Select to expand quotedecrepit said..
I've always thought it would be better to attach harness lines as high as possible, that's why I was perplexed by advice to lower the boom when overpowered.
So we'd need sails with a cut out about 2.5m up for the attachments??? I've some old sails, I could experiment. But I can foresee loose ropes whipping around being a problem, and securing them before a gybe tricky. Maybe they clip into normal lines, then you can transfer from one to the other as needed?
Is this worth pursuing?
It's going to alter sail dynamics of course, mast bend will be altered significantly, may not get maximum benefit unless a sail maker is also involved.
Yes, and what USA46 has brought to our attention is how the height of the connection can change the moment that a sailor at a given orientation hiking out board can generate! I suppose we were locked into the thinking behind dinghies with rigid stayed masts where the method of supporting the hiking sailor doesn't matter. It's all the same whether supported by a trapeze wire or a hiking plank. That's because it's a rigid triangulation and a small change in mast lean results in the exact same small angle of ballast movement.
Trouble with windsurfers is that the mast, boom, mast foot , sailor and harness lines form a pentagon with only one angle, boom to mast fixed. A small movement in theta in the diagram causes a small movement in phi behind the sailors feet. But the ratio of the angular movements is not constant, it depends on the height of the harness hook. Unlike a triangle that whole pentagon distorts, and the sides are all different lengths.
This is where I thought it easier to look at it from an energy conservation aspect rather than try and analyse the forces. Force times distance against the wind = force times distance lost to gravity. But the trigonometry is still horrendous. But I'm pretty convinced, just looking at the diagram, that there will be an optimum hook height given that the boom height is constrained by being at a comfortable height for the sailor to do the fine control but leaving the bulk of the righting component to the harness lines. Maybe we'd get used to chest harnesses again if we could show they had advantages if we tolerated the discomfort?
Maybe USA46 can do the trigonometry for us and use a bit of differentiation to find the optimum?
Maybe it's not that hard. USA46's trigonometry should give the same solution for the optimum harness hook position. Just have to replace the 60 degree angle of the brown line to horizontal with an angle theta. The harness lines won't necessarily pass through the C of G now, and the unknown theta can be carried all the way thru the calculations to step 3 and the result differentiated and solved for zero.
Way beyond my abilities, I'm a suck it an see man. Intuition guided trial and error. Can be slow and painful, but also fun if you learn something. Even better if it's something useful
Select to expand quoteIan K said..USA46 said..
answer on picture
Mebers will go crazy ,
Yes we won't get any green thumbs. At least the mebers can't give us red ones any more.
After a few false starts I now agree with your answer on the picture. In which case, assuming we are both correct, why do we insist on attaching our harness lines to the boom? (Apart from convenience)
If you conect rope to sail you will have two big problems:
1)every sail has, lets call it " pivot axis" which conect center of pressure(COE) and mast base,sail will rotate around this pivot axis every time if projected harness line dont cross pivot axis(this happend every time when you put harness line in wrong place),so you must conect this rope somewhere at this pivot axis.
So if you want high conection ,than rope conection will be to far back in the sail.sail handling like this will be imposible.
2)high conection on sail will deform sail profile,sail is to soft/flexibile to conenct so much force at one point..also it will kill sail ability to twist.
Of course, I just realised that this morning! How could we miss such a basic point? It's all to do with 2d diagrams leading us astray I think.
(Ian K, how do you mean optimal hook height,for what optimal?
low hook(seat harness) will increase Fsail but decrease vertical force on feet to board,so you will lose control in high wind condition
high hook(chest harness)-opposite)
similar as low vs high boom,changing boom height we change angle of harness line
I will use this equation on picture below for calculate Fsail for my original task...
here is result:
Not sure what I mean by optimal? Still thinking about it. But thanks US46, your diagram with the exaggerated low hook shows that as you go lower the leverage on the sail increases. (67 vs 62). Not by all that much though. So I suppose if you get lifted by the sail while the board remains flat you could lower the hook. If the board lifts on the leeward rail and you maintain position raise the hook for more comfort. Is the sweet spot where both happen at once?
Select to expand quoteIan K said..
But thanks US46, your diagram with the exaggerated low hook shows that as you go lower the leverage on the sail increases. (67 vs 62).
Not by all that much though.
67kg vs 62kg is not low vs high hook,it is "wrong" vs correct calculation.
My configuration is the same as original task,I just want to show when use "wrong" equation we get 62kg and with correct calculation is 67kg.
Sorry USA46. I read your post too quickly. Jim Drake's method is not incorrect. He is analysing forces on the system as a whole.
Jim states this " A useful method for the analysis of forces on a body (in this case the body is the combination of board, rig and sailor)"
It doesn't matter whether the internal forces are rigid or articulated the analysis is valid.
Jim Drakes's method is all you need if you just want to analyse the external forces, the fin, the sail. He just estimated the sailor's C of G relative to the board, the height of the sail's centre of lift and away he went.
If you need to wonder about the forces on the universal, the harnesses you use your method.
Another advantage of your method is that by starting with the requirement that the sailor can balance the rig you can then consider what is needed to balance the system with the sail force already independently calculated.
As well as considering the length of the fin you can also determine where the centre of hydrodynamic lift is under the hull. Jim has assumed it is on the centreline of the hull. But that was in the days of narrow boards not so much of an issue. With wide boards the centre of hydrodynamic lift can move off centre towards the leeward rail to make sure the system is balanced. We all know this, you can sheet in more on a wide board if you get the windward rail up a bit. (red arrow, first post figure 6)
(Anyway trigonometric wiz that you undoubtedly are if you apply your method and Jim's method to your scenario and don't come up with the exact same answer, you've made a minor slip up somewhere. Though you did use your hydrodynamic lift centre, when you checked using Jim's method? Jim assumed it on the centreline. That's not incorrect he makes it clear,that he's done it. It's an approximation. Both methods have plenty of approximations )
Select to expand quoteIan K said..
Jim Drakes's method is all you need if you just want to analyse the external forces, the fin, the sail.
Ok than just find Fsail, easy task,using "Jim" formula Fxh=WxL
25 kg. Assuming hydrodynamic lift from the hull is on the centreline and vertical.
( How the fellow can stand vertically while holding the boom is a problem internal to the system. Maybe he's made of titanium and his feet are welded to the deck. Not something we are asked to solve)
Select to expand quoteIan K said..
25 kg. Assuming hydrodynamic lift from the hull is on the centreline and vertical.
( How the fellow can stand vertically while holding the boom is a problem internal to the system. Maybe he's made of titanium and his feet are welded to the deck. Not something we are asked to solve)
wrong answer..
Sailor stand 100% vertical his- pull is zero,so Fsail= 0, system is not in balance,sailor will sink windward rail..
(we know that sailor dont have titanium feet,we also know that feet are not welded to board and we also know that sailor can not transfer moment to sail via board)
(if mast is conected to board with rigid conection,than Fsail=25kg ,system is in balance)
Select to expand quoteUSA46 said..Ian K said..
25 kg. Assuming hydrodynamic lift from the hull is on the centreline and vertical.
( How the fellow can stand vertically while holding the boom is a problem internal to the system. Maybe he's made of titanium and his feet are welded to the deck. Not something we are asked to solve)
wrong answer..
Sailor stand 100% vertical his- pull is zero,so Fsail= 0, system is not in balance,sailor will sink windward rail..
(we know that sailor dont have titanium feet,we also know that feet are not welded to board and we also know that sailor can not transfer moment to sail via board)
(if mast is conected to board with rigid conection,than Fsail=25kg ,system is in balance)
"wrong answer.."
"we know that sailor dont have titanium feet,"
We also know you'll never be able to find a photo like that to base your calculations on, But if you did find such a photo you would have to assume something like that is going on.. Shrouds made of invisible dark matter maybe?
I when put all my calculation in one formula,get this.
I added distance (d)=from feet to sailor C.G. now we can play with hook height(L) as well.
As you can see sail angle (C) dont exist in formula,because Fsail dont change when sail angle change.
If harness line is 100% horizontal then B=0,but if you use angle below horizontal than you must have minus sign to formula works correct,
anyway, this is never case in reality because harness line allways stay slightly upright...
If sailor stay at 100% upright,then A=90, cos90=0 ,as you know multiplication by zero gives zero, so Fsail =0 ,what is also correct becuase if sailor stay 100% upright his pull is zero so Fsail is zero too....
get picture,take protractor to messure angle A and B,from board width you know what is (X),(h) is ussualy about 40% of mast and you can find Fsail..for this picutre and this numbers I get Fsail=33kg
below explanation why sail angle(C) dont change Fsail..
As far as sailing goes, angle "C" does have an effect. It changes the direction of F so the component of vertical lift changes. This will affect board trim.
Select to expand quotedecrepit said..
As far as sailing goes, angle "C" does have an effect. It changes the direction of F so the component of vertical lift changes. This will affect board trim.
Yes for sure.Angle C will not change magnitude of Fsail but it will change direction.When sail is angled then works less efficient , thrust component is reduced,so for speed sail must be upright as possible..
Not sure about that, yes less forward thrust, but the sail is taking some of the vertical lift off the hull, this may help to get a more ideal hull trim. And reduce hull drag
I thought that was why that very fast foil boat (sorry ancient brain forgets names) had the sail heavily canted to windward.
Yes you are right, but I think for windsurifng at +50knots so small area of board is in water,that we dont need so much lift from sail.
Select to expand quoteUSA46 said..
Yes you are right, but I think for windsurifng at +50knots so small area of board is in water,that we dont need so much lift from sail.
That's OK, I haven't managed 40kts yet, but would like to optimise trim, to give me a chance of getting there.
