Radar Physics Simulator

Watch a single radar pulse race out at the speed of light, bounce off a target, and return — its timing gives range, its strength gives detectability, its frequency shift gives speed.

Starting values — Radar Physics Simulator

MTI clutter filter	off

Change any control above and the results recompute from these.

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Halving the detection range costs a 16-fold cut in cross-section 🖖

Received power falls as 1/R⁴ — the pulse spreads on the way out and the echo spreads on the way back — so detection range scales as the fourth root of radar cross-section. That fourth root is brutal in both directions. This tool's own RCS presets run from 0.01 m² for a stealth aircraft to 1000 m² for a ship, a span of 100,000×, yet the ranges they produce differ by only 100000^¼ ≈ 17.8×. Against the 5 m² fighter, a stealth design cuts cross-section 500-fold and buys a range reduction of just 500^¼ ≈ 4.7×. Every further halving of the range at which you are seen costs another factor of sixteen.

One echo, three answers 🖖

Radar fires a single pulse of radio waves and reads the returning echo three ways at once. The delay before it comes back gives distance (R = c·Δt/2, halved because the pulse makes a round trip); the strength of the echo tells you whether the target is even detectable; and any shift in its frequency reveals how fast the target is closing or receding. Takeaway: light is so quick that a target 150 km away answers in under a millisecond.

How a plane can hide at the right speed 🖖

MTI radar rejects ground clutter by ignoring echoes whose frequency doesn't change from pulse to pulse. But a target flying at a blind speed — where its Doppler shift lands on an exact multiple of the PRF — looks frozen between pulses and gets filtered out along with the hills. These speeds follow v = n·λ·PRF/2; with λ = 0.1 m and PRF = 1000 Hz the first is 50 m/s. Real radars hop their PRF to close the gaps.

Problems solved in full

A fighter 90 km out with a radar cross-section of 5 m² 7 steps
A fighter 90 km out, with a radar cross-section of 5 m². The set transmits 100 kW at 3 GHz through an antenna of gain 1000, and its receiver calls a detection at -120 dBm. Work out what comes back, and then how far this radar could see if the target kept flying away.
1. The pulse crosses the gap twice, so the clock is timing 180 km of travel, not 90. Halve before you divide by c, or every range you report comes out doubled: 600 µs here.
  
  $Δ t = \frac{2 R}{c} = \frac{2 \times 9 \times 1 0 ^{4}}{3 \times 1 0 ^{8}} = 6 \times 1 0^{- 4} s = 600 μ s$
2. Frequency fixes the wavelength, and the wavelength earns its place twice — it sets the Doppler sensitivity, and it sets how much of the returning wave the dish can gather. At 3 GHz, λ = 0.1 m.
  
  $λ = \frac{c}{f} = \frac{3 \times 1 0 ^{8}}{3 \times 1 0 ^{9}} = 0.1 m$
3. Now the part that makes radar hard. On the way out the power spreads over a sphere, so the density arriving at the target falls as 1/R². The target intercepts σ of that and re-radiates it, and the re-radiated power spreads over a sphere as well — a second, independent 1/R². The dish then collects from an effective area A_e = Gλ²/4π.
  
  $S_{t} = \frac{P _{t} G}{4 π R ^{2}}, S_{r} = \frac{S _{t} σ}{4 π R ^{2}}, A_{e} = \frac{G λ ^{2}}{4 π}$
4. Multiply the three and the R² factors square. Received power falls as R⁻⁴, so a radar echo is not merely faint but catastrophically faint: double the range and the echo drops by 16, not by 4.
  
  $P_{r} = S_{r} A_{e} = \frac{P _{t} G ^{2} λ ^{2} σ}{( 4 π ) ^{3} R ^{4}}$
5. Put the numbers in. The numerator is 5×10⁹; the denominator is (4π)³ = 1984 times R⁴ = 6.56×10¹⁹, so 1.30×10²³. The echo is 3.84×10⁻¹⁴ W. That is 38 femtowatts returning from a 100 kW transmitter, and the ratio between those two is the whole problem.
  
  $P_{r} = \frac{1 0 ^{5} \cdot 1 0 ^{6} \cdot 1 0 ^{- 2} \cdot 5}{1984 \times ( 9 \times 1 0 ^{4} ) ^{4}} = \frac{5 \times 1 0 ^{9}}{1.30 \times 1 0 ^{23}} = 3.84 \times 1 0^{- 14} W$
6. Radar powers span 20 orders of magnitude, so they are quoted in dB relative to a milliwatt. 3.84×10⁻¹⁴ W is -104 dBm, comfortably clear of the -120 dBm the receiver needs.
  
  $10 lo g_{10} (\frac{3.84 \times 1 0 ^{- 14}}{1 0 ^{- 3}}) = 10 lo g_{10} (3.84 \times 1 0^{- 11}) = - 104 dBm$
7. That margin converts into range, but not generously. In watts the echo is 38.4 times the threshold, and power goes as R⁻⁴, so the range can grow only by the fourth root of 38.4. That is 2.489, and 90 km × 2.489 = 224 km.
  
  $\frac{P _{r}}{P _{m i n}} = \frac{3.84 \times 1 0 ^{- 14}}{1 0 ^{- 15}} = 38.4 ⟹ R_{m a x} = 90 \times 38. 4^{1/4} = 90 \times 2.489 = 224 km$
Answer

224 km — and it took an echo 38 times stronger than the receiver needs to reach it. There is a second limit hiding inside that number, and nothing on the panel shows it. An echo from 224 km takes 2 × 224 km / c = 1.49 ms to come home. Fire the next pulse any sooner and the late echo lands after a fresh pulse has already left, and the set cannot tell which pulse it belongs to — it will report a 224 km target as a near one. So the detection range you just derived fixes a pulse repetition frequency ceiling of about 670 Hz. How far a radar can see and how often it can look are traded against each other by the speed of light alone, and no amount of transmit power touches that trade.

$\frac{2 R _{m a x}}{c} = 1.49 ms ⟹ PRF < \frac{1}{1.49 ms} = 670 Hz$
A target with σ = 0.01 m² at 70 km 5 steps
The same radar, still 100 kW at 3 GHz, now looking at a target with σ = 0.01 m² at 70 km. That is 500 times less radar cross-section than the fighter, and it is 20 km closer. Find the echo, and then find how close it would have to come to be seen at all.
1. Only one factor in the numerator changed: 10⁵ × 10⁶ × 10⁻² × 10⁻² = 10⁷. With R⁴ at 70 km giving a denominator of 4.76×10²², the echo is 2.1×10⁻¹⁶ W — 0.21 femtowatts.
  
  $P_{r} = \frac{1 0 ^{5} \cdot 1 0 ^{6} \cdot 1 0 ^{- 2} \cdot 1 0 ^{- 2}}{1984 \times ( 7 \times 1 0 ^{4} ) ^{4}} = \frac{1 0 ^{7}}{4.76 \times 1 0 ^{22}} = 2.1 \times 1 0^{- 16} W$
2. Notice how badly the two changes are matched. Closing from 90 km to 70 km multiplies the echo by (90/70)⁴ = 2.73; the cross-section divides it by 500. The geometry gain is swamped.
  
  $(\frac{90}{70})^{4} = 2.73 ≪ \frac{5}{0.01} = 500$
3. In dBm that is -127, against a receiver floor of -120. The echo is under the floor, so at 70 km this target does not exist as far as the set is concerned.
  
  $10 lo g_{10} (2.1 \times 1 0^{- 13}) = - 127 dBm < - 120 dBm$
4. Range is the only variable left. In watts the echo is short by a factor of 10⁻¹⁵ / 2.1×10⁻¹⁶ = 4.76, and it rises as R⁻⁴, so the range has to fall by the fourth root of 4.76 — a factor of 1.477. 70 km / 1.477 = 47.4 km.
  
  $\frac{P _{m i n}}{P _{r}} = \frac{1 0 ^{- 15}}{2.1 \times 1 0 ^{- 16}} = 4.76 ⟹ R = \frac{70}{4.7 6 ^{1/4}} = \frac{70}{1.477} = 47.4 km$
5. Reach the same number from the other end, starting at the fighter's 224 km. Detection range scales as σ to the power 1/4, so dividing σ by 500 divides the range by 500 to the power 1/4, which is 4.73. 224 / 4.73 = 47.4 km. One law, read in both directions.
  
  $R_{m a x} \propto σ^{1/4} : \frac{224}{50 0 ^{1/4}} = \frac{224}{4.73} = 47.4 km$
Answer

Cutting the radar cross-section by 500 buys a factor of 4.73 in detection range, and that fourth root is the entire economics of low observability. It also prices the alternative. To put this target back at 224 km with power alone you have to restore the product P_tσ: σ fell by 500, so P_t must rise by 500 — from 100 kW to 50 MW. That is the argument for shaping the target rather than enlarging the radar. The exponent charges the shaper at the same rate, though: every further halving of the detection range costs another factor of 16 in σ.

$R_{m a x} \propto (P_{t} σ)^{1/4} ⟹ P_{t} \to 500 \times 100 kW = 50 MW$

References (2)

The radar range equation and the R⁴ two-way loss: M. I. Skolnik, Introduction to Radar Systems, 3rd ed., ch. 1–2. McGraw-Hill, 2001. ISBN 978-0-07-288138-7.
MTI, blind speeds and PRF staggering, behind the third block: M. A. Richards, Fundamentals of Radar Signal Processing, 2nd ed., ch. 5. McGraw-Hill, 2014. ISBN 978-0-07-179832-7.

By the 42Math team · Updated 2026-07-17

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

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The R⁴ wall
Stealth - Stealth RCS=0.01 m² at 120 km — detection range only 47 km
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