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NASA's published specs for the Curiosity rover cameras says that the image resolution is in milliradians per pixel. Which is odd because image resolution is measured in physical size units, i.e. millimetres.

Could it be referring to angular resolution?

To check I tried to see if the 0.82 milliradians could be used to accurately get the field-of-view (FoV) angle.

So the Navcam has 45 degree FoV and has 1028 pixel resolution.

0.82 milliradians is 0.047 degrees, so one pixel is 0.047 degrees.

0.047 degrees x 1028 pixel is 48.29 degrees, not 45 degrees.

It doesn't really match up, but maybe I messed something up, because I am not confident in math.

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    $\begingroup$ Good question, although just to point out that in the Wikipedia article that you linked to it states, "Resolution units can be tied to physical sizes (e.g. lines per mm, lines per inch), to the overall size of a picture (lines per picture height, also known simply as lines, TV lines, or TVL), or to angular subtense." $\endgroup$ Oct 29, 2023 at 12:44
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    $\begingroup$ Resolution of a camera is not generally measured in "physical size units" like millimeters. An exception in Space would be satellites, where "ground resolution" or "pixel size" are often assigned a physical size. Here, one can say that one pixel represents, say, a 3 meter square on the ground, and this is possible because the satellite is typically at a constant distance from the target (the ground). $\endgroup$
    – Dragongeek
    Oct 30, 2023 at 8:17

3 Answers 3

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Let's have a look at a more technical and detailed description:

https://an.rsl.wustl.edu/help/Content/About%20the%20mission/MSL/Instruments/MSL%20Navcam.htm

The Navcam optics are f-theta fisheye lenses with 45 degree x 45 degree horizontal/vertical field of view and a 67 degree diagonal field of view. They have an angular resolution at the center of the picture of 0.82 milliradians/ pixel.

So, the lens used is a fisheye which means that it's expected that the image gets distorted towards the edges and even more so towards the corners. This is evident from the numbers of the field of view given: A perfect 45° square would give you a diagonal field of view of only 63.6°, but the actual is 67° due to distortion.

Consequently the actual resolution varies from the center towards the edges and the stated resolution of 0.82 mrad/px is only valid in the central region.

As Hobbes already wrote, you can state the resolution of the pixel sensor itself using units of length or area (here: "12 micron-squared pixels", not sure whether that is supposed to mean 12µm² or 12µm*12µm (source)).

You can equally well state the spatial resolution of a photographed object, e.g. "1 pixel corresponds to 8mm on the object". But in this case you also need to state the distance to the object - twice the distance gives half the resolution.

The number that stays constant is the angular resolution which is the difference in direction between two pixels. If you know the distance you can easily convert this number to the actual spatial resolution. The angular resolution is the typical measure used e.g. for telescopic cameras.

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  • $\begingroup$ Tangent: You seem to have calculated the ideal diagonal FoV for a rectilinear lens as $\sqrt2\cdot45°≈63.6°$, but that shortcut only works for small angles where $\tan(x) ≈ x$. (In particular, for large enough angles it can give results larger than $180°$, which is clearly impossible for a rectilinear lens.) For the correct result you need to convert the view angle to the side length of the view square (at some arbitrary distance, which we may take to equal 1), then multiply by $\sqrt2$ and convert back, giving $2\arctan(\sqrt2\cdot\tan(45°/2))≈60.7°$. That's even further from $67°$, though. $\endgroup$ Oct 30, 2023 at 9:44
  • $\begingroup$ @IlmariKaronen Right, at 22.5° the tan is already off from the tan(x)=x approximation by 5%. But I'm also not fully sure your calculation is correct after a complicated lens assembly that changes incident angles. $\endgroup$
    – asdfex
    Oct 30, 2023 at 10:56
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When you specify resolution as a linear size, it becomes dependent on how far away the subject is. Curiosity's cameras take photos at varying distances, so you'd have to have a table with resolutions at every distance.

When the resolution is specified as an angle, you can calculate the linear resolution for any distance.

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  • $\begingroup$ Is it possible to clarify the delta between the calculated field of view of 48 and the published 45 degree? Is this a maths problem or a rounding for publication thing? $\endgroup$ Oct 29, 2023 at 8:12
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    $\begingroup$ The 48 vs. 45 disagreement is likely rounding for editorial purposes. Also, the angular resolution may not be the same over the entire sensor depending on the optics in the system. $\endgroup$ Oct 29, 2023 at 9:53
  • $\begingroup$ If the 45 fov is just rounding, is the 48 fov likely to the actual fov? $\endgroup$ Oct 29, 2023 at 10:28
  • $\begingroup$ @spaceamoeba1010 I can' see how any rounding rounds 48 to 45. 48 to 50 yes unless the actual figure is 47.5 and you are usinbg very different rounding methods to find the nearest. $\endgroup$
    – mmmmmm
    Oct 30, 2023 at 14:02
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Field of view

45 degree doesn't seem a rounding error but just the effect of trigonometry. Simplifying the camera as projecting the image though a point (the center of the lens), knowing the angular resolution allows to compute the focal distance:

> (f <- 1/tan(0.82e-3))
[1] 1219.512

The focal distance is 1219.5 pixels, that is, the focal distance is 1219.5 times larger than the size of one pixel in the sensor. With the focal distance and the size of the sensor we can know the field of view in radians (514 is just half of the sensor, 1028/2):

> atan(514/f)*2
[1] 0.797771

Or in degrees:

> atan(514/f)*2*180/pi
[1] 45.70891

Of course, lens and sensor geometry can be more complex, and some rounding might be involved, but using the old rules for perspective gives the same results that have been published.

I'd like to add a diagram but it's nearly midnight at my place, and Wikipedia already has an explanation of how to compute the field of view angle.

Resolution: an example

And about the angular resolution 0.82 milliradians per pixel, other answers already explain its meaning and why it is used but I'd like to add an example of how it can be used. Since for small angles the angle in radians is approximately equal to its sine and its tangent, 0.82 milliradians per pixel means that the size of the object covered by one pixel is 0.00082 times the distance from the camera to the object. If the object is 1 meter away, each pixel covers .82 mm. If the object is 10 meters away, each pixel covers 8.2 mm. If the object is 100 m away, each pixel covers 82 mm, and so on.

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