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Clarification on YZ Ceti, please
It has been pointed out to me that wikipedia gives the distance between Tau Ceti and YZ Ceti as 0.72 ly, not the much larger distance I gave (Solstation gives the distance as 1.6 ly). Can anyone out there clarify which value is correct?

If the low value is the right one, YZ Ceti exerts about 40% more force on Tau Ceti's comets than Alpha Centauri does on ours.

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I seem to have forgotten all of my highschool geometry.

How do I go from this

Star              RA                DEC          Distance 

Tau Ceti          01 44 04.1        -15 56 15    3.64
YZ Ceti           01 12 30.6        -16 59 57    3.72

to calculating the distance between those two sets of coordinates?

Re: Yes, I am an idiot

Complete algorithm care of Winchell Chung:

Re: Yes, I am an idiot

You want to convert RA and DEC to a vector in spherical coordinates.
Next, convert the spherical vector to a cartesian vector. Finally,
Determine the magnitude of the difference between the two vectors.

If you think of the Earth's rotational axis as the z-axis (with
positive z in the northward direction), then RA is a rotation about
the z-axis. We'll call this angle phi. There are 15 degrees per

The next angle, theta, is measured from +z toward the xy plane. DEC
is measured above or below the xy plane, so theta = 90 degrees - DEC.

The final spherical component, r, is just the distance. You'll want
to convert parsecs to light years.

So the spherical vectors for your two stars should be (in degrees
and light years):

Star phi theta r
Tau Ceti 26.017 105.938 11.881
YZ Ceti 18.128 106.999 12.126

To convert from spherical to cartesian:

x = r*sin (theta)*cos (phi)
y = r*sin (theta)*sin (phi)
z = r*cos (theta)

This gives you (in light years):

Star x y z
Tau Ceti 10.267 5.011 -3.262
YZ Ceti 11.021 3.608 -3.545

To get the vector from one star to the other, subtract
the vector components. Since you only want the magnitude
of the vector, it doesn't matter which way you subtract.

(YZ Ceti - Tau Ceti) -0.754 1.403 0.283

To get the magnitude of the resulting vector, take the
square root of the sum of the squares of the vector

And the answer is: 1.6 light years!

Joe Groene

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