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Asteroid Still Not movingLast night I recorded

Ned Smith
2022-01-13
2022-01-14
  • Ned Smith

    Ned Smith - 2022-01-13

    Last night I recorded over 800 fits frames of a predicted occultation. The predicted error in time was 4 sec. I started capturing frames 2 min. before and 2 min after the predicted time. I did not see an occultaion but still I should see the asteroid reresenttion move past the target star. I don't see that. Solving the first and last frame and annotating any asteroid onto the image shows the asteroid marker in roughly the same location. It should be seen to approach the target and pass on.
    The only way I can see that this a problem at my side is if I am captureing the same image 800 times.
    I would be glad to sent you the whole folder or any part of it.

     
  • han.k

    han.k - 2022-01-13

    No problem, send me the whole set.

    How many pixels will the asteroid move in the 4 minutes?

    Han

     
  • han.k

    han.k - 2022-01-13

    I can create an log output of the calculated position in fractions of an pixel.

     
  • Ned Smith

    Ned Smith - 2022-01-13

    "How many pixels will the asteroid move in the 4 minutes?"
    Real answer, I don't know. If its position is correctly represented in the first frame, I would guess the asteroid should move roughly 600 pixels. The predictions I use are not perfect but very good.
    Scale is .594.
    How can I send you the folder. It is 4.12G. Too big for my free WeTransfer account.

     
  • han.k

    han.k - 2022-01-14

    Send me a link the first and last image only. Then I will analyse the calculated movement and actual movement of the asteroid.

     
  • Ned Smith

    Ned Smith - 2022-01-14

    Can I attach more than one frame to tjis post?
    What is your emaill address?

     
  • Ned Smith

    Ned Smith - 2022-01-14

    Here is WeTransfer link
    https://we.tl/t-TWNfXCU7sd

     
  • han.k

    han.k - 2022-01-14

    Here are the two calculated positions of the asteroid. Both the RA and DEC are in degrees. Looking to the x,y position difference is difficult due to the seeing.

    Asteroid position at :2022-01-13T01:35:25.8719596
    RA 82.8477390200269
    DEC 35.9949895469369

    Asteroid position at :2022-01-13T01:39:25.6642019
    RA 82.8472219462482
    DEC 35.9948410276734

    Looks to the deltas and compare it with the deltas you can get from e.g JPL horizon. They should be about the same.

    I would expect only see a significant move after maybe 20 minutes. These object take a few years to go 360 degrees around in the sky.

    JPL horizon
    https://ssd.jpl.nasa.gov/horizons/app.html#/

    Han

     
  • Ned Smith

    Ned Smith - 2022-01-14

    Perhaps I am misunderstanding the graphic I am seeing. Let me go through the steps i use to get there.
    Plate solvethe frame. From finder charts i believe the target is near the center of the image.
    Add marker at. RA and Dec. This puts a square at the specified point and I see it surrounds the star I thought was the target, vefiying my guess.
    Asteroid annotation: This puts the name and position of the expected asteroid on the image. I believe this is the position of the asteroid at the exposed time from the FITS header. The asteroid name is given and the right angle marks the current position, I believe.
    If the abouve is true then the last FITS frame in the series should show the asteroid name and position marker at an equal distance on the other side of the target star.
    But it dosen,t So my assumptions must be wrong.

     
  • han.k

    han.k - 2022-01-14

    Your assumptions are correct except that the travel distance is very small. If either side of the target is equal to a 1/2 pixel then you will not see it.

    I think my values in previous posting are based on an older ephemeride (dated 2021-7) I had default. So absolute values are not so accurate but it doesn't matter for the deltas.

    The delta in RA is (82.84773902 - 82.8472219462482) x 3600=1.8 arcseconds. Too small to be detectable .

    The 1.8 arcsecond travel in 4 minutes, equals 68 degrees/year. so it requires a few years to do 360 degrees. So that looks very correct for a minor planet.

     

    Last edit: han.k 2022-01-14

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