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Posted by: manishm1020 Jun 11 2010, 03:23 AM

Dear HiRISE experts,

I have obtained .tiff files of HiRISE images taken from PDS. They are all 10-bit data. I would just like to obtain the I/F for each of these images. I basically calculate the I/F from equation: I/F = (gain*DN) + offset. I use the gain and offset provided in the .LBL within PDS for each image. Now the question is - can I use the 10-bit data in the I/F calibration equation? Do I need to convert this 10-bit data to 14-bit data? If I do that, can I basically multiply each DN value by 2^4? If this is not possible, how can I obtain the original 14-bit data? If it is 10-bit data, the dynamic range of I/F for my images in the southern polar dunes (PSP_003386_1080) only spans from 0.09 to 0.2. I have read that frost/ice can approach much higher reflectance values. There must be something that I am missing. If it is 14-bit data, the dynamic range would span 0.09 to 1.5.

Any advice that you could provide, would signficantly help. This has been a bit confusing for me.

Thank you so very much.

-manish mehta

Posted by: Nirgal Jun 11 2010, 04:10 PM

The real dynamic range is actually only 8 bits. This is because of a fixed on-board lookup-table (LUT) reducing the original 14 bit to 8 bit *before* the data is sent to earth, i.e. it is lost forever.
On the other hand, the 10 bit range of the JP2 PDS products is a result of the postprocessing like photometric calibration and map projection that involves, among other things, resampling of the data, thereby kind of "smoothing" the dynamic range from 8 to 10 bit again. But still there is no more of 8 bit "real" dynamic range in a HiRISE image.

As far as the I/F conversion is concerned: it should be ok to just use the D/N values of the PDS product JP2 with the "factor" and "offset" parameters from the corresponding label file.

EDIT: yes, the low I/F value of 0.2 you reported for clear frost/ice does seem rather low ... it should be higher at least for exactly sun facing slopes... maybe it's the dust attenuation.

Hope this helps
Bernhard

Posted by: elakdawalla Jun 11 2010, 04:31 PM

Would the low I/F values you are getting result from the low angle of the Sun? The HiRISE web page for that image says the sun's only 24 degrees above the horizon.

Posted by: JohnVV Jun 11 2010, 04:34 PM

0.09 to 0.2

that is looking like that tiff is a 32bit float .I get those when i import a isis cub into openev
well a 16 bit image values in a 32 bit format
for PSP_003386_1080
http://hirise.lpl.arizona.edu/PSP_003386_1080
the jp2 remapped should have the correct values

Posted by: Nirgal Jun 11 2010, 07:34 PM

yes, with 66 degrees incident angle, a slope would have to be as steep as 66 degrees (!) in order to directly facing the sun for maximum reflectance.
whereas in reality the slopes on mars in this kind of dune landscapes rarely exceed, say 20-30 degrees so its quite plausible to see far less than maximum reflectance even in high-albedo regions when there are both shallow slopes and low sun incidence angles.