NavList:

From: hanno ix <hannoix@att.net>
Subject: Fw: [NavList] Re: Ix Diagram and Pub 208
To: "Gary LaPook" <garylapook@pacbell.net>
Date: Thursday, December 13, 2012, 10:13 AM

Greg:

I first understood your remark about the "interesting" set of values

as a hint there might be a problem. However, 

I don't see any problem whatsoever - or I am blind!

The diagram delivers for  dec = 30; LHA = 40; Hc = 50

the azimuth Z = 60 which is correct.

The diagram delivers for  dec = 40; LHA = 30; Hc = 50

the azimuth Z = 36 which is close 36.58 - the correct value.

Do you agree?

h

----- Forwarded Message -----
From: hanno ix <hannoix@att.net>
To: "NavList@fer3.com" <NavList@fer3.com>
Sent: Wednesday, December 12, 2012 10:26 AM
Subject: [NavList] Re: Ix Diagram and Pub 208

Greg,

you are correct. 

I believe there is at least one practical way to deal with this.

The question then is, what is the most practical way?

I have to ponder this - unfortunately I am pressed for time.

h.

---

From: Gary LaPook <garylapook@pacbell.net>
To: NavList@fer3.com
Sent: Tuesday, December 11, 2012 9:35 PM
Subject: [NavList] Re: Ix Diagram and Pub 208

"While trying some different examples, I discovered an interesting set of values to use. Let dec = 30, LHA = 40, Hc = 50 and then the Z  comes out to exactly 60. There must be some trig reason for this but I haven't figured out why. " --------------------------------------------------------------------------------- The answer just came to me in a flash. The diagram solves this equation: sin Z cos Hc = sin LHA cos dec = C      ( I wrote it wrong in my first post) In my example: sin 60 cos 50  = sin 40  cos 30 = C since cos 50 = sin 40 and also sin 60 = cos 30 so Z ends up coming out exactly as 60 because it is arc sin (cos 30) gl --- On Tue, 12/11/12, hanno ix <hannoix@att.net> wrote: From: hanno ix <hannoix@att.net> Subject: [NavList] Re: Ix Diagram and Pub 208 To: "NavList@fer3.com" <NavList@fer3.com> Date: Tuesday, December 11, 2012, 6:33 PM Greg, you found it out! Yes, one can use the diagram for multiplying trig functions as you describe but I think the accuracy of this little diagram is rather limited for most  other CelNav purposes, no? As to the sample you are bringing up - let me think about it. h From: Gary LaPook <garylapook@pacbell.net> To: NavList@fer3.com Sent: Tuesday, December 11, 2012 2:01 PM Subject: [NavList] Re: Ix Diagram and Pub 208 Addendum: sin Z cos Hc = cos LHA sin dec = C gl --- On Tue, 12/11/12, Gary LaPook <garylapook@pacbell.net> wrote: From: Gary LaPook <garylapook@pacbell.net> Subject: [NavList] Re: Ix Diagram and Pub 208 To: NavList@fer3.com Date: Tuesday, December 11, 2012, 1:57 PM I like your diagram, it is pretty slick. I was trying to figure out how it works and I noticed that your "C" value is the product of sin LHA times cos dec. So this comes from the standard formula for azimuth, sin Z = sin LHA cos dec /cos Hc. You then rearrange this formula to  sin Z cos Hc = sin LHA cos dec. Using your diagram, moving from the outside scales to an intersection multiplies the two values so moving out to one scale from an intersection is the same as dividing. So after finding "C" (which is cos LHA sin dec and also sin Z cos Hc) starting with the value of "C" and the intersection with cos Hc and moving out to the azimuth scale your are actually dividing "C" by cos Hc and thereby  finding Z. The diagram could also be used for other trig problems involving multiplying and dividing sines and cosines. While trying some different examples, I discovered an interesting set of values to use. Let dec = 30, LHA = 40, Hc = 50 and then the Z  comes out to exactly 60. There must be some trig reason for this but I haven't figured out why. gl --- On Mon, 12/10/12, hanno ix <hannoix@att.net> wrote: From: hanno ix <hannoix@att.net> Subject: [NavList] Re: Ix Diagram and Pub 208 To: "NavList@fer3.com" <NavList@fer3.com> Date: Monday, December 10, 2012, 10:18 AM Thanks for,the kudos, Greg! In the meantime, I have developed a version that transforms the diagram into polar coordinates.  Perhaps that will be still easier to use? After Xmas I will send out more about that. Also, there is a novel sight-reduction-by-hand in the works. It works with a single 2-page table and covers angles from 0 to 90 degrees without special rules. The calculation uses only  integer numbers of just 4 digits. The error is +/- 1 sm throughout the full range of angles. The simple process will be contained on the worksheet. Sounds too good to be true? You guys will be the judges... Merry X-mas H From: Greg Rudzinski <gregrudzinski@yahoo.com> To: NavList@fer3.com Sent: Sunday, December 9, 2012 1:51 PM Subject: [NavList] Ix Diagram and Pub 208 Hanno Ix's rendition of the azimuth diagram has really delivered. After a plethora of trials it has proven remarkably fast and error free. I highly recommend it's use with Pub 208 or Weem's Line of Position in lieu of the Rust diagram. http://fer3.com/arc/m2.aspx/Azimuth-Diagram-example-use-HannoIx-oct-2012-g20982 Attached is an example of an octant bubble Sun observation sight reduction using Pub 208 and the Ix Diagram. The NA look-up and sight reduction fit easily on one side of an index card. Greg Rudzinski ---------------------------------------------------------------- NavList message boards and member settings: www.fer3.com/NavList Members may optionally receive posts by email. To cancel email delivery, send a message to NoMail[at]fer3.com ----------------------------------------------------------------