A field team faces an unresolved physical question: How can one antenna setting reshape a coverage model? They must answer it before changing gain 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 gain. The middle card applies this page's relationship. The green card is linear gain. 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 gain is 8.
- 2
Name the relationship. G=10^(8/10)=6.31x Pt=20-8=12 dBm=15.8 mW ohm=4π/6.31=1.99 sr=15.8% of the sphere Fixed-power range factor=√6.31=2.51x
- 3
Substitute the chapter fixture. Set gain to 8. The page ledger gives linear gain as 6.31 times.
- 4
Read the result. Keep times beside the value. Use it only inside the technical boundary on this page.
Predict, then change gain
Try Predict the direction of linear gain. Move one control, calculate, then check your prediction.
Observe The same gain can either reshape an EIRP-limited pattern or extend in-beam range when conducted power is held fixed; those are different scenarios. Reset the control to 8 and compare linear gain.
Explain Only gain 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 the whole sphere
An isotropic source is not a neutral placeholder. It spreads power equally across 4π steradians. A sector antenna concentrates energy into a smaller part of that sphere.
2. Name every algebra move
Undo decibelsG=10^(GdBi/10).
Narrow the sphereΩ≈4π/G and sphere fraction=1/G.
Respect the ceilingPt(dBm)=EIRP−GdBi.
Convert powerPt(mW)=10^(Pt(dBm)/10).
Compare fixed-power rangeRange factor=√G.
3. Reproduce the 8 dBi case
Pt=20−8=12 dBm=15.8 mW
Ω=4π/6.31=1.99 sr=15.8% of the sphere
Fixed-power range factor=√6.31=2.51×
The isotropic ceiling case is 20 dBm=100 mW across 12.6 sr. The sector case uses less conducted power at the same EIRP while changing where energy goes.
4. Try the antenna gain
TryMove the sector gain while the 20 dBm EIRP ceiling stays fixed.
ObserveMore gain narrows the ideal lobe and reduces allowed conducted power at a fixed EIRP ceiling.
ExplainThe same gain can either reshape an EIRP-limited pattern or extend in-beam range when conducted power is held fixed; those are different scenarios.
This is one ideal main-lobe ledger, not an antenna pattern solver.
- Pattern
- Real antennas have beamwidth, sidelobes, nulls, polarization, and mounting effects
- EIRP
- The permitted ceiling depends on band, region, channel, and equipment rules
- Range
- √G assumes fixed conducted power and the same required power density
Configure the measured pattern and validate it with the installed antenna and an RF walk test.
5. Test the assumption that matters
Compare the isotropic model with the intended antenna file, azimuth, tilt, cable loss, and mounting. Walk both the illuminated dock and the predicted off-axis areas.
6. Record the evidence state
Store model version, antenna pattern, gain reference, EIRP rule, conducted power, orientation, field points, and the mismatch that forces a rerun.
7. Check yourself
Why is 8 dBi equal to 6.31× rather than 8×?
Why does conducted power fall to 12 dBm?
Does 1.99 sr describe a real sector exactly?
The arithmetic reproduces the chapter's catalog-typical 20 dBm and 8 dBi case.
- 6.31×
- Linear gain implied by 8 dBi
- 15.8%
- Ideal sphere fraction, not a measured beam footprint
- 2.51×
- Fixed-power in-beam comparison, not a deployment range promise
Correct, not complete: this antenna ledger does not qualify loading-dock coverage.
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