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jonbosley
professor emeritus
Reged: 10/19/05
Posts: 540
Loc: Texas
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Hi,
I was hoping someone could help me work out a conversion.
With Mira Pro I measure a large white spot on jupiter to be 1.76348 arcseconds. How do I convert arcseconds to miles? After googling it I could not find any clear answers.
Thanks,
Jon
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C11 XLT Celestron
Atlas Goto Mount on Concrete Pier
WO 66 SD Doublet APO Refractor
Philips SNC900NC Webcam
Faymax FC-1000m Webcam with cooling fan
Orion Star Shoot Deep-Space CCD
POD Observatory
www.whitedwarfobservatory.com
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Charlie B
super member
Reged: 03/22/08
Posts: 164
Loc: Virginia
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Use the small angle approximation tan (theta)= sin(theta) = theta; (in radians)= D/d, where d is the distance to Jupiter and D is the distance measured. So D=d*theta; or D = d·theta(arcsec) / 206265.
Of course you need the distance to Jupiter in miles at the time you observed. See http://cfa-www.harvard.edu/astro16/docs/trig_pilman.pdf
Charlie B
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Meade SN-8, DS-90, AT-66
DSI Pro II (Schuler Filters), DSI-C, LPI, Canon XTi
AIP4WIN, Nebulosity, DSS, Registrax, GIMP
running on Dell 1420 with Vista
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jonbosley
professor emeritus
Reged: 10/19/05
Posts: 540
Loc: Texas
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I was hoping they would be a simple solution for the mathematically inept like me. Why would I need to know the distance to Jupiter at the time observed? I know that it was 44 arcseconds in size and that at the equator it is 71,492km. So why can’t I just divide its size by its arcseconds (which would give me 1624.8 km per arcsecond)? This would mean that the white spot storm was 2859km across. I may be missing the plot here but that seems a lot easier.
Jon
--------------------
C11 XLT Celestron
Atlas Goto Mount on Concrete Pier
WO 66 SD Doublet APO Refractor
Philips SNC900NC Webcam
Faymax FC-1000m Webcam with cooling fan
Orion Star Shoot Deep-Space CCD
POD Observatory
www.whitedwarfobservatory.com
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jonbosley
professor emeritus
Reged: 10/19/05
Posts: 540
Loc: Texas
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Looks like I should be using the diameter of 142,984 km but it seems I am right with the arcseconds to find feature sizes without Trig. (Thanks Lusy for the link).Thanks Charlie you got me to the right track by making me think more on distance and arcsecond size.
Jon
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C11 XLT Celestron
Atlas Goto Mount on Concrete Pier
WO 66 SD Doublet APO Refractor
Philips SNC900NC Webcam
Faymax FC-1000m Webcam with cooling fan
Orion Star Shoot Deep-Space CCD
POD Observatory
www.whitedwarfobservatory.com
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Centaur
Pooh-Bah
Reged: 07/12/04
Posts: 1123
Loc: Chicago
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I’ll try to help, Jon. First though, I can’t imagine how you made a measurement as precise as 1.76348 arcseconds. Naturally, based on the information you provided in your initial post, responders are going to give you trigonometric formulae. However, once you provided Jupiter’s diameter in both arcseconds and kilometers, you are right that simple arithmetic is sufficient. Since you want the result to be in miles, divide the 142,984 km by 1.609344 to get 88,846 miles. Then multiply that by 1.76348” and divide by 44” to get 3561 miles.
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For astronomical graphics, including
monthly wallpaper calendar, visit:
www.CurtRenz.com/astronomical.html
Curt Renz - "Centaur"
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Kobayashi
sage
Reged: 07/10/08
Posts: 291
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Quote:
Why would I need to know the distance to Jupiter at the time observed? I know that it was 44 arcseconds in size and that at the equator it is 71,492km.
If you know the apparent size and actual size, you already know the distance. Though of course you don't have to convert it to distance - the method you described is perfectly fine.
You should pay a bit more attention to significant digits though. I don't think you can measure the spot size to 4 digit accuracy...
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-- Ken Kobayashi
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jonbosley
professor emeritus
Reged: 10/19/05
Posts: 540
Loc: Texas
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Thanks -
About the spot size accuracy, Mira Pro has some very accurate measuring tools. The art is to get the correct focal length to the CCD chip with imaging setup using the plate scale, this results in the correct arc/pixel for the image. Mira Pro will measure ADU output per pixel and it is clear on a feature such as a white spot where on the graph it starts and stops (it is much brighter then the surrounding clouds). Any number of readings can be taken and then average to four digit accuracy (remember this is only for the image being measured and we are counting/measuring pixels so its easy to see how to get a four digit reading.I do believe that the result is reasonable). The function if the software is to do a lot of the more complicated mathematics for the user. Hope this helps a little.
Jon
--------------------
C11 XLT Celestron
Atlas Goto Mount on Concrete Pier
WO 66 SD Doublet APO Refractor
Philips SNC900NC Webcam
Faymax FC-1000m Webcam with cooling fan
Orion Star Shoot Deep-Space CCD
POD Observatory
www.whitedwarfobservatory.com
Edited by jonbosley (08/29/08 02:29 PM)
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Charlie B
super member
Reged: 03/22/08
Posts: 164
Loc: Virginia
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Quote:
I may be missing the plot here but that seems a lot easier.
Yes it is, but as been pointed out, the question was how to convert arcseconds to distance. However, I should have thought about the question a little more.
Charlie B
--------------------
Meade SN-8, DS-90, AT-66
DSI Pro II (Schuler Filters), DSI-C, LPI, Canon XTi
AIP4WIN, Nebulosity, DSS, Registrax, GIMP
running on Dell 1420 with Vista
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Kobayashi
sage
Reged: 07/10/08
Posts: 291
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Quote:
Any number of readings can be taken and then average to four digit accuracy (remember this is only for the image being measured and we are counting/measuring pixels so its easy to see how to get a four digit reading.I do believe that the result is reasonable).
Nevertheless, I don't believe you can measure the size to 0.001" accuracy using a telescope that has 0.3" resolution.
Assuming you are operating at about 0.1"/pixel plate scale, the size of this spot is 17~18 pixels, and the edge of the feature has 2 or 3 pixels of fuzzyness at best. Your software may calculate the size to 0.001" precision, but I'd guess the error bar would be somewhere between 0.1" and 0.3". You can test it by making multiple measurements of the object and seeing how consistent the measurements are.
Of course since you're doing this as a hobby (I presume) you don't have to care about all this detail. But if you are interested in making accurate measurements, do keep in mind that the error bar is an important part of the measurement.
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-- Ken Kobayashi
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canopus56
sage
Reged: 05/01/05
Posts: 298
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Google's search bar can be used to make a quick rough estimate the physical size of a small angle using the small angle formula.
First obtain the distance to Jupiter at the time of observation in either astronomical units or kilometers, either from your planetarium program or off the web from the NASA-JPL Solar System Simulator:
http://space.jpl.nasa.gov/
In the NASA Solar System Simulator, display the view of Jupiter as seen from Earth on the UTC date-time of your observation. The NASA JPL Solar System Simulator prints the distance to the object below the object in the simulated view display.
Assuming the observation occured at the transit of 8-28-2008 at 9UTC, the NASA Solar System Simulator gives the distance to Jupiter as 672,406,000 kilometers.
Second, at a Google page type in the formula in the Google search bar:
sin( ( [your object's axis in arcsecs] / 3600 degrees ) ) * [a.u.s to target] * 149597871 [kilometers per au]
If you know the distance to Jupiter in kilometers (e.g. the value obtained from the NASA JPL Solar System Simulator), then just use:
sin( ( [your object's axis in arcsecs] / 3600 degrees ) ) * [distance to target in kilometers]
For example for your axis of 1.76348 arcsecs, assuming the observation occured at the transit of 8-28-2008 at 9UTC:
sin( ( 1.76348 / 3600 degrees ) ) * 672,406,000
or just:
sin( ( 1.76348 arcseconds ) ) * 672,406,000
Google's built-in calculator returns: 5 748.79716 kilometers
To convert kilometers to miles using Google's built-in calculator, type the following in the Google search bar:
5748.6 kilometer in miles
Google's built-in calculator returns: 3 572.01444 miles
This estimate has a rough hobbyist level precision. On the safe side, I'd quote it as 5700 kilometers +- 15% or about 860 kilometers.
- Canopus56
Google calculator reference guides:
http://www.soople.com/soople_intcalchome.php
http://www.googleguide.com/calculator.html
Some other useful conversion examples:
Degrees to radians type:
45 degrees in radians
Arcseconds to radians type:
1.78 arcsecs in radians
Radians to degrees type:
0.785398163 radians in degrees
Clickable link to run example small angle formulae above:
http://www.google.com/search?hl=en&q=sin%28+%28+1.76348++arcseconds+%29+%29++*++672%2C406%2C000+
Clickable link to run kilometers to miles conversion:
http://www.google.com/search?hl=en&q=5748.6+kilometer+in+miles
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canopus56
sage
Reged: 05/01/05
Posts: 298
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Another simple, low precision method to take your image of Jupiter in Mira Pro and measure the number of pixels of the equatorial diameter of Jupiter as the base reference distance. Jupiter has an equatorial diameter of diameter of 142,984 kilometers and a polar diameter of 133,708 kilometers. Determine the number of kilometers per pixel for each dimension and average. Then use that metric - km per pixel - and times it by the number of pixels that transect the minor and major axis of Jovian spot. In this method, the size of features are estimated using the proportional method using the Jovian equator as the reference distance.
Neither this or the basic estimation method I described upthread is the "real" way to measure a feature size on Jupiter, Mars, Saturn or any other planet from Earth.
Neither basic estimation method corrects for the significant effect of foreshortening. Foreshortening of spots is most easily seen on lunar craters. Craters at high latitudes or at longitudes close to the limb are circular, but look like ellipses due to foreshortening.
The more mathematically intensive method to determine the size of a feature on Jupiter is to determine its apparent latitude and longitude in degrees. Then the the great circle distance formulae can be applied to estimate its size. See -
http://en.wikipedia.org/wiki/Great-circle_distance
http://www.movable-type.co.uk/scripts/gis-faq-5.1.html
http://mathworld.wolfram.com/GreatCircle.html
That is probably more math than you want to get involved with. If your spot is near the Jovian equator and is near transit, then using the proportional method is probably sufficiently accurate for your hobby purpose.
Fortunately, there is freeware that will do the work for you. Although I haven't used it, winjupos contains extensive features to perform this math for you by clicking on your image.
http://www.grischa-hahn.homepage.t-online.de/astro/winjupos/index.htm
- Canopus56
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