<p>The expression in the question would provide the physical aperture, but the value of interest is the effective aperture which accounts for non-uniform <a href="https://en.wikipedia.org/wiki/Parabolic_antenna#Feed_pattern" rel="nofollow noreferrer">illumination</a> of the dish reflectors by the feed.</p>
<p><span class="math-container">$$A_e = n \eta \pi r^2$$</span></p>
<p>Where <span class="math-container">$A_e$</span> is the effective aperture, <span class="math-container">$n$</span> is the number of antennas, <span class="math-container">$\eta$</span> is the <a href="https://en.wikipedia.org/wiki/Aperture_(antenna)#Aperture_efficiency" rel="nofollow noreferrer">aperture efficiency</a>, and <span class="math-container">$r$</span> is the dish diameter.</p>
<p>Each of the 27 <a href="https://en.wikipedia.org/wiki/Very_Large_Array" rel="nofollow noreferrer">Very Large Array</a> (VLA) antennas has a diameter of 25 meters, so <span class="math-container">$r = 12.5m$</span>.  According to <a href="https://www.ursi.org/proceedings/procGA05/pdf/J01-P.10(0734).pdf" rel="nofollow noreferrer">Efficiency at L-band on the Expanded Very Large Array Synthesis Telescope</a> the aperture efficiency at 1425 MHz (1.425 GHz) is 0.43.</p>
<p>Combining all of these:</p>
<p><span class="math-container">$$A_e = 27 \cdot 0.43 \cdot \pi \cdot (12.5m)^2 = 6.20\times10^3 m^2$$</span></p>

CC BY-SA 4.0
2023-07-31T11:26:45.333Z
Okay that looks so right. However, also I cannot catch where i used to this A_e thing to calculate Flux from Jansky, then Luminosity by that Flux. I know this A_e is having a huge part and effect on Flux. But I'm not a radio astronomer so my knowledge is the limiting thing.
Ege Tunç
CC BY-SA 4.0
2023-07-31T11:28:52.853Z
Is it just L = S * 4π * D^2 * ν, where S is the flux in Janskies. Or L = Jansky * A_e * 4π * D^2 * ν is it what should be? But units cannot match.
Ege Tunç
CC BY-SA 4.0
2023-07-31T00:31:12.043Z
2023-08-03T01:05:36.553Z
2023-08-03T01:05:36.553Z
GrapefruitIsAwesome
<p>I need to calculate NRAO VLA Sky Surveys effective area that's been held in 1998. (The observation of 1.4GHz NRAO VLA Sky Survey (NVSS) (Condon+ 1998)). I couldn't find the exact values, tried many times. But I cannot find a decent source to confirm the area of each antennas' surface. So, basically i know that <span class="math-container">$$\pi*r^2*n_{antenna}$$</span> will be the effective area of VLASS. Where <span class="math-container">$$n_{antenna}$$</span> is 27. So i just need the <span class="math-container">$$r$$</span> of them but couldn't find it.</p>

<radio-astronomy>
What is the effective area of VLASS?
Ege Tunç
# What is the effective area of VLASS?

(Asked by: username_0 on 2023-07-30)

I need to calculate NRAO VLA Sky Surveys effective area that's been held in 1998. (The observation of 1.4GHz NRAO VLA Sky Survey (NVSS) (Condon+ 1998)). I couldn't find the exact values, tried many times. But I cannot find a decent source to confirm the area of each antennas' surface. So, basically i know that $$\pi\*r^2\*n\_{antenna}$$ will be the effective area of VLASS. Where $$n\_{antenna}$$ is 27. So i just need the $$r$$ of them but couldn't find it.

**Tags:** radio-astronomy

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## 1 answers

### Answer by username_1 on 2023-07-31. Score: 1 | Accepted answer

The expression in the question would provide the physical aperture, but the value of interest is the effective aperture which accounts for non-uniform [illumination](https://en.wikipedia.org/wiki/Parabolic_antenna#Feed_pattern) of the dish reflectors by the feed.

$$A\_e = n \eta \pi r^2$$

Where $A\_e$ is the effective aperture, $n$ is the number of antennas, $\eta$ is the [aperture efficiency](https://en.wikipedia.org/wiki/Aperture_(antenna)#Aperture_efficiency), and $r$ is the dish diameter.

Each of the 27 [Very Large Array](https://en.wikipedia.org/wiki/Very_Large_Array) (VLA) antennas has a diameter of 25 meters, so $r = 12.5m$. According to [Efficiency at L-band on the Expanded Very Large Array Synthesis Telescope](https://www.ursi.org/proceedings/procGA05/pdf/J01-P.10(0734).pdf) the aperture efficiency at 1425 MHz (1.425 GHz) is 0.43.

Combining all of these:

$$A\_e = 27 \cdot 0.43 \cdot \pi \cdot (12.5m)^2 = 6.20\times10^3 m^2$$

> **Comments:**
> **username_0**: Okay that looks so right. However, also I cannot catch where i used to this A_e thing to calculate Flux from Jansky, then Luminosity by that Flux. I know this A_e is having a huge part and effect on Flux. But I'm not a radio astronomer so my knowledge is the limiting thing.  
> **username_0**: Is it just L = S * 4π * D^2 * ν, where S is the flux in Janskies. Or L = Jansky * A_e * 4π * D^2 * ν is it what should be? But units cannot match.  

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