SME
Surveying · Problem type
Range from signal travel time
Use this for “A GNSS signal takes 0.0723 s to arrive. How far is the satellite, and what pseudorange does a receiver clock 1 μs off measure?”
This is an example. Type your own values to start a new problem.
ρ (km)
=
299,792.458
km/s
×
Δt (s)
Δρ (m)
=
299,792.458
km/s
×
δt (μs)
P (km)
=
ρ (km)
+
Δρ (m)
Formulas
ρ
=
299,792.458
×
Δt
21,674.9947 = 299,792.458 × 0.0723
Δρ = 299,792.458 × δt
0.2998
=
299,792.458
×
(1
×
10
-6
)
P
=
ρ
+
Δρ
21,675.2945 = 21,674.9947 + 0.2998
Formulas are worked in s, km. The step-by-step shows the conversions.
Assumptions
•
The signal travels at the speed of light, c = 299,792.458 km/s; the atmosphere’s delay is left out.
•
The receiver times the signal with its own clock, so a clock error δt adds cδt to every range.
•
A range that carries the clock error is a pseudorange; a fourth satellite lets the receiver solve for δt.
Step-by-step
WE KNOW
Δt = 0.0723 s, δt = 1 μs
FIND
range (ρ), range error (Δρ), pseudorange (P)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to s, km).
δt = 1 μs = 1 × 10⁻⁶ s (1 s = 10⁶ μs)
Step 2 · Find range (ρ)
Use ρ = 299,792.458 × Δt
Distance is speed times time: c in km/s times the travel time in seconds.
ρ
=
299,792.458
×
0.0723
ρ
=
21,674.99
km
Step 3 · Find range error (Δρ)
Use Δρ = 299,792.458 × δt
Light covers c × δt during the clock error, so the range is off by that much (c in km/s).
Δρ
=
299,792.458 ×
10
-6
Δρ = 0.2998 km
Step 4 · Find pseudorange (P)
Use P = ρ + Δρ
The receiver measures the true range plus the clock’s range error.
P
=
21,674.99
+
0.2998
P
=
21,675.29
km
Step 5 · Convert the answers
Change each answer back to the unit you chose.
Δρ = 0.2998 km = 299.7925 m (1 km = 1000 m)
Check
Put the numbers back in. Both sides should match.
21,674.99 = 299,792.458 × 0.0723 ✓
0.2998
=
299,792.458 ×
10
-6
✓
21,675.29 = 21,674.99 + 0.2998 ✓
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