<p><a href="https://en.wikipedia.org/wiki/Spherical_trigonometry#Excess" rel="nofollow noreferrer">Wikipedia</a> has formulas for the area of a celestial triangle in steradians (which they call the &quot;excess&quot; and write <span class="math-container">$E$</span>) in terms of the angles, the side lengths, and side-angle-side. The angle formula is the simplest, but calculating the angles is tricky. It may be easier to calculate the side lengths and use the side-side-side formula.</p>
<p>You can convert right ascension and declination into a unit vector as shown <a href="/a/34428">here</a>. The distance between points represented by unit vectors <span class="math-container">$v$</span> and <span class="math-container">$w$</span> is <span class="math-container">$cos^{-1}\, v\cdot w$</span>. Plugging the three distances into the formula that Wikipedia attributes to L'Huilier will give you the area.</p>
<p>For a polygon with more than 3 vertices you'll have to use the angle formula since it's the only one that generalizes. The angle between <span class="math-container">$u$</span> and <span class="math-container">$w$</span> at <span class="math-container">$v$</span> is <span class="math-container">$cos^{-1}\, \frac{v\times u}{||v\times u||}\cdot\frac{v\times w}{||v\times w||}$</span>, but there may be a simpler way to calculate it.</p>

CC BY-SA 4.0
2020-10-08T19:43:11.790Z
2020-10-08T19:43:11.790Z
benrg
<p>The area of a triangle enclosed by 3 stars on the celestial sphere, in square degrees, is given by:</p>
<p><span class="math-container">$$ A = \frac{180}{\pi}\times E$$</span></p>
<p>Where E is the <a href="https://en.wikipedia.org/wiki/Spherical_trigonometry#Area_and_spherical_excess" rel="noreferrer">spherical excess</a> and is equal to the sum of all angles of the triangle minus 180°. Problem is, we don't know the angles of the Summer Triangle and instead we have to use an alternate equation for the spherical excess given by:</p>
<p><span class="math-container">$$ E = 4\tan^{-1}\left({\sqrt{\tan\left(\frac{s}{2}\right)\times\tan\left(\frac{s-a}{2}\right)\times\tan\left(\frac{s-b}{2}\right)\times\tan\left(\frac{s-c}{2}\right)}} \right) $$</span></p>
<p>Where a, b, and c are the angular distances between 2 stars and thus the sides of the Summer Triangle in degrees, and <span class="math-container">$s = \frac{a+b+c}{2}$</span>.</p>
<p>I highly recommend you just go to <a href="https://www.wolframalpha.com/input/?i=Angular%20distance%20between%20Vega%20and%20Deneb" rel="noreferrer">Wolfram Alpha</a> and just type &quot;Angular distance between Star 1 and Star 2&quot; and get all values for a, b, and c and then you could calculate the spherical excess and subsequently the area.</p>
<p>But if you insist on giving yourself a headache and finding the values yourself then check the next section. Otherwise skip to the bottom.</p>
<hr />
<p>The angular distance between 2 stars on the celestial sphere is given by:</p>
<p><span class="math-container">$$\cos^{-1} \left ( \sin(\text{D} _{1})\sin(\text{D} _{2})+\cos(\text{D} _{1})\cos(\text{D} _{2}) \cos(\text{R} _{1}-\text{R} _{2})\right ) $$</span></p>
<p>Where:</p>
<ul>
<li><p><span class="math-container">$\text{D}_{1}$</span> and <span class="math-container">$\text{D}_{2}$</span> are the declination values of both stars in degrees.</p>
</li>
<li><p><span class="math-container">$\text{R}_{1}$</span> and <span class="math-container">$\text{R}_{2}$</span> are the right ascension values for both stars, also in degrees.</p>
</li>
</ul>
<p>To convert a star's declination from DMS to degrees, divide the arcminutes by 60 and arcseconds by 3600. For example Vega's declination is 38° 47' 1.28&quot;. Converted to degrees only, that becomes:</p>
<p><span class="math-container">$$ 38 + \frac{47}{60} + \frac{1.28}{3600} = 38.7836° $$</span></p>
<p>The exact same formula is used to convert RA to degrees, only difference is you multiply it by 15 at the end. Vega's right ascension is 18h 36m 56.33s. Converted to degrees only, that becomes:</p>
<p><span class="math-container">$$ \left(18 + \frac{36}{60} + \frac{56.33}{3600}\right) \times 15 = 279.2347° $$</span></p>
<p>You could just spare yourself all this mess by directly getting the angular distances from Wolfram Alpha or you could calculate them yourself. I recommend the former.</p>
<hr />
<p>Once you plug in all the numbers, you get the spherical excess <span class="math-container">$E = 7.296508$</span> and the area of the Summer Triangle:</p>
<p><span class="math-container">$$ \bbox[5px,border:2px solid #C0A000]{A = 418.059 \text{deg}^2} $$</span></p>
<p>It would cover 1.0134% of the celestial sphere and be the 40th largest constellation in the sky, bigger than Capricornus but smaller than Aries.</p>

CC BY-SA 4.0
2020-10-08T22:07:24.280Z
2020-10-08T22:07:24.280Z
ChristieToWin
<p>I'm pretty sure spherical trigonometry comes into use here but I don't know how to calculate the area. I'm pretty sure Vega, Deneb, and Altair's declination and right ascension figures need to be converted to coordinates, but again, I have no idea how.</p>
<p>Is there a formula for a general case like this? Not necessarily just the Summer Triangle, but a formula for the area enclosed by any 3 (or more) stars on the celestial sphere?</p>

<declination><right-ascension>
What is the area of the Summer Triangle?
user177107
# What is the area of the Summer Triangle?

(Asked by: username_0 on 2020-10-08)

I'm pretty sure spherical trigonometry comes into use here but I don't know how to calculate the area. I'm pretty sure Vega, Deneb, and Altair's declination and right ascension figures need to be converted to coordinates, but again, I have no idea how.

Is there a formula for a general case like this? Not necessarily just the Summer Triangle, but a formula for the area enclosed by any 3 (or more) stars on the celestial sphere?

**Tags:** declination, right-ascension


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## 2 answers

### Answer by username_2 on 2020-10-08. Score: 7 | Accepted answer

The area of a triangle enclosed by 3 stars on the celestial sphere, in square degrees, is given by:

$$ A = \frac{180}{\pi}\times E$$

Where E is the [spherical excess](https://en.wikipedia.org/wiki/Spherical_trigonometry#Area_and_spherical_excess) and is equal to the sum of all angles of the triangle minus 180°. Problem is, we don't know the angles of the Summer Triangle and instead we have to use an alternate equation for the spherical excess given by:

$$ E = 4\tan^{-1}\left({\sqrt{\tan\left(\frac{s}{2}\right)\times\tan\left(\frac{s-a}{2}\right)\times\tan\left(\frac{s-b}{2}\right)\times\tan\left(\frac{s-c}{2}\right)}} \right) $$

Where a, b, and c are the angular distances between 2 stars and thus the sides of the Summer Triangle in degrees, and $s = \frac{a+b+c}{2}$.

I highly recommend you just go to [Wolfram Alpha](https://www.wolframalpha.com/input/?i=Angular%20distance%20between%20Vega%20and%20Deneb) and just type "Angular distance between Star 1 and Star 2" and get all values for a, b, and c and then you could calculate the spherical excess and subsequently the area.

But if you insist on giving yourself a headache and finding the values yourself then check the next section. Otherwise skip to the bottom.

---

The angular distance between 2 stars on the celestial sphere is given by:

$$\cos^{-1} \left ( \sin(\text{D} \_{1})\sin(\text{D} \_{2})+\cos(\text{D} \_{1})\cos(\text{D} \_{2}) \cos(\text{R} \_{1}-\text{R} \_{2})\right ) $$

Where:

* $\text{D}\_{1}$ and $\text{D}\_{2}$ are the declination values of both stars in degrees.
* $\text{R}\_{1}$ and $\text{R}\_{2}$ are the right ascension values for both stars, also in degrees.

To convert a star's declination from DMS to degrees, divide the arcminutes by 60 and arcseconds by 3600. For example Vega's declination is 38° 47' 1.28". Converted to degrees only, that becomes:

$$ 38 + \frac{47}{60} + \frac{1.28}{3600} = 38.7836° $$

The exact same formula is used to convert RA to degrees, only difference is you multiply it by 15 at the end. Vega's right ascension is 18h 36m 56.33s. Converted to degrees only, that becomes:

$$ \left(18 + \frac{36}{60} + \frac{56.33}{3600}\right) \times 15 = 279.2347° $$

You could just spare yourself all this mess by directly getting the angular distances from Wolfram Alpha or you could calculate them yourself. I recommend the former.

---

Once you plug in all the numbers, you get the spherical excess $E = 7.296508$ and the area of the Summer Triangle:

$$ \bbox[5px,border:2px solid #C0A000]{A = 418.059 \text{deg}^2} $$

It would cover 1.0134% of the celestial sphere and be the 40th largest constellation in the sky, bigger than Capricornus but smaller than Aries.


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### Answer by username_1 on 2020-10-08. Score: 2

[Wikipedia](https://en.wikipedia.org/wiki/Spherical_trigonometry#Excess) has formulas for the area of a celestial triangle in steradians (which they call the "excess" and write $E$) in terms of the angles, the side lengths, and side-angle-side. The angle formula is the simplest, but calculating the angles is tricky. It may be easier to calculate the side lengths and use the side-side-side formula.

You can convert right ascension and declination into a unit vector as shown [here](/a/34428). The distance between points represented by unit vectors $v$ and $w$ is $cos^{-1}\, v\cdot w$. Plugging the three distances into the formula that Wikipedia attributes to L'Huilier will give you the area.

For a polygon with more than 3 vertices you'll have to use the angle formula since it's the only one that generalizes. The angle between $u$ and $w$ at $v$ is $cos^{-1}\, \frac{v\times u}{||v\times u||}\cdot\frac{v\times w}{||v\times w||}$, but there may be a simpler way to calculate it.


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