I Physics of sailing/windsurfing systems

[jbriggs444]

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..

Let me be clear. I am focused on the part of the problem where we take a toy model and use it to generate predictions. The part where we decide which toy model matches the real world situation best, I make no claims. The frustrating bit for me has been trying to figure out which exact toy model is being considered.

 

[John Mcrain]

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.

You just explain what I am talking about last 50 posts specifily for "horizontal line case".

 

[John Mcrain]

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

Yes your idea of sliding teflon track is very good.
In reality you can stop and just low the boom and try new settings in second run..

(As I said before,in reality line allways has some downward angle,and you don't change boom height when sailing so line attachment point remain the same during sailing..
But I deliberately use horizontal line case to prove that you can't press nose down with this settings..)

 

[jbriggs444]

Yes your idea of sliding teflon track is very good.
In reality you can stop and just low the boom and try new settings in second run..

(As I said before,in reality line allways has some downward angle,and you don't change boom height when sailing so line attachment point remain the same during sailing..
But I deliberately use horizontal line case to prove that you can't press nose down with this settings..)

I am not sure that I understand what you are saying. That the toy model is never correct, but that it proves the thing that you want it to prove so that it must be right anyway?

 

[John Mcrain]

I am not sure that I understand what you are saying. That the toy model is never correct, but that it proves the thing that you want it to prove so that it must be right anyway?

As you see ,my english is not so good,so I have very hard time to explain so complex system with words.
I didnt understand your question.Can you simplfy?

 

[John Mcrain]

One more thing about lines angle

If feet and mast lay in same line ,than for given sailor lean out angle ,all three angles (blue rope) give same sail force.It doesn metter where is attachment point at mast

If feet is out from center line,than for given lean out angle, sail force change with different lines angle.
lowest angle give small sail force,higher you choose angle sail force is higer..
also the more distance is from feet to center line,grather sail force will be..

all of this I can prove with math

 

[pbuk]

If feet and mast lay in same line ,than for given sailor lean out angle ,all three angles (blue rope) give same sail force.It doesn metter where is attachment point at mast

This is wrong, but perhaps this time it is due to a problem of language. All three angles give the same righting moment, but not the same force. Also in this case the mast step joint is free to rotate laterally and so none of the righting moment can be transferred to the board; unless the righting moment is exactly equal to the heeling moment provided by the force of the wind on the sail then the rig and sailor will fall into the water.

If feet is out from center line,than for given lean out angle, sail force change with different lines angle.
lowest angle give small sail force,higher you choose angle sail force is higer..

The total righting moment is the same wherever the wishbone is mounted but now it has two components: the rig righting moment and the [component forcing the windward rail into the water - I can't remember what this is called right now]. And yes, the higher up the rig the wishbone is mounted the greater proportion of the righting moment goes into the rig.

also the more distance is from feet to center line,grather sail force will be..

No, this just increases the [component forcing the windward rail into the water].

all of this I can prove with math

I don't think attempting that would be a good use of anybody's time. It would be more profitable to get out on the water and do some sailing.

 

[A.T.]

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?

yes..

You don't understand why a triangular truss is rigid?

 

[John Mcrain]

You don't understand why a triangular truss is rigid?

Yes I understand that but that has nothing to do with wsurf system.
If sailor-sail-board is rigid system like you suggest,they will allways "rotate" togeather..

You can see at video that sail and sailor are " moving" separatley from board.Board is almost flat and sail and sailor moving all the time...


If you really make wsurf system rigid,so "welded" sail to board, borad will be moving in pitch and roll direction all the time ,it will be completely out of control..

 

[pbuk]

Yes of course a real windsurfer is moving all the time, but you have not taken this into account either - you have only considered the simplified situation assuming that the position is completely stable.

Lets look at an different (but still dynamic) system: a tiller on a monohull dinghy attached to the rudder with a universal joint like this (best seen from about 1 minute in):


See how the sailor is constantly moving the rudder by applying a torque about its pivot point? The only way the sailor can apply this torque is through the universal joint attaching the tiller.

This clearly demonstrates that you can apply a torque through a UJ so you need to stop believing that you can't.

Once you have stopped focussing on this you can start looking at a better free body diagram like the ones on the site I linked a few posts ago.

 

[A.T.]

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?

Yes I understand that .but that has nothing to do with wsurf system.

It has a lot more to with the wsurf system than the single joint you keep showing:

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

View attachment 279075

Do you understand that multiple universal joints attached at different locations can transfer a torque?

 

[John Mcrain]

Lets look at an different (but still dynamic) system: a tiller on a monohull dinghy attached to the rudder with a universal joint like this (best seen from about 1 minute in):


See how the sailor is constantly moving the rudder by applying a torque about its pivot point? The only way the sailor can apply this torque is through the universal joint attaching the tiller.

This clearly demonstrates that you can apply a torque through a UJ so you need to stop believing that you can't.

This is wrong discription of physics.

Universal joint transfer side force from tiller-extension to tiller.Conseqeunce of side force at tiller is rotating rudder around his pivot point.
So universal joint at the end of tiller does not trasfer torque to tiller,it transfer side force, and consequence of this side force is rotating rudder..

I never said that universal joint can not transfer forces...

Does any member agree with me about this what I just explain in this post?

 

[russ_watters]

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
[separate post]
Does any member agree with me about this what I just explain in this post?

@John Mcrain you need to start doing better, because at this point we are starting to doubt your sincerity. You are stating false things and misrepresenting what people are telling you. The key issue is so simple it is difficult for us to believe that you are really having this much trouble with it. So I need you to be crystal clear in your understanding of the following:

1. In scenario 1, the mast has a force/torque applied by force F and falls over. The board has no torque applied due to fore F.
2. In scenario 2, the mast does not fall over. The entire structure acts like one rigid object and force F pushes forward and applies a clockwise torque to the entire system. The entire object rotates clockwise unless there are other forces.

So:
3. A single universal joint cannot transfer a torque around itself.
4. A connected series of universal joints becomes a rigid object. Torques can be transferred via internal forces. E.G., in scenario 2, the torque is not transferred through the universal joint connecting the mast to the board, it is transferred by linear forces through the brace and its connection points.

Do you understand and agree with these statements?

 

[John Mcrain]

@John Mcrain you need to start doing better, because at this point we are starting to doubt your sincerity. You are stating false things and misrepresenting what people are telling you. The key issue is so simple it is difficult for us to believe that you are really having this much trouble with it. So I need you to be crystal clear in your understanding of the following:

View attachment 279132

1. In scenario 1, the mast has a force/torque applied by force F and falls over. The board has no torque applied due to fore F.
2. In scenario 2, the mast does not fall over. The entire structure acts like one rigid object and force F pushes forward and applies a clockwise torque to the entire system. The entire object rotates clockwise unless there are other forces.

So:
3. A single universal joint cannot transfer a torque around itself.
4. A connected series of universal joints becomes a rigid object. Torques can be transferred via internal forces. E.G., in scenario 2, the torque is not transferred through the universal joint connecting the mast to the board, it is transferred by linear forces through the brace and its connection points.

Do you understand and agree with these statements?

Yes I understand evertything you said and 100% agree with all what you said.

Now you tell me one think,do you understand that sailor do not hold sail like on picture below?

Also quote my words that you don't agree with me ,what I said wrong...?

 

[russ_watters]

Yes I understand evertything you said and 100% agree with all what you said.

Ok, good.

Now you tell me one think, do you understand that sailor do not hold sail like on picture below?

The internal geometry is irrelevant to the points I was making, so I don't understand the point of that question. E.G.; he could do that. Whether he does or doesn't is a new scenario that hasn't been described.

Also quote my words that you don't agree with me ,what I said wrong...?

I did. And you repeated it several times.
[edit] Here it is again:

here is no any torque from the wind force, because joint can't transfer torque to board.

That's wrong. There *is* a torque on the board from the wind force because a series of universal joints *can* (and does) transfer torque to the board.

 

[sophiecentaur]

Universal Joint: Is there a bit of cross-purpose going on here? The use of this term in this thread is not a common one. A so-called Universal Joint is normally used for actually transferring a torque between a shaft and an object (or another shaft).This applies in car transmissions and some socket spanner sets. In that context it is not truly 'universal' as a ball and socket would be. Otoh the rubber 'finger' and ball that's used on a sailboard transmits no torque; the sailor doesn't grip the mast like a screwdriver and when the sail changes direction, the joint will move round and allow the sail angle to be set in any direction the sailor sets by her body / arm / feet positions. There is no torque applied to that UJ because it will move however you want it to, with no reaction.
The only explicit torque that can be applied here is by the sailor's feet (as a pair) push / pulling laterally on the surface of the deck.

 

[John Mcrain]

The internal geometry is irrelevant to the points I was making, so I don't understand the point of that question.

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?

 

[russ_watters]

Universal Joint: Is there a bit of cross-purpose going on here? The use of this term in this thread is not a common one. A so-called Universal Joint is normally used for actually transferring a torque between a shaft and an object (or another shaft).This applies in car transmissions and some socket spanner sets. In that context it is not truly 'universal' as a ball and socket would be.

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.

 

[russ_watters]

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

Certainly not. You must be adding assumptions/constraints you aren't saying (or you're just wrong). The bad grammar doesn't help either. I'll not speculate about an under-defined scenario. Draw all the forces and label all the lengths, and solve for your scenario.

You're jumping around so much here that it is difficult to know what your ultimate goal is. That makes it look like your ultimate goal is a never-ending argument.

 

[russ_watters]

To put a finer point on it, this was your original question:

Sailing boat has pitching moment caused by sail drive force and hydrodnamic drag froce in the water,distance between them is lever arm.
It cause nose of boat to push down..

Does windsurfing has pitching moment and if yes how it is transfer to the board ,if sail is connected to board with universal joint and sailor feet also act as joint. But joint can not transfer moment..?
I don't understand physics of this..

Do you now understand that a series of members connected with non-rigid joints forms a rigid object that enables the sail force to apply a torque to the board (entire system)? You replied to my post #113 in a way that suggests this misunderstanding has been cleared-up and you now understand, but your other recent posts imply you are still holding on to the misunderstanding.

 

[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??