Physics of sailing/windsurfing systems

[russ_watters]

I don't get it,what is your results for case at post #131?

If none of the forces change, then none of the forces change. But I'm not sure you had the right answer to begin with; I'm getting a different answer now (I only checked the first case and didn't realize it they weren't really in agreement because they were close).

With a 20 degree angle, the horizontal component of weight is 1000 cos(20) = 940 N.
The height of the attachment point of the harness is sin (20) = 0.342 m
The torque about the root of the mast is .342 * 940 = 321 N-m
The propulsion force is 321 / 2 = 161 N.

 

[pbuk]

No ,you can see in my calculations, it is attached at sin20x1=0.342m

Yes, you've shortened the lever arm which provides the torque about the mast foot from the sailor and as a result you have had to reduce the wind force to keep everything in balance under the sailors feet.

 

[John Mcrain]

Yes, you've shortened the lever arm which provides the torque about the mast foot from the sailor and as a result you have had to reduce the wind force to keep everything in balance under the sailors feet.

I use sin20x1=0.342m for 20degress lean out angle
and sin70x1=0.939m for 70degrees case

What is wrong with this?

 

[John Mcrain]

If none of the forces change, then none of the forces change. But I'm not sure you had the right answer to begin with; I'm getting a different answer now (I only checked the first case and didn't realize it they weren't really in agreement because they were close).

With a 20 degree angle, the horizontal component of weight is 1000 cos(20) = 940 N.
The height of the attachment point of the harness is sin (20) = 0.342 m
The torque about the root of the mast is .342 * 940 = 321 N-m
The propulsion force is 321 / 2 = 161 N.

You must know that attachment point,and lean angle determine magnitude of mast force
Differnece between fixed connection and joint is almost double it is not rounding error..

 

[John Mcrain]

once again two cases
Task is find mast force,lift position and magnitude,for given geometry and weight at picture

20degress with joint
case 20degrees
1) find tension in rope T
calculate from moment at feet Mfeet=0
100kgxcos20 -Txcos70=0
T=274.4kg

2)calculate mast force
Mjoint=0
-Fx2m + 274.4kg x0.342m=0
F=46.9kg...460N

3)find Lift position and magnitude
magnitude is same as sailor weight...100kg=1000N clear as glass

Mjoint=0
-46,9kgx2m +100kgx1,9396m -100kg x X= 0
X=1m ...1meter right from joint is sailor feet..

mast with joint,20degress
results:
Mast force= 460N
Lift position under sailor feet

fixed mast ,20 degrees
mast force will be
100kgx1,939m -Fx2m=0
F=96,98kg...951N

And lift magnitude will be 100kg..1000N and lift position will be under mast
fixed mast ,20degress
results:
mast force=951N
lift position under mast

What is wrong with this calculations??

 

[russ_watters]

Nevermind on my check on the calcs...

Solve for tension. $t = mg \cot \theta$. Multiply by sailor height above deck ($h \sin \theta$) and divide by mast height (H) to get force at top of mast. If you use g=10 m/s^2, h=1m, H=2m the numbers check out.

...I entered cos instead of cot.

In any case, this doesn't change the point we're at now.

 

[russ_watters]

Differnece between fixed connection and joint is almost double it is not rounding error..

A fixed connection doesn't on its own change anything. The forces and torques still have to balance. If the forces and torques are all balanced and the connection point has zero torque on it, then changing it to a fixed connection doesn't change the torque. Or to put it another way; since all joints are pins, none of them have torques around them. Changing them to fixed doesn't create a torque where none exists anyway.

Specifically:

[Case 1:]
1) find tension in rope T
calculate from moment at feet Mfeet=0
100kgxcos20 -Txcos70=0
T=274.4kg

-Fx2m + 274.4kg x0.342m=0
F=46.9kg...460N

[Case 2:]
100kgx1,939m -Fx2m=0
F=96,98kg...951N

For case 1, where does the 100 kg x 1.939m come from? And what happened to the tension on the harness (274.4*10=2744 N)?

[edit] Oh, I see it now; the 1.939 is the distance from the rider's COM to the mast. What does that have to do with anything? What do you think that is/is doing? It looks like you arbitrarily chose to increase the driving force and moved the center of lift to compensate. Why?

 

[John Mcrain]

A fixed connection doesn't on its own change anything. The forces and torques still have to balance. If the forces and torques are all balanced and the connection point has zero torque on it, then changing it to a fixed connection doesn't change the torque. Or to put it another way; since all joints are pins, none of them have torques around them. Changing them to fixed doesn't create a torque where none exists anyway.

Specifically:

For case 1, where does the 100 kg x 1.939m come from? And what happened to the tension on the harness (274.4*10=2744 N)?

[edit] Oh, I see it now; the 1.939 is the distance from the rider's COM to the mast. What does that have to do with anything? What do you think that is/is doing?

You mean on this part in case 1?
"Mjoint=0
-46,9kgx2m +100kgx1,9396m -100kg x X= 0
X=1m ...1meter right from joint is sailor feet.."

Once when I calculate external mast force using internal geometry from picture,I am using external forces to find position of lift force.

 

[russ_watters]

You mean on this part in case 1?
Mjoint=0
-46,9kgx2m +100kgx1,9396m -100kg x X= 0
X=1m ...1meter right from joint is sailor feet..

Once when I calculate external mast force using internal geometry from picture,I am using external forces to find position of lift force.?

I see it now. So you've changed a lot more than just fixing the mast. Why did you move the lift force and increase the driving force? You didn't have to.

 

[John Mcrain]

Why did you move the lift force and increase the driving force? You didn't have to.

Are you kidding me or what?

I have given geometry and sailor weight at picture for these two caseses ,I must find/CALCULATE mast force,lift force position and magnitude..

So I didnt move lift force position and increase mast force by mayself deliberlty..
They end up like this when I calculate...Math did it...They are results,not something that I can choose..I can't believe what question you ask me!
At how many topics you are in same time??

 

[russ_watters]

Are you kidding me or what?

I have given geometry and sailor weight at picture for these two caseses ,I must find/CALCULATE mast force,lift force position and magnitude..

So I didnt move lift force position and increase mast force by mayself deliberlty..
They end up like this when I calculate...Math did it...They are results,not something that I can choose..I can't believe what question you ask me!
At how many topics you are in same time??

No, you made the choice. You chose the lift force location when you chose the axis to zero the moments about. It's not a coincidence or result of the calculations. Put the lift force back at his feet and re-calculate.

 

[John Mcrain]

No, you made the choice. You chose the lift force location when you chose the axis to zero the moments about. It's not a coincidence or result of the calculations. Put the lift force back at his feet and re-calculate.

Yes if use lift force fixed at feet position,then mast force will be same as in case with joint...But position of lift force is my "indicator" of wetted surface,if lift force is shift toward nose ,that mean nose is press down and wetted surface increase..so I put lift force position as unknown,to see what happened with board wetted surface

So even for second case,sail force increase almost double,wetted surface is increased so drag will go up..

 

[russ_watters]

Yes if use lift force fixed at feet position,then mast force will be same as in case with joint...

Good. I feel like that's a gigantic revelation for you...

But position of lift force is my "indicator" of wetted surface,if lift force is shift toward nose ,that mean nose is press down and wetted surface increase..so I put lift force position as unknown,to see what happened with board wetted surface

I don't really know what you are trying to do here or why. The board has buoyancy based on the volume of water displaced (edit; well, more complicated, it is at least partially a planing hull when moving fast). That value is constant in the case of a horizontal driving force. It doesn't matter where you put it. If it's aft, the tail end has to sink. If it's centered the board has to sit more level, but higher. Putting it all the way aft is of course impossible, but it's your model and I wasn't quibbling with that. What this has to do with the discussion though, I don't know.

So even for second case,sail force increase almost double,wetted surface is increased so drag will go up..

Again, these haven't been part of the model, so I don't know why you are bringing them in now/what the point is. Frankly, I really don't know where any of this is going.

 

[jbriggs444]

Nevermind on my check on the calcs...

...I entered cos instead of cot.

I did $\cot$ instead of $\cos$ on my first attempt to reproduce "mast force". Then figured out what quantity "mast force" was supposed to represent.

In any case, this doesn't change the point we're at now.

I know, I know... Right back where we started.

 

[John Mcrain]

I did $\cot$ instead of $\cos$ on my first attempt to reproduce "mast force". Then figured out what quantity "mast force" was supposed to represent.

I know, I know... Right back where we started.

If I now must have fixed lift force under mast for both case,and must find mast force ,system must be in eqlibrium
For fixed case F= (100x1.939) /2m= 96.95kh x 9.81=951N

keep geometry the same,just change fixed mast with joint connection
it seems this configuration can not be in equlibrium,what prevent that right side of board don't flip down?

Where I am wrong again?

 

[jbriggs444]

it seems this configuration can not be in equlibrium,what prevent that right side of board don't flip down?

Where I am wrong again?

You've over-constrained the set up.

With two supports and a scale at each one, there was something to calculate. No problem.
With one support there is a constraint instead -- a thing that the calculations must match. And they don't.

 

[russ_watters]

If I now must have fixed lift force under mast for both case,and must find mast force...

Why is that a "must" now? That's oddly worded. We're still on the same page that you've chosen to move it, right?

In any case, I agree with the second case not being in equilibrium. Now what?

Listen; the other moderators and some members are pressuring me to close the thread because it's unproductive and a waste of time. So you need to get to a point, rapidly. As far as I can tell, your original misunderstanding has been corrected, but I'm not even completely sure of that. Can you confirm that? So, what is the point of the discussion now?

 

[John Mcrain]

In any case, I agree with the second case not being in equilibrium. Now what?

Imagine joint has pin,so when pin is inside ,joint is locked(fixed conection) and when you remove pin it work like joint.

first pin is inside,joint is locked(fixed conection) and everything is balance.mast force is 951N..
now I just remove pin from joint,system become out of balance ,right side of board will flip down

So how you can say that I can consider system with joints like one rigid thing if I just remove pin(switch from fixed connection to joint/"only" change internal conection) and system is suddenly out of balance.?

That is all I want to say all the time...
When you explain this ,topic is finished..

 

[russ_watters]

Imagine joint has pin,so when pin is inside ,joint is locked(fixed conection) and when you remove pin it work like joint.

first pin is inside,joint is locked(fixed conection) and everything is balance.mast force is 951N..
now I just remove pin from joint,system become out of balance ,right side of board will flip down

So how you can say that I can consider system with joints like one rigid thing if I just remove pin(switch from fixed connection to joint/"only" change internal conection) and system is suddenly out of balance.?

OMG. The system we were talking about yesterday had no internal torques around the attachment points. It was in equilibrium/balanced. That's what a pin joint is for. This setup has internal torques. They behave differently because they are different.

Yes, a triangle of members is intrinsically rigid and a four member structure is not.

Again, where is this going/what does it have to do with the original question?

 

[John Mcrain]

They behave differently because they are different.

That I was talking from my first post...
Wait a minute,didnt you say that internal arranging can not change net moment of system?

 

[pbuk]

Moving the support from under the feet to under the mast is not an internal rearrangement.

 

[John Mcrain]

Moving the support from under the feet to under the mast is not an internal rearrangement.

Read my post #158

I just change fixed conection into joint connection,lift is under mast in both case.

So I just remove pin / unclok the joint,and system "crash",as expected..

 

[russ_watters]

Wait a minute,didnt you say that internal arranging can not change net moment of system?

If the system is internally rigid, the shape doesn't matter. The human body is a complex machine and different rules may apply in different situations/configurations. You keep changing the situations and then get confused or upset when it changes the rules.

 

[russ_watters]

The human body can act like a one member system, two member system or ten member system. It can be internally rigid or not. It can be dynamically adjusted to act internally rigid. It can have two members and behave rigid, turning your four member, movable system into a three member, rigid system.

 

[pbuk]

I just change fixed conection into joint connection,lift is under mast in both case.

But nobody said that removing a pin does not change anything, this is what @russ_watters said:

since all joints are pins, none of them have torques around them. Changing them to fixed doesn't create a torque where none exists anyway.

Here there is a net torque at the mast foot because the force from the sailor on the rope attaching him to the mast does not balance the torque from the wind.

 

[pbuk]

Wait a minute,didnt you say that internal arranging can not change net moment of system?

Where did anyone say that? When you misquote people like that it is very difficult to believe that you are genuine and are not simply trying to create an argument.

 

[hutchphd]

 

[meekerdb]

Sailing boat has pitching moment caused by sail drive force and hydrodnamic drag froce in the water,distance between them is lever arm.
It cause nose of boat to push down..

Does windsurfing has pitching moment and if yes how it is transfer to the board ,if sail is connected to board with universal joint and sailor feet also act as joint. But joint can not transfer moment..?
I don't understand physics of this..View attachment 278922

A joint can transmit torque if it has muscles and ligaments attached to it. But what's going on here is that the surfer is holding the boom down and keeps the mast from tipping forward, just like the back stay on a sailboat. You know it wouldn't matter if the back were fastened via a ball joint.

 

[John Mcrain]

You've over-constrained the set up.

With two supports and a scale at each one, there was something to calculate. No problem.
With one support there is a constraint instead -- a thing that the calculations must match. And they don't.

I am at beach and windsurfer sailing towards me,so I see his front view.
What I know:
I see board is flat,that mean lift from board act at board centerline.
I know he is using sail that has center of pressure at 2m above board.(assume sail is upright for easier calculation)
Sailor has 100kg,lean out angle 20degrees,board is wide 1m(feet to centerline=0.5m),horizontal harness line,sailor c.g. is same spot where line is attached,from feet to c.g. is 1m

I have case a) and b) for hydro side force position..

Task is find sail side force.

case a) I assume that side force is in board level,let say he don't have fin so board rail produce all hydro side force..

Here I can not calculate sail side force like wsurf has fixed sail to board-rigid thing(F=100kg x 1.43m / 2m =71.5kg=701N)
I see from logic that system can not be in balance if sail and board is conected with joint,because joint can not transfer moment from sail force to keep right side of board UP,so right side of board will flip down...

I also see that if sail and mast is fixed(sailing boat configuration) this system will be in balance.
So I understand that members suggest me that I can allways look at system like it is one rigid thing(fixed connection),but if I do this at this case I will make mistake..

So this is my example why you need to care about type of connection between board and sail when do calculation,that is all I want to say from my first post,but obviusly I am really bed in explaining what I want to say..
case b) I know that he use fin that has center of pressure 0.35m below board

in this case only"hydro side force" is at different position and suddley I can look at system as it has fixed connection(even it don't has) and still get correct result,that fascinates me.
F=100kg x 1.43m / 2.35m= 60,85kg = 596N
Conclusion question:
We see that case a) is only possible if sail and mast is fixed.So if we use fixed conection we have benfits that hydro side fore can be in board level,at the end that has consequnece to offer greater righting moment=greater sail force=greater speed..
Am I right?

In other words,can fixed type of connection improve wsurf-system maximum speed potential,theoretically ?
(I say theoretically ,because fixed connection will never work for regular wsurf for other reasons..)

 

[pbuk]

I am at beach and windsurfer sailing towards me,so I see his front view.

So everyone is clear I will point out that the previous 168 messages in this thread have been talking about a side view, now you want to look at a front view.

I see board is flat,that mean lift from board act at board centerline.

If the sailor is standing on the rail like that the board will not be flat: they would have to have their feet nearer the centre line.

...
...

The 'attachment point to the mast' (which does not exist on a real windsurf board of course, the sailor holds on to the boom) is far to low.

I see from logic that system can not be in balance if sail and board is conected with joint,because joint can not transfer moment from sail force to keep right side of board UP,so right side of board will flip down...

It won't 'flip down' because it is supported by the hydrostatic/hydrodynamic buoyancy/planing forces, but it certainly won't be horizontal as I noted above. With 100kg on the rail unless it is a very buoyant board with a very stiff UJ it is likely to capsize.

So I understand that members suggest me that I can allways look at system like it is one rigid thing(fixed connection),but if I do this at this case I will make mistake..

Nobody has ever said that.

So this is my example why you need to care about type of connection between board and sail when do calculation,that is all I want to say from my first post,but obviusly I am really bed in explaining what I want to say..

No, you are not bad at explaining you are bad at listening. You need to listen to what has been said: you need to look at the whole system to see what forces and torques apply, you cannot simply say 'all systems are rigid' or 'all joints are flexible' and nobody has said that.

case b) I know that he use fin that has center of pressure 0.35m below board
... [some more dubious fin calculations]

The fin is not there to keep the board upright, it is there to stop it (i) slipping sideways and (ii) turning away from the wind.

Conclusion question:
We see that case a) is only possible if sail and mast is fixed.So if we use fixed conection we have benfits that hydro side fore can be in board level,at the end that has consequnece to offer greater righting moment=greater sail force=greater speed..

Yes you can hike out harder on a dinghy, however what you can't do on a dinghy is bring the centre of effort back and to windward as shown in any video of a windsurfer sailing fast. There is also much greater weight and (except for foiling dinghies) wetted surface area in a dinghy and the advantages of a larger righting moment are generally outweighed by these disadvantages so in general windsurfers can sail faster than dinghies.

Am I right?

No.

In other words,can fixed type of connection improve wsurf-system maximum speed potential,theoretically ?
(I say theoretically ,because fixed connection will never work for regular wsurf for other reasons..)

No, because you would lose the ability to move the centre of effort and as stated above this is key to the speed advantage of a windsurfer. Note however that all UJs have some degree of stiffness and depending on wind and wave conditions, board dimensions and ability a stiff power or 'boge' joint may be faster than a flexible tendon joint, or vice versa.

 

[jedishrfu]

Now seems like a good time to close this thread. The OPs question has been answered in numerous ways from a variety of perspectives with @pbuk s excellent commentary we can now conclude this topic.

Thank you all for contributing here.

Jedi