BYTE SHIFT
HELLO WORLD

 

Part2

SIGNAL SPEEDS

Han Erim

November 30, 2015 

Copyright 2015 © Han Erim. All right reserved

 

 

In the first part of the article, we learnt how Byte Shift situation occurred on the electromagnetic signals because of Doppler Shift when wireless connection is done with the moving frames (reference systems). In this second part, benefiting from the Byte Shift event, I will tell the topic of signal speed going towards a moving frame.

 

I did not tell the things here told in the first text to not to repeat. Because of this, if you did not read the first part, I request you to start by reading it firstly.

 

Our case here consists of two planes flying in the opposite directions, a mountain and signal transmitter (Figure 1) when the planes fly towards meeting point, the message "HELLO WORLD" departs. We will investigate the moment at the "Meeting Point" where the signal receivers on the two planes and the mountain are at equal distance to the transmitter.

 

 

 

Letís edit the case like this because we will calculate the signal speed: at the moment arrived to the meeting point, the signal receiver on the mountain receives 44th Bit of the message "HELLO WORLD". 44th Bit falls to the midpoint of the message because the message "Hello World" consists of 88 Bit. Now, letís find which Bit of the message the planes received by calculating the necessary Byte Shift Calculations.

 

 

On the Figure 2 we see above, signals come from the same transmitter but the transmitter has not been added to the figure because it is at long distance. The figure was prepared by taking the Excel Table below as basement. You can download the Excel Table from here for inspection.

 

We see on the figure that, because of Byte Shift, the receiver on the mountain receives the character " " (Space), while the receiver of the plane moving away takes the message "L" and the receiver of the plane coming closer takes the message "O". But our topic is signal speed, so we should talk here on Bits instead of Bytes. We see on the calculations that; the message chain carrying the message "HELLO WORLD" shifted, +1,4176 meters for the plane departing from the receiver, and -1,4176 meters for the one approaching to the receiver. By benefiting from this distances, the position of 44th Bit of the message "HELLO WORLD" with respect to the planesí frames has been signed on the figure. In the next level where we will find the signal speed, we will benefit from 44th Bitís position places.

 

As a note: On the figure, we see that the signals continue to their ways after arrival point. In reality such a kind of thing does not happen, a signalís travel end when it reaches its target. Consequently, actually the travel of the signals reaching the targets has been completed. The reason of the figure to be prepared this way is for necessary calculations to be done.

 

 


BYTE SHIFT CALCULATIONS

 

 

 

Plane Speeds (Jet Speed) (as MACH)

2,5

Signal Frequency (GHz)

3,18

Transmitter distance (kilometers)

500

 

 

Speed of light (c)   (meters/second)

299792458

Plane Speeds (v) (MACH Value x (340 Meters/second))

850

Transmitter distance (Converted to millimeters)

500000000

 

 

DOPPLER SHIFT ON THE SIGNAL

 

We assumed that transmitter is stationary. All λ values are millimeters

 

Wavelength of the signal for the mountain: 

λ0 = speed of light/(Signal Frequency*100000)

94,27435786

Wavelength of the signal for the outgoing plane:  λ10 x (c+v)/c

94,27462516

Wavelength of the signal for the incoming plane:  λ20 x (c-v)/c

94,27409057

 

 

METING POINT

 

How many bits fill the distance between transmitter and receivers? 

(Length of 1 Bit  = λ)

 

n0 bits for the mountain  (n0=distance/λ0)

5303669,114

n1 bits for outgoing plane  (n1=distance/λ1)

5303654,076

n2 bits for incoming plane  (n2=distance/λ2)

5303684,151

 

 

BYTE SHIFT  

How many Bits shift between mountain and outgoing plane? (n0-n1)

15,0374

How many Bits shift between mountain and incoming plane? (n0-n2)

-15,0375

How many Bits shift between incoming and outgoing planes?  (n2-n1)

30,0749

 

 

SIGNAL "HELLO WORLD"

 

Numbers of Char

11

Numbers of Bytes

11

Numbers of Bits (1 Byte = 8 Bits)

88

Total signal length for mountain in the sky

= λ0 * Numbers of Bits /1000 (meters)

8,29614

Total signal length for outgoing plane in the sky 

= λ1 * Numbers of Bits /1000 (meters)

8,29617

Total signal length for in coming plane in the sky 

= λ2 * Numbers of Bits /1000 (meters)

8,29612

   

The receiver on the mountain is receiving 44th Bit (middle point) of the message.

 

How many meters shift the signal for outgoing plane? 

= (n0-n1)*λ1/1000 (meters)

1,417647405

How many meters shift the signal for incoming plane?

= (n0-n2)*λ2/1000 (meters)

-1,417647405

 

 

CALCULATION OF THE SIGNAL SPEED

Figure 3 has been prepared similarly taking the values in the Excel Table as base. For all 3 frames, as we know where 44th Bit is, let us write at what distances the 44th Bit is with respect to signal transmitter.

44th Bitís distance to itself with respect to signal transmitter (in meters):
For the mountain:                             = 5000000
For the outgoing plane : 500000+1,4176 =500001,4176
For the incoming plane : 5000000-1,4176 = 499998,5854

Letís find the arrival time of the signal to the receiver on the mountain leaving the transmitter:
Δt = 500000/299792458 = 0,00166782 seconds.

As we consider the 44th Bitís positions, in Δt time (0,00166782 seconds); 
The signal going to the outgoing plane has done 500001,4176 meters way,
The signal going to the incoming plane has done 499998,5824 meters way. 


In this stage, the equations below has occurred between the signal speed and distances;


For the outgoing plane:
(Speed of light + plane's speed) x time = (c+v) . Δ
= (299792458 + 850) x 0,00166782 = 500001,4176 meters

For the incoming plane:
(Speed of light - planeís speed) x time = (c-v) . Δ
= (299792458 - 850) x 0,00166782 = 499998,5824 meters

So, the speed of the signalís going to the planes and mountain with respect to transmitterís reference system are like that:
Speed of the signal going to the mountain = c = 299792458 meters / second
Speed of the signal going to the outgoing plane = c+v = 299792458 + 850 = 299793308 meters / second
Speed of the signal going to the incoming plane = c-v = 299792458 - 850 = 299791608 meters / second


The table below includes the part in the Excel file doing the calculations about the signal speed. (The part about the things told just above). You can reach the Excel file from here.

FINDING OF  SIGNAL SPEEDS

 

 

 

Where is the 44th Bit relative to transmitter? (via sum of distances)

 

For mountain (distance+0) (meters)

500000

For Outgoing Plane (distance + shift amount)  (meters)

500001,4176

For Incoming Plane (distance - shift amount) (meters)

499998,5824

 

 

Travelling time of the signal (from transmitter to the mountain)

(Δt = distance / c)  (seconds)

0,00166782

 

 

Where is the 44th Bit relative to transmitter? (via signal speeds)

 

For mountain:  c . Δt (meters)

500000

For outgoing plane: (c+v) . Δt (meters)

500001,4176

For incoming plane: (c-v) . Δt (meters)

499998,5824

 

 

Relative To Transmitter's Frame Speed Of Signals

 

Speed of the signal for mountain = c (meters/second)  

299792458

Speed of the signal for outgoing plane =  (c+v) (meters/second)

299793308

Speed of the signal for incoming plane =  (c-v) (meters/second)

299791608

 
 

(c+v)(c-v)  MATHEMATICS FOR THE ELECTROMAGNETIC THEORY 

Without using any theoretical explanations here, based on Doppler Shift Equations I stated that (c+v)(c-v) mathematic is valid for Electromagnetic Theory. I should point out with momentously, going of the electromagnetic waves exiting from the same resource to the moving objects with different speeds does not mean infraction of constant "c". In our example here we can easily see that the light speed constant "c" is preserved: 

The speed of the signal coming to itself with respect to outgoing plane = 299793308 - 850 = 299792458 = c
The speed of the signal coming to itself with respect to incoming plane = 299791608 + 850= 299792458 = c

Consequently, the mistake in the current Electromagnetic Theory is not in the constant "c", it is about how the constant "c" should be interpreted. As a result, an electromagnetic wave takes the reference system of the arrival target as base, goes with the speed c with respect to the targeting reference system, this is the basic of the job. The example used in this page is telling this situation; each of the electromagnetic waves taking the road at the same time and carrying the same message value has taken its own reference system of arrival target as base and went with speed "c" with respect to that reference system. Mathematics of Byte Shift shows this to us. 

I want to mention that, the speed of an electromagnetic wave going to a moving target has not been measured up to now. So, Electromagnetic Theory without a such a kind of measurement result from which it can benefit has been constructed with a wrong comment of constant c, as the result of this, it is normal, natural to be some mistakes, absences in the theory. If this measurement has been done in the stage of construction of Electromagnetic Theory, this situation told in this page would be a situation known by everybody. The (c+v)(c-v) mathematics is extending the existing mathematics of Electromagnetic Theory considering Electromagnetic interaction between the moving frames and removing its imperfections.

I hope that the scientists will consider my this article which I prepared with a simple mathematical reality. Now the important thing is to make the measurement of Byte Shift and to verify the truth of the things told. This measurement will give (c+v)(c-v) mathematics to the Electromagnetic Theory and will maintain a big improvement in science. 

Thanks for reading.
Han Erim

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