Physics of sailing/windsurfing systems

[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),becuase 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...

 

[jbriggs444]

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 etc etc

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...

View attachment 279068

Let us fill in some numbers. The line is 1 meter above the board. The wind force is centered $H$ meters high and has magnitude $F$. The person has his center of mass sitting $R$ meters back from the mast. The line is at height $h$ above the deck. The surfer's center of mass is also at this same height above the desk. Just to keep everything symbolic, let us use $m$ for the surfer's mass, even though we are told that it is 100 kg.

We assume that the person adjusts his lean angle as needed (without adjusting his center of mass) to provide enough tension to keep the mast from falling over.

We can begin by calculating the required tension ($t$) in the line. Based on a torque balance on the mast about its base we can immediately write down an equation:$$ht=HF$$ We can easily solve this for t and deduce that:$$t=F\frac{H}{h}$$

Now let solve another force balance to determine the offset of the surfer's feet. We can do a torque balance about the surfer's feet. His center of gravity is $r$ meters back from the feet. And the rope tension is $h$ meters above the feet. So we can write down:$$mgr = ht$$ Solving for r, we get: $$r=\frac{ht}{mg}$$ But we already know that $t=F\frac{H}{h}$. And substituting that in for t, we get that: $$r=\frac{FH}{mg}$$
You should be seeing what is going on by now. With this model, as the wind force is increasing, the surfer's feet are moving toward the mast.

The position of the 100 kg (force) load from surfer on board is moving from stern to bow as the wind force increases. Accordingly, the scale forces skew more and more toward the forward scale.

We could keep going and, given an assumption that the right hand scale is directly under the surfer's center of mass, and the distance between the two scales, determine the resulting scale readings, but surely there is no need?

 

[John Mcrain]

Let us fill in some numbers. The line is 1 meter above the board. The wind force is centered $H$ meters high and has magnitude $F$. The person has his center of mass sitting $R$ meters back from the mast. The line is at height $h$ above the deck. The surfer's center of mass is also at this same height above the desk. Just to keep everything symbolic, let us use $m$ for the surfer's mass, even though we are told that it is 100 kg.

We assume that the person adjusts his lean angle as needed (without adjusting his center of mass) to provide enough tension to keep the mast from falling over.

We can begin by calculating the required tension ($t$) in the line. Based on a torque balance on the mast about its base we can immediately write down an equation:$$ht=HF$$ We can easily solve this for t and deduce that:$$t=F\frac{H}{h}$$

Now let solve another force balance to determine the offset of the surfer's feet. We can do a torque balance about the surfer's feet. His center of gravity is $r$ meters back from the feet. And the rope tension is $h$ meters above the feet. So we can write down:$$mgr = ht$$ Solving for r, we get: $$r=\frac{ht}{mg}$$ But we already know that $t=F\frac{H}{h}$. And substituting that in for t, we get that: $$r=\frac{FH}{mg}$$
You should be seeing what is going on by now. With this model, as the wind force is increasing, the surfer's feet are moving toward the mast.

The position of the 100 kg (force) load from surfer on board is moving from stern to bow as the wind force increases. Accordingly, the scale forces skew more and more toward the forward scale.

You can't move sailor feet toward mast,feet is stuck in footstraps all the time ,they are allways above right weight scale.

 

[jbriggs444]

You can't move sailor feet toward mast,feet is stuck in footstraps all the time ,they are allways above right weight scale.

Again, you need to spell out what is held fixed and what is allowed to change.

It seems that you now want to fix the position of the sailors feet, allow the position of his center of mass to vary and retain a demand that the mast be in equilibrium.

So the sailor must extend his legs (or lean back and lift his arms to maintain a horizontal line) in this case, changing the external torque of gravity on the system thereby balancing with the changing external torque of wind on the system.

With that model, changing the wind force does not change the scale readings.

The attachment point of the line remains irrelevant to the scale readings. Only the external forces enter in.

 

[John Mcrain]

Again, you need to spell out what is held fixed and what is allowed to change.

It seems that you now want to fix the position of the sailors feet, allow the position of his center of mass to vary and retain a demand that the mast be in equilibrium.

So the sailor must extend his legs (or lean back and lift his arms to maintain a horizontal line) in this case, changing the external torque of gravity on the system thereby balancing with the changing external torque of wind on the system.

With that model, changing the wind force does not change the scale readings.

The attachment point of the line remains irrelevant to the scale readings. Only the external forces enter in.

I am talking about windsurfing all the time,board has footstraps at tail and when you planning, both feet are in footstraps all the time..Boom is sailor "throttle", he adjust how much sail power he need with sheet in(increase angle of attack) and sheet out(decrease AoA).
Also sailor must lean out to balance sail torque..If gust is to strong and sailor don't open sail(sheet out) at time, he will be catapulted.

look at video



1)So my question is,when sailor extend his legs to lean out more to compensate increase in sail power,why is nose of board press down?
2) Does downward angle of harness line/hands are needed to press nose down or this can be achieved with horizontal lines/hands too?

NOTE:(horizontal lines/hands are never case is real windsurfing,I choose this deliberately,to easeir find out why nose is press down when sailor sheet in..In real windsurfing there is allways some downward angle of lines/hands..)

 

[jbriggs444]

Pick a model and we can calculate. Fail to pick a model and we can't. It is that simple.

 

[John Mcrain]

Pick a model and we can calculate. Fail to pick a model and we can't. It is that simple.

Are you still with your state that angle of harness line/hands is irrelevant for press nose down("board pitching moment")?

 

[jbriggs444]

Are you still with your state that angle of harness line/hands is irrelevant for press nose down("board pitching moment")?

Yes. Given a fixed set of external forces, it is definitely irrelevant. But you have not clarified if a fixed set of external forces is a characteristic of a model that you can accept.

 

[John Mcrain]

Yes. Given a fixed set of external forces, it is definitely irrelevant. But you have not clarified if a fixed set of external forces is a characteristic of a model that you can accept.

Do you aware that with downward angle lines sailor can hold more sail force than horizontal line for same sailor lean out angle?

 

[jbriggs444]

Do you aware that with downward angle lines sailor can hold more sail force than horizontal line for same sailor lean out angle?

So what? If the external forces are fixed, that is irrelevant. Please decide what question you are asking.

 

[John Mcrain]

So what? If the external forces are fixed, that is irrelevant. Please decide what question you are asking.

External forces are not fixed,when sailor hold sail at 4 degrees AoA(relativly open sail-sheet out) sail force is way smaller compare when he sheet in and make sail at 18 degrees AoA..

So I am going from low sail power to high sail power ,and question is why when I sheet in(increase sail power) nose goes down?