Achieving Mars-Earth comunication High-Bandwidth

[MaximT]

Base on my post in computer engineering, about chatplace technics:

ORBITAL SPACE ARRAY

 

With the same kind of device, we could expect to reach Mars by telecommunication system with great bandwidth by only few step, the result will be an array of satellites (at least 6, for redundancy) to be placed in orbit around the Sun, our star, between Earth and Mars. My first estimation gives to me, with an array of 9 X 18-packed coloured LASER, distanced by 115 km. The power will be produced in every single devices, because the power requirement is low, by the detection rate that need only about less than 1E-12 J. We could achieved that without the use of massive lens, that could cost us billions of USD.

 

We will aim at the target with the best positioning device, and make the all thing oscillating both axis range, with a medium frequency, very low amplitude vibrator, to find within an ante-selected range obtain by Lambert algorithm, and with two synchronized clocks, send a signal when the target is hit... after we will follow the moving target by the lost of points in the signal.

 

Some calculations:

 

mirrors(4 m diameter), that will concentrate the diluted laser: theta = M² * λ / (pi * diameter)

1,1 * 325 E-9 m / (3,14159 * 0,0021 m) = 0,000 054 188°

distance between the mirrors = (1000 / 5) E9 m * Tan(0,000 054 188°) = 190 km

 

CATS AND RABBITS LINEAR SUPERPOSITION

 

 

Base on adding wave one to another we “may” achieved (18!)⁹. But it's stupid, because we can't have enough memory to translate it In that scenario, we could reduced the array to 2 X 2 or simply a triangle of 3: (18!)³ ...

 

USUAL IMAGERY DEVICES

IMX294CJK

Diagonal 21.63 mm (Type 4/3) Approx. 10.71M-Effective Pixel Color CMOS Image Sensor

• Large-size optical format (Type 4/3)

• Supports 4K output at 120 frame/s

• High sensitivity (SNR1s = 0.14 lx)

• High-speed interfaces (CSI-2/SLVS-EC*2)

• Supports Quad Bayer Coding HDR

 

It's clearly not what we need, but I need to get some number: 0,14 lux, it's going to be better

 

lux = lumen/m²

lumen = related to power consumption by specific device: 1/683 J/s

 

The LASER should have data like this: 230 mJ at 0,4 W, that gives 0,23 Joule / 15E-9 s = 15E6 J/s

That gives 10,2 E9 lumen, let's expect 2 mm of area, gives 3,25 E15 lux

At 200E9 meters, gives 3,25E9 / 1,13E11 = 0,03 lux

Magnify by a mirror of 10 X = 0,3 lux

 

Now, we only need the smallest telescope of best buy company

 

ELECTRICITY

 

A solar array like one in the ISS, is about 25 W per m². The consumption will be in the order of 10 W, without the communication device. Be provided to a Hall effect thruster is about 1 kW to achieved 83 mN at 3000 second of efficiency. But we could accumulated energy in a capacitor to achieved that...

8,3E-3 m/s², will be enough, the tank of fuel, with 10 kg, can hold at least 50 years...

 

 

[swansont]

Is this based on the topic you didn’t explain in another thread, and didn’t link to?

why use lux and lumen? Those are units that assume the human eye is involved?

why switch fonts in the middle, making it hard to read your already incomprehensible math? (Throwing an equation into a post without context is pretty useless)

[Strange]

11 hours ago, MaximT said:

With the same kind of device

Same kind of device as what?

11 hours ago, MaximT said:

we could expect to reach Mars by telecommunication system with great bandwidth by only few step

We can already do that? You may have missed the last few decades, but there have been robots on the surface, spacecraft in orbit and we have returned vast amounts of data about Mars.

11 hours ago, MaximT said:

the result will be an array of satellites (at least 6, for redundancy) to be placed in orbit around the Sun, our star, between Earth and Mars.

What for?

11 hours ago, MaximT said:

My first estimation gives to me, with an array of 9 X 18-packed coloured LASER, distanced by 115 km.

Where are these lasers and what are they for?

How did you estimate this?

11 hours ago, MaximT said:

We will aim at the target with the best positioning device, and make the all thing oscillating

Aim what?

At what target?

And what is the "thing" that is oscillating? And why?

11 hours ago, MaximT said:

CATS AND RABBITS LINEAR SUPERPOSITION

What do cats and rabbits have to do with it?

 

[Bufofrog]

11 hours ago, MaximT said:

CATS AND RABBITS LINEAR SUPERPOSITION

 

13 minutes ago, Strange said:

What do cats and rabbits have to do with it?

Clearly it was a typo, he meant to say ducks and walruses.

[MaximT]

3 hours ago, swansont said:

Is this based on the topic you didn’t explain in another thread, and didn’t link to?

Yes, I'm sorry, I should have wait for more time before posting again about a subject I didn't link, I will correct that as soon as possible.

 

3 hours ago, swansont said:

why use lux and lumen? Those are units that assume the human eye is involved?

It's because the sensor sensibility is given in those units.

 

3 hours ago, swansont said:

why switch fonts in the middle, making it hard to read your already incomprehensible math? (Throwing an equation into a post without context is pretty useless)

It's an image, I don't control the font, and I will implement more about that math, making it more comprehensible, as soon as possible.

 

1 hour ago, Strange said:

Same kind of device as what?

It's all about 2D compression ration, that I called CHATPLACE TECHNICS, with material memory compression.

 

1 hour ago, Strange said:

We can already do that? You may have missed the last few decades, but there have been robots on the surface, spacecraft in orbit and we have returned vast amounts of data about Mars.

We are limited to 5 MBytes/s, that's not enough for Mars colonization.

 

1 hour ago, Strange said:

What do cats and rabbits have to do with it?

It's a joke name built to be funny.

 

1 hour ago, Strange said:

Aim what?

At what target?

And what is the "thing" that is oscillating? And why?

One satellite array to the other next to it, to complete the link between Mars and Earth, the oscillating thing is the aiming direction of the LASER.

[swansont]

2 hours ago, MaximT said:

We are limited to 5 MBytes/s, that's not enough for Mars colonization.

Evidence?

Quote

It’s an image

No, it’s not. Your equation for the mirror was typed

15 hours ago, MaximT said:

mirrors(4 m diameter), that will concentrate the diluted laser: theta = M² * λ / (pi * diameter)

1,1 * 325 E-9 m / (3,14159 * 0,0021 m) = 0,000 054 188°

distance between the mirrors = (1000 / 5) E9 m * Tan(0,000 054 188°) = 190 km

See? (I copy-pasted it and removed the formatting)

Now all you have to do is explain what you think you’re doing with it

[Strange]

3 hours ago, MaximT said:

We are limited to 5 MBytes/s, that's not enough for Mars colonization.

By the time we are able to send people to Mars, the data rate will be higher. 

What data rate do you believe is required? And why?

3 hours ago, MaximT said:

It's a joke name built to be funny.

It is incomprehensible and not funny.

3 hours ago, MaximT said:

One satellite array to the other next to it, to complete the link between Mars and Earth, the oscillating thing is the aiming direction of the LASER.

What evidence do you have that higher data rates can be achieved with lasers than current technology (X- and Ka-band)?

Why is the laser oscillating instead of being consistently aimed?

How many real high-speed communication systems have you worked on? Are you familiar with information theory?

[MaximT]

1 hour ago, Strange said:

By the time we are able to send people to Mars, the data rate will be higher. 

What data rate do you believe is required? And why?

This amount of data is saturated by the use of all the wave length scale, we only could achieved few more than 5 M bytes / s, and that for all nation of Earth.

NASA:

Quote

Communications with Earth

 

 

Mars Reconnaissance Orbiter communicates with the Deep Space Network antennas on Earth using two different kinds of radio waves:

X-band:	the current standard in communications, which, when amplified, allows the orbiter to send data back to Earth more than 10 times faster than previous missions.
Ka-band:	a previously untested radio frequency four times higher than X-band, which allows scientists to bring data back even faster.

From the viewpoint of a Deep Space Network antenna on Earth, the orbiter spends about one-third of its time behind Mars during each orbit. During these times, the orbiter is "occulted from the Earth." During occultations, Mars Reconnaissance Orbiter cannot usefully send or receive radio signals.

So, out of 16 hours of daily Deep Space Network tracking, Mars Reconnaissance Orbiter sends data to Earth for 10 to 11 hours, and does that for about 700 days. The data rate is about 0.5 to 4 megabits per second.

 

For, the landing of the first astronautes and cosmonautes… I'm expecting, as citizen of Earth, at least one 4K size image data rate by landing site for every nation that will take part of this landing. So at least 20 nations X 6 M Bytes/s = 200 X the actual limit by those bands.

1 hour ago, Strange said:

Why is the laser oscillating instead of being consistently aimed?

It is oscillating, to find it in space at first, after we are going to aim more precisely to it, and apply a small oscillation, to make sure it hit it, because it is very far away: 120 E9 meters from each other. That because, it is almost improbable to find another way to achieved such precision.

 

1 hour ago, Strange said:

How many real high-speed communication systems have you worked on? Are you familiar with information theory?

Three theoretical devices. I studied in computer engineering, but it's more than a hobby to me.

 

2 hours ago, swansont said:

See? (I copy-pasted it and removed the formatting)

Now all you have to do is explain what you think you’re doing with it

Yeah, your are right, next time I will format my text evenly.

Those equation give the dispersion for a gaussian LASER, so an array of 3 X 3 or simply 3 in a triangle, will gives a huge data rate by that fact:

Where the summation of dot point = n,   data possible by frame = 2ⁿ

[Strange]

Just now, MaximT said:

This amount of data is saturated by the use of all the wave length scale, we only could achieved few more than 5 M bytes / s, and that for all nation of Earth.

OK. Don't answer the question. I was just wondering if you could. Obviously not.

1 minute ago, MaximT said:

For, the landing of the first astronautes and cosmonautes… I'm expecting, as citizen of Earth, at least one 4K size image data rate by landing site for every nation that will take part of this landing. So at least 20 nations X 6 M Bytes/s = 200 X the actual limit by those bands.

Do I really need to explain why this is complete nonsense?

2 minutes ago, MaximT said:

That because, it is almost improbable to find another way to achieved such precision.

That would be because you don't know what you are talking about.

What is the dispersion of a laser at that distance? In other words, how large an area would the signal be spread over?

And, again, why lasers?

 

[MaximT]

7 minutes ago, Strange said:

What is the dispersion of a laser at that distance? In other words, how large an area would the signal be spread over?

And, again, why lasers?

 

18 hours ago, MaximT said:

theta = M² * λ / (pi * diameter)

LASER allow us to trig a 2D array, so achieved by 2ⁿ, where n

 

14 minutes ago, MaximT said:

Where the summation of dot point = n,   data possible by frame = 2ⁿ

Actualy

 

15 minutes ago, MaximT said:

This amount of data is saturated by the use of all the wave length scale, we only could achieved few more than 5 M bytes / s, and that for all nation of Earth.

 

[swansont]

MRO mission was > 10 years ago, and the technology for missions is locked in earlier. So it’s not exactly new. Satellites have one difficulty that a surface station would not: the Doppler shift will have a wider range and change more rapidly. This might have an impact on the bandwidth 

Ka band has been used in Gigabit satellite communications, so the number for MRO is limited by something else. Which you have not identified.

25 minutes ago, Strange said:

What is the dispersion of a laser at that distance? In other words, how large an area would the signal be spread over?

I will give a hint to help with Strange’s question: research the moon corner cube retroreflectors, which they use to measure the distance to the moon. 

[MaximT]

14 minutes ago, swansont said:

Ka band has been used in Gigabit satellite communications, so the number for MRO is limited by something else. Which you have not identified.

Quote

Noise

Apart the Sun and the problem of distance, two other noise sources interfere with telecommunications : cosmic rays and thermal noise generated by the receiver.

The signal strength or noise level estimation, also known as the "dB below W" or dBW, is a measurement of the absolute power expressed in watts, and no more a power ratio like could be the decibel.

Knowing the signal power and the noise level at the source, at the distance of the Orbiter, we can estimate the signal-to-noise ratio (S/N) according to the bandwidth used.

Like in radioastronomy, in space communications, engineers estimate that a noise level of -215 dBW/Hz at 10 GHz is acceptable for the large ears of the DSN network.

For a bandwidth of 100 kHz and a signal close to 2x10^-16 W or -157 dB (-157 dBW) at reception, the S/N is only 8 dB. It can be twice as higher if the bandwidth is ten times shorter but this configuration is almost unusable in practice excepted in some digital transmission modes.

But 8 dB means that the DSN can theoretically receive such a signal without using error correction protocols, DSP systems or any BPSK or alike mode (although it does). In such conditions the transmission rate is relatively fast, up to 21 KB/s (166 kbit/s). It is this kind of "small budget" configuration that was used until 2005 by space probes like MGS and other Cassini.

That may help us to understand.

Quote

Signal loss calculation

(in free space)

L_dB = 92.4 + 20 Log (F_GHz x d_km)

Example. At 56 millions km from Earth, a signal transmitted from Mars in X-band at 8.4 GHz displays a loss of 266 dB, a power ratio >10²⁴.

steradian, looks to be the answer...

[swansont]

Please get into the habit of providing links to information, if it’s not your work. Such as things you quote.

4 minutes ago, MaximT said:

 steradian, looks to be the answer...

Snorkel. Albuquerque.

I can do it, too.

[MaximT]

6 hours ago, swansont said:

Snorkel. Albuquerque.

I can do it, too.

I looked into dictionary, I'm not sure if it's an insult.

[Strange]

8 hours ago, MaximT said:

LASER allow us to trig a 2D array, so achieved by 2ⁿ, where n

I have no idea what that means. 

Why lasers?

8 hours ago, MaximT said:

steradian, looks to be the answer...

That doesn’t answer anything

[Sensei]

The first of all 4K image, like on normal 4K TV, has 3840 x 2160 pixels with at least 24 bits per pixel RGB (3 bytes per pixel). So it's 3840 * 2160 * 3 bytes = 24,883,200 bytes = 23.73 MB of raw data. With 30 FPS it's 711.9 MB of raw data for video. Number of possible combinations of single frame is 2^ ( 3840 * 2160 * 3 * 8 ) = 2^ 199,065,600...

Because of the real (but lossy) compression, not the nonsense that you're writing in other thread, it can be squeezed to fraction of this value. Especially if new frame of video is very similar to previous frame which has been already sent, you can reduce amount of data sending just difference between frames, instead of full frame each time.

NASA should be more interested in receiving HDRI pictures, not RGB/RGBA. HDRI often uses 16 bits per channel or 32 bits per channel.

Edited July 29, 2019 by Sensei

[swansont]

3 hours ago, MaximT said:

I looked into dictionary, I'm not sure if it's an insult.

It's from National Treasure. Meaningless words, in the context of a conversation.

Steradian is the unit of solid angle. It makes as much sense to say "kilometer, looks to be the answer"