A Conversation for Parsecs

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Post 1

Researcher 93445

Correct me if I'm wrong, but isn't the illustration for this entry inaccurate? I was under the impression that the second of arc was the change seen in position against the "fixed" (distant) stars from two observations taken six months apart. The illustration shows the angle encompassing the radius of the earth's orbit rather than the diameter.

You might also want to mention that an arc-second is 1/60 of 1/60 of 1/360 of a circle...


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Post 2

Jan^

I agree. The change in angle of 1 arcsecond is measured from the earth when it is either side of the sun (6 months apart, as ffmike said), not from the centre of the sun (too hot, the theodolite melts before you can take a reading smiley - smiley).


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Post 3

Jim diGriz

I thought it was wrong at first.

I think it's just an unconventional way of drawing it. Normally the parallax is shown from the point of view of Earth only. But 1 arcsecond parallax shift is the same regardless of the locations from which you measure it (General Relativity ignored conveniently smiley - smiley.

So I think it's right. The distance works out the same.

But it is a sleepy Tuesday morning.


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Post 4

Lupa Mirabilis, Serious Inquisitor

It's not the same. You have to measure it from opposite ends of the Earth's orbit because that's the baseline. I just did a lab on this.


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Post 5

Researcher 93445

Of course it's not the same. Parallax depends on the movement of the observer. If the observer moves half as far, then the same parallax translates to half the distance. So the figure is off by a factor of two.


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Post 6

Jim diGriz

Aaaarrrrrggghhhh!

Of course, you're right!

It's true that 1 second parallax is the same regardless of the locations you measure it from. But that's not the point.

I'd completely forgotten that in order for the parallax to be able to define a distance then the baseline has to be a known distance.

(Just sketched it on the back-of-an-envelope and then it was bloody obvious!)

D'oh! smiley - smiley


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Post 7

Lupa Mirabilis, Serious Inquisitor

Everything is obvious in retrospect....smiley - smiley


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Post 8

Jan^

Just checked - by my calculation a parsec is 6.52 light years.


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Post 9

Researcher 93445

Oh damn. Of course, the correct figure is 3.26 light years. So, I dug into this a bit further. It turns out that, although the measurements are taken six months apart, those craft astronomers then divide the angle by two to get a right triangle to work with. There's a pretty good discussion here: http://www.ccm.net/~jrsmith/parallax.html.

The original figure is *still* wrong...it's just wrong in a different fashion than I originally believed. It should show a right triangle with the right angle in the middle of the sun and the parallax angle at the far, pointy tip.


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Post 10

Jan^

Well done, that researcher. I had a feeling 6.52 ly was a bit big. I'll flag up the mistake with Anna and she can talk to the graphics department.


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Post 11

Spirit

tee hee hee...we're listening to you smiley - winkeye


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Post 12

Spirit

Is the new illustration any better? Distances have been left out because of space and artistic license. Thanks for pointing it out smiley - smiley


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Post 13

Researcher 93445

Good job. Thanks for listening.


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Post 14

Jan^

Outstanding service. I take back all those things I said about you smiley - winkeye
LOL smiley - fish


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Post 15

Lupa Mirabilis, Serious Inquisitor

Just one other thing: maybe it would be a little clearer if you defined "parallax" for those (like me, until recently) who don't know what it is.


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Post 16

Spirit

I'd love to, however that's an issue to take up with our wonderful editors, rather than the artists xxx ferociously yours...fierce smiley - smiley


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Post 17

Lupa Mirabilis, Serious Inquisitor

So point me toward 'em.


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Post 18

Jan^

click on the editor's name on the right hand side of the page.


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Post 19

SetupWeasel

The figure is right. A parallax is calculated by taking the total angular displacement over a diameter and dividing by two so that it fits into a right triangle with a leg that is an AU long. The right triangle thing makes the math much easier.


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