21 replies to this topic

[rkinnett]

I managed to spot JWST last night and imaged it for awhile.  My image is nothing special compared to others', but I wanted to share that I was able to estimate range to JWST with decent accuracy... and you can too!

 

I imaged from 1/4/22 08:31 UTC through 09:47 UTC, from Los Angeles, using a C9.25 at f/7.5 and asi1600mm with pixel scale of 0.47".  Initial azimuth was 220 deg (yeah, I got a late start), 30 deg west of local meridian.

 

During that 1hr 16min period, JWST appeared to move 578 px across my camera view, relative to background sky.  That's 275 arcsec (0.00133 rad) of apparent motion.  In that period, the Earth rotated 19 degrees, shifting Los Angeles 1307 km eastward (straight line distance).  Accounting for my viewing vector being 30 degrees west of local meridian, the distance between my initial view vector and final view vector was 1132 km.  Range to target is approximated (small angle theorem) as the separation distance between the parallel viewing vectors divided by the angle of the target's apparent motion, in this case 0.00133 rad.

 

I calculate 850,000 km range.  JPL Horizons estimated 867,000 km.  Pretty darn close!   (EDIT:  see update below, in post #6, correcting several errors, with an end result not quite as good as this initial result seemed)

 

 

Edited by rkinnett, 05 January 2022 - 05:11 PM.

[happylimpet]

Brilliant! It hadnt occurred to me that the offset was largely due to the observers movement.

[matt_astro_tx]

Nice job.

[Sarciness]

I managed to spot JWST last night and imaged it for awhile. My image is nothing special compared to others', but I wanted to share that I was able to estimate range to JWST with decent accuracy... and you can too!

I imaged from 1/4/22 08:31 UTC through 09:47 UTC, from Los Angeles, using a C9.25 at f/7.5 and asi1600mm with pixel scale of 0.47". Initial azimuth was 220 deg (yeah, I got a late start), 30 deg west of local meridian.

During that 1hr 16min period, JWST appeared to move 578 px across my camera view, relative to background sky. That's 275 arcsec (0.00133 rad) of apparent motion. In that period, the Earth rotated 19 degrees, shifting Los Angeles 1307 km eastward (straight line distance). Accounting for my viewing vector being 30 degrees west of local meridian, the distance between my initial view vector and final view vector was 1132 km. Range to target is approximated (small angle theorem) as the separation distance between the parallel viewing vectors divided by the angle of the target's apparent motion, in this case 0.00133 rad.

I calculate 850,000 km range. JPL Horizons estimated 867,000 km. Pretty darn close!

jwst_parallax.jpg

jwst_parallax_calc.png

Quickly looking, the technique looks sound, but isn't the radius of Earth around 6400km?

[RedLionNJ]

Brilliant! It hadnt occurred to me that the offset was largely due to the observers movement.

Correct. This puzzled me when I was figuring out how visible it might be, a few weeks prior launch, but it turned out by day10 or so the velocity (mostly toward L2) would be around 1750 km/hr. The sideways (visible) component of this is relatively small.  So even for the purposes of Ryan's excellent math above, the 2000 or so km the JWST moved within the imaging period is still well within the margin of rounding.

 

Great work, Ryan!

[rkinnett]

I jumped the gun here.  Turns out my initial seemingly accurate result was largely coincidental.

 

Quickly looking, the technique looks sound, but isn't the radius of Earth around 6400km?

Yeah, you're right, it's 6400 km.  The 3960 figure I used is in miles.   (Can't we just be done with imperial units already?!)

 

I also neglected to account for latitude and a few other things which I incorrectly assumed were insignificant.  The additional range to JWST from high latitude compared to an equatorial observer is insignificant for these rough calculations.  However, the observer's latitude influences the straight line distance between the observer's position during the first and last observations, and that baseline distance strongly influences the range estimate, so latitude must be taken into account.  D2 (orthogonal baseline distance between initial and final viewing vectors) must also account for rotation of the Earth during the observation, not just the initial view azimuth.  This makes a small difference.

 

Here's a more detailed diagram and outline of the calculations:

 

 

This representation still makes several simplifying assumptions, including neglecting tilt of the Earth relative to the ecliptic which is okay near winter and summer solstices but potentially more influential around vernal and autumnal equinoxes.  I really don't want to work out the geometry to account for that generally.  I think I've captured the most significant factors, though I'm not too confident in it yet.  With these corrections, my estimate overshoots the official ephemeris estimates by 17%.  Not as good as I previously thought, but it's in the ballpark.

 

Example calcs:

• The Earth rotated 0.33 rad (19 deg) during this 1hr 16min observation.
• My latitude is +34 deg, so my radial distance from Earth's rotation axis ("R_O") is 5306 km.
• My initial pointing azimuth was 220 deg wrt North, or 30 deg west of local meridian.
• Since Earth rotated 19 deg during this observation, my final pointing direction was roughly 49 deg, and my mid-observation center azimuth was about 40 deg (0.69 rad) west of due south.
• My straight-line distance ("D₁") from initial topo position to final was 2*5306km*sin(0.33/2 rad) = 1751 km.
• I tracked sidereal (autoguided) during this observation, maintaining constant celestial pointing.
• The orthogonal baseline distance ("D₂") between my initial viewing vector and final was:  1751km * cos(0.69 rad) = 1351 km.
• JWST started near the center of my camera view at the start of the observation and moved 578 pixels during the observation.
• With plate scale 0.47"/px, the apparent motion of JWST relative to background sky was 578px * 0.47"/px = 275"  (0.00133 rad).
• JWST range is approximated (small angle theorem) as 1351km / 0.00133 rad = 1,015,000 km.

[rkinnett]

Correct. This puzzled me when I was figuring out how visible it might be, a few weeks prior launch, but it turned out by day10 or so the velocity (mostly toward L2) would be around 1750 km/hr. The sideways (visible) component of this is relatively small.  So even for the purposes of Ryan's excellent math above, the 2000 or so km the JWST moved within the imaging period is still well within the margin of rounding.

 

Great work, Ryan!

Thanks!  Excellent point about the inertial speed of JWST on its way out to L2 being insignificant to these calculations.  Thanks for adding that.

[freestar8n]

I used to do asteroid occultation timings and associated studies and similar geometry applies.  It involves a tilted and rotating sphere (Earth) and a distant point that may or may not be moving much relative to the center of the Earth - so different observers on the Earth may see very different speeds and directions of the asteroid as it passes over the star.

 

It is described in this pdf with 3D renderings:

 

https://smallstarspo...rdDirection.pdf

 

For the JWST I think it's motion is mostly away from the Earth, so it would correspond to the diagrams with little x, y motion rate.

 

Oh - and JWST is pretty much in the plane of the equator, so there wouldn't be much Earth tilt involved.

 

Frank

Edited by freestar8n, 05 January 2022 - 08:55 PM.

[rkinnett]

Thanks Frank for the insight.  Slight correction though, JWST is in (near) the ecliptic plane, not Earth's equatorial plane.  You are nonetheless correct that Earth's tilt does not significantly factor into these rough projective approximations.

 

My comment about accounting for Earth's tilt had to do with motion of an observer relative to the ecliptic across the duration of the observation.  An observer's motion is roughly parallel to the ecliptic around winter and summer solstices, whereas in spring and fall, the observer moves north or south relative to the ecliptic.  Looking back at Earth from L2, this is roughly what the observer's apparent motion might look like at each time of year:

 

For some reason, I thought that it might be necessary to project the viewing geometry onto the ecliptic plane, in which case the apparent N/S component of the observer's apparent motion in spring and fall would reduce the projected baseline distance by 8%.  However, I was mistaken, there is no need to project the viewing geometry into the ecliptic plane, at least not in this particular application with a single observer and relatively short duration between observations.

[freestar8n]

Thanks Frank for the insight.  Slight correction though, JWST is in (near) the ecliptic plane, not Earth's equatorial plane.  You are nonetheless correct that Earth's tilt does not significantly factor into these rough projective approximations.

 

My comment about accounting for Earth's tilt had to do with motion of an observer relative to the ecliptic across the duration of the observation.  An observer's motion is roughly parallel to the ecliptic around winter and summer solstices, whereas in spring and fall, the observer moves north or south relative to the ecliptic.  Looking back at Earth from L2, this is roughly what the observer's apparent motion might look like at each time of year:

earth_rotation_viewed_from_L2.png

 

For some reason, I thought that it might be necessary to project the viewing geometry onto the ecliptic plane, in which case the apparent N/S component of the observer's apparent motion in spring and fall would reduce the projected baseline distance by 8%.  However, I was mistaken, there is no need to project the viewing geometry into the ecliptic plane, at least not in this particular application with a single observer and relatively short duration between observations.

Thanks - that's a good point about the ecliptic.  For me right now it is near the equator at dec. around +2 degrees, but I guess it will range a fair amount from there - particularly when it is in actual orbit around L2.

 

Frank

[RedLionNJ]

Thanks - that's a good point about the ecliptic.  For me right now it is near the equator at dec. around +2 degrees, but I guess it will range a fair amount from there - particularly when it is in actual orbit around L2.

 

Frank

Given the orbits of existing probes (e.g. Planck, Herschel) around the Sun-Earth L2 point, it could reasonably be expected to vary from +23 to -23 dec over the course of a few months.

[swisstars]

Good to see all this analysis. I think there are some neat opportunities to observe the JWST. I got it on January 2 and wondered about the east west proper motion. So, this is the quick calculation that convinced me that it was moving the right direction and distance.

[RedLionNJ]

Good to see all this analysis. I think there are some neat opportunities to observe the JWST. I got it on January 2 and wondered about the east west proper motion. So, this is the quick calculation that convinced me that it was moving the right direction and distance.

Or simply, from Horizons:

 

 

Where Sky_Motion is in arcsec/minute.

[freestar8n]

I guess it makes sense that it was near the celestial equator after launch - because by launching from the Earth's equator when L2 is near the celestial equator would fling it out directly and save fuel.

 

And the 23 degree tilt of the orbit lets the scope cover the entire sky every 6 months despite being able to tilt only +/- 25 degrees.

 

Here is a nice thread on how sky coverage works during the year:

 

https://space.stacke...an-the-jwst-see

 

Frank

[matt_astro_tx]

So it won't necessarily be a reactive instrument... able to jump on targets of opportunity.  That is, unless the target is within the current FOV.  Interesting.

[freestar8n]

So it won't necessarily be a reactive instrument... able to jump on targets of opportunity.  That is, unless the target is within the current FOV.  Interesting.

From the diagrams it has a very wide view of the sky at a given time - just not all of the sky.  Tilting +/- 25 degrees and spinning 360 covers a lot.  I don't know how fast it can move, though.  And I don't know how quickly you could get a surprise target into the pipeline.

 

Frank

[dcaponeii]

From the diagrams it has a very wide view of the sky at a given time - just not all of the sky. Tilting +/- 25 degrees and spinning 360 covers a lot. I don't know how fast it can move, though. And I don't know how quickly you could get a surprise target into the pipeline.

Frank

I’m not sure that solar radiation incident on the sun shade allows for 360 degrees of rotation. If you mean rotation around the scopes pointing axis +/- 25 degree tilt then I agree but that doesn’t give you that much of the sky. That sun shield has to stay between the instruments and the sun if I understand that sun shade correctly.

[freestar8n]

I’m not sure that solar radiation incident on the sun shade allows for 360 degrees of rotation. If you mean rotation around the scopes pointing axis +/- 25 degree tilt then I agree but that doesn’t give you that much of the sky. That sun shield has to stay between the instruments and the sun if I understand that sun shade correctly.

The craft can tilt +/- 25 degrees relative to the sun, but in the other axis it can spin a full 360 degrees.  This lets it see a wide band of the sky at any given time, while keeping the shield nearly perpendicular to the sun.  And during a year it has 2 6-month periods during which it can cover the entire sky.

 

Frank

[Psion]

Two of my colleagues took JWST images 300 km apart, 10.1.2022, 21:50 UT. I use THIS algorithm. My result was 1 134 159 km +/- 2% and right distance was 1 111 000 km.

Attached Thumbnails

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[Tapio]

Cool.
But not an easy calculation to do.
Wish there was some website, or at least (Python)script to make it easier.

[rkinnett]

Cool.
But not an easy calculation to do.
Wish there was some website, or at least (Python)script to make it easier.

That would be cool.  I'll put it on my to-do list to grok the algorithm Psion mentioned and see if I can code that into a javascript web app.

[jfgout]

Cool.
But not an easy calculation to do.
Wish there was some website, or at least (Python)script to make it easier.

You can use my R code for this: https://github.com/jfgout/parallax

 

jf