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4NEC2? |

4NEC2?Brian Reay wrote:
On 24/10/2018 08:35, brian wrote: In message , Spike writes On 19/10/2018 06:15, Jeff Liebermann wrote: Spike contended that a distant station cannot tell the difference between a sending station that has been tuned up using a torch bulb and one that has used an expensive VNA for that purpose. Via some other topics, the discussion on the torch bulb vs VNA issue has now reached this point: Let's start with an RF powered light at some brightness level.ﾂ* Next to it, I take a brighter bulb, where I know the brightness.ﾂ* This light is NOT adjustable and is always the same known brightness.ﾂ* Now, I move this bulb farther away until it appears to be exactly the same brightness as the RF powered light bulb.ﾂ* At this point, I know: 1.ﾂ* The distance between the observer (me) and the RF powered light bulb which I'll call A. 2.ﾂ* The distance between the observer (me) and the reference light bulb which I'll call B. 3.ﾂ* The brightness of the reference light bulb which I'll call C. 4.ﾂ* I'll call the unknown brightness of the RF powered bulb as D. Let's say that the observer is 2 meters away from the RF powered light, and that the reference light is the same brightness as the RF powered light at a distance of 5 meters.ﾂ* I'll assume the reference light produced 1000 lux.ﾂ* Therefore, the brightness of the RF powered light is: ﾂ*ﾂ* 1000 / (5/2)^0.5 = 1000 / 1.58 = 632 lux Presumably, the reference light was calibrated for brightness at a given RF level.ﾂ* Let's say it's 50 watts for 1000 lux.ﾂ* Therefore, the RF power of the RF powered light would be: ﾂ*ﾂ* 632 / 1000 * 50 = 32 watts A lamp of power or light output P has a light intensity at a point at a distance r from it that is a function of P/4pi*r^2 Two lamps of differing power or light output, P1 and P2, and spaced apart will have a point somewhere between them where the light intensities are equal. At this point the distance from P1 to the point of equal intensity is given by r1, and for P2 that distance is r2 and we thus have the equality given by: P1/(4pi*r1^2) = P2/(4pi*r2^2) Simplifying: P1/r1^2 = P2/r2^2 From which it can be seen that, if the power or light output of P1 is known then: P2 = P1(r2^2)/(r1^2) If P1 = 50 units of power or light output, and r1 and r2 are 5 and 2 units of distance respectively then: P2 = 50 * 2^2/5^2 = 50 * 4/25 = 8 units of power or light output, Now take your equationﾂ* above: 1000 / (5/2)^0.5 = 1000 / 1.58 = 632 lux Using my notation as above, this becomes P1/((r1/r2)^0.5) = P2 Rearranging this and separating the terms gives P1/(r1)^0.5 = P2/(r2)^0.5 You seem to have invented the Lieberman Law of Inverse Square Roots... If you're using incandescentﾂ* bulbs , the two bulbs have different colour temperatures . The dimmer one radiates more power in the red and infra red or "heat bands" My good friend Mr Planck sussed this out. If you try to use your eyes to judge brightness, the human eye spectral response causes the redder one to look disproportionally dimmer. I've had these sort of problems trying to simulate sunlight using tungsten lamps. There's a further problem introduced by the area of the lamp, which [email protected]@ers up the inverse square law. You might not have that problem with a "pea" bulb. Spike has enough rope to play with already. Jeff, like myself, tends to give idiots just enough to hang themselves. He (Jeff), really is very good at it. It窶冱 like that time Jeff tore a new arsehole into Frank Hunter GI4NKB over his appalling lack of knowledge of basic scripture. -- STC / M0TEY / http://twitter.com/ukradioamateur |

4NEC2?On 24/10/2018 07:35, brian wrote:
Spike writes On 19/10/2018 06:15, Jeff Liebermann wrote: Spike contended that a distant station cannot tell the difference between a sending station that has been tuned up using a torch bulb and one that has used an expensive VNA for that purpose. Via some other topics, the discussion on the torch bulb vs VNA issue has now reached this point: Let's start with an RF powered light at some brightness level. Next to it, I take a brighter bulb, where I know the brightness. This light is NOT adjustable and is always the same known brightness. Now, I move this bulb farther away until it appears to be exactly the same brightness as the RF powered light bulb. At this point, I know: 1. The distance between the observer (me) and the RF powered light bulb which I'll call A. 2. The distance between the observer (me) and the reference light bulb which I'll call B. 3. The brightness of the reference light bulb which I'll call C. 4. I'll call the unknown brightness of the RF powered bulb as D. Let's say that the observer is 2 meters away from the RF powered light, and that the reference light is the same brightness as the RF powered light at a distance of 5 meters. I'll assume the reference light produced 1000 lux. Therefore, the brightness of the RF powered light is: 1000 / (5/2)^0.5 = 1000 / 1.58 = 632 lux Presumably, the reference light was calibrated for brightness at a given RF level. Let's say it's 50 watts for 1000 lux. Therefore, the RF power of the RF powered light would be: 632 / 1000 * 50 = 32 watts A lamp of power or light output P has a light intensity at a point at a distance r from it that is a function of P/4pi*r^2 Two lamps of differing power or light output, P1 and P2, and spaced apart will have a point somewhere between them where the light intensities are equal. At this point the distance from P1 to the point of equal intensity is given by r1, and for P2 that distance is r2 and we thus have the equality given by: P1/(4pi*r1^2) = P2/(4pi*r2^2) Simplifying: P1/r1^2 = P2/r2^2 From which it can be seen that, if the power or light output of P1 is known then: P2 = P1(r2^2)/(r1^2) If P1 = 50 units of power or light output, and r1 and r2 are 5 and 2 units of distance respectively then: P2 = 50 * 2^2/5^2 = 50 * 4/25 = 8 units of power or light output, Now take your equation above: 1000 / (5/2)^0.5 = 1000 / 1.58 = 632 lux Using my notation as above, this becomes P1/((r1/r2)^0.5) = P2 Rearranging this and separating the terms gives P1/(r1)^0.5 = P2/(r2)^0.5 You seem to have invented the Lieberman Law of Inverse Square Roots... If you're using incandescent bulbs , the two bulbs have different colour temperatures . The dimmer one radiates more power in the red and infra red or "heat bands" My good friend Mr Planck sussed this out. If you try to use your eyes to judge brightness, the human eye spectral response causes the redder one to look disproportionally dimmer. I've had these sort of problems trying to simulate sunlight using tungsten lamps. There's a further problem introduced by the area of the lamp, which [email protected]@ers up the inverse square law. You might not have that problem with a "pea" bulb. Before moving on to the technological side of the issue, I thought it best to get the physics right first. Having done that one could match lamps for all sorts of things and improve the methodology, but if the premise of the physics is wrong, there's little point in doing that... In any case, the intensity indicating device will also bring limitations of its own to the method, but it may be that these are second-order issues that have only a minor effect on the system's overall accuracy, bearing in mind the original reason for using the set-up. -- Spike "Nearly all men can stand adversity, but if you want to test a man's character, give him an internet group to manage" |

4NEC2?On 24/10/2018 09:27, Brian Reay wrote:
On 24/10/2018 08:35, brian wrote: Spike writes On 19/10/2018 06:15, Jeff Liebermann wrote: Spike contended that a distant station cannot tell the difference between a sending station that has been tuned up using a torch bulb and one that has used an expensive VNA for that purpose. Via some other topics, the discussion on the torch bulb vs VNA issue has now reached this point: Let's start with an RF powered light at some brightness level.ﾃつ* Next to it, I take a brighter bulb, where I know the brightness.ﾃつ* This light is NOT adjustable and is always the same known brightness.ﾃつ* Now, I move this bulb farther away until it appears to be exactly the same brightness as the RF powered light bulb.ﾃつ* At this point, I know: 1.ﾃつ* The distance between the observer (me) and the RF powered light bulb which I'll call A. 2.ﾃつ* The distance between the observer (me) and the reference light bulb which I'll call B. 3.ﾃつ* The brightness of the reference light bulb which I'll call C. 4.ﾃつ* I'll call the unknown brightness of the RF powered bulb as D. Let's say that the observer is 2 meters away from the RF powered light, and that the reference light is the same brightness as the RF powered light at a distance of 5 meters.ﾃつ* I'll assume the reference light produced 1000 lux.ﾃつ* Therefore, the brightness of the RF powered light is: ﾃつ*ﾃつ* 1000 / (5/2)^0.5 = 1000 / 1.58 = 632 lux Presumably, the reference light was calibrated for brightness at a given RF level.ﾃつ* Let's say it's 50 watts for 1000 lux.ﾃつ* Therefore, the RF power of the RF powered light would be: ﾃつ*ﾃつ* 632 / 1000 * 50 = 32 watts A lamp of power or light output P has a light intensity at a point at a distance r from it that is a function of P/4pi*r^2 Two lamps of differing power or light output, P1 and P2, and spaced apart will have a point somewhere between them where the light intensities are equal. At this point the distance from P1 to the point of equal intensity is given by r1, and for P2 that distance is r2 and we thus have the equality given by: P1/(4pi*r1^2) = P2/(4pi*r2^2) Simplifying: P1/r1^2 = P2/r2^2 From which it can be seen that, if the power or light output of P1 is known then: P2 = P1(r2^2)/(r1^2) If P1 = 50 units of power or light output, and r1 and r2 are 5 and 2 units of distance respectively then: P2 = 50 * 2^2/5^2 = 50 * 4/25 = 8 units of power or light output, Now take your equationﾃつ* above: 1000 / (5/2)^0.5 = 1000 / 1.58 = 632 lux Using my notation as above, this becomes P1/((r1/r2)^0.5) = P2 Rearranging this and separating the terms gives P1/(r1)^0.5 = P2/(r2)^0.5 You seem to have invented the Lieberman Law of Inverse Square Roots... If you're using incandescentﾃつ* bulbs , the two bulbs have different colour temperatures . The dimmer one radiates more power in the red and infra red or "heat bands" My good friend Mr Planck sussed this out. If you try to use your eyes to judge brightness, the human eye spectral response causes the redder one to look disproportionally dimmer. I've had these sort of problems trying to simulate sunlight using tungsten lamps. There's a further problem introduced by the area of the lamp, which [email protected]@ers up the inverse square law. You might not have that problem with a "pea" bulb. Spike has enough rope to play with already. Excellent! Unable to attack the science, technology, or methodology, go for the man instead! One could not have a better seal of approval from Brian Reay, should anyone think such a thing had any value. Jeff, like myself, tends to give idiots just enough to hang themselves. He (Jeff), really is very good at it. Jeff's problems began when his nuclear-weapons handbook didn't appear to explain what the purpose was of the test under discussion. As a result he limited himself to EMP in a test that used a weapon that wasn't designed to produce it, and which was in fact being tested for different effects[1] so had to have this explained to him. Anyway this is all far above any pay grade you ever held so I would not have expected you to understand the difference, either then or now, but it was generous of you to remind us all once more of your shortcomings. [1] Jeff wrote "The Russian K program was there to investigate EMP just as were the Starfish and Fish-bowl US programmes, They were high altitude tests and produced no ground effects other than EMP, not good for blowing thing up on the ground", which of course ignores what the K tests were actually designed to test - which wasn't EMP. -- Spike "Nearly all men can stand adversity, but if you want to test a man's character, give him an internet group to manage" |

4NEC2?Spike wrote:
On 24/10/2018 07:35, brian wrote: Spike writes On 19/10/2018 06:15, Jeff Liebermann wrote: Spike contended that a distant station cannot tell the difference between a sending station that has been tuned up using a torch bulb and one that has used an expensive VNA for that purpose. Via some other topics, the discussion on the torch bulb vs VNA issue has now reached this point: Let's start with an RF powered light at some brightness level. Next to it, I take a brighter bulb, where I know the brightness. This light is NOT adjustable and is always the same known brightness. Now, I move this bulb farther away until it appears to be exactly the same brightness as the RF powered light bulb. At this point, I know: 1. The distance between the observer (me) and the RF powered light bulb which I'll call A. 2. The distance between the observer (me) and the reference light bulb which I'll call B. 3. The brightness of the reference light bulb which I'll call C. 4. I'll call the unknown brightness of the RF powered bulb as D. Let's say that the observer is 2 meters away from the RF powered light, and that the reference light is the same brightness as the RF powered light at a distance of 5 meters. I'll assume the reference light produced 1000 lux. Therefore, the brightness of the RF powered light is: 1000 / (5/2)^0.5 = 1000 / 1.58 = 632 lux Presumably, the reference light was calibrated for brightness at a given RF level. Let's say it's 50 watts for 1000 lux. Therefore, the RF power of the RF powered light would be: 632 / 1000 * 50 = 32 watts A lamp of power or light output P has a light intensity at a point at a distance r from it that is a function of P/4pi*r^2 Two lamps of differing power or light output, P1 and P2, and spaced apart will have a point somewhere between them where the light intensities are equal. At this point the distance from P1 to the point of equal intensity is given by r1, and for P2 that distance is r2 and we thus have the equality given by: P1/(4pi*r1^2) = P2/(4pi*r2^2) Simplifying: P1/r1^2 = P2/r2^2 From which it can be seen that, if the power or light output of P1 is known then: P2 = P1(r2^2)/(r1^2) If P1 = 50 units of power or light output, and r1 and r2 are 5 and 2 units of distance respectively then: P2 = 50 * 2^2/5^2 = 50 * 4/25 = 8 units of power or light output, Now take your equation above: 1000 / (5/2)^0.5 = 1000 / 1.58 = 632 lux Using my notation as above, this becomes P1/((r1/r2)^0.5) = P2 Rearranging this and separating the terms gives P1/(r1)^0.5 = P2/(r2)^0.5 You seem to have invented the Lieberman Law of Inverse Square Roots... If you're using incandescent bulbs , the two bulbs have different colour temperatures . The dimmer one radiates more power in the red and infra red or "heat bands" My good friend Mr Planck sussed this out. If you try to use your eyes to judge brightness, the human eye spectral response causes the redder one to look disproportionally dimmer. I've had these sort of problems trying to simulate sunlight using tungsten lamps. There's a further problem introduced by the area of the lamp, which [email protected]@ers up the inverse square law. You might not have that problem with a "pea" bulb. Before moving on to the technological side of the issue, I thought it best to get the physics right first. Having done that one could match lamps for all sorts of things and improve the methodology, but if the premise of the physics is wrong, there's little point in doing that... In any case, the intensity indicating device will also bring limitations of its own to the method, but it may be that these are second-order issues that have only a minor effect on the system's overall accuracy, bearing in mind the original reason for using the set-up. So many FURIOUS backpedals. -- STC / M0TEY / http://twitter.com/ukradioamateur |

4NEC2?On 25/10/2018 07:06, Jeff wrote:
Jeff's problems began when his nuclear-weapons handbook didn't appear to explain what the purpose was of the test under discussion. As a result he limited himself to EMP in a test that used a weapon that wasn't designed to produce it, and which was in fact being tested for different effects[1] so had to have this explained to him. Anyway this is all far above any pay grade you ever held so I would not have expected you to understand the difference, either then or now, but it was generous of you to remind us all once more of your shortcomings. [1] Jeff wrote "The Russian K program was there to investigate EMP just as were the Starfish and Fish-bowl US programmes, They were high altitude tests and produced no ground effects other than EMP, not good for blowing thing up on the ground", which of course ignores what the K tests were actually designed to test - which wasn't EMP. You were trying to deflect the argument, which was that low level bursts do no produce wide area EMP. To do that requires a high altitude burst, in the region of 50km +, which is outside the capabilities of most terrorist groups, which was what was bing discussed (although looking at the behaviour of some states these days the definition might be a little fudged). IIRC that's what Reay has claimed as well, but his first contribution to the thread was an attack on Cummins, one of several of his in that thread. Note that the Trinity test in 1945 had screened/hardened test equipment because of the EMP threat, and some of it failed due to the EMP. In any case Gareth's original post that started the thread makes no mention of the burst height, so there was nothing to 'fudge', as you put it, although it should be noted that you did say "Low altitude or ground level bursts produce little in the way of EMP, this is due to the mechanism that produces the pulse which requires a large mean free path between electrons in the atmosphere that only exists at very high altitudes.". 'Little' does not equal 'none', as they found out at Alamogordo. -- Spike "Nearly all men can stand adversity, but if you want to test a man's character, give him an internet group to manage" |

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