Satellite internet · Session 1 of 8
What a message actually is
What is the smallest honest unit of a picture, and what does reliability cost?
Zoom into any photograph far enough and the curves break into squares of flat colour. Now try to describe one of those pictures to someone in the next room using only the words “white” and “black”, square by square. A small 8 × 8 picture takes 64 spoken words, takes about a minute, and fails completely if either of you loses count by one — every square after the slip lands in the wrong place. The problem is not the picture. It is that you never agreed how to read it.
By the end
- Measure a picture in bits, and calculate how many distinct pictures a grid of that size can hold.
- Compare a parity check with a triple-repetition code on rate, detection and correction.
- Name what the device, router, terminal, satellite, ground station and destination server each do, and trace a request and its reply through all six.
The idea
A screen image is a grid of pixels, each one a coloured square. In this lesson every pixel is black or white, so describing one takes a single yes-or-no choice. That choice is a bit: the smallest unit of information there is. An 8 × 8 grid therefore holds 64 bits, which is 8 bytes, a byte being 8 bits.
Bits multiply. Each extra bit doubles the number of distinct messages, so n bits can describe 2ⁿ different things. Sixty-four bits is 2⁶⁴ ≈ 1.8 × 10¹⁹ — about 18 quintillion possible pictures on a grid you can draw in under a minute. Real images add bit depth: 8 bits per colour channel gives 256 levels each, and 24-bit colour gives 2²⁴ = 16,777,216 colours per pixel.
That is why compression exists rather than being a nicety. One uncompressed 1920 × 1080 frame at 24 bits per pixel is 49,766,400 bits ≈ 6.2 MB; thirty frames a second is about 1.5 Gbit/s. A video call actually streams at a few Mbit/s, so the codec is discarding roughly 300 parts in 301 and reconstructing the rest — it sends what changed and what your eye will not miss.
Bits are a description. Nothing shaped like a 1 leaves your house. Equipment represents each bit as a controlled change in a physical quantity — a voltage on copper, a brightness in fibre, the phase of a radio wave — and the receiver measures that quantity and decides which symbol was meant. Physics carries the signal; the protocol is the shared agreement that turns a measured signal back into the intended bits.
Every real protocol spends some of its capacity on being wrong less often. A parity bit is one extra bit chosen to make the number of 1s even: nine bits carry eight, 89% efficient, and any single flipped bit shows up as an odd count — but parity can only detect, never locate. A repetition code sends each bit three times and the receiver takes the majority, which corrects one error per triple but is only 33% efficient. Neither is free, and choosing between them is an engineering decision, not a right answer.
The route matters as much as the code. A router moves traffic between devices on a local network. A satellite terminal is a radio that tracks spacecraft. A ground station (also called a gateway) joins the space segment to the terrestrial internet. None of these substitutes for another, and a message from your room to a distant server passes through all of them — then does the whole trip again in reverse, because almost everything you do online is a request expecting a reply.
Gather
Graph paper, pencils, six labelled cards, string and a calculator.
Build and investigate
- Draw an 8 × 8 grid and colour a simple symbol on it without showing your partner. Agree the reading order before you start: rows left to right, top to bottom.
- Write the 64 bits as eight rows of eight. Hand your partner the bits, the grid dimensions and the reading order, and nothing else. Have them reconstruct the picture, then count exact matches out of 64.
- Run the same 64 bits as a single line and have a second person read them back as four rows of sixteen. Compare the two reconstructions. Record what was lost — and note that no bit was altered.
- Take one 8-bit row. Append a parity bit so the number of 1s is even. Have your partner flip exactly one bit in secret, then check parity. Repeat with two flipped bits and record what parity reports.
- Encode the same row with triple repetition — 24 bits — and have your partner flip one bit per triple. Decode by majority. Record the bits sent, the errors introduced and the errors survived, for both codes.
- Lay out six cards — device, router, terminal, satellite, ground station, destination server — and link a route with string. Trace a request outward and its reply back, naming what each node does to the message. Add a second satellite as an inter-satellite relay and say where the message must still reach the ground.
Run the numbers
Information has units, and every protection scheme has a price. Work in bits and keep the rate explicit.
| Relation | Expression | What it means |
|---|---|---|
| Message size | N = w × h × d bits | w × h pixels at d bits of depth each |
| Distinct messages | M = 2ⁿ | n bits distinguish 2ⁿ possibilities |
| Bits needed | n = log₂ M | to label M things you need log₂ M bits, rounded up |
| Code rate | R = data bits ÷ total bits | the fraction of the channel doing useful work |
Core · Grade 7–8
- Your 8 × 8 grid: confirm N = 64 bits = 8 bytes. How many bits would a 16 × 16 grid take? The side doubled — by what factor did the message grow, and why is it not 2?
- A message is one of 64 numbered squares. How many bits does it take to say which square? Check your answer by writing 63 in binary.
- Calculate the rate of both codes you tested: 8 data bits in 9 sent, and 8 data bits in 24 sent. Express each as a percentage, then state in one sentence what you bought with the difference.
Deeper · Grade 9–12
- An uncompressed 1920 × 1080 frame at 24-bit colour: calculate the bits per frame, then the raw bit rate at 30 frames per second. Compare with a 5 Mbit/s stream and state the compression ratio.
- Triple repetition corrects one error per triple. What happens if two of the three copies are corrupted? Write the condition under which majority decoding returns the wrong bit, and explain why a code that never fails is impossible rather than merely difficult.
- A 4 × 4 grid has 2¹⁶ = 65,536 possible pictures; an 8 × 8 grid has 2⁶⁴. The grid has 4 times the squares. By what factor did the number of possible pictures grow? Express the answer as a power of 2 and explain why exponential growth makes brute-force guessing useless as a decoding strategy.
Measure and record
Record matched pixels out of 64 for each reconstruction. For each code, record the data bits, the total bits sent, the rate as a percentage, the errors introduced and whether the receiver detected, corrected or missed them.
Worked example
Take a finished picture's eight rows and run them together into one line of 64 bits. Read that line back as four rows of sixteen and you get a picture — a different one, from exactly the same bits, with nothing lost in transit. The receiver simply applied a different rule. So 64 bits is not the message; 64 bits plus “8 wide, rows top to bottom, 1 = black” is the message. Now price the protection on that frame: parity adds 1 bit to 8, a rate of 8 ÷ 9 = 88.9%, and flags any single error without saying where it is. Triple repetition sends 24 bits for the same 8, a rate of 8 ÷ 24 = 33.3%, and fixes one error per triple. Tripling the cost did not triple the protection — it bought a different kind.
Try the idea
Draw a message in bits
Toggle squares to make a picture. Copy the eight rows to send it, or paste eight rows of eight 0s and 1s to decode one. The status line reports the message size in bits and in bytes, and the parity of each row.
Pause and explain
Your partner receives all 64 bits perfectly, with no errors at all, but no grid dimensions and no reading order. Can they recover the intended picture? Justify your answer with a number.
Check your reasoning after your attempt
No — and the number says why. Sixty-four bits can be arranged as 1 × 64, 2 × 32, 4 × 16, 8 × 8, 16 × 4, 32 × 2 or 64 × 1, and each reading order and symbol convention multiplies the possibilities again. A perfect channel does not supply meaning. The protocol does: dimensions, reading order and the meaning of each symbol have to be agreed in advance or sent alongside, which is why real formats begin with a header before any pixel data.
Save in your mission notebook
Both reconstructions and what differed; bits, rate and outcome for each code; the six-node route with the return path marked; one sentence on what a protocol has to supply beyond the bits.
Parent / teacher notes
The one-line-read-back in step 3 is the load-bearing moment: learners who meet it stop treating a bitstream as self-describing. Have the parity check fail in front of them with two flipped bits — the silence is more instructive than the catch. For a solo learner, flip bits with a die and keep the original face down. Grade 7–8 can stop after the rate percentages; Grade 9–12 should do the video bit-rate calculation, which is where the orders of magnitude land.
Suggested timing: 10 minutes predict, 15 read and explain, 30 build and investigate, 25 measure and calculate, 10 independent explanation. Adapt the activities rather than rush unfinished reasoning.
Take it further
Build a (7,4) Hamming code: 4 data bits, 3 parity bits, and a decoder that not only detects a single error but locates it. Compare its rate with triple repetition and explain how it corrects with so much less overhead.