Physics of sailing/windsurfing systems

[John Mcrain]

The problem here is that you have an active human with many joints as part of the system, so biomechanics is part of it. Any change of the vessels geometry affects the pose the human will take, and where his center of mass will end up.

Yes that is right.
What do you think about " mast foot pressure theory",does it failed like I state with my three counterquestions in post 29?

 

[A.T.]

What do you think about " mast foot pressure theory",

I don't see much theory there, just things that were empirically found to work. To verify how they work exactly, you would need motion analysis and force measurements, which is difficult to do on water.

 

[John Mcrain]

I don't see much theory there, just things that were empirically found to work. To verify how they work exactly, you would need motion analysis and force measurements, which is difficult to do on water.

But mast foot pressure is internal force,internal force can not change net moment of system..?
So you don't have explanation how sail power press nose down?

 

[willem2]

But mast foot pressure is internal force,internal force can not change net moment of system..?
So you don't have explanation how sail power press nose down?

It's hard to see what the problem is here.
If you look at the complete surfer/plank/system, there's a couple that would press the nose down, from the wind force and the drag, and there will have to be a couple from gravity + buoyant force in the opposite direction.

If you look at only the plank, there can't be a large net torque on it from the sail and the surfer for two reasons.

1. there is indeed no way for such a torque to be exerted on the plank. (unless you stand before the mast in a tailwind and push against the sail).

2. The only way to counteract such a torque, would be for the nose to go deeper in the water and for the buoyant force to act further forward. This is undesirable. You'd flip forward at higher speeds.

So the surfer will have to stand or lean backwards to counteract the torque from the sail.

 

[A.T.]

But mast foot pressure is internal force,internal force can not change net moment of system..?
So you don't have explanation how sail power press nose down?

As I told you in post #3 already, you have to decide if you are analyzing the moments on the whole system or just on the board.

 

[John Mcrain]

So the surfer will have to stand or lean backwards to counteract the torque from the sail.

For sure rider must lean back more when sail power increase,otherwise rider will be catupulted with sail...
When catapult happened rider and sail are catapuled but board not rotate with them,because joint can't transfer moment to board..



If sail has fixed connection to board ,when catapult happened board will be also rotate in crash,like sailing boat ..
This is main difference in joint and fixed connection..

 

[John Mcrain]

As I told you in post #3 already, you have to decide if you are analyzing the moments on the whole system or just on the board.

Today I make some calculations ,it seems that is impossible to press nose down if harness lines/hands are 100% horizontal.That math say..

Sailor must transfer some of his weight to the sail/mast foot to press nose down,but that is only possible if harness lines /hand have some downward angle...
So it seems popular "mast foot pressure" explanation is correct..
http://www.guycribb.com/userfiles/documents/downforce.pdf

first for 70degress lean back angle

sail power increas,sailor lean back at 20degress.

I get it in both case board lift in same position,below rider feet,1m from joint,so nose is not press down even drive force from sail increase

 

[jbriggs444]

Sailor must transfer some of his weight to the sail/mast foot to press nose down,but that is only possible if harness lines /hand have some downward angle...

The angle of the hands/line is irrelevant in any case to the torque exerted on the rest of the system by the sailor. The only controls the sailor has in that regard are:

1. Where he places his center of mass.
2. How he manipulates control surfaces to affect the pressure of wind and wave.

 

[John Mcrain]

The angle of the hands/line is irrelevant in any case to the torque exerted on the rest of the system by the sailor. The only controls the sailor has in that regard are:

1. Where he places his center of mass.
2. How he manipulates control surfaces to affect the pressure of wind and wave.

I do math and there is no mast foot pressure if lines /hands are 100% horizontal.

If you put lines more vertical,you can hold more sail power ,so drive force increase which has effect on system..

In same manner it is easier to hold falling tree if you attached rope at top instead at bottom of tree..
isnt it?

 

[jbriggs444]

I do math and there is no mast foot pressure if lines /hands are 100% horizontal.

If you put lines more vertical,you can hold more sail power ,so drive force increase which has effect on system..

In same manner it is easier to hold falling tree if you attached rope at top instead at bottom of tree..
isnt it?

None of that negates what I just finished saying.

Also, if you work the math, you should find that holding a falling tree is (for a fixed distance from the tree) is pretty much just as easy with a moderately high attachment point as a much higher attachment point. It is the fact that a higher attachment point allows for a more remote standing point that makes the opposite hold true.

It would help if you decided what you want to hold fixed and what you want to vary.

 

[John Mcrain]

Also, if you work the math, you should find that holding a falling tree is (for a fixed distance from the tree) easier with the rope lower.

It would help if you decided what you want to hold fixed and what you want to vary.

You want to say that if I am 10m meters from tree,it is easier to hold tree with rope attached low than at top of tree?
That can't be true,if you want I can prove with with math?

rider has fixed feets position because he is in footstraps all the time,that are at the back of board.
I am interest why when sail power increase(sheet in) board of nose is press down and when he sheet out ,nose goes up...

So sail power is vary but also rider lean angle because he must lean down more when sail power increase ,otherwise he will be catapult..

 

[jbriggs444]

You want to say that if I am 10m meters from tree,it is easier to hold tree with rope attached low than at top of tree?
That can't be true,if you want I can prove with with math?

I'd updated the claim before you posted to "nearly as easy". But I can do the math.

With an ideal rope pulling from ground level a fixed distance from the tree with a fixed tension and an attachment point up the tree at an angle of $\theta$ from the horizontal, you get a torque proportional to $\sin \theta \cos \theta$. It is a well known trig identity that this is equal to $\frac{sin \ 2 \theta}{2}$. So in this idealized case, your best attachment point is at a 45 degree angle above the horizontal.

 

[A.T.]

rider has fixed feets position because he is in footstraps all the time

But he has two feet, and can transfer the weight between them. So the effective center of pressure of the feet combined is not fixed.

 

[John Mcrain]

The angle of the hands/line is irrelevant in any case to the torque exerted on the rest of the system by the sailor.

Are you sure in that?

if you have massless beam,and man with 100kg lean back like on picture,and horizontal drive force.
In case lines are horizontal ,left weight scale will allways show 0 kg,right weight scale will allways show 100kg
(left weight scale showing 0kg is "indicator "that nose is not press down)

If lines has downward angle ,left weight scale will allways show weight greather than 0kg,right scale will show less than 100kg

 

[A.T.]

Are you sure in that?

if you have massless beam,and man with 100kg lean back like on picture,and horizontal drive force.
In case lines are horizontal ,left weight scale will allways show 0 kg,right weight scale will allways show 100kg
(left weight scale showing 0kg is "indicator "that nose is not press down)

If lines has downward angle ,left weight scale will allways show weight greather than 0kg,right scale will show less than 100kgView attachment 279038

If the force on the upper mast and the center of mass of the man are the same, the readings of the scales will also be the same.

 

[DrStupid]

If the force on the upper mast and the center of mass of the man are the same, the readings of the scales will also be the same.

Sure? If the mast is placed over the left scale and the vertical force acting on the man cancels his weight, than the left scale should read 100 kg and the right 0.

 

[A.T.]

Sure? If the mast is placed over the left scale and the vertical force acting on the man cancels his weight, than the left scale should read 100 kg and the right 0.

There are 4 external forces on the whole system. If sail force and weight don't change (same vectors and points of applicaton), then neither will the two scale forces change.

 

[DrStupid]

There are 4 external forces on the whole system. If sail force and weight don't change (same vectors and points of applicaton), then neither will the two scale forces change.

I still don't see it. If the center of mass remains fixed then weight must be compensated by the sum of the two scale forces. But why need the individual scale forces to be the same?

 

[jbriggs444]

I still don't see it. If the center of mass remains fixed then weight must be compensated by the sum of the two scale forces. But why need the individual scale forces to be the same?

Both net force and net torque remain fixed. That is two parameters that must be matched. Two equations. Two unknowns (the scale forces). The system can be solved.

 

[DrStupid]

Both net force and net torque remain fixed.

Do they? Of course the net torque changes if the scale forces change (while their sum remains constant). But that's what this thread is all about.

 

[A.T.]

Do they? Of course the net torque changes if the scale forces change (while their sum remains constant). But that's what this thread is all about.

I'm assuming both setups in post #44 are supposed to be in equilibrium: The net external force and the net external torque on the whole system are zero.

 

[DrStupid]

I'm assuming both setups in post #44 are supposed to be in equilibrium: The net external force and the net external torque on the whole system are zero.

Than we were talking at cross-purposes. I was assuming that the pitching moment may change. Maybe @John Mcrain need to clarify what he had in mind.

 

[A.T.]

Than we were talking at cross-purposes. I was assuming that the pitching moment may change. Maybe @John Mcrain need to clarify what he had in mind.

I was replying to post #44. Nothing in that post suggests that the "horizontal drive force" (on the mast) is different for the two cases, or that the center of mass has moved. That's all you need to know to tell that the scale forces remain the same.

 

[hutchphd]

Rigid body: So long as nothing massive moves (rope has no mass) the body (man+sail+board) is a rigid body. Rigid bodies subject to external forces always obey Newton's Laws. The same forces always produce the same result.

 

[John Mcrain]

Than we were talking at cross-purposes. I was assuming that the pitching moment may change. Maybe @John Mcrain need to clarify what he had in mind.

In post #44 I just want to prove that @jbriggs444 is wrong when say that harness lines/hands angle is irrelvant .

If lines are horizontal,left scale will allways show 0kg,that mean you can't press nose down with horizontal lines/hands.

 

[John Mcrain]

If the force on the upper mast and the center of mass of the man are the same, the readings of the scales will also be the same.

What are you compare ,left and right picture?

Force at mast at right picture,must be greater then on left(if everything else/geometery and sailor mass are the same,except lines angle),because lines are attached higher at mast..

So forces at mast can't be same in left and right picture..

 

[John Mcrain]

I was replying to post #44. Nothing in that post suggests that the "horizontal drive force" (on the mast) is different for the two cases, or that the center of mass has moved.

You must see without any calculation,that "horizontal drive forces" at masts must be different,becasue line on right picture is connect at higher point,so right drive force is greater then left...

 

[A.T.]

Force at mast at right picture,must be greater then on left(if everything else/geometery and sailor mass are the same,except lines angle)

If the force at mast changes, while the center of mass remains the same, the scale forces will change as well.

 

[jbriggs444]

You must see without any calculation,that "horizontal drive forces" at masts must be different,becasue line on right picture is connect at higher point,so right drive force is greater then left...

You need to specify what you are holding fixed and what you are allowing to vary.

If you hold the wind force on the sail constant and hold the position of the surfer's center of mass constant but vary the angle of the line while requiring the system to remain in equilibrium then the line tension must change. The resulting scale readings will not change in this case. [The line tension can change if the surfer adjusts posture and foot position while leaving his center of mass stationary]

If you hold the line tension and the position of the surfer's center of mass constant but vary the angle of the line while requiring the system to remain in equilibrium then you must be varying the force of the wind. The resulting scale readings will change in this case.

 

[John Mcrain]

You need to specify what you are holding fixed and what you are allowing to vary.

If you hold the wind force on the sail constant and hold the position of the surfer's center of mass constant but vary the angle of the line while requiring the system to remain in equilibrium then the line tension must change. The resulting scale readings will not change in this case. [The line tension can change if the surfer adjusts posture and foot position while leaving his center of mass stationary]

If you hold the line tension and the position of the surfer's center of mass constant but vary the angle of the line while requiring the system to remain in equilibrium then you must be varying the force of the wind. The resulting scale readings will change in this case.

Lets firs solve this case.

Force at mast and harness line(blue rope) must be horizontal and system must be in equlibrium.
(catapult is not equlibrium)

Everything else you can change if you want,lean back angle,distance from joint to feet,weight but provided that force at mast and harness line are horizontal...

Tell what will left weight scale show? If you think that left weight scale can show number that is not 0kg,explain how..

I tell left weight scale will allways show 0 kg...