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Old 15-06-2021, 07:59   #1
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Trig Help

Hi Friends,
Need a little help with Trigonometry
It's been a long time since High School
I'm trying to figure out how much heel angle I would
need to induce to get a 60 foot stick under a 54 foot bridge ?
And what width of channel would be required.
I know one has to use the Pythagorean Theory,
but to tell you the truth just looking at it on Google
made my head spin.
So Id appreciate any help from anybody good with sums
Cheers
Neil
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Old 15-06-2021, 08:05   #2
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Re: Trig Help

Quote:
Originally Posted by Time2Go View Post
Hi Friends,
Need a little help with Trigonometry
It's been a long time since High School
I'm trying to figure out how much heel angle I would
need to induce to get a 60 foot stick under a 54 foot bridge ?
And what width of channel would be required.
I know one has to use the Pythagorean Theory,
but to tell you the truth just looking at it on Google
made my head spin.
So Id appreciate any help from anybody good with sums
Cheers
Neil

At least 26°


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Old 15-06-2021, 08:33   #3
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Re: Trig Help

You want cos.

Adjacent length being air draft vessel

Hi the long word being air draft bridge.

Is stick 60 feet from rotation bouyansy of vessel?

54/60=0.9

Cos (30 degrees) =0.86 or close enough which is lower or in basic terms allowance for a bit of a gap.

54/0.86= 62.7 or 2.7 feet extra at 30 degree heel.
Pitch alters air draft, might be a passing vessel.

I might have mathematics wrong. Been a while. Pretty sure that true though.
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Old 15-06-2021, 08:39   #4
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Re: Trig Help

How wide your boat?
Although mast air draft reducing, mast step rising.
You need allow for bouyansy of heel that keeps our wind clean. Or you ain't under bridge although trig angle suggests above.

0100 hours. I'll mislead like normal
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Old 15-06-2021, 08:40   #5
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Re: Trig Help

Quote:
Originally Posted by Time2Go View Post
.........need to induce to get a 60 foot stick under a 54 foot bridge? And what width of channel would be required..........
Solving for width by the Pythagorean theorem based on 60 ft mast plus 4 feet from mast base to the waterline and 54 ft clearance under the bridge to the waterline and assuming 15 ft beam:

width = √ ([60+4]^2 - 54^2) + 15/2 = 42 feet minimum
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Old 15-06-2021, 08:59   #6
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Re: Trig Help

Time2go:

Why make your life complicated???

Just grab a piece of paper and draw a horizontal line across it near the bottom. Then draw a vertical line about an inch in from the right hand side. Then measure up 5" from the first horizontal line you drew, and draw another horizontal line. Now pick up 6" in your compasses and with the compass point at the crossing of the bottom horizontal and the vertical, draw a part of a circle so it crosses the top horizontal.

A line drawn from the point where the point sits to the intersect of the top horizontal and the circle segment will have the slope you seek, and therefore if you measure the angle between that slopey line and the vertical with a protractor, you can read directly the number of degrees you need to heel the boat.

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Old 15-06-2021, 09:34   #7
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Re: Trig Help

I did not have a protractor.........tossed it when I retired......but using the formula in this link came up 57.54 degrees based on a = 54 ft, b= 34.5 ft, c= 64 ft

https://www.omnicalculator.com/math/triangle-angle
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Old 15-06-2021, 10:40   #8
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Re: Trig Help

Assuming the top of your masthead is 60 feet above your center of buoyancy. If your mast is 60 feet this number will be bigger. Also note that the center of buoyancy will be located BELOW the waterline.

Heel = ACOS((54-clearance)/60), where clearance is the vertical distance in feet from the masthead to the underside of the bridge.

So, at the threshold where the mast would just touch the bridge (zero clearance), ACOS(54/60) = ACOS(0.9) = 25.8⁰
For three feet of clearance, ACOS((54-3)/60) = ACOS(0.85) = 31.8⁰, etc.

You will need to verify that clearance is indeed 54' (sometimes bridges get modified and data might not get updated, e.g. utilities or signage could change clearance for example) at the specified water level.

Bridge clearance should be based at MHW so you will gain clearance at lower water levels.
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Old 15-06-2021, 14:22   #9
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Re: Trig Help

We don't need no stinkin' trig! Or maybe we really do.....
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Old 15-06-2021, 15:06   #10
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Re: Trig Help

Good job, Spot, That's EXACTLY how you do it. :-)!

Now, I went for 50 feet under the bridge structure just to give a little extra heel to be on the safe side.

Now, to help out our OP, how would you suggest he induce a 40º heel on a 10 or 15 ton boat? I'll bet you my limit (35¢ Can.) that his boat doesn't have a running topping lift that's up to the job. And that his masthead sheaves for main and jib halyards don't swivel. :-)

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Old 15-06-2021, 15:52   #11
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Re: Trig Help

Hmmm...

As a once upon a time physicist I understand the trig and the associated theory full well.

As a practical sailor who knows just how badly things can go wrong, I suggest the following non-theoretical approach:

Make up a line that is as long as the clearance under the bridge plus whatever safety factor you are comfy with. Pput a substantial weight on one end of it. Attach the other end to the masthead. As the boat heels, the line will swing out athwartship. If the weight is touching or under the surface of the water, the masthead will clear the bridge.

With such a setup you can go out and practice heeling and maintaining the boat to the required angle out in open water. You may find that a bit challenging!

Jim
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Old 15-06-2021, 16:07   #12
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Re: Trig Help

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Hmmm...

Make up a line that is as long as the clearance under the bridge plus whatever safety factor you are comfy with.

Jim
Good plan, but I think that should read “Make up a line that is as long as the clearance under the bridge MINUS whatever safety factor you are comfy with.”

i.e. If the bridge is 50 feet, you want the “clearance line” to be 45 feet or so.
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Old 15-06-2021, 16:10   #13
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Re: Trig Help

Quote:
Originally Posted by Jim Cate View Post
Hmmm...

As a once upon a time physicist I understand the trig and the associated theory full well.

As a practical sailor who knows just how badly things can go wrong, I suggest the following non-theoretical approach:

Make up a line that is as long as the clearance under the bridge plus whatever safety factor you are comfy with. Pput a substantial weight on one end of it. Attach the other end to the masthead. As the boat heels, the line will swing out athwartship. If the weight is touching or under the surface of the water, the masthead will clear the bridge.

With such a setup you can go out and practice heeling and maintaining the boat to the required angle out in open water. You may find that a bit challenging!

Jim
There you go! The practical empirical approach!
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Old 15-06-2021, 16:21   #14
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Re: Trig Help

Quote:
Originally Posted by Jim Cate View Post
Make up a line that is as long as the clearance under the bridge plus whatever safety factor you are comfy with. Pput a substantial weight on one end of it. Attach the other end to the masthead. As the boat heels, the line will swing out athwartship. If the weight is touching or under the surface of the water, the masthead will clear the bridge.
Suggest the "substantial weight" is a barrel(s) full of water, so that when it hits the water, it stops pulling down.
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Old 15-06-2021, 16:45   #15
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Re: Trig Help

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Originally Posted by ChrisJHC View Post
Good plan, but I think that should read “Make up a line that is as long as the clearance under the bridge MINUS whatever safety factor you are comfy with.”

i.e. If the bridge is 50 feet, you want the “clearance line” to be 45 feet or so.
Well, you are absolutely correct! Sorry about the unclear statement, and thanks for pointing it out, Chris!

Jim
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