# Revisit time estimation for a SAR constellation

I am studying satellite coverage analysis and I have read something about revisit time I did not understand.

you can increase and increase the number of satellites to further improve the revisit time, but there is a limit if you use one plane constellation. The maximum revisit time you can achieve is 1 day, because you have to wait for the earth to rotate. Even if you put infinite satellites, you have to wait one day for EO sensors or 12 hours for SAR.

I do not understand or rather I cannot visualize how the increase in the number of satellites on the same orbital plane cannot bring about an improvement in the revisit time. On the other hand I have also read that the maximum number of satellites for which there is an improvement in the revisit time is 6 (always on the same orbital plane). Could someone clarify this concept for me, perhaps using animations or images? Thanks in advance

• What is the source of the quote? Commented Aug 22, 2022 at 17:25
• @OrganicMarble notes I took in class Commented Aug 22, 2022 at 17:27
• Commented Aug 22, 2022 at 17:52
• tangentially relevant; I assume EO implies imaging using daylight, so unless it's Sun-synchronous it's once a day. SAR doesn't need direct sunlight, so twice a day can work, but that assumes it has big batteries because SAR uses a lot of electrical power. Often SAR is done mostly in daylight and only sparingly at night due to power constraints. As far as your question, I think "revisits" imply equally spaced in time. If you have 42 satellites in one plane (about one pass every two minutes) you might get a burst of three closely-spaced "visits" but then wait 12 or 24 hours for the next burst.
– uhoh
Commented Aug 23, 2022 at 7:22

The ground track of a low orbiting satellite looks like a single period of the sine function. Here is a recent ISS ground track for reference. If you want to observe a specific city, say at 30°N latitude, you will notice that our ground track crosses 30°N exactly twice per orbit. One crossing is in daylight and the other at night. Here, the daylight crossing of 30°N is approximately in Houston and the nighttime crossing is in Shanghai.

The next orbit appears to be shifted west, but it is actually earth that slowly turns eastward. If you want to observe Cairo, also on 30°N, you have to wait until earth turned enough to bring Cairo under the ground track. In 6 hours, Cairo will be under ISS's nighttime pass over 30°N, shown with a red arrow. In 16 hours, Cairo will be under the daylight pass. Each grid square is two hours.

More satellites on the same orbit won't help you because the orbit comes nowhere near the middle east until earth has turned a quarter revolution and that takes 6 hours.

One plane, no matter how many sensors you place in it, only observes part of the Earth at a time. The observed section is a band like a ribbon or a belt, covering the orbit's ground track and a bit to either side, which is sometimes called the sensor's "swath". As the Earth turns under the satellites, parts of it go out of view and new parts come in. The time it takes for any point on the surface to rotate from crossing the orbit plane once to crossing it again takes half a day. In order to observe a spot continuously, you have to put up enough planes so that the swath width of one plane touches the swath width of its neighbor. If you use a high inclination, you will revisit the poles faster than the equator; for a sun-synchronous weather satellite, you will see the pole on every rev, but there will be gaps along the equator, and a gradient of increasing coverage with latitude. If you use too low an inclination, you will never see the poles at all.

As far as the maximum useful number of spacecraft in a single plane, 6 is not always the answer. The most well-known example is geostationary, in which case 3 suffices. As the orbit altitude drops, the maximum viewable region drops with it, and the number of satellites needed to maintain continuous coverage of the ground track increases accordingly, with a maximum well beyond 6. To explain why, I have to draw a few pictures and then do some trigonometry. In the diagrams and equations below, capital letters indicate points, lowercase Greek letters are angle measures, and lowercase Latin letters are distances.

In Figure 1, the satellite is at point S, the center of the Earth is at point E, and point N is the nadir, the point on Earth's surface "directly below" the satellite, also known as the subpoint. Points P and Q may be on the edge of the satellite sensor's field of view (FOV), but they may not. There is some disagreement on exactly how FOV is defined, which brings us to the next two images.

Figure 2 labels the distance from E to N as $$n$$, and from E to P as $$p$$. These are both radii of the Earth, which are equal if the Earth is assumed to be circular in the plane of the diagram, or you can have different ones that depend on angle to express non-spherical terms. The distance from the satellite to its subpoint I have called $$h$$, for height; it is sometimes called altitude, but semimajor axis has a better claim on the letter $$a$$. The distance from the satellite to some point, P, other than the nadir, N, I have called $$s$$, which stands for slant range.

The yellow line is perpendicular to the Earth's radius at P; this is the local horizontal, tangent to a smooth surface and ignoring terrain for this discussion. $$\alpha$$ is the cone angle of the view, and $$\varepsilon$$ is the elevation angle. We often measure $$\alpha$$ in "spacecraft degrees", to distinguish it from the Earth angle $$\theta$$, which may be latitude, longitude, or some combination of the two, depending on how the plane of the diagram is oriented in 3-D space. From the perspective of the spacecraft, that rotation is often called the clock angle.

The largest possible spot on the ground for a given orbital height is determined by the set of points with elevation $$\varepsilon=0$$, which I'll call the horizon, as seen in Figure 3.

This would be easy to calculate if the Earth were a perfectly smooth, round ball (as in the image), but the presence of terrain makes calculating its exact position very messy. It's conventional to ignore terrain when drawing the boundary, both because it is easier and because it is very difficult to obtain good measurements when looking sideways through much more atmosphere than if you were looking straight down. Even if you did get more data than you expected back from that edge region, your sensing algorithms would probably have significantly greater difficulty in processing it, unless they are specifically designed for grazing incidence. For some sensors, the practical horizon is at $$\varepsilon=5^\circ$$ or $$10^\circ$$, or even above $$30^\circ$$ for something like SAR, since the accessable area depends not on where you can send radar signals, but on where you can get enough echo back to detect; forward scattering doesn't help.

Some people use one or both of the terms "field of view" and "footprint" to mean everything the satellite could possibly see without having to look through the solid Earth. I have worked with a variety of scanning and steerable sensors, so I prefer to restrict those terms to only the instantaneous area the sensor is currently collecting, so some call that the "IFOV". I use "field of regard" (FOR) to mean everywhere the automatically scanning sensor is going to point, or everywhere the steerable sensor could be commanded to point, which has no necessary relation to the geometrical horizon. For example, a small diameter antenna at a low radio frequency might have its FOV include not only the visible Earth but also large amounts of empty space around Earth, which has consequences for antenna noise temperature, and various other things. With reference to Figure 2, I consider the cone angle $$\alpha$$ to be a property of the spacecraft hardware and software, such as its antenna pattern and the amount of acceptable loss before the SNR drops too low to collect confidently.

Now, we can finally use some equations that actually get to a numerical answer for your question.

In the case of the $$\varepsilon=0$$ horizon, Pythagoras says $$p^2 + s^2 = (n + h)^2$$, and we also know $$\cos\theta = p/(n+h) = \sin\alpha$$. For a spherical Earth, $$n=p$$, which for purposes of my calculations will equal 6371 km (the global average, not the equatorial maximum). At geosynchronous altitude, $$\alpha = \arcsin(6371/42164) = 8.7^\circ$$, so $$\theta=81.3^\circ$$. This means that a GEO with inclination less than $$8.7^\circ$$ never sees the poles, but a GEO with inclination more than that sees each pole for at least a short time every day (a time which grows larger as the inclination does). Turning the plane $$90^\circ$$, so that $$\theta$$ measures longitude instead of latitude, that means each GEO's geometric horizon is 162 degrees across, so two of them don't quite cover the entire equator, but three of them are more than enough (alternately, give complete, continuous visibility to a wider band around the equator, as the number increases). If on the other hand we look down from the ISS, $$h=420$$ km means $$\theta = \arccos(6371/6791) =20.3^\circ$$, so you would need at least 9 (360 / 40.5, rounded up) vehicles at that altitude to keep the entire ground track continuously observed. At 120 km ($$\theta=11^\circ$$), you'd need more than 16.

If on the other hand we take a fixed sensor spot beam size of constant $$\alpha$$, which is a better model for many kinds of sensing systems, we need to be a little trickier. Looking at Figure 2, the length $$c$$ (named for the chord of the arc from P to Q), we find is a side shared in common by two right triangles. On the left, $$p \sin \theta = c$$, and on the right, $$s \sin \alpha = c$$. The line segment from S to E, orthogonal to $$c$$, is $$n+h$$, which equals the sum of another side of those same two triangles, namely $$p \cos \theta + s \cos \alpha$$. We can also note in passing that $$\alpha + \theta + \varepsilon = \pi/2$$, but we don't actually need that for this computation.

Instead, taking $$n$$, $$p$$, $$h$$, and $$\alpha$$ as known, we eliminate $$\theta$$ and solve for $$s$$. We have $$p \sin \theta = s \sin \alpha \\ p^2 \sin^2 \theta = s^2 \sin^2 \alpha \\ p \cos \theta = n+h-s \cos\alpha \\ p^2 \cos^2 \theta = (n+h-s \cos\alpha)^2 \\ p^2 \sin^2 \theta + p^2 \cos^2 \theta = s^2 \sin^2 \alpha + (n+h)^2 - 2(n+h)s \cos\alpha + s^2\cos^2 \alpha \\ p^2 = s^2 - 2s(n+h)\cos\alpha + (n+h)^2 \\ 0 = s^2 + [-2(n+h)\cos\alpha] s + [(n+h)^2-p^2]$$ and we use the quadratic formula to obtain $$s=(n+h)\cos\alpha-\sqrt{(n+h)^2\cos^2\alpha-(n+h)^2+p^2}\\ s = (n+h)(\cos\alpha - \sqrt{\cos^2\alpha - (1 - p^2/(n+h)^2)})$$ Note that we keep only the $$-b-\sqrt{b^2-4ac}$$ answer, and throw away the $$+$$, because the $$+$$ represents what happens when our line of sight pierces through the solid Earth and sticks out on the opposite side.

Armed with this equation, we discover that if we have a fixed sensor viewing cone of 5 degrees from center to edge, when that sensor is mounted on a GEO, $$\theta=30^\circ$$, so six of them could keep the ground track observed continuously. If instead that same sensor was mounted on something orbiting at the altitude of the ISS, $$\theta$$ is only two-thirds of a degree, so you would need 270 satellites in one plane to keep that one plane continuously observed.

• Every time I read Ryan's answers, I get a sense of good and true mentorship. I am not the OP, but Thank you for your rigorous explanation. What a stellar community. Commented Aug 26, 2022 at 23:12