modelling Level 4

Satellite Overpasses and Visibility

When is a satellite visible from your location? You need line of sight — the satellite must be above your horizon and not blocked by Earth itself. This model derives the geometry of satellite visibility, calculates elevation and azimuth angles, and predicts overpass times for any ground station.

Prerequisites: 3d geometry, line of sight, elevation angle, azimuth, vector geometry

26 Feb 2026 Updated 12 Mar 2026 23 min read

1. The Question

When will the International Space Station pass over Denver?

A satellite orbits Earth in a known path. A ground observer at a specific latitude and longitude wants to know:

When is the satellite above my horizon?
Where in the sky should I look (elevation angle, compass direction)?
How long will it be visible?

The mathematical question: Given the satellite’s position and the observer’s position, calculate the line-of-sight geometry.

2. The Conceptual Model

Line of Sight

The satellite is visible if:

Above the local horizon: Elevation angle > 0°
Not blocked by Earth: The straight line from observer to satellite doesn’t intersect Earth’s surface
Sufficient elevation: Practical visibility often requires elevation > 10° (atmospheric effects, obstructions)

Horizon Range

Even if a satellite is “overhead” in terms of ground track, it may be beyond the horizon if it’s too far away laterally.

Maximum range to horizon from altitude $h$:

\[d_{\text{max}} = \sqrt{2R_{\oplus}h + h^2}\]

For LEO satellites (~400 km), this is roughly 2300 km surface distance.

Elevation and Azimuth

From the observer’s perspective:

Elevation angle ($E$): Angle above the horizon (0° = horizon, 90° = zenith)
Azimuth ($Az$): Compass bearing (0° = north, 90° = east, 180° = south, 270° = west)

These are the look angles — where to point your antenna or telescope.

3. Building the Mathematical Model

Step 1: Position Vectors

Observer position (lat, lon, altitude ≈ 0):

\[\mathbf{r}_{\text{obs}} = R_{\oplus} \begin{pmatrix} \cos(\text{lat}) \cos(\text{lon}) \\ \cos(\text{lat}) \sin(\text{lon}) \\ \sin(\text{lat}) \end{pmatrix}\]

Satellite position at time $t$:

\[\mathbf{r}_{\text{sat}}(t) = \begin{pmatrix} x_{\text{sat}}(t) \\ y_{\text{sat}}(t) \\ z_{\text{sat}}(t) \end{pmatrix}\]

(Calculated from orbital parameters using ground track methods from Model 14)

Range vector (from observer to satellite):

\[\mathbf{\rho} = \mathbf{r}_{\text{sat}} - \mathbf{r}_{\text{obs}}\]

Step 2: Local Horizon Frame

Define a coordinate system at the observer’s location:

East ($\mathbf{\hat{e}}$): Points east (tangent to latitude circle)
North ($\mathbf{\hat{n}}$): Points north (tangent to meridian)
Up ($\mathbf{\hat{u}}$): Points away from Earth’s center (local vertical)

Up direction (unit radial vector):

\[\mathbf{\hat{u}} = \frac{\mathbf{r}_{\text{obs}}}{|\mathbf{r}_{\text{obs}}|} = \begin{pmatrix} \cos(\text{lat}) \cos(\text{lon}) \\ \cos(\text{lat}) \sin(\text{lon}) \\ \sin(\text{lat}) \end{pmatrix}\]

East direction:

\[\mathbf{\hat{e}} = \begin{pmatrix} -\sin(\text{lon}) \\ \cos(\text{lon}) \\ 0 \end{pmatrix}\]

North direction:

\[\mathbf{\hat{n}} = \begin{pmatrix} -\sin(\text{lat}) \cos(\text{lon}) \\ -\sin(\text{lat}) \sin(\text{lon}) \\ \cos(\text{lat}) \end{pmatrix}\]

These form an orthonormal basis (all unit length, all perpendicular).

Step 3: Transform Range Vector to Local Frame

Project the range vector $\mathbf{\rho}$ onto the local East-North-Up frame:

$\rho_e = \mathbf{\rho} \cdot \mathbf{\hat{e}}$ $\rho_n = \mathbf{\rho} \cdot \mathbf{\hat{n}}$ $\rho_u = \mathbf{\rho} \cdot \mathbf{\hat{u}}$

These are the topocentric coordinates — position relative to the observer’s local horizon.

Step 4: Calculate Elevation Angle

The elevation angle is the angle above the horizontal plane:

\[E = \arcsin\left(\frac{\rho_u}{|\mathbf{\rho}|}\right)\]

Where $

\mathbf{\rho}

= \sqrt{\rho_e^2 + \rho_n^2 + \rho_u^2}$ is the slant range (direct distance from observer to satellite).

Alternative form:

\[E = \arctan\left(\frac{\rho_u}{\sqrt{\rho_e^2 + \rho_n^2}}\right)\]

Visibility condition: $E > 0°$ (satellite is above horizon)

Step 5: Calculate Azimuth

The azimuth is the compass bearing in the horizontal plane:

\[Az = \arctan2(\rho_e, \rho_n)\]

(Note: atan2(y, x) gives the angle from the positive x-axis; here we use atan2(east, north) to measure clockwise from north)

Converting to compass degrees:

0° = North
90° = East
180° = South
270° = West

If the result from atan2 is negative, add 360°.

4. Worked Example by Hand

Problem: An observer is at Denver (39.7°N, 105.0°W). A satellite is at position:

\[\mathbf{r}_{\text{sat}} = \begin{pmatrix} -2000 \\ 3000 \\ 5500 \end{pmatrix} \text{ km}\]

(a) Calculate the range vector $\mathbf{\rho}$.
(b) Transform to local East-North-Up coordinates.
(c) Calculate elevation angle and azimuth.
(d) Is the satellite visible?

Solution

(a) Observer and range vector

Denver position ($R_{\oplus} = 6371$ km):

\[\mathbf{r}_{\text{obs}} = 6371 \begin{pmatrix} \cos(39.7°) \cos(-105.0°) \\ \cos(39.7°) \sin(-105.0°) \\ \sin(39.7°) \end{pmatrix}\] \[= 6371 \begin{pmatrix} 0.766 \times (-0.259) \\ 0.766 \times (-0.966) \\ 0.640 \end{pmatrix} = \begin{pmatrix} -1264 \\ -4718 \\ 4077 \end{pmatrix} \text{ km}\]

Range vector:

\[\mathbf{\rho} = \mathbf{r}_{\text{sat}} - \mathbf{r}_{\text{obs}} = \begin{pmatrix} -2000 - (-1264) \\ 3000 - (-4718) \\ 5500 - 4077 \end{pmatrix} = \begin{pmatrix} -736 \\ 7718 \\ 1423 \end{pmatrix} \text{ km}\]

(b) Local frame basis vectors

Up:

\[\mathbf{\hat{u}} = \begin{pmatrix} \cos(39.7°) \cos(-105.0°) \\ \cos(39.7°) \sin(-105.0°) \\ \sin(39.7°) \end{pmatrix} = \begin{pmatrix} -0.198 \\ -0.740 \\ 0.640 \end{pmatrix}\]

East:

\[\mathbf{\hat{e}} = \begin{pmatrix} -\sin(-105.0°) \\ \cos(-105.0°) \\ 0 \end{pmatrix} = \begin{pmatrix} 0.966 \\ -0.259 \\ 0 \end{pmatrix}\]

North:

\[\mathbf{\hat{n}} = \begin{pmatrix} -\sin(39.7°) \cos(-105.0°) \\ -\sin(39.7°) \sin(-105.0°) \\ \cos(39.7°) \end{pmatrix} = \begin{pmatrix} 0.166 \\ 0.618 \\ 0.766 \end{pmatrix}\]

Project range onto local frame:

\[\rho_e = \mathbf{\rho} \cdot \mathbf{\hat{e}} = (-736)(0.966) + (7718)(-0.259) + (1423)(0) = -711 - 2000 = -2711 \text{ km}\] \[\rho_n = \mathbf{\rho} \cdot \mathbf{\hat{n}} = (-736)(0.166) + (7718)(0.618) + (1423)(0.766) = -122 + 4770 + 1090 = 5738 \text{ km}\] \[\rho_u = \mathbf{\rho} \cdot \mathbf{\hat{u}} = (-736)(-0.198) + (7718)(-0.740) + (1423)(0.640) = 146 - 5711 + 911 = -4654 \text{ km}\]

(c) Elevation and azimuth

Slant range:

\[|\mathbf{\rho}| = \sqrt{(-2711)^2 + (5738)^2 + (-4654)^2} = \sqrt{7.35 \times 10^6 + 3.29 \times 10^7 + 2.17 \times 10^7} = \sqrt{6.36 \times 10^7} = 7975 \text{ km}\]

Elevation angle:

\[E = \arcsin\left(\frac{-4654}{7975}\right) = \arcsin(-0.584) = -35.7°\]

Azimuth:

\[Az = \arctan2(-2711, 5738) = \arctan2(-2711, 5738) \times \frac{180}{\pi} \approx -25.3°\]

Convert to compass bearing: $-25.3° + 360° = 334.7°$ (NNW)

(d) Visibility

Elevation angle is −35.7° → Satellite is below the horizon (not visible).

The negative elevation means the satellite is on the opposite side of Earth from Denver.

5. Computational Implementation

Below is an interactive satellite visibility calculator.

Observer Latitude: 40°N Observer Longitude: -105°E Satellite Altitude (km): 400 km Time (hours since epoch): 0.0 hours

Current Satellite Position:

Latitude:

Longitude:

Slant Range: km

Elevation Angle: °

Azimuth: ° ()

Visibility:

Try this:

Animate Time: Watch the satellite rise and set
Find Next Pass: Jump to the next time the satellite is above 10° elevation
Move the observer’s location: Different latitudes see different passes
Notice: Peaks in the elevation plot correspond to satellite overpasses
The chart shows when the satellite is visible (elevation > 0°) and when it’s usefully above the horizon (>10°)

Key insight: Visibility is periodic — the satellite passes overhead multiple times per day at predictable intervals determined by the orbital period and Earth’s rotation.

6. Polar Overpass Diagram

Below is a polar plot showing satellite passes from the observer’s perspective — a local sky map with zenith at center and horizon at the perimeter.

Observer Latitude: 40°N Satellite Altitude (km): 600 km Sensor Swath Width (km): 185 km Time Window (hours): 24 hours

Reading the plot: Center = zenith (directly overhead), outer ring = horizon. Each colored arc is one satellite pass. Wider bands show sensor swath coverage.

Try this:

Change latitude: Higher latitudes see more polar orbit passes (inclination 98°)
Adjust swath width: Set to 185 km (Landsat), 290 km (Sentinel-2), or 500 km (MODIS)
Time window: 24 hours shows daily coverage pattern; 48 hours reveals repeat geometry
Toggle swath: Turn off swath display to see just the satellite ground tracks
Notice how passes at different azimuths provide multi-directional coverage

Key insight: The polar plot reveals the viewing geometry from the observer’s perspective. Multiple passes from different directions provide stereoscopic potential and reduce shadow/illumination bias in imagery.

7. Interpretation

Reading the Polar Plot

The polar overpass diagram is a local sky map from the observer’s viewpoint:

Center: Zenith (directly overhead, 90° elevation)
Outer ring: Horizon (0° elevation)
Radial distance: Zenith angle (90° - elevation)
Angular position: Azimuth (compass bearing)

Each colored arc represents one satellite pass. The arc shows the satellite’s path across the sky as seen from the ground station.

Swath coverage: The dashed lines flanking each pass show the sensor’s ground coverage footprint. Points on the ground within this swath are imaged during the pass.

Applications:

Mission planning: Which passes provide coverage of a target area?
Stereo imaging: Passes from different azimuths enable 3D reconstruction
Illumination analysis: Morning passes (eastern azimuths) vs. evening passes (western azimuths) have different sun angles
Antenna scheduling: Determine optimal times for data downlink based on elevation angle

Swath Width and Coverage

The sensor swath is the width of the strip of Earth’s surface imaged in a single pass.

Wide swath (e.g., MODIS 2330 km):

Fewer passes needed for global coverage
Lower spatial resolution (250 m–1 km)
Daily global coverage possible

Narrow swath (e.g., Landsat 185 km):

Many passes needed for global coverage
Higher spatial resolution (15–30 m)
16-day revisit cycle for exact repeat

Swath geometry depends on:

Sensor field of view (FOV): Angular width of detector array
Pointing capability: Can the sensor look sideways (off-nadir)?
Altitude: Higher satellites see wider swaths but with coarser resolution

Look angle from swath width:

For a nadir-pointing sensor at altitude $h$ with ground swath width $W$:

\[\text{Half-swath angle} \approx \arctan\left(\frac{W/2}{R_{\oplus} + h}\right)\]

This is the angular extent from the satellite’s perspective.

Coverage Gaps and Overlaps

From the polar plot, you can identify:

Gaps: Regions of sky with no passes (equatorial zones for polar orbits from high latitudes)
Overlap: Multiple passes covering the same area from different angles
Coverage uniformity: Are passes evenly distributed in azimuth?

Example: At 40°N with a sun-synchronous orbit (98° inclination):

Passes cluster in north-south directions
East-west coverage comes from orbital precession over days
No truly equatorial passes (inclination limits)

Maximum Elevation Angle

The highest point in the sky the satellite reaches depends on:

How close the ground track passes to the observer
Satellite altitude (higher satellites appear lower in the sky from the same ground track distance)

Zenith pass ($E = 90°$): Satellite directly overhead — ground track passes through observer’s location.

Low pass ($E < 30°$): Satellite is near the horizon — ground track is far from observer.

Visibility Duration

For LEO satellites:

Typical pass duration: 5–10 minutes
Maximum elevation depends on orbital geometry
Number of visible passes per day: varies with inclination

For geostationary satellites:

Continuously visible from the same location (if above horizon)
Elevation angle is constant
Only visible from latitudes within ~±81° of equator

Antenna Tracking

Ground stations with directional antennas must track the satellite:

Azimuth changes as satellite moves across the sky
Elevation changes from rise to set
Tracking systems use predicted ephemerides (orbital predictions) to point the antenna

Two-line element sets (TLEs): Standard format for distributing orbital parameters. Used to generate predictions.

7. What Could Go Wrong?

Forgetting to Check Both Conditions

A satellite can be:

Above the horizon from your location but behind Earth (blocked by the planet)
Close in slant range but below the horizon (on the other side of Earth)

Always check: elevation angle > 0° AND not blocked by Earth’s sphere.

Assuming Visibility = Ground Track Proximity

The satellite’s ground track is where it’s directly overhead. But you can see satellites whose ground track is hundreds of kilometers away — they just appear lower in the sky.

Horizon distance for 400 km altitude satellite: ~2300 km surface distance.

Neglecting Atmospheric Refraction

Near the horizon, Earth’s atmosphere bends light rays, making satellites appear slightly higher than their geometric position.

Refraction correction: ~0.5° at horizon, negligible above 10° elevation.

Mixing Up Azimuth Conventions

Mathematical convention: Angle from positive x-axis, counterclockwise.
Compass convention: Angle from north, clockwise.

Always use atan2(east, north) for compass azimuth, not atan2(north, east).

8. Extension: Access Time and Revisit Frequency

Access time: Total time per day a satellite is visible above a minimum elevation (e.g., 10°).

For a polar orbit at 800 km:

Each pass: ~10 minutes above 10° elevation
Passes per day: ~14
Total access time: ~140 minutes/day (if all passes are visible)

Revisit frequency: How often a satellite can image the same ground location.

Factors:

Orbital period
Swath width (sensor coverage width)
Off-nadir pointing capability (can the satellite look sideways?)

Example: Landsat 8 has a 16-day exact repeat cycle (same ground track every 16 days) but can image any location every ~8 days due to overlapping swaths.

9. Math Refresher: Dot Product for Projections

Projecting a Vector onto a Direction

Given vector $\mathbf{v}$ and unit vector $\mathbf{\hat{u}}$, the component of $\mathbf{v}$ in the direction of $\mathbf{\hat{u}}$ is:

\[v_{\parallel} = \mathbf{v} \cdot \mathbf{\hat{u}}\]

Geometric meaning: The length of $\mathbf{v}$’s shadow when projected onto $\mathbf{\hat{u}}$.

Example: Range vector $\mathbf{\rho}$ projected onto “up” direction gives the vertical component (used to compute elevation angle).

Orthonormal Basis

Three unit vectors $\mathbf{\hat{e}}$, $\mathbf{\hat{n}}$, $\mathbf{\hat{u}}$ form an orthonormal basis if:

Each has length 1: $ \mathbf{\hat{e}} = \mathbf{\hat{n}} = \mathbf{\hat{u}} = 1$
All are perpendicular: $\mathbf{\hat{e}} \cdot \mathbf{\hat{n}} = 0$, etc.

Any vector can be expressed as:

\[\mathbf{v} = (v \cdot \mathbf{\hat{e}})\mathbf{\hat{e}} + (v \cdot \mathbf{\hat{n}})\mathbf{\hat{n}} + (v \cdot \mathbf{\hat{u}})\mathbf{\hat{u}}\]

This is what we do when converting the range vector to topocentric (East-North-Up) coordinates.

Summary

Visibility requires satellite above local horizon: elevation angle $E > 0°$
Elevation angle: $E = \arcsin(\rho_u / \mathbf{\rho} )$ where $\rho_u$ is vertical component
Azimuth: $Az = \arctan2(\rho_e, \rho_n)$ compass bearing (0° = N, 90° = E)
Local horizon frame: East-North-Up basis vectors at observer’s location
Maximum slant range for LEO satellites: ~2000–2500 km
Pass duration: typically 5–10 minutes for LEO
Multiple passes per day, predictable from orbital parameters
Geostationary satellites appear fixed in the sky

Phase II Complete: Models 13–15 provide the foundation for orbital mechanics and satellite tracking. Future extensions could cover:

Sensor swath geometry and coverage calculations
Off-nadir pointing and agile satellites
Constellation design for continuous coverage
Doppler shift and radio tracking
Two-line element set interpretation