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To see that you can use hour angle and LHA equally well, you just have 
to look at H.O. 249. Look at the LHA columns at each edge of the page 
and you will see that you get the same Hc for two values of LHA. Look at 
LHA 10 and LHA 350 on this example page:

http://www.fer3.com/arc/img/106576.2007-page%20152.pdf

LHA 10 is the same as hour angle 10 west., LHA 350 is the same as hour 
angle 10 east.

This relationship is not so evident if you just look at H.O 229 due to 
the arrangement of the tables.
gl


Gary LaPook wrote:
> Remember, LHA is in the range of 0� to 360� and is always measured to 
> the west from the assumed longitude (ALon). It is the angle measured 
> westward from the meridian of the ALon to the meridian containing the 
> body's grographic position (GP). There is no such thing as easterly LHA.
>
> In the past, various computations methods and tables (e.g H.O. 214) used 
> "hour angle," (abbreviated "H.A." or "H" or "t"), which is the angle 
> measured between the meridian of the ALon and the meridian containing 
> the body's GP measured the shortest way, either west or east. Using this 
> notation, hour angle ends up in the range of 0� to 180� only and is 
> denoted "east" or "west." Because of the the way the trig formulas work, 
> using either method computes the same Hc and the same "azimuth angle" 
> ("Az" or "Z"). The only thing affected by choice of notation is the 
> method used for the final determination of Zn,(azimuth used for plotting 
> the LOP.)
>
> The original Bygrave used hour angle, not LHA, and the scales were 
> marked from 0� to 90� and then back the other way, 90� to 180�. My 
> implementation of the Bygrave eliminated the second set of markings on 
> the scale, 90�-180�, to eliminate clutter so I added an extra step to 
> bring hour angle into the range of 0� to 90� only and provided the 
> necessary rules for the final computation of Zn. This is what is 
> happening on the top of the form I provided. If LHA is less than 90�, H 
> = LHA; if LHA is greater than 90� but less than 180�, H = 180� - LHA; if 
> LHA is greater than 180� but less than 270� then H = LHA - 180�; and if 
> LHA is greater than 270� but less than 360�, H = 360 �- LHA. 
> Conceptually, this is the smallest angle measured from either the upper 
> branch or from the lower branch of the observer's meridian to the 
> meridian containing the body's GP.
>
> See the revised form at:
>
> http://www.fer3.com/arc/img/108719.revised%20form%206-18-09.pdf
>
> The formulas for calculating LHA are:
>
> If your AP is in west longitude: LHA = GHA - ALon. (If necessary, add 
> 360 �to GHA prior to subtracting ALon.)
>
> If your AP is in east longitude: LHA = GHA + ALan. (if LHA then exceeds 
> 360�, subtract 360� from the result.)
>
> Using the first formula for your first two examples, GHA (55� + 360�) - 
> 77� = LHA = 338�.
>
> GHA 95� - 77� = LHA = 18�
>
> The third example you bring up makes no sense since GHA is never 
> measured to the east, it is always measured west from Greenwich.
>
> gl
>
>
> Andrew Corl wrote:
>   
>> All,
>>  
>> I need some help.  I am attempting to work the problem in Ocean 
>> Navigator using the Lapook-Bygrave Slide Rule.  I am uncertain how to 
>> compute the Local Hour Angle (LHA).
>>  
>> In the following cases I am assuming my longitude to be 77 degrees 
>> west of Greenwich.
>>  
>> 1. The GHA of the Sun is 55 degrees west of Greenwich
>> 2. The GHA of the Sun is 95 degrees west of Greenwich
>>  
>>  
>> In the following cases I am assuming my longitude to be 120 degrees 
>> west of Greenwich
>>  
>> 1. The GHA of the sun is 170 degrees east of Greenwich.
>>  
>> Also what is the formula if my position is east of Greenwich and the 
>> sun has a GHA of more than 180 degrees?
>>  
>> Thanks
>> Andrew
>>
>>     
>
>
> >
>
>   


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