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Old 01-04-2015, 09:31   #46
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Re: Trigonometry question...

A squared +B squared =C squared
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Old 01-04-2015, 12:38   #47
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Re: Trigonometry question...

Quote:
Originally Posted by andres View Post
I think your formula can be simplified using some trigonometric formulas:

sin(A-B) = sin(A) cos(B) - cos(A) sin(B)
sin(A+B) = sin(A) cos(B) + cos(A) sin(B)

...

[sin(TA-WA) + sin(TA+WA)/[sin(180-2A)] = cos(WA)/cos(TA)
Very elegant!

I didn't even think of looking for a simplification of my formula.

It is certainly not intuitive that it would reduce to such a simple solution.
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Old 01-04-2015, 13:27   #48
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Re: Trigonometry question...

Quote:
Originally Posted by StuM View Post
........
The Law of Sines :
a b c
---- = ---- = ----
sin(A) sin(B) sin(C)

.........
and the formula for Cell B5:

=(SIN(RADIANS(B1-B2))+SIN(RADIANS(B1+B2)))/SIN(RADIANS(180-2*B1))*B3

and not a single right angled triangle used
It would have helped had I remembered the Law of Sines .
Been too many decades since anything I needed anything other than the simplest trig . My version was much clumsier.
Nicely done.

Quote:
Originally Posted by andres View Post
Hi Stum,

I think your formula can be simplified using some trigonometric formulas:

sin(A-B) = sin(A) cos(B) - cos(A) sin(B)
sin(A+B) = sin(A) cos(B) + cos(A) sin(B)

then

sin(A-B) + sin(A+B) = 2 sin(A) cos(B)

and

sin(180-2A) = sin(2A)
sin(2A) = sin(A+A) = 2 sin(A) cos(A)

so

sin(180-2A) = 2 sin(A) cos(A).

then

[sin(A-B) + sin(A+B)/[sin(180-2A)] = [2 sin(A) cos(B)]/[2 sin(A) cos(A)]
= cos(B)/cos(A).

Taking A= TA and B = WA, we get

[sin(TA-WA) + sin(TA+WA)/[sin(180-2A)] = cos(WA)/cos(TA)
Andres, that was just beautiful!
I would never have guessed it could be reduced so elegantly!

SWL
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Old 01-04-2015, 13:48   #49
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Re: Trigonometry question...

Thanks Stum and Seaworhy Lass!

But I teach mathematics, so I use these formulas every now and then.
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Old 01-04-2015, 14:00   #50
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Re: Trigonometry question...

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Thanks Stum and Seaworhy Lass!

But I teach mathematics, so I use these formulas every now and then.
This was actually the first time I have seem them in action since high school exercises.
To someone who finds these "puzzles" a pleasure that was just so satisfying to see the final result reduce so neatly
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Old 01-04-2015, 19:52   #51
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Re: Trigonometry question...

Nice in theory! In practice if my tacks look like right angles, I assume I'm going to sail 1.5x the distance, if they look like, equilateral triangles against the rhumb line, I assume I'm going twice the distance and roughly calculate eta on those gustimates. Anyway, wind, sea and tides all play too many games with my predictions but some rules of thumb come in handy..
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Old 02-04-2015, 03:13   #52
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Re: Trigonometry question...

Sitting here in the hotel again for my last evening on site, I decided to extend the concept slightly:



You can get the spreadsheet here:
http://camcopng.com/download/Windage.xlsx

(Don't look too closely at the formulae in E6 & 7. I'm not at all proud of them. I'm sure there is a much neater way to decide the preferred first tack )
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Old 02-04-2015, 03:33   #53
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Re: Trigonometry question...

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I'm sure there is a much neater way to decide the preferred first tack )
For me overwhelmingly the decision on which tack to first take depends on how I think the true wind will alter (due to either ground wind shift or current change). Local effects along the way are also considered (true wind may shift with land/sea features).

If no clue and no crystal ball, I would take the leg that keeps me closest to my course first.

I wouldn't need a table for any of those decisions. What are I missing?

SWL
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Old 02-04-2015, 04:23   #54
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Re: Trigonometry question...

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Originally Posted by Seaworthy Lass View Post
For me overwhelmingly the decision on which tack to first take depends on how I think the true wind will alter (due to either ground wind shift or current change). Local effects along the way are also considered (true wind may shift with land/sea features).

If no clue and no crystal ball, I would take the leg that keeps me closest to my course first.

I wouldn't need a table for any of those decisions. What are I missing?

SWL
You're not missing anything. It is obvious which is the preferred leg to any of us. It is not obvious to a dumb spreadsheet.

It was working out the logic of "which leg keeps me closest" (which is obviously also the longest leg) so that I could display the distance and heading of that leg first that got convoluted - especially when you have to handle passing through 0/360 degrees in the calcs)

Having it tell you the distance and bearing to the end of the optimum first tack (all other considerations being equal) may be useful for planning. You can easily plot the planned legs and tacking waypoint on your chart.
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Old 02-04-2015, 08:12   #55
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Re: Trigonometry question...

One thing I found interesting while going through this exercise was the relative impact of changes in tacking angle vs. changes in wind direction.

For example, a 20% change in wind direction might result in a 10% change in distance sailed while a 20% change in tacking angle (pointing ability) can result in a 40% change in distance sailed.

I guess the moral to the story, if there is one, is to rig one's boat to point as high as possible; especially critical for racers.
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Old 02-04-2015, 09:24   #56
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Re: Trigonometry question...

For great distances you will need to use spherical trig because you are now dealing with a third dimension....the shape of the Earth and the fact that it is also an oblate spheroid. The earth bulges slightly in the middle from the centripetal effect of it's rotation. You will need to use tables whether paper or electronic to determine this.

With spherical trig the lines are great circle routes and the sum of the angles of a spherical triangle is greater than 180 degrees.

Planar trig is only good for two dimensions.
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Old 02-04-2015, 12:16   #57
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Re: Trigonometry question...

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For great distances you will need to use spherical trig because you are now dealing with a third dimension....the shape of the Earth and the fact that it is also an oblate spheroid. The earth bulges slightly in the middle from the centripetal effect of it's rotation. You will need to use tables whether paper or electronic to determine this.

With spherical trig the lines are great circle routes and the sum of the angles of a spherical triangle is greater than 180 degrees.

Planar trig is only good for two dimensions.
So what's would the difference be on a say a 200NM run, given that the radius of the earth is nearly 3500NM?
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Old 02-04-2015, 12:20   #58
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Re: Trigonometry question...

Quote:
Originally Posted by Seaworthy Lass View Post
For me overwhelmingly the decision on which tack to first take depends on how I think the true wind will alter (due to either ground wind shift or current change). Local effects along the way are also considered (true wind may shift with land/sea features).

If no clue and no crystal ball, I would take the leg that keeps me closest to my course first.

I wouldn't need a table for any of those decisions. What are I missing?

SWL
Now don't go introducing logic into this!
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Old 02-04-2015, 12:35   #59
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Re: Trigonometry question...

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Now don't go introducing logic into this!
Oops, sorry .

Let's get back to how imprecise pi is for cruisers
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Old 02-04-2015, 12:35   #60
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Re: Trigonometry question...

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Originally Posted by OldFrog75 View Post
I read it. Little beyond my comprehension. Was hoping for something simple for determining the lengths of the sides of an isosceles triangle given the length of the base and angles of the legs off the base...
eg

Hypotenuse = sqrt(adjacent^2 + opposite^2)

Adjacent = cosine(angle) * hypotenuse

Simple trig. Sqrt of the sum of the squares if you have two lengths. Otherwise rearrange cosine, tangent or sine of the angle to get the length of interest.

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