Undergrad Physics of sailing/windsurfing systems

[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"