The 12.5 kbit/s Loop That Is Already Too Slow
A regional turboprop fleet still uses ARINC 429 low-speed links at 12.5 kbit/s. On the air-data bus the ADC transmits 38 words in one fixed repeating loop; each word is 32 bits and the transmitter uses the minimum 4-bit inter-word gap. The captain's airspeed indicator (label at position 34 of the loop) must be updated at least 10 times per second, and the autopilot uses the same label. A modification proposes appending two new labels to the loop, and the modifier argues that two extra words out of forty is roughly a 5% increase and cannot possibly matter. The bus has no other transmitters, and every receiver on the pair is certified for 12.5 kbit/s only.
1. Calculate the worst-case refresh interval of the airspeed label in the current build and state whether the 10 Hz requirement is met. Explain what the label's position in the loop does and does not affect.
Read the worked answer
Word slot time. 32 data bits + 4 gap bits = 36 bit times:
36 / 12 500 = 2.88 ms per word
Refresh interval. In a fixed loop each label reappears exactly once per cycle, so the worst case is the full cycle, not the label's position:
38 x 2.88 ms = 109.4 ms -> 1 / 0.1094 = 9.1 Hz
Verdict. The requirement is at least 10 Hz (a 100 ms budget). The existing installation already delivers only about 9.1 Hz, so it does not meet the requirement before any modification is considered.
Position 34 affects only latency: how long after power-up, or after a corrupted word, before the first valid update arrives. It does not change the steady-state rate, which is set by the total loop length.
2. Evaluate the modifier's 5% argument. Support your answer with numbers, including the maximum number of words the loop can contain at 10 Hz.
Read the worked answer
Limit first. 100 ms / 2.88 ms = 34.7, so the loop may contain at most 34 words and still meet 10 Hz.
With the two added labels: 40 x 2.88 ms = 115.2 ms -> 8.7 Hz.
Assessment of the claim
- The arithmetic of 2/40 is right; the conclusion is wrong. The requirement is an absolute 100 ms budget, not a percentage, and there is zero slack to spend. Any increase takes a marginal 9.1 Hz further below the limit.
- A 5% increase in words is a 5% increase in cycle time, i.e. a 5% reduction in update rate - the opposite of what the aircraft needs.
- Duty cycle: 38 words x 36 bits over 109.4 ms is 12.5 kbit/s, i.e. the loop is effectively at 100% of the bus at minimum gap. Real transmitters often exceed the minimum gap for processing, so the true rate is likely slightly below 9.1 Hz. Measure the actual gap with a bus analyser before accepting any calculated figure.
3. Recommend a technically defensible fix, including why duplicating the airspeed label inside the same loop does not work.
Read the worked answer
Why duplication fails. Sending the label twice per loop adds one word, so the cycle grows by exactly the 2.88 ms the second transmission was meant to save: 39 x 2.88 = 112.3 ms -> 8.9 Hz. Refresh rate depends on total loop time, not on how often a label appears.
Viable options
- Raise the link to 100 kbit/s. Word slot becomes 36 / 100 000 = 0.36 ms; 40 words -> 14.4 ms -> about 69 Hz, with large margin. Requires the ADC and every receiver to be released for high speed, and re-verification of the wiring and terminations. Never mix 12.5 kbit/s and 100 kbit/s equipment on the same pair.
- Split the load. Keep airspeed and the other autopilot-critical labels on a short loop of 34 words or fewer (97.9 ms, 10.2 Hz) and move non-critical labels to a second, parallel transmitter/bus. Same arithmetic benefit as option 1 with no new LRU certification.
- Reduce word count. Slow or delete non-critical labels until the loop is 34 words or fewer. This meets the letter of the requirement with almost no margin and is normally rejected as insufficient for a growth-tolerant design.
Whichever is chosen, verify the final timing by measurement, not calculation, and record the result.