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Mars Global Surveyor Image Analysis

Height of sand dunes near Ophir Chasma on Mars

by Ashley Hall

May, 2004

Abstract:

In this lab I analyzed a photo taken by the Mars Global Surveyor of an area of Mars that contained sand dunes. I found the angle that the Mars Global Surveyor was compared to the Sun and the surface. That angle was 26.89°. I used NIH Image 1.62 to slice the picture across the dune field at the angle 26.89°. I then got a brightness vs. distance graph and used it to find the average spacing of the dunes and the length of the shadow of a typical dune in this area. I then used this information, basic trigonometry, and the information given to me by the Mars Global Surveyor to determine the height of a typical dune in this area.

Analysis of the MGS photo information

Explanations and values from the MGS Ancillary Information explanation file.  

M = Mars Global Surveyor Position

N = Normal to surface at Target

S = Sun

T = center of image (Target)

s = angle between normal and spacecraft (emission angle)  = 18.05 degrees

m = angle between sun and the normal (incidence angle) = 90-solar altitude = 55.94 degrees

n = angle between spacecraft and sun (phase angle) =71.76 degrees

*Note that one pixel is equal to 4.76 meters of distance*

The main objective of this lab was to use a picture taken of Mars by the Mars Global Surveyor of an area that contained sand dunes and use the information given to me by MGS and the actual photo to slice the photo and find the height and spacing of the sand dunes.

This is the photo I analyzed. I concentrated on the upper dune field. This is located near a crack in the edge of an unnamed crater in Ophir Chasma on Mars.

Part I: Determining the direction of the sunlight in the photo.

  • First, I obtained a picture from the Mars Global Surveyor website that was taken for my teacher Mr. Adkins. The picture is called R10-04234 and the SCET time at the start of the image was 17:34:48.70 on October 24, 2003. The picture has a center longitude of 4.56 and a center aerographic latitude of 26.2°.

  • I needed to find angle M in order to know the direction of the sunlight so I know the angle my slice would have to be, so I considered this figure as seen from above:

The lines s, n, and m are arc lengths (not to scale)

  • In order to find the direction of the sunlight, I used spherical trigonometry. The relationship in spherical triangles can be used to determine an unknown angle when the arc lengths are all known as

cos (a) = cos(b)cos(c) + (sin(b)sin(c))cos (A)

for a triangle with corner angles A, B, and C and arc lengths opposite the angles labeled a, b, and c.

  • The only angle in the equation is A, so I know that is the angle I need to solve for, so I substituted the letter A with M. Since I also know that the arc lengths (lower case letters) are opposite the corners (capitol letters) that are labeled with the same letters, I can then change cos (a) to cos (m). I also substituted in the lower case letters s and n into the equation for b and c. It does not matter which letter they are substituted in for, because multiplication is commutative. I then got

cos (m) = cos(s)cos(n) + (sin(s)sin(n))cos (M)

  • My objective here is to solve for angle M, so I solved the above equation for M so I just had to plug in numbers later. I got

  • I then converted the arc lengths s, n, and m to radians.

s = .31503193 radians

m = .976337184 radians

n = 1.252448271 radians

  • After that, I plugged them into the above equation set equal to cos (M). I got .469269092 radians, but I wanted degrees. I then converted the answer using

         

 

      I concluded that angle M equals 26.89°.

 

Part II: Slicing the picture.  Slicing makes a brightness versus distance graph from the pixels along a line you draw on the picture.

  • I sliced my original picture at angle M (26.89°).

  • I used NIH Image 1.62 to do the photo analysis and slice. I then used the angle button in NIH Image to measure the angle I needed through the center of the dune field. After I had that angle, I visualized where the line was and then I clicked on the line button and drew a line through the picture where I thought the line was. I then used the angle tool to measure my line to see if the line I drew was drawn at the correct angle. I had to redo the line two or three times until I was satisfied with my result.

  • Next, I used the slice tool. I drew the slice parallel to my original line.

  • After I sliced the photo, I got a brightness vs. distance graph:

 


 

Next, I used the graph by counting the number of pixels from the top of the first dune I used to the top of the last dune I used. I counted a total of ten dunes. They were 68 pixels apart. I then divided this number by ten and found the average of these ten numbers to be 6.8 pixels or 32.37 meters. (This step is just another question, and is not needed to solve for the height of the dune)

  • Next, I used the graph and the picture to find the length of the shadow of a typical dune measured in this direction. I used my own judgment to determine the start and end of each shadow. To do this I looked for the point that the dark coloring of the shadow changed to the lighter coloring of the dunes. I did this ten times and came up with: 5, 6, 4, 3, 6, 8, 8, 7, 5, and 5. The average of these numbers is 5.7 pixels or 27.132 meters.

Part III: Finding the height of a dune.

 

This diagram shows the relationship between the observed shadow length from the perspective of the spacecraft and the actual shadow length. The line M lies on is the line of sight. Length-observed is the projection of the shadow onto a plane that is perpendicular to the line of sight. We can then conclude that angle D is a right angle. We can also conclude that the point the line angle M lies on intersects the Length-observed is also a right angle.

 

  • I used trigonometry to determine the height of the dune.

First, I substituted in what I knew. I plugged in 18.05° for angle s, 34.06° for

            90-m and 5.7 pixels for Length-observed.

I then used the complementary angle theorem to solve for angle B which is 71.95°. After that, I used the fact that the sum of all the angles inside of a triangle is 180° to solve for angle t. Angle t is 73.99°.

Next, I used the rule of supplementary angles to find the angle above t. The angle is in fact equal to t which is 73.99°. The triangle that this angle lies in is a right triangle due to similar triangles. Since I already knew two angles and I know that there are 180° inside of a triangle, I can solve for angle a. Angle a is 16.01°. I then looked at the larger triangle containing angle a. I used tangent (opposite/adjacent) to solve for length Z. Z equals 5.93 pixels. Now that I know what Z is, I can use the law of sines or sine to find the height of the dune. I concluded that the height of the dune is 3.32 pixels or 15.81 meters of distance

Conclusion: I have found that angle M is 26.89°. I then found that the average spacing of the dunes in the center of the dune field to be 6.8 pixels or 32.37 meters of distance. I then concluded that the length of the shadow of a typical dune measured in this direction is 5.7 pixels or 27.13 meters of distance. I then used this information to conclude that the height of a typical dune in this field is 3.32 pixels or 15.81 meters of distance.

Next Step: This is my major project for my junior year. If I carry it on into my senior year, I would like to try and find a relationship between the size of a crater and the number of, spacing of, and height of the dunes inside of it.

References:

  • The Mars Global Surveyor photo analysis was performed on a Macintosh computer using the public domain NIH Image program (developed at the U.S. National Institutes of Health and available on the Internet at http://rsb.info.nih.gov/nih-image/).

  • Mars dune picture courtesy of NASA/JPL/Malin Space Science Systems.

  • The picture R10-04235 can be seen at the URL http://www.msss.com/mars_images/moc/publicresults/R10data.html

Notes:

In this document, the symbol ° is degrees.


Jeff Adkins, Director
astronomyteacher@mac.com

Cheryl Domenichelli, Assistant Director
cheryldomenichelli@antioch.k12.ca.us

4700 Lone Tree Way
Antioch, CA 94531

The ESPACE Academy is sponsored in part by a grant from the California Department of Education's Specialized Secondary Program.