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Re: Basic question regarding Napier's triangles
From: Peter Hakel
Date: 2009 May 28, 16:04 -0700
From: "NavList@fer3.com" <NavList@fer3.com>
To: NavList@fer3.com
Sent: Thursday, May 28, 2009 12:16:23 PM
Subject: [NavList 8432] Basic question regarding Napier's triangles
Hello my name is Christian Scheele and I live in Cape Town. I am by no means a professional navigator or mathematician and hope that I am not gatecrashing any party by making myself heard in here without much ado.
I would much appreciate it if somebody could help me understand the application of Napier's triangles to specific scenarios. I have obviously run afoul of some very basic premise governing Napier's rules.
Consider a spherical triangle superimposed on the following parts of an ideally spherical rather than spheroid globe, where capitals denote angles and small letters denote sides: A = 90 degrees, c = 70 degrees , a = 40 degrees.
Applying Napier's rules, (90~A) is the "middle part" which is opposite b and c.
Sin (90~A) = Cos b. Cos C
removing complement: Cos a = Cos b. Cos c
transposing formula: Cos b = Cos a/ Cos c
substituting values: Cos b = Cos 40/ Cos 70
Cos b = 0.766.../ 0.342...
Cos b = 2.24 ( 2 d.p.)
But b is not defined for values > 1. It would appear that certain right-angles spherical triangles, namely such yielding Cos values > 1, cannot be solved as illustrated in the above problem and application.
I have obviously made a blunder somewhere...
Either way, could somebody please help me approach this problem? I would be most grateful.
(This problem can be complicated should one assume that the triangle lies in the plane of a spheroid, such as the earth, rather than a sphere. For example, one could assume that b lies on 20 degrees N and a lies on 0 degrees longitude. This implies that c can be broken up into a component of 20 N and 50 S. In that case corrections must be made to any results arrived at by having assumed perfectly spherical properties initially. Perhaps Inman's Tables could come in handy here...)
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From: Peter Hakel
Date: 2009 May 28, 16:04 -0700
Hi Christian,
From the spherical law of sines we have
sin a / sin A = sin c / sin C, or,
sin C = sin c * sin A / sin a
Your values are A = 90 (i.e. sin A = 1) and c > a, which leads to sin C > 1.
This triangle does not exist.
Peter
From the spherical law of sines we have
sin a / sin A = sin c / sin C, or,
sin C = sin c * sin A / sin a
Your values are A = 90 (i.e. sin A = 1) and c > a, which leads to sin C > 1.
This triangle does not exist.
Peter
From: "NavList@fer3.com" <NavList@fer3.com>
To: NavList@fer3.com
Sent: Thursday, May 28, 2009 12:16:23 PM
Subject: [NavList 8432] Basic question regarding Napier's triangles
Hello my name is Christian Scheele and I live in Cape Town. I am by no means a professional navigator or mathematician and hope that I am not gatecrashing any party by making myself heard in here without much ado.
I would much appreciate it if somebody could help me understand the application of Napier's triangles to specific scenarios. I have obviously run afoul of some very basic premise governing Napier's rules.
Consider a spherical triangle superimposed on the following parts of an ideally spherical rather than spheroid globe, where capitals denote angles and small letters denote sides: A = 90 degrees, c = 70 degrees , a = 40 degrees.
Applying Napier's rules, (90~A) is the "middle part" which is opposite b and c.
Sin (90~A) = Cos b. Cos C
removing complement: Cos a = Cos b. Cos c
transposing formula: Cos b = Cos a/ Cos c
substituting values: Cos b = Cos 40/ Cos 70
Cos b = 0.766.../ 0.342...
Cos b = 2.24 ( 2 d.p.)
But b is not defined for values > 1. It would appear that certain right-angles spherical triangles, namely such yielding Cos values > 1, cannot be solved as illustrated in the above problem and application.
I have obviously made a blunder somewhere...
Either way, could somebody please help me approach this problem? I would be most grateful.
(This problem can be complicated should one assume that the triangle lies in the plane of a spheroid, such as the earth, rather than a sphere. For example, one could assume that b lies on 20 degrees N and a lies on 0 degrees longitude. This implies that c can be broken up into a component of 20 N and 50 S. In that case corrections must be made to any results arrived at by having assumed perfectly spherical properties initially. Perhaps Inman's Tables could come in handy here...)
-----------------------------------------------
[Sent from archive by: scheele-AT-telkomsa.net]
--~--~---------~--~----~------------~-------~--~----~
Navigation List archive: www.fer3.com/arc
To post, email NavList@fer3.com
To unsubscribe, email NavList-unsubscribe@fer3.com
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