Jupiter and GRS size

[astroman001]

Here is an image from the UK on 30th June. Hard to find a good focus, but managed to extract some detail. Maybe (blur)  a dark flake on following end of GRS. The whole GRS well area looks very different to a month ago.

As per the guidance on the BAA website I estimated the length of the GRS with WINjupos.

CM2 longitude was 306.8 to 321.5 degrees = 14.7 degrees long x 1187 km/deg = 17449 km

Peter

[geoflewis]

Well done Peter, that’s a very nice image - shows what’s still possible from the UK.

I saw the BAA request for GRS measurements, but don’t know how to use WinJupos to do it. Were there instructions on the BAA site, if so I missed them? Presumably it’s easier to measure when the GRS is face on transiting the meridian as in your excellent image.

Cheers, Geof

[si@nite]

Some good detail showing through Peter, well done!

[astroman001]

Hi Geoff,

There were no instructions on the BAA website. Yes it is easier to do when the GRS is near the CM.

You need to use the correct file naming format from Firecapture or similar YYYY-MM-DD-TTTT for winJUPOS. Check Jupiter is the selected planet under program / celestial body 

Basically you have to load your image into the recording/image measurement tab. Align the outline with the planet in the adj tab. Use automatic detection under the outline frame tab. Then  go to the pos tab and click the cursor on either end of the GRS. The longitude will be displayed at top left.

Here is a link to some winJUPOS tutorials

http://www.grischa-hahn.homepage.t-online.de/astro/winjupos/tutorials.htm

Peter

[geoflewis]

1 hour ago, astroman001 said:

Hi Geoff,

There were no instructions on the BAA website. Yes it is easier to do when the GRS is near the CM.

You need to use the correct file naming format from Firecapture or similar YYYY-MM-DD-TTTT for winJUPOS. Check Jupiter is the selected planet under program / celestial body 

Basically you have to load your image into the recording/image measurement tab. Align the outline with the planet in the adj tab. Use automatic detection under the outline frame tab. Then  go to the pos tab and click the cursor on either end of the GRS. The longitude will be displayed at top left.

Here is a link to some winJUPOS tutorials

http://www.grischa-hahn.homepage.t-online.de/astro/winjupos/tutorials.htm

Peter

Thanks Peter,

I took a look at the tutorials, especially on measuring the GRS, but couldn't work out how to instal and use the latest CMT and Measurement files. Ignoring those I used the Pos tab under Image Measurement to measured the GRS using the earliest of my images from 28 June, i.e. when the GRS was closest to the CM and it also measured at 14.7 deg (slightly different L2s 306.5 and 321.2), but from where did you get, or how did you derive the length per degree at 1187km? The equatorial circumference of Jupiter is 439,263.8 km (NASA), hence ~1220km/deg, so your figure looks to be right given the GRS latitude, but wondering how/where you got it.

14 hours ago, astroman001 said:

CM2 longitude was 306.8 to 321.5 degrees = 14.7 degrees long x 1187 km/deg = 17449 km

Again many thanks,

Geof

[astroman001]

I got the figure from the BAA website observing challenge on their home page upper left. It covers the methodology.

Interestingly one of the images is credited to you!!

Peter

[geoflewis]

Thanks again Peter,

I need to pay more attention to what I read and yes, that thumbnail on the BAA challenge page is my image from a couple days prior to Jupiter's 2018 opposition. I saw your's posted and have just uploaded mine too with my estimated measurement of 17,449km (same as yours 👌), even though it was when the GRS was far from the CM.

Cheers, Geof