<p>I don't have enough room in the comments for this, so I'm writing here, although it's probably not a true answer since I know nothing about that particular telescope.</p>

<p>Anyway, if you look at many SCT systems, and their derivatives such as Ritchey-Chretien, Dall-Kirkham, etc, the distance between primary and secondary is often not too big.</p>

<p>If the secondary has negative curvature, that means one of its focal points is behind it. That's where the focal point of the primary also is. So the focal length of the primary is bigger than it seems just by looking at the picture. There's a focal point geometrically located in front of the telescope; the distance between it and the secondary or the primary depends on the design parameters of the system. It can't be too close to the secondary, or else the secondary would have to have extraordinary amounts of curvature.</p>

<p>Also keep in mind that there must be a gap between the edge of the primary and the inner surface of the OTA, so the primary is smaller than the visual estimate of the hole.</p>

<p>We also don't see the bottom of the instrument, so we don't know how far the primary mirror cell is protruding out the back of the instrument - though that's limited by the mount's big fork.</p>

<p>Also, look at a small SCT, such as the Celestron EdgeHD8 OTA. Visually, the tube seems a bit longer, compared to its diameter. However, the EdgeHD is a much smaller instrument. The relative amount "wasted" at the bottom by the primary cell is different. Also, the secondary is "buried" into the OTA, whereas the 0.65m scope has its secondary sticking out of the OTA.</p>

<p>Anyway, this is a bunch of handwaving based on visual estimates. It is possible that this is a system with a primary somewhat more strongly curved than usual.</p>

CC BY-SA 4.0
2019-09-12T22:30:59.240Z
Thanks for your thoughts! I'm not well versed in astro-acronymology; can you mention what SCT represents?
uhoh
CC BY-SA 4.0
2019-09-13T12:28:41.243Z
@uhoh SCT = Schmidt-Cassegrain telescope, OTA = optical tube assembly
Mike G
CC BY-SA 4.0
2019-09-13T16:16:23.147Z
In this [forum topic](https://astronomy.ru/forum/index.php/topic,175643.0.html) Borisov appears to call it an f/1.5 Hamiltonian.
Mike G
CC BY-SA 4.0
2020-11-05T22:45:47.273Z
@MikeG I've just asked [What (the heck) is a Hamiltonian telescope? Is this one?](https://astronomy.stackexchange.com/q/39702/7982)
uhoh
CC BY-SA 4.0
2019-09-12T16:54:12.490Z
2019-09-12T16:54:12.490Z
Florin Andrei
<p>The recently discovered object <a href="https://en.wikipedia.org/wiki/C/2019_Q4_(Borisov)" rel="nofollow noreferrer">C/2019 Q4 (Borisov)</a> is in the news because it might have an eccentricity greater than one, which means it <em>might be</em> from outside the solar system, though not necessarily.
<a href="https://minorplanetcenter.net/mpec/K19/K19RA6.html" rel="nofollow noreferrer">MPEC 2019-R106</a> estimates e=3.08.</p>

<p>I looked at Sky and Telescope's <a href="https://www.skyandtelescope.com/astronomy-news/possible-interstellar-comet-headed-our-way/" rel="nofollow noreferrer">Is Another Interstellar Visitor Headed Our Way?</a> which shows the image and caption below.</p>

<p>Although it is impossible to be sure from only this photograph, the 0.65 m telescope looks like the distance from primary to the small secondary is only about 1 meter, maybe 1.2 at most. </p>

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

<ol>
<li>Why is this telescope so short? </li>
<li>How hard is it to make such a fast (and therefore deep) primary?</li>
</ol>

<p><img src="https://i.sstatic.net/eCnwb.jpg" alt="enter image description here"></p>

<blockquote>
  <p>Gennady Borisov with the 0.65-meter telescope he built and used to discover the new comet. G. Borisov</p>
</blockquote>

CC BY-SA 4.0
2019-09-12T11:26:13.470Z
There's a [real MPEC](https://minorplanetcenter.net/mpec/K19/K19RA6.html) now, e=3.08.
Mike G
CC BY-SA 4.0
2019-09-12T11:31:41.933Z
@MikeG wow thanks! That's quite a switch from 1.0718 +/- 0.0009 Feel free to edit/update my question if so inclined.
uhoh
CC BY-SA 4.0
2019-09-12T12:20:03.923Z
Maybe barycentric vs. heliocentric?
Mike G
CC BY-SA 4.0
2019-09-16T17:01:47.677Z
D = 65cm, F / 1.5.  FLI ML16803,
The field of view with this matrix is 128x128 arc minutes, penetration of 20 magnitude in 60 seconds of exposure
A. Rumlin
CC BY-SA 4.0
2020-11-05T22:45:25.317Z
@A.Rumlin I've just asked [What (the heck) is a Hamiltonian telescope? Is this one?](https://astronomy.stackexchange.com/q/39702/7982)
uhoh
<telescope><amateur-observing><telescope-making><mirror-making>
Why is this telescope so short? How hard is it to make such a fast primary?
uhoh
# Why is this telescope so short? How hard is it to make such a fast primary?

(Asked by: username_0 on 2019-09-12)

The recently discovered object [C/2019 Q4 (Borisov)](https://en.wikipedia.org/wiki/C/2019_Q4_(Borisov)) is in the news because it might have an eccentricity greater than one, which means it *might be* from outside the solar system, though not necessarily.
[MPEC 2019-R106](https://minorplanetcenter.net/mpec/K19/K19RA6.html) estimates e=3.08.

I looked at Sky and Telescope's [Is Another Interstellar Visitor Headed Our Way?](https://www.skyandtelescope.com/astronomy-news/possible-interstellar-comet-headed-our-way/) which shows the image and caption below.

Although it is impossible to be sure from only this photograph, the 0.65 m telescope looks like the distance from primary to the small secondary is only about 1 meter, maybe 1.2 at most.

**Questions:**

1. Why is this telescope so short?
2. How hard is it to make such a fast (and therefore deep) primary?

![enter image description here](https://i.sstatic.net/eCnwb.jpg)

> Gennady Borisov with the 0.65-meter telescope he built and used to discover the new comet. G. Borisov

**Tags:** telescope, amateur-observing, telescope-making, mirror-making

> **Comments:**
> **username_1**: There's a [real MPEC](https://minorplanetcenter.net/mpec/K19/K19RA6.html) now, e=3.08.  
> **username_0**: @username_1 wow thanks! That's quite a switch from 1.0718 +/- 0.0009 Feel free to edit/update my question if so inclined.  
> **username_1**: Maybe barycentric vs. heliocentric?  
> **username_2**: D = 65cm, F / 1.5.  FLI ML16803, The field of view with this matrix is 128x128 arc minutes, penetration of 20 magnitude in 60 seconds of exposure  
> **username_0**: @A.username_2 I've just asked [What (the heck) is a Hamiltonian telescope? Is this one?](https://astronomy.stackexchange.com/q/39702/7982)  

---

## 1 answers

### Answer by username_3 on 2019-09-12. Score: 2 | Accepted answer

I don't have enough room in the comments for this, so I'm writing here, although it's probably not a true answer since I know nothing about that particular telescope.

Anyway, if you look at many SCT systems, and their derivatives such as Ritchey-Chretien, Dall-Kirkham, etc, the distance between primary and secondary is often not too big.

If the secondary has negative curvature, that means one of its focal points is behind it. That's where the focal point of the primary also is. So the focal length of the primary is bigger than it seems just by looking at the picture. There's a focal point geometrically located in front of the telescope; the distance between it and the secondary or the primary depends on the design parameters of the system. It can't be too close to the secondary, or else the secondary would have to have extraordinary amounts of curvature.

Also keep in mind that there must be a gap between the edge of the primary and the inner surface of the OTA, so the primary is smaller than the visual estimate of the hole.

We also don't see the bottom of the instrument, so we don't know how far the primary mirror cell is protruding out the back of the instrument - though that's limited by the mount's big fork.

Also, look at a small SCT, such as the Celestron EdgeHD8 OTA. Visually, the tube seems a bit longer, compared to its diameter. However, the EdgeHD is a much smaller instrument. The relative amount "wasted" at the bottom by the primary cell is different. Also, the secondary is "buried" into the OTA, whereas the 0.65m scope has its secondary sticking out of the OTA.

Anyway, this is a bunch of handwaving based on visual estimates. It is possible that this is a system with a primary somewhat more strongly curved than usual.

> **Comments:**
> **username_0**: Thanks for your thoughts! I'm not well versed in astro-acronymology; can you mention what SCT represents?  
> **username_1**: @username_0 SCT = Schmidt-Cassegrain telescope, OTA = optical tube assembly  
> **username_1**: In this [forum topic](https://astronomy.ru/forum/index.php/topic,175643.0.html) Borisov appears to call it an f/1.5 Hamiltonian.  
> **username_0**: @username_1 I've just asked [What (the heck) is a Hamiltonian telescope? Is this one?](https://astronomy.stackexchange.com/q/39702/7982)  

---

