I Dawn dead in Ceres orbit, ran out of fuel Oct 2018

[marcus]

That's right! There is what appears to be an element of luck. Ceres orbit period is 4.6 years. That means its seasons, like e.g. Winter, last on the order of a year. Dawn might have arrived e.g. in the middle of Winter when the north polar regions are in arctic darkness. The probe's limited supplies only allow it to operate for a few months. So it would be unable to map those regions in darkness. I don't know if the planet's axis inclination was determined earlier by Hubble space telescope and this went into planning the mission, or if it was just luck. Maybe the inclination is a small angle anyway, or we simply happen to be in an "equinoctial" season just now (a Ceres spring or autumn.)

Om's annotated diagram showing the approach timeline from two different perspectives is useful for reference, and since we have turned a page I will bring it forward for convenient reference. Thanks to Marc Rayman for posting the original un-annotated approach trajectories in his April 2014 Dawn Journal.

The lower diagram (looking down on Ceres north pole, sun to the left, solar orbit motion "up" in the figure) shows the probe having deficient solar orbit speed and falling behind until around 24 February when Ceres gravity causes it to start catching up. By 23 April, probe is in a circular polar orbit and beginning to pass under Ceres south pole. The orbit would not look circular in the lower diagram because we are viewing it somewhat edge-on, from the north pole direction.

According to current status page http://neo.jpl.nasa.gov/orbits/fullview2.jpg the distance to Ceres is now (as of 6 February 9AM pacific ) 125.92 thousand km.
2 arcsin(.475 /125.92) = 0.432... deg, about 86% angular size of full moon .

 

[mfb]

Ceres' axial tilt is just 3 degrees, no matter when you arrive you can map nearly everything. The Wikipedia article cites a paper from September 2005, two years before Dawn was started. Hubble could see the bright spot, so it was possible to measure its axis of rotation directly.

 

[lpetrich]

Dawn got complete coverage of Vesta just before it departed for Ceres: Spring arrives to Vesta's north pole, as Dawn departs, plus a request for citizen scientists | The Planetary Society, Dawn Mission: News & Events > Shape of Vesta

For other Solar-System objects, one can get a clue from published maps -- where do they lack detail? Planetary Visions: Texture maps

Mariner 10 flew by Mercury 3 times, but saw only half of that planet's surface. That's from a Mercury solar day being two Mercury years and Mariner 10 being in an orbit with a period of about 2 Mercury years. However, the MESSENGER Mercury orbiter has succeeded in seeing all of that planet's surface (MESSENGER: Global Mosaics of Mercury).

Venus has 98% coverage from Magellan's radar, all except near its south pole. Likewise, the Moon and Mars have nearly complete coverage.

Jupiter's and Saturn's big moons are well-covered, except for near their poles. However, Uranus's moons only had their southern hemispheres visible. That is because the only spacecraft to visit them, Voyager 2, flew by near Uranus's southern summer solstice.

 

[lpetrich]

Here are some figures on Ceres, from its Wikipedia article.

Stable orbits around an object are all inside its "Hill sphere", and Ceres's Hill sphere has a radius of about 220,000 km.

Surface-satellite orbital velocity and orbital period: 360 m/s and 2.3 hours.

Dawn is currently 122,000 km away from Ceres and traveling around 100 m/s toward it. Escape velocity from Ceres at that distance is about 32 m/s. But it's a month before Dawn has slowed down enough relative to be Ceres to be captured.

We should be getting much better values of Ceres's mass and size before long, however.

 

[mfb]

At 105000km Ceres will look as large as the moon (using its semi-major axis). 2 days to go.

Ceres' mass is known to better than 1%, getting a better measurement know looks complicated - it would need a very good distance measurement to see the small effect of gravity (relative to thrust) with an uncertainty below 1%.

 

[Dotini]

Rayed craters and possible maria make this look like our familiar Moon, IMO.
"Here be dragons -- but not for much longer!"
http://www.planetary.org/blogs/emily-lakdawalla/2015/02061616-ceres-coming-into-focus.html

NASA / JPL / UCLA / MPS / DLR / IDA / Emily Lakdawalla
Dawn's view of Ceres on February 4, 2015 (animation)

This animation consists of 20 individual frames shot as part of Dawn's third optical navigation campaign on approach to Ceres. The images have been enlarged from the original size by a factor of two, and contrast enhanced to bring out more detail.

 

[marcus]

...
...
http://www.planetary.org/blogs/emily-lakdawalla/2015/02061616-ceres-coming-into-focus.html

NASA / JPL / UCLA / MPS / DLR / IDA / Emily Lakdawalla

Beautiful job! I like Emily Lakdawalla's comments at her blog.
Dotini, could you please explain how you transferred the animation from her blog (at the link you gave) to your post here? Did you use the "media" button, with the icon that looks like some 35mm film frames. And how exactly does one proceed?

 

[Dotini]

You're going to roll your eyes when I tell you. I have a 20 year old Mac that is so obsolete I don't have cursor control even typing in this reply block, tho I do gain a cursor when I edit.

Anyway, I just highlight, copy and paste.

 

[mheslep]

I see not just scant, but zero mention of Dawn's arrival at Ceres in the mainstream newspapers, which is disappointing. Though, perhaps they're waiting for closest approach. I don't really expect a front page piece every day between now and April.

 

[marcus]

...
http://www.planetary.org/blogs/emily-lakdawalla/2015/02061616-ceres-coming-into-focus.html
Emily says Dawn is approaching Ceres "from the south". That means we should be able to see more small-circle rotation detail at the bottom of the Ceres image. I think I do. This agrees with the upper Om-figure of post #62. Going by that figure, Dawn crosses the equatorial plane of Ceres around 15 Feb.
Or let's say the orbit plane determined by direction to sun (left) and orbit direction (into the page).
But the axial tilt is only about 3 deg, so the planes are nearly parallel. Until mid-February Dawn is looking slightly from the south and sees more of the south polar region. Increasingly after mid-Feb it will be looking more from the north.
Please correct if I'm missing something. In the animation, lighting also seems stronger near the north pole, with northern regions blanked out in glare. We are told the planet's axial tilt is only 3 degrees, could the current tilt (small as it is) be towards the sun?

Current angular size: 2 arcsin(.475 / 110.67) = 0.491832996 deg (as of 9AM pacific, on 8 Feb)

Om has some more figures tabulated in post #55 but here is the table of optical navigation photo shoots (with a few extra rows added) from 29 January D.J.
http://dawnblog.jpl.nasa.gov/2015/01/29/dawn-journal-january-29/

Jan 25    (237,000)    43    (22)     1.3    96%
Feb 1     (168,000)
Feb 3     (146,000)    70    (14)     2.2    97%
Feb 10    (100,000)
Feb 12    (83,000)    121    (7.8)    3.8    98%
Feb 19    (46,000)    221    (4.3)    7.0    87%
Feb 25    (40,000)    253    (3.7)    8.0    44%
Mar 1     (49,000)    207    (4.6)    6.5    22%
Apr 10    (33,000)    304    (3.1)    9.6    18%
Apr 15    (22,000)    455    (2.1)    14     50%
Apr 23    (14,000)

Relevant column headings as given in the original table:
Distance from Dawn to Ceres in (kilometers)
Ceres diameter in pixels
Resolution in (kilometers) per pixel
Resolution compared to Hubble
Illuminated portion of disk

 

[OmCheeto]

...
http://www.planetary.org/blogs/emily-lakdawalla/2015/02061616-ceres-coming-into-focus.html
Emily says Dawn is approaching Ceres "from the south". That means we should be able to see more small-circle rotation detail at the bottom of the Ceres image. I think I do. This agrees with the upper Om-figure of post #62. Going by that figure, Dawn crosses the equatorial plane of Ceres around 15 Feb.
Or let's say the orbit plane determined by direction to sun (left) and orbit direction (into the page).
But the axial tilt is only about 3 deg, so the planes are nearly parallel. Until mid-February Dawn is looking slightly from the south and sees more of the south polar region. Increasingly after mid-Feb it will be looking more from the north.
Please correct if I'm missing something. In the animation, lighting also seems stronger near the north pole, with northern regions blanked out in glare. We are told the planet's axial tilt is only 3 degrees, could the current tilt (small as it is) be towards the sun?

Dr. Rayman, on Friday stated:

Marc Rayman says:
February 6, 2015 at 10:49 am
Tom,

You’re right that the direction of Ceres’ axis is not precisely known (yet!), but it is tipped only a few degrees from the plane of its orbit. (This might help others understand a bit more about the axial tilt.) It seems to me then that in one practical (albeit unconventional) sense, Ceres is always near equinox. That is, the sun is always near the equator, so, yes indeed, most of the surface is lit during a Cerean day. That would also mean the poles are never more than “just barely illuminated.” Astronomers have several estimates of the pole location but using one popular one, a colleague calculated for me that the sun right now is about 4 degrees south latitude and near the southern summer solstice.

I should add that Dawn is not over Ceres’ equator, so we are not seeing the northern and southern hemispheres equally. It is approaching over the southern hemisphere, and it was above (roughly) 22 degrees south latitude for OpNav 3. Dawn will cross the equator between RC1 and RC2.

Marc

My guess is that the washing out of the polar regions may be due to the lack of available light.

Current angular size: 2 arcsin(.475 / 110.67) = 0.491832996 deg (as of 9AM pacific, on 8 Feb)

Om has some more figures tabulated in post #55 but here is the table of optical navigation photo shoots (with a few extra rows added) from 29 January D.J.
http://dawnblog.jpl.nasa.gov/2015/01/29/dawn-journal-january-29/

Jan 25    (237,000)    43    (22)     1.3    96%
Feb 1     (168,000)
Feb 3     (146,000)    70    (14)     2.2    97%
Feb 10    (100,000)
Feb 12    (83,000)    121    (7.8)    3.8    98%
Feb 19    (46,000)    221    (4.3)    7.0    87%
Feb 25    (40,000)    253    (3.7)    8.0    44%
Mar 1     (49,000)    207    (4.6)    6.5    22%
Apr 10    (33,000)    304    (3.1)    9.6    18%
Apr 15    (22,000)    455    (2.1)    14     50%
Apr 23    (14,000)

Relevant column headings as given in the original table:
Distance from Dawn to Ceres in (kilometers)
Ceres diameter in pixels
Resolution in (kilometers) per pixel
Resolution compared to Hubble
Illuminated portion of disk

In an effort to figure out what "Capture by Ceres' gravity" means, I spent the last few days doing some "orbital mechanics" studying.
The only thing I've concluded, is that Newton was a FREAK!
Even with spreadsheets, the internet, and one of the worlds most powerful laptops, I have not a clue what I'm doing.

But in my feeble attempt to solve the capture problem, I discovered lots of weird and wonderful things, which is always a good thing.

About the only useful thing I can add, is my z-y plot of Dawn, as it approaches Ceres, as viewed from the Sun. (z becoming the x-axis in the 2D plot below)

Ceres is at the origin.
Position labels are date and distance in kkm from my digitization.
3-6 & 3-20 are extractions from one of Dr. Rayman's journals.

 

[marcus]

My guess is that the washing out of the polar regions may be due to the lack of available light.

Good guess! I'm glad you got a response from Dr. Rayman about the axial tilt. Just a few degrees (Wikipedium on Ceres say about 3 deg) and it is currently Southern summer--around the solstice in fact.

Om your extra interest and energetic research makes all the difference! Keeps me feeling optimistic and excited by what we are learning in this thread.

I would guess the most important criterion for "capture" is that the speed V be less than the escape velocity from the distance R.
so V < sqrt(2GM/R) would be the condition. That could very well be the equation you are working with and just don't have a closed formula for the speed V, because we don't know the thrusting program and it would be hairy to project what the speed is going to be at various distances of approach.

 

[marcus]

About capture, I measured on your annotated diagram and it looks to me like in either diagram the distance from Ceres at time of capture is 40% more than the distance on 1 March which is 49000 km. So around 6 March, capture, distance must be R = 68000 or 69000 something like that.
So we can calculate the upper limit on speed (2GM/68000 km)^.5
remembering that the mass of our planet is M = 943 billion billion kg.
Just paste this into google: (2G*943e18 kg/68000 km)^.5
When I do that, google comes back with 43 m/s
Actually it says "((2 * G * (943e18 kg)) / (68 000 km))^.5 = 43.0233559 m / s" :D

Current status says the current speed is about 100 m/s. So if there is going to be capture on 6 March then by this rough calculation the speed relative to Ceres has to come down to below 43 m/s.

I didn't measure very accurately just held a plastic ruler up to the diagrams on the screen. It was about 40% farther on 6 March than on 1 March.
Current angular size: 2 arcsin(.475 / 108.83) = 0.500148519 deg (as of 3PM pacific, on 8 Feb)

BINGO, half a degree. Full moon size!

 

[marcus]

It's always fun to see better and better pictures, but I would say that the next big landmark day for me will be FEBRUARY 24 WHEN DAWN IS 28 kkm directly North and

28 kkm directly behind Ceres in their race around the sun.

so that the hypoteneuse distance works out to 40 kkm (28²+28²=40²).

The idea is that 28 thousand km lag should be the FARTHEST BEHIND CERES THAT DAWN EVER TRAILS.

It has to cling fiercely to Ceres' coat-tails (if Grain&Fertility godesses can be said to have coat tails).
It has to stop drifting behind at that point and from that day forward begin catching up.

The combined x,y,z distance will generally be greater than that but from then on the behind distance along the Ceres orbit track has to be gradually reduced until they are neck and neck and
Dawn is safely in a circular polar orbit around the planet.

 

[OmCheeto]

Good guess! I'm glad you got a response from Dr. Rayman about the axial tilt. Just a few degrees (Wikipedium on Ceres say about 3 deg) and it is currently Southern summer--around the solstice in fact.

Om your extra interest and energetic research makes all the difference! Keeps me feeling optimistic and excited by what we are learning in this thread.

I would guess the most important criterion for "capture" is that the speed V be less than the escape velocity from the distance R.
so V < sqrt(2GM/R) would be the condition. That could very well be the equation you are working with and just don't have a closed formula for the speed V, because we don't know the thrusting program and it would be hairy to project what the speed is going to be at various distances of approach.

Well, I figured the speed had to be somewhere between the orbital and escape velocities. But one of the only relevant equations I found was for a Kepler orbit, which required a "semi-major axis", which, given my complete geometrical incompetence, is completely useless. How on Earth do you extract a "semi-major axis" from a curve segment?

I think I'll just sit back now, and wait for the pretty pictures.

 

[mfb]

Every speed below the escape velocity will give an orbit. We don't know the planned 3D motion so it is hard to estimate the speed of Dawn. I guess the speed relative to Ceres will drop below the escape velocity at that point. It will make Dawn the first object to have been in orbit around two different celestial objects (not counting Earth).

 

[marcus]

Om, However it happened we got fairly close agreement for the speed V around "capture" on 6 March. I was just roughly estimating and got 43 m/s and you got 45 m/s I think.
Whatever the speed actually is, if capture occurs that day then the speed must be just slightly LESS than escape speed (43 or 45, something around there).

 

[lpetrich]

I would guess the most important criterion for "capture" is that the speed V be less than the escape velocity from the distance R.
so V < sqrt(2GM/R) would be the condition.

Yes, that's a reasonably good approximation, though it's a spherical-cow sort of formula. Or, in this case, a spherically-symmetric asteroid all alone.

I'll estimate the sizes of the two main sources of departure from that ideal state: Ceres's equatorial bulge and the Sun. Ceres should have an equatorial bulge from its rotation, and one can estimate its size from the hypotheses of hydrostatic equilibrium and constant density.

Ceres's equatorial bulge:
$$ \text{Oblateness gravitational coefficient } J_2 \sim \text{flattening } f \sim \left( \frac{T_{SS}}{T_{rotation}} \right)^2 $$
SS = surface satellite

For Ceres, it should be about 0.06 or 1/17. To do a better job, one should use the formulas for Maclaurin spheroids, as they are called.
$$ \text{EB's relative effect } \sim J_2 \left( \frac{r_{equatorial}}{r} \right)^2 $$

The Sun:
$$ \text{Sun's relative effect} \sim \frac{M_{Sun}}{M} \left( \frac{r}{r_{to\ Sun}} \right)^3 \sim \left( \frac{T_{SS}}{T_{Sun\ orbit}} \right)^2 \left( \frac{r}{r_{equatorial}} \right)^3 $$

At 100,000 km (100 megameters):
EB: 1.4*10^(-6)
Sun: 0.026

That could very well be the equation you are working with and just don't have a closed formula for the speed V, because we don't know the thrusting program and it would be hairy to project what the speed is going to be at various distances of approach.

A good approximation ought to be constant acceleration, since Dawn's ion engines have *very* low thrust, and since they are typically run almost continuously.

 

[mfb]

The uncertainty on the mass of Ceres is much larger than the effect of the equatorial bulge.

Dawn has a constant acceleration as measured by Dawn, but not in the system of Ceres.
103400km now (90m/s relative velocity), which means Ceres appears larger to Dawn than the moon to us (using the semimajor axis).

Oh, and Ceres is a very good approximation to a spherical cow in frictionless vacuum. It does not give milk, however.

 

[OmCheeto]

Every speed below the escape velocity will give an orbit. We don't know the planned 3D motion

I kind of do.

The x-axis would be a line from the sun to Ceres.
The y-axis would be through the north pole.
The z-axis is along the line of motion of Ceres.

I derived this from plotting out the points from the published images.
Much massaging has been done, and I do not recommend piloting crafts in this manner.
The original y and z values jumped around so badly, that it looked like pinball.
Then I found a couple of buttons on my spreadsheet: Trendlines & Polynomial

so it is hard to estimate the speed of Dawn.

It will be fun to see how close I got, just by eyeballing the graphs.

I guess the speed relative to Ceres will drop below the escape velocity at that point.

I'm afraid I still don't understand how this works.
But as wiki states; "The consequences of the rules of orbital mechanics are sometimes counter-intuitive."
I have found this to be quite true.
When I first saw stated that PE tends towards zero at infinite distances, I scoffed.
Everyone here one Earth knows the equation is PE = mgh, and therefore, as h approaches infinity, so should PE.
Things are different, out in space.

It will make Dawn the first object to have been in orbit around two different celestial objects (not counting Earth).

Yay!

 

[marcus]

Since we just turned a page I will bring forward the essentials. Here's Emily Lakdawalla's version of the latest shots.
...
http://www.planetary.org/blogs/emily-lakdawalla/2015/02061616-ceres-coming-into-focus.html

NASA / JPL / UCLA / MPS / DLR / IDA / Emily Lakdawalla

Here's Om's date/distance labeled version of Marc Rayman's approach trajectory diagrams. In upper, the trajectory path is projected onto the plane normal to Ceres orbit motion. So Ceres' motion is into the page. The upper diagram does not show how the probe falls behind Ceres (almost 30 thousand km) until 25 Feb, when it begins to catch up. Because that falling behind would be out of the page. The lower diagram (which Dawn entered on the left edge today, 10 Feb) shows the path projected down on Ceres orbit PLANE, with the sun off to the left and the planet's orbit motion directly up on the page. This is the same path projected on a different plane. It shows how the probe falls behind at first by nearly 30 thousand km--I estimate 28,000 km but that remains to be seen.

Here's the link to current status:
http://neo.jpl.nasa.gov/orbits/fullview2.jpg

 

[marcus]

25 Feb is an important upcoming date because it is when the spacecraft will hopefully stop falling behind Ceres and (with Ceres' gravity helping) begin to catch up. At that time, I estimate that Dawn will be 28 kkm behind Ceres (measured on the orbit plane projection) and 28 kkm up measured on the upper diagram---the projection on the plane orthogonal to planet motion. Combining those two distances at right angles to each other gives 40 kkm, which agrees with Marc Rayman's figure for 25 Feb.

http://dawnblog.jpl.nasa.gov/2015/01/29/dawn-journal-january-29/

Jan 25    (237,000)    43    (22)     1.3    96%
Feb 1     (168,000)
Feb 3     (146,000)    70    (14)     2.2    97%
Feb 10    (100,000)
Feb 12    (83,000)    121    (7.8)    3.8    98%
Feb 19    (46,000)    221    (4.3)    7.0    87%
Feb 25    (40,000)    253    (3.7)    8.0    44%
Mar 1     (49,000)    207    (4.6)    6.5    22%
Apr 10    (33,000)    304    (3.1)    9.6    18%
Apr 15    (22,000)    455    (2.1)    14     50%
Apr 23    (14,000)

Relevant column headings as given in the original table, which listed planned navigational photo shoots. (I added three rows, without photo data, to help connect with the trajectory diagrams.)
Distance from Dawn to Ceres in (kilometers)
Ceres diameter in pixels
Resolution in (kilometers) per pixel
Resolution compared to Hubble
Illuminated portion of disk

 

[marcus]

Current status is for 11 Feb at 4AM UTC, which is 10 Feb at 8 PM pacific time. (8 hours earlier than UTC)
The distance given is 93.02 kkm (thousand kilometers).
http://neo.jpl.nasa.gov/orbits/fullview2.jpg
2 arcsin(.475 / 93.02) = 0.585... deg
117% the size of a full moon.

Those estimates I gave for 25 Feb are just rough approximations. I said 28 kkm up (off the orbit plane) and trailing 28 kkm behind (for a combined distance of 40 kkm.
But it could, for instance be a bit less up and a bit more back.
It could be say 27.1 kkm up and 29.4 kkm back

The main thing is that around 25 Feb it stops falling behind and begins to catch up to Ceres in their race around the sun, and the farthest it falls behind is something short of 30 thousand kilometers.

If you have been periodically checking the current status page, with the simulated view of Ceres seen from Dawn, you may be enjoying seeing that brown dot in the middle of the frame grow large.
The white dot you get as a cursor, on that page, is full moon size (it's a 30 degree field of view and the dot is 1/60 of the frame width). So it gives something to compare with.

 

[marcus]

Current status ( http://neo.jpl.nasa.gov/orbits/fullview2.jpg ) now shows the Dawn spacecraft in picture-taking mode--having temporarily stopped thrusting.
And having rotated so as to change orientation--point the camera I guess.
So we can expect a bunch of new photographs (perhaps in a day or two, after processing) unless they are just calibrating instruments this time.

This particular photo-shoot is designated RC1. There should be a noticeable improvement in resolution over last time (assuming we get to see the results.)

Current status says distance as of 7PM pacific on 11 Feb (3h 12Feb UTC) is 86.43 kkm
2 arcsin(.475 / 86.43) = 0.6298 deg
126% of full moon
It gives the approach speed as 90 m/s. That's around 8 thousand km a day. So it is reasonable to suppose that some of the pictures, maybe all, will be shot at the nominal range of 83 kkm .
==========
Looking ahead, on 19 Feb, and 25 Feb, range will be 46 kkm and 40 kkm so angular size will be
2arcsin(.475/46) and 2arcsin(.475/40) respectively
237% and 272% of full moon size, respectively.

 

[marcus]

Om, I recall you checked out the Deep Space Network (DSN) to see if it was receiving images from Dawn, at one point, during a past photo shoot.
I got curious and tried to do this. I saw three locations---Goldstone, Madrid, Canberra. Each location had several antennas.
http://eyes.nasa.gov/dsn/dsn.html
I see! you click on the number under each antenna and it tells you which spacecraft the antenna is receiving data from. At the moment none of the antennas were taking Dawn data.

EDIT: I checked more recently (9PM pacific time) and saw that Dawn was talking to Canberra. Also an antenna at Madrid appeared to be standing ready to take over.

Checked current status as of 5PM pacific 12 Feb, distance 80.03 thousand km
2 arcsin(.475 / 80.03) = 0.680136325 deg
136% of moon-size :D

 

[OmCheeto]

Om, I recall you checked out the Deep Space Network (DSN) to see if it was receiving images from Dawn, at one point, during a past photo shoot.
I got curious and tried to do this. I saw three locations---Goldstone, Madrid, Canberra. Each location had several antennas.
http://eyes.nasa.gov/dsn/dsn.html
I see! you click on the number under each antenna and it tells you which spacecraft the antenna is receiving data from. At the moment none of the antennas were taking Dawn data.

EDIT: I checked more recently (9PM pacific time) and saw that Dawn was talking to Canberra. Also an antenna at Madrid appeared to be standing ready to take over.

Checked current status as of 5PM pacific 12 Feb, distance 80.03 thousand km
2 arcsin(.475 / 80.03) = 0.680136325 deg
136% of moon-size :D

I checked at around 5:30 pm PST yesterday, and communication had started.
This morning, around 2 am PST, communication had by that time, switched to Madrid.
Currently, there is no communication, and I see that Dawn's ion drive is back on. (Duh!)

I should mention, that my mathematical skills have become quite rusty over the last 30 years, from dis-use. But the plotting, and mathematical magic, of Dawn's trajectory have really re-sparked my interest. Not to mention, that I've discovered 3D rendering software on my laptop.

I discovered that it can even create an animation of the above. But at 20 megabytes, it's a bit above my storage limit.

 

[mfb]

Two images from different viewing angles, switching back and forth, would be sufficient to get a nice 3D impression.

 

[OmCheeto]

Two images from different viewing angles, switching back and forth, would be sufficient to get a nice 3D impression.

Never mind. I just checked my account. I'm not even close to my limit. :)

But my latest attempt came out to 34 megabytes! It took me 8 minutes to upload.

3D Animation

 

[lpetrich]

I've measured the positions off of marcus's most recent diagram. They are as a tab-delimited spreadsheet text file that I've attached to this message.

Columns:
Date
Distance in kkm, where given
X1 raw -- first picture horizontal
Y1 raw -- first picture vertical
X2 raw -- second picture horizontal
Z2 raw -- second picture vertical

The second row is my measurement of Ceres's position in the picture

Scaling my measured positions to Dawn's distances from Ceres has proved more tricky than I expected, so I'll use Mathematica for that.

BTW, OmCheeto, what software did you use for:
- Measuring Dawn's position off of marcus's picture
- Plotting Dawn's position in 3D
?

I wrote an image measurer for myself since I couldn't find a good one that enters a position with each click on the picture being measured. It's OSX-native, so to port it to Windows or Linux, you'll need GNUstep.

 

[marcus]

Om,
Impressive bit of animation, thanks for sharing it. The figure in your post #87 rotates! I'm curious: what type of account puts a storage limit on this? Is it a PF limit, or one connected with your ISP (internet service provider)? Or has it to do with some "cloud" thing that you upload to, analogous to YouTube for videos or SoundCloud for music.