<p>Answer over in Photography:
<a href="https://photo.stackexchange.com/questions/119761/sensor-size-of-sony-hdr-hc7e/119764#119764">https://photo.stackexchange.com/questions/119761/sensor-size-of-sony-hdr-hc7e/119764#119764</a>
It turns out to be Vertical: 3.74 horizontal 4.98 and diagonal 6.23</p>

CC BY-SA 4.0
2020-09-22T18:46:11.310Z
2022-10-13T11:53:02.927Z
2022-10-13T11:53:02.927Z
Phönix 64
<p>I want to motion track a shot from my Sony HDR-HC7e camcorder and I am not quite sure about the horizontal sensor size(width).
I set the focal length to minimum and the recording to HDV 16:9</p>
<p><a href="https://www.sony.com/electronics/support/res/manuals/2319/23198061M.pdf" rel="nofollow noreferrer">This</a> manual (HC7e and HC7 are basically the same) states it is a 1/2.9&quot; sensor with a diagonal of 6.3mm (page 101). But if I do my calculations and multiply (1/2.9) with 25.4 to convert it to mm I get 8.75862069?</p>
<p>The camcorder outputs a 1440*1080 interlaced .m2t file. The image in this file is squeezed by a factor of 1.33. as seen in this <a href="https://i.sstatic.net/x2AvN.jpg" rel="nofollow noreferrer">example</a>.
So I suppose that the sensor has an aspect ratio of 4/3(1.33) So I did some math:</p>
<hr />
<p><span class="math-container">$$a^2 = b^2 +c^2$$</span></p>
<p><span class="math-container">$$
\begin{matrix}
a_1 &amp;=&amp; 6.3\\
a_2 &amp;=&amp; 8.75862069\\
b &amp;=&amp; x\\
c &amp;=&amp; 1.33x\\
\end{matrix}
$$</span></p>
<hr />
<p><span class="math-container">$$6.3^2 = x^2 + (1.33x)^2$$</span></p>
<p><span class="math-container">$$
\begin{matrix}
x &amp;=&amp; 3.78605\\
1.33x &amp;=&amp; 5.0354465\\
\textrm{horizontal 1} &amp;=&amp; 5.0354465
\end{matrix}
$$</span></p>
<hr />
<p><span class="math-container">$$8.75862069^2 = x^2 + (1.33x)^2$$</span></p>
<p><span class="math-container">$$
\begin{matrix}
x &amp;=&amp; 5.26359\\
1.33x &amp;=&amp; 7.0005747\\
\textrm{horizontal 2} &amp;=&amp; 7.0005747\\
\end{matrix}
$$</span></p>
<p>So I thought because there always is a relation between sensor size and focal length, I just take the sensor size horizontal 35mm and the corresponding 35mm focal length equivalent of 49mm (manual page 101) and get the ration: 35 / 49 = 0.714285714 then multiply the true focal length of 5.4mm by that and get a horizontal sensor size of 3.857142856</p>
<p>So now I have got 3 possible horizontal sensor sizes. Now I want to calibrate my lens with a set of images like the linked one. Should I leave them with the 1.33 aspect ratio and where would I specify that in the OpenCV camera calibration or do I unsqueze them and then calibrate? And do I even need the sensor size and focal length for OpenCV camera calibration?</p>

CC BY-SA 4.0
2020-09-22T17:51:46.280Z
You might have a better time asking about this on [Photography.SE](photo.stackexchange.com), I'd guess they will know more about camera sensor geometry. This is a bit out of our bailiwick here at CGSE.
Nathan Reed
<camera><opencv>
Sensor size of Sony HDR-HC7e
Phönix 64
# Sensor size of Sony HDR-HC7e

(Asked by: username_0 on 2020-09-21)

I want to motion track a shot from my Sony HDR-HC7e camcorder and I am not quite sure about the horizontal sensor size(width).
I set the focal length to minimum and the recording to HDV 16:9

[This](https://www.sony.com/electronics/support/res/manuals/2319/23198061M.pdf) manual (HC7e and HC7 are basically the same) states it is a 1/2.9" sensor with a diagonal of 6.3mm (page 101). But if I do my calculations and multiply (1/2.9) with 25.4 to convert it to mm I get 8.75862069?

The camcorder outputs a 1440\*1080 interlaced .m2t file. The image in this file is squeezed by a factor of 1.33. as seen in this [example](https://i.sstatic.net/x2AvN.jpg).
So I suppose that the sensor has an aspect ratio of 4/3(1.33) So I did some math:

---

$$a^2 = b^2 +c^2$$

$$
\begin{matrix}
a\_1 &=& 6.3\\
a\_2 &=& 8.75862069\\
b &=& x\\
c &=& 1.33x\\
\end{matrix}
$$

---

$$6.3^2 = x^2 + (1.33x)^2$$

$$
\begin{matrix}
x &=& 3.78605\\
1.33x &=& 5.0354465\\
\textrm{horizontal 1} &=& 5.0354465
\end{matrix}
$$

---

$$8.75862069^2 = x^2 + (1.33x)^2$$

$$
\begin{matrix}
x &=& 5.26359\\
1.33x &=& 7.0005747\\
\textrm{horizontal 2} &=& 7.0005747\\
\end{matrix}
$$

So I thought because there always is a relation between sensor size and focal length, I just take the sensor size horizontal 35mm and the corresponding 35mm focal length equivalent of 49mm (manual page 101) and get the ration: 35 / 49 = 0.714285714 then multiply the true focal length of 5.4mm by that and get a horizontal sensor size of 3.857142856

So now I have got 3 possible horizontal sensor sizes. Now I want to calibrate my lens with a set of images like the linked one. Should I leave them with the 1.33 aspect ratio and where would I specify that in the OpenCV camera calibration or do I unsqueze them and then calibrate? And do I even need the sensor size and focal length for OpenCV camera calibration?

**Tags:** camera, opencv

> **Comments:**
> **username_1**: You might have a better time asking about this on [Photography.SE](photo.stackexchange.com), I'd guess they will know more about camera sensor geometry. This is a bit out of our bailiwick here at CGSE.  

---

## 1 answers

### Answer by username_0 on 2020-09-22. Score: 0

Answer over in Photography:
<https://photo.stackexchange.com/questions/119761/sensor-size-of-sony-hdr-hc7e/119764#119764>
It turns out to be Vertical: 3.74 horizontal 4.98 and diagonal 6.23


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