<p>If the grid over the first image is reliable, then it's surely not a rectilinear projection.</p>
<p><strong>EDIT: The above conclusion is wrong, as the rings don't have a 10-degrees step, but only 5 degrees.</strong> I'll keep the answer as is, maybe it's helpful to someone else.</p>
<p>It covers 150 degrees over the diagonal, and if it were rectilinear, the grid cells should appear much longer at the image border than in the center. But the distance between your 10-degree &quot;rings&quot; stays rather constant all over the image, which looks more like a fisheye.</p>
<p>Why that?</p>
<p>In a rectilinear projection, let's say we have a point that's away some angle α from the center of the image. Then it gets projected onto the image at a metric distance of</p>
<pre><code>d = K tan α
</code></pre>
<p>(with K being a constant dependent on focal length, image sensor size etc.). Now let's look at a table of the tan function:</p>
<pre><code> α      tan α
--------------
 0°     0
10°     0.176
20°     0.364
30°     0.577
40°     0.839
50°     1.191
60°     1.732
70°     2.747
80°     5.671
</code></pre>
<p>So, on the diagonal of a rectilinear image covering roughly 150 degrees, the outermost 10-degrees step (from 60° to 70°) should be 5 times larger than the innermost one (from 0° to 10°), namely 1.015 compared to 0.176.</p>
<p>In your image, the rings aren't centered exactly around the image center (you didn't point at Polaris). But the growing-distance effect should still be very obvious if the projection were rectilinear.</p>
<p>To definitely find out about the projection (and any imperfections), take a photograph of some grid structure, using the same camera settings. If, on the image, the grid still consists of mainly straight lines, you have a photo-typical rectilinear projection. If the lines are significantly curved, you need a different projection model.</p>

CC BY-SA 4.0
2022-03-15T16:12:51.250Z
The grid over the first image is mostly accurate. I drew it referencing the stars. Only at the extreme corners that stars brightness were not good and I might have used some imagination.Could you please expand on why the ring sizes should be variable?
amr gameel
CC BY-SA 4.0
2022-03-15T17:05:51.690Z
I'll add to my answer...
Ralf Kleberhoff
CC BY-SA 4.0
2022-03-16T13:57:07.223Z
Thanks a lot for the clarification. I think I know where the confusion is coming from. These rings are 5° rings rather than 10°. And they are actually enlarging as you go peripherally, even though it doesn’t appear so to the eye. But if you measure the pixels you’ll see it.
amr gameel
CC BY-SA 4.0
2022-03-16T14:48:25.057Z
@amrgameel okay, then I completely misread the image based on a 10-degrees assumption.
Ralf Kleberhoff
CC BY-SA 4.0
2022-03-15T12:47:37.180Z
2022-03-16T14:50:39.783Z
2022-03-16T14:50:39.783Z
Ralf Kleberhoff
<p>The iphone camera, in common with most other lenses, produces a rectilinear projection - also known as a gnomonic or perspective projection.</p>
<p>The main feature of this projection is that straight lines (ie great circles on the sky) map to straight lines.  For normal photography this is desirable: it avoids the &quot;fisheye&quot; effect of a curvilinear projection.</p>
<p>It does create lots of distortion far from the centre of the projection, which is why very wide field lenses tend to use &quot;fisheye&quot;.  But it is very acceptable for the relatively narrow field of view of a regular camera (about 60 degrees)</p>

CC BY-SA 4.0
2022-03-15T09:07:39.080Z
Really helpful. Thanks a lot.
amr gameel
CC BY-SA 4.0
2022-03-14T21:35:56.733Z
2022-03-14T21:35:56.733Z
James K
<p>I’m trying to develop software that overlays sky coordinates over 2d images of the sky taken by an iphone. I’ve done the sky coordinates part, but I’m trying to figure out what kind of perspective projection I could assume my iphone lens uses. Could you tell from these 2 images attached (that I have manually labelled) what kind of projection it is?</p>
<p><a href="https://i.sstatic.net/ylJb7.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ylJb7.jpg" alt="sky image 1" /></a></p>
<p><a href="https://i.sstatic.net/RfIt4.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/RfIt4.jpg" alt="sky image 2" /></a></p>

<observational-astronomy><mathematics><projection>
What kind of perspective projection is shown in this image?
amr gameel
# What kind of perspective projection is shown in this image?

(Asked by: username_0 on 2022-03-14)

I’m trying to develop software that overlays sky coordinates over 2d images of the sky taken by an iphone. I’ve done the sky coordinates part, but I’m trying to figure out what kind of perspective projection I could assume my iphone lens uses. Could you tell from these 2 images attached (that I have manually labelled) what kind of projection it is?

[![sky image 1](https://i.sstatic.net/ylJb7.jpg)](https://i.sstatic.net/ylJb7.jpg)

[![sky image 2](https://i.sstatic.net/RfIt4.jpg)](https://i.sstatic.net/RfIt4.jpg)

**Tags:** observational-astronomy, mathematics, projection

---

## 2 answers

### Answer by username_2 on 2022-03-14. Score: 9 | Accepted answer

The iphone camera, in common with most other lenses, produces a rectilinear projection - also known as a gnomonic or perspective projection.

The main feature of this projection is that straight lines (ie great circles on the sky) map to straight lines. For normal photography this is desirable: it avoids the "fisheye" effect of a curvilinear projection.

It does create lots of distortion far from the centre of the projection, which is why very wide field lenses tend to use "fisheye". But it is very acceptable for the relatively narrow field of view of a regular camera (about 60 degrees)

> **Comments:**
> **username_0**: Really helpful. Thanks a lot.  

---

### Answer by username_1 on 2022-03-15. Score: 3

If the grid over the first image is reliable, then it's surely not a rectilinear projection.
EDIT: The above conclusion is wrong, as the rings don't have a 10-degrees step, but only 5 degrees. I'll keep the answer as is, maybe it's helpful to someone else.
It covers 150 degrees over the diagonal, and if it were rectilinear, the grid cells should appear much longer at the image border than in the center. But the distance between your 10-degree "rings" stays rather constant all over the image, which looks more like a fisheye.
Why that?
In a rectilinear projection, let's say we have a point that's away some angle α from the center of the image. Then it gets projected onto the image at a metric distance of
d = K tan α

(with K being a constant dependent on focal length, image sensor size etc.). Now let's look at a table of the tan function:
 α      tan α
--------------
 0°     0
10°     0.176
20°     0.364
30°     0.577
40°     0.839
50°     1.191
60°     1.732
70°     2.747
80°     5.671

So, on the diagonal of a rectilinear image covering roughly 150 degrees, the outermost 10-degrees step (from 60° to 70°) should be 5 times larger than the innermost one (from 0° to 10°), namely 1.015 compared to 0.176.
In your image, the rings aren't centered exactly around the image center (you didn't point at Polaris). But the growing-distance effect should still be very obvious if the projection were rectilinear.
To definitely find out about the projection (and any imperfections), take a photograph of some grid structure, using the same camera settings. If, on the image, the grid still consists of mainly straight lines, you have a photo-typical rectilinear projection. If the lines are significantly curved, you need a different projection model.

> **Comments:**
> **username_0**: The grid over the first image is mostly accurate. I drew it referencing the stars. Only at the extreme corners that stars brightness were not good and I might have used some imagination.Could you please expand on why the ring sizes should be variable?  
> **username_1**: I'll add to my answer...  
> **username_0**: Thanks a lot for the clarification. I think I know where the confusion is coming from. These rings are 5° rings rather than 10°. And they are actually enlarging as you go peripherally, even though it doesn’t appear so to the eye. But if you measure the pixels you’ll see it.  
> **username_1**: @amrgameel okay, then I completely misread the image based on a 10-degrees assumption.  

---

