<p>For a single dish radio telescope we define the primary beam as the telescope's response on the sky as a function of angle. What this means is that half way between the centre and edge of the primary beam, a radio source with a flux of 1 Jy will be observed to have a flux of 0.5 Jy. This response is circular i.e. it is a function of angular separation, and is not dependent on altitude/azimuth. The full-width of this response is typically written as:</p>
<p><img src="https://latex.codecogs.com/gif.latex?%5Ctheta=%5Cfrac%7B%5Clambda%7D%7BD%7D" alt="Primary beam width" /></p>
<p>where <img src="https://latex.codecogs.com/gif.latex?%5Ctheta,%5Clambda,%5Cmathrm%7Band%7D,D" alt="Primary beam width" /> are the primary beam width in radians, observing wavelength and dish diameter.</p>
<p>As for the big ear telescope, it is obviously not a parabolic reflector or Gregorian offset like the Lovell or Greenbank telescopes (<a href="http://www.naapo.org/W8JK/Images/JDK097l.jpg" rel="nofollow noreferrer">www.naapo.org/W8JK/Images/JDK097l.jpg</a>). Since the secondary reflector is the important one in the determination of resolution, we shall use its dimensions, which are 103m long by 21m high.</p>
<p>The Wow! signal was received at 1420MHz which has a wavelength of 0.21m. Using the above equation, the response would not be circular, but 'fan-shaped'. The response would thereby have approximate dimensions on the sky of <strong>0.12deg by 0.58deg</strong> (in azimuth and altitude respectively).</p>
<p>Caveats of this calculation are the non-circular reflectors, which means the actual shape of the telescope's response on the sky would not be Gaussian/symmetrical for that matter. But as for an approximate, angular resolution, 0.12deg by 0.58deg will suffice.</p>
<p>Further to this, if you can find the LST (local sidereal time) of when the Wow! signal was received, and assuming that the big ear telescope is pointing at the zenith (i.e. straight up), the spot the telescope was pointing to would be easy enough to derive based upon telescope latitude and LST. Enjoy!</p>

CC BY-SA 4.0
2020-09-24T14:52:48.423Z
Thank you for your answer! Yep, if we had to use a simple model and assume the source was on the meridian at the time, then maybe a single-slit diffraction pattern is a pretty good approximation, and as you point out, better than an Airy disk for this antenna. If we knew the declination that it was pointed at, then the angular full width resolution could be compared to the 72 seconds (duration in time) mentioned in the other answer.
uhoh
CC BY-SA 4.0
2020-09-24T14:54:21.133Z
btw are 0.12 and 058 degrees switched? I'd expect azimuth to have a narrower beam  than altitude, since the aperture was a lot longer than it was tall.
uhoh
CC BY-SA 4.0
2020-09-24T15:59:36.577Z
Ah, yes, I switched them by accident. Editing my answer to correct.
simonp2207
CC BY-SA 4.0
2020-09-24T16:02:58.813Z
With regards to where the telescope was pointing... I'm guessing with the fixed receiver building, only small adjustments in RA/Dec were possible, if at all. I wouldn't be surprised if the technology of the time simply used earth rotation alone to change the pointing centre of the telescope.
simonp2207
CC BY-SA 4.0
2020-09-24T17:01:48.120Z
[The flat reflector tilts to adjust the elevation (declination, if you are an astronomer) of the beam.](http://www.bigear.org/w1ghz/w1ghz.htm) It can be pointed up and down to different places *along the local meridian.* There's a little more [here](http://www.bigear.org/guinness.htm) (mentioned [here](https://chat.stackexchange.com/transcript/message/55526589#55526589))
uhoh
CC BY-SA 4.0
2020-09-24T14:43:19.123Z
2020-09-24T16:00:53.830Z
2020-09-24T16:00:53.830Z
simonp2207
<p>@uhoh, I hope this answers your first question.</p>
<p>I sourced the following 2 paragraphs from an NBC article by Jesse Emspak, <a href="https://www.nbcnews.com/mach/science/has-mysterious-signal-space-finally-been-explained-ncna771671" rel="nofollow noreferrer">&quot;Has Mysterious Signal From Space Finally Been Explained?&quot;</a>:</p>
<blockquote>
<p>Two big issues are that the signal didn't repeat, and it appeared for such a short time. Ehman noted that the Big Ear telescope had two &quot;feed horns,&quot; each of which provides a slightly different field of view for a radio telescope.</p>
</blockquote>
<blockquote>
<p>&quot;We should have seen the source come through twice in about 3 minutes: one response lasting 72 seconds and a second response for 72 seconds following within about a minute and a half,&quot; Ehman told Live Science. &quot;We didn't see the second one.&quot;</p>
</blockquote>
<p>The first para clearly states that the 'Big Ear' had no single field of view. I also tried to search for this in many ways, but with no luck, the telescope has also been disassembled.</p>
<p>For your second question though, it requires some thinking as the 2 'feed horns' of the telescope give slightly different field of views. I don't know the answer to this yet, but I will attempt to find it.</p>

CC BY-SA 4.0
2020-09-13T08:12:55.003Z
Hmm... interesting, thanks! Well 72 seconds would be 0.3 degrees if the telescope were pointed towards the celestial equator (assume 360 degrees in 24 hours) and you can multiply that by the cosine of the declination, so I think you are almost there!
uhoh
CC BY-SA 4.0
2020-09-13T08:42:09.247Z
@uhoh, I haven't really ventured deep into Astronomical Math, but thanks for the info!
Infinity Milestone
CC BY-SA 4.0
2020-09-13T06:45:49.307Z
2020-09-18T12:49:21.697Z
2020-09-18T12:49:21.697Z
Infinity Milestone
<p>24h x 60 min = 1440 mins/day = 86400 sec
360° / 86400 = 0.0041° per second
72 seconds = 0.3°</p>
<p>An arcminute (denoted by the symbol '), is an angular measurement equal to 1/60 of a degree or 60 arcseconds.
To convert a degree measurement to a minute of arc measurement, multiply the angle by the conversion ratio. The angle in minutes of arc is equal to the degrees multiplied by 60.</p>
<p>0.3 x 60 = 18 arcminutes?</p>
<p>As seen from the Earth, the Sun and Moon both have angular diameters of about 30 arcminutes. The full moon's average apparent size is about 31 arcminutes (or 0.52°).</p>
<p>In other words, the Wow! signal spanned an area of about half the size of the Sun or the Moon, in the sky. <a href="https://i.sstatic.net/04qQs.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/04qQs.jpg" alt="enter image description here" /></a></p>
<p>That is a rather large area in astronomy.</p>
<p>Best, Eric, for <a href="https://contactproject.org" rel="nofollow noreferrer">https://contactproject.org</a></p>

CC BY-SA 4.0
2022-01-15T19:07:11.163Z
2022-01-15T19:07:11.163Z
Erich Habich-Traut
<p><a href="https://astronomy.stackexchange.com/a/35090/7982">This answer</a> to <em>Did comets 266P/Christensen or P/2008 Y2 (Gibbs) cause the Wow! signal?</em> points out that the comets in question were nowhere near where the radio telescope was pointed.</p>

<p><a href="https://en.wikipedia.org/wiki/Ohio_State_University_Radio_Observatory" rel="nofollow noreferrer">Wikipedia says</a> that the Ohio State University Radio Observatory (or Big Ear) was a <a href="https://en.wikipedia.org/wiki/Kraus-type_radio_telescope" rel="nofollow noreferrer">Kraus-type radio telescope</a> but doesn't provide enough information to deduce a field of view.</p>

<p><strong>Questions:</strong> </p>

<ol>
<li>What was the field of view of the Ohio State University Radio Observatory of Wow! signal fame?</li>
<li>What would be the typical rise and fall times of a steady source passing through the fixed telescope's field of view as the Earth rotated?</li>
</ol>

<observational-astronomy><radio-astronomy><history><radio-telescope>
What was the field of view of the Ohio State University Radio Observatory of Wow! signal fame?
uhoh
# What was the field of view of the Ohio State University Radio Observatory of Wow! signal fame?

(Asked by: username_0 on 2020-02-15)

[This answer](https://astronomy.stackexchange.com/a/35090/7982) to *Did comets 266P/Christensen or P/2008 Y2 (Gibbs) cause the Wow! signal?* points out that the comets in question were nowhere near where the radio telescope was pointed.

[Wikipedia says](https://en.wikipedia.org/wiki/Ohio_State_University_Radio_Observatory) that the Ohio State University Radio Observatory (or Big Ear) was a [Kraus-type radio telescope](https://en.wikipedia.org/wiki/Kraus-type_radio_telescope) but doesn't provide enough information to deduce a field of view.

**Questions:**

1. What was the field of view of the Ohio State University Radio Observatory of Wow! signal fame?
2. What would be the typical rise and fall times of a steady source passing through the fixed telescope's field of view as the Earth rotated?

**Tags:** observational-astronomy, radio-astronomy, history, radio-telescope

---

## 3 answers

### Answer by username_1 on 2020-09-24. Score: 3 | Accepted answer

For a single dish radio telescope we define the primary beam as the telescope's response on the sky as a function of angle. What this means is that half way between the centre and edge of the primary beam, a radio source with a flux of 1 Jy will be observed to have a flux of 0.5 Jy. This response is circular i.e. it is a function of angular separation, and is not dependent on altitude/azimuth. The full-width of this response is typically written as:

![Primary beam width](https://latex.codecogs.com/gif.latex?%5Ctheta=%5Cfrac%7B%5Clambda%7D%7BD%7D)

where ![Primary beam width](https://latex.codecogs.com/gif.latex?%5Ctheta,%5Clambda,%5Cmathrm%7Band%7D,D) are the primary beam width in radians, observing wavelength and dish diameter.

As for the big ear telescope, it is obviously not a parabolic reflector or Gregorian offset like the Lovell or Greenbank telescopes ([www.naapo.org/W8JK/Images/JDK097l.jpg](http://www.naapo.org/W8JK/Images/JDK097l.jpg)). Since the secondary reflector is the important one in the determination of resolution, we shall use its dimensions, which are 103m long by 21m high.

The Wow! signal was received at 1420MHz which has a wavelength of 0.21m. Using the above equation, the response would not be circular, but 'fan-shaped'. The response would thereby have approximate dimensions on the sky of **0.12deg by 0.58deg** (in azimuth and altitude respectively).

Caveats of this calculation are the non-circular reflectors, which means the actual shape of the telescope's response on the sky would not be Gaussian/symmetrical for that matter. But as for an approximate, angular resolution, 0.12deg by 0.58deg will suffice.

Further to this, if you can find the LST (local sidereal time) of when the Wow! signal was received, and assuming that the big ear telescope is pointing at the zenith (i.e. straight up), the spot the telescope was pointing to would be easy enough to derive based upon telescope latitude and LST. Enjoy!

> **Comments:**
> **username_0**: Thank you for your answer! Yep, if we had to use a simple model and assume the source was on the meridian at the time, then maybe a single-slit diffraction pattern is a pretty good approximation, and as you point out, better than an Airy disk for this antenna. If we knew the declination that it was pointed at, then the angular full width resolution could be compared to the 72 seconds (duration in time) mentioned in the other answer.  
> **username_0**: btw are 0.12 and 058 degrees switched? I'd expect azimuth to have a narrower beam  than altitude, since the aperture was a lot longer than it was tall.  
> **username_1**: Ah, yes, I switched them by accident. Editing my answer to correct.  
> **username_1**: With regards to where the telescope was pointing... I'm guessing with the fixed receiver building, only small adjustments in RA/Dec were possible, if at all. I wouldn't be surprised if the technology of the time simply used earth rotation alone to change the pointing centre of the telescope.  
> **username_0**: [The flat reflector tilts to adjust the elevation (declination, if you are an astronomer) of the beam.](http://www.bigear.org/w1ghz/w1ghz.htm) It can be pointed up and down to different places *along the local meridian.* There's a little more [here](http://www.bigear.org/guinness.htm) (mentioned [here](https://chat.stackexchange.com/transcript/message/55526589#55526589))  

---

### Answer by username_2 on 2020-09-13. Score: 1

@uhoh, I hope this answers your first question.

I sourced the following 2 paragraphs from an NBC article by Jesse Emspak, ["Has Mysterious Signal From Space Finally Been Explained?"](https://www.nbcnews.com/mach/science/has-mysterious-signal-space-finally-been-explained-ncna771671):

> Two big issues are that the signal didn't repeat, and it appeared for such a short time. Ehman noted that the Big Ear telescope had two "feed horns," each of which provides a slightly different field of view for a radio telescope.

> "We should have seen the source come through twice in about 3 minutes: one response lasting 72 seconds and a second response for 72 seconds following within about a minute and a half," Ehman told Live Science. "We didn't see the second one."

The first para clearly states that the 'Big Ear' had no single field of view. I also tried to search for this in many ways, but with no luck, the telescope has also been disassembled.

For your second question though, it requires some thinking as the 2 'feed horns' of the telescope give slightly different field of views. I don't know the answer to this yet, but I will attempt to find it.

> **Comments:**
> **username_0**: Hmm... interesting, thanks! Well 72 seconds would be 0.3 degrees if the telescope were pointed towards the celestial equator (assume 360 degrees in 24 hours) and you can multiply that by the cosine of the declination, so I think you are almost there!  
> **username_2**: @uhoh, I haven't really ventured deep into Astronomical Math, but thanks for the info!  

---

### Answer by username_3 on 2022-01-15. Score: 0

24h x 60 min = 1440 mins/day = 86400 sec
360° / 86400 = 0.0041° per second
72 seconds = 0.3°

An arcminute (denoted by the symbol '), is an angular measurement equal to 1/60 of a degree or 60 arcseconds.
To convert a degree measurement to a minute of arc measurement, multiply the angle by the conversion ratio. The angle in minutes of arc is equal to the degrees multiplied by 60.

0.3 x 60 = 18 arcminutes?

As seen from the Earth, the Sun and Moon both have angular diameters of about 30 arcminutes. The full moon's average apparent size is about 31 arcminutes (or 0.52°).

In other words, the Wow! signal spanned an area of about half the size of the Sun or the Moon, in the sky. [![enter image description here](https://i.sstatic.net/04qQs.jpg)](https://i.sstatic.net/04qQs.jpg)

That is a rather large area in astronomy.

Best, Eric, for <https://contactproject.org>


---

