Home |
Search |
Today's Posts |
#12
|
|||
|
|||
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.Â* Next to it, I take a brighter bulb, where I know the brightness.Â* This light is NOT adjustable and is always the same known brightness.Â* Now, I move this bulb farther away until it appears to be exactly the same brightness as the RF powered light bulb.Â* At this point, I know: 1.Â* The distance between the observer (me) and the RF powered light bulb which I'll call A. 2.Â* The distance between the observer (me) and the reference light bulb which I'll call B. 3.Â* The brightness of the reference light bulb which I'll call C. 4.Â* 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.Â* I'll assume the reference light produced 1000 lux.Â* Therefore, the brightness of the RF powered light is: Â*Â* 1000 / (5/2)^0.5 = 1000 / 1.58 = 632 lux Presumably, the reference light was calibrated for brightness at a given RF level.Â* Let's say it's 50 watts for 1000 lux.Â* Therefore, the RF power of the RF powered light would be: Â*Â* 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 equationÂ* 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 incandescentÂ* 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 bu@@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’s 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 |
Thread Tools | Search this Thread |
Display Modes | |
|
|
Similar Threads | ||||
Thread | Forum | |||
a little 4nec2 help? | Antenna | |||
Anybody tried 4nec2 on Vista ? | Antenna | |||
New 4nec2 version | Antenna | |||
4nec2 and linux ?? | Antenna | |||
4nec2 question | Antenna |