Angular width of Astronomy Picture of the Day

[W6CWJ]

I would like to see the angular width of the APOD in arc seconds posted along with the photo.

Alooooha

Jerry
Maui

[apodman]

I would like to see the angular width of the APOD in arc seconds posted along with the photo.

Jerry
Maui

For example:

M76 Above and Below (Little Dumbbell Nebula) - APOD 2008 November 21

http://antwrp.gsfc.nasa.gov/apod/ap081121.html

Distance estimates place M76 about 3 to 5 thousand light-years away toward the heroic constellation Perseus, making the nebula over a light-year in diameter.

Okay, until someone else starts doing my homework for me, I whip out my calculator from my handy desk drawer.

1 ly divided by 3,000 ly is 1/3000.
Take the arctan or arcsin (close enough) of 1/3000 in degrees.
Multiply by 3600 to convert degrees to arc seconds.

1 ly divided by 5,000 ly is 1/5000.
Take the arctan or arcsin (close enough) of 1/5000 in degrees.
Multiply by 3600 to convert degrees to arc seconds.

So M76 is about 40 to 70 arc seconds in diameter.

Not every APOD description provides enough information to do this calculation, either, but if it's there it's easy enough to use. If your calculator is not solar and the battery is dead, I recommend referring to the Messier place mats (the ones with the column for angular size) available at the Dollar Store in Roswell, NM (but they close at 5:00). Sorry, Jerry (and welcome to the forum, too) - that's for mainlanders - you can get the same thing (and the matching shower curtain) at the Keck gift shop.

Of course, angular size is measurable while the other two variables (actual size and distance) are educated guesses. So it's backwards to calculate a measurable quantity from the guesses (actually one guess determines the other - distance vs. size - once you have the angular measurement), but in this case we could only work with what we were given - hence an answer ranging from 40 to 70 arc seconds instead of a single number.

[Chris Peterson]

Okay, until someone else starts doing my homework for me, I whip out my calculator from my handy desk drawer.
...
So M76 is about 40 to 70 arc seconds in diameter.

Or, working from the image itself: the plate scale is 1.4"/pixel. So the image is 18.7 arcmin by 14 arcmin, and the nebula is 4.7 arcmin by 2.5 arcmin- quite a bit larger than you get doing the calculation from sizes and distances.

[apodman]

So here's what I find in a fast search:

42.85"

2.7' x 1.8' (= 162" x 108" - apodman)

2.7' x 1.8' (= 162" x 108" - apodman)

and:

The bright bar-shaped main body (measuring 42" × 87"), is probably a slightly elliptical ring seen edge-on from only a few degrees off its equatorial plane. ... Along the axis perpendicular to the ring plane, the gas is moving out more rapidly to form lower surface-brightness wings (157" × 87"). Finally, there is a faint halo covering a region about 290".

That's quite a range of numbers. I'm not sure what to compare with what.

---

Anyway, here's the Messier angular size chart without having to buy place mats or a shower curtain:

http://www.astronomy-pictures.com/Messi ... s.htm#data

[Chris Peterson]

That's quite a range of numbers. I'm not sure what to compare with what.

The short answer is you can't compare these values. Most deep sky objects are irregular and diffuse. Many of the published values for size go back to visual estimates, but images show a far greater extent. Whoever came up with a set of numbers made an arbitrary decision on both where the edges were, and what direction or directions to measure.

The rather deep image in the APOD, and the rectangular shape, makes it pretty easy to measure this object without requiring a lot of subjective decisions. M76 actually is object-like. There are other nebular targets that just seen to get larger as you look deeper, and defining a reasonable size for them is nearly impossible. But for any given image, you can usually come up with a reasonable size- as imaged- by just measuring the number of pixels across it and multiplying by the plate scale.

[NoelC]

I would like to see the angular width of the APOD in arc seconds posted along with the photo.

The angular width of an image is unambiguous. It has nothing to do with the size of the object. It has to do with the field of view of the camera. The sky is divided up into a coordinate system. The angular width and height can be determined by finding the distance from the coordinates of one corner to the coordinates of the opposite corner.

The angular width, for example, of the APOD shown on http://antwrp.gsfc.nasa.gov/apod/ap081121.html is just about 14 x 19 arc-minutes.

I use a freeware sky chart program called Cartes du Ciel that allows me to overlay a camera field of view. It's a simple matter to define a test field of view, compare it with an image, and iterate until one has determined the angular size - i.e. when the red outline projected on the chart matches the image.

There's also a neat site out there called Astrometry.net to which you can submit a photo, and it will find it on its sky charts and return info to you.

-Noel