Fresnel Clearance: Radius, Antenna Height, and Deployment Cost

Fresnel Clearance: Radius, Antenna Height, and Deployment Cost

Ada re-derives the chapter’s Fresnel radius, required clearance, mast height, and the deployment cost the geometry unlocks

foundations
math-foundations
wireless-propagation
fresnel-zone
intermediate
Ada ADA · CALCULATION AUDIT

Fresnel Clearance: Radius, Antenna Height, and Deployment Cost

Fresnel planning is geometry with a radio consequence: first calculate the invisible clearance envelope, then ask whether the antenna height really leaves enough of it open. The chapter’s own examples run a 100 m Wi-Fi bridge at 2.4 GHz and a 5 km LoRa link at 915 MHz, then turn the required clearance into a mast height and a deployment cost.

Companion to the chapter Fresnel Zones and Deployment — every number here comes from that chapter.

See the relationship before changing it

The figure reads from left to right. The blue card is wi-fi path length. The middle card applies this page's rule. The green card is required fresnel clearance. 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 wi-fi path length, so the numeric fixture does not switch without explanation.

Wi-Fi path length changes required fresnel clearance An input card leads through the rule clearance = 0.6 x 17.3 x sqrt((distance / 1,000) / (4 x 2.4)) to the required fresnel clearance result. INPUT PAGE INPUT APPLY THE RULE predict calculate check units OUTPUT RESULT
Walk the arrows. A longer path widens the first Fresnel zone when frequency and the midpoint split stay fixed.

Derive the baseline in four named moves

  1. 1

    Name the input. The chapter baseline is 100 m.

  2. 2

    Name the relationship. clearance = 0.6 x 17.3 x sqrt((distance / 1,000) / (4 x 2.4))

  3. 3

    Substitute with units. 0.6 x 17.3 x sqrt(0.100 km / (4 x 2.4 GHz)) = 1.06 m

  4. 4

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

Predict, then change wi-fi path length

Try Predict the direction of clearance = 0.6 x 17.3 x sqrt((distance / 1,000) / (4 x 2.4)). Test another wi-fi path length, then compare required fresnel clearance.

100 m
Chapter baseline
Required Fresnel clearance

Observe A longer path widens the first Fresnel zone when frequency and the midpoint split stay fixed. Reset wi-fi path length to 100 and compare required fresnel clearance.

Explain A longer path widens the first Fresnel zone when frequency and the midpoint split stay fixed.

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 wi-fi path length moves here. Field effects named in the technical boundary stay fixed.
TryUse Calculate on the 100 m, 2.4 GHz bridge, then compare the 5 km, 915 MHz LoRa path.
ObserveThe required-clearance readout grows when the 100 m, 2.4 GHz bridge is replaced by the 5 km, 915 MHz path.
ExplainFirst-Fresnel radius follows the square root of wavelength and split path lengths, so 915 MHz and 5 km enlarge the clearance envelope.

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

The clearance ledger

r1 = 17.3 * sqrt(d / (4f)), with d in km and f in GHz; required clearance = 0.6 * r1
Check Arithmetic shown Result Design meaning
100 m Wi-Fi bridge at 2.4 GHz 17.3 * sqrt(0.1 / (4 * 2.4)) = 17.3 * sqrt(0.0104) r1 = 1.77 m; 0.6 * 1.77 = 1.06 m The short Wi-Fi link only needs about one meter of midpoint clearance before installation margin.
5 km LoRa link at 915 MHz 17.3 * sqrt(5 / (4 * 0.915)) = 17.3 * sqrt(1.366) r1 = 20.22 m; 0.6 * 20.22 = 12.13 m The long low-frequency link needs a much taller clearance envelope.
Gateway above 2 m crops 2 + 12.13 = 14.13 m; round to a 15 m mast (15 - 2) / 20.22 = 64.3% clear The 15 m gateway clears the 60% rule; a 3 m gateway gives only (3 - 2) / 20.22 = 4.9%.
Infrastructure comparison Ground: 39 * $1,500 = $58,500; tower: 4 * $1,500 + 4 * $3,000 = $18,000 ($58,500 - $18,000) / $58,500 = 69.2% The physics calculation explains why the taller design is cheaper despite the tower cost.

Audit rule: do not copy a range claim forward unless the distance, frequency units, obstacle height, and clearance percentage are written beside the decision.

Every number above is taken from the chapter’s own Fresnel-zone worked examples and re-derived step by step.

Technical boundaries. Terrain diffraction, foliage motion, atmospheric refraction, antenna patterns, Earth curvature, and installation tolerances lie beyond this ideal clearance geometry.