16. THE
FORTY-FORTH BIT (The 44th BIT)
I’d like to conclude the topic
of Byte Shift with an example using real values. The results we will
get here are important because the things we will see here will make it
a lot easier to understand the topics in the queue. On the other hand,
the use of real values while covering the topic will precisely show us
how the event of Byte Shift occurs.
The set-up of
the event closely resembles the previous ones that we have seen. While
the transmitter is sending the message “Hello World”, the planes are
moving towards “Meeting Line” (see the figure above).
We will deal with a special
situation where the distances of all three signal receivers to the
transmitter are equal when the planes are at “Meeting Line” (see the
figure below).
Let’s assume
that, at this moment, the station on the mountain receives the 44th bit
of the message. In this case, which Bit of the message that is coming
to them will the planes be receiving? This is our topic. We will do the
necessary calculations to answer this question.
The 44th Bit
falls in the middle of the message that consists of 88 Bits.
Let’s see the situation in a
figure. We are going to find the locations of the 44th bit shown with a
“question mark” relative to the meeting line in the figure below for
the message going to the planes.
We use the following table for base data and
calculations:
| Description | Symbol / Calculation Method | Value | Unit |
|---|---|---|---|
| Transmitter factory setting | |||
| Frequency | f0 | 3.18 | GHz |
| Wavelength | λ0 = c / f0 | 0.09427435786 | m |
| Constants | |||
| Speed of light constant | c | 299792458 | m/sn |
| Speed of sound | u | 340 | m/sn |
| Aircraft speed (Mach) | m. | 2.5 | Mach |
| Aircraft speed (meters per second) | v = m. . u | 850 | m/sn |
| Distance to meeting line | d0 | 50000 | m |
| Calculation of signal wavelengths to aircraft | |||
| For departing aircraft | λ1 = λ0 . (c+v) / c | 0.09427462516 | m |
| For approaching aircraft | λ2 = λ0 . (c-v) / c | 0.09427409057 | m |
Information on message groups:
| Description | Symbol / Calculation Method | Value | Unit |
|---|---|---|---|
| The number of characters, bytes, and bits that make up the phrase "Hello World." | |||
| Number of characters | 11 | Number | |
| Number of Bytes | 11 | Number | |
| Number of Bits | Number of Bytes x 8 | 88 | Number |
| Length of message groups in the sky | |||
| c Group | l1 = λ0 . 88 | 8,296143 | m |
| c+v Group | l1 = λ1 . 88 | 8,296167 | m |
| c-v Group | l2 = λ2 . 88 | 8,296120 | m |
We can use two methods to calculate where the 44th Bit is for the
planes.
- Calculation method by using the number of wavelengths
“Let’s assume that the message “Hello World” is a message that starts as “Hello World. Welcome to … ………………..” and has infinite length. As each wavelength belonging to the message carries data of 1 Bit, we can do calculations by making use of the number of wavelengths number between the receiver and the transmitter. - Calculation method by making use of signal speeds
We can do calculations by using the travel time of the signal that goes from the transmitter to the station on the mountain and the speeds of signal groups.
We see the calculations done by each method in the table below.
| Description | Symbol / Calculation Method | Value | Unit |
|---|---|---|---|
| Calculation using wavelength number | |||
| Wavelength number (for λ₀) | n0 = d0 / λ0 | 5303669,14 | Number |
| c Group | d0 = λ0 . n0 | 500000 | m |
| c+v Group | d1 = λ1 . n0 | 500001,4176 | m |
| c-v Group | d2 = λ2 . n0 | 499998,5824 | m |
| Calculation of the 44th bit positions using signal speeds | |||
| Signal arrival time | tΔ = d0 /c | 0,00166782 | sn |
| c Group | d0 = c . tΔ | 500000 | m |
| c+v Group | d1 = (c+v) . tΔ | 500001,4176 | m |
| c-v Group | d2 = (c-v) . tΔ | 499998,5824 | m |
Let’s try to complete the previous figure with the information we
obtained.
In this way, we
see how “Byte Shift” takes place. Although the receivers are at the
same distance from the transmitter, the station on the mountain, the
plane going away and the plane approaching receive a different part of
the message at that moment. While the transmitter on the mountain gets
the character “ ” (Space), the plane going away gets “O” character and
the plane approaching gets “L” character.
I’m going to end this part
about Byte Shift with a reminder. It is crucial to remember that (c+v)
(c-v) mathematics takes place between two objects. At the moment the
signal emitted reaches its target, its journey is over. The previous
figures were illustrated to cover the topic and do the calculations
and, in those illustrations, it seems as if the signals keep traveling
after they reach their targets. Of course, this doesn’t happen. The
illustration reflecting the reality is below. The journey is over for
the signal parts that reach the receiver.
We can directly
find which Bit of the message the planes receive at “Meeting Line” with
a third calculation method based on wavelength differences.
| Description | Symbol / Calculation Method | Value | Unit |
|---|---|---|---|
| Calculation using wavelength differences | |||
| Wavelength number (for λ₀) | n0 = d0 / λ0 | 5303669,114 | Piece |
| Signal shift amount (c+v) | x1 = n0 . ( λ1 - λ0 ) | +1,4176 | m |
| Signal shift amount (c-v) | x2 = n0 . ( λ2 - λ0 ) | -1,4176 | m |
| Number of shifting bits | |||
| For departing aircraft | x1 / λ1 | ≈ +15 | Bit |
| For approaching aircraft | x2 / λ2 | ≈ -15 | Bit |
| Which bit do receivers on the meeting line receive? | |||
| Ground station | 44 | ||
| Departing aircraft | 44 + 15 =59 | ||
| Approaching aircraft | 44 - 15 = 29 | ||
