Line of Sight Calculator — Radio Horizon & Earth Bulge

Radio horizon of each antenna, maximum line-of-sight distance between two masts, Earth bulge and first Fresnel-zone radius at any point along the path, and the clearance a link needs there, over a smooth Earth with the k-factor of your choice.

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Formula

d_{los} = \sqrt{2kR}\left(\sqrt{h_1} + \sqrt{h_2}\right),\quad b = \frac{d_1 d_2}{2kR},\quad r_1 = \sqrt{\frac{\lambda d_1 d_2}{d}},\quad h_{req} = b + p\,r_1

Reference: ITU-R P.526-15, Eqs. (2), (21) and (22); ITU-R P.530-17, §2.2.2; ITU-R P.834-9, §2

h_1, h_2— Antenna heights above the datum (m)

kR— Effective Earth radius, R = 6371 km (m)

d_1, d_2— Distances from the point to each antenna (m)

b— Earth bulge at the point (m)

r_1— First Fresnel-zone radius at the point (m)

p— Required fraction of the first Fresnel zone

How It Works

Two antennas can see each other only until the Earth's surface gets in the way. A radio ray does not travel quite straight: the atmosphere's refractive index falls with height, which bends rays gently back toward the ground. The standard way to account for it is to keep the ray straight and enlarge the Earth instead, to an effective radius $a_e = kR$ with $R = 6371$ km. For the standard atmosphere $k = 4/3$ (ITU-R P.834-9); $k = 1$ is the geometric case with no refraction, and $k$ below 1, sub-refraction, makes the Earth look more curved.

Radio horizon and line-of-sight distance

An antenna at height $h$ sees the smooth Earth out to its radio horizon, and two antennas see each other while the path is shorter than the sum of their horizons (ITU-R P.526-15, equation 21):

d_h = \sqrt{2 a_e h}, \qquad d_{los} = \sqrt{2 a_e}\left(\sqrt{h_1} + \sqrt{h_2}\right)

A 10 m antenna has a radio horizon of 13.03 km at $k = 4/3$ ; 30 m and 10 m antennas see each other out to 35.61 km.

Earth bulge

Line of sight alone is not enough: the ray also needs room around it. At a point $d_1$ from one end and $d_2$ from the other on a path of length $d$ , the Earth rises above the straight chord between the antenna sites by the Earth bulge, and the first Fresnel zone has its radius (ITU-R P.526-15, equation 2):

b = \frac{d_1 d_2}{2 a_e}, \qquad r_1 = \sqrt{\frac{\lambda d_1 d_2}{d}}

At the middle of a 50 km path the bulge is 36.79 m at $k = 4/3$ .

Clearance

A ray that clears every obstacle by 60% of the first Fresnel zone propagates essentially as in free space. ITU-R P.530-17, section 2.2.2, asks for 1.0 F1 over the highest obstacle at the median $k$ (4/3 by default), and for 0.0 to 0.6 F1 at the low $k$ exceeded 99.9% of the worst month. The calculator gives the clearance a straight ray needs above a flat datum at the chosen point, $b + p\,r_1$ , the height of the ray above the smooth Earth there (ITU-R P.526-15, equation 22), and the highest obstacle the path tolerates at that point:

h = \frac{h_1 d_2 + h_2 d_1}{d} - b, \qquad h_{obs} = h - p\,r_1

Validity

The Earth is a smooth sphere with one $k$ along the whole path; terrain, buildings, vegetation and ducting are not modelled. The geometry is the small-angle form P.526-15 uses, which holds for paths within about 5° of the horizontal (section 4.4), and the spherical-Earth method applies from 10 MHz up (section 3.2). Outside either, the results are shown with a warning. A point beyond either end of the path is an error.

Worked Example

Problem: A 6 GHz point-to-point link runs 30 km between a 40 m and a 30 m mast over open, level ground. Does it clear the Earth at mid-path, and how tall an obstacle can it tolerate there?

Step 1 - Effective Earth radius at k = 4/3: a_e = 4/3 × 6371 km = 8494.7 km

Step 2 - Line of sight: d_h1 = √(2 × 8494667 × 40) = 26.07 km d_h2 = √(2 × 8494667 × 30) = 22.58 km d_los = 48.64 km, so the 30 km path is within line of sight.

Step 3 - Earth bulge at mid-path: b = 15000 × 15000 / (2 × 8494667) = 13.24 m

Step 4 - First Fresnel zone: λ = 299792458 / 6e9 = 0.04997 m r_1 = √(0.04997 × 15000 × 15000 / 30000) = 19.36 m

Step 5 - Required clearance for 0.6 F1: b + 0.6 r_1 = 13.24 + 11.62 = 24.86 m

Step 6 - What the ray actually has: h = (40 + 30)/2 − 13.24 = 21.76 m above the smooth Earth Highest obstacle at mid-path: 21.76 − 11.62 = 10.14 m

Step 7 - The ITU-R P.530-17 checks: At k = 4/3 with 1.0 F1, the obstacle allowance falls to 2.40 m. At k = 2/3 with 0.3 F1, the bulge doubles to 26.49 m and the allowance is 2.71 m.

The smooth Earth alone clears, but the tighter of the two P.530 conditions leaves only 2.4 m for trees or buildings at mid-path, so the masts need to go higher if anything stands there.

Practical Tips

✓Evaluate the clearance at the highest obstacle, not only at mid-path; the bulge is largest at mid-path, but the obstacle may be elsewhere.
✓Run the same path at k = 4/3 with 1.0 F1 and at a low k with 0.3 F1, as ITU-R P.530 does, and design to the more demanding of the two.
✓Raising the lower antenna is cheaper than raising both: the ray height at a point is weighted toward the nearer antenna.
✓For the Fresnel zone alone at any frequency, the Fresnel zone calculator gives the radius and the 60% clearance directly.

Common Mistakes

✗Checking only geometric line of sight. A path inside the radio horizon can still lose many decibels to diffraction if the Earth or an obstacle intrudes into the first Fresnel zone.
✗Using k = 4/3 alone. Sub-refractive conditions lower k toward 2/3 or less for a small percentage of the time, doubling the bulge; ITU-R P.530 plans clearance at both the median and the low k.
✗Adding the bulge to the obstacle heights and then also subtracting it from the ray. The bulge is counted once: either raise the terrain profile by it or lower the ray by it.
✗Mixing heights above sea level for one antenna and above ground for the other. Both heights, and any obstacle heights, must be measured from the same datum.

Frequently Asked Questions

How far is the radio horizon?

The square root of twice the effective Earth radius times the antenna height. At the standard k of 4/3 a 10 m antenna has a radio horizon of about 13 km; the visual, geometric horizon with k equal to 1 is about 11.3 km for the same height.

What is Earth bulge?

The height by which the curved Earth rises above a straight line drawn between two points on its surface. At a point d1 from one end and d2 from the other it is d1 times d2 divided by twice the effective Earth radius, so it is largest at mid-path.

What k-factor should I use?

4/3 for the standard atmosphere, which ITU-R P.834 gives as the median. For planning clearance, ITU-R P.530 also checks the lower value exceeded 99.9 percent of the worst month, read from its Figure 2 for the path length; it is below 1 and lowest on short paths.

How much Fresnel-zone clearance does a link need?

At least 60 percent of the first Fresnel zone for essentially free-space propagation. ITU-R P.530 asks for the full first zone over the highest obstacle at the median k, and between grazing and 60 percent at the low k, depending on climate and the kind of obstruction.

Does this calculator use terrain data?

No. It assumes a smooth spherical Earth. Enter obstacle heights against the highest-obstacle output at the point you are checking, or use a terrain-profile tool for a full path study.

Methodology & References

References

Propagation by diffraction — Recommendation ITU-R P.526-15 (10/2019), §2.1 Eq. (2), §3.2 Eqs. (21)–(22), §4.4 — Fresnel radius, line-of-sight distance, ray clearance, a_e = k × 6371 km
Propagation data and prediction methods required for the design of terrestrial line-of-sight systems — Recommendation ITU-R P.530-17 (12/2017), §2.2.2 — planning criteria for path clearance
Effects of tropospheric refraction on radiowave propagation — Recommendation ITU-R P.834-9 (12/2017), §2 — effective Earth radius, k = 4/3

The horizon, line-of-sight and bulge figures are P.526-15's own equations; as k grows the bulge falls to zero and the clearance reduces to the flat-Earth chord.

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