catamaran sailing test equipment

[Fabius3D]

hello to all
I propose an experiment that I tried to carry out but I am not sure of the final result.
consists in the realization of a test equipment for the sailing of a catamaran.

calculate the effort on the shaft knowing that this is 8 m tall, the boma b=2.5 m, assuming a linear speed profile with the height, vm=20 km/h to 0.5 m from the ground, boma insertion quota, vm=30 km/h in the head of the tree.
in the next image we see the tree with the sail and the boma.

to be more precise piece 1 is the boma, the 2 the tree:

I calculated the average speed: v(x)= vm + (vm-vm)/hthe force is equal to: f(x)= 1/2· edNewsv^2=1/2NewsOldNewsv(x)^2·b(x)where b(x) is the length of the boma.
After that I integrated f(x) (\t 0 e 7,5)as a result I got ≈830 nwhile the right one is 238 n
after aI realized there was something wrong!ma non so cosa! I'll take you back. What do you think? Is that the wrong way?

 

[Onda]

the sail is overwhelmed, and its fat changes from below to above.
so its bearing coefficient varies a lot and is not a constant.
Moreover the formulas you have put are not clear.
b(x) should be the rope of the sail, which varies linearly along the height of the tree.
the average speed formula is wrong. but if you do not need the average speed but the local speed at the given height.
vh=vm+(vm-vm)*(h/hmax)) which varies linearly between vm to h=0 and vm to h=hmax.
b(h) is instead the length (cord) of the sail function h.
now you can integrate the whole (between h=0 and h=hmax) and get a useless result, as already said the carrier of the sail varies linearly according to its depth and angle of incidence.
then I didn't understand what you want to do with the portanza once you got it.

 

[Fabius3D]

the sail is overwhelmed, and its fat changes from below to above.
so its bearing coefficient varies a lot and is not a constant.
Moreover the formulas you have put are not clear.
b(x) should be the rope of the sail............... .

Thank you for the answer and for the time. very clear, though. .
...I believe that the problem requires a more elementary reading than this (in fact I didn't think it might seem a question of fluid dynamics). the sail understood as a rigid body.
the problem is that I can't get a strategy that suits the performance!
In practice it is as if we had a distributed force on the sail and I have to find the point where this is stronger. and then the force that is distributed on the tree. like this:

 

[Onda]

provided that there are so many regulations and explanations that do what you ask, so I don't understand why you leave alone. .. in a dubious direction.

what I explained to you is correct, you can integrate strength with the given formula, then you have to calculate the center of strength by making the average weighed of each horizontal strip of sail.
and in this way find the moment applied to the tree.
the road followed by all the designers is however different.
because sailing is not a rigid body and can be adjusted in camber, twist and incidence, it is too difficult to calculate a tree starting from the wind force.
starts from what keeps the tree straight and that is the straightening moment. then balances the tree knowing the watering moment of the boat.
all the regulations that verify the trees, as well as the design texts start from this point of view.
if then the wind is higher than the straightening moment can happen that a monohull bands more and therefore reduces the surface exposed to the wind.
A multi-hull shakes.
basic never exceeds the straightening of the boat.
if then the frog is flat slipped or other affects little. (relatively little) as the lever arm of a slipped frog lowers and therefore the load on the tree increase, same speech for the thirdroles, but at first approximation is not considered.
the point of application of force depends on the shape of the frog, normally not triangular. it is considered, by memory, from 41% of the inferior for the conventional ranks (roach to 20%) up to 47% of the ranks square top. these values already introduce the difference of pressure of the wind between high and low and have been verified (by others) in the wind tunnel.

 

[Fabius3D]

provided that there are so many regulations and explanations that do what you ask, so I don't understand why you leave alone.. .

Sorry I didn't want to be insistent. I'll try again. Sooner or later something will come out. read so "integrate strength with the given formula, then you have to calculate the center of strength by making the average weighed of each horizontal strip of sail" seems easier than expected.
Thanks again

 

[Fabius3D]

Nothing to do. I can't connect the speed profile of the wind with force. are unbelievable values!

 

[Onda]

try to take a look at the attached accounts.
I think it's true.

 

[Fabius3D]

this result is very close to what in theory should be (f=238 n). I got a f=236,3 n yesterday but it doesn't totally match. I looked for the center of gravity, made a weighed average of speeds at that point, as you had suggested, then without integrating I also found mf and mt and also these values approach the results. But there's always something that doesn't convince me. Perhaps it is better if I pass to other because there are 3 days above and the test is next Wednesday!
Thank you very much for everything. availability has no price!

 

[Onda]

here I also calculated the thrust center, which turns out a little low since the hypothesized frog is perfectly triangular.
keep in mind that the result is hypothesized as the length of the boma was not a supplied data and neither the profile c.p.
we say that from a didactic point of view it is correct, then in reality it is far light years from what happens on a frog.
I used a summary rather than an integral, but the concept is identical.
Moreover, while being the sum of 1 to 1000 and therefore the strips in very small consideration, there is a small error since the sail was considered as rectangle strips and not trapezoidal, and for the center of the strip I did not take the center but one end, but adding on 1000 strips changes very little. making an iteration on 2000 strips changes the third decimal digit.
As for your result, perhaps depends precisely on the cp and the boma size. If you give me the correct values, let's see if it comes back

 

[Fabius3D]

the assigned values are:
♪
♪
♪

 

[Onda]

e chi sono mf e mt?

 

[Onda]

I recalculated to return what is required.
Even if I don't understand the goal of this?
an engineering problem?
Moreover, I think the position of the carrier on the sail at 50% of the rope. really unrealistic position, being this at 25% for a symmetric profile and a little over 30-35% for an asymmetric profile.
all this assuming that mt is the twisting moment around the tree.

 

[Fabius3D]

exact, mt and mf respectively torque and stinging moment. the end is to find the efforts that undergoes a square plate, to which the tree is connected, connected to the frame through screws and welding, the latter finally to be dimensional based on f, mf and mt, which is rather simple, at least regarding the search for those damn forces! As you say, it's not very realistic because the problem, in my opinion, is just a cover. it was about calculating the whole having already the force (carrying) as given would already be another thing! But obviously there's a rotten man!

 

[Fulvio Romano]

is not the dairy ham, it is the problem approached by the wrong side.

If I'm not mistaken, post #4 suggests a different approach, engineeringly much more valid.. .

 

[Fabius3D]

Yes absolutely, in fact I have tried several times but I do not return the results, even if recently...

 

[Onda]

But you take out the curiosity of all this?
Is it a job?
an engineering exercise?
a plum?

from what the data you have, which by the way are partial, not having given us the length of the boma.

Why do you say the results are almost accurate? who determined the exact results and how?

Thank you.

 

[Fabius3D]

It's a test. The measurements are what I put at the beginning. the results derive from an aleatory explanation of the teacher. the way, the method used is the one described in post #1. the boma measures, from text, 2.5 m.
thank you anyway. I didn't want to cheat on anyone.