A field team faces an unresolved physical question: How do fixed radios and long current loops survive the real site? They must answer it before changing cable length 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 cable length. The middle card applies this page's relationship. The green card is gateway eirp. 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 cable length is 500.
- 2
Name the relationship. Gateway EIRP=14+6=20 dBm Wearable EIRP=0+0=0 dBm Ideal range ratio=10^(20/20)=10.0x Rcable=0.0842x500x2=84.2 ohm Vneeded=0.020x(84.2+250)=6.68 V
- 3
Substitute the chapter fixture. Set cable length to 500. The page ledger gives gateway eirp as 20.00 dBm.
- 4
Read the result. Keep dBm beside the value. Use it only inside the technical boundary on this page.
Predict, then change cable length
Try Predict the direction of gateway eirp. Move one control, calculate, then check your prediction.
Observe Both designs survive the site by budgeting a physical carrier, but RF orientation and wired compliance are different mechanisms. Reset the control to 500 and compare gateway eirp.
Explain Only cable length 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. Begin with what stays fixed
A bolted gateway can keep its antenna pointed; a moving wearable cannot. A series current loop keeps the same current through each element, but its supply must still provide enough voltage for the cable and burden.
2. Name every algebra move
Combine radio termsEIRP=Pt+G.
Compare ideal ranged2/d1=10^(ΔEIRP/20) when all other link terms match.
Count both conductorsRcable=rL×2.
Pay for cable and burdenVneeded=I(Rcable+Rburden).
Check headroomVheadroom=Vsupply−Vneeded.
3. Reproduce the 500 m case
Wearable EIRP=0+0=0 dBm
Ideal range ratio=10^(20/20)=10.0×
Rcable=0.0842×500×2=84.2 Ω
Vneeded=0.020×(84.2+250)=6.68 V
The 24 V loop has 17.32 V of teaching headroom. A plain voltage signal carrying the same 20 mA would lose 1.68 V in that cable; the current loop keeps the measured current but still needs compliance voltage.
4. Try the cable length
TryExtend the one-way cable while wire gauge, loop current, burden, and supply stay fixed.
ObserveCable length changes loop resistance and voltage headroom, but it does not alter the radio ledger.
ExplainBoth designs survive the site by budgeting a physical carrier, but RF orientation and wired compliance are different mechanisms.
This combines two ideal ledgers for comparison; it does not model one shared signal path.
- RF range
- The 10× value holds only if receiver, frequency, propagation, loss, and orientation assumptions match
- Loop
- The transmitter also needs its own compliance voltage and protection allowance
- Cable
- Temperature, joins, barriers, shielding, capacitance, and installation rules matter
Measure the installed RF link and verify the exact transmitter's compliance budget over worst-case cable resistance.
5. Test the installation assumptions
Rotate the wearable and walk the body-shadow cases. Aim and detune-check the fixed gateway. For the loop, test full-scale current at minimum supply and maximum expected cable resistance.
6. Record the evidence state
Store radio power, antenna gain and pattern, orientation, frequency, receiver, site points, wire gauge, one-way length, loop current, burden, transmitter compliance, supply tolerance, and measured headroom.
7. Check yourself
Why does a 20 dB link difference give a 10× ideal range ratio?
Why is the 500 m cable counted twice?
Does a current loop ignore cable resistance completely?
The arithmetic uses the chapter's catalog-typical radio, antenna, 24 AWG cable, 250 Ω burden, and 24 V supply examples.
- 10.0×
- An ideal matched-link ratio, not a site range promise
- 6.68 V
- Cable plus burden only, not the transmitter's full requirement
- 17.32 V
- Teaching headroom before other loop drops and tolerances
Correct, not complete: these budgets do not qualify an industrial radio or field loop installation.
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