Duna's Angular Size

[KrakenNinja]

I'm trying to find Duna's angular size. I found a calculator for angular size online but it is a few arcmilliseconds off. I want to find Duna's angular size at it's closest point to kerbin.

[AHHans]

Well, according to the KSP Wiki Duna's relevant parameters are:

• Periapsis: 19 669 121 365 m

• Orbital inclination: 0.06 °

• Equatorial radius: 320 000 m

For Kerbin we have:

• Apoapsis: 13 599 840 256 m

The low orbital inclination means that we can ignore it. (Feel free to do the math yourself.)(*) So we can do for a theoretical closest approach:

• Minimal distance Kerbin-Duna: 19669121365 m - 13599840256 m = 6069281109 m
• Diameter of Duna: 2 * 320000 m = 640000 m
• Angular size in radians(**): 640000 m / 6069281109 m = 0.00010544906200685917
• Angular size seconds of arc:  0.00010544906200685917 / pi * 180 / 60 /60 = 21.75043030347386

I don't know if Duna ever comes this close to Kerbin, but if it does: there it is.

P.S. (*) Not to self: typing while in transit over Eve's Explodium Sea is O.K. Typing while on approach for landing is discouraged. (Don't worry, no Kerbals were harmed while typing this message, but it was a close call. )
(**) sin(x) = x for all relevant x.

 

[KrakenNinja]

Thanks.

[paul23]

14 hours ago, AHHans said:

Well, according to the KSP Wiki Duna's relevant parameters are:

• Periapsis: 19 669 121 365 m

• Orbital inclination: 0.06 °

• Equatorial radius: 320 000 m

For Kerbin we have:

• Apoapsis: 13 599 840 256 m

The low orbital inclination means that we can ignore it. (Feel free to do the math yourself.)(*) So we can do for a theoretical closest approach:

• Minimal distance Kerbin-Duna: 19669121365 m - 13599840256 m = 6069281109 m
• Diameter of Duna: 2 * 320000 m = 640000 m
• Angular size in radians(**): 640000 m / 6069281109 m = 0.00010544906200685917
• Angular size seconds of arc:  0.00010544906200685917 / pi * 180 / 60 /60 = 21.75043030347386

I don't know if Duna ever comes this close to Kerbin, but if it does: there it is.

P.S. (*) Not to self: typing while in transit over Eve's Explodium Sea is O.K. Typing while on approach for landing is discouraged. (Don't worry, no Kerbals were harmed while typing this message, but it was a close call. )
(**) sin(x) = x for all relevant x.

 

You can't do it quite like this: this is only correct if apoapsis and periapsis of duna and kerbin align perfectly. If, say, the apoapsis is on the other side of kerbol compared to duna's periapsis the closes point is much further and probably somewhere in the middle of the curve.

 

If there's inclinations (difference) involved it even gets more complex. The equations get so complex that I'd strongly suggest solving it numerically.

[AHHans]

10 hours ago, paul23 said:

You can't do it quite like this: this is only correct if apoapsis and periapsis of duna and kerbin align perfectly. If, say, the apoapsis is on the other side of kerbol compared to duna's periapsis the closes point is much further and probably somewhere in the middle of the curve.

Which is why I wrote: "theoretical closest approach"! Obviously (well, to me at least, and I'd hope to any astronomer) the actual angular size will change over time with the distance of Duna to Kerbin. But having shown the calculations here, it shouldn't be too hard to compute it for other situations.

10 hours ago, paul23 said:

If there's inclinations (difference) involved it even gets more complex. The equations get so complex that I'd strongly suggest solving it numerically.

You did see that I commented about that? How many significant digits will be unaffected by ignoring the inclination of Duna's orbit?

Yes, I gave way to many "significant" digits in what I wrote. And, yes, in a test I would probably have deducted points from a student who wrote down that many digits. But copy&paste is easy and actually switching on your brain is hard. In general the rule-of-thumb is that anything beyond the 3rd or 4th significant digit can be safely ignored.

Edited October 13, 2019 by AHHans
fixed typo

[KrakenNinja]

How many pixels would it be on a 1280 x 1024 display?

Edited October 13, 2019 by KrakenNinja

[Curveball Anders]

3 hours ago, AHHans said:

 In general the rule-of-thumb is that anything beyond the 3rd or 4th significant digit can be safely ignored.

For an engineer yes, for a mathematician it would be heresy

 

[AHHans]

2 minutes ago, Curveball Anders said:

For an engineer yes, for a mathematician it would be heresy

Well, show me a "true"(TM) mathematician who cares about actual numbers. 

Edit: And it's not just engineers, everyone who actually measures something only cares about the first few significant digits. (Well, nearly everyone, some physicists make a sport out of measuring some values extra precise.)

13 minutes ago, KrakenNinja said:

How many pixels would it be on a 1280 x 1024 display?

Please think about that question, and why there is no answer to it. (At least not without additional information.)

Edited October 13, 2019 by AHHans

[Curveball Anders]

3 minutes ago, AHHans said:

Well, show me a "true"(TM) mathematician who cares about actual numbers. 

True, but they stick to the principle, usability be damned

 

[AHHans]

Just now, Curveball Anders said:

True, but they stick to the principle, usability be damned

Which is why they don't like actual numbers. Once you start writing them down, there is not enough space on the blackboard for all the digits of Pi.

[paul23]

7 hours ago, AHHans said:

Which is why I wrote: "theoretical closest approach"! Obviously (well, to me at least, and I'd hope to any astronomer) the actual angular size will change over time with the distance of Duna to Kerbin. But having shown the calculations here, it shouldn't be too hard to compute it for other situations.

You did see that I commented about that? How many significant digits will be unaffected by ignoring the inclination of Duna's orbit?

Yes, I gave way to many "significant" digits in what I wrote. And, yes, in a test I would probably have deducted points from a student who wrote down that many digits. But copy&paste is easy and actually switching on your brain is hard. In general the rule-of-thumb is that anything beyond the 3rd or 4th significant digit can be safely ignored.

The number of digits ignored depends entirely on where the closest point actually is. If inclination difference is 90 degrees, and/or very high eccentric orbits your calculations might be off by an order of magnitude (or more). So really "ignoring inclination/longitude of ascending node" is not a simplification that is allowed from an engineering point, only if orbits look similar that is allowed.

[AHHans]

16 hours ago, paul23 said:

The number of digits ignored depends entirely on where the closest point actually is. If inclination difference is 90 degrees, and/or very high eccentric orbits your calculations might be off by an order of magnitude (or more).

I wasn't doing the calculation for any general case, I was doing the calculation for Duna being viewed from Kerbin. Of course you cannot generally ignore the relative inclination of the orbits, but for Duna and Kerbin you can. I though that was clear from what I wrote. I even quoted all the relevant values in my message. (Well, except perhaps the fact that Kerbins's orbit is the definition of the equatorial ecliptic(*) plane and thus Kerbin's orbital inclination is zero.)

In a very similar fashion of course "sin(x)" isn't in general equal to "x". But in astronomy the angles are usually so small that the error you introduce by just assuming "sin(x) = x" is insignificant compared to the measurement error.

Edit: (*) Relevant here is the ecliptic plane, not the equatorial plane. For Kerbin both are identical, but for e.g. Earth they aren't.

Edited October 14, 2019 by AHHans

[Snark]

On 10/12/2019 at 5:22 PM, paul23 said:

You can't do it quite like this: this is only correct if apoapsis and periapsis of duna and kerbin align perfectly. If, say, the apoapsis is on the other side of kerbol compared to duna's periapsis the closes point is much further and probably somewhere in the middle of the curve.

If there's inclinations (difference) involved it even gets more complex. The equations get so complex that I'd strongly suggest solving it numerically.

In general that's true, but those concerns are greatly mitigated in this particular case.  Kerbin's orbit happens to be perfectly circular, which simplifies the concern about periapsis / apoapsis; and Duna's inclination relative to Kerbin is so tiny (0.06 degrees) that it might as well be zero-- the cosine of that is over 0.999999.

The only concern about "closest approach" distance is how close Duna is to periapsis when it's at opposition to Kerbin, which will change on every orbit. But the easy way to deal with this is to just provide a range-- i.e. its angular size at opposition when it's at Pe versus Ap.  That will give the min/max sizes involved.

12 hours ago, KrakenNinja said:

How many pixels would it be on a 1280 x 1024 display?

Depends on what the angular field of view (FOV) is of your camera (which is an adjustable setting in KSP).  But to give you a ballpark idea, suppose the FOV is 60 degrees (I have no idea if that's even close to being the default, but it's probably somewhere between 45 and 90, so let's go with it for the moment and see where that takes us.)

If Duna is 22 arcseconds, that means Duna is 22 / 3600 = 0.0061 degrees in size.  One degree in this case would be 1280/60 = 21.33 pixels.

So if your FOV is 60 degrees, Duna's size on your 1280-pixel monitor would be about 0.13 pixels, i.e. just barely over 1/8th of a pixel.  It would be a bit larger than that if you have a narrower FOV, a bit smaller if you have a wider one.

And that is why you don't see Duna when you look up at Kerbin's night skies. 

[AHHans]

7 hours ago, Snark said:

In general that's true, but those concerns are greatly mitigated in this particular case.  Kerbin's orbit happens to be perfectly circular, which simplifies the concern about periapsis / apoapsis; and Duna's inclination relative to Kerbin is so tiny (0.06 degrees) that it might as well be zero-- the cosine of that is over 0.999999.

Well, that's also doing it a bit too simple. If the orbits of Duna and Kerbin were otherwise close, then even this small inclination could have a significant effect. What is important is the absolute distance to the ecliptic plane compared to the distance in the ecliptic plane.

As a quick check to see if I need to bother, I computed the largest possible distance to the ecliptic at that inclination ("sin(inlcination)*apoapsis") and compared that to the minimum distance in the ecliptic (Duna_Pe - Kerbin_Ap). The result was as expected: I don't need to bother taking the inclination into account.

7 hours ago, Snark said:

Depends on what the angular field of view (FOV) is of your camera (which is an adjustable setting in KSP).

Oh, that is (sort of) fixed? (I expected it to change when you zoom out.)

Hmmmm... I don't see any setting that looks like it adjusts the camera FOV to me in the settings menu. There are some settings of which I don't know / understand what they do though.

7 hours ago, Snark said:

And that is why you don't see Duna when you look up at Kerbin's night skies.

Well, there is also the question if KSP implements a render distance for planets, a distance beyond which it doesn't bother to display objects even if they would be visible.

[paul23]

10 hours ago, Snark said:

In general that's true, but those concerns are greatly mitigated in this particular case.  Kerbin's orbit happens to be perfectly circular, which simplifies the concern about periapsis / apoapsis; and Duna's inclination relative to Kerbin is so tiny (0.06 degrees) that it might as well be zero-- the cosine of that is over 0.999999.

The only concern about "closest approach" distance is how close Duna is to periapsis when it's at opposition to Kerbin, which will change on every orbit. But the easy way to deal with this is to just provide a range-- i.e. its angular size at opposition when it's at Pe versus Ap.  That will give the min/max sizes involved.

I agree that this is the case, however just blindly ignoring it is wrong. You have to state you ignore it... (And I'm not even talking about resonance orbits, which could mean kerbin and duna are always in a certain angle from each other when duna is at periapsis).

 

[Snark]

2 hours ago, AHHans said:

 Oh, that is (sort of) fixed? (I expected it to change when you zoom out.

Basically yeah. The default thumbwheel zoom in KSP doesn't change the FOV the way an IRL camera does; rather, it physically moves the camera closer or farther away. In photography terms, it's a dolly, not a zoom. (This becomes pretty obvious if you watch for it when you play with it it.)

2 hours ago, AHHans said:

Well, there is also the question if KSP implements a render distance for planets, a distance beyond which it doesn't bother to display objects even if they would be visible.

Pretty sure it doesn't, at least not at interplanetary scales. The sun is a visible disk even viewed from Jool, after all. Plus, when approaching a planet, it becomes visible as soon as it's even one pixel in size, if you know where to look (can generally distinguish it from skybox stars because of "twinkling" from rendering artifacts).

37 minutes ago, paul23 said:

I agree that this is the case, however just blindly ignoring it is wrong. You have to state you ignore it... (And I'm not even talking about resonance orbits, which could mean kerbin and duna are always in a certain angle from each other when duna is at periapsis).

It's not "blindly ignoring" if it's perfectly obvious that it's ignorable in the case under discussion. If you ask me how long it takes me to travel between two cities by train, for example, my calculations will not include relativistic time dilation. I'm not "blindly ignoring" Einstein, nor would I bother explicitly stating that I'm neglecting relativistic effects when giving my answer, as I'd assume it's obvious.

In this case, the question wasn't presented as "solve the general problem of closest approach of two bodies in arbitrary elliptical orbits", which would indeed be very gnarly (I know, because I've had to solve it myself for BetterBurnTime). Rather, the problem was explicitly stated as involving Duna and Kerbin specifically, which enormously simplifies the problem, since one is perfectly circular and they're nearly perfectly coplanar. Nor did the OP request all the details and a fully perfectly rigorous explanation of all the math-- he just wanted an answer.

So when a kindly soul like AHHans takes the trouble of providing a correct answer, and shows enough of his math to get an idea of how he got it, in a way that's perfectly appropriate for the context of what the OP actually wants to know, then I for one am not inclined to quibble.