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Co-Latitude - c_davidson - 03-24-2016
Would someone please explain to me what Co-Latitude is?
C. D.
RE: Co-Latitude - Fred_B - 03-24-2016
Co-latitude, from wikipedia, https://en.wikipedia.org/wiki/Colatitude
The colatitude is most useful in astronomy because it refers to the zenith distance of the celestial poles. For example, at latitude 42°N, Polaris (approximately on the North celestial pole) has an altitude of 42°, so the distance from the zenith (overhead point) to Polaris is 90 − 42 = 48°.
RE: Co-Latitude - jeremyparker - 03-27-2016
(03-24-2016, 01:21 AM)c_davidson Wrote: Would someone please explain to me what Co-Latitude is?
C. D.
CoLat is simply the complementary angle of the Latitude, (as CoDec is the complement of the Declination). A complementary angle is one that adds to 90º: that is to say, if our Latitude is 30º then the CoLat is 60º.
So where does this come into CelNav?
Conventionally, we express our North/South position as the angular distance between the Equator and our Latitude, which means that the CoLat is the angular distance between the Elevated Pole (the pole in our own hemisphere) and our Latitude: if we are at 25ºN then our CoLat is 90-25=65º.
When we solve the PZX Triangle - usually to determine the Calculated Altitude (Hc) in the intercept method - we take as our arguments the three known values: Latitude of the EP, Declination of the body, and LHA, which is the difference in Longitude between EP and GP. We can solve any triangle if we know the length of two sides and an included angle, but Lat and Dec are not technically the two sides - Lat and Dec lie outside the triangle. The sides we need are CoLat and CoDec.
All this is purely academic, however, when we actually come to solve the triangle in practice, because the Sine of an angle is equal to the Cosine of its complement: Sin 30º=0.5; Cos 60º=0.5 and so on!
If we use the Sin/Cosine formula to solve PZX we normally express it as:
Sin Hc=Cos LHA x Cos Dec x Cos Lat + Sin Dec x Sin Lat
but what we are really calculating is:
Cos ZD = Cos LHA x Sin CoDec x Sin CoLat + Cos CoDec x Cos CoLat
So we can see that although we really need CoLat and CoDec to solve the triangle - and the solution actually gives us the Cosine of the Zenith Distance (the third side) - we don't have to actually calculate these values, we simply apply the complementary term (Cos for Sin etc.) And the aCos of the ZD is the same as the aSin of the Hc!
I hope this makes sense. I've tried to keep is as unmathematical as possible. The short answer to your question is in the first paragraph; the rest is an explanation of its relevance to CelNav!
cheers
Jeremy