Talk:Hour angle

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Looking at the definition of the meridian it seems to be a great circle, therefore any object is going to pass through the meridian twice in a 24 hour period. I assume the hour angle is measured from when the object crosses the meridian below the celestial pole rather than above it? If this is the case the article needs to clarify this point. Htaccess 13:06, 9 May 2006 (UTC)Reply

It looks like Culmination makes this distinction, so the question become is the hour angle measured from when the lower or upper culmination occurs? Htaccess 13:32, 9 May 2006 (UTC)Reply

The article states: "For example, if an object has an hour angle of 2.5 hours, it transited across the local meridian 2.5 hours ago (measured using sidereal time), and is currently 37.5 degrees west of the meridian." Should this not be east rather than west as in the northern hemisphere objects rotate the pole in a counterclockkwise direction therefore two hours after crossing the meridian it will be to the east of the meridian and not west. Likewise in the south stars rotate clockwise round the pole therefore an object will be to the east of the meridian. In fact should compass points be used at all as the direction will be changing over time, would it not be better to use clockwise and counter clockwise to describe rotation? Htaccess 13:06, 9 May 2006 (UTC)Reply

Short answer... no. Think about it for a moment. You know that the sun, moon, and stars rise in the east, and set in the west -- so they're always moving west when they're above the horizon. Of course, there are the stars near the poles that never set, or go for months without setting, but since we're talking about the time period 2.5 hours after transiting the meridian, we can also say for sure that one of those objects is traveling west at the time. 67.95.66.69 21:59, 28 November 2006 (UTC)Reply

Isn't the hour angle defined as the angle between the local meridian plane and the plane passing through the polar axis and the star? And, as such, it is an alternative to the right ascension in the equatorial coordinate system? And then the sideral time is defined as the hour angle of the vernal point? Rlupsa 18:53, 3 October 2007 (UTC)Reply

Yes, those statements are all correct. Sidereal time is the hour angle of the vernal equinox - Jthorstensen (talk) 00:24, 8 December 2007 (UTC)Reply

Some graphical illustration might be extremely helpful. — Preceding unsigned comment added by 46.126.159.162 (talk) 10:45, 28 January 2012 (UTC)Reply

The article hour circle seems very similar to the hour angle. Should they be merged?

Sam-2727 (talk) 22:19, 30 December 2018 (UTC)Reply

In the article the formula for hour angle (HA) is given as HA = LST (local siderial time) - RA (right ascension). This formula is only correct if LST - RA is a positive number. If LST - RA is a negative number then the formula should be (if measuring in hours:min:sec): HA = LST - RA + 24. This is assuming that HA is defined as the number of hours:min:sec that the celestial object is west of the local meridian (from 0 hours to < 24 hours, and alway positive). Thus, an object having a HA of 23 hours would pass the meridian in one hour. An object having an HA of 1 hour passed the meridian an hour previously. Programs such as Stellarium define HA in this manner (always positive). Objects with HAs of >0 to <12 are west of the meridian and objects with HAs of >12 to <24 are east of the meridian. The formula presented in the article gives inconsistent and confusing results. For example, if LST = 23, and the RA of the object = 2, then from the formula in the article HA = 23 - 2 = 21, and the object will cross the meridian in 3 hours, which is correct. However, say LST = 1 and RA = 4. Then the formula in the article gives: 1 - 4 = -3, yet this should be the same HA as the previous example: ie. the object crossing the meridian in 3 hours. One might say that both answers (-3 and 21) show the same thing, but it is more consistent and less confusing to add 24 when LST - RA is negative so that the HAs are always positive and increasing in the westward direction. Defining HA this way (alway positive and increasing toward the west) is also more consistent with the graphic in the article showing the relationships of RA, LST and HA looking down from above the north pole. Let me know if this makes sense (or doesn't). — Preceding unsigned comment added by 199.91.228.217 (talk) 21:15, 10 November 2021 (UTC)Reply

Where is the Spherical Trigonometry? What you have here is kindergarten stuff. Useless! 213.233.108.137 (talk) 09:31, 17 February 2023 (UTC)Reply