

30032015, 17:14

#31

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Re: Trigonometry question...
For any practical tacking course in a sailboat, plane geometry is close enough.
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30032015, 23:12

#32

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Posts: 12,825

Re: Trigonometry question...
Quote:
Originally Posted by barnakiel
That's only when the WPTs are close by. For distant WPTs you will be using spherical geometry and cos of the angle will not apply. See orthodrome distance and spherical triangle formulas.
b

It is close enough for several dozen nm.
It is primarily an academic exercise anyway  the true wind cannot be expected to stay constant in direction, so a huge error is introduced there.
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31032015, 01:56

#33

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Join Date: Aug 2013
Location: Cowherding down on the campo
Boat: 39' Westerly Sealord
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Re: Trigonometry question...
Quote:
Originally Posted by OldFrog75
I'm trying to compute the distance to a waypoint for a 1 degree change in heading off the wind.
I know 1 degree off course translates to 1 mile off course for every 60 miles sailed. I also know that a boat trying to sail to a waypoint directly into the wind with only one tack will sail a course more or less resembling an isosceles triangle.
My question is: how does one determine the length of the two legs given different degrees off the wind, which when added together will give total distance sailed. ...........

Why not just pick up a second hand copy of Nories Nautical Tables and use the Traverse tables that you shall find therein? What you are trying to do is a very simple example of the "Day's Work" problem....



31032015, 08:11

#34

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Join Date: Feb 2013
Location: Santa Monica, CA
Boat: Club Sailor; various
Posts: 922

Re: Trigonometry question...
Quote:
Originally Posted by El Pinguino
Why not just pick up a second hand copy of Nories Nautical Tables and use the Traverse tables that you shall find therein? What you are trying to do is a very simple example of the "Day's Work" problem....

I might ultimately do that very thing but for now I'm having too much fun playing with my spreadsheets. I'm getting to that age where it's always good to exercise the mind anyway I can and creating a table vs. buying one helps with that.
Now that I've gotten some familiarity with the Sin, Radian, and COS functions in Excel, I'm working on Trigonometry question #2  total distance sailed to a way point various degrees off the wind given different tacking angles.
Perhaps a totally meaningless endeavor but it keeps me busy. SWL has her knots (among other things); I have my numbers.
Thanks again to everyone who helped me yesterday.
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31032015, 23:41

#35

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Posts: 1,403

Re: Trigonometry question...
The value of a radian in degrees is 360/2Pi I think although in theory it's never going to be exact.
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01042015, 00:27

#36

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Re: Trigonometry question...
Quote:
Originally Posted by RaymondR
The value of a radian in degrees is 360/2Pi I think although in theory it's never going to be exact.

In theory it will always be exact .
Explanation:
The definition of pi is the ratio between a circle's circumference and its diameter (ie circumference/diameter). The diameter is twice the radius (r)
This leads to pi = circumference/2r
So circumference = 2 pi r
The definition of a radian is the angle subtended by an arc that is equal in length to its radius. As a circle's circumference is 2 pi r, this means there are 2 pi radians in a circle.
So given there are 360 degrees in a circle and 2 pi radians in a circle, one radian equals 360/2 pi.
All of this is exact in theory .
Where the precision breaks down in practice is that the value of pi goes on forever without repeating (mind boggling isn't it!).
I think excel spreadsheet use a value up to 15 digits. Close enough to being precise for even the most exacting sailor's needs .
SWL
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01042015, 00:53

#37

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Join Date: Nov 2013
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Re: Trigonometry question...
Quote:
Originally Posted by Seaworthy Lass
In theory it will always be exact .
..
All of this is exact in theory .
Where the precision breaks down in practice is that the value of pi goes on forever without repeating (mind boggling isn't it!).
...
I think excel spreadsheet use a value up to 15 digits. Close enough to being precise for even the most exacting sailor's needs

In theory, there is no difference between theory and practice.
In practice, there is.
(not Yogi Bera!)
But 15 digits precision means that you could potentially by out by 0.04 nanometers in a circumnavigation of the earth (40,000 Km)
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01042015, 01:10

#38

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Re: Trigonometry question...
Quote:
Originally Posted by StuM
But 15 digits precision means that you could potentially by out by 0.04 nanometers in a circumnavigation of the earth (40,000 Km)

I was wrong then. It is not "close enough to being precise for even the most exacting sailor's needs". We have some very exacting sailors here on CF .
SWL
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01042015, 01:11

#39

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Join Date: Oct 2008
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Re: Trigonometry question...
Quote:
Originally Posted by OldFrog75
Now that I've gotten some familiarity with the Sin, Radian, and COS functions in Excel, I'm working on Trigonometry question #2  total distance sailed to a way point various degrees off the wind given different tacking angles.
Perhaps a totally meaningless endeavor but it keeps me busy. SWL has her knots (among other things); I have my numbers.

OldFrog, we have something in common  I love numbers too .
And trigonometry is just a puzzle, not unlike crossword puzzles or sudoku . Not meaningless endeavours at all.
The puzzle you have set yourself is more challenging than the first exercise where your destination was directly into wind.
It still only needs the most basic principles applied, but more triangles need to be solved, so the whole thing becomes more complex.
If you haven't done much trigonometry before, here are the basics:
 The three angles inside any triangle add up to 180 degrees.
 If you have a right angled triangle (ie one side is perpendicular to another), then the sum of the squares of the two short sides equals the square of the long side.
 The relationship between the angles and the sides is given by the sin, cos, tan functions (found on many even basic calculators nowadays).
sin angle = length side opposite to the angle/ length of the longest side
cos angle = length side adjacent to the angle/ length of the longest side
tan angle = length side opposite to the angle/ length side adjacent to the angle
The angle can be specified in degrees or radians, but as you have found out excel only takes radians, so you need to convert (or tell excel to convert, as Stu showed you how).
To solve any problem like the one you have set yourself, you need to draw a diagram of the situation and then find how you can draw in right angled triangles so you can solve the problem using the above tools (setting this up is usually the key to solving the problem).
You just need two bits of info to solve any right angled triangle  one of the two smaller angles and the length one of the sides, or the length of two of its sides. From this you can determine all three angles and the length of all three sides.
If you are keen to pursue this, just yell if you need any help. This is a fun puzzle .
SWL
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01042015, 01:59

#40

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Join Date: Apr 2007
Location: Australia
Boat: Island Packet 40
Posts: 1,403

Re: Trigonometry question...
Quote:
In theory it will always be exact

Oh good, someone has finally worked out Pi to a sufficient number of decimal places to get an absolutely exact value. I was not aware of that (either that or they finally reached infinity and consider that good enough?)
(This thread could rapidly go the way of the Arctic (or was it Antarctic) ice cover one)
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01042015, 02:06

#41

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Join Date: Oct 2008
Boat: Aluminium cutter rigged sloop
Posts: 12,825

Re: Trigonometry question...
Quote:
Originally Posted by Seaworthy Lass
If you haven't done much trigonometry before, here are the basics:
 The three angles inside any triangle add up to 180 degrees.
 If you have a right angled triangle (ie one side is perpendicular to another), then the sum of the squares of the two short sides equals the square of the long side.
 The relationship between the angles and the sides is given by the sin, cos, tan functions (found on many even basic calculators nowadays).
sin angle = length side opposite to the angle/ length of the longest side
cos angle = length side adjacent to the angle/ length of the longest side
tan angle = length side opposite to the angle/ length side adjacent to the angle
The angle can be specified in degrees or radians, but as you have found out excel only takes radians, so you need to convert (or tell excel to convert, as Stu showed you how).
To solve any problem like the one you have set yourself, you need to draw a diagram of the situation and then find how you can draw in right angled triangles so you can solve the problem using the above tools (setting this up is usually the key to solving the problem).
You just need two bits of info to solve any right angled triangle  one of the two smaller angles and the length one of the sides, or the length of two of its sides. From this you can determine all three angles and the length of all three sides.

Pictures tell a thousand words, so here is a diagram of all the trig basics you need to know (nothing else needed). This will make it clearer:
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01042015, 03:55

#42

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Join Date: Nov 2013
Location: Port Moresby,Papua New Guinea
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Re: Trigonometry question...
OK, I'm sitting in a hotel room in Buka with nothing else to do so:
Vertical lines are Wind direction
CD is at 90 degrees to Wind Direction
Code:
DAC = TA (tacking Angle)
DAB = WA (Wind angle)
Since wind lines are parallel,
ACD = 90  TA
and BCA = 90  TA
A = TA  WA
C = 180  2TA
The three angles of a triangle add up to 180, so
B = 180  (180 2TA) (TA  WA)
= 2TA  TA + WA
= TA + WA
Let c = 1
The Law of Sines :
a b c
 =  = 
sin(A) sin(B) sin(C)
a = sin(A) and b = sin(B)
 
sin(C) sin(C)
so DISTANCE SAILED (b + c)
sin(A) + sin(B)
= 
sin(C)
sin(TA  WA) + sin(TA + WA)
= 
sin (180  2TA)
Here's an Excel spreadsheet that does it:
and the formula for Cell B5:
=(SIN(RADIANS(B1B2))+SIN(RADIANS(B1+B2)))/SIN(RADIANS(1802*B1))*B3
and not a single right angled triangle used
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01042015, 08:48

#43

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Join Date: May 2009
Location: Montevideo, Uruguay
Boat: Catamaran, Wharram Tiki 26, Polynesia
Posts: 21

Re: Trigonometry question...
Hi Stum,
I think your formula can be simplified using some trigonometric formulas:
sin(AB) = sin(A) cos(B)  cos(A) sin(B)
sin(A+B) = sin(A) cos(B) + cos(A) sin(B)
then
sin(AB) + sin(A+B) = 2 sin(A) cos(B)
and
sin(1802A) = sin(2A)
sin(2A) = sin(A+A) = 2 sin(A) cos(A)
so
sin(1802A) = 2 sin(A) cos(A).
then
[sin(AB) + sin(A+B)/[sin(1802A)] = [2 sin(A) cos(B)]/[2 sin(A) cos(A)]
= cos(B)/cos(A).
Taking A= TA and B = WA, we get
[sin(TAWA) + sin(TA+WA)/[sin(1802A)] = cos(WA)/cos(TA)
When your destination is exactly in the direction of the wind, then WA=0 and you get 1/[cos(TA)]; this is the case first proposed by OldFrog75.
In this last situation your triangle ABC is isosceles and DCA is a rectangular triangle.
But your formula is more general.
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01042015, 09:02

#44

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Join Date: Feb 2013
Location: Santa Monica, CA
Boat: Club Sailor; various
Posts: 922

Re: Trigonometry question...
Quote:
Originally Posted by Seaworthy Lass
OldFrog, we have something in common  I love numbers too .
And trigonometry is just a puzzle, not unlike crossword puzzles or sudoku . Not meaningless endeavours at all.
The puzzle you have set yourself is more challenging than the first exercise where your destination was directly into wind.
It still only needs the most basic principles applied, but more triangles need to be solved, so the whole thing becomes more complex.
You just need two bits of info to solve any right angled triangle  one of the two smaller angles and the length one of the sides, or the length of two of its sides. From this you can determine all three angles and the length of all three sides.
If you are keen to pursue this, just yell if you need any help. This is a fun puzzle .
SWL

Thanks for the encouragement. Spent a few hours working on it yesterday and finally got the spreadsheet finished using some of the logic you expressed but primarily SIN functions.
Set it up so I can adjust for various degrees of pointing ability and wind shift (independent variables) to determine how that affects total distance sailed to the waypoint.
Although "finished", I'm sure I'll continue tweaking it until absolutely certain my formulas and assumptions are correct, the most challenging for me being the angle at the first tack for a given degree of wind shift. Might never use it but it sure is pretty.
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01042015, 09:23

#45

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Join Date: Feb 2013
Location: Santa Monica, CA
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Posts: 922

Re: Trigonometry question...
Quote:
Originally Posted by StuM
OK, I'm sitting in a hotel room in Buka with nothing else to do so:
Vertical lines are Wind direction
CD is at 90 degrees to Wind Direction
Code:
DAC = TA (tacking Angle)
DAB = WA (Wind angle)
Since wind lines are parallel,
ACD = 90  TA
and BCA = 90  TA
A = TA  WA
C = 180  2TA
The three angles of a triangle add up to 180, so
B = 180  (180 2TA) (TA  WA)
= 2TA  TA + WA
= TA + WA
Let c = 1
The Law of Sines :
a b c
 =  = 
sin(A) sin(B) sin(C)
a = sin(A) and b = sin(B)
 
sin(C) sin(C)
so DISTANCE SAILED (b + c)
sin(A) + sin(B)
= 
sin(C)
sin(TA  WA) + sin(TA + WA)
= 
sin (180  2TA)
Here's an Excel spreadsheet that does it:
and the formula for Cell B5:
=(SIN(RADIANS(B1B2))+SIN(RADIANS(B1+B2)))/SIN(RADIANS(1802*B1))*B3
and not a single right angled triangle used

Hallelujah! I set up my spreadsheet formulas differently than yours but when I entered the independent variables (TA, Wind, Distance) in your example and hit the enter button I got the exact same result for distance sailed.
My biggest skepticism was trusting that I was correct in my assumption that (C = 180 2TA).
Now that a second party has confirmed that to be true, I feel much more confident in my results.
Thanks for playing!
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