Apollo 11

[numbernu...@gmail.com]

The Caltech-MIT lunar laser ranging experiment is described in Smullin paper "Optical Echoes from the Moon" (1962). The Caltech-MIT lunar reflector experiment is an attempt at measuring the distance from the earth to the moon using a mirrored reflector that was install on the surface of the moon during the Apollo 11 mission. At the Lick observatory, a 2.3 W laser beam is pointed at the lunar reflector that is 238,900 miles from the earth. The laser beam's intensity propagates to the moon and is reflected by the lunar reflector then propagates back to the earth and detected by the Lick observatory. Using A/B = cos θ where A is the diameter of an object on the surface of the moon, B is the distance to the moon and θ is the telescopic resolution that is represented with A/B = θ when A/B → 0. The minimum diameter of a illuminated object that can be viewed from the earth by the Lick (.6 arcsec) is calculated,





A = θ x B = [(.6) / 3600] x (3.84 x 108 m) = 64 km...........................................................................................82





For the Lick to detected an illuminated object on the surface of the moon, the object must have the minimum diameter of 64 km yet the Lick telescope is detecting the intensity of the Caltech-MIT lunar reflector that has a surface area of one square meter. The Apollo 11 mission did not land on the moon since the intensity of a radio signal is dependent on the inverse of the second order of the distance I = A cos2(kr)/r2 . After propagating the distance of 238,000 miles (3.83 x 10^8 m) the intensity of a 1 kW radio signal would be reduced to 10-13 W. Radio telescopes are used to justify the communication with the Apollo mission but the sensitivity of a radio telescope is estimated at 1 x 10-9 W which would not be able to detect the 10-13 Watt radio signal that originates from the Apollo 11 mission. It is questionable how NASA communicated with the Apollo missions, Voyagers, and Mars probes using radio waves.

[Vaughn Fleck]

numbernumber1964 wrote:

> For the Lick to detected an illuminated object on the surface of the
> moon, the object must have the minimum diameter of 64 km yet the Lick
> telescope is detecting the intensity of the Caltech-MIT lunar reflector
> that has a surface area of one square meter.

This looks lika serious limitation to me. It makes the entire program
questionable.

[Steve BH]

=======

We have tried to explain high-gain (directional) antennas to you, but you're too stupid to understand the idea. So, you're stuck.

https://en.wikipedia.org/wiki/Directional_antenna

[intuit...@gmail.com]

I hope that you guys do realise that you don't actually need to land on the moon to place reflective mirrors there? You can just drop them onto the surface from an orbiting craft. The precise orientation in which they land won't even matter if the geometry of the reflectors are such that they consist of three mutually perpendicular mirrors like the inside of the corner of a cube, as that geometry will reflect any signal back in the direction from which it came.

[Zandra Sobczak]

intuitionist1 wrote:

>> https://en.wikipedia.org/wiki/Directional_antenna
>
> I hope that you guys do realise that you don't actually need to land on
> the moon to place reflective mirrors there? You can just drop them onto
> the surface from an orbiting craft.

No, they can't. This is impossible.

[dlzc]

Dear Zandra Sobczak:

https://en.wikipedia.org/wiki/List_of_retroreflectors_on_the_Moon
... 2 placed by Russians, who still have not landed men on the Moon.

But of course, you mean they cannot just be dropped from altitude at orbital speed, but requires a lander of some sort.

David A. Smith

[Zandra Sobczak]

My good old friend, so we should not trust too much that "manned" moon
landing? If it is this what you say. I feel depressed.

[intuit...@gmail.com]

I honestly don't see why not. You can embed lots of them into the surface of a large springy sponge or rubbery material which you can toss out of the orbiting craft. It will bounce around for a while and then settle. Perhaps a chemical inside the sponge can then be released which dissolves the foam, or some mechanical contraption inside the rubber which dismantles it, leaving the reflectors lying happily on the ground. Or perhaps... etc.

Or you could actually just pad the reflectors out and throw them out and hope for the best. They don't have to be bulky at all, and given the very simple mechanism by which they work they would probably still be functional even if they did happened to take quite a bit of damage on impact.

[Zandra Sobczak]

intuitionist1 wrote:

>> But of course, you mean they cannot just be dropped from altitude at
>> orbital speed, but requires a lander of some sort. David A. Smith
>
> I honestly don't see why not. You can embed lots of them into the
> surface of a large springy sponge or rubbery material which you can toss
> out of the orbiting craft. It will bounce around for a while and then
> settle. Perhaps a chemical inside the sponge can then be released which
> dissolves the foam, or some mechanical contraption inside the rubber
> which dismantles it, leaving the reflectors lying happily on the ground.
> Or perhaps... etc.

You are not easy to fool.

[intuit...@gmail.com]

Despite all of the other obvious problems, I personally do not understand how anyone with any ounce of common sense can possibly believe that they were able to successfully launch back off the moon and dock with the orbiter when it is not even possible in principle to test this rather complex process here on Earth due to the inability to recreate the moon's gravitational field and lack of atmosphere in any realistic way.

Would this not constitute some kind of an engineering miracle? Actually, 'miracle' is putting it mildly. It is an engineering impossibility and it simply did not happen.

It is a shame that this kind of mass delusion/blatant disregard for common sense can take place on such a large scale, but alas these things are commonplace, and we see it all the time, not least within the theoretical physics community.

Well, not that it particularly matters any more, but anyway. Just saying.

[Steve BH]

Orbital pursuit and docking was thoroughly tested in the Gemini program as one of the prime reasons to have it. It’s harder to do in Earth orbit where speed is 3.5 mile/sec, than Lunar orbit at one mile/sec.

In any case Newton’s equations (which you will never understand) predict the differences.

[numbernu...@gmail.com]

How did Nasa communicate with the Apollo 11 mission----since the intensity of a radio signal is dependent on the inverse of the second order of the distance I = A cos2(kr)/r2 . After propagating the distance of 238,000 miles (3.83 x 10^8 m) the intensity of a 1 kW radio signal would be reduced to 10-13 W. Radio telescopes are used to justify the communication with the Apollo mission but the sensitivity of a radio telescope is estimated at 1 x 10-9 W which would not be able to detect the 10-13 Watt radio signal that originates from the Apollo 11 mission.

[Odd Bodkin]

<numbernu...@gmail.com> wrote:
> How did Nasa communicate with the Apollo 11 mission----since the
> intensity of a radio signal is dependent on the inverse of the second
> order of the distance I = A cos2(kr)/r2 .

Wrong formula to use for a focused dish transmitter. The formula you’re
using is for a pole transmitter that an FM radio station might use.

> After propagating the distance of 238,000 miles (3.83 x 10^8 m) the
> intensity of a 1 kW radio signal would be reduced to 10-13 W. Radio
> telescopes are used to justify the communication with the Apollo mission
> but the sensitivity of a radio telescope is estimated at 1 x 10-9 W which
> would not be able to detect the 10-13 Watt radio signal that originates
> from the Apollo 11 mission.
>

--
Odd Bodkin — Maker of fine toys, tools, tables

[Volney]

On 6/7/2018 1:12 PM, numbernu...@gmail.com wrote:
> How did Nasa communicate with the Apollo 11 mission----since the intensity of a radio signal is dependent on the inverse of the second order of the distance I = A cos2(kr)/r2 . After propagating the distance of 238,000 miles (3.83 x 10^8 m) the intensity of a 1 kW radio signal would be reduced to 10-13 W. Radio telescopes are used to justify the communication with the Apollo mission but the sensitivity of a radio telescope is estimated at 1 x 10-9 W which would not be able to detect the 10-13 Watt radio signal that originates from the Apollo 11 mission.

> You were already told about high-gain antennas.

1kW? Why would NASA, with access to super dish antennas, use such a
wimpy ERP? Heck, the actual transmitter itself may have been in the
megawatt range. (although hams regularly bounce radio signals off the
moon using transmitters limited to 1000 watts input power. You should
see the homemade high-gain antennas many of them came up with)

[Paparios]

El jueves, 7 de junio de 2018, 13:12:58 (UTC-4), numbernu...@gmail.com escribió:
> How did Nasa communicate with the Apollo 11 mission----since the intensity of a radio signal is dependent on the inverse of the second order of the distance I = A cos2(kr)/r2 . After propagating the distance of 238,000 miles (3.83 x 10^8 m) the intensity of a 1 kW radio signal would be reduced to 10-13 W. Radio telescopes are used to justify the communication with the Apollo mission but the sensitivity of a radio telescope is estimated at 1 x 10-9 W which would not be able to detect the 10-13 Watt radio signal that originates from the Apollo 11 mission.

You are quite wrong!!

For typical radio applications, the FSPL equation becomes

FSPL(dB) = 20 log ( d ) + 20 log⁡ ( f ) + 92.45 db

At f = 1 GHz, d = 383000 km, FSPL = 148.28 dB

A 1.5 m radio antenna of the Apollo has a gain of 21.8 db at 1 GHz

a 34 m radio antenna receiver has a gain of 48.9 db at 1 GHz

Then if the Apollo transmitted at Pt = 10 W (10 dBW), the received power
was Pr = Pt + Gt - FSPL + Gr = 10+21.8-148.28+48.9 = -67.6 dBW,
which is easily detected.

[numbernu...@gmail.com]

Wrong formula to use for a focused dish transmitter. The formula you’re
using is for a pole transmitter that an FM radio station might use.

________________________________________________________


So what is the dispersion rate of a radio wave?

[numbernu...@gmail.com]

1kW? Why would NASA, with access to super dish antennas, use such a
wimpy ERP?

_________________________________________________________


https://www.google.com/search?q=apollo+11+radio+communication+system&rlz=1C1GCEA_enUS800US800&source=lnms&tbm=isch&sa=X&ved=0ahUKEwjn9fCrosLbAhV1OX0KHSfmAqUQ_AUICigB&biw=1600&bih=769#imgrc=Js_C60xzJzeOYM:


This looks like about 1 k W?

[numbernu...@gmail.com]

Then if the Apollo transmitted at Pt = 10 W (10 dBW), the received power
was Pr = Pt + Gt - FSPL + Gr = 10+21.8-148.28+48.9 = -67.6 dBW,
which is easily detected.

___________________________________


10 W is going to give you 100 mile range and microwaves does not increase the range.

[Volney]

With those nice dishes on the lunar rover and next to the lander? Those
will produce a wonderful ERP if fed with any nontrivial power!

(since you seem to have highlighted what's apparently a transmitter, I
assume you have no clue what I mean by the ERP and only care about the
transmitter power. So, go back and reread the post where antenna gain is
discussed)

(and you deleted the part about hams bouncing radio signals off the moon
with a 1 kW transmitter [actual input power, not ERP]. I have seen that
done _in person_. US law for amateur radio restricted transmitter power
to 1kW input, meaning even less than that actual output power. But no US
restriction on ERP! Just the laws of physics!)

[Steve BH]

How far away can you see a 10 watt (not milliwatt) green laser pointed at you, vs. a standard 100 W bulb that radiates 10 watts of vivible light (in all directions). 10 watts directed vs. 10 watts not.

It's the same question. Ten watts directed and focused is not all the same thing as 10 watts radiated in all directions.

Microwaves help because the focusing power of a dish antenna is in terns of number of wavelengths across the dish is.

[Steve BH]

On Thursday, June 7, 2018 at 12:22:21 PM UTC-7, numbernu...@gmail.com wrote:

No, a camera on the moon can see a 10 watt laser directed at it from the dark side of the Earth. That experiment was done with a Surveyor in the 1960's.

[Edward Prochak]

Never heard of directional antennae?
It is a directed beam.
Here is a nice slide show of all the antennae on the Apollo.
Some omni directional, some unidirectional.

https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20090015392.pdf

You know questions like this are often easily answered by
going to the source, in this case, NASA. Of course I assume
that you are interested in learning and not in promoting
some anti moon landing agenda.

Enjoy.
ed

[Edward Prochak]

Thanks for the calculations

FYI, the larger antenna used
Mode 1 = Frequency Modulation, High Power (20 watts)
included TV
Mode 2 = Phase Modulation, Low Power (.75 watts)
without TV

Transmit (moon to earth) was 2282.5 MHz
Receive (earth to moon) was 2101.8 MHz

I assume mode 2 was voice only. All from the
link I just posted a little bit ago.

Enjoy,
Ed

[Odd Bodkin]

Depends on the shape of the transmitter, obviously.
Note also that the dispersion rate of a 10W light bulb is different than
the dispersion rate of a 10W flashlight, which is different than the
dispersion rate of a 10W laser.

[Paparios]

Nonsense. You do not know what you are talking about!!

All long range radio (like the one of Cassini at Saturn) do operate at
relatively low power (43 dBm at the satellite in the Cassini case, that is 13
dBW or 20 Watts of transmitter power).

See https://descanso.jpl.nasa.gov/DPSummary/Descanso3--Cassini2.pdf for details.

The Cassini antenna had a 47.2 dB gain, which together with all losses, gives an
EIRP of 88.1 dBm (see table 5-2).

The free space loss from Saturn orbit is a whooping 294.4 dB.

The gain of the receiver antennas (DSN) is 68.24 dB.

So the total receiver power is -137.87 dBm

And the final result is that the receiver Pt/No is 45.19 dB-Hz, and the received
pictures prove that the signals from Saturn are easily received!!!

[intuit...@gmail.com]

On Thursday, June 7, 2018 at 5:35:48 PM UTC+3, Steve BH wrote:
> Orbital pursuit and docking was thoroughly tested in the Gemini program as one of the prime reasons to have it. It’s harder to do in Earth orbit where speed is 3.5 mile/sec, than Lunar orbit at one mile/sec.
>
> In any case Newton’s equations (which you will never understand) predict the differences.

I see, and this logic of yours' also applies to the perfect take-off from the moon first time without the ability to test beforehand as well does it? For some reason you don't say anything about that.

BTW, I have a PhD in theoretical physics from MIT.

[Steve BH]

The omnidirectional Moon antennas of course all compensated for with highly directional dish receivers. on Earth.

[furthermo...@gmail.com]

With those nice dishes on the lunar rover and next to the lander? Those

will produce a wonderful ERP if fed with any nontrivial power! ..

_____________________________________________________________________


A microwave is a radio wave. After leaving the dish a radio wave disperses dependent on the inverse of the second order of the distance I = A cos2(kr)/r2 . After propagating the distance of 238,000 miles (3.83 x 10^8 m) the intensity of a 1 kW radio signal would be reduced to 10-13 W which is not detectable.

[furthermo...@gmail.com]

How far away can you see a 10 watt (not milliwatt) green laser pointed at you, vs. a standard 100 W bulb that radiates 10 watts of vivible light (in all directions)

_____________________________________________________


A laser does not produce a radio signal.

[furthermo...@gmail.com]

Depends on the shape of the transmitter, obviously.
Note also that the dispersion rate of a 10W light bulb is different than
the dispersion rate of a 10W flashlight, which is different than the

dispersion rate of a 10W laser.....


_________________________________________


Again, laser or flashlight does not produce a radio signal.

[Odd Bodkin]

Laser light and radio waves are both electromagnetic radiation. The
dispersion properties for both depend on the shape of the transmitter. A
radio dish has the same shape as the parabolic reflector in a household
flashlight for a reason.

Is there some rational reason why everyday examples do not dissuade you
from repeating simple errors?

[Zandra Sobczak]

Paparios wrote:

>> 10 W is going to give you 100 mile range and microwaves does not
>> increase the range.
>
> Nonsense. You do not know what you are talking about!!
> All long range radio (like the one of Cassini at Saturn) do operate at
> relatively low power (43 dBm at the satellite in the Cassini case, that
> is 13 dBW or 20 Watts of transmitter power).

my friendos, the radio amateurs are been using the moon as reference from
the very beginning. No "mirror" is required. The Moon itself is the mirror.

[furthermo...@gmail.com]

Laser light and radio waves are both electromagnetic radiation.. The
dispersion properties for both depend on the shape of the transmitter.. A

radio dish has the same shape as the parabolic reflector in a household

flashlight for a reason..

_______________________________________________________


A parabolic reflector does not change the fact that a laser does not produce a radio signal.

[Odd Bodkin]

Alright. But it is STILL true that the dispersion rate for a radio wave
depends on the shape of the transmitter, and you cited the wrong dispersion
rate for the shape of the transmitter used by NASA for Apollo.

[furthermo...@gmail.com]

Alright. But it is STILL true that the dispersion. rate for a radio wave
depends on the shape of the transmitter, and you .cited the wrong dispersion
rate for the shape of the transmitter used by NASA. for Apollo.


___________________________________


The shape of the dish concentrates the radio wave and does not change the inherent structure of a radio wave that is based on spherical waves.

[furthermo...@gmail.com]

Nonsense. You do not know what you are talking about!!

All long range radio (like the one of Cassini at Saturn) do operate at
relatively low power (43 dBm at the satellite in the Cassini case, that is 13
dBW or 20 Watts of transmitter power).

See https://descanso.jpl.nasa.gov/DPSummary/Descanso3--Cassini2.pdf for details.

The Cassini antenna had a 47.2 dB gain, which together with all losses, gives an
EIRP of 88.1 dBm (see table 5-2).

The free space loss from Saturn orbit is a whooping 294.4 dB.

___________________________________________________________________


Which is closer the moon or Saturn?

[Edward Prochak]

On Friday, June 8, 2018 at 8:56:58 AM UTC-4, intuit...@gmail.com wrote:
>
> BTW, I have a PhD in theoretical physics from MIT.

yeah, right.

[Edward Prochak]

On Friday, June 8, 2018 at 11:36:06 AM UTC-4, Steve BH wrote:
> The omnidirectional Moon antennas of course all compensated
> for with highly directional dish receivers. on Earth.

Steve, the omni antennas were for local communications.
Astronaut - lander on the moon
LEM - Command Module in orbit

I already posted the link for the NASA presentation about
all the antennas on the LEM and CM. Here it is again

https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20090015392.pdf

HTH,
Ed

[Edward Prochak]

Which just shows you are incapable of understanding
the examples presented to you.
ed

[Odd Bodkin]

<furthermo...@gmail.com> wrote:
>
> Alright. But it is STILL true that the dispersion. rate for a radio wave
> depends on the shape of the transmitter, and you .cited the wrong dispersion
> rate for the shape of the transmitter used by NASA. for Apollo.
>
>
> ___________________________________
>
>
> The shape of the dish concentrates the radio wave

That is, it changes the dispersion.

> and does not change the inherent structure of a radio wave that is based
> on spherical waves.
>

There is no “inherent structure” to radio waves based on spherical waves.
What on earth are you talking about?

[Paparios]

Saturn is over 3000 times farther than the Moon and yet we received the HD
pictures of Cassini perfectly well...

[Steve BH]

But an antenna DOES produce coherent EM radiation, just like a laser, so the comparison is apt. What is difficult to do with visible light is the norm in radio and microwave generation. Like a laser, this longer wave EM radiation can then easily be focused into a tight beam, with divergence limited only by the width of the collector (actually a ratio of wavelength to antenna diameter).

With a big dish we can not only communicate with Voyager probes outside the solar system (which each have a small dish), but two radio dishes the size of Arecibo, with its 2.5 megawatt transmitter, it could in theory communicate with each other across the width of our galaxy. Closer to home, the Arecibo can hear amateur HAM messages bounced off the Moon. This doesn't even require an antenna on the Moon-- the radar is bounced off Lunar soil.

https://en.wikipedia.org/wiki/Arecibo_Observatory

https://en.wikipedia.org/wiki/Earth%E2%80%93Moon%E2%80%93Earth_communication

[Steve BH]

It doesn't need to. A dish concentrates the POWER of a radio wave in one direction (solid angle) at the expense of its power in other directions. The same net power goes out, but mostly in one direction, with the power mostly confined to a narrow solid angle. Within that angle the power still falls as 1/r^2 (conservation of energy) but the point is that it STARTS at a much high intensity because it is directed.

The gain of Acecibo is about 80 dBi at 3 GHz, which is 90 million or so. That mean's its 90 million times brighter in the direction of its beam than a spherical source that wasn't focused by the dish.

https://en.wikipedia.org/wiki/Parabolic_antenna#Gain

[Zandra Sobczak]

Steve BH wrote:

>> The shape of the dish concentrates the radio wave and does not change
>> the inherent structure of a radio wave that is based on spherical
>> waves.
>
> It doesn't need to. A dish concentrates the POWER of a radio wave in one
> direction (solid angle) at the expense of its power in other directions.

Of course not. Only a dish as a receiver. As a sender, there is not much
"concentration" but rather _directionality_, something having a power
distribution shape.

[Zandra Sobczak]

Paparios wrote:

>> The free space loss from Saturn orbit is a whooping 294.4 dB.
>> ___________________________________________________________________
>> Which is closer the moon or Saturn?
>
> Saturn is over 3000 times farther than the Moon and yet we received the
> HD pictures of Cassini perfectly well...

So you don't know how pictures are transmitted over large distances.

[intuit...@gmail.com]

https://arxiv.org/search/?query=sabbir+rahman&searchtype=all&source=header

[Edward Prochak]

What point are you trying to make?
just an insult? Why not add constructive information
or just stay quiet.

Ed

[Paparios]

Do not waste your time with the nymshifting troll. He is not interested
in anything but to write nonsensical contradictory stuff and irrelevancies.

[Steve BH]

Uh, either you're just trying to make the same point I am, or else you're wrong. <g>. ;'p

If you take a dipole antenna radiating 100 watts, it goes out in two large lobes in opposite directions, 50 watts into each. Now, put a simple (very large) metal "mirror" just behind the antenna (very near it-- inside 1/4 wave), in a position to cut off one lobe. What happens to the other? The mirror reflects the back lobe and it reinforces the front one. Which now looks much the same but has 100 watts going into it. It now it has twice the power in the same volume. Power has been CONCENTRATED (by a factor of two) into a smaller volume. There is no other word for it. Yagi arrays do a similar thing, but they use a bunch of elements placed like a diffraction grating to get the same effect without all that solid metal.

A dish antenna (which can be made of a thin net of wire, so long as it's smaller than quarter wave) has a radiation pattern consisting of very many small lobes and even a tiny back lobe, but nearly all the power is concentrated into the forward lobe.

That power from a dish is not only concentrated by removing it from other lobes (or from a theoretical radiation pattern which is uniform in all directions, never seen in practice), but also because the forward lobe has a very narrow radial distribution in both directions (theta and azimuth). This is far narrower than the distribution of one dipole lobe. The theoretic gain (after conversion from dB) is is the fraction of the solid angle getting power from the antenna, at the expense of the rest of the 4pi that sees little or nothing. It can be as high as a 100 million (Arecebo and 3 cm), and in practice can be nearly 70% as much as theory. Due to the reciprocity theorem the dish antenna as receiver is that much better at picking up radio power from THAT solid angle than it is from any other.

[furthermo...@gmail.com]

There is no “inherent structure” to radio waves based on spherical waves...
What on earth are you talking about?..

____________________________________________________________________


Can I communicate with the Apollo 11 mission using a cell phone?

[furthermo...@gmail.com]

With a big dish .we can not only communicate with Voyager probes outside the solar system (which each have a small dish), but two radio dishes the size of Arecibo, with its 2.5. megawatt transmitter, it could in theory communicate with each other across. the width of our galaxy. Closer to home, the Arecibo can hear amateur HAM. messages bounced off the Moon. This doesn't even require an antenna on the Moon-- the radar is bounced off Lunar soil..

__________________________________________________________________


What is the dispersion rate of a spherical wave?

[furthermo...@gmail.com]

Of course not. Only a dish as a receiver. As a sender, there is not much
"concentration" but rather _directionality_, something having a power
distribution shape.

________________________________________________


Again, what is the dispersion rate of a radio wave?

[furthermo...@gmail.com]

>> The free space loss from Saturn orbit is a whooping 294.4 dB..
>> ___________________________________________________________________

>> Which is closer the moon or Saturn?.

>
> Saturn is over 3000 times farther than the Moon and yet we received the

> HD pictures of Cassini perfectly well....

So you don't know how pictures are transmitted over large distances.

_____________________________________________________


What is the dispersion rate of a radio wave transmitted over large distances?

[furthermo...@gmail.com]

That power from a dish is not only concentrated. by removing it from other lobes (or from a theoretical radiation pattern which is uniform in all directions, never seen in practice), but also. because the forward lobe has a very narrow radial distribution in both directions (theta and azimuth). This is far narrower than the distribution of. one dipole lobe. The theoretic gain (after conversion from dB) is is the fraction of the solid angle getting power from the antenna, at the. expense of the rest of the 4pi that sees little or nothing. It can be as high as a 100 million (Arecebo and 3 cm), and in practice can be nearly 70% as. much as theory. Due to the reciprocity theorem the dish antenna as receiver. is that much better at picking up radio power from THAT solid angle than. it is from any other.

_______________________________________________________________


What is the dispression rate of the radio wave so I can make the calculations.

[Odd Bodkin]

No, of course not, because the antenna design for a cell phone is
completely different than the antenna design NASA used to communicate with
Apollo.

[Odd Bodkin]

Depends on the antenna design. There is no intrinsic dispersion rate
formula for radio waves. It all depends on the antenna.

[furthermo...@gmail.com]

No, of course not., because the antenna design for a cell phone is
completely different. than the antenna design NASA used to communicate with
Apollo. .


______________________________________________________


Is the NASA apollo mission using radio waves?

[furthermo...@gmail.com]

Depends on the antenna design.. There is no intrinsic dispersion rate
formula for radio waves.. It all depends on the antenna.

_____________________________________________________


Gives use an approximation dispersion rate of the best one.

[Odd Bodkin]

Yes, with a completely different antenna design used for other radio
applications.

You have used the wrong dispersion rate expression for the radio waves used
in the Apollo application. You used the dispersion rate expression for a
completely different application.

[Odd Bodkin]

This you should learn how to look up on your own. Your posts are filled
with factual errors, which you must correct by doing better online
research.

[Paparios]

https://en.wikipedia.org/wiki/Link_budget

[furthermo...@gmail.com]

https://en.wikipedia.org/wiki/Link_budget


This does not give the structural representation of the dispersion rate of a propagating radio wave.

[furthermo...@gmail.com]

- show quoted text -
This you should learn how to look up on your own. Your posts are filled.
with factual errors, which you must correct by doing better online.
research..

____________________________________________________


Really, then please correct me if I am wrong.

[numbernu...@gmail.com]

The photographic images of the Apollo 11 lander does not depict a blast zone beneath the exhaust nozzle of the lander caused by the 3,000 lb rocket thrust during the final descent. The argument that the 3,000 lb thrust is not significant enough to produce a blast zone beneath the lander is used to justify the non-existence of the blast zone. Using an analogy, a Lear jet engine is rated at 3,500 lb thrust. The 3,000 lb rocket thrust would result in a blast zone beneath the Apollo 11 lunar lander yet the Apollo 11 photographs (fig 22) show absolutely no disturbance of the fine particle matter in the area beneath the lunar lander.

[numbernu...@gmail.com]

The original film of Neil Armstrong stepping on the surface of the moon has been lost and the only record of the event was taken indirectly from a NASA monitor by a network television camera. NASA lost the original film of the most important engineering and scientific achievement in the history of man. There is also a problem regarding the distance of the camera and Armstrong making his first step on the moon's surface since according to NASA a video camera extends from the side of the lander is approximately three meters from the lander stairs but the film of Armstrong's first step onto the moon's surface depicts a camera that is more than six meters away from the lander since Armstrong takes four steps towards the camera yet 50% of Armstrong's image appears in the film which proves that the camera was more than five meters from the ladder and that the filming of Armstrong's first step on the lunar surface was taken by a camera separate from the lander in a movie studio. The opening eye of the camera is pointing at a position .5 feet below the lunar ladder and does not correspond with the angle of the image represented in the film.

[numbernu...@gmail.com]

That's with the addition of the fact that radio waves do not function as a commuication system from the moon.

[Odd Bodkin]

You are wrong in dozens of places. You even got the fuel composition of the
Lunar Lander wrong and that is an EASY fact to look up).

Posting something that is riddled with dozens of errors and asking people
to correct each of your mistakes doesn’t accomplish anything, because
you’ll have learned nothing about how to check facts and you’ll just go on
to make more dozens of mistakes. Instead, you have to be taught how to
research facts on your own, so that you’ll correct dozens of mistakes FIRST
before you post them.

[numbernu...@gmail.com]

Posting something that is riddled with dozens of errors. and asking people
to correct each of your mistakes doesn’t accomplish. anything, because
you’ll have learned nothing about how to check facts. and you’ll just go on
to make more dozens of mistakes. Instead, you have. to be taught how to
research facts on your own, so that you’ll correct. dozens of mistakes FIRST
before you post them..

____________________________________________________________

Please, elaborate. Also, don't forget to include the dispersion rate of a radio wave, while your at it which shouldn't be hard since there are dozens (30) of errors.

[numbernu...@gmail.com]

Rocket propellant - Wikipedia
https://en.wikipedia.org/wiki/Rocket_propellant
Rocket propellant is a material used either directly by a rocket as the reaction mass (propulsive .... Thus, the Saturn V first stage used kerosene-liquid oxygen rather than the liquid hydrogen-liquid oxygen used on its upper stages. Similarly, the ...


RP-1 (alternately, Rocket Propellant-1 or Refined Petroleum-1) is a highly refined form of kerosene outwardly similar to jet fuel, used as rocket fuel. RP-1 has a lower specific impulse than liquid hydrogen (LH2), but is cheaper, stable at room temperature, far less of an explosion hazard, and far denser. RP-1 is significantly more powerful than LH2 by volume. RP-1 also has a fraction of the toxicity and carcinogenic hazards of hydrazine, another room-temperature liquid fuel. RP-1 is most commonly burned with LOX (liquid oxygen) as the oxidizer, though other oxidizers have also been used. RP-1 is a fuel in the first-stage boosters of the Soyuz-FG, Zenit, Delta I-III, Atlas, Falcon 9, Antares and Tronador II rockets. It also powered the first stages of the Energia, Titan I, Saturn I and IB, and Saturn V. ISRO is also developing a RP-1 fueled engine for its future rockets.[2]

____________________________________________________



Calling rocket fuel kerosene is close enough to show how much rocket smoke is produced but according to you. Rocket fuel does not produce smoke and radio waves can propagate to the moon. Right

[Odd Bodkin]

I’ve already explained why I’m not going to do that.

There’s a famous saying that the lazy man’s way of getting answers on the
internet is to post the wrong answer and wait for others to correct it.

[Odd Bodkin]

No it is not close enough. The specific fuel used in the Lunar Lander
contains no kerosene whatsoever. And the fuel that was in fact used does
not produce smoke. So it is not close enough at all.

[numbernu...@gmail.com]

https://www.youtube.com/watch?v=xO-CE6SW7eo


Aerozine 50 is a 50/50 mix by weight of hydrazine and unsymmetrical dimethylhydrazine (UDMH),[1][2] originally developed in the late 1950s by Aerojet General Corporation as a storable, high-energy, hypergolic fuel for the Titan II ICBM rocket engines.


________________________________________________



Is that enough smoke for yous???

[Paparios]

El lunes, 11 de junio de 2018, 15:24:00 (UTC-4), furthermo...@gmail.com escribió:
> https://en.wikipedia.org/wiki/Link_budget
>
>
> This does not give the structural representation of the dispersion rate of a propagating radio wave.

You are a troll right?

Equation Lfs in the above link does exactly that, but you are too dumb to
recognize it.

PLONK!

[numbernu...@gmail.com]

.https://en.wikipedia.org/wiki/Link_budg
.This does not give the structural representation of the dispersion rate of a propagating radio wave..

You are a troll right?.

Equation Lfs in the above link does exactly that, but you are too dumb to

recognize it. .

__________________________________________________


According to your link a cell phone could communicate with the Apollo mission.



https://www.google.com/search?q=girl+troll&tbm=isch&tbs=rimg:CUWkO4vKACLXIjjZ7LE3ewSFZ9agV2n41G--m5j7S8R0idEll2rG91_18ktJo8in-RmOC4q_1zvjwtNbJQcITzkhdDYCoSCdnssTd7BIVnETD7n5c8yiIAKhIJ1qBXafjUb74RptCqszHkzCkqEgmbmPtLxHSJ0RGDAXPvvE1PQyoSCSWXasb3X_1ySEZFakbOWbwY-KhIJ0mjyKf5GY4IRdDn2d-yqZ64qEgnir_1O-PC01shHFXLcozjAzbyoSCVBwhPOSF0NgETrcch6Ff7gA&tbo=u&sa=X&ved=2ahUKEwjfkMihzczbAhUqi1QKHYYRCLIQ9C96BAgBEBs&biw=1920&bih=1014&dpr=1#imgrc=m5j7S8R0idEsdM:

[numbernu...@gmail.com]

[numbernu...@gmail.com]

The lunar surface during the day reaches the temperature of 400o Fahrenheit which would result in the Apollo astronaut's space suit to explode and the plastic magnetic tape within the video camera to melt after more than four hours on the lunar surface. The argument that the space suit reflects the Sun's intensity is insufficient since the lunar heat would conduct through the astronaut's boots and the interior gas within the space suit which would result in the space suit to exploding. The near side of the moon where the lander descended is continuously illuminated by the Sun's maximum intensity for more than a thousand years. The moon near side is always illuminate and forms a full moon perpetually except during a lunar eclipse.

[numbernu...@gmail.com]

In a film of an Apollo astronaut walking on the surface of the moon shows the placement of the American flag on the surface of the moon but in the film, the flag appears to be flapping similar to a flag blowing in the wind yet the surface of the moon has no atmosphere that could form the waving of the flag with the intensity depicted in the Apollo 11 film. It is argued that the vibration of the flag created by the astronaut placing the flag onto the moon causes the flag to wave but the intensity of the flag waving in the horizontal direction suggests that the Apollo 11 mission flag waving was created by a cooling fan within a movie studio

[Zandra Sobczak]

Steve BH wrote:

>> > It doesn't need to. A dish concentrates the POWER of a radio wave in
>> > one direction (solid angle) at the expense of its power in other
>> > directions.

>>
>> Of course not. Only a dish as a receiver. As a sender, there is not
>> much "concentration" but rather _directionality_, something having a
>> power distribution shape.
>
>

> Uh, either you're just trying to make the same point I am, or else
> you're wrong. <g>. ;'p
> If you take a dipole antenna radiating 100 watts, it goes out in two
> large

I don't know what you are talking about. If don't understand "power
distribution" from a sender, and NO "directionality" I can't help.

As a receiver, yes, a parabola may concentrate the power in one focal
point. But not the other way around.

[Zandra Sobczak]

Edward Prochak wrote:

> On Saturday, June 9, 2018 at 11:13:39 AM UTC-4, Zandra Sobczak wrote:
>> Paparios wrote:
>>
>> >> The free space loss from Saturn orbit is a whooping 294.4 dB.
>> >> ___________________________________________________________________
>> >> Which is closer the moon or Saturn?
>> >
>> > Saturn is over 3000 times farther than the Moon and yet we received
>> > the HD pictures of Cassini perfectly well...
>>
>> So you don't know how pictures are transmitted over large distances.
>
> What point are you trying to make?
> just an insult? Why not add constructive information or just stay quiet.

What's the point?? Can't you read, the poster implies analog modulated EM
waves. Over large distances this is not possible due to the week signal.
Hence digital modulation has to be used. Which is slow but accurate to a
given resolution. Kindergarten stuff, where have you been.

[Zandra Sobczak]

Paparios wrote:

>> > >> The free space loss from Saturn orbit is a whooping 294.4 dB.
>> > >> ___________________________________________________________________
>> > >> Which is closer the moon or Saturn?
>> > >
>> > > Saturn is over 3000 times farther than the Moon and yet we received
>> > > the HD pictures of Cassini perfectly well...
>> >
>> > So you don't know how pictures are transmitted over large distances.
>>
>> What point are you trying to make?
>> just an insult? Why not add constructive information or just stay

>> quiet. Ed
>
> Do not waste your time with the nymshifting troll. He is not interested
> in anything but to write nonsensical contradictory stuff and
> irrelevancies.

It appears you just been caught stupid once again. I can't believe this is
the first time.

[numbernu...@gmail.com]

The point is that immediately after leaving the transmission antenna, no matter what the shape the the anatenna the intensity of a radio signal is dependent on the inverse of the second order of the distance I = A cos^2(kr)/r^2 . After propagating the distance of 238,000 miles (3.83 x 10^8 m) the intensity of a 1 kW radio signal would be reduced to 10^-13 W that is not detectable. In another words, there is in fact a limit to the range of a radio wave that cannot reach the moon which prove NASA is a hoax.

[Odd Bodkin]

<numbernu...@gmail.com> wrote:
> The point is that immediately after leaving the transmission antenna, no
> matter what the shape the the anatenna the intensity of a radio signal is
> dependent on the inverse of the second order of the distance I = A cos^2(kr)/r^2 .

That is an incorrect statement, one of many incorrect statements you make.
Then, when the mistake is pointed out, you persist in repeating the same
mistake.

You clearly have an agenda, one that is compromising your integrity. What
is that motive?

> After propagating the distance of 238,000 miles (3.83 x 10^8 m) the
> intensity of a 1 kW radio signal would be reduced to 10^-13 W that is not
> detectable. In another words, there is in fact a limit to the range of a
> radio wave that cannot reach the moon which prove NASA is a hoax.
>

[numbernu...@gmail.com]

That is an incorrect statement,. one of many incorrect statements you make.
Then, when the mistake is pointed out,. you persist in repeating the same
mistake..

You clearly have an agenda, one that is compromising your integrity.. What
is that motive?,
_________________________________________________________________________


You always say this. Why don't you point out what is incorrect in your statement to save time.

[Steve BH]

On Tuesday, June 12, 2018 at 1:50:58 PM UTC-7, numbernu...@gmail.com wrote:
> The point is that immediately after leaving the transmission antenna, no matter what the shape the the anatenna the intensity of a radio signal is dependent on the inverse of the second order of the distance I = A cos^2(kr)/r^2 . After propagating the distance of 238,000 miles (3.83 x 10^8 m) the intensity of a 1 kW radio signal would be reduced to 10^-13 W that is not detectable. In another words, there is in fact a limit to the range of a radio wave that cannot reach the moon which prove NASA is a hoax.

A radiowave cannot reach the Moon? I suppose then we can't control spacecraft beyond the Moon? Or receive signals from them?

How do we get radiowaves from SigA* in the center of our galaxy? Or from other galaxies, pray tell? Radioastronomy is a hoax, you think?

Okay, here's your clue: Although your equation is correct, you don't know what "A" is. It's the initial power or amplitude of the radio beam, but that's not the power to the radio transmitter. The whole point of a high gain antenna, is that it takes the power of the radio transmitter and amplifies it in one direction (which removing it from others, due to conservation of energy).

Just a 2 m dish antenna, the kind satellite crazies have in their back yards, can give you a gain of 44 dB, which is 10^4.4 = 25,000 at the wavelength used. If you feed such an antenna with 100 watts of power, along the beam direction it's putting out 25,000 times the isotropic power, which is something like 2.5 million watts (if it were going in all directions, which it is not. It works like a magnifying glass with sunlight.

So 25 million watts is your "A". And it drops by 1/r^2, but it starts out very large.

http://www.astrosurf.com/luxorion/qsl-antenna5.htm

SBH

[furthermo...@gmail.com]

How do we get radiowaves from SigA* in the center of our galaxy? Or from other galaxies, pray tell? Radioastronomy is a hoax, you think?

____________________________________________________________________


Try to find the sensitivity of a radio telescope, in watts. When you attempt to find the sensitivity they give your a range of frequencies. Now if radioastronomy was not a hoax they would give you the sensitivity but the sensitivity would prove that radioastronomy is in fact a hoax which is the reason they are concealing the sensitivity in watts.. Jackson.

[furthermo...@gmail.com]

ust a 2 m dish antenna, the kind satellite crazies have in their back yards, can give you a gain of 44 dB, which is 10^4.4 = 25,000 at the wavelength used. If you feed such an antenna with 100 watts of power, along the beam direction it's putting out 25,000 times the isotropic power, which is something like 2.5 million watts (if it were going in all directions, which it is not. It works like a magnifying glass with sunlight.

So 25 million watts is your "A". And it drops by 1/r^2, but it starts out very large.

___________________________________________________________________



25 million watts is alot power for a radio wave transmitter. Ah old chap.

[Odd Bodkin]

I just pointed out which of your statements is incorrect (or at least one
of them). You apparently want me to provide a corrected statement to save
you time. I think you need to spend the time and effort to find the correct
answer, rather than just babbling one incorrect statement after another.
Being wrong is easy and a baboon can do it. Being right takes work. Don’t
be a lazy baboon, m’k?

[Steve BH]

Yes it would be. But you're getting 100 watts into 1/25000 of the solid angle, in this case just 1/2000 steradian. So it looks like 2.5 million watts along that solid angle.

The intensity of a radio beam A is not in watts but watts PER AREA.

How do you think an ant feels under a magnifying glass? Is the amplitude and intensity of the light the same? Hardly.

[furthermo...@gmail.com]

I just pointed out which of your statements is incorrect (or at least one
of them). You apparently want me to provide a corrected statement to save
you time. I think you need to spend the time and effort to find the correct
answer, rather than just babbling one incorrect statement after another.
Being wrong is easy and a baboon can do it.

______________________________________________________



I cann't remember which one could you repeat it since there are so many posts.

[furthermo...@gmail.com]

es it would be. But you're getting 100 watts into 1/25000 of the solid angle, in this case just 1/2000 steradian.. So it looks like 2.5 million watts along that solid angle..

The intensity of a radio beam A is not in watts but watts PER AREA..

How do you think an ant feels under a magnifying glass? Is the amplitude and intensity of the light the same? Hardly..

______________________________________________________


What is your definition of the dispersion rate of a radio wave? Example, 1/r^2 or 1/r^3 or 1/r^4 which is the closest to your rate so that I can further analysis the problem. Jackson.

[furthermo...@gmail.com]

The intensity of a radio beam A is not in watts but watts PER AREA.

__________________________________________________________________


That might be so but the original power of the source is also important. Example, the power of an FM radio station when represented with 3 W as compared to 20 kW would give an approximation of the signal strength or weakness. Say it Jackson.

[Odd Bodkin]

So let’s see. You post in the internet but you can’t use the internet to
check facts. You use Usenet but you don’t know how to follow a conversation
thread to remind yourself what was said yesterday. Hmmmm.

[numbernu...@gmail.com]

The point is that immediately after leaving the transmission antenna,, no matter what the shape the the anatenna the intensity of a radio signal is dependent on the inverse of the second order of the distance I = A cos^2(kr)/r^2 ., After propagating the distance of 238,000 miles (3.83 x 10^8 m) the intensity of a 1 kW radio signal would be reduced to 10^-13 W that is not detectable.. In another words, there is in fact a limit to the range of a radio wave that cannot reach the moon which prove NASA is a hoax..

____________________________________________________________________



What is incorrect about this?

[Steve BH]

Only because the radiating antenna is the same. With a dish antenna 3 watts in the right direction can look exactly like 20 kW. Or more. The energy is directionally concentrated into a narrow beam. What about that do you not understand? Apparently, everything.

[Steve BH]

I = A cos^2(kr)/r^2 . I don't know where you got this formula, but it's certainly not universally applicable.

You can't just assume that I and A will be the same at any r around an antenna. The 1/r^2 dependence for example only describes the far field (EMR), which begins at a dish at 2D^2/lambda. If D is 100 times lambda, that's 200 lambda. For a 2 m dish and and 2 cm waves you need 200 * 2cm = 4 m before you have a decent far field, and your 1/r^2 dependence appears there and farther. And it will indeed be thousands of times brighter in that direction than the field from a simple 5 mm dipole antenna connected to the same source and power, and seen from 4 m away.

Look at the hot filament of an auto headlamp from various angles in front of the car, with your eye a foot from the bulb face. When you get your eye exactly in front, suddenly the image of that hot filament fills every part of the reflector behind it, and it's dazzlingly brighter. I and A have jumped by a factor of several hundred. If you could see microwaves, you'd get the same effect from dish antenna.

And yet the headlight beam from an automobile exhibits the same 1/r^2 intensity (intensity = power density = power/area) dependence.

[Edward Prochak]

Okay students, how many factual errors can you find in num's post today?

On Tuesday, June 12, 2018 at 12:36:23 PM UTC-4, numbernu...@gmail.com wrote:
> The lunar surface during the day reaches the temperature of 400o Fahrenheit which would result in the Apollo astronaut's space suit to explode and the plastic magnetic tape within the video camera to melt after more than four hours on the lunar surface. The argument that the space suit reflects the Sun's intensity is insufficient since the lunar heat would conduct through the astronaut's boots and the interior gas within the space suit which would result in the space suit to exploding. The near side of the moon where the lander descended is continuously illuminated by the Sun's maximum intensity for more than a thousand years. The moon near side is always illuminate and forms a full moon perpetually except during a lunar eclipse.

Num, you are hilarious!
Thanks for the laugh.
ed

[Edward Prochak]

On Tuesday, June 12, 2018 at 12:37:08 PM UTC-4, numbernu...@gmail.com wrote:
> In a film of an Apollo astronaut walking on the surface of the moon shows the placement of the American flag on the surface of the moon but in the film, the flag appears to be flapping similar to a flag blowing in the wind yet the surface of the moon has no atmosphere that could form the waving of the flag with the intensity depicted in the Apollo 11 film. It is argued that the vibration of the flag created by the astronaut placing the flag onto the moon causes the flag to wave but the intensity of the flag waving in the horizontal direction suggests that the Apollo 11 mission flag waving was created by a cooling fan within a movie studio

boy even the non-scientists of Myth Busters show this claim is false!
ed

[Edward Prochak]

I do not see the implied analog claim. In fact I thought the mention
of HD implied digital transmission. But at least now you made your
comment clear. No need to add insults.

Ed

[Edward Prochak]

that should give you a clue about the accuracy of your posts!

[furthermo...@gmail.com]

Only because the radiating antenna is the same. With a dish antenna 3 watts in the right direction can look exactly like 20 kW. Or more. The energy is directionally concentrated into a narrow beam. What about that do you not understand? Apparently, everything.

__________________________________________________


Yes, but what is the dispersion rate of this concentrated beam?