Ethernet Margins: Attenuation, PoE Budget, Camera Capacity, and Deterministic Traffic

Ethernet Margins: Attenuation, PoE Budget, Camera Capacity, and Deterministic Traffic

Ada rechecks the chapter’s Ethernet cable-loss margin, camera oversubscription, PoE headroom, robot timing, and VLAN field size

ADA · CALCULATION AUDIT

Ethernet Margins: Attenuation, PoE Budget, Camera Capacity, and Deterministic Traffic

Ethernet feels simple because the cable plugs in, but the physics still sets margins: insertion loss, power budget, shared uplinks, and scheduled traffic all need arithmetic before the link is called reliable. The chapter’s deployment records supply the numbers — a 72 m Cat6 run, 50 cameras at 25 Mbps each, a 370 W PoE switch, and a fleet of 50 robots. This audit rechecks every one of those margins.

Companion to the chapter Wired Access: Ethernet — every number here comes from that chapter.

See the relationship before changing it

The figure reads from left to right. The blue card is camera count. The middle card applies this page's rule. The green card is aggregate video rate. Walk the arrows once: set the input, apply the rule, then read the result with its unit.

The retained audit below checks several chapter fixtures. This model keeps those stated values fixed and changes only camera count, so the numeric fixture does not switch without explanation.

Walk the arrows. Each identical camera adds another 25 megabits per second to the shared aggregation path.

Derive the baseline in four named moves

1. 1

   Name the input. The chapter baseline is 50 cameras.

2. 2

   Name the relationship. aggregate rate = cameras x 25 Mbit/s

3. 3

   Substitute with units. 50 cameras x 25 Mbit/s = 1,250 Mbit/s

4. 4

   Read the result. Keep the unit beside the value. Use it only inside the technical boundary on this page.

Predict, then change camera count

Try Predict the direction of aggregate rate = cameras x 25 Mbit/s. Test another camera count, then compare aggregate video rate.

Camera count

50 cameras

Chapter baseline

Aggregate video rate

Observe Each identical camera adds another 25 megabits per second to the shared aggregation path. Reset camera count to 50 and compare aggregate video rate.

Explain Each identical camera adds another 25 megabits per second to the shared aggregation path.

Check yourself

What should you do before trusting a moved-control result?

Answer: Predict its direction, apply the shown relationship, keep the units, and reset to the worked baseline.

What does this small model leave out?

Answer: Only camera count moves here. Field effects named in the technical boundary stay fixed.

Honesty note. This teaching model changes only camera count. It is not a field prediction.

TryRun Calculate for the 72 m Cat6 channel, 50 cameras at 25 Mbps, and 24 PoE loads at 15 W.

ObserveThe page exposes cable-loss headroom, a 1.25 Gbps camera aggregate, and only 10 W left on the 370 W PoE budget.

ExplainLength scales insertion loss, device count multiplies shared traffic and power, and deterministic robot timing requires bounded queues rather than unused raw bandwidth alone.

Ready: use the stated baseline inputs, then compare each displayed result.

Rebuild the Cable-Loss Margin

1. Rebuild the cable-loss margin. The chapter gives a 21 dB channel-loss budget at 100 MHz and Cat6 attenuation of 19.8 dB per 100 m. For the 72 m run in the deployment record:

loss = 19.8 dB/100 m x 72 m = 14.256 dB; margin = 21 dB - 14.256 dB = 6.744 dB

At the full 100 m channel limit, the same calculation leaves only 21 - 19.8 = 1.2 dB of insertion-loss margin, which is why termination quality and cable testing matter near the limit.

Audit the Camera Bandwidth Claim

2. Audit the camera bandwidth claim. The warehouse scenario gives 50 cameras at 25 Mbps each:

aggregate video = 50 x 25 Mbps = 1250 Mbps = 1.25 Gbps

Check	Arithmetic shown	Audit result
One camera on a 1 Gbps access port	25 Mbps / 1000 Mbps = 0.025 = 2.5%	Matches the chapter's per-camera headroom claim.
All 50 cameras on one 1 Gbps uplink	1250 Mbps / 1000 Mbps = 1.25 = 125%	The access ports are fine, but a single 1 Gbps uplink would be oversubscribed.
All 50 cameras on a 10 Gbps aggregation path	1250 Mbps / 10000 Mbps = 0.125 = 12.5%	A 10 Gbps aggregation path leaves operational headroom.

Check the PoE Switch Headroom

3. Check the PoE switch headroom. The chapter's PoE example uses a 370 W switch and 24 devices at 15 W each:

peak draw = 24 x 15 W = 360 W; utilization = 360 / 370 = 0.97297 = 97.3%

80% planning limit = 370 W x 0.80 = 296 W; 360 W - 296 W = 64 W over the recommended headroom target

Separate Bandwidth From Timing for Robot Traffic

4. Separate bandwidth from timing for robot traffic. The robot scenario gives 50 robots, 500 bytes per update, one update every 100 ms, and a 10 ms latency requirement:

per robot = 500 bytes x 8 / 0.1 s = 40000 bps = 40 kbps; fleet = 50 x 40 kbps = 2 Mbps

2 Mbps / 1000 Mbps = 0.002 = 0.2% of a 1 Gbps link; 0.069 ms / 10 ms = 0.0069 = 0.69% of the latency budget

The math explains the physics claim: the hard part is not raw throughput, but bounded timing, queue configuration, and a wired path that preserves the schedule.

Verify the VLAN Field Count

5. Verify the VLAN field count. The under-the-hood section says an 802.1Q tag has a 12-bit VLAN identifier:

possible VLAN ID codes = 2¹² = 4096

Every number above is taken from the chapter’s own Ethernet deployment examples and re-derived step by step.

Technical boundaries. Cable faults, connector ageing, PoE conversion loss, traffic bursts, switch queues, and thermal derating are outside these fixed attenuation, load, and timing checks.