Physics of sailing/windsurfing systems

[jbriggs444]

Do you understand that left weight scale will show 0kg for every scenario where blue line and mast force is 100% horizontal,feet must stay above right scale,and system must be in equlibrium?

Do you understand that the wind force still exerts a torque in spite of the correctness of the assertion above?

Do you understand that you yourself have said that this toy model is unrealistic?

 

[russ_watters]

Do you understand that the wind force still exerts a torque in spite of the correctness of the assertion above?

Wait, is this about net torque vs individual torque components? Yeah; if the net torque on the system is zero, that doesn't mean that the individual torques don't exist, it just means they sum to zero.

 

[A.T.]

once again
Do you understand that left weight scale will show 0kg for every scenario where blue line and mast force is 100% horizontal,feet must stay above right scale,and system must be in equlibrium?

View attachment 279138

The single fixed foot is not a realistic model for two feet with variable load sharing. In wind surfing, the center of pressure of the feet combined and the center of mass of the rider are not fixed.

 

[John Mcrain]

Certainly not. You must be adding assumptions/constraints you aren't saying (or you're just wrong). . Draw all the forces and label all the lengths, and solve for your scenario.

Find position and magnitude of lift force,system must be in balance..
first for sailor lean out angle 70 then for 20 degress

My results : lift position is for both case(70deg and 20 deg) below sailor feet and magnitude for both case is 1000N
mast force for 70deg case= 171N
mast force for 20deg case=469N
(I use 1kg=10N)70degress

20degress

 

[russ_watters]

Find position and magnitude of lift force,system must be in balance..
first for sailor lean out angle 70 then for 20 degress

My results : lift position is for both case(70deg and 20 deg) below sailor feet and magnitude for both case is 1000N
mast force for 70deg case= 171N
mast force for 20deg case=469N
(I use 1kg=10N)70degress
View attachment 27914220degress
View attachment 279143

Edit; Sorry, I do see now that this is a complete answer. Ok, yes, I agree. Now what?

 

[jbriggs444]

My results : lift position is for both case(70deg and 20 deg) below sailor feet and magnitude for both case is 1000N
mast force for 70deg case= 171N
mast force for 20deg case=469N
70degress

So these are figures for the 2-meter-high wind force that will be balanced by the surfer's tension on a horizontal line at 70 degrees and at 20 degrees. Yes, I agree. In both cases, the result is $\frac{mgh \cos \theta}{H}$

We have the board+mast+sailor all balanced on a point support located at the sailor's feet. The farther back the sailor leans, the more clockwise torque from gravity about the support point. And the more counter-clockwise torque from wind that can be resisted.

In the toy model where the line is horizontal, it all boils down to a simple balance between sailor position and wind force.

You show me what answers you have calculated and how you calculated them.

In a bit more detail, do a torque balance on the sailor. 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.

 

[russ_watters]

In a bit more detail, do a torque balance on the sailor. 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.

...I edited my reply above. It's enough of the answers for that step, to me. Now I'd like to know what his point is.

 

[sophiecentaur]

You're right that the actual joint here is a ball-and-socket, which is free to rotate in all three axes, un-constrained. I don't think that's in dispute here. I don't think calling it a universal joint is a point of confusion.

You are probably right but using the right term avoids the wrong implications and some of the arguments above have been a bit loose. Frankly, I am always uneasy with arguments which are based on numerical values instead of symbols. The numbers become anonymous in the end result and the pattern is lost.

Also, why do we see a 'force' of 100kg in diagrams? And, although I never got beyond a very wobbly pose with backside sticking out, I have noticed that windsurf masts are always leaning to windward and not to leeward as with other sailing craft. I think the sailboard should only be studied when the basics of a 'simple' fixed mast, monosail has been sorted.

 

[John Mcrain]

Edit; Sorry, I do see now that this is a complete answer. Ok, yes, I agree. Now what?

You ask me to prove math..here you are..

First, there is now need for any calculation for this easy case where line and mast force is horizontal,it is pure logic that lift force must be under sailor feet for both case for two reason.

1) line is horizontal so sailor don't "loose" any weight under his feet and don't produce any internal vertical force at joint
2)external mast force is horizontal so no any internal vertical force at 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...469N

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..

double check with internal forces
Mjoint=0

100kgx1m -100kgx X=0
X=1m

(If line has some downard angle and mast force is horizontal,lift position will be somewhere left from sailor feet...that I can say just from logic,but in this case,I can't say accurate position without calculation..)

 

[russ_watters]

You ask me to prove math..here you are..

First, there is now need for any calculation for this easy case where line and mast force is horizontal, it is pure logic that lift force must be under sailor feet for both case for two reason.

It's dangerous to assume you don't need math. That's an easy path to making mistakes.

In any case, this model doesn't take into account the weight of the board and sail/mast/boom. The force on the left scale isn't zero (and the force on the right scale does not equal the riders weight), but they are the same between the two scenarios. Maybe that detail isn't important to whatever point you are trying to make, though, I don't know.

(If line has some downard angle and mast force is horizontal, lift position will be somewhere left from sailor feet...

I agree with that as well. So now what?

 

[John Mcrain]

If evertyting is same just mast has fixed conection:
mast force will be
100kgx1,939m -Fx2m=0
F=96,98kg...969,8N

And lift magnitude will be 100kg..1000N and lift position will be under mast

Am I right?

 

[russ_watters]

If evertyting is same just mast has fixed conection:
mast force will be
100kgx1,939m -Fx2m=0
F=96,98kg...969,8N

And lift magnitude will be 100kg..1000N and lift position will be under mast

Am I right?
View attachment 279164

...[edit] dang, multi-tasking again. Hang on.

(thanks @pbuk )

 

[John Mcrain]

Yes.

So you see that mast connection make difference.
So I can not look at first case with joint like it is rigid(welded mast)

What does it mean in practice,if I have fixed mast I can increase righting moment=increase sail power =go faster??

 

[russ_watters]

So you see that mast connection make difference.

Sorry, no, I misread while multi-tasking. It was confusing when you said "everything is the same" and then started changing things.

If the only physical setup change is that you've added guy-wires to the mast, then nothing about the force balance changes.

 

[John Mcrain]

Sorry, no, I misread while multi-tasking. It was confusing when you said "everything is the same" and then started changing things.

If the only physical setup change is that you've added guy-wires to the mast, then nothing about the force balance changes.

But what is then wrong with my calculation,for joint case mast force is 469N and for fixed conection mast force is 969N.What did I wrong ?

 

[russ_watters]

But what is then wrong with my calculation, for joint case mast force is 469N and for fixed conection mast force is 969N.What did I wrong ?

I think you assumed that the fixed mass applies a torque at the attachment point.

 

[pbuk]

No, this does not make any sense.

You said that you 'keep everything the same', just fix the mast but you have NOT kept everything the same, you have changed the wind force on the mast.

In the diagram where the mast is 'not fixed', the lift point is only directly below the sailors feet because you have fixed it that way: you have assumed that there is no other torque on the board (for example due to buoyancy or tail fin drag) and so in order for it not to rotate the centre of lift has to act at the feet!

You have proved nothing about your 'model' of the side view of a windsurfer, but it is pointless discussing this model anyway, it is nothing like a real windsurfer which looks like this:

 

[John Mcrain]

I think you assumed that the fixed mass applies a torque at the attachment point.

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

 

[russ_watters]

Find position and magnitude of lift force,system must be in balance..
first for sailor lean out angle 70 then for 20 degress

My results : lift position is for both case(70deg and 20 deg) below sailor feet and magnitude for both case is 1000N
mast force for 70deg case= 171N
mast force for 20deg case=469N
(I use 1kg=10N)70degress
View attachment 27914220degress
View attachment 279143

...also, I think I found an error here I didn't notice before. I only checked the 70 degree case, and my answer was 160 N. I figured there might be a rounding error difference, but maybe not: are you assuming the attachment point of the harness to the mast stays at 1m high? This may not matter to what you are really after though, so we can set it aside for now.

 

[John Mcrain]

...also, I think I found an error here I didn't notice before. I only checked the 70 degree case, and my answer was 160 N. I figured there might be a rounding error difference, but maybe not: are you assuming the attachment point of the harness to the mast stays at 1m high? This may not matter to what you are really after though, so we can set it aside for now.

No ,you can see in my calculations, it is attached at sin20x1=0.342m (for 20degress case)
and sin70x1=0.939m for(70deg case..)

 

[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??