# Units of "pixels" in a research paper on digital streak formation for optical payloads

I am trying to wrap my head around how a single bright event might spill over and streak across pixels in a sensing array (CMOS) if your exposure time is high enough at some orbital velocity.

In the 2023 paper Design of an optical system for a Multi-Cubesats debris surveillance mission (similar here), Pineau and Felicetti state the following equation:

$$L_{st} =\frac{f \ v_{t, a} \ \tau}{\cos(\theta_{obj}) \ d \ p}$$

where $$\tau$$ is the exposure time, $$f$$ is the focal length, $$v_{t,a}$$ is the averaged value of the object's tangential velocity in the CubeSat frame over $$\tau$$, $$p$$ is the size of each pixel of the sensor, the cosine of the angle is apparent from the image (see below), and $$d$$ is the distance to the object in real life.

What is the unit of "pixel?" Why do they refer to it as the size of each pixel of the sensor, when that implies dimensions such as square meters [m^2]. If it is square meters or even just meters do you not get either inverse meters or a dimensionless number - neither of which make any sense?

It must be that pixel is just a pure number of pixels on the sensing array, correct?

(Apologies for not using proper math formatting.)

• google "mathjax stackexchange" to find MathJax basic tutorial and quick reference (I think it's available in a help link somewhere on the page also, but I don't remember where)
– uhoh
Commented May 1 at 22:50

$$L_{st} =\frac{f \ v_{t, a} \ \tau}{\cos(\theta_{obj}) \ d \ p}$$

where tau is the exposure time, $$f$$ is the focal length, $$v_{t,a}$$ is the averaged value of the object's tangential velocity in the CubeSat frame over $$\tau$$, $$p$$ is the size of each pixel of the sensor, the cosine of the angle is apparent from the image (see below), and $$d$$ is the distance to the object in real life.

Let's do dimensional analysis of the right side.

$$\frac{m \cdot m/s \cdot s}{m \cdot m}$$

The result is unit-less, which is exactly what it needs to be for number of pixels.

So for a focal length $$f$$ of 100 mm, a relative tangential velocity between observing and tracked satellites $$v_{t, a}/cos(\theta_{obj})$$ of say 200 m/s, a real distance $$d$$ of 1000 km, a pixel size $$p$$ of 10 microns, and a time $$\tau$$ of 10 seconds, we get $$L = 20$$ pixels.

That sounds about right.

I haven't read the paper yet, I'll leave that as an exercise for the reader.

• Just to clarify further, the equation measures the number of pixels along a given direction (horizontal in the picture), that's why here the size of the pixel is the size along that direction, rather than the surface. Commented May 3 at 8:52
• @jcaron I did not dig in to the paper at all. I see, it's the length of the trail on the sensor surface, measured in units of $L$. If the trail is horizontal or vertical then it is also length in number of pixels. If the trail is in any other direction, we can't really define number of pixels in a clear way. Am I understanding it correctly? Feel free to edit this answer or add another.
– uhoh
Commented May 3 at 12:24
• I haven’t looked at the paper, but the diagram is clearly for the case of a trail which is parallel to one of the sensor axes. For other directions it would be more complex to count exactly how many pixels are impacted, but one could find the enclosing rectangle size by using the above formula in each of the two dimensions. Commented May 3 at 12:30

Per the first link uhoh provided in their edit, $$p$$ appears to be the size of one side of a pixel, with units of length/pixel. I draw this conclusion from:

• Figure 6 in the paper, where p is labeled along one side of a square on the image plane

• Table 1 "Balor digital sensor parameters" in the paper, where the "Pixel size $$p$$" is $$12 \mu\text{m}$$

• Equations 34 and 35, where dimensional analysis show that the width, in pixels, of a streak is supposed to be the twice the Airy radius (a length) divided by $$p$$, so

$${\text{pixel}} = {\text{length}\over{\text{length}/\text{pixel}}}$$

uhoh's answer, which popped up while I was typing this, covers the equation in the question (equation 8), which is consistent with the above.