Spike has enough rope to play with already.


https://www.photobox.co.uk/my/photo?...d=501183701461

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

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 bu@@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"

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.ÂÂ* Next
to it, I take a brighter bulb, where I know the brightness.ÂÂ* This
light is NOT adjustable and is always the same known brightness.ÂÂ* Now,
I move this bulb farther away until it appears to be exactly the same
brightness as the RF powered light bulb.ÂÂ* At this point, I know:

1.ÂÂ* The distance between the observer (me) and the RF powered light
bulb which I'll call A.
2.ÂÂ* The distance between the observer (me) and the reference light
bulb which I'll call B.
3.ÂÂ* The brightness of the reference light bulb which I'll call C.
4.ÂÂ* 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.ÂÂ* I'll assume the reference
light produced 1000 lux.ÂÂ* Therefore, the brightness of the RF powered
light is:
ÂÂ*ÂÂ* 1000 / (5/2)^0.5 = 1000 / 1.58 = 632 lux

Presumably, the reference light was calibrated for brightness at a
given RF level.ÂÂ* Let's say it's 50 watts for 1000 lux.ÂÂ* Therefore, the
RF power of the RF powered light would be:
ÂÂ*ÂÂ* 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 equationÂÂ* 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 incandescentÂÂ* 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.

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"

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 bu@@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

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"