/2014 1:36 PM, wrote: gareth wrote: Quoting from Electromagnetism By F.N.H.Robinson in the Oxford Physics Series 1973 edition ISBN 0 19 8518913 Chapter 11, Radiation, page 102 Formula 11.11 Has in the equation for radiated power the term (2*PI*L/LAMBDA)**2 where L is the antenna length and LAMBDA the wavelength, thereby showing that the radiated power decreases when the antenna length decreases. I will read up further and report further... You do that and while you are at it take note of the fact that the expression you give is unitless and can not be power. You will also find that the total power radiated by an antenna is the surface integral of the average Poynting vector over a surface enclosing the antenna. The surface usually chosen is a sphere in the far field to keep the equations "simple". He is taking a portion of the equation and presenting it out of context assuming that this is a valid way to consider what he wishes to show. I would like to see the full equation. The devil is in the details. Actually, there is no "the" equation for the power radiated by an antenna other than the surface integral of the average Poynting vector over a surface enclosing the antenna. There are some approximate rules for specific cases and limiting conditions, but this isn't one of them. What he presented is for a 1/2 wavelegth antenna 9.87 and a full wave antenna 39.48. WTF is that?? I have no idea what you are talking about. Where did you get these numbers? 9.87 what? (2 * 3.14 * 5 meters / 10 meters) ^ 2 9.87 nothing; the expression is unitless, i.e. a pure number without units. In the expression you have a length divided by a length, which cancels into a unitles number. As my old physics professor used to say, always check the units of your answer; the arithmatic may be correct but it is meaningless unless the units are correct. Sample problem: You drove 100 miles and used 5 gallons of gas. What was your mileage? 5 gallons * 100 miles = 500 gallon-miles --- wrong units. 5 gallons / 100 miles = .05 gal/mile --- wrong units. 100 miles / 5 gallons = 20 miles/gal --- correct units and correct answer. -- Jim Pennino |
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#4
October 14th 14, 09:05 AM
posted to rec.radio.amateur.antenna
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Short antenna = reduced power
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#5
October 14th 14, 09:10 AM
posted to rec.radio.amateur.antenna
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Short antenna = reduced power
On 10/14/2014 1:21 AM, wrote:
rickman wrote: On 10/13/2014 11:45 PM, wrote: rickman wrote: On 10/13/2014 1:36 PM, wrote: gareth wrote: Quoting from Electromagnetism By F.N.H.Robinson in the Oxford Physics Series 1973 edition ISBN 0 19 8518913 Chapter 11, Radiation, page 102 Formula 11.11 Has in the equation for radiated power the term (2*PI*L/LAMBDA)**2 where L is the antenna length and LAMBDA the wavelength, thereby showing that the radiated power decreases when the antenna length decreases. I will read up further and report further... You do that and while you are at it take note of the fact that the expression you give is unitless and can not be power. You will also find that the total power radiated by an antenna is the surface integral of the average Poynting vector over a surface enclosing the antenna. The surface usually chosen is a sphere in the far field to keep the equations "simple". He is taking a portion of the equation and presenting it out of context assuming that this is a valid way to consider what he wishes to show. I would like to see the full equation. The devil is in the details. Actually, there is no "the" equation for the power radiated by an antenna other than the surface integral of the average Poynting vector over a surface enclosing the antenna. There are some approximate rules for specific cases and limiting conditions, but this isn't one of them. What he presented is for a 1/2 wavelegth antenna 9.87 and a full wave antenna 39.48. WTF is that?? I have no idea what you are talking about. Where did you get these numbers? 9.87 what? (2 * 3.14 * 5 meters / 10 meters) ^ 2 9.87 nothing; the expression is unitless, i.e. a pure number without units. In the expression you have a length divided by a length, which cancels into a unitles number. As my old physics professor used to say, always check the units of your answer; the arithmatic may be correct but it is meaningless unless the units are correct. Sample problem: You drove 100 miles and used 5 gallons of gas. What was your mileage? 5 gallons * 100 miles = 500 gallon-miles --- wrong units. 5 gallons / 100 miles = .05 gal/mile --- wrong units. 100 miles / 5 gallons = 20 miles/gal --- correct units and correct answer. I have no idea what you are going on about. Ok, 9.87 is a unitless number. So is 33.043. Now what? -- Rick |
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#6
October 14th 14, 05:58 PM
posted to rec.radio.amateur.antenna
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Short antenna = reduced power
rickman wrote:
On 10/14/2014 1:21 AM, wrote: rickman wrote: On 10/13/2014 11:45 PM, wrote: rickman wrote: On 10/13/2014 1:36 PM, wrote: gareth wrote: Quoting from Electromagnetism By F.N.H.Robinson in the Oxford Physics Series 1973 edition ISBN 0 19 8518913 Chapter 11, Radiation, page 102 Formula 11.11 Has in the equation for radiated power the term (2*PI*L/LAMBDA)**2 where L is the antenna length and LAMBDA the wavelength, thereby showing that the radiated power decreases when the antenna length decreases. I will read up further and report further... You do that and while you are at it take note of the fact that the expression you give is unitless and can not be power. You will also find that the total power radiated by an antenna is the surface integral of the average Poynting vector over a surface enclosing the antenna. The surface usually chosen is a sphere in the far field to keep the equations "simple". He is taking a portion of the equation and presenting it out of context assuming that this is a valid way to consider what he wishes to show. I would like to see the full equation. The devil is in the details. Actually, there is no "the" equation for the power radiated by an antenna other than the surface integral of the average Poynting vector over a surface enclosing the antenna. There are some approximate rules for specific cases and limiting conditions, but this isn't one of them. What he presented is for a 1/2 wavelegth antenna 9.87 and a full wave antenna 39.48. WTF is that?? I have no idea what you are talking about. Where did you get these numbers? 9.87 what? (2 * 3.14 * 5 meters / 10 meters) ^ 2 9.87 nothing; the expression is unitless, i.e. a pure number without units. In the expression you have a length divided by a length, which cancels into a unitles number. As my old physics professor used to say, always check the units of your answer; the arithmatic may be correct but it is meaningless unless the units are correct. Sample problem: You drove 100 miles and used 5 gallons of gas. What was your mileage? 5 gallons * 100 miles = 500 gallon-miles --- wrong units. 5 gallons / 100 miles = .05 gal/mile --- wrong units. 100 miles / 5 gallons = 20 miles/gal --- correct units and correct answer. I have no idea what you are going on about. Ok, 9.87 is a unitless number. So is 33.043. Now what? Exactly. Since it is a unitless number, it can not be power as claimed. Nor can it be a rule of thumb for power versus antenna length as a full wave antenna does not radiate 4 (39.48/9.87) more power than a 1/2 wave antenna. It is just nonsense. -- Jim Pennino |
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#7
October 14th 14, 06:29 PM
posted to rec.radio.amateur.antenna
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Short antenna = reduced power
On 10/14/2014 12:58 PM, wrote:
rickman wrote: On 10/14/2014 1:21 AM, wrote: rickman wrote: On 10/13/2014 11:45 PM, wrote: rickman wrote: On 10/13/2014 1:36 PM, wrote: gareth wrote: Quoting from Electromagnetism By F.N.H.Robinson in the Oxford Physics Series 1973 edition ISBN 0 19 8518913 Chapter 11, Radiation, page 102 Formula 11.11 Has in the equation for radiated power the term (2*PI*L/LAMBDA)**2 where L is the antenna length and LAMBDA the wavelength, thereby showing that the radiated power decreases when the antenna length decreases. I will read up further and report further... You do that and while you are at it take note of the fact that the expression you give is unitless and can not be power. You will also find that the total power radiated by an antenna is the surface integral of the average Poynting vector over a surface enclosing the antenna. The surface usually chosen is a sphere in the far field to keep the equations "simple". He is taking a portion of the equation and presenting it out of context assuming that this is a valid way to consider what he wishes to show. I would like to see the full equation. The devil is in the details. Actually, there is no "the" equation for the power radiated by an antenna other than the surface integral of the average Poynting vector over a surface enclosing the antenna. There are some approximate rules for specific cases and limiting conditions, but this isn't one of them. What he presented is for a 1/2 wavelegth antenna 9.87 and a full wave antenna 39.48. WTF is that?? I have no idea what you are talking about. Where did you get these numbers? 9.87 what? (2 * 3.14 * 5 meters / 10 meters) ^ 2 9.87 nothing; the expression is unitless, i.e. a pure number without units. In the expression you have a length divided by a length, which cancels into a unitles number. As my old physics professor used to say, always check the units of your answer; the arithmatic may be correct but it is meaningless unless the units are correct. Sample problem: You drove 100 miles and used 5 gallons of gas. What was your mileage? 5 gallons * 100 miles = 500 gallon-miles --- wrong units. 5 gallons / 100 miles = .05 gal/mile --- wrong units. 100 miles / 5 gallons = 20 miles/gal --- correct units and correct answer. I have no idea what you are going on about. Ok, 9.87 is a unitless number. So is 33.043. Now what? Exactly. Since it is a unitless number, it can not be power as claimed. Nor can it be a rule of thumb for power versus antenna length as a full wave antenna does not radiate 4 (39.48/9.87) more power than a 1/2 wave antenna. It is just nonsense. Did you read the OP? He says: "Has in the equation for radiated power the term" He is just giving us a portion of the equation to show the dependence on wavelength vs antenna length. But without the full equation we can't know if there are mitigating factors. Nothing that you have posted makes any sense in this context. -- Rick |
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#8
October 14th 14, 07:09 PM
posted to rec.radio.amateur.antenna
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Short antenna = reduced power
rickman wrote:
On 10/14/2014 12:58 PM, wrote: rickman wrote: On 10/14/2014 1:21 AM, wrote: rickman wrote: On 10/13/2014 11:45 PM, wrote: snip 9.87 nothing; the expression is unitless, i.e. a pure number without units. In the expression you have a length divided by a length, which cancels into a unitles number. As my old physics professor used to say, always check the units of your answer; the arithmatic may be correct but it is meaningless unless the units are correct. Sample problem: You drove 100 miles and used 5 gallons of gas. What was your mileage? 5 gallons * 100 miles = 500 gallon-miles --- wrong units. 5 gallons / 100 miles = .05 gal/mile --- wrong units. 100 miles / 5 gallons = 20 miles/gal --- correct units and correct answer. I have no idea what you are going on about. Ok, 9.87 is a unitless number. So is 33.043. Now what? Exactly. Since it is a unitless number, it can not be power as claimed. Nor can it be a rule of thumb for power versus antenna length as a full wave antenna does not radiate 4 (39.48/9.87) more power than a 1/2 wave antenna. It is just nonsense. Did you read the OP? He says: "Has in the equation for radiated power the term" He is just giving us a portion of the equation to show the dependence on wavelength vs antenna length. But without the full equation we can't know if there are mitigating factors. And it is utter nonsense. There is NOTHING about a 1 wavelength antenna that is 4 times that of a 1/2 wave antenna, or 16 times than of a 1/4 wave antenna, which is what the expression evaluates to. The part L/LAMBDA is the fractional size of the antenna, and the rest just numbers and I assume you have a calculator of some kind. It has already been shown by others that, neglecting loss, all power is radiated by an antenna no matter what the size. -- Jim Pennino |
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#9
October 14th 14, 09:19 AM
posted to rec.radio.amateur.antenna
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Short antenna = reduced power
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#10
October 14th 14, 05:59 PM
posted to rec.radio.amateur.antenna
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Short antenna = reduced power
Lostgallifreyan wrote:
wrote in : 5 gallons / 100 miles = .05 gal/mile --- wrong units. Context is everything.  There are cases where this reciprocated form gets used to good effect. Maybe convention demands 1/20 gal/mile just to indicate reciprocation of a normal convention. True but it is the answer to a different question. -- Jim Pennino |
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