Astronomy · Distance measurement

Stellar parallax

Q: Why is distance in parsecs simply the reciprocal?

Today one parsec is exactly $\frac{648000}{\pi}$ astronomical units. The idealized right triangle first gives $d=\frac{1\,\mathrm{au}}{\tan p}$. At very small angles, tangent can be replaced by the angle in radians. Converting arcseconds introduces the same factor as the parsec definition, giving the numerical relation $d_{\mathrm{pc}}\approx\frac1{p_{\mathrm{arcsec}}}$. This extremely accurate astronomical approximation is not an exact tangent identity. For milliarcseconds, use 1000 in the numerator instead: 50 mas corresponds to approximately 20 pc. Degrees or radians need different conversion factors. Also check half-angle, measurement uncertainty and catalogue conventions before interpreting the result as a distance.

Trigonometric parallax gives a star’s distance from its apparent annual position shift relative to very distant background stars.

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Formula

Calculate

d [pc] = \frac{1}{p [ arcsec ]}

d in parsecs (pc); p in arcseconds (arcsec)

Variables & units

Variables & units
Symbol	Meaning	Unit
$d$	Distance to the star	pc
$p$	Annual parallax half-angle, not the full position swing	arcsec

LaTeX

d\,[\mathrm{pc}] = \frac{1}{p\,[\mathrm{arcsec}]}

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Two Earth positions above and below the Sun form two right triangles with the star. Each half-baseline is 1 au. At the star, p is marked between the Sun sightline and the upper Earth sightline. — Two opposite Earth positions are approximately 2 au apart. The half-baseline from the Sun is 1 au; p is the half-angle. Not to scale.

Stellar parallax distance calculator

Find the distance of a star from its annual parallax or the other way round, using the small-angle approximation d in pc = 1 / p in arcseconds.

Calculation model

Find

Parallax p (half-angle)

Annual parallax as the half-angle, not the full shift, e.g. 0.768″ for Proxima Centauri.

Filled with example values. Type your own and the result updates straight away.

Working · rearrange for d, substitute, evaluate

1. Rearrange for d
$d\approx\frac{1\,\mathrm{au}}{p_{\rm rad}}$
2. Substitute values (in SI units)
- p = 0.768 arcsec = 3.72337 · 10⁻⁶ rad
$\frac{149597870700}{0.00000372337}=40178093509000000$
3. Result
d = 1.302 pc

Geometric small-angle approximation; measurement uncertainty, proper motion, catalogue corrections and statistical priors are not automatically accounted for.

Result: Distance d

d = 1.302 pc

With p = 0.768 arcsec, d = 1.302 pc.

Quick tips

Reciprocal in your head: 0.1″ gives 10 pc, 0.01″ gives 100 pc. In milliarcseconds d in pc = 1000 / p in mas, so 8 mas gives 125 pc.
Twice the parallax, half the distance. Watch the half-angle: using the full shift over half a year as p gives only half the distance.

Assumptions and limits

Annual trigonometric parallax p is the half-angle for a 1-au baseline, not the full two-position shift. Positive reliable geometric input; d≈1 au/p(rad) is the stated small-angle approximation, not an exact tangent identity. Not automatically a statistical catalogue distance; negative/zero noisy estimates are not inverted. Parsec is exactly 648000/π au; light-year uses a 365.25-day Julian year, with rounded stored double factors.

Derivation & background

A parsec is exactly 648000/π au. In the astronomical small-angle approximation, a parallax of one arcsecond corresponds to one parsec. The apparent annual shift is determined from observations across the year; p is the angle between the mean line of sight and the line of sight from an observing position displaced by one AU. In milliarcseconds, d[pc] = 1000/p[mas]. Relative measurement uncertainty grows as p becomes small.

Exam blueprint

Validity range

Annual parallax half-angle p in arcseconds or mas. The reciprocal uses the small-angle approximation. Large relative uncertainty or nonpositive measured parallax requires a statistical distance model.

Derivation steps

A right triangle and the small-angle approximation connect half the orbital baseline to the parallax angle.

1Specify half-baseline $b=1\,\mathrm{au}$ and parallax half-angle p; the full idealised angular difference is 2p.
2The triangle gives $\tan(p_{\mathrm{rad}})=\frac bd$ , hence $d=\frac b{\tan(p_{\mathrm{rad}})}$ .
3For the very small angle, $\tan(p_{\mathrm{rad}})\approx p_{\mathrm{rad}}$ , giving $d\approx\frac b{p_{\mathrm{rad}}}$ .
4Convert arcseconds: $p_{\mathrm{rad}}=p_{\mathrm{arcsec}}\frac\pi{648000}$ . One parsec is exactly $\frac{648000}\pi\,\mathrm{au}$ .
5The conversion factor cancels: $d_{\mathrm{pc}}\approx\frac1{p_{\mathrm{arcsec}}}$ or $d_{\mathrm{pc}}\approx\frac{1000}{p_{\mathrm{mas}}}$ .
6Check units, half-angle and measurement uncertainty. At large relative uncertainty, the reciprocal is not a reliable statistical distance estimate.

Rearrangements

Parallax

p\,[\mathrm{arcsec}] = \frac{1}{d\,[\mathrm{pc}]}

This numerical shortcut works when d is expressed in pc.

Distance with milliarcseconds

d\,[\mathrm{pc}] = \frac{1000}{p\,[\mathrm{mas}]}

1″ = 1000 mas.

Task variant

The full idealised displacement between opposite Earth positions is 0.080″. Find d.

p = 0.040″ is half the displacement. d = 1/0.040 pc = 25 pc.

Common mistakes

Substituting the full angular difference 2p as p.

Halve the full displacement in the simplified geometry.

Using milliarcseconds without a conversion factor.

For mas, d_pc ≈ 1000/p_mas; for arcseconds, d_pc ≈ 1/p_arcsec.

Inferring negative physical distance from negative measured parallax.

Noisy angular estimators can be negative; estimate distance with a suitable inference model.

Exam context

Apply angle conventions and units to geometric stellar distances.
Use σd/d ≈ σp/p for small relative errors and explain its limits.

Formula cluster

Geometry and stellar distance

Parallax gives geometric distance. Brightness methods need additional luminosity and, where relevant, extinction information.

mV − MV = 5 log₁₀(d/10 pc) + AV

Astronomical distance modulus

F = L/(4πd²)

Luminosity, flux and distance

Mᵥ = a log₁₀(P/d) + b; d aus mᵥ−Mᵥ−Aᵥ

Cepheids: Period–Luminosity Relation

Worked example

For p = 0.10″, d = 1/0.10 = 10 pc. This is about 32.6 light-years; that conversion is an additional approximation, not part of the parallax formula.

Applications

Measure nearby stellar distances geometrically and provide an independent calibration for larger astronomical distance scales.

How do you calculate with Stellar parallax and distance?

Here is how to work through a typical Stellar parallax and distance (d [pc] = 1 / p [″]) task step by step:

1
Task
The full idealised displacement between opposite Earth positions is 0.080″. Find d.
Solution path
p = 0.040″ is half the displacement. d = 1/0.040 pc = 25 pc.

Understand and apply the equation

Annual parallax turns Earth’s orbit into a geometric baseline for stellar distances. The small angle p belongs to half the baseline. This guide develops the triangle, calculations with arcseconds and milliarcseconds, and the limits imposed by measurement uncertainty.

Why is annual parallax the half-angle?

A nearby star appears in different directions against a more distant background when observed from different positions. This is a change in the line of sight; it does not imply that the star travels along the apparent track. Earth’s orbit makes the geometric part repeat annually.[1]

In the simplified right-triangle construction, use two suitable observations from opposite sides of Earth’s orbit. The separation is approximately 2 au. Parallax p is the half-angle, referenced to 1 au from the mean line of sight. The full angular difference in this idealised arrangement is 2p.[2]

A nearer star has a larger parallax; a more distant star has a smaller one.
Several observations model the annual displacement. Two images alone need not distinguish all other motion components.
The diagram exaggerates the small angle to make the geometry legible.

How do you convert parallax in mas or arcseconds into parsecs?

1^\circ=3600^{\prime\prime},\qquad1^{\prime\prime}=1000\,\mathrm{mas}=\frac{\pi}{648000}\,\mathrm{rad}

An arcsecond is 1/3 600 of a degree. A milliarcsecond is one thousandth of an arcsecond.

1\,\mathrm{pc}=\frac{648000}{\pi}\,\mathrm{au}\approx3.085677581\times10^{16}\,\mathrm m

The modern definition is exactly 648 000/π au; the rounded metre value follows from the exact astronomical unit.[3]

A parsec is a length, not a time. For ordinary exercises, 1 pc ≈ 3.26156 light-years. A light-year is also a length: the distance light travels in vacuum during a Julian year of 365.25 days.

Derive the relation from a right triangle

In the idealised triangle, the half-baseline b = 1 au lies opposite angle p. The adjacent side is distance d along the mean line of sight. The definition of tangent gives the geometric relationship.[2]

\tan(p_{\mathrm{rad}})=\frac{b}{d}\quad\Longrightarrow\quad d=\frac{b}{\tan(p_{\mathrm{rad}})}

Solve for d and use a consistent angle convention for tangent.

\tan(p_{\mathrm{rad}})\approx p_{\mathrm{rad}},\qquad d\approx\frac{1\,\mathrm{au}}{p_{\mathrm{rad}}}

For very small angles, tangent can be replaced by the angle in radians.

p_{\mathrm{rad}}=p_{\mathrm{arcsec}}\frac{\pi}{648000}\quad\Longrightarrow\quad d\approx\frac{648000}{\pi}\frac{\mathrm{au}}{p_{\mathrm{arcsec}}}

Convert the numerical arcsecond value to radians and substitute.

d_{\mathrm{pc}}\approx\frac{1}{p_{\mathrm{arcsec}}},\qquad d_{\mathrm{pc}}\approx\frac{1000}{p_{\mathrm{mas}}}

The parsec definition cancels the conversion factor. The usual reference formula, often written with an equals sign, uses the extremely accurate astronomical small-angle approximation.

Distinguish numerical values from physical quantities: 1/p directly gives a numerical distance in pc only when p is the numerical value in arcseconds. Substituting degrees, radians or mas without the corresponding factor gives the wrong distance.

Three calculations with angle and uncertainty checks

An angular difference between two observations

Given: The full displacement between opposite observing positions is 0.080″ in the simplified geometry.
Find: Distance in pc and approximately in light-years.

Approach: The angle between the two extreme observations is twice the parallax. $p=\frac{\Delta\theta}{2},\quad d_{\mathrm{pc}}=\frac1{p_{\mathrm{arcsec}}}$ . Halve the total angle: p = 0.040″.
$p=\frac{0.080^{\prime\prime}}2=0.040^{\prime\prime}$
Take the reciprocal of the numerical parallax.
$d_{\mathrm{pc}}=\frac1{0.040}=25$
Convert parsecs to light-years separately.
$d\approx25\times3.26156\,\mathrm{ly}\approx81.5\,\mathrm{ly}$

Result: d ≈ 25 pc, or approximately 82 light-years at the precision of the input.

Check: $p=\frac1{25}\,\mathrm{arcsec}=0.040\,\mathrm{arcsec}$ and $2p=0.080\,\mathrm{arcsec}$ agree with the input. Substituting 0.080″ directly would incorrectly halve the distance.

A catalogue value in milliarcseconds

Given: A reliable parallax estimate is p = 8.0 mas with standard uncertainty σp = 0.2 mas.
Find: Estimate distance and its small relative uncertainty.

Approach: Account for milliarcseconds before taking reciprocal distance. $d_{\mathrm{pc}}=\frac{1000}{p_{\mathrm{mas}}}$ . Use the mas conversion factor.
$d_{\mathrm{pc}}=\frac{1000}{8.0}=125$
Calculate relative parallax uncertainty: $\frac{0.2}{8.0}=0.025$ .
For this small uncertainty, use first-order propagation.
$\sigma_d\approx\frac{1000\sigma_p}{p_{\mathrm{mas}}^2}=3.125\,\mathrm{pc}$

Result: d ≈ 125 pc with an approximate 3 pc standard uncertainty from this measurement error alone.

Check: The relative distance uncertainty is approximately 2.5%. Account separately for systematic offsets; this linear calculation is not a general interval estimate.

Invert a positive angular interval correctly

Given: In the simplified geometric model, let $p=5.0\,\mathrm{mas}$ , with reported positive interval endpoints at $4.5$ and $5.5\,\mathrm{mas}$ .
Find: Find the central value and geometrically corresponding distance endpoints; no additional probabilistic interpretation is assigned to this interval.

Approach: for positive parallax, reciprocal distance decreases strictly. The larger angular endpoint therefore gives the smaller distance endpoint.
$d_{\mathrm{pc}}\approx\frac{1000}{p_{\mathrm{mas}}}$
Invert the central numerical value using the mas factor.
$d_0\approx\frac{1000}{5}=200\,\mathrm{pc}$
Compute both endpoints separately and round afterwards.
$d_{\min}\approx\frac{1000}{5.5}=181.818\,\mathrm{pc},\qquad d_{\max}\approx\frac{1000}{4.5}=222.222\,\mathrm{pc}$

Result: The geometric interval is approximately $[181.8,222.2]\,\mathrm{pc}$ , around central value $200\,\mathrm{pc}$ .

Check: The offsets are approximately $-18.2$ and $+22.2\,\mathrm{pc}$ . Symmetric angular offsets thus do not give symmetric distance offsets; statistical estimation additionally requires a measurement model.

Why small measured angles need careful statistics

\left|\frac{\mathrm{d}d}{\mathrm{d}p}\right|=\frac1{p^2},\qquad\frac{\sigma_d}{d}\approx\frac{\sigma_p}{p}

The reciprocal derivative gives first-order propagation for small relative uncertainty, provided the numerical units are consistent.

At larger relative parallax uncertainty, inversion is nonlinear. Symmetric errors in p do not become symmetric errors in d. A statistical distance estimate then needs a measurement model and explicit prior assumptions; a reciprocal with symmetric plus-or-minus errors can mislead.[4]

A negative measured parallax is not a negative physical distance. It can arise from scatter in a noisy angular estimator. Do not reinterpret it as a negative stellar distance using 1/p. Likewise, p = 0 has no finite reciprocal.[4]

Show the interval curve

Reciprocation transforms an angular interval

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For positive parallax, d≈1000/p decreases nonlinearly. The example interval 4.5–5.5 mas therefore gives unequal offsets around 200 pc. This curve is a geometric conversion, not a measured or statistically inferred distance distribution.[2][4]

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Parallax and other distance methods

Parallax provides a geometric entry into the astronomical distance scale. A brightness-based method needs a known or estimated absolute brightness and, where relevant, a correction for intervening absorption. The distance modulus then connects apparent and absolute brightness. These methods require different assumptions and are not interchangeable without checking them. Finding a stellar diameter from angular size and distance also requires the appropriate angular size and its geometric convention. A magnitude difference from a flux ratio instead compares measured brightness and alone does not provide a geometric distance.

Proper motion describes an ongoing position change; annual parallax describes a periodic geometric component.
Orbital geometry and observation times determine the apparent pattern. The sketch is a simplified measurement arrangement.
A catalogue distance may already be a statistical estimate. Check column definitions and units before taking your own reciprocal.

Five exercises with solutions

Calculate · Foundation

1. A star has p = 25 mas. What is its distance in pc?

Show hint

Use the formula with 1 000.

Learning goal: Calculate distance from milliarcseconds.

Show solution and calculation

Approach: $d_{\mathrm{pc}}\approx\frac{1000}{p_{\mathrm{mas}}}$ .
$d=\frac{1000}{25}\,\mathrm{pc}=40\,\mathrm{pc}$ .
Converting to $p=0.025\,\mathrm{arcsec}$ gives the same numerical value in pc from $\frac1{0.025}=40$ .

d_{\mathrm{pc}}=\frac{1000}{25}=40

Calculate · Application

2. A star is 80 pc away. What annual parallax is approximately expected?

Show hint

Take the reciprocal in arcseconds, then convert to mas.

Learning goal: Distinguish parallax from full baseline shift.

Show solution and calculation

Approach: $p_{\mathrm{mas}}\approx\frac{1000}{d_{\mathrm{pc}}}$ .
$p=\frac1{80}\,\mathrm{arcsec}=0.0125\,\mathrm{arcsec}=12.5\,\mathrm{mas}$ .
The full idealised displacement between opposite positions would be 25 mas.

p_{\mathrm{mas}}=\frac{1000}{80}=12.5

Interpret · Application

3. A measurement gives p = 0.3 ± 0.4 mas. Is 3 333 pc with a small symmetric error a reliable statement?

Show hint

Assess relative uncertainty and the region near p = 0.

Learning goal: Assess noisy parallax statistically.

Show solution and calculation

Approach: $\frac{\sigma_p}{p}$ .
No. σp/p ≈ 1.33 is not small; the measurement range includes zero and negative values.
A defensible distance requires statistical inference with stated model and prior assumptions.

\frac{\sigma_p}{p}=\frac{0.4}{0.3}\approx1.33

Interpret · Application

A positive parallax is reported as an interval from 8 to 12 mas, central value 10 mas. Use geometric inversion only; this is not a posterior probability interval. Find the corresponding distance interval and offsets from the central value.

Show hint

Taking reciprocals reverses endpoint order.

Learning goal: Interpret nonlinear uncertainty intervals under inversion.

Show solution and calculation

Approach: Invert the two positive endpoints in reversed order.
The central value gives 100 pc; the interval spans about 83.3333 to 125 pc.
Offsets are −16.6667 and +25 pc, hence asymmetric.
Geometric interval mapping is not automatically a statistical distance distribution or a credibility statement.

\begin{aligned}d_0&=\frac{1000}{10}\,\mathrm{pc}=100\,\mathrm{pc}\\{}d&\in\left[\frac{1000}{12},\frac{1000}{8}\right]\,\mathrm{pc}\end{aligned}

[2][4]

Error analysis · Application

In an idealized geometry the line of sight is normal to the orbital plane. The full angular shift between opposite Earth positions is 40 mas. Someone inserts 40 mas as annual parallax and reports 25 pc. Correct the distance.

Show hint

Annual parallax belongs to the half baseline, not full displacement.

Learning goal: Distinguish half and full parallax baselines.

Show solution and calculation

Approach: Under the stated geometry the parallax angle is half the full shift, namely 20 mas.
Distance is 50 pc.
The wrong substitution doubles angle and halves distance.
Outside this ideal geometry one must model the projected annual orbit rather than halve an arbitrary two-epoch displacement.

\begin{aligned}p&=\frac{40\,\mathrm{mas}}2=20\,\mathrm{mas}\\{}d&=\frac{1000}{20}\,\mathrm{pc}=50\,\mathrm{pc}\end{aligned}

[2]

Check the calculation in four steps

Check whether the angle is p or the full displacement 2p.
Identify units: arcsec → 1/p, mas → 1 000/p, with distance as a numerical value in pc.
Check using p = 1/d and round to the precision of the input.
Assess relative measurement uncertainty and systematics before treating the distance as reliable.

Check half-angle and angular unit before taking the reciprocal. Small relative parallax uncertainty permits approximate first-order propagation. Large uncertainty and negative measurements require statistical distance estimation.

Sources for this explanation

Quanta · Updated October 2, 2026

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This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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October 2, 2026Refined question headings and relevant paragraph links; preserved existing scientific explanations, worked examples and exercises.
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October 1, 2026Reviewed the learning sequence; added a third example, an original quantitative figure, symbolic exercise approaches and a short summary.

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Quanta. Stellar parallax and distance. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/stellarparallaxe

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Frequently asked questions about Stellar parallax and distance

What is stellar parallax?+

Stellar parallax is the apparent shift of a nearby star relative to distant background stars as the observing position changes over a year. Earth’s orbit around the Sun supplies a baseline. The star does not need to move back and forth in the same way; the apparent shift comes from the changed line of sight. The small angle gives a geometric distance. The nearer the star, the larger its parallax. This makes parallax especially useful for nearby stars. The annual pattern is a perspective effect and should be modeled separately from the star’s proper motion.

Why is distance in parsecs simply the reciprocal?+

Today one parsec is exactly $\frac{648000}{\pi}$ astronomical units. The idealized right triangle first gives $d=\frac{1\,\mathrm{au}}{\tan p}$ . At very small angles, tangent can be replaced by the angle in radians. Converting arcseconds introduces the same factor as the parsec definition, giving the numerical relation $d_{\mathrm{pc}}\approx\frac1{p_{\mathrm{arcsec}}}$ . This extremely accurate astronomical approximation is not an exact tangent identity. For milliarcseconds, use 1000 in the numerator instead: 50 mas corresponds to approximately 20 pc. Degrees or radians need different conversion factors. Also check half-angle, measurement uncertainty and catalogue conventions before interpreting the result as a distance.

Is p the full yearly motion across the sky?+

No. In the standard definition, p is the parallax half-angle: the angle between the mean line of sight and the line of sight from one position of Earth in its orbit. Between opposite observing dates, the full apparent span can be about 2p. This distinction prevents a factor-of-two error in distance. Catalogues usually report parallax as a single angular value. Follow the catalogue definition rather than simply measuring the full motion shown in a sketch. An astrometry catalogue usually reports p directly, though the measurement method still affects uncertainty.

What should I do with parallax in milliarcseconds?+

A milliarcsecond is one thousandth of an arcsecond. You can first divide by 1000 and then apply d=1/p in pc, or directly compute d[pc]=1000/p[mas]. For example, p=20 mas gives 1000/20=50 pc. Check that the measurement is positive and well constrained. For very small or uncertain parallaxes, taking the reciprocal can be biased; uncertainty distributions and other distance information may matter instead of simply inverting the central measured value. For Gaia data, consider measurement uncertainty before interpreting the reciprocal as a distance.

Why does a parallax distance become less precise at large distances?+

Parallax decreases inversely with distance. For distant stars, p becomes very small, so the same absolute measurement error is a larger fraction of the value. Since d=1/p is nonlinear, a symmetric error in p also becomes asymmetric in d. This is why careful astrometry, calibration and uncertainty models matter. A simple reciprocal remains useful as a first estimate, but it does not replace statistical distance inference when the signal is weak. The formula gives a geometric distance, not a photometric estimate based on brightness.

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Stellar parallax

Trigonometric parallax gives a star’s distance from its apparent annual position shift relative to very distant background stars.

BasicUpper secondaryUniversity/collegeBachelor foundations

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

d

Distance to the star

p

Annual parallax half-angle, not the full position swing

arcsec

Derivation & background

Exam blueprint

Validity range

Derivation steps

A right triangle and the small-angle approximation connect half the orbital baseline to the parallax angle.

1Specify half-baseline $b=1\,\mathrm{au}$ and parallax half-angle p; the full idealised angular difference is 2p.
2The triangle gives $\tan(p_{\mathrm{rad}})=\frac bd$ , hence $d=\frac b{\tan(p_{\mathrm{rad}})}$ .
3For the very small angle, $\tan(p_{\mathrm{rad}})\approx p_{\mathrm{rad}}$ , giving $d\approx\frac b{p_{\mathrm{rad}}}$ .
4Convert arcseconds: $p_{\mathrm{rad}}=p_{\mathrm{arcsec}}\frac\pi{648000}$ . One parsec is exactly $\frac{648000}\pi\,\mathrm{au}$ .
5The conversion factor cancels: $d_{\mathrm{pc}}\approx\frac1{p_{\mathrm{arcsec}}}$ or $d_{\mathrm{pc}}\approx\frac{1000}{p_{\mathrm{mas}}}$ .
6Check units, half-angle and measurement uncertainty. At large relative uncertainty, the reciprocal is not a reliable statistical distance estimate.

Why is annual parallax the half-angle?

A nearer star has a larger parallax; a more distant star has a smaller one.
Several observations model the annual displacement. Two images alone need not distinguish all other motion components.
The diagram exaggerates the small angle to make the geometry legible.

How do you convert parallax in mas or arcseconds into parsecs?

1^\circ=3600^{\prime\prime},\qquad1^{\prime\prime}=1000\,\mathrm{mas}=\frac{\pi}{648000}\,\mathrm{rad}

An arcsecond is 1/3 600 of a degree. A milliarcsecond is one thousandth of an arcsecond.

1\,\mathrm{pc}=\frac{648000}{\pi}\,\mathrm{au}\approx3.085677581\times10^{16}\,\mathrm m

The modern definition is exactly 648 000/π au; the rounded metre value follows from the exact astronomical unit.[3]

A parsec is a length, not a time. For ordinary exercises, 1 pc ≈ 3.26156 light-years. A light-year is also a length: the distance light travels in vacuum during a Julian year of 365.25 days.

Derive the relation from a right triangle

\tan(p_{\mathrm{rad}})=\frac{b}{d}\quad\Longrightarrow\quad d=\frac{b}{\tan(p_{\mathrm{rad}})}

Solve for d and use a consistent angle convention for tangent.

\tan(p_{\mathrm{rad}})\approx p_{\mathrm{rad}},\qquad d\approx\frac{1\,\mathrm{au}}{p_{\mathrm{rad}}}

For very small angles, tangent can be replaced by the angle in radians.

p_{\mathrm{rad}}=p_{\mathrm{arcsec}}\frac{\pi}{648000}\quad\Longrightarrow\quad d\approx\frac{648000}{\pi}\frac{\mathrm{au}}{p_{\mathrm{arcsec}}}

Convert the numerical arcsecond value to radians and substitute.

d_{\mathrm{pc}}\approx\frac{1}{p_{\mathrm{arcsec}}},\qquad d_{\mathrm{pc}}\approx\frac{1000}{p_{\mathrm{mas}}}

The parsec definition cancels the conversion factor. The usual reference formula, often written with an equals sign, uses the extremely accurate astronomical small-angle approximation.

Three calculations with angle and uncertainty checks

An angular difference between two observations

Given: The full displacement between opposite observing positions is 0.080″ in the simplified geometry.
Find: Distance in pc and approximately in light-years.

Approach: The angle between the two extreme observations is twice the parallax. $p=\frac{\Delta\theta}{2},\quad d_{\mathrm{pc}}=\frac1{p_{\mathrm{arcsec}}}$ . Halve the total angle: p = 0.040″.
$p=\frac{0.080^{\prime\prime}}2=0.040^{\prime\prime}$
Take the reciprocal of the numerical parallax.
$d_{\mathrm{pc}}=\frac1{0.040}=25$
Convert parsecs to light-years separately.
$d\approx25\times3.26156\,\mathrm{ly}\approx81.5\,\mathrm{ly}$

Result: d ≈ 25 pc, or approximately 82 light-years at the precision of the input.

Check: $p=\frac1{25}\,\mathrm{arcsec}=0.040\,\mathrm{arcsec}$ and $2p=0.080\,\mathrm{arcsec}$ agree with the input. Substituting 0.080″ directly would incorrectly halve the distance.

A catalogue value in milliarcseconds

Given: A reliable parallax estimate is p = 8.0 mas with standard uncertainty σp = 0.2 mas.
Find: Estimate distance and its small relative uncertainty.

Approach: Account for milliarcseconds before taking reciprocal distance. $d_{\mathrm{pc}}=\frac{1000}{p_{\mathrm{mas}}}$ . Use the mas conversion factor.
$d_{\mathrm{pc}}=\frac{1000}{8.0}=125$
Calculate relative parallax uncertainty: $\frac{0.2}{8.0}=0.025$ .
For this small uncertainty, use first-order propagation.
$\sigma_d\approx\frac{1000\sigma_p}{p_{\mathrm{mas}}^2}=3.125\,\mathrm{pc}$

Result: d ≈ 125 pc with an approximate 3 pc standard uncertainty from this measurement error alone.

Check: The relative distance uncertainty is approximately 2.5%. Account separately for systematic offsets; this linear calculation is not a general interval estimate.

Invert a positive angular interval correctly

Given: In the simplified geometric model, let $p=5.0\,\mathrm{mas}$ , with reported positive interval endpoints at $4.5$ and $5.5\,\mathrm{mas}$ .
Find: Find the central value and geometrically corresponding distance endpoints; no additional probabilistic interpretation is assigned to this interval.

Approach: for positive parallax, reciprocal distance decreases strictly. The larger angular endpoint therefore gives the smaller distance endpoint.
$d_{\mathrm{pc}}\approx\frac{1000}{p_{\mathrm{mas}}}$
Invert the central numerical value using the mas factor.
$d_0\approx\frac{1000}{5}=200\,\mathrm{pc}$
Compute both endpoints separately and round afterwards.
$d_{\min}\approx\frac{1000}{5.5}=181.818\,\mathrm{pc},\qquad d_{\max}\approx\frac{1000}{4.5}=222.222\,\mathrm{pc}$

Result: The geometric interval is approximately $[181.8,222.2]\,\mathrm{pc}$ , around central value $200\,\mathrm{pc}$ .

Why small measured angles need careful statistics

\left|\frac{\mathrm{d}d}{\mathrm{d}p}\right|=\frac1{p^2},\qquad\frac{\sigma_d}{d}\approx\frac{\sigma_p}{p}

The reciprocal derivative gives first-order propagation for small relative uncertainty, provided the numerical units are consistent.

Show the interval curve

Reciprocation transforms an angular interval

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Parallax and other distance methods

Proper motion describes an ongoing position change; annual parallax describes a periodic geometric component.
Orbital geometry and observation times determine the apparent pattern. The sketch is a simplified measurement arrangement.
A catalogue distance may already be a statistical estimate. Check column definitions and units before taking your own reciprocal.

Five exercises with solutions

Calculate · Foundation

1. A star has p = 25 mas. What is its distance in pc?

Show hint

Use the formula with 1 000.

Learning goal: Calculate distance from milliarcseconds.

Show solution and calculation

Approach: $d_{\mathrm{pc}}\approx\frac{1000}{p_{\mathrm{mas}}}$ .
$d=\frac{1000}{25}\,\mathrm{pc}=40\,\mathrm{pc}$ .
Converting to $p=0.025\,\mathrm{arcsec}$ gives the same numerical value in pc from $\frac1{0.025}=40$ .

d_{\mathrm{pc}}=\frac{1000}{25}=40

Calculate · Application

2. A star is 80 pc away. What annual parallax is approximately expected?

Show hint

Take the reciprocal in arcseconds, then convert to mas.

Learning goal: Distinguish parallax from full baseline shift.

Show solution and calculation

Approach: $p_{\mathrm{mas}}\approx\frac{1000}{d_{\mathrm{pc}}}$ .
$p=\frac1{80}\,\mathrm{arcsec}=0.0125\,\mathrm{arcsec}=12.5\,\mathrm{mas}$ .
The full idealised displacement between opposite positions would be 25 mas.

p_{\mathrm{mas}}=\frac{1000}{80}=12.5

Interpret · Application

3. A measurement gives p = 0.3 ± 0.4 mas. Is 3 333 pc with a small symmetric error a reliable statement?

Show hint

Assess relative uncertainty and the region near p = 0.

Learning goal: Assess noisy parallax statistically.

Show solution and calculation

Approach: $\frac{\sigma_p}{p}$ .
No. σp/p ≈ 1.33 is not small; the measurement range includes zero and negative values.
A defensible distance requires statistical inference with stated model and prior assumptions.

\frac{\sigma_p}{p}=\frac{0.4}{0.3}\approx1.33

Interpret · Application

A positive parallax is reported as an interval from 8 to 12 mas, central value 10 mas. Use geometric inversion only; this is not a posterior probability interval. Find the corresponding distance interval and offsets from the central value.

Show hint

Taking reciprocals reverses endpoint order.

Learning goal: Interpret nonlinear uncertainty intervals under inversion.

Show solution and calculation

Approach: Invert the two positive endpoints in reversed order.
The central value gives 100 pc; the interval spans about 83.3333 to 125 pc.
Offsets are −16.6667 and +25 pc, hence asymmetric.
Geometric interval mapping is not automatically a statistical distance distribution or a credibility statement.

\begin{aligned}d_0&=\frac{1000}{10}\,\mathrm{pc}=100\,\mathrm{pc}\\{}d&\in\left[\frac{1000}{12},\frac{1000}{8}\right]\,\mathrm{pc}\end{aligned}

[2][4]

Error analysis · Application

In an idealized geometry the line of sight is normal to the orbital plane. The full angular shift between opposite Earth positions is 40 mas. Someone inserts 40 mas as annual parallax and reports 25 pc. Correct the distance.

Show hint

Annual parallax belongs to the half baseline, not full displacement.

Learning goal: Distinguish half and full parallax baselines.

Show solution and calculation

Approach: Under the stated geometry the parallax angle is half the full shift, namely 20 mas.
Distance is 50 pc.
The wrong substitution doubles angle and halves distance.
Outside this ideal geometry one must model the projected annual orbit rather than halve an arbitrary two-epoch displacement.

\begin{aligned}p&=\frac{40\,\mathrm{mas}}2=20\,\mathrm{mas}\\{}d&=\frac{1000}{20}\,\mathrm{pc}=50\,\mathrm{pc}\end{aligned}

[2]

Check the calculation in four steps

Check whether the angle is p or the full displacement 2p.
Identify units: arcsec → 1/p, mas → 1 000/p, with distance as a numerical value in pc.
Check using p = 1/d and round to the precision of the input.
Assess relative measurement uncertainty and systematics before treating the distance as reliable.

Sources for this explanation

Quanta · Updated October 2, 2026

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Change history

October 2, 2026Refined question headings and relevant paragraph links; preserved existing scientific explanations, worked examples and exercises.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Reviewed the learning sequence; added a third example, an original quantitative figure, symbolic exercise approaches and a short summary.

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Quanta. Stellar parallax and distance. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/stellarparallaxe

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