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.

2 hours ago, Strange said:

Is it? Perhaps you can explain how taking 100 times more space to store data is "compression"?

In a cell phone with a ratio of 2048X, it will takes (1 mm X (10 mm X 10 mm)) / (7.5 mm X 200 mm X 100 mm) X 100 = less than 0.1%

3 hours ago, Strange said:

What is your compression algorithm?

Your right, it's not really a compression algorithm, it's material, with the ratio 2^m/2 / m

3 hours ago, Strange said:

Is it it lossless or lossy

It's lossless 100%

3 hours ago, Strange said:

How does it compare with other compression algorithms?

It can't be compare, it's incomparable!

3 hours ago, Strange said:

What data sources have you tested it with?

It's completely theoretical, the biggest issue is the time in nano second of the memory chip, to complete a cycle of comparaison, but with that ratio, it does not matter! 

    I imagined a huge net of copper wires or optic fiber, depending on ratio cost/bandwidth you want to achieved, place at the bottom of the sea. At every kilometer of wire, there will be a wire that goes upward with a communication device on it. When the submarine, or the aircraft carrier or other, want a secure and very fast connexion, they send a probe with a wire, to the bottom of the sea, to get a connexion with the nodal network.

    So, I don't know what are the odd: could I develop such a system, in the interest of Human kind?

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...

1 minute ago, swansont said:

Are you ever going to detail what this “thing” is?

Yes, of course, there are two option available, one with radio wave, and the best one with a system that consist in a camera and a small screen, at both end, to encode the data with a memory compressed system. The bandwidth could be a lot more impressive, in the preferable option.

2 hours ago, swansont said:

They use ELF signals to communicate with submerged subs. Slow, but it exists. How is sending this “thing” - that you have not explained - to the bottom and back faster?

Quote

Submarine communications[edit]

 

A ground dipole antenna used for transmitting ELF waves, similar to the U.S. Navy Clam Lake antennas, showing how it works. It functions as a huge loop antenna, with the alternating current I from the transmitter P passing through an overhead transmission line, then deep in the earth from one ground connection G to the other, then through another transmission line back to the transmitter. This creates an alternating magnetic field H which radiates ELF waves. The alternating current is shown flowing in one direction only through the loop for clarity.

The United States Navy utilized extremely low frequencies (ELFs) as radio band and radio communications. The Submarine Integrated Antenna System (SIAS) was a research and development effort to communicate with submerged submarines.^[26] The Soviet/Russian Navy also utilized ELFs for submarine communications system, ZEVS.^[27] The Indian Navy has an operational ELF communication facility at the INS Kattabomman naval base to communicate with its Arihant class and Akula class submarines.^[28][29]

Explanation[edit]

Because of its electrical conductivity, seawater shields submarines from most higher frequency radio waves, making radio communication with submerged submarines at ordinary frequencies impossible. Signals in the ELF frequency range, however, can penetrate much deeper. Two factors limit the usefulness of ELF communications channels: the low data transmission rate of a few characters per minute and, to a lesser extent, the one-way nature due to the impracticality of installing an antenna of the required size on a submarine (the antenna needs to be of an exceptional size in order to achieve successful communication). Generally, ELF signals have been used to order a submarine to rise to a shallow depth where it could receive some other form of communication.

Difficulties of ELF communication[edit]

One of the difficulties posed when broadcasting in the ELF frequency range is antenna size, because the length of the antenna must be at least a substantial fraction of the length of the waves. Simply put, a 3 Hz (cycle per second) signal would have a wavelength equal to the distance EM waves travel through a given medium in one third of a second. Taking account of refractive index, ELF waves propagate slightly slower than the speed of light in a vacuum. As used in military applications, the wavelength is 299,792 km (186,282 mi) per second divided by 50–85 Hz, which equals around 3,500 to 6,000 km (2,200 to 3,700 mi) long. This is comparable to the Earth's diameter of around 12,742 km (7,918 mi). Because of this huge size requirement, to transmit internationally using ELF frequencies, the Earth itself forms a significant part of the antenna, and extremely long leads are necessary into the ground. Various means, such as electrical lengthening, are used to construct practical radio stations with smaller sizes.

The US maintained two sites, in the Chequamegon-Nicolet National Forest, Wisconsin and in the Escanaba River State Forest, Michigan (originally named Project Sanguine, then downsized and rechristened Project ELF prior to construction), until they were dismantled, beginning in late September 2004. Both sites used long power lines, so-called ground dipoles, as leads. These leads were in multiple strands ranging from 22.5 to 45 kilometres (14.0 to 28.0 mi) long. Because of the inefficiency of this method, considerable amounts of electrical power were required to operate the system.

Because of the inefficiency of this method: ELF

We could achieved Gbit/s by my "thing".

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.

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

I'm talking about, the fact that a nuclear nation won't be allowed to change it's mind about a large scale attack, or have to wait a "lag" time period of many hour, staying submerge for detection purpose, because there is no transmission technique under the sea.

1 hour ago, Strange said:

In summary: I have absolutely no idea what you are jabbering on about.

 

Maybe you should start from the beginning and say what you are truing to achieve.

It's a data compression technics, that use a lot of memory. That memory data, will be throw in the data line on the reception of the integer address where it is stored, I'm using imaging to try explain all that. Sorry about being unable to achieved it easily, but I will complete today...

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.

Is this concept acceptable for a non-nuclear atrocity believer?

HIGH BANDWIDTH CONVENTIONAL SUBMARINES CLOSE TRANSFER WITH THE GROUND

Maybe, we could achieved an international network, dedicated to nuclear submarines efficiency, that way reducing the response time, so the risk associated with such lag responses :() :() :()

The formidable investment, will be rewarded by a new era of stability :()

My first estimation, for a complete network mapping of all oceanic floor, in case of an alien invasion :

• 361 132 000 km² of oceanic floor

• 361 132 000 / 25 = 14 Millions km

• 14E6 km * 132 USD/km = 1,8 Billions USD

• Lets double this price to include receptor every km

• The insignificant cost of the submarines modification (all the fleet? Yes, but this time the new submarine concept, I proposed, in case of aliens invasion...)

The principle is simple, the ship carry all the thing, and when she want a high debit connection, it send the thing at the bottom of the sea, and after the thing going back up with the link, and the connection is establish

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...

THE CHATPLACE TECHNICS

In honor of Laplace, I'm building this imaginary space that will be a complete screen of symbol to be read and encode by the two end computer that will transfer the images. In the imaginary space of 4K, there is the place for all characters of this planet in 20X20 format pixels, of about 20,736 completed characters. All the screen possibility will be encoded materially into a memory read only device, that will trig the data stored into by the call of the transfer screen number. In the case of Mars, that should be a 7 X 7 low dispersion laser array.

This system is now improved, by the use of single pixel (bit) in different screen space (memory available), for the different use.

PARENTHESIS ON THE POWER GENERATED BY THE CHATPLACE TECHNICS

The list of device that follow doesn't include the CHATPLACE device, because it does not exist actually

• SDRAM:

  • 4 G bytes / 10 ns : 1,5 W

• EEPROM:

  • 512 k bytes / 70 ns : 0,1 W

  • 1 M bytes / 70 ns : 0,1 W

  • 1 M bytes / 200 ns : 0,05 W

• LPDDR4:

  • 32 G bytes / 0,25 ns : 1 W N.B.: This 0,25 ns could theoretically be increase by burst data to 0,032 ns by adding a memory complex to compare with the packet within a 4096 Bus width, all that modifiable

My approximate evaluation for the power consumption, will include the others parts of the device, that will be estimated by doubling the power consumption of the memory. For a cell phone, with a factor CHATS-LAPINS of 2048X, it should be about 2 W, for a 4 G bytes memory, remembering the reader that the ratio shall be calculated with bits values...

I must remember the reader, that this ratio is applied on the real bandwidth, so, without parallelism of some devices, you won't achieved any increase in effective data rate in the phone, but only reduced the cost of the transmission by 2048

With a 5 M bytes transmission rate device, the cost of decreasing the rate and maintaining the total phone data rate, should be:

• With a 32 G bytes memory unit

  • A ratio of 2048, that mean 64 CHATPLACE unit parallel

  • 64/44 of the data: 1,6X

• With 4 X 32 G bytes memory unit

  • A ratio of 2048, that mean 256 CHATPLACE unit parallel

  • 256/40 of the data: 6,4X = 32 M bytes / second At the cost of 8 W and 4 mm X (10 X 10) mm, that will be restore to the phone by lowering the WiFi bandwidth by 320X, even more

• With a 32 G bytes memory unit

  • A ratio of 1092, 256 parallel

  • 256/21: 12,2X, but this time 160X on the WiFi, and 2 W, and 1 mm X (10 X 10) mm

SOME ELECTRONICS

To achieved better efficiency, there are some designs specifications to be described:

• The two different Bus of the single chip IC, that will included memory, are as usual address and data bus.

• Since that scanning comparison procedure will be incremented one by one in the same way every time.

• There is no need to input the addresses, but only to trigger the process, for synchronization purpose.

• The total memory size of the chip IC, will be (2^m/2 + 2) * (2^m/2) bits. For storing the compared data packet.

• That storage capacity is justify, by the use of very huge internal Bus width, for speed (overall time).

• The comparison could be achieved by multi “Equal” transistorized parallel system, for # of cycles too.

• Multi signal (cables or frequency) merge inside a separate chip IC (or the same one: heat consideration)

The final cycle time (to produced a “m” width “integer”) : Max ( 0,25 ns X 2^m/2 X 2^m/2 / Internal Bus Width, 2^m/2 (from more than one signal) / achievable input data Bus width (physical chip width) / 100 sub memory packets (0,002 ns transistor switching time, on a pin synchronization, within the physical input data Bus: theoretical) )

The maximum theoretical input: 5X 100 G bit/s input, but we won't achieved that The device looks to be limited only by the comparison process...

At the end: 0,25 ns X 2^m/2 seconds per cycle, with an internal Bus width of 2^m/2 = 61 kHz per parallel device

With the 100 G bits/s cable: 51,230 devices in parallel, to saturate it... with an astonishing 1,64E15 bytes of memory

3 minutes ago, Strange said:

I don't know where you get that idea from. It is wrong.

Misunderstanding there, when the moon is at the opposite of the sun, the moon attraction retrieved from the force of the Sun on the Earth. I mean both attract the earth, so the sum of force will be less on the Sun side, by that fact. But I understand that it's not modifying the gravity...

3 minutes ago, Strange said:

The Moon doesn't reduce the mass of the Earth.

In fact, yes, when it's at the opposite of the Sun, our star. I'm asking to myself, if that phenomena's happening at galaxy level? 

After some thinking, I'm expecting that the galaxy, the milky way, is not orbiting around it's Black Hole, but around it's own mass. A mass that is One Million time bigger than it's suppose Black Hole, at Sagittarius. The density of the BH, is extract from the star S2 motion, and has nothing to do with the whole rotation of the Galaxy...

Quote

In the Milky Way

 

Inferred orbits of 6 stars around supermassive black hole candidate Sagittarius A* at the Milky Way galactic center^[44]

Astronomers are very confident that the Milky Way galaxy has a supermassive black hole at its center, 26,000 light-years from the Solar System, in a region called Sagittarius A*^[45] because:

• The star S2 follows an elliptical orbit with a period of 15.2 years and a pericenter (closest distance) of 17 light-hours (1.8×10¹³ m or 120 AU) from the center of the central object.^[46]
• From the motion of star S2, the object's mass can be estimated as 4.1 million M_☉,^[47][48] or about 8.2×10³⁶ kg.
• The radius of the central object must be less than 17 light-hours, because otherwise S2 would collide with it. Observations of the star S14^[49] indicate that the radius is no more than 6.25 light-hours, about the diameter of Uranus' orbit.
• No known astronomical object other than a black hole can contain 4.1 million M_☉ in this volume of space.

Is a sum of mass orbiting a Black Hole, could contribute to the gravity force maintaining the rotation in orbit around the center of the galaxy, to make that galaxy objects spin around the center?

Is that sum of mass (all orbiting objects) could reduced the theoretical mass of the Black Hole itself, creating a empty, cannot be seen, zone around it, at the place of dilating it's volume to achieved the same density?

3 minutes ago, chrisnhowell said:

1. Would he need to travel constantly away from earth or could he just orbit for the relative time difference to occur?

Orbiting earth at such speed, can't, but if you are constantly pushing inside, why not. To me, it will suffice to achieve a time differential base on relativity.

5 minutes ago, chrisnhowell said:

2. What would would we hear/see from earth if we were in radio/video contact with him?

Time dilation should be constant, the signal won't reach him normally, it will have to be corrected every second... After a year of continuous correction, it will be the same, at correct time, the differential don't add over time.

Is that because we analysed the mass of rotating matter around them, and extracted that they shall weight enough to attracted it, at a certain rotation speed? And after deduced from their looking size, by our observation, what density they should have? Where is the Science on that, and the part of relativity to it ?

1 hour ago, awaterpon said:

Gravitational waves propagate with the speed of light c.

How do we measure speed of gravity, how to create waves of gravity, to get their speed?

21 hours ago, sandokhan said:

"The scalar portion of the original Maxwell equations expressed in quaternions was discarded (by Oliver Heaviside) to form "modern" EM theory; thus also the unified field interaction between electromagnetics and gravitation was discarded as well.

The quaternion scalar expression has, in fact, captured the local stress due to the forces acting one on the other. It is focused on the local stress, and the abstract vector space, adding a higher dimension to it.

One sees that, if we would capture gravitation in a vector mathematics theory of EM, we must again restore the scalar term and convert the vector to a quaternion, so that one captures the quaternionically infolded stresses. These infolded stresses actually represent curvature effects in the abstract vector space itself. Changing to quaternions changes the abstract vector space, adding higher dimensions to it.

Quaternions have a vector and a scalar part and have a higher topology than vector and tensor analysis."

I disagree, I revised Maxwell's work recently, and there is nothing like that. But Maxwell was searching for a reasoning to achieved a medium for propagation of light, but without confirming it. At the last chapter of his work, he made clear that he didn't know, with putting this in relation with the work of an Italian, Betti, a well known one. If you have some clues about quaternions, where are they? Give us more...

8 minutes ago, sandokhan said:

The Sagnac effect is a non-relativistic effect.

Personally, I agree, but I don't get what you are meaning. And, all the sciences, is against me.

You could refer to his book: Maxwell book tome II

16 minutes ago, sandokhan said:

A Sagnac light interferometer (MGX, RLGs) can detect BOTH EFFECTS: one is a physical effect proportional to the area - the CORIOLIS EFFECT; the other one is an electromagnetic effect proportional to the radius of rotation - the SAGNAC EFFECT.

I must agree with you, but not the actual sciences, Sagnac effect results had been explained by relativity, you may look on Wikipedia, the paper from India is wrong, to me. But what is behind you tough, about Aether? Are you sustaining that ether is not fix about all the Universe? Because all that, can only be explained by a non uniform repartition of the Aether, if there is...

This theoretical image show how to create matter, with this concept of Aether.

The Aether theory is not death yet:

Louis de Broglie

6 hours ago, swansont said:

your equation can’t work, since it has the wrong units.

In fact, you are right, My mistake, sorry:

In celestial mechanics, the standard gravitational parameter μ of a celestial body is the product of the gravitational constant G and the mass M of the body.

μ=GM

For several objects in the Solar System, the value of μ is known to greater accuracy than either G or M.^[11] The SI units of the standard gravitational parameter are m³s⁻². However, units of km³s⁻² are frequently used in the scientific literature and in spacecraft navigation

15 hours ago, MaximT said:

Basically, the speed on the surface will have to follow this rule: (where u = 4.905E12 m³/s²)

This equation is derived from the circular orbit speed, as described in literacy:

15 hours ago, MaximT said:

The limitation, for the length of the cable is 4,500 km high...

It's simply the breaking strength, in the spreadsheet, related to the mass of the cable, In it's most simple expression, but with centripetal force applied.