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.

Let’s
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 44^{th}
Bit of the message "HELLO WORLD". 44^{th}
Bit falls to the midpoint of the message because the message "Hello
World" consists of 88 Bit. Now, let’s 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
44^{th} Bit of the message "HELLO WORLD" with respect to the planes’ frames has been signed on the figure. In the next level where we will find the signal speed, we will benefit from
44^{th} Bit’s 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 signal’s 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: λ1=λ0
x (c+v)/c

94,27462516

Wavelength
of the signal for the incoming plane:
λ2=λ0
x (cv)/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? (n0n1)

15,0374

How
many Bits shift between mountain and incoming plane? (n0n2)

15,0375

How
many Bits shift between incoming and outgoing planes?
(n2n1)

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 44^{th}
Bit (middle point) of the message.


How many meters shift the signal for outgoing plane?
= (n0n1)*λ1/1000
(meters)

1,417647405

How many meters shift the signal for incoming plane?
= (n0n2)*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
44^{th} Bit is, let us write at what distances the 44^{th} Bit is with respect to signal transmitter.
44^{th} Bit’s 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 : 50000001,4176 = 499998,5854
Let’s find the arrival time of the signal to the receiver on the mountain leaving the transmitter:
_{Δ}t = 500000/299792458 = 0,00166782 seconds.
As we consider the 44^{th} Bit’s positions, in _{Δ}t 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}t
= (299792458 + 850) x 0,00166782 = 500001,4176 meters
For the incoming plane:
(Speed of light  plane’s speed) x time = (cv) . _{Δ}t
= (299792458  850) x 0,00166782 = 499998,5824 meters
So, the speed of the signal’s going to the planes and mountain with respect to transmitter’s 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 = cv = 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 44^{th}
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)
(_{Δ}t = distance / c) (seconds)

0,00166782



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


For
mountain: c .
_{Δ}t
(meters)

500000

For
outgoing plane: (c+v) . _{Δ}t (meters)

500001,4176

For
incoming plane: (cv) . _{Δ}t (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 =
(cv) (meters/second)

299791608




(c+v)(cv) MATHEMATICS FOR THE ELECTROMAGNETIC THEORY
Without using any theoretical explanations here, based on Doppler Shift Equations I stated that (c+v)(cv) 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)(cv) 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)(cv) mathematics to the Electromagnetic Theory and will maintain a big improvement in science.
Thanks for reading.
Han Erim


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