Phoebe guidesA field team faces an unresolved physical question: The Five-Year Cost of One Extra Byte They must answer it before changing deployment life in years on the real device. Predict the direction first.
See the relationship before changing it
The figure reads from left to right. The blue card is deployment life in years. The middle card applies this page's relationship. The green card is energy per reading. 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 added model holds every other chapter fixture fixed, so the numeric fixture does not switch without explanation.
Derive the baseline in four named moves
- 1
Name the input. The chapter baseline for deployment life in years is 5.
- 2
Name the relationship. t_extra = bits/R; E_reading = IVt_extra; E_total = E_reading x readings/day x 365 x years
- 3
Substitute the chapter fixture. Set deployment life in years to 5. The page ledger gives energy per reading as 0.483 mJ.
- 4
Read the result. Keep mJ beside the value. Use it only inside the technical boundary on this page.
Predict, then change deployment life in years
Try Predict the direction of energy per reading. Move one control, calculate, then check your prediction.
Observe The outputs use t = bits/R, P = IV, E = Pt, and the chapter's cadence multiplier, exactly the derivation above. Reset the control to 5 and compare energy per reading.
Explain Only deployment life in years moves here. The other chapter fixtures remain fixed.
Check yourself
What should you do before trusting a moved-control result?
What does this small model leave out?
1. A byte keeps the radio awake
Energy is power multiplied by time. An extra payload byte adds eight bits. At a fixed data rate, those bits require extra radio-on time. Repeating that small cost for every reading and every day turns a format choice into a lifetime budget line.
2. Keep units visible
| Quantity | Calculation | Unit |
|---|---|---|
| radio power | P = IV = 0.100 A × 3.3 V | 0.33 W |
| extra time | t = bits/R | seconds |
| energy per reading | E = Pt | joules |
| lifetime energy | E × readings/day × days | joules, then watt-hours |
3. Derive the lifetime cost
Turn a byte into bitsOne byte = 8 bits.
Turn bits into secondsAt R bits/s, t_extra = 8/R.
Find radio powerP = IV = 0.100×3.3 = 0.33 W.
Find energy per readingE = Pt_extra.
Multiply by cadence and timeUse 96 readings/day and 365 days/year, then divide joules by 3,600 to get Wh.
4. Reproduce the chapter's numbers
At 5,470 bit/s, eight bits take 8/5470 = 1.46 ms. A 0.33 W radio therefore spends 0.33×0.00146 = 0.000483 J = 0.483 mJ per reading. Every 15 minutes means 96 readings/day, so the extra byte costs 46.3 mJ/day = 1.29×10⁻⁵ Wh/day. Over 1,825 days, that is 0.0235 Wh = 23.5 mWh. The chapter also shows why a 16-bit path buys 98.1 dB versus 74.0 dB for 12 bits—24.1 dB that may not be useful if the sensor noise already dominates.
5. Try the same formula
TryMove deployment life from one to five years while the chapter's extra byte, radio, and 15-minute cadence stay fixed.
ObserveTime and energy per reading stay fixed; lifetime energy grows in direct proportion to years.
ExplainThe outputs use t = bits/R, P = IV, E = Pt, and the chapter's cadence multiplier, exactly the derivation above.
This isolates payload airtime for one extra byte.
- preamble
- Needs separate evidence
- headers
- Needs separate evidence
- acknowledgements
- Needs separate evidence
- retries
- Needs separate evidence
- startup
- Needs separate evidence
- receive windows
- Needs separate evidence
- sleep current
- Needs separate evidence
- data-rate changes
- Needs separate evidence
Use field evidence or a deeper model before release.
6. What the result buys you
The 23.5 mWh result is small beside many batteries, but it exposes the mechanism. Multiply it across more bytes, more frequent reports, lower data rates, retries, and a large fleet. Keep precision only when measured noise and the application requirement can use it.
7. Check yourself
1. How long do eight bits take at 5,470 bit/s?
Answer: 8/5470 = 0.00146 s = 1.46 ms.
2. Why are there 96 readings per day?
Answer: Four 15-minute intervals per hour × 24 hours = 96.
3. What important energy is excluded?
Answer: Transaction overhead, retries, startup, receive windows, sleep current, and other device loads.
These are the chapter inputs, worked results, and named teaching assumptions.
- 12/16-bit
- Digital resolution or converter setting
- 74.0/98.1 dB
- Gain, loss, margin, or level ratio
- 24.1 dB
- Gain, loss, margin, or level ratio
- 24-byte
- Device, payload, or sample count
- 15-minute
- Time, interval, or service-life value
- 100 mA
- Current or responsivity value
- 3.3 V
- Voltage or voltage-step value
- 0.33 W
- Power or power-loss value
- 5,470 bit/s
- Time, interval, or service-life value
- 1.46 ms
- Time, interval, or service-life value
- 0.483 mJ
- Charge or energy value
- 96/day
- Time, interval, or service-life value
- 46.3 mJ/day
- Time, interval, or service-life value
- 1.29×10⁻⁵ Wh/day
- Time, interval, or service-life value
- 1,825-day
- Time, interval, or service-life value
- 0.0235 Wh
- Charge or energy value
- 23.5 mWh
- Charge or energy value
This correct payload-only calculation is not a complete battery model; use the chapter's whole-transaction and delivery evidence for design.