I Physics of sailing/windsurfing systems

[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?

 

[A.T.]

board has footstraps at tail and when you planning, both feet are in footstraps all the time..

What is your point here? Did you read my post #43?

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.

This is just like having a foot that can move forward and backward along the board.

 

[jbriggs444]

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?

Why would it not? You've increased an external downward pitching torque. Surely one would expect the craft to nose down unless/until countered by another external torque. The details of the internal bracing do not change that.

If the craft is stable (not a certainty - it could be maintained in an unstable equilibrium by active controls -- e.g. a unicycle), one would expect a pitch down change to result in a restoring upward pitching torque.

 

[John Mcrain]

What is your point here? Did you read my post #43?

This is just like having a foot that can move forward and backward along the board.

Yes I am...

So only way to press nose down is thorugh sailor feet weight distribution?

What about sail "pitching moment"?

 

[A.T.]

So only way to press nose down is thorugh sailor feet weight distribution?

Who said that?

 

[John Mcrain]

Why would it not? You've increased an external downward pitching torque.

Because my brain can not understand how universal joint can transfer torque to the board?

If mast and board is connected with fixed conection(like all sailboat) than it is very easy to understand boat pitching caused by sail force..

If push at mast B,beam A will not rotate...let say joint allows 360 degress rotation

 

[pbuk]

Because my brain can not understand how universal joint can transfer torque to the board?

Because when the mast is held in place by the sailor the mast step is not acting as a universal joint. If the sailor let's go of the wishbone then the mast will rotate about the universal joint, in exactly the same way as the mast on a sailing boat will fall if any of the stays under tension breaks (or the gate collapses for an unstayed mast).

 

[John Mcrain]

Because when the mast is held in place by the sailor the mast step is not acting as a universal joint. If the sailor let's go of the wishbone then the mast will rotate about the universal joint, in exactly the same way as the mast on a sailing boat will fall if any of the stays under tension breaks (or the gate collapses for an unstayed mast).

So I can treat mast like is welded to the board and use only external forces for calculating pitch torque?

 

[pbuk]

Why are you picking on the pitching force: do you think that there is no heeling force either?

 

[John Mcrain]

Why are you picking on the pitching force: do you think that there is no heeling force either?

Because pitch torque put nose down...and this is theme of this topic..

 

[pbuk]

OK, let's leave heel out of this.

So I can treat mast like is welded to the board and use only external forces for calculating pitch torque?

Yes, as long as the board/sailor/rig system is entirely rigid (which it is not in real life of course, the sailor is continuously pumping the rig).

 

[A.T.]

Because my brain can not understand how universal joint can transfer torque to the board?

This was explained in post #8 already:

That's not the only connection between the sail and the board. The surfer also connects them. Two universal joints in combination can transfer moments.

 

[John Mcrain]

OK, let's leave heel out of this.

Yes, as long as the board/sailor/rig system is entirely rigid (which it is not in real life of course, the sailor is continuously pumping the rig).

Yes but in windsurf ,internal angles of lines and geometry determine how much will be magnitude of sail force,which at the end determine how much will be pitch torque..
So I can't neglect internal "geometry"..

 

[A.T.]

So I can treat mast like is welded to the board and use only external forces for calculating pitch torque?

When you connect three beams via universal joints to a triangle, you have a rigid shape. A force applied to one corner will create a torque on the opposite beam. Do you have a problem with two universal joints transferring a torque here?

 

[pbuk]

Yes but in windsurf ,internal angles of lines and geometry determine how much will be magnitude of sail force,which at the end determine how much will be pitch torque..
So I can't neglect internal "geometry"..

No, the flow of the wind over the sail determines the magnitude and direction of the sail force. You must ignore internal geometry.

 

[John Mcrain]

No, the flow of the wind over the sail determines the magnitude and direction of the sail force. You must ignore internal geometry.

Yes wind increase sail force,but useing optimal "internal geometry" allow sailor to hold more sail power,increase his "righting moment".
For example if you put boom at 1m height,your righting moment is very small..

harness line angle only don't change sailor righting moment if sail feet is at centerline,in line with mast.
But sailor feet can be even 0.5m out of center line ,which increase righting moment

 

[John Mcrain]

Do you have a problem with two universal joints transferring a torque here?

yes..

 

[John Mcrain]

You must ignore internal geometry.

Do you see as long as harness line(blue rope) is 100% horizontal,you can not press left weight scale?
Scale will read 0kg..

 

[jbriggs444]

Do you see as long as harness line(blue rope) is 100% horizontal,you can not press left weight scale?

View attachment 279086

Again, what are you holding fixed and what are you allowing to vary? If you hold the line angle fixed at directly horizontal, that constrains the relationship that can exist between center of mass, foot position and tension when the sail force is allowed to vary and the mast is required not to fall over.

 

[pbuk]

Do you see as long as harness line(blue rope) is 100% horizontal,you can not press left weight scale?
Scale will read 0kg..

View attachment 279086

No, the left scale will read > 0kg due to the torque from the wind force [edit] and the mass of the board and the rig. But this diagram is useless because (i) unless this sailor is on a dead run they will fall into leeward and (ii) the board is supported by a buoyant force not a pair of scales.

 

[John Mcrain]

Again, what are you holding fixed and what are you allowing to vary? If you hold the line angle fixed at directly horizontal, that constrains the relationship that can exist between center of mass, foot position and tension when the sail force is allowed to vary and the mast is required not to fall over.

feet are fixed above right weight scale,just like in wsurf,where feet can not move ,they are allways in footstraps
line is fixed in horizontal position as I said

sail force can very ,so sailor must lean back more and change line attached point to balance sail torque...

again as long as lines stay horizontal ,hydrodnamic lift will be allways under sailor feet..

But if you put lines at downward angle,you can now press left weight scale,so hydrodynamic lift is shift somewhere forward,nose is " press down"

 

[pbuk]

yes..

Well you need to get rid of that problem. How do you think that the rig knows that it has a flexible joint at the foot when it is held in place? Do you think that it is impossible for a rudder to turn a boat because it is attached by a flexible joint?

 

[pbuk]

feet are fixed above right weight scale,just like in wsurf,where feet can not move ,they are allways in footstraps
line is fixed in horizontal position as I said

sail force can very ,so sailor must lean back more and change line attached point to balance sail torque...

again as long as lines stay horizontal ,hydrodnamic lift will be allways under sailor feet..

In order to keep the harness line horizontal you would have to be 12 feet tall.

But if you put lines at downward angle,you can now press left weight scale,so hydrodynamic lift is shift somewhere forward,nose is " press down"

The only thing that changes the hydrodynamic lift is the amount of the board that is in the water when in displacement mode (which you never are of course) or the speed and angle of attack when in planing mode.

 

[John Mcrain]

No, the left scale will read > 0kg due to the torque from the wind force [edit] and the mass of the board and the rig. But this diagram is useless because (i) unless this sailor is on a dead run they will fall into leeward and (ii) the board is supported by a buoyant force not a pair of scales.

Board and rig has very little weight compare to sailor,so I use approximation as they are massless..I thought that was clear.. You are wrong, there is no any torque from the wind force, because joint can't transfer torque to board.
Line must be at angle to press left weight scale.

I really don't understand how you don't see that without any calcualtion,this must be crystal clear

 

[jbriggs444]

sail force can very ,so sailor must lean back more and change line attached point to balance sail torque...

So now the attachment point of line to mast is in a greased Teflon track so that it is free to slide up and down the mast as the sailor leans back. That is an interesting model. Let us put some numbers on it.

Let us make the simplifying assumption that the surfer's center of mass is at the same level as his hands (a distance h from his feet). We can do a force balance on the sailor and see that for a lean angle of $\theta$ (measured from vertical = 0) that the torque from the line must match the torque from gravity: $h t \cos \theta = h mg \sin \theta$.

The torque on the mast is also equal to $h t \cos \theta$. So the maximum torque on the mast is mgh which is approached in the limit as the sailor reaches the horizontal.

If the force from the wind exceeds this, the mast falls over. The only useful effect of the surfer's lean is that it puts his center of gravity farther back. And there is a limit on how far he can move his center of gravity with his feet stuck in the cleats.

 

[pbuk]

because joint can't transfer torque to board.

So what stops the mast from rotating forwards?

 

[John Mcrain]

So now the attachment point of line to sail is in a greased Teflon track so that it is free to slide up and down the mast as the sailor leans back. That is an interesting model.

Yes it can be like this if you want.
Or you can stop, lower the boom(because line is conected to boom) try this new geometry in second run and see what is board doing at water..

For sure sail force is limited by sailor max "righting moment",when sail torque is greather than that,sailor and sail is catapulted...

 

[pbuk]

48 hours ago you posted this:

I don't understand physics of this..

Why have you now decided that you understand it so well that you are going to tell everyone else that they are wrong?

 

[jbriggs444]

Yes it can be like this if you want.
Or you can stop, lower the boom(because line is conected to boom) try this new geometry in second run and see what is board doing at water..

For sure sail force is limited by sailor max "righting moment",when sail torque is greather than that,sailor and sail is catapulted...

Not what I want; what you have demanded. You want the line to be horizontal. So I am making the line horizontal.

Using this model, the scale readings are both unchanged until the moment when the maximum wind force is exceeded. That is because the increased down pitch from the wind torque is matched by the increased up pitch from the surfer's center of mass moving back.

If you change your mind about the model and use a line with a fixed attachment point on the mast, the situation changes. There is a vertical down force from line on mast and then from mast on board. The scale readings change as wind force increases because the mast force increases.

 

[John Mcrain]

48 hours ago you posted this:

Why have you now decided that you understand it so well that you are going to tell everyone else that they are wrong?

No I just tell you ,that you are wrong only for my let's call it "horizontal line case".
@jbriggs444 just confrim that I am correct for this specific case,read his posts and explanation..

 

[pbuk]

I suggest you take a look at this page http://joewindsurfer.blogspot.com/2008/04/jim-drakes-windsurf-physics.html and come back with any questions.