How to measure planet disk position in an image?

[municher]

Hello,

I have taken series images of jupiter and its moons, to document the orbits of the moons for some young students science project.

Now i would like to measure the positions in the image. The moons are easily detected as stars in the image but jupiter ist a 150px elliptical white blob. How can i measure the center of brightness / center of that planetary disk with Astroart?

Best regards,
Martin

[fabdev]

Hello,
I cannot imagine an automatic procedure to do that, but if it's sufficient to work on 5-10 images it could be done manually.
For example:
1) Stretch the visualization to make the edges of the planet clear, but be careful with the dark side.
2) Zoom 2X.
3) Create a rectangular or an elliptic Annotation.
4) Click "Coordinates" in the Annotation Window and place the rectangle around the planet.
5) Read the coordinates from the "Center X-Y" field.
Here is an example:

Best regards, Fabio.

[municher]

Hello Fabio,

Thanks for the proposal, but then i would directly determine the value with my borders and no calculation would happen.

Can we not add a "center of brighness" calculation to the statistics view? We already can work with rectangles as input to that view. We have the background there (to subtract during the calculation). That would seem like a likely place.

Best regards,
Martin

[fabdev]

Hello Martin,
it could be done with a simple script, see for example the second example here:
viewtopic.php?f=5&t=586

for example:

Code: Select all

x1 = Image.Rectangle.X1
y1 = Image.Rectangle.Y1
x2 = Image.Rectangle.X2
y2 = Image.Rectangle.Y2
if x1 = x2 then
  Message("Please select a rectangle")
  end
end if
bkg = Image.Background
sum = 0 : sumx = 0 : sumy = 0
for y = y1 to y2
 for x = x1 to x2
   v = Image.GetPixel(x,y) - bkg
   if v < 0 then continue
   sum = sum + v
   sumx = sumx + v*x
   sumy = sumy + v*y
 next x
next y
xc = sumx / sum
yc = sumy / sum
print "Centroid =" ; xc ; yc

By the way, since the planet has a variable brightness, this centroid will not be the geometric center of the disk (but you could "saturate" it someway, e.g. if v > 2000 then v = 2000).