Calculating a ParSec

I have come across several references, including the definition in
Webster's, that a parsec is about 3.26 light years. I also understand
that a parsec is the distance of an object that shows a displacement
due to parallax of one arc second when observed from opposite sides
of the Earth's orbit.

My problem is this. When I do the trig calculation using an orbital
diameter of 186E6 miles I get a value of about 3.8E13 miles, or about
6.54 light years. Why do I get twice the expected value?

Just Curious,

Mike Hogan

You have to use radius, not diameter.

My problem is this. When I do the trig calculation using an orbital
diameter of 186E6 miles I get a value of about 3.8E13 miles, or about
6.54 light years. Why do I get twice the expected value?

In article , Mike Hogan wrote:

Tom Kerr wrote:

In article , Mike Hogan
wrote:
I have come across several references, including the definition in
Webster's, that a parsec is about 3.26 light years. I also understand
that a parsec is the distance of an object that shows a displacement
due to parallax of one arc second when observed from opposite sides
of the Earth's orbit.

My problem is this. When I do the trig calculation using an orbital
diameter of 186E6 miles I get a value of about 3.8E13 miles, or about
6.54 light years. Why do I get twice the expected value?

You're using an incorrect definition for a parsec. A parsec is defined as
the distance at which 1 AU subtends an angle of 1 arcsec. You're using a
distance of 2 AU.

I agree that using 1 AU gives the expected answer, but both the
descriptions on the following page say from opposite sides of the
orbit. Are both these guys wrong?
http://www.physlink.com/Education/AskExperts/ae239.cfm

Yes, they're wrong, as your own calculation has shown. If you do a google
search for the definition of a parsec, you'll find plenty of correct
definitions.

e.g.

http://www.site.uottawa.ca:4321/astr...ex.html#parsec
http://www.twcac.org/Tutorials/Stellar%20Distance.htm
http://www.wikipedia.org/wiki/Parsec
http://sirtf.caltech.edu/EPO/Field/distance.html

Thanks for the references. It is interesting that pretty much
everyone agrees on the value, but not on the derivation. One of
the definitions in your first reference is:

"has definition The distance at which one astronomical
unit subtends an angle of one second of arc;
equivalently, the distance to an object having an annual
parallax of one second of arc. (abbreviation for parallax
second)"

I think "annual parallax" means that which is observed during one
year, or a complete revolution around the Sun. That would involve
2 AU, the diameter of the orbit. So, this definition actually is
internally conflicting. Anyway, thanks again for the clarification.

Always Curious,
Mike


Tom Kerr wrote:


I agree that using 1 AU gives the expected answer, but both the
descriptions on the following page say from opposite sides of the
orbit. Are both these guys wrong?
http://www.physlink.com/Education/AskExperts/ae239.cfm

Yes, they're wrong, as your own calculation has shown. If you do a google
search for the definition of a parsec, you'll find plenty of correct
definitions.

e.g.

http://www.site.uottawa.ca:4321/astr...ex.html#parsec
http://www.twcac.org/Tutorials/Stellar%20Distance.htm
http://www.wikipedia.org/wiki/Parsec
http://sirtf.caltech.edu/EPO/Field/distance.html

Scratch that last reply. I found a reference for "annual parallax"
and it refers to "semi-major axis" of the orbit so that is 1 AU.
It's morning, so I guess the egg on my face is appropriate.

Curious,
Mike

HI there. You posted:

I have come across several references, including the definition in
Webster's, that a parsec is about 3.26 light years. I also understand
that a parsec is the distance of an object that shows a displacement
due to parallax of one arc second when observed from opposite sides
of the Earth's orbit.

A parsec is the distance an object will have when the angular change in
its position when viewed from two different positions (at right angles
to the direction of the object) one astronomical unit apart is one arc
second (a distance of 206,264.8 Astronomical Units is one parsec or
3.26163 light years). The Astronomical Unit is roughly the mean
distance between the Earth and the sun (the more precise definition of
the A.U. involves the mean radius of an orbit which has a period of one
year).

My problem is this. When I do the trig calculation using an orbital
diameter of 186E6 miles I get a value of about 3.8E13 miles, or about
6.54 light years. Why do I get twice the expected value?

Well, you don't use the diameter of the Earth's orbit, you must use the
*radius* of the Earth's orbit to get the distance. Think of
constructing a right triangle with the sun on the corner with the right
angle, the Earth on the second one, and the distant object on the
third. Parallax figures are generally arrived at by measuring the
position of the star first on one side of the Earth's orbit and then on
the other (to get a bit larger number from the bigger baseline), and
this value is *then* divided by two. This is then the trigonometric
parallax figure listed in most catalogs. Forget about the use of miles
here, as they are much too small a unit. If you want to know the
distance to a star, take its parallax figure and invert it (ie: take the
number one and divide it by the parallax in arc seconds). The result is
the distance in parsecs (that's the beauty of this simple formula). If
an object has a listed trigonometric parallax of 0.50 arc seconds, it
will be 2.0 parsecs away (6.52 light years). Clear skies to you.
--
David W. Knisely
Prairie Astronomy Club: http://www.prairieastronomyclub.org
Hyde Memorial Observatory: http://www.hydeobservatory.info/

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"Mike Hogan" wrote...
in message ...

Scratch that last reply. I found a reference for "annual parallax"
and it refers to "semi-major axis" of the orbit so that is 1 AU.
It's morning, so I guess the egg on my face is appropriate.

Curious,
Mike

It's a good thing that eggs are tasty, for we've all had some on our
faces at one time or another, Mike.

In our imagination, we may see a giant isosceles triangle with a base
that is the change in Earth's position over six months. The top point
of the triangle is, say, a target star. Now we drop a vertical line from
the target star back to the Sun. The two identical triangles (actually
an infinite number of triangles when seen in three dimensions and
across the period of a year) each have a short leg of 1 AU.

Back up at the top of the triangles, at the target star, a larger angle
is sliced in half by the vertical line we dropped down to our Sun.
So the "parallax angle" is one-half of the total angular shift. Knowing
this parallax angle and also given the length of the short leg, 1 AU, we
can find the length of the vertical line we dropped.

And of course, if we give the parallax angle a certain value, say we
give it a value of 1 arc second, then we can use the 1AU of the short
leg to solve for the vertical line we dropped and get 206,265 AU.

As you can see, that's a rather high number of astronomical units. And
yet this is a very useful distance to astronomers, so they gave it the new
name of "one parsec." Now since people like me who are not scientists,
but who love reading about science, usually visualize astronomical
distances in "light years," most of your really good astronomy websites
will give distances in both parsecs *and* light years.

Here is yet another reference for you that uses images to explain the
trig used to find the parsec...

http://www.astronomynotes.com/starprop/s2.htm

hth

happy days and...
starry starry nights!

--
Kiss the girls and make them cry
Just like old Georgie Porgy--
In math base-2 brings Love to mind,
Base-10 turns to an Orgy!

Paine Ellsworth