Relativistic Velocity Addition Calculator

Throw a ball forwards from a fast rocket and the speeds do not simply add. This page does the real sum, shows the wrong one beside it, and draws the gap between them — because the gap is the whole point, and it is the clearest thing in relativity to show a child.

Last updated 11 Nov 2026

Two speeds → the speed the ground actually measures

w = (u′ + v)/(1 + u′v/c²), with the classical sum beside it

Speed of the object, as the rocket measures it (u′)This is the speed in the MOVING frame — what someone on the rocket would read off their own speedometer for the ball they threw. It may be negative: type −0.8 and the ball has been thrown backwards, out of the rear of the rocket, and the diagram turns its arrow round. The backwards case is worth trying, because it shows that the formula also refuses to let anything come out slower than −c.

Speed of the rocket, as the ground measures it (v)The speed of the moving frame itself, measured from the ground. Kept positive here so that “forwards” has one meaning throughout; if your rocket is going the other way, flip the sign of everything and read the answer the same way round.

Both speeds above are given asBoth boxes use the same unit, because adding speeds in different units is the one way to get this wrong that has nothing to do with relativity.

Four arrows on one scale, with the red line at c. The top two are the speeds you typed. Below the divider, the dashed red arrow is what you get by simply adding them and the solid dark arrow is the real answer. Push both speeds up and watch the dashed arrow sail past the c line while the solid one crowds up against it and never passes it. Feed it c itself and it returns exactly c, which is the one case where the solid arrow reaches the line. The short red bar between the two tips is the gap, and it is the whole of special relativity in one picture.

0.975610cExample

A rocket at 0.8c fires a probe forwards at 0.8c relative to itself

The sum that keeps c the same for everybody

w = (u′ + v) / (1 + u′v/c²) · in units of c: β_w = (β_u + β_v) / (1 + β_uβ_v) · classical: w = u′ + v

u′: the object’s speed IN THE MOVING FRAME — the rocket’s own measurement of the ball it threw. The prime is not decoration; putting the ground-frame speed here is the one mistake that matters
v: the speed of that moving frame relative to the ground. Swapping u′ and v gives the same answer, because the formula is symmetric in them — which is a useful check and also a trap, because it makes a wrong substitution look right
w: the object’s speed measured from the ground. Always smaller in magnitude than u′ + v when both are positive, and always strictly below c
1 + u′v/c²: the entire difference from Galileo. At 30 m/s and 300 m/s it is 1.000 000 000 000 0001, so the classical sum is right to more figures than anyone can measure; at 0.8c and 0.8c it is 1.64

Worked example

A rocket at 0.8c fires a probe forwards at 0.8c relative to itself
WRITE DOWN WHAT THE WRONG ANSWER WOULD BE. 0.8c + 0.8c = 1.6c. Do this deliberately: the question is almost always SET so that the naive answer exceeds c, and noticing that is half the marks.
IDENTIFY THE FRAMES. v = 0.8c is the rocket relative to the ground. u′ = 0.8c is the probe relative to the ROCKET. w is what we want: the probe relative to the ground.
NUMERATOR. u′ + v = 0.8c + 0.8c = 1.6c. Yes, really — the numerator is allowed to be bigger than c, because it is not the answer.
DENOMINATOR. 1 + u′v/c² = 1 + (0.8)(0.8) = 1.64. Working in units of c makes this trivial; working in m/s means squaring 2.4×10⁸ and dividing by 8.99×10¹⁶, which is the same number by a longer road.
DIVIDE. w = 1.6c / 1.64 = 0.975 610c, which is 2.925×10⁸ m/s. Below c, as it must be.
CHECK THE LIMIT. Put u′ = c into the formula: w = (c + v)/(1 + v/c) = c(c + v)/(c + v) = c, for ANY v. Light fired forwards from a fast rocket is measured at c from the ground. That is not a consequence of the formula — it is the postulate the formula was built to satisfy.

What the two sums give, side by side

The last two rows are the ones to look at twice. Light fired forwards from anything, at any speed, is still measured at exactly c — that is not a result of the formula so much as the requirement the formula was reverse-engineered from. And the final row shows the formula behaving perfectly sensibly in the ordinary direction: throw the ball backwards at exactly the rocket’s speed and it hangs motionless in the air, which is also what Galileo would have told you.
u′	v	Simply added	The real answer	Gap
30 m/s	300 m/s	330 m/s	330 m/s	3×10⁻²² — nothing you could measure
0.01c	0.01c	0.02c	0.019998c	0.01%
0.1c	0.1c	0.2c	0.198020c	1%
0.5c	0.5c	1.0c	0.8c	the first row where the naive sum reaches c exactly
0.8c	0.8c	1.6c	0.975610c	the naive sum is 64% over
0.9c	0.9c	1.8c	0.994475c
0.99c	0.99c	1.98c	0.999950c	two things at 99% of c give 99.995%
c	0.9c	1.9c	c exactly	light stays at c in every frame — the postulate
−0.8c	0.8c	0	0 exactly	a ball thrown backwards at the rocket’s own speed stands still

Speeds do not add, and the denominator is the reason

Start with the wrong answer on purpose. If a rocket travelling at 0.8c fires a probe forwards at 0.8c, the obvious answer is 1.6c, and the obvious answer is wrong. It is wrong not because something intervenes to slow the probe down, but because adding velocities was never a law of physics — it was an assumption about how time and space work, and that assumption fails. The real rule divides by 1 + u′v/c², and 1.6c becomes 0.9756c.

Why there is a denominator at all. Two observers moving relative to each other disagree about which events are simultaneous. A speed is a distance divided by a time, and if two frames disagree about both distances and times — and about which pairs of events count as “at the same moment” — then there is no reason a speed in one should be a speed in the other plus a constant. The denominator is exactly what falls out of the Lorentz transformation when you work out dx/dt in the new frame, and nothing was added by hand to make the answer come out below c.

c is a ceiling, not a speed limit sign. There is no point at which the formula clips the answer. Feed it 0.999c and 0.999c and it returns 0.9999995c; feed it 0.999999c twice and it returns closer still. The result approaches c and never arrives, because the denominator grows at exactly the rate needed. The one input that produces c is c itself, and then the answer is c no matter what the other speed is — which is the invariance of the speed of light, dropping out of the algebra rather than being imposed on it.

Where this goes wrong. Putting the ground-frame speed in the u′ box. The formula needs the speed as measured in the MOVING frame, and the two differ by exactly the thing you are trying to compute, so getting it wrong is circular rather than merely inaccurate. Next, mixing units: u′ in m/s with v in fractions of c gives nonsense, which is why this page uses one selector for both. Then forgetting the sign when the object goes backwards. And finally, using this formula for an acceleration or a closing speed between two objects seen from a third frame — that last one is a different question, and in the ground frame two things approaching each other at 0.9c each really do close at 1.8c, because a closing rate is not anybody’s velocity.

Frequently asked questions

Why can’t I just add the speeds?

Because adding speeds assumes everyone agrees about distances and times, and that assumption is what relativity drops. The honest way to put it is that the Galilean sum is an approximation to this one, accurate when u′v/c² is negligible. At 30 m/s and 300 m/s the correction is in the twenty-second decimal place, so for the whole of ordinary life the simple sum is right.

What if I throw the object backwards?

Type a negative u′. The formula handles it with no change: the denominator becomes less than 1 and the magnitude of the answer is bigger than the simple difference would be. A ball thrown backwards at exactly the rocket’s own speed comes out at exactly zero, which is the ordinary result and a good check that the formula is not doing anything strange in the familiar regime.

What happens if one of the speeds is c?

You get exactly c back, for any value of the other speed. Substituting u′ = c gives w = (c + v)/(1 + v/c) = c. That is the second postulate of special relativity appearing as an algebraic identity. On this page typing c clamps the input to 0.999 999 999c, so you will see 0.999 999 999 rather than a clean 1 — the clamp exists to keep the arithmetic finite and the diagram drawn, and the exact statement is in this answer rather than in the number.

Two spaceships approach each other at 0.9c each. Do they close at 1.8c?

In the ground frame, yes, and that is not a violation of anything. The gap between them shrinks at 1.8c as measured by ground clocks and rulers, but 1.8c is not the speed of any object in any frame — no single thing is moving at 1.8c. If you ask what ONE ship measures for the OTHER, that is this formula’s question, and the answer is (0.9 + 0.9)/(1 + 0.81) = 0.9945c.

Is this on my syllabus?

Not usually, and it is worth knowing that. AQA’s special relativity section 3.12.3 lists time dilation, length contraction and mass-energy, and does not require velocity addition. The CBSE Class XII curriculum has no relativity at all. This page is here because it is the easiest piece of relativity to understand and the hardest to forget, not because a board will examine it.

Does the order of the two speeds matter?

No. The formula is symmetric: swapping u′ and v gives the same w. That is handy as a check, and dangerous as a habit — it means a wrong assignment of the two frames produces a number that looks fine. The discipline is to say out loud which frame each speed was measured in before you substitute.

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References

CODATA Internationally Recommended 2022 Values of the Fundamental Physical Constants, NIST Standard Reference Database 121. Read on 11 November 2026 from the machine-readable table at physics.nist.gov/cuu/Constants/Table/allascii.txt, which prints “speed of light in vacuum 299 792 458 (exact)” — exact because the 1983 definition of the metre fixes it, so c² = 8.9875517873681764×10¹⁶ m²/s² carries no uncertainty either. The same table gives G = 6.674 30(15)×10⁻¹¹ m³kg⁻¹s⁻², which is the least precisely known of the constants on this page by a factor of ten thousand, standard acceleration of gravity 9.806 65 (exact), and the rest energies used here: electron 0.510 998 950 69(16) MeV, proton 938.272 089 43(29) MeV, muon 105.658 3755(23) MeV.
A. Einstein, “Zur Elektrodynamik bewegter Körper”, Annalen der Physik 17 (1905) 891–921 — the paper in which the two postulates, the Lorentz transformation, time dilation and length contraction all appear together, and the companion three-page note “Ist die Trägheit eines Körpers von seinem Energieinhalt abhängig?”, Annalen der Physik 18 (1905) 639–641, which adds the mass-energy result. Cited for the result, not for a number: neither paper was read for this build, and nothing on this page rests on a figure taken from them.
AQA, AS and A-level Physics (7407, 7408) specification, section 3.12.3 Special relativity, inside the option 3.12 Turning points in physics (A-level only). Read from aqa.org.uk on 11 November 2026. It is worth knowing exactly what a board asks for: 3.12.3.3 requires proper time and time dilation in the form t = t₀ / √(1 − v²/c²), with evidence from muon decay; 3.12.3.4 requires length contraction as l = l₀√(1 − v²/c²); and 3.12.3.5 requires mass-energy equivalence, the form E = m₀c²/√(1 − v²/c²), graphs of mass and kinetic energy against speed, and Bertozzi’s experiment as direct evidence. Note what that implies: AQA still teaches relativistic mass. This page cites the specification; it does not claim the board endorses it.
CBSE / NCERT, Physics, Class XII, Curriculum 2025–26, read on 11 November 2026 from the government-hosted syllabus at edustud.nic.in (SYLLABUS_2025_26/12/12_Physics_EM.pdf). The nine units are Electrostatics, Current Electricity, Magnetic Effects of Current and Magnetism, Electromagnetic Induction and Alternating Currents, Electromagnetic Waves, Optics, Dual Nature of Radiation and Matter, Atoms and Nuclei, and Electronic Devices. Special relativity, time dilation, length contraction and the Lorentz factor appear nowhere in it. The only trace of the subject is “mass-energy relation, mass defect” in Chapter 13, Nuclei — and the document does not even write E = mc² out. So if you are on this syllabus, this topic is reading beyond it rather than revision for it, which is a useful thing to know the night before a paper.
Particle Data Group, Review of Particle Physics — Summary Table of Leptons, read on 11 November 2026 from pdg.lbl.gov/2025/tables/rpp2025-sum-leptons.pdf. Muon mean life τ = (2.196 981 1 ± 0.000 002 2)×10⁻⁶ s; muon mass 105.658 3755 ± 0.000 0023 MeV. It is a MEAN life, not a half-life: the half-life is τ ln 2 = 1.5230 µs, and quoting one where the other is meant is the commonest error in a muon question. cτ = 658.64 m, which is the distance a muon would cover in one mean lifetime if there were no dilation at all.