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

[marcus]

Judging from fullview2, the distance to Ceres is going down markedly. It was well over 78 and is now 76-something. Speed rel Ceres was as low as 32 and is now up to 36 mph.
The craft is thrusting towards Ceres, speeding up its descent. A couple of posts back I was estimating that by around 4 April it would have to start flipping around to point thruster at Ceres in order to slow down (so as not to arrive too fast).
When the distance is down to 54 kkm (more precisely 54.43) the angular size will be 1 degree. ...<SNIP>...
2arcsin(.475/54.43) = 1.000 degree
EDIT: Decided I should take back suggestion to change headline format.

 

[marcus]

Altitude of the closest planned orbit is 375 km which means radius of 850 km. Angular size from there is 68 degrees, so that will more than fill the 30 degree simulated view frame. I wonder if they will keep up the simulated views.
Angular widths in this earlier post were calculated according to Om's suggestion.

. The orbit altitudes are 13500, 4400, 1740, and 375 km. so adding the average radius of around 475 km, we get the orbit radii.
13975 km, 4875 km, 2215 km, and 850 km
...
2arcsin(475/13975) in degrees = 3.9 degrees
2arcsin(475/4875) in degrees = 11.2 degrees
2arcsin(475/2215) in degrees = 24.8 degrees
2arcsin(475/850) in degrees = 67.9 degrees

 

[OmCheeto]

...Anybody know what day in April they will flip it around?

...

I've been working on this, off and on, since you asked the question 7 days ago.
I got dreadfully stuck on how to add acceleration to velocity, vectorially.
I finally figured it out yesterday, but am afraid Dawn will be in RC3 orbit before I'm able to crunch all the numbers.

lpetrich's numbers are still tracking exceptionally well, and I was going to create a daily vector arrow thrust map based on the difference between his numbers, and the gravitational pull of Ceres.

 

[marcus]

Clever idea! I can see that to actually carry it out might involve too much calculation, but the concept is nice: to figure out what the thruster is doing by comparing Petrich's daily trajectory with a dead fall.

According to that "artist conception" animation flipping around is not quite the right word. Reorienting the thruster is spread over several days. So there isn't even a well defined answer to when it happens, since it is so gradual. I'm guessing they start the pivot in about a week---like 3 April.

The whole thing seems to be going nicely (to the extent that we can trust the simulated views).

If they took the trouble to put Ceres in this view it would be sitting right above the "t" in "to Sun 17.19..." In Dawn's view the planetto forms an isosceles triangle with the two shoulder stars of Orion. Speed is now mainly towards Ceres, at about 2000 km per day.

I guess if I were able to update the thread-line I would say
73 thousand km to go, speed 2000 km/day, increasing

 

[Jimster41]

I had no idea Ceres was so interesting... what's the theory as to why it's so warm (-38 F)?

 

[mfb]

lpetrich's numbers are still tracking exceptionally well, and I was going to create a daily vector arrow thrust map based on the difference between his numbers, and the gravitational pull of Ceres.

I did that about a month ago and attached the excel file in my post back then. Thrust and motion are not well aligned right now - they still have to change the orbital plane a bit (and the detour allowed to do so very well as Dawn is so slow). Starting April 4th according to the numbers, thrust will get more and more against the flight direction, one week later its acceleration is well aligned with the flight course (in opposite direction of course).

Distance 71010km, speed 52 mph (84km/h, 23m/s) in three hours.

 

[OmCheeto]

I did that about a month ago and attached the excel file in my post back then. Thrust and motion are not well aligned right now - they still have to change the orbital plane a bit (and the detour allowed to do so very well as Dawn is so slow). Starting April 4th according to the numbers, thrust will get more and more against the flight direction, one week later its acceleration is well aligned with the flight course (in opposite direction of course).

Distance 71010km, speed 52 mph (84km/h, 23m/s) in three hours.

Oh. Ummm...

I used your numbers to calculate accelerations from gravity and from Dawn.

Ok...

I see now that you did.

Ummmm...

Look over there! Squirrel!



ps. My laptop says this is the second time I've downloaded your spreadsheet.

I had no idea Ceres was so interesting... what's the theory as to why it's so warm (-38 F)?

"it"?

Ceres Surface Properties by High-Resolution Imaging from Earth
The measured temperature of the warmest area is 235 ± 4 K

And that was an old paper. October 1993?

And I've heard that interstellar space has a temperature equivalent to that of the sun.
I didn't believe it when I saw it. (It was here at PF)
I googled it.
It's true.

Temperature "classifications" are one of the billion "interesting" things that make me glad I have ADD.

 

[marcus]

I had no idea Ceres was so interesting... what's the theory as to why it's so warm (-38 F)?

Is it unexpectedly warm for 2.8 AU? I had just assumed that its surface would be at the equilibrium temperature.
Albedo is less than 10% (a dark surface)

Surface has to radiate away 1/2.8^2 of the power so I'd think the temperature would be 1/2.8^.5 of what it is at our distance.
That is about 0.6 of what it is at our distance.

I looked up http://en.wikipedia.org/wiki/Ceres_(dwarf_planet)
and it said mean surface temp was 168K

At our distance from sun (AU) without atmosphere greenhouse etc. I think the equilibrium temp is about 255, I could be wrong. So roughly, 60% of that is 153 K. around 168 K. Sort of what one would expect. Maybe the fact that it is a bit warmer has to do with the difference in albedo. Ceres absorbs over 90% of the incident sunlight.

 

[mfb]

And I've heard that interstellar space has a temperature equivalent to that of the sun.
I didn't believe it when I saw it. (It was here at PF)
I googled it.
It's true.

The equilibrium temperature for a black body as far out as Ceres has an equilibrium temperature that is significantly lower than the temperature of the sun (no matter where you measure it).

 

[Jimster41]

I just saw this line below in the wiki. I hadn't noticed that the 235K was for the max. I just thought liquid water was thought to only exist on Earth and probably on Europa (not IO, sorry), and I thought Europa was thought to be possible because of geologic activity. I guess I was just expecting utterly frozen rock - but they think it may have water?

Almost twice as far as Mars and so much smaller, less massive, I guess I thought it would be about as cold as space except for maybe the sunlit surface.

Wiki:

"The Cererian surface is relatively warm. The maximum temperature with the Sun overhead was estimated from measurements to be 235 K (about −38 °C, −36 °F) on 5 May 1991.[18] Ice is unstable at this temperature. Material left behind by the sublimation of surface ice could explain the dark surface of Ceres compared to the icy moons of the outer Solar System."
Ceres:
Surface temp. min mean max
Kelvin ? ≈ 168 K[17] 235 K[18]
Aphelion 2.9773 AU
(445410000 km)

Mars:
Surface temp. min mean max
Kelvin 130 K 210 K[6] 308 K
Celsius −143 °C[9] −63 °C 35 °C[10]
Aphelion 1.6660 AU
249.2 million km

 

[Jimster41]

At our distance from sun (AU) without atmosphere greenhouse etc. I think the equilibrium temp is about 255, I could be wrong. So roughly, 60% of that is 153 K. around 168 K. Sort of what one would expect. Maybe the fact that it is a bit warmer has to do with the difference in albedo. Ceres absorbs over 90% of the incident sunlight.

I see.

Maybe it's chock full of water, so it holds a lot of heat?

 

[OmCheeto]

I did that about a month ago and attached the excel file in my post back then. Thrust and motion are not well aligned right now - they still have to change the orbital plane a bit (and the detour allowed to do so very well as Dawn is so slow). Starting April 4th according to the numbers, thrust will get more and more against the flight direction, one week later its acceleration is well aligned with the flight course (in opposite direction of course).

Distance 71010km, speed 52 mph (84km/h, 23m/s) in three hours.

I've a Mac, and am curious if your Excel spreadsheet had more than one "tabelle".
I only got one tabelle, and the plot for the x velocity was very dirty.

The curve fit though, which I derived from the data from tab 1, shows an x direction switch on April 6th, which closely matches your April 4th remark, and Marcus's observation from the latest video.

 

[mfb]

The velocity values are from leptrich.
What is special about the x direction?

 

[OmCheeto]

The velocity values are from leptrich.
What is special about the x direction?

I think I chose the x direction, because I assumed it had the least amount of error.

Also, I'd like to convert all of the coordinates to polar notation, and rotate the entire system, as it looks as though Dawn will enter the orbital plane around April 2nd.

Which makes me very suspect of the "what a waste of xenon!" z thrust velocity graph:

Day zero = Feb 20_____________________________________________ Day 60 = April 20
​

The smoothed line crosses the y zero on ≈February 27th and ≈April 12th.I suppose the evil Dr. Rayman planned this from the very beginning, as he somehow knows I haven't translated a 3D coordinate system in my life, and it's about time I learned how.
His "Dawn Journals" are also currently offline. He posted several new comments on March 26th. One of which was an explanation of why they were powering in, rather than just coasting.
I check his journal daily.

[edit - 18 minutes later] His journal is back up. I told you he was evil...

 

[OmCheeto]

The comment I was referring to:

Tim G says:
March 26, 2015 at 6:36 am
Marc, the MYSTIC simulator indicates that Dawn’s
specific orbital energy, v^2/2 -GM/r, has been creeping up.
I noticed that it dipped below zero on March 6th when the mission
planners celebrated Dawn’s capture.
Is this related to saving time because saving xenon propellant isn’t the only issue?

Marc Rayman says:
March 26, 2015 at 11:46 am
Hi Tim,
...Rest assured, however, that Dawn is securely in orbit and is not going to leave. (So its orbital energy will remain negative.)

You are quite right that the energy is increasing... To summarize, Dawn would naturally fall toward Ceres after reaching the high point of its elliptical orbit ..., but we use the ion engine to accelerate toward Ceres even faster. We have plenty of xenon and there is no need to conserve it this late in the mission. So we reduce the time it takes to travel to lower altitude (which is part of what we have to do to reshape the orbit into RC3) by propelling the spacecraft toward it. You can also see this in the animation above of the approach trajectory. Note how the thrust does not reverse quite when Dawn reaches that maximum orbital altitude. Rather, the ion beam is still directed away from Ceres, meaning the ship is still pushing itself toward Ceres. Only after it has accelerated its descent does it thrust in the reverse direction to brake smoothly into its low, circular orbit.

Edited for brevity, and bolding for cheer factor.

 

[lpetrich]

As to OmCheeto's noisy graph, that's a problem with calculating numerical derivatives. Doing so magnifies errors. That's why I used a filter function that uses 4 neighbors instead of 2 neighbors. Another way of reducing noise is to fit the numbers to some curve, like a cubic spline, and then take derivatives of that curve.

 

[OmCheeto]

As to OmCheeto's noisy graph, that's a problem with calculating numerical derivatives. Doing so magnifies errors. That's why I used a filter function that uses 4 neighbors instead of 2 neighbors. Another way of reducing noise is to fit the numbers to some curve, like a cubic spline, and then take derivatives of that curve.

I have no idea what you just said.
But, I kind of enjoy the "schmootzig" data.

Looking at a spreadsheet of numbers doesn't really do much for me, so I convert everything into graphs, and try and figure out what the "further" smoothed graph numbers mean.

For instance, I've been plotting Dawn's speed in relation to Ceres since March 9th. The graph is very linear.

About all this tells me though, is that the acceleration of Dawn, directly towards Ceres, has been virtually constant.

Boring! (Ok. Only after I thought about it for a few hours.)

So I plotted the y intercept numbers since March 20th. There was a notable change in slope on March 24th.

I have no idea how to interpret this, other than; "Om. It's time, once again, to engage your brain".

 

[wabbit]

And I've heard that interstellar space has a temperature equivalent to that of the sun.
didn't believe it when I saw it. (It was here at PF)
I googled it.
It's true.

I found that surprising so I googled it too. Weird as it sounds, according to https://en.wikipedia.org/wiki/Interstellar_medium it's true - within the Milky Way most of the interstellar medium is at >6000K, and about half is coronal gas at >10⁶K. Clearly not at equilibrium with the CMB...
Now within the solar system in our neighborhood I suppose this translates to :
- an extremely dilute coronal gas at >10⁶K
- a gas of photons flowing out of the sun, at ~250K
And the effective temperature from averaging both would be ~250K
Is that right?

Still sounds weird that in the ISM the coronal gas would win over the photons. I suspect that the ISM temperature reported is just its matter temperature, and its average effective temperature is actually close to 3K.

 

[Jimster41]

I hope this doesn't come off as too off topic. I have been enjoying thinking about Dawn all week, just picturing that machine out there. And I for one have been mulling over the temperature problem. I'm hung up on the idea that they could possibly find liquid water and life friendly chemistry if not even life itself on an... asteroid. I hate to get excited about aliens, but my first thought was, "I hope they find an old freaking fort"

But more seriously, it's exasperating because there is my intuition, which I trust not one bit, or maybe one bit, but not enough, and I'm trying to study other topics, and realize I don't even understand... Temperature in outer space. And for me learning is about answering questions, when they occur.

So thermal energy in "space" (not including the Unruh effect etc) is driven by EM radiation from any EM sources visible. Plus if there is a gas or plasma, like the "solar wind" or the Interstellar Medium add that? BTW, I thought the Interstellar Medium didn't come into the picture until well past Pluto, at the "heliopause"? Is that wrong? So then the object's temp is a function of its reflectivity/absorptivity at the the EMR wavelength, then some degree of thermal conductivity or kinetic heat transfer efficiency at the interface to the passing warm mass (gas/plasma) and the surface area, but then also it's "specific heat" at equilibrium, which is going to correlate with density and mass to some degree, material composition, then also (and this is where I really don't trust my setup) there is the objects own "Black Body" emissivity or power. But what is the primary variable in final equilibrium? Is it only the uneven-ness of exposure to the sources (daylight/night, kinetic flux windward/leeward) or is it just going to match the sources eventually, carrying some bit of constant thermal mass. Is there a strong way to estimate final equilibrium temp?

There is that one innocuous sentence in the wiki about it being "relatively warm", but to a real planetary scientist, is Ceres Temperature interesting from the physics? Or as Marcus says, it's around expectation. Are other potential sources, like geologic activity (in an asteroid?) or some exothermic chemistry ruled out? Could the somewhat rare latent heat carrying cap of liquid water explain higher equilibrium or lower day/night cycle variance? I know that cryo-volcanism is a thing with comets, but is it common in objects with relatively stable orbits, without geologic activity. I guess I don't even know if Ceres orbit is that round.

Sorry that's a lot o questions. Just wanted to share...

 

[OmCheeto]

I found that surprising so I googled it too. Weird as it sounds, according to https://en.wikipedia.org/wiki/Interstellar_medium it's true - within the Milky Way most of the interstellar medium is at >6000K, and about half is coronal gas at >10⁶K. Clearly not at equilibrium with the CMB...
Now within the solar system in our neighborhood I suppose this translates to :
- an extremely dilute coronal gas at >10⁶K
- a gas of photons flowing out of the sun, at ~250K
And the effective temperature from averaging both would be ~250K
Is that right?

Still sounds weird that in the ISM the coronal gas would win over the photons. I suspect that the ISM temperature reported is just its matter temperature, and its average effective temperature is actually close to 3K.

This might be a counterexample to sophiecentaur's quote; "The enemy of understanding is classification."
Temperature in the ISM may need a different classification.
I learned long ago that temperature was defined as; "The average internal translational kinetic energy", or something like that.
For solids, liquids, and gasses, this kind of makes sense to me.
It's the average jigglinesss.

But in both the interstellar, and intergalactic mediums, it makes less sense to me, as it strikes me more as particles moving at ungodly speeds.

And it appears we are getting a bit off topic. I would imagine @marcus could direct us to the appropriate thread for such a discussion.
[edit]I found one. D H's comment kind of looks like what I was trying to say.

I'm currently interested in learning more about the surface temperature variations on Ceres, and if my theory* holds up.

*I haven't posted it, as it's pure speculation. It's more based on my observations, and head scratching, regarding comet 67P. But it may be relevant to Ceres.

 

[wabbit]

It's the average jigglinesss.

I think your image shows yet another kind of temperature, related to vibrations within a molecule. ISM temp is probably just ordinary gas temperature, related to the kinetic energy of the gas molecules (or atoms, or ions).

As to Ceres, is its temperature different from the one quoted by Marcus from equilibrium with solar light?

 

[OmCheeto]

I think your image shows yet another kind of temperature, related to vibrations within a molecule. ISM temp is probably just ordinary gas temperature, related to the kinetic energy of the gas molecules (or atoms, or ions).

As to Ceres, is its temperature different from the one quoted by Marcus from equilibrium with solar light?

Not sure if you noticed that I added an edit to my above post. Very interesting thread.
Marcus at one point shoved 400,000,000 photons into a box, and said that was a temperature also.
I think.

...To recap what was said earlier in the thread, space has about 400 million CMB photons per cubic meter. And those photons have a near-perfect thermal distribution, with a temperature of 2.728 kelvin.

As to your question, I'll have to look at that later. Chat trivia is starting in about 11 minutes.

 

[marcus]

According to current status (fullview2) Dawn is already beginning to flip around. Fullview shows it's ionized tail pointing more down today (down meaning Ceres south, i.e. solar system orbit plane south.)

After a few days of more "upwards" thrust then according to Om's timeline trajectory, and the somewhat impressionistic animated movie we saw, it will turn some more and have the thruster tail pointed more sunward and towards Ceres, to slow its fall.

Current status says distance is now 62 kkm and speed about 2.8 kkm per day.
Angular size must be getting close to one degree (if I remember right that occurs at 54 kkm.)

 

[marcus]

Marc Rayman has a new Dawn Journal
http://dawnblog.jpl.nasa.gov/?p=2437
dated 31 March 2015

In it he gives distance to Ceres as 57 kkm, this is closer than the estimate we got from the current status simulated views. He also notes that distance peaked out (apoapsis) at around 75 kkm on 18 March, which is also nearer Ceres than what we were seeing (which was around 78 kkm).

So we might want to mentally deduct some 3 kkm from the current status figure for a while.

An interesting note on the first circular orbit (called "RC3", after that come "survey" "HAMO" and "LAMO" successively closer in). The altitude of RC3, 13,500 km, was chosen because at that height Ceres will just fill the camera's field of view.

From the nearer orbits, Dawn's camera will not see the full disk, but only limited patches at higher resolution.

BTW RC stands for "rotation characterization". By photographing the full disk over an extended period of a few hours they can determine Ceres rotation axis. The intent is to have Dawn orbiting right over the poles.
And also to have its orbit plane veered a few degrees off the sunwards direction (clockwise seen from above north pole.) The clockwise skew is to make sure that Dawn never passes through Ceres shadow because then it would lose electric power.

The simulated views already show Dawn has stopped thrusting *towards* Ceres and is beginning to adjust orbit plane, preparing for eventual circular polar orbit, angled slightly off the sun. It is thrusting "north" and slightly aft. Sort of in an 11 o'clock direction in the simulated view frames.
http://neo.jpl.nasa.gov/orbits/fullview1.jpg
http://neo.jpl.nasa.gov/orbits/fullview2.jpg
If Ceres were added to the Sun view frame (fullview1) it would be about where the letter "a" is in "relative to Sun" at the bottom edge.
Only a sliver of the upper rim of Ceres is illuminated, from Dawn's perspective. The sun is behind and a few degrees (about 20 degrees) above Ceres.

 

[OmCheeto]

Marc Rayman has a new Dawn Journal
http://dawnblog.jpl.nasa.gov/?p=2437
dated 31 March 2015
...

It's the first thing I check for, every morning.

A couple of things I did after reading it, was go back to my spreadsheet and add the orbital revolution times for the 4 upcoming orbits, and double check if I was using "center of mass" vs "surface" for my distances for the orbits.

I immediately noticed that my HAMO numbers were off from those in today's journal.
Since January, it looks as though we've had the altitude listed as 1740 km, and it should be 1470.
Not that big a deal, as it will be the end of July before Dawn gets to that orbit.

And I'm not sure if the following differences in numbers are a change of plans, or just non-nerd PR.

April 30, 2014 Journal

Dawn will follow a tighter spiral down to the (misleadingly named) high altitude mapping orbit (HAMO) at 910 miles (1,470 kilometers).​

In todays journal, it is listed at 900 miles (1,450 km)​

 

[OmCheeto]

I have an overwhelming urge to send Dr. Rayman a bunch of questions.
But he seems to be a bit busy, and want to pick the best one.
What does everyone think?

1. Apodemeter

I don't mean to sound like Sgt. Friday, but why was Dawn communicating with the DSN between the hours of approximately:

Mar 17, 2015, 06:25 UTC: Start
Mar 18, 2015, 11: 00 UTC: Stop​

28.5 hours!
Parallax?
Doppler shift in carrier signal?​

2. HAMO altitude changed from 1470 to 1450 km in this blog.

Is this a change of plans, or just non-nerd number smoothing?​

3. Orbital plane.

When is Dawn predicted to enter it?​

4. Ice Carrots.

Is there a possibility that the headlights are "Ice Carrots"?
(Yes, we've all googled "complex impact crater morphology", but Easter is coming up!)​

 

[marcus]

... want to pick the best one.
What does everyone think?
...
3. Orbital plane.

When is Dawn predicted to enter it?​

...

I'd be most interested in knowing more about the time-table and manner of Dawn getting into the right orbit plane. the plane can't be be exactly in line with the sun direction (else Dawn would fall in shadow during orbit). It has to be turned "clockwise" (looking down on Ceres north pole) by some 5 to 7 degrees. It looks to me like this will involve a complicated maneuver---a slightly "S-shaped" approach.

As it stands, if the probe comes straight in, without that kind of weaving or zig-zag maneuver, it would get into an orbit plane that was actually turned slightly COUNTER-clockwise relative to sun direction, seen looking down along Ceres rotation axis.

BTW I see that current status (fullview2) shows the probe in communication mode this morning with thruster turned off. But when I checked DSN I did not see any antenna talking with Dawn at that time.
You mentioned doppler measurement of the carrier wave. I think that is what Rayman means by "radio-navigation". It's remarkable that they can derive useful info about where the probe is and where it's going by keeping track of radial velocity relative to Earth stations, by doppler.

Loathe to seem nitpicky about spelling but we shouldn't offend adherents of Greek&Roman pagan religion by misspelling the goddess Demeter's name. Better edit and change "dometer" to "demeter" lest you incur the wrath of the goddess.

(Yes, we've all googled "complex impact crater morphology", ...)
Nice link. The Wippykidia referenced a book
http://www.lpi.usra.edu/publications/books/CB-954/CB-954.intro.html
and I found interesting diagrams and discussion of the central uplift formation in large craters---chapter 3 page 26
http://www.lpi.usra.edu/publications/books/CB-954/chapter3.pdf

 

[marcus]

I see from current status that distance is 54 kkm which corresponds to angualar size of 1 degree.

 

[OmCheeto]

I'd be most interested in knowing more about the time-table and manner of Dawn getting into the right orbit plane. the plane can't be be exactly in line with the sun direction (else Dawn would fall in shadow during orbit). It has to be turned "clockwise" (looking down on Ceres north pole) by some 5 to 7 degrees. It looks to me like this will involve a complicated maneuver---a slightly "S-shaped" approach.
As it stands, if the probe comes straight in, without that kind of weaving or zig-zag maneuver, it would get into an orbit plane that was actually turned slightly COUNTER-clockwise relative to sun direction, seen looking down along Ceres rotation axis.

Along with capturing distance, speed, thrust status, and communication status, two to three times a day, I've also been capturing the images.
Using lpetrich's digitizer, I should be able to determine when Dawn enters the orbital plane, if it hasn't already.

BTW I see that current status (fullview2) shows the probe in communication mode this morning with thruster turned off. But when I checked DSN I did not see any antenna talking with Dawn at that time.

As of this moment, 16:11 UTC (19:11 PDT), Canberra #34 and Goldstone #15 indicate carrier signal. When I checked about an hour ago, 15:03 UTC, Canberra #34 was in standby.

You mentioned doppler measurement of the carrier wave. I think that is what Rayman means by "radio-navigation". It's remarkable that they can derive useful info about where the probe is and where it's going by keeping track of radial velocity relative to Earth stations, by doppler.

I'd like to see an article on "radio-navigation". Perhaps I'll google that later.

Loathe to seem nitpicky about spelling but we shouldn't offend adherents of Greek&Roman pagan religion by misspelling the goddess Demeter's name. Better edit and change "dometer" to "demeter" lest you incur the wrath of the goddess.

Unfortunately, my timeframe for editing has expired. I wonder, if I'm very polite, and offer them cookies, that one of the mentors would edit my post:

1. Apodometer ( <-- -1sp. Bad Om!) Apodemeter​

(Yes, we've all googled "complex impact crater morphology", ...)
Nice link. The Wippykidia referenced a book
http://www.lpi.usra.edu/publications/books/CB-954/CB-954.intro.html
and I found interesting diagrams and discussion of the central uplift formation in large craters---chapter 3 page 26
http://www.lpi.usra.edu/publications/books/CB-954/chapter3.pdf

I looked at a few "complex crater" references, but chose the wippykidia link as general reference.

I think I steered away from the others, as I think the craters on Ceres will need to have a new book written about them. To my knowledge, nobody has seen "headlights" before.

How does the central peak in moon craters form? (re: Tycho Crater on Earth's moon)

Kepler Crater - Central Peak (NASA, Earth's moon)
...
Complex craters occur above a certain diameter crater, the cutoff diameter is dependent on gravity, so it varies from planet to planet (or moon to moon).
...
Despite the label "central peak," a central peak is not always exactly in the center of a crater, nor is it always symmetrically shaped; Kepler crater is an example. Instead of having a nice central peak, Kepler crater has an irregular off-center peak. This form is most likely due to the crater being close to the boundary diameter between a simple and complex crater. Larger craters, such as King crater, can also display oddly shaped central peaks that are likely the result of an oblique impact.

Update! 16:47 UTC
Canberra #34 and Dawn are now in two way communication mode.
Goldstone #15 is in Dawn standby.

[update edit] 16:52 UTC
Goldstone #15 is no longer assigned to Dawn[/edit]

[update edit] 17:12 UTC
Just checked my spreadsheet from the last time I did "OmNav" triangulation calculations, and saw this note:

"forget it. this is impossibly hard."​

I interpret this as; "Don't hold your collective breathes, for me to figure out the entry time into orbital plane." [/edit]

[update edit] 20:32 UTC
Canberra #34 and Dawn are still in two way communication mode.
Und, viele dank, mfb.
Mein schwester wurd in Deutschland in sechs wochen. She will gladly give the cookies, for which, I will pay her for, on Dienstag.
[/edit]

 

[mfb]

I wonder, if I'm very polite, and offer them cookies, that one of the mentors would edit my post:

Cookies!
210% size of moon now (full moon in 15 hours, and the moon is close to apogee).