Z makes kitefoils now, including both the women's and the men's national champs foils in kitefoiling.
He still has 2 local riders who are very competitive in windfoil racing in speed and angle, but one is now 72 years old, the other over 65.
He's been thinking of making ws foils now for 2 years, but has not committed yet.
Select to expand quoteIan K said..
ie. I'm thinking, still using using your calculations and trigonometry, that when looking at lateral forces acting on the board, ( there's only four of them : two feet, mast base and fin ). That the fin side force = Ffin = 47.7 - 10.4 = 37.3 kg.
That just means that the mast base and the feet act in opposite directions, ie. to compress the board, and that if the mast base breaks it will flick up and hit you in the knees.
(unless of course your sail plan is symmetric about the boom, like that of a wind wing which doesn't need a mast base at all!)
Feet and mast foot are only two conections which transfer sail force to the board,so sum of horizontal forces at feet and mast foot must be allways equal to sail horizontal component (Fh).
47.7+10.4=58.1
In my case ,this is very little side force (+10.4kg),almost neutral.But with different configuration(angles,distances..etc) it can change magnitude and direction as well.
(Here is vector calculator where you can check - must adapt my angles from original picture to cordinate system,so input will be 67.09 / 30degrees, 95.4 / -120 degrees, then see sum column for x components...
link:
www.mathsisfun.com/algebra/vector-calculator.html )
vector diagram at mast foot
Since we are talking about side force at mast foot, Starboard used this side force for easier railing...
Phantom race starboard use slanted mast track,it has raised deck where mast foot is placed.That gives aditional lever arm (h) where mast foot side force (directed to leeward) make torque which lift windward rail up,that help board get on its rail even in lighter winds/slower speed...
In fig 4 right diagram of your original post are you balancing the forces on the rig? The force on the board at the mast base will be equal and opposite. But first you have to get the horizontal forces on the rig to balance. 10.4 kg is small in the scheme of things but if you add it rather than subtract it you'll get 21 kg less force on the fin. Then you'll be in the ballpark of the calculations by Jim Drake.
Interesting point about the race board with the raised mast track. You've got me puzzling about it?
(not that we should necessarily assume Jim Drake got it right. We've got to question everything.)
Select to expand quoteSubsonic said..
Force/"moment"of force
A force moment is another name for torque, is that correct ? (Unit: Nm)
Select to expand quotejn1 said..Subsonic said..
Force/"moment"of force
A force moment is another name for torque, is that correct ? (Unit: Nm)
Yes. force times perpendicular distance from an axis you might be considering rotation about.
( should say the component of force acting at right angles to both the axis and the perpendicular distance, i.e.. how you use a spanner )
It's useful because it gives a rotational equivalent of F = ma
T = I X angular acceleration.
I is the moment of inertia ie. add up all the masses times the square of each distance from the axis of rotation.
If something is going in a straight line at constant speed with a constant rate of tumble then both the torques and forces must balance.
If speed and rotation isn't constant and you want to put numbers on it all, use newtons and newton metres. If it's a stable situation you can use kg, lbs and any unit of length you like
Ok, thanks Ian. As Mikey100 said - I got stuck at paragraph 1 
. The intro makes sense now.
Select to expand quotejn1 said..
Silly question: why isn't wind force inputted into the analysis ?
Not a silly question!
It's because it's the force on the sail that's relevant, this is a combo of sail size and wind strength. So it's sort of included, but combined with sail size.
Does the zero sum force & torque assumptions at the start of the OP post and the "0" on the end of the calculation in step 3 relate to not needing to specify a wind force input into the calculations ??
Select to expand quotejn1 said..
Does the zero sum force & torque assumptions at the start of the OP post and the "0" on the end of the calculation in step 3 relate to not needing to specify a wind force input into the calculations ??
The wind force on the sail is indirectly deduced by calculating the tension in the harness lines/arms based on sailor's centre-of-mass distance from the fulcrum at the sailor's feet. There is an assumption that the centre of lift of the sail is 2 metres above the deck. Moments were then calculated around the mast base to deduce the force of the wind. You can calculate moments about any location and they will give the same answer, but the mast base was chosen because then the forces at the mast base drop out and make the calculation simpler.
The 2 metre estimate for the sail's centre of lift sounds about right. I suppose if it wasn't you could add more downhaul until it is.
Lil off track here. But still to do with forces. What forces are involved in a catapult? Im 100kg travelling at 45kmh. What level of force do i feel when i hit the water hard enough to snap a new harness line?
Select to expand quoteolskool said..
>What forces are involved in a catapult? Im 100kg travelling at 45kmh. What level of force do i feel when i hit the water hard enough to snap a new harness line?
After a very bad catapult one of my mates came up, eyes spinning, with 60kts on his GPS. So catapult acceleration can be very high. Not only have I snapped harness lines, but harness webbing and boom in two places as well. No idea how you'd calculate those forces. You could do a destructive test of your equipment and see what sort of force they fail at.
Select to expand quoteIan K said..
The wind force on the sail is indirectly deduced by calculating the tension in the harness lines/arms based on sailor's centre-of-mass distance from the fulcrum at the sailor's feet. There is an assumption that the centre of lift of the sail is 2 metres above the deck. Moments were then calculated around the mast base to deduce the force of the wind. You can calculate moments about any location and they will give the same answer, but the mast base was chosen because then the forces at the mast base drop out and make the calculation simpler.
Fwind = -Fsail ?
Select to expand quotejn1 said..Ian K said..
The wind force on the sail is indirectly deduced by calculating the tension in the harness lines/arms based on sailor's centre-of-mass distance from the fulcrum at the sailor's feet. There is an assumption that the centre of lift of the sail is 2 metres above the deck. Moments were then calculated around the mast base to deduce the force of the wind. You can calculate moments about any location and they will give the same answer, but the mast base was chosen because then the forces at the mast base drop out and make the calculation simpler.
Fwind = -Fsail ?
Going on the first diagram after the photo, yes. The sail exerts an equal and opposite force on the wind.
Now you just need to give me the numbers I need to make my jibes work, I find that baffling enough, let alone the mathematics.
Select to expand quoteolskool said..
Lil off track here. But still to do with forces. What forces are involved in a catapult? Im 100kg travelling at 45kmh. What level of force do i feel when i hit the water hard enough to snap a new harness line?
Just had a look at this, my 5hz gps track shows deceleration of my last catapult
24 m/s2 is about 2.5gs, rigged up I'm about 70kg so that's about a force of 175Kg.
I think 10hz would give a more accurate reading of acceleration, but this gives a good idea.
This didn't break anything, but ripped the motion off my arm, which then disappeared into the weed.
Select to expand quoteIan K said..jn1 said..
Fwind = -Fsail ?
Going on the first diagram after the photo, yes. The sail exerts an equal and opposite force on the wind.
I'm not a mechanical Engineer, but now have a basic understanding of the OP's proof. This is a really good post actually. Probably bread and butter for professionals here ?. I remember when I was a beginner going through a basic calculation like this, but I lacked the mechanical background. Thanks all for helping me out with my dumb questions . Nothing wrong exerting a bit of mental effort to understand the physics of our sport.
Select to expand quoteIan K said..
In fig 4 right diagram of your original post are you balancing the forces on the rig? The force on the board at the mast base will be equal and opposite. But first you have to get the horizontal forces on the rig to balance. 10.4 kg is small in the scheme of things but if you add it rather than subtract it you'll get 21 kg less force on the fin. Then you'll be in the ballpark of the calculations by Jim Drake.
Interesting point about the race board with the raised mast track. You've got me puzzling about it?
(not that we should necessarily assume Jim Drake got it right. We've got to question everything.)
For sure Jim know physics,but he use general Fsail xh=WxL formula,because dont want bother people with sinus,cosinus,moments,force components etc..Also less complex equation is ,it is more understandable for most people.
Yes this formula is correct ones everything is put in balance..so you can use it at the end of calculaiton for one more check system balance..
Equation Fsail x h= Weight x L dont include sailor lean out angle ,and harness line angle...Sailor weight ,lean out angle and harness line angle(T1) in combination with sailor distance from centerline,deterimne what sail force (Fsail) will be.But when I put everything in balance, I can use this formula to check system balance(You can see when I find that my first "fin" (0.35m) will lift windward rail up ,then I put smaller fin (0.14m) to make system balance,now I can use this simple formula to once again check system balance..(Try it ,input fin (0.14m) into this simple formula Fsail x h= Weight x L and you will get ,Fsail= 67kg again..)
(Statics is the branch of mechanics,I didnt inveted this rules,rules for calculation for universal joints exist for hundreds of years..)
Listen at 6:40-7:30,he said low hook increase railing, high hook-put rail down-more control..
"2m tall sailor - lowest boom position - waist harness = harness line almost horizontal"
Also rider at 50 degree angle and mast at 70 degree
For distance L , looks like you should use the boom height from the mast base?
Select to expand quoteUSA46 said..Ian K said..
Yes this formula is correct ones everything is put in balance..so you can use it at the end of calculaiton for one more check system balance..
Yes OK it is the correct one, but you had me head scratching for almost an hour when you balanced the forces on the rig in figure 4 right. Where you gave the horizontal force on the rig at the mast foot as +10.4. Ie. three horizontal forces acting on the rig +10.4 +58.1 - 47.7, Numbers that don't add up to 0! More confusing to me, because intuitively I was thinking that that would leave an anticlockwise moment about the point 1.68 metres up the mast to sort of balance the clockwise moment of the sail at 2 metres. however! I'd forgot, there can still be anticlockwise moment to counteract from the vertical component of force acting on the mast base. But then you magically didn't reverse the sign, (hidden it up your sleeve?) when analysing the forces on the board so in the end it all adds up. Well done.
It is reassuring that by drilling down through all the internal forces, feet, mast foot, etc that you ended up in the same place as the big picture approach i.e.. the horizontal force on the sail balances the horizontal force on the fin .
Anyway passes the time in lockdown, Revising what I last did in year 12 physics in 1970. 50 years ago! Do they still even do high school physics these days? How old are you?
Select to expand quoteIan K said..
Good job. No wonder fins bend. I would have estimated the sailor in the photos C of G makes an angle of 30 degrees with the vertical from where his feet contact the deck. Are you sure that isn't -10.4 kg for the mast's horizontal component in fig 4 ?
Ian K, I think that you mixed up with signs of action and reactions forces..
If make moments equation for side forces at feet and mast foot ,you will get -47,7kg and -10.4kg, but this is signs for reaction forces.
I need signs for action forces for later calculating board balance
Got no idea what the difference is between an action force and a reaction force. You'll have to define them for me.
Select to expand quoteIan K said..
It is reassuring that by drilling down through all the internal forces, feet, mast foot, etc that you ended up in the same place as the big picture approach i.e.. the horizontal force on the sail balances the horizontal force on the fin .
Just tell me two things.
1)Do you understand that for picture below you can not use formula Fxh=WxL ?
2)Do you understand if you conected with rope T1,that force F will be greater than if you conected with rope T2 ,yet the formula implies it makes no difference?
(these two questions are essence of all my "drilling ")
Yes I understand those two points.
W X the horizontal distance from his C of G to his heels = F X h (as long as T1 and T2 are ropes as indicated)
No not quite that was wrong, even though the tension in the lower rope will be higher the high connection will allow for a greater F
A question for you. What is the optimum height of the rope for maximum F ?
(we're both a bit mad, my Mum used to play internet scrabble to pass the time..... She always won)
Select to expand quoteIan K said..
Yes I understand those two points.
W X the horizontal distance from his C of G to his heels = F X h (as long as T1 and T2 are ropes as indicated)
No not quite that was wrong, even though the tension in the lower rope will be higher the high connection will allow for a greater F
A question for you. What is the optimum height of the rope for maximum F ?
(we're both a bit mad, my Mum used to play internet scrabble to pass the time..... She always won)
"W X the horizontal distance from his C of G to his heels = F X h (as long as T1 and T2 are ropes as indicated)"
you can use this formula for T2(but only because rope is 100% horizontal,so with changing distance L ,height of conection on mast will remain the same).For T1 you must use my calculation from original task..
(Also you can use your formula if feet are at universal joint,because then changing angle of rope will not change force F)
"A question for you. What is the optimum height of the rope for maximum F ?"
answer on picture
I thing we are going to deep,too complex.Mebers will go crazy ,but they have right..
Forget all this numbers,sinus ,cosinus,etc ,I just want that people understand that equation Fxh=WxL dont "see" that left rider is at 60 degree angle and right rider at 20 degree.
All very well, but can someone calculate the probability (as a ratio) of me using this thread to convince my wife to let me buy some new gear.
cause I'm sure now my fins are both too long and short, and the F forces in my old sails are totally F'd
Select to expand quoteUSA46 said..
answer on picture
Mebers will go crazy ,
Yes we won't get any green thumbs. At least the mebers can't give us red ones any more.
After a few false starts I now agree with your answer on the picture. In which case, assuming we are both correct, why do we insist on attaching our harness lines to the boom? (Apart from convenience)
I've always thought it would be better to attach harness lines as high as possible, that's why I was perplexed by advice to lower the boom when overpowered.
So we'd need sails with a cut out about 2.5m up for the attachments??? I've some old sails, I could experiment. But I can foresee loose ropes whipping around being a problem, and securing them before a gybe tricky. Maybe they clip into normal lines, then you can transfer from one to the other as needed?
Is this worth pursuing?
It's going to alter sail dynamics of course, mast bend will be altered significantly, may not get maximum benefit unless a sail maker is also involved.
Select to expand quotedecrepit said..
I've always thought it would be better to attach harness lines as high as possible, that's why I was perplexed by advice to lower the boom when overpowered.
So we'd need sails with a cut out about 2.5m up for the attachments??? I've some old sails, I could experiment. But I can foresee loose ropes whipping around being a problem, and securing them before a gybe tricky. Maybe they clip into normal lines, then you can transfer from one to the other as needed?
Is this worth pursuing?
It's going to alter sail dynamics of course, mast bend will be altered significantly, may not get maximum benefit unless a sail maker is also involved.
Securing them isn't a large problem. Easy enough to adapt an elasticated system similar to whats used on sailing dinghy trapeze systems, but off the boom.
i would think the problems would be more to do with how much it will affect mast bend curve and sail handling. You'd get better leverage, but you'll lose your steady sheeting angle and be muscling the sail with the back hand.
Select to expand quoteSubsonic said..
>>> You'd get better leverage, but you'll lose your steady sheeting angle and be muscling the sail with the back hand.
Yes, that's my worry, and why I thought a sail designer would need to be involved, they may see a way around any negatives.
