Last night I took this photo where the shadow was pretty good on Mons Huygens, the highest mountain in the Montes Apenninus Mountain Range at 18,046 feet. So, the question comes to mind how elevations are referenced on the moon. I found a few Internet sites, but nothing that seems to really spell it out clearly where the Zero elevation is referenced.
Mean Sea (of Tranquility) Level
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Here's a description of NASA's lunar ellipsoid. It's all Geek to me...
ask Ralph. I assume he issued instructions
(cool photo)
Nice brief document.
Latitude longitude are not different than what we use. Prime meridian is a little funny. "0?ø Longitude is the "mean sub-earth point", which does not pass through any single prominent feature."
As for height/elevation the only reference I saw was height above mass center, and it's called radius. "If necessary, a reference sphere radius of 1737.4 Km may be subtracted to avoid large number rounding error."
Something like that. It appears that the LRO project is the one that has most data, and therefore most input. But. .. the data is ascii so they can adapt if required.
Check out this page: https://astrogeology.usgs.gov/search/map/Moon/LRO/LOLA/Lunar_LRO_LOLA_Global_LDEM_118m_Mar2014
paden cash, post: 444232, member: 20 wrote: Here's a description of NASA's lunar ellipsoid. It's all Geek to me...
And there's no apparent reference to an ellipsoid. The term used is sphere. If it's close enough to sphere, go with sphere.
Larry Scott, post: 444275, member: 8766 wrote: And there's no apparent reference to an ellipsoid. The term used is sphere. If it's close enough to sphere, go with sphere.
There's a link to a rather large pdf buried in that pub:
http://ssd.jpl.nasa.gov/dat/lunar_cmd_2005_jpl_d32296.pdf
On the doc page (about) 18...printed page 4, I think, there are models mentioned with the Lunar Geoid LP150Q with references to the variation of geoid to what they consider the "mean lunar surface".
edit: these are things we may need to know to successfully complete LOMRs on the moon....;)
paden cash, post: 444280, member: 20 wrote: There's a link to a rather large pdf buried in that pub:
http://ssd.jpl.nasa.gov/dat/lunar_cmd_2005_jpl_d32296.pdf
On the doc page (about) 18...printed page 4, I think, there are models mentioned with the Lunar Geoid LP150Q with references to the variation of geoid to what they consider the "mean lunar surface".
edit: these are things we may need to know to successfully complete LOMRs on the moon....;)
They list polar and equatorial radii being equal, a f: = 0.00. Sphere not spheroid. There are significant gravity anomalies, so running spirit levels may be difficult. And that would make for a distinct geoid as we think of it. But, there's not a lot of gravity drainage to worry about. Without gps inertial navigation isn't as accurate.
The first doc listed in the orig post has a brief overview. But not different than the doc you listed, just no detail.
The doc you listed ..,
Section 2.3, pg 22:
RMoon-Equator = 1737.4 km
RMoon-Pole fMoon= 1737.4 km
Flattening = 0.0
(Radius of Moon equator from IAU/IAG)
(Radius of Moon pole from IAU/IAG)
(Moon flattening factor, derived from IAU/IAG values ( (RMoon-Equator - RMoon-Pole)/ RMoon-Equator) )
As with all things, everything is a compromise.
Doesn't mean there isn't a "geoid", just that it's either for future, or higher level science.
I cut and paste...
2.3 Moon Shape Parameters
The general shape of the Moon is very nearly a perfect sphere, excluding local topography variations. In fact, the magnitude of the local topography variations are much larger than the overall flattening of the lunar poles or any ellipticity of the lunar equator. Based upon the data presented in Reference 7, the magnitude of the lunar flattening relative to the equatorial radius is about 2 km. This memo, however, recommends representing the overall shape of the Moon for any mission design and navigation analyses as a perfect sphere for the following reasons.
First, the IAU/IAG 2000 Report recommends using the same radius value for both the lunar equator and the lunar pole (i.e. a sphere). Second the lunar topography data, discussed in Section 2.6.1, are expressed relative to a sphere with the same radius as recommended by the IAU/IAG 2000 Report. The values in the IAU/IAG 2000 Report are listed below
I realize the term "geoid" and "ellipsoid" are terrestrial terms, not lunar. I was merely relating their numbers to terms us earthlings are familiar with. And as for our scientific community, their sphere definition will suffice for now. At least until one of us surveyors sets foot there and turns all their good terms and numbers on their heads. 😉
paden cash, post: 444292, member: 20 wrote: I realize the term "geoid" and "ellipsoid" are terrestrial terms, not lunar. I was merely relating their numbers to terms us earthlings are familiar with. And as for our scientific community, their sphere definition will suffice for now. At least until one of us surveyors sets foot there and turns all their good terms and numbers on their heads. 😉
Ellipsoid describes all the of the planets.
Wiki ....
Because of the combined effects of https://en.m.wikipedia.org/wiki/Gravity&apos ;">gravity and https://en.m.wikipedia.org/wiki/Rotation_of_the_Earth&apos ;">rotation, the https://en.m.wikipedia.org/wiki/Earth%27s_shape&apos ;">shape of the Earth, and of all https://en.m.wikipedia.org/wiki/Planet&apos ;">planets, is not quite a sphere but instead is slightly flattened in the direction of its axis of rotation.
Larry Scott, post: 444295, member: 8766 wrote: Ellipsoid describes all the of the planets.
Wiki ....
Because of the combined effects of https://en.m.wikipedia.org/wiki/Gravity&apos ;">gravity and https://en.m.wikipedia.org/wiki/Rotation_of_the_Earth&apos ;">rotation, the https://en.m.wikipedia.org/wiki/Earth%27s_shape&apos ;">shape of the Earth, and of all https://en.m.wikipedia.org/wiki/Planet&apos ;">planets, is not quite a sphere but instead is slightly flattened in the direction of its axis of rotation.
To make matters worse, the Earth also tugs on ol' La Luna and actually gives it an egg shaped quality..with the large end of the "egg"facing, of course, the Earth.
paden cash, post: 444297, member: 20 wrote: To make matters worse, the Earth also tugs on ol' La Luna and actually gives it an egg shaped quality..with the large end of the "egg"facing, of course, the Earth.
The gravity (geoid) of the moon is quite irregular due to a solid core and meteor impacts, and tidal locked with the earth. The term used in the docs is 'masscon' (mass concentrations) in proximity to major craters.
I guess if it had a molten core, and not tidal locked, it would smooth out.
I took these photos last night of the sun striking the tops of the peaks of the Montes Jura range which makes them appear to the left of the shadow line. The images were taken about 2 hours apart.
Beautiful photos. What kind of camera are you using?
I read an article about a similar issue comparing the largest volcano in the solar system (on Mars) with earth based mountains. It is about three times higher than Everest but where do you measure from? The plain below, some Martian datum?
Dave Karoly, post: 444558, member: 94 wrote: I read an article about a similar issue comparing the largest volcano in the solar system (on Mars) with earth based mountains. It is about three times higher than Everest but where do you measure from? The plain below, some Martian datum?
The old question: tall or high?
Mars elevation datum:
"...On Earth, the zero elevation datum is based on https://en.m.wikipedia.org/wiki/Sea_level&apos ;">sea level. Since Mars has no oceans and hence no 'sea level', it is convenient to define an arbitrary zero-elevation level or "https://en.m.wikipedia.org/wiki/Datum_(geodesy)&apos ;">datum" for mapping the surface. The datum for Mars is arbitrarily defined in terms of a constant atmospheric pressure. 610.5 Pa
In 2001, https://en.m.wikipedia.org/wiki/Mars_Orbiter_Laser_Altimeter&apos ;">Mars Orbiter Laser Altimeter data led to a new convention of zero elevation defined as the https://en.m.wikipedia.org/wiki/Equipotential_surface&apos ;">equipotential surface (gravitational plus rotational) whose average value at the equator is equal to the mean radius of the planet.https://en.m.wikipedia.org/wiki/Geography_of_Mars#cite_note-5&apos ;">[5] ..."
Peter Ehlert, post: 444544, member: 60 wrote: Beautiful photos. What kind of camera are you using?
Canon Powershot SX530 HS
J. Penry, post: 444628, member: 321 wrote: Canon Powershot SX530 HS
you got me thinking... I would love to capture many things that my old cameras and cell phone can't grab.
I searched eBay, Amazon, etc. and that is now an affordable camera. That surprised me, a lot.
Question: If you were needed to replace it would you get another the same, or some other brand or model?
... price point being less than $300.