Satellite internet · Session 6 of 8
Where the delay actually comes from
What makes a connection feel slow, and why is that not the same as slow?
Send a photo and watch the progress bar. It does not creep smoothly — it jumps, stalls, jumps again. The photo never travelled as one object. It left in hundreds of numbered pieces, down whatever paths happened to be free, and arrived out of order to be reassembled at the far end. The stalls are not the network being lazy. They are pieces waiting in a queue somewhere, and queuing is a different thing from distance.
By the end
- Separate the three contributions to delay — propagation, serialisation and queuing — and identify which one a given change affects.
- Measure first-packet latency and throughput independently and show that one does not determine the other.
- Predict when a vacuum inter-satellite laser route beats a terrestrial fibre route over the same ground distance.
The idea
Networks do not send messages; they send packets. Each carries a slice of data plus a header with addresses and a sequence number. Packets from one message may take different routes, arrive out of order, or not arrive at all; the receiver reorders them by sequence number and asks for any that are missing. This is packet switching, and it is why a network can keep working when a link fails.
Delay is not one thing. Propagation delay is distance ÷ signal speed and nothing can reduce it but a shorter path. Serialisation delay is the time to clock the bits of a packet onto the wire, equal to packet size ÷ link rate. Queuing delay is time spent waiting behind other packets, and unlike the other two it depends on how busy the link is, not on its physics. Measured delay is the sum, and improving the wrong term achieves nothing.
Latency and throughput are independent measurements. Latency is how long the first bit takes; throughput is how much arrives per second. A link can deliver a great deal per second and still make you wait a long time for the first thing — a lorry full of hard drives has superb throughput and dreadful latency. Everyday speech has one word, “fast”, for both, and that single word causes more confusion in this subject than any equation.
Here is where orbital altitude earns its keep. Propagation to 550 km takes 550 ÷ 299,792 = 1.83 ms; up and down is 3.67 ms, and a request plus its reply crosses four times for about 7.3 ms. The same arithmetic at geostationary altitude gives 119 ms one way and roughly 477 ms for a request and reply — approaching half a second before any equipment has done anything at all. That is the difference between a usable video call and an unusable one, and it is pure geometry.
Over long distances, orbit can beat the ground. Light in glass travels at c / n with n ≈ 1.47, so fibre carries signals at about 204,000 km/s — just 68% of c — and real cable routes run perhaps 20% longer than the great-circle distance. A vacuum inter-satellite laser link travels at the full c over a nearly direct path. London to New York is about 5,570 km: roughly 6,700 km of fibre at 204,000 km/s is 32.8 ms one way, while a laser relay covering 5,570 km plus 1,100 km of climb and descent at 299,792 km/s is 22.2 ms. Over 20 ms saved on the round trip, which is why these links exist on long routes and not on short ones.
Congestion is the term you can actually change. A link's capacity is fixed, so when the offered load approaches it, queues grow — and near saturation they grow sharply rather than gradually. Little's law, L = λW, ties average queue length to arrival rate and waiting time. Throughput is also not the same as goodput: a 1,500-byte IP packet carries at most 1,460 bytes of payload after headers, about 97%, and retransmitted packets consume capacity while delivering nothing new.
Three delays, one measurement
Gather
20 numbered index cards, string, a timer, node labels and a calculator. A single learner can move counters on a drawn network.
Build and investigate
- Lay out A → B → C → D: A the user terminal, B a satellite, C a ground station, D the destination server. Give A all 20 numbered cards; D reassembles them in order.
- Fix the forwarding rule and write it down: at each 2-second tick, every node forwards at most one card that it held at the start of that tick; cards received during a tick wait for the next. All nodes act simultaneously, so no card crosses two hops in one tick.
- Run ten ticks. Record the first arrival time and the number of cards delivered by 20 s. Repeat the run to confirm your procedure is reproducible before you change anything.
- Add a second sender at B with five cards, alternating between the two queues when both are occupied. Re-run and compare the original message's delivery count. The link length has not changed — identify which term of the delay grew.
- Add the route B → E → C as an inter-satellite relay. Block B → C and re-run. Record the new first-arrival time and delivery count, then explain the change in terms of hop count rather than distance.
- Now do the real arithmetic. Calculate propagation delay for a 550 km hop and for a geostationary hop. Then calculate the fibre-versus-laser comparison for a 5,570 km route, and state the distance below which the satellite path stops winning.
Run the numbers
Three delays add. Work out which one each experiment changed, and the counter-intuitive results stop being counter-intuitive.
| Relation | Expression | What it means |
|---|---|---|
| Propagation delay | t = d / v | v = 299,792 km/s in vacuum, ≈ 204,000 km/s in fibre |
| Serialisation delay | t = packet bits / link rate | time to clock a packet onto the link |
| Total one-way delay | t = t_prop + t_serial + t_queue | only the last depends on traffic |
| Throughput | packets (or bits) ÷ observation time | state the window; an average hides bursts |
| Little's law | L = λW | queue length = arrival rate × waiting time |
| Bandwidth–delay product | bits in flight = rate × round-trip time | how much data is unacknowledged en route |
Core · Grade 7–8
- Calculate the one-way propagation delay to 550 km and to 35,786 km at 299,792 km/s. Multiply each by four for a request and its reply, and state which is usable for a live conversation.
- From your ten-tick run, calculate throughput as delivered packets ÷ 20 s. Then state the first-arrival time separately. Explain why quoting only one of these numbers would mislead someone.
- In the congested run, the distance between nodes never changed and neither did the tick length. Which term of t_prop + t_serial + t_queue grew, and by how much?
Deeper · Grade 9–12
- Compare routes for a 5,570 km great-circle distance. Fibre: about 6,700 km of cable at 204,000 km/s. Laser relay: 5,570 km across plus 1,100 km of climb and descent at 299,792 km/s. Compute both one-way delays and the round-trip saving.
- Find the break-even distance. Write both delays as functions of great-circle distance d, assuming fibre runs 1.2d and the relay runs d + 1,100 km, and solve for the d at which they are equal. Below that distance, which wins, and why does the constant term decide it?
- A 100 Mbit/s link has a 40 ms round-trip time. Calculate the bandwidth–delay product in kilobytes. This is data already sent but not yet acknowledged — explain why a protocol that waits for an acknowledgement after every packet would waste almost all of this link's capacity.
Measure and record
For each run record the route, the forwarding rule, whether competing traffic was present, the first arrival in seconds, packets delivered by 20 s and the rate in packets per second. Use packet/s here, since model packets carry no defined bit length.
Worked example
With three links and the stated rule, packet 1 is forwarded by A on tick 1, by B on tick 2 and by C on tick 3, arriving at 6 s. By tick 10 — 20 s — eight packets have reached D, so throughput is 8 ÷ 20 = 0.4 packet/s. Switch to the four-link relay route and the first arrival moves to 8 s with seven delivered: one extra hop cost one tick of latency and one packet of throughput, and no distance changed at all. Now the real numbers, which run the other way: a 550 km hop is 550 ÷ 299,792 = 1.83 ms, so a request and reply through one satellite is about 7.3 ms of propagation, against 477 ms through a geostationary satellite. The model shows hops costing delay; the arithmetic shows distance costing far more. Both are true, and a real latency budget has to add them.
Try the idea
Route twenty packets
Each tick represents two model seconds. A node forwards at most one packet per tick, and a packet received during a tick waits for the next one.
With competing traffic, B sends the original message first and then alternates while both queues hold packets. Counts track the original 20 packets only. The tick stands in for serialisation and processing delay; it is not a radio travel time, and none of these are Starlink measurements.
Pause and explain
Two satellite links are measured. Both deliver 100 Mbit/s of throughput. One is a 550 km low-orbit link, the other geostationary. A user reports that one is fine for watching films but unusable for a video call. Which is which, and which measurement explains it?
Check your reasoning after your attempt
The geostationary link is the one that fails for calls. Throughput is identical, so throughput cannot be the explanation; latency is. A request and its reply cross 35,786 km four times, about 477 ms of pure propagation, and a conversation with half a second of lag each way collapses into people talking over each other. A film is buffered — it is a throughput problem, and a delay before it starts is barely noticed. This is the clearest case in the course of two systems being identical on one measurement and unusable apart on another.
Save in your mission notebook
The route diagram; results for the normal, congested and failed-route runs with first arrival and throughput recorded separately; the propagation arithmetic for both shells; the fibre-versus-laser comparison; one sentence naming which delay term each experiment changed.
Parent / teacher notes
Walk two ticks slowly before timing anything, and use two piles per node — held and newly received — so the simultaneity rule survives contact with enthusiasm. The congestion run in step 4 is the point of the lesson: the link did not get longer, yet the delay grew, and that observation is what separates queuing from propagation. Grade 7–8 can stop after the propagation comparison between the two shells. Grade 9–12 should do the fibre-versus-laser break-even, which needs solving a linear equation, and the bandwidth–delay product.
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
Give the links different capacities — say B → C can carry two packets per tick while A → B carries one — and find the bottleneck. Show that end-to-end throughput is set by the slowest link alone, so upgrading any other link changes nothing, and connect this to why the route with the fewest hops is often not the fastest.