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Weight equalisation

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    Posted: 03 Jan 22 at 4:34pm
Originally posted by iGRF

Originally posted by tink



I give up if you are not prepared to accept basic laws of physics I not going to try and change that


Humans wrote the basic laws of physics and they have frequently been proved wrong.

That's the trouble with engineers, they only do as they're told and never question anything.

Rules are and have always been, made to be broken.

?? The whole focus of Engineering is to question everything, that is what we do. I’m dyslexic and look at problems with very fresh eyes. PS  not sure my IQ but it is in the 99.99 percentile from full educational physiologist evaluation. I’m not questioning your intelligence either, hope I haven’t come across that I am. The physics we are talking is the same as a see-saw uses it’s not going to disproved 
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Post Options Post Options   Quote iGRF Quote  Post ReplyReply Direct Link To This Post Posted: 03 Jan 22 at 4:34pm
Had to revert back to schoolboy maths and moments, now this is not a statement, it's still a question.

So lets say racks, and our helm is 90 kgs and the racks are 2.5 metres out from the pivot point which we'll call the mast base. So Distance from pivot times mass to get moment is 2.5 x 90kg equals 225Nm yes?

So now with a trapeze do we not have to accept that the pivot distance changes from the foot of the mast to the end of the rack, to the foot of the mast to the point at which the weight is acting on the mast which is the new lever at say 4 mtrs to the hounds? So our 90 kilo mass now supported flat wiring from the mast must be 4 x 90 kgs and 360 Nm?

This doesn't asnwer my question but it does give some maths to start with.

Or, do we totally discount the mast lever suggestion because it's neutralised by the sail sideforce and the calculation remains directly from the pivot point to the point at which the harness is attached or the height of the sailor?

If it's the harness hook attachment point then my suggestion holds water, if it's the height of the sailor then it doesn't.

Those holding the racks trapeze makes no difference answer will hold with the view that wether racks or trap the c.o.g of the sailor distance prevails, i.e. 250 so 225 Nm righting moment either way.

I maintain that there is an element of mast leverage and probable weight reduction as a result, but it's an interesting thing to continue to work on and I hate maths, but there has to be a formula there somewhere to support what I'm sure I've read somewhere along the path.

Edited by iGRF - 03 Jan 22 at 4:41pm
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Post Options Post Options   Quote Old bloke Quote  Post ReplyReply Direct Link To This Post Posted: 03 Jan 22 at 4:38pm
I know its only because its Christmas and you're bored.
However, 2 simple proofs
1, picture the telltale on your main. They flow horizontally and the stronger the wind, the more horizontally they flow. So with an upright mast and horizontal flow there can't be a vertical force for the trapeze hand to counteract
2, go out to the garage, loosely screw 2 pieces of wood to form a L. Then keeping the angle the same use a piece of line as your trapeze wire, attach it to various places on the "mast" and the "trapezing sailor" and see if you can feel a difference.
It nay or may not convince you, but it will get you out from under the long suffering Mrs iGRF's feet
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Post Options Post Options   Quote tink Quote  Post ReplyReply Direct Link To This Post Posted: 03 Jan 22 at 4:42pm
Originally posted by A2Z

Interesting isn’t it?  Lots of people have told you that you are wrong about moving the trapeze attachment outboard increasing righting moment but no one has really explained why.  

I have explained 3 times, and giving up now 
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Post Options Post Options   Quote iGRF Quote  Post ReplyReply Direct Link To This Post Posted: 03 Jan 22 at 4:50pm
So why does the windward shroud go loose when I'm hanging on the wire?

I'm not suggesting a vertical force, is it not a sideways force?

And if the wind suddenly dropped without me getting back in the boat the weight of the wire will pull the boat over.
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Post Options Post Options   Quote Oatsandbeans Quote  Post ReplyReply Direct Link To This Post Posted: 03 Jan 22 at 4:59pm
The mast has nothing to do with it. As I said it is two moments which are balanced and composed of two forces and it the magnitude of these force and their separation that is important.
The heeling moment is composed of the heeling force (sideways) from the rig and the lift (sideways) from the foils. This is balanced by the righting moment, which is composed of the crew weight and the buoyancy of the hull.
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Post Options Post Options   Quote iGRF Quote  Post ReplyReply Direct Link To This Post Posted: 03 Jan 22 at 5:11pm
I've worked it out but you'll have to wait.

Mrs iGRF is not happy, I've tried to explain to her but she won't listen.


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Post Options Post Options   Quote Oatsandbeans Quote  Post ReplyReply Direct Link To This Post Posted: 03 Jan 22 at 5:18pm
Grf -I tell the youngsters that I work with that if they can’t explain something to their mum they don’t understand it ( in your case it would be your wife!)
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Post Options Post Options   Quote tink Quote  Post ReplyReply Direct Link To This Post Posted: 03 Jan 22 at 5:19pm
Originally posted by iGRF

So why does the windward shroud go loose when I'm hanging on the wire?

I'm not suggesting a vertical force, is it not a sideways force?

And if the wind suddenly dropped without me getting back in the boat the weight of the wire will pull the boat over.
It goes slack because the trapeze wire is acting as a stay and taking the load off the shroud. The centre of gravity of the whole boat has not changed, the thing that is counter acting the sail force is the righting moment: that is the horizontal distance between the centre of gravity and the centre of buoyancy multiplied by the whole weight of the boat and crew.

Ignore dinghies think of a deep keel boat. When the mast is vertical the keel is actually doing nothing. Now imagine a heavy person steps on the gunwale. He will force the boat to tip towards him, as he does the the underwater shape changes and the center of buoyancy moves towards him, as this happens the keel and the centre of gravity moves in the opposite direction - at some point the movement of the keel will counter act the tipping momentum of the man on the gunwale. 
If the man on the gunwale pulled with all his might on the halyard nothing, absolutely nothing would change. His weight is still in exactly the same position and he weighs the same amount. 
If you took your Farr 3.7 on its trailer to a public weigh bridge and stood roughly where you trapeze from then actually trapezed or just sat in the boat there would be no change in the weight of the boat. 

It is very different for windsurfers as I have said because the mast pivots. 

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Post Options Post Options   Quote tink Quote  Post ReplyReply Direct Link To This Post Posted: 03 Jan 22 at 5:23pm
Originally posted by iGRF

I've worked it out but you'll have to wait.

Mrs iGRF is not happy, I've tried to explain to her but she won't 



Please tell me Mrs iGRF is a professor of pure physics 
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