MSL data in the PDS and the Analyst's Notebook, Working with the archived science & engineering data |
MSL data in the PDS and the Analyst's Notebook, Working with the archived science & engineering data |
Feb 27 2013, 07:22 PM
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Solar System Cartographer Group: Members Posts: 10229 Joined: 5-April 05 From: Canada Member No.: 227 |
"February 27, 2013. MSL Release 1, part 1, Sols 0-89.
The first release of MSL data takes place in two parts. Part 1, February 27, 2013, includes raw data products (EDRs) acquired on Sols 0 through 89, August 6 through November 5, 2012, for these instruments: APXS, ChemCam, DAN, Hazcam, Navcam, and REMS, along with SPICE data. Part 2, March 20, 2013, will include the derived data products (RDRs) for Sols 0 though 89 for the APXS, ChemCam, DAN, Hazcam, Navcam, and REMS instruments, along with both the EDRs and RDRs for the CheMin and RAD instruments, and the RDRs for the SAM instrument. Release 1 does not include data from the MAHLI, MARDI, or Mastcam instruments. These instrument teams have not yet delivered data products to PDS. Some documents in the MSL archives are awaiting clearance by JPL Document Review and/or the JPL Import/Export Control Office. They will be posted online as soon as clearance has been received, and announced on this web site." Phil -------------------- ... because the Solar System ain't gonna map itself.
Also to be found posting similar content on https://mastodon.social/@PhilStooke Maps for download (free PDF: https://upload.wikimedia.org/wikipedia/comm...Cartography.pdf NOTE: everything created by me which I post on UMSF is considered to be in the public domain (NOT CC, public domain) |
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Mar 27 2013, 05:02 PM
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Senior Member Group: Members Posts: 2346 Joined: 7-December 12 Member No.: 6780 |
The paper can be read online here.
Now I understand better how the thermopiles work internally. But I didn't find an answer, how to estimate the emissivity; there is mentioned an estimate of somewhere between 0.9 and 1.0, depending on the composition of the ground. A 10% uncertainty means up to about 27K near 0°C. The influence of reflected sun light is estimated to be below 0.5%. As quotient of the Planck function for Kelvin temperature T, and two different (nu/mu)^3 x exp(4.8 x 10^-11 x [(mu-nu)/T] s K), s = second, K = Kelvin, 4.8 x 10^-11 is Planck's constant h divided by Boltzmann's constant k. (no errors assumed) This avoids emissivity by cancelling out, if it is assumed to be the same for both By taking 11 um for mean wavelength thermopile A and 18 um for mean wavelength thermopile B, one could try to avoid calculus in a first attempt. Using the brightness temperatures instead of the radiance sounds good, especially because they will already be calibrated. I'll need to think a bit, how to do it exactly. EDIT: The above formula just shows, how emissivity cancels out, but to be useful, integrals of the Planck function over the respective wavelengths have to be divided, with the same cancellation effect. Hard to avoid calculus or a numerical solution at the end. With the very rough assumptions and the formula above, and without calculus, I get T = 509 K / ln (0.228 q), with q = I(18 um) / I(11 um). Still to solve is, how to calculate I(18 um) and I(11 um) from the brightness temperatures. P.S.: In the meanwhile I found the calculus version here. (The paper uses the wavelength version of Planck's law.) |
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