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Fight181

An orbit that allows you to map the entirety of kerbin

Question

Hi

I'm wondering if it is possible to have a single orbit in which a camera inside a spacecraft could take pictures/map the entire world of kerbin. The cameras field of view would be the exact same as the field of view seen if you were looking from the locked view from the point on the ship that is closest to kerbin . Daytime/nighttime would not be an issue. I think it is
implied in the post up to this point but to clarify this can be done over many orbits of kerbin but the orbit must not be changed and it must be done in a single spacecraft. However, that being said I want an orbit that would complete this task in a reasonable period of time like under 35 years although it is preferable to have an orbit which could complete this in a faster period of time. Also it is preferable to have an orbit that is as close to kerbin as possible.

I would imagine that this would be some kind of polar orbit/ an orbit close to a polar orbit.

Parameters

- Single spacecraft

-The camera must be able to map the entire world

-Field of view of the camera is the same as the field of view from the spacecraft's closest part to earth in the locked view

-The orbital path must remain the same

-The time to map must be under 35 years (Preferably in the fastest period of time)

-Attempt to get an orbit that is as close to kerbin as possible.

Edited by Fight181

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This is just going to be a straight-up polar orbit. Let Kerbin rotate beneath you, and before too long, the entirety of the planet has been mapped. Play with the ScanSAT mod, and you'll see how it goes.

In theory, if your field of view is sufficiently wide, you can get it in half a sidereal day. If the FoV is more narrow, you'll get stripes that are covered and stripes that aren't, and you'll have to wait to fill in the gaps.

Now, for a bit of Fun (TM), try out Venus (Real Solar System) or other mod planets with very slow sidereal rotations. Half a Venusian sidereal day is 120 Earth days.

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While it's not a big enough improvement for me to bother, I think the optimal inclination would be 90 degrees plus half your field of view, such that you just barely graze the poles with each pass. This gives you a bit of additional rotation under your scanner as you pass the equator, letting you scan slightly more of the equator with each pass.

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I can't remember why but the advice I had some time ago was for a few degrees off a polar orbit. I typically pick 85°.  Someone might be able to explain why or why not this might be better than 90°.

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1 hour ago, Foxster said:

I can't remember why but the advice I had some time ago was for a few degrees off a polar orbit. I typically pick 85°.  Someone might be able to explain why or why not this might be better than 90°.

I'd guess half of your field of view should be your tilt off of 90 degrees, to decrease what you are guaranteed to see every pass of each pole, thereby increasing the chance of seeing new stuff.

Edited by 5thHorseman

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5 hours ago, Foxster said:

Someone might be able to explain why or why not this might be better than 90°.

You'll be less efficient with a true polar orbit, as you'll pass over the exact same spot each orbit (the poles).  Let's say your camera can see a 10' field of view, so an inclination of something like 85' will still see the poles as you pass, but you will cover more new ground with each pass.     

As for efficiency, there will be some orbits that might fall into some resonance with the rotation of the planet, and you'll end up cover the same, or almost the same, ground with each orbit.   You'd want to setup your orbit so each pass covers mainly new ground. 

I have no idea as to the actual maths, but there would be an equation that would reveal the optimum orbital inclination for a given altitude and camera field of view.

From playing with scansat in the past, I would think with a halfway decent field of view (10-15'), you should be able to cover the entire planet of Kerbin within a few weeks.  

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7 hours ago, Foxster said:

I can't remember why but the advice I had some time ago was for a few degrees off a polar orbit. I typically pick 85°.  Someone might be able to explain why or why not this might be better than 90°.

In the real world Sun synchronous orbits are usually slightly retrograde (around 98°) and set up so that the orbit precesses eastwards to keep them aligned to the Sun throughout the year but I don't think that applies in KSP.

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Does anyone know an optimal altitude to do this at where it would minimize time to map but still not compromise on mapping the entire planet?

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2 hours ago, Fight181 said:

Does anyone know an optimal altitude to do this at where it would minimize time to map but still not compromise on mapping the entire planet?

A 150km polar orbit will give you about a 36 min orbit and your field of view is about 36° so you should get a reasonable coverage looking at only the daylight side in 10 orbits.

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12 hours ago, Reactordrone said:

A 150km polar orbit will give you about a 36 min orbit and your field of view is about 36° so you should get a reasonable coverage looking at only the daylight side in 10 orbits.

You mean five orbits. Remember, you scan 36° on one hemisphere, then another 36° on the opposite hemisphere.

Incidentally, it's a curious optimization issue: a low orbit restricts your field of view, but also gives you more north-south angular velocity.

EDIT: Just read the "daylight passes" bit. In that case, either 10, or maybe 5 if you orbit on the day-night terminator.

Edited by Starman4308
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It's just a polar orbit - when you vertically, and the planet rotates horizontally, then soon you have covered the whole body. That's why you need a polar orbit to map for resources. It's the same for doing a rescue mission - after a while, the whole planet is covered and at some point, you will be directly over your target.

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8 hours ago, Duck McFuddle said:

It's just a polar orbit

A polar orbit may or may not give you full coverage of the planet.   Given a planets radius,  it's rotational period, and your satellite's orbital period, you might end up with your scanner only covering the same spots over and over.   You could easily be left with patches that you will never see due to a resonance of the variables. 

You will have to be within one field of view of the poles though.  So if your camera can only see 15' of the planet at at time, any inclination less than 75' will never see the poles.    So in this case, the correct answer lies between 75' and 90' (inclusive). 

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You'd have to be incredibly unlucky to accidentally get an orbit with a perfect resonance to the rotational frame of reference, though. It's hard enough to do that intentionally.

 

Anything that has the pole in field of view should work; in real life orbital altitude for imaging satellites is a trade-off between field of view and ground tracking ability Vs resolution and clarity, but that applies less here.

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13 hours ago, Espatie said:

You'd have to be incredibly unlucky to accidentally get an orbit with a perfect resonance to the rotational frame of reference, though. It's hard enough to do that intentionally.

 

Anything that has the pole in field of view should work; in real life orbital altitude for imaging satellites is a trade-off between field of view and ground tracking ability Vs resolution and clarity, but that applies less here.

While a perfect resonance generally doesn't happen, approximate resonances do. If, for example, you have an orbital period of 1.01 local sidereal days, it will take nearly 50 local days to map the body in question. This can be a factor if time is a significant factor.

Edited by Starman4308

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4 hours ago, Starman4308 said:

This can be a factor if time is a significant factor.

Which is the basis of my argument.   If we're looking  for the most efficient orbit, it's not polar. 

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I'd call it marginal with thinner ground tracks ... the change in the polar mapping isn't going to matter much, they'll get covered completely the earliest anyway.  What MAY make a difference is the way the changed angle increases equatorial coverage in a retrograde orbit.

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Is it possible to do it with a very small field of vision say 2 degrees?

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3 hours ago, Fight181 said:

Is it possible to do it with a very small field of vision say 2 degrees?

It's possible with any field of view or even a constant scan-width (eg; 1m).  Obviously the smaller the coverage of each orbit the more orbits it will take but the basic approach of an almost-polar orbit remains the same.

For orbit altitude consider how far the planet rotates beneath you each orbit.  A Kerbin day is 6hrs (and 360-degrees by definition) and you scan two strips per orbit (one going N, then going S).  You want the lowest (fastest) orbit possible while avoiding a resonant orbit that repeats the same strips before covering the whole planet.  Most orbits won't scan 'in sequence' but will gradually fill-in the gaps between initial widely-spread scans.

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10 hours ago, Pecan said:

It's possible with any field of view or even a constant scan-width (eg; 1m).  Obviously the smaller the coverage of each orbit the more orbits it will take but the basic approach of an almost-polar orbit remains the same.

For orbit altitude consider how far the planet rotates beneath you each orbit.  A Kerbin day is 6hrs (and 360-degrees by definition) and you scan two strips per orbit (one going N, then going S).  You want the lowest (fastest) orbit possible while avoiding a resonant orbit that repeats the same strips before covering the whole planet.  Most orbits won't scan 'in sequence' but will gradually fill-in the gaps between initial widely-spread scans.

Not necessarily true. There are two effects going on that partially work against that.

First, as you increase altitude, a fixed scan angle will cover a wider strip of terrain.

Second, With SCANSat, the primary mod implementing any sort of terrain scanning, there's a mechanic where your scan angle is narrowed if you're at less than the optimal altitude. A similar effect goes on for "look down and take screenshots"-type mapping: as you go up, you see more of the planet at once.

The actual optimal altitude varies on a lot of factors: scan angle, whether the scan angle is fixed with altitude, planet radius, planet mass (and thus orbital speed at X altitude), and sidereal rotational period.

With slowly rotating Venus, for example, the optimal is "high as you can get without compromising scan angle". With almost any orbit and scan angle, you're spending around a hundred days twiddling your thumbs waiting for Venus to rotate under you. With a very small body such as Gilly, however, a high orbit might mean it's rotating under you plenty fast; the limiting factor is how fast you can go from pole to pole.

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19 hours ago, Starman4308 said:

Not necessarily true

True, of course, but he did say 2 degrees, which probably means the wider strip covered with a higher orbit would be more than offset by the longer period, almost anywhere.  Again though, you're correct that it does depend on how fast the planet is rotating, so if a planet is tide-locked to its sun it'll take a local year to fully rotate (and therefore 6 months to scan).  Ultimately if it has a day the same length as its year it would have no apparent rotation and you'd have to do a hemisphere at once or have no chance, since precession isn't a thing in KSP.  The question then is whether you could find an orbit far enough out to give you such a wide view while staying within the body's SOI.

Apart from that if we're talking about ScanSat or anything else with specified orbit ranges then there's no real choice but to use those (and avoid resonance within the given bounds).

Edited by Pecan
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