Error in a wikipedia illustration?

[aphh]

<p>I thought I found an obvious error in this wiki-image depicting difference in sidereal and solar time:</p><p>http://upload.wikimedia.org/wikipedia/commons/1/1d/Tiempo_sid%C3%A9reo.en.png</p><p>At the rightmost icon sun should be on the local meridian again, as 24 hrs and thus one solar day has passed. But they have accidentally used the wrong icon, that depicts one sidereal day, where the sun has not yet made a full circle.</p><p>I'm not a member of wiki, so I don't know who to inform about this error. Maybe somebody here does wiki, and could inform them.</p><p>The image belongs to wiki article about sidereal time: http://en.wikipedia.org/wiki/Sidereal_time</p><p>Edit: it turned out the image was not in error, so slight editing has been done to this post.&nbsp;</p>

 

[derekmcd]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I found an obvious error in this wiki-image depicting difference in sidereal and solar time:http://upload.wikimedia.org/wikipedia/commons/1/1d/Tiempo_sid%C3%A9reo.en.pngAt the rightmost icon sun should be on the local meridian again, as 24 hrs and thus one solar day has passed. But they have accidentally used the wrong icon, that depicts one sidereal day, where the sun has not yet made a full circle.I'm not a member of wiki, so I don't know who to inform about this error. Maybe somebody here does wiki, and could inform them.The image belongs to wiki article about sidereal time: http://en.wikipedia.org/wiki/Sidereal_timeI hi-lighted the wrong icon: &nbsp;&nbsp;&nbsp; <br /> Posted by aphh</DIV></p><p>The illustration is correct.&nbsp; It is slighly exaggerated to make a point, but it is correct nontheless.&nbsp; A sidereal day is the Earth rotating about it's axis 360 degrees.&nbsp; One full rotation.&nbsp; A Solar day is just a bit over one full rotation because the Earth is moving to the right (anti-clockwise from a top down view) in orbit around the sun.&nbsp; In order for the sun to return to it's same position in the sky, the Earth has to over-rotate slightly to compensate for the movement.&nbsp; The red dot in the illustration represents a distant star and will, essentially, remain in a fixed position despite the Earth's orbit around the sun.&nbsp; In other words, no being able to "eyeball" the parallax.</p><p>&nbsp;&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>

 

[aphh]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>The illustration is correct.&nbsp; It is slighly exaggerated to make a point, but it is correct nontheless.&nbsp; A sidereal day is the Earth rotating about it's axis 360 degrees.&nbsp; One full rotation.&nbsp; A Solar day is just a bit over one full rotation because the Earth is moving to the right (anti-clockwise from a top down view) in orbit around the sun.&nbsp; In order for the sun to return to it's same position in the sky, the Earth has to over-rotate slightly to compensate for the movement.&nbsp; The red dot in the illustration represents a distant star and will, essentially, remain in a fixed position despite the Earth's orbit around the sun.&nbsp; In other words, no being able to "eyeball" the parallax.&nbsp;&nbsp; <br /> Posted by derekmcd</DIV></p><p>Thanks for the clarification, the sun is indeed on the local meridian on the rightmost icon, whereas the star has moved on, which is factually correct and illustrates the relationship between different durations of sidereal and solar day.</p><p>Maybe admin could remove this post. </p>

 

[derekmcd]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Thanks for the clarification, the sun is indeed on the local meridian on the rightmost icon, whereas the star has moved on, which is factually correct and illustrates the relationship between different durations of sidereal and solar day.Maybe admin could remove this post. <br /> Posted by aphh</DIV></p><p>From our reference frame, the star doesn't move... It's a fixed point in the sky.&nbsp; If we start our rotation about our axis with the star at zero degrees and rotate 360 degrees, that star will be in the same spot in the sky after 360 degrees.&nbsp; That's not so with the sun.&nbsp; The sun will be off to the left a bit as we are, essentially, travelling from left to right as we move in our orbit.&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>

 

[aphh]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>From our reference frame, the star doesn't move... It's a fixed point in the sky.&nbsp; If we start our rotation about our axis with the star at zero degrees and rotate 360 degrees, that star will be in the same spot in the sky after 360 degrees.&nbsp; That's not so with the sun.&nbsp; The sun will be off to the left a bit as we are, essentially, travelling from left to right as we move in our orbit.&nbsp; <br /> Posted by derekmcd</DIV></p><p>Exactly. And this is why the Hour Angle grows as the world turns on it's axis counter-clockwise.</p><p>If I looked at a star exactly in my South, the Hour Angle would be 0H (for that moment in time). Now that the world turns counter-clockwise, the star looks like it's heading westward, but instead we're turning eastward hence making the Hour Angle grow, until one full rotation has occurred in some 23H 56' 04" duration.&nbsp;</p>

 

[DrRocket]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>From our reference frame, the star doesn't move... It's a fixed point in the sky.&nbsp; If we start our rotation about our axis with the star at zero degrees and rotate 360 degrees, that star will be in the same spot in the sky after 360 degrees.&nbsp; That's not so with the sun.&nbsp; The sun will be off to the left a bit as we are, essentially, travelling from left to right as we move in our orbit.&nbsp; <br />Posted by derekmcd</DIV></p><p>The star doesn't move -- very much.</p><p>That is the problem with trying to define a universal "inertial reference frame".&nbsp; It does move a little bit and a reference frame that is "stationary with respect to the fixed stars" is only approximately inertial.&nbsp; This comment applies to any other reference point that you might pick that is tied to anything visible.</p><p>But it is philosophically one way that some people get tripped up and start believing the myth about the existence of "absolute rest".&nbsp; There is simply no way to determine absolute rest even if it were a viable concept, which it is not.&nbsp; The biggest problem with inertial reference frames is that, so far as I can tell, there is no&nbsp;truly inertial frame.&nbsp; <br /></p> <div class="Discussion_UserSignature"> </div>

 

[aphh]

<p>If I planned a long flight from, say, Helsinki to Tokyo, would I need to account for the extra distance caused by Tokyo moving 15 degrees eastward every hour of the flight? Or is the plane moving too in the same reference frame?</p><p>I'm guessing the athmosphere, in which the plane flies, turns the same as the earth, but there may be slight difference in speeds between the ground and the upper athmosphere hence causing a bit more distance to fly when flying eastward, speed difference between the ground and the upper athmosphere growing towards the equator?&nbsp;</p>

 

[derekmcd]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>The star doesn't move -- very much.That is the problem with trying to define a universal "inertial reference frame".&nbsp; It does move a little bit and a reference frame that is "stationary with respect to the fixed stars" is only approximately inertial.&nbsp; This comment applies to any other reference point that you might pick that is tied to anything visible.But it is philosophically one way that some people get tripped up and start believing the myth about the existence of "absolute rest".&nbsp; There is simply no way to determine absolute rest even if it were a viable concept, which it is not.&nbsp; The biggest problem with inertial reference frames is that, so far as I can tell, there is no&nbsp;truly inertial frame.&nbsp; <br /> Posted by DrRocket</DIV></p><p>I completely agree hence the last sentence in my first post about not being able to "eyeball" the parallax.&nbsp; Didn't feel the need to repeat it in the second post.&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>

 

[derekmcd]

<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>If I planned a long flight from, say, Helsinki to Tokyo, would I need to account for the extra distance caused by Tokyo moving 15 degrees eastward every hour of the flight? Or is the plane moving too in the same reference frame?I'm guessing the athmosphere, in which the plane flies, turns the same as the earth, but there may be slight difference in speeds between the ground and the upper athmosphere hence causing a bit more distance to fly when flying eastward, speed difference between the ground and the upper athmosphere growing towards the equator?&nbsp; <br /> Posted by aphh</DIV></p><p>The plane is, indeed, in the same reference frame as the ground/earth.&nbsp; No accounting necessary.&nbsp; However, depending on your latitude and the direction of the overall weather system's pattern, you fly more efficiently.&nbsp; If I'm not mistaken (not a weather guy), the atmosphere as a whole does maintain it's inertial frame with the rotation of the earth, but there are variations at different latitudes where planes can take advantage of weather patterns, jet streams and such.&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>