Sentinel incidence angle and pass

Hi, I have examined the product from 26th August and as shown in the image below and using position of 12deg 38’16" 77deg 26’ 22" and terrain height of 695m shows a point target corresponding to your target CR1 (at pixel coordinate 10200, 12658 in IW2). This and the fainter target CR2 are both not visible in the VH polarisation as expected for trihedral corner reflectors.

Dear Peter,

thanks for the very useful information about orientation of CRs for Sentinel. I have just a doubt regarding the elevation angle. In the IWS acquisition mode, incidence angle of the SAR sensor is different in the scene depending on the position of your target of interest (i.e. steeper in near range, IW1, and shallower in far range, IW3).

https://sentinel.esa.int/web/sentinel/user-guides/sentinel-1-sar/acquisition-modes/interferometric-wide-swath

Do we have to adapt the elevation angle obtained from heavens-above accordingly? The subtraction of 35 degrees you suggested is to maximise the CR surface exposed to SAR waves? Thanks in advance

Hi, The elevation angle or satellite altitude derived from heavens-above is specific to the location of your corner reflector and the relative orbit of your acquisition. If you move your corner reflector to a different location, then you will need to use heavens-above with your new lat/long. Similarly if an acquisition is obtained over your site with a different relative orbit, then you need to use heavens-above to find a new elevation angle.

The image below shows our 2.5m corner reflector (left) and a diagram showing the corner reflector base angle (shown in black) and the satellite altitude/elevation angle (shown in red). For a triheadral corner reflector is it much easier to orientate the corner reflector in elevation using the base plate. The red line is the line through the centre of the corner reflector which is 35 deg with respect to the base plate (actually arctan(1/sqrt(2)) or 35.26 deg), as shown in blue. For example if the satellite altitude/elevation from heavens-above is 55 deg, then the base angle is 55 - 35 = 20 deg.

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Thanks a lot for the additional info!

Dear Peter,
I have done my deployment of target in the site and collected IW data of sentinel. Please let me know , is Snap tool suitable to find out the power value of pixels?
I am using matlab to process the tiff file of GRDH image, so that i could extract the power from each pixels using interpolation. Using SNAP Tool i could able to locate my reflector in the GRDH image but when i tried to open the same tiff file (file/measurement/aaa.tiff) using geotiffread it apears as a complete dark screen…
Please kindly guide me to locate the target using matlab…

Thanking you

Why don’t you locate it with the SNAP GUI and then export a subset to Matlab?

yes I could locate with snap tool. Pls find below the attachment where i ve marked my target .


But please tell me how to subset it and export to matlab without compromising in intensity?

Convert the intensity from Virtual to real band, then export that band only. You can use subset from view to cut the area you want.

Thank you for your response and your time. I am really new to image processing. Kindly clarify me, in detail about converting from virtual to real and export to matlab
Thanks in advance

When i convert slc to grd from the S1 TOPS it says can not construct data buffer. What it means? how to convert the SLC to grd data

Please consult the S1TBX and SNAP tutorials and try them out yourself. Also use the search-function in the forum to see whether your questions have already been discussed.

Hi,
Can someone, tell me how to make reflector for subsidence monitoring (i am interested only in vertical displacement), or post some pictures and link for proper literature?
Thanks

https://www.google.com/search?q=SAR+corner+reflector&tbm=isch

Hello @peter.meadows

For the orientation of azimuth, what is my reference? if i want to oriente my corner refector to this parameters:

Satellite elevation: 85 °
Satellite azimuth: 281 ° (ONO)

I need to substract 85 °-35 °= 50° for elevation but to orientate azimuth, can i use a magnetic compass to find the azimuth?

Thank you.

I found this:

would you please to share the source literature if that possible.

Hi, azimuth is with respect to true north (o deg is north, 90 deg east etc). Yes a magnetic compass can be used provided the correction for magnetic declination (https://en.wikipedia.org/wiki/Magnetic_declination) is taken into account the difference between magnetic and true/geographic north for your location.

Sure @falahfakhri

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Hello @peter.meadows

I need to orientate my corner reflector in the position (19.4731° N, 98.9890° W). I saw in Heavens Above the azimuth for my position is 260°(O), according to the table, I need to do measure 260° from W or 350° from true north, It is correct?.

For elevation, I have 35° then, I don’t need to extract anything.

Hi @CFEgildan71, Your azimuth angle of 260 deg is measured from true north (i.e. this angle is 10 deg south of due west). The (O) just indicates that 260 deg is close to west (I guess O indicates west in Spanish).
In elevation you need to subtract 35 deg from the altitude in HA for the angle of the base of your trihedral CR. So for this example, your base will be level (angle of 0 deg).