Need 3D functions for Uhden Boom Vang

[john.uhden]

I haven't seen one of these contraptions anywhere, but I am imagining it.

The goal is that by tightening the vang (from either side of the boat), not only do you pull the boom down but also pull it forward to optimize the sail shape and outward position mainly on a run.  Conventional vangs attached to the base of the mast don't pull the boom forward.

Rather than trial and error by drawing the thing a number of times with the boom in or out, I want to calculate the height above the deck.  I am pretty sure that the XY force vector is the bisector of the two legs of the vang, but calculating the Z will indicate to me the relative downward pull that the vang will exert on the boom at its point of attachment to the boom.  With this daydream design, I am assuming that the closer hauled the sail (boom being almost midships) that the downward pull will be the minimum, which is what you want when sailing upwind because the mainsheet should control the downward pull in that condition.

Anyone got any functions that will help me calculate the Z for various boom angles and tightness on the vang?

My premonition is that the force directions are simple (x y z) vectors and that the Z value is just a matter of applying the sine of the vertical part of the vector.  But I could spend days on this and spin myself in circles (just like many sailors do on the water after they have fouled another boat.)  I would like to build and use it (if it mathematically works) before the end of the racing season, just three weeks away.  Even better for the next races this coming Friday.

Now it's probably not that important because I lapped the rest of the fleet twice in the last race.  I'm not a sore loser, but winning is fun.

For those of you who have read this far and are familiar with yacht racing rules, for fun I suggested to the committee that the 360° penalty is too severe and should be cut in half.😆

[Sea-Haven]

Interesting 3d xyz calcs required, sailed a 16; mosquito cat with a spinnaker put the hull 2' under a wave and shot out other side gear busting speed. Not sure any direct answer it may be a 3d style answer like a surface with 4 variables.

Have you seen the windsurfer guys with hydrofoils unbelieve speed 2' above the water.

Back to the problem, the boom can swing L&R and go up down so the 1st block would have a spherical answer, hor angle, ver angle. then take that point and work to block on deck as its fixed.

There may be some 3d cable calcs yachting software out there.

[diagodose2009]

I think , in my mind, i found a few steps for engine-math inside Model-Space

1)You share here your-drawing.dwg

2)You convert' all entities to polar coordinates.(lines/arcs)

You convert the equations from Physics to AutoLisp with this program 

3) https://www.youtube.com/watch?v=qLqzfF88UjU 

[john.uhden]

Those foils are incredible!
Speaking of blasts, there was the time my buddy and I were sailing my M
Scow around before the start. The wind was blowing about 30mph and we were
on a screaming reach just planing from wave top to wave top. I had a
wooden mast without any spreaders. The mast kept going sproing sproing
until it snapped in two. OMG, it was so much fun (in spite of throwing
away any chance of a 1st place finish in the series).
Since you know what you're talking about, I guess I'm on my own, but I've
been there before.
I was thinking I need a polar3D function. But I'm guessing I can break
down the vector into an XY component and an XZ component. Does that sound
right?

[CodeDing]

@john.uhden ,

Not a direct answer to your question, but maybe check out the CAL command and its many uses. I was looking into these last night when I stumbled upon your question.

Best,

~DD

[john.uhden]

I googled and found what appears to be a very helpful website...

https://chortle.ccsu.edu/VectorLessons/vch04/vch04_8.html

It starts at the beginning and reveals more and more of the things I never learned.

Why should I have; AutoCAD wasn't around yet when I was in high school and college.

For example:

| (x, y, z)^T |  =  √( x² + y² + z²)

[marko_ribar]

This with forces reminds me on topic about trusses, though I can't be perfectly sure what are inside forces and yes I broke forces onto components, but still not sure if that was correct approach... If someone interested here is topic at the swamp under show stuff, so you must be active member to have access and be able to download files...

http://www.theswamp.org/index.php?topic=50403.0 

[john.uhden]

Thanks, Marko. Maybe I will join.

[dbroad]

A boom vang is a block and tackle between the boom and the base of the mast. Think 2d geometry here. The connection point to the mast should be in line vertically with the gooseneck.  It's only role is to flatten the sail.  To adjust the sail port/starboard, you use another block and tackle called a sheet between the boom and either the center of the boat near the center of lateral resistance and cleated with a jam cleat so that it can be popped out for adjustment by the skipper. If far enough aft, the sheet can be connected to a traveller instead of a fixed center point.

[john.uhden]

Hi, Doug. I am honored to receive your input.
You have described conventional rigging, none of which forces a boom down
AND OUT. You can't push a main sheet.
On my Barnegat Bay Sneakbox I had added "boom hauls" (whatever) that pulled
the boom toward a block attached to the side stay tangs. But there was one
on each side, and God forbid you forget to pop the tight one before
gybing. But you always sailed with crew on a Sneakbox. Sunfish are
single-handed and don't have side stays with tangs. Thus my twisted
mind started thinking. The smoke fumes are still pervading my computer
area.

[Sea-Haven]

A 3d spatial analysis tool where you build a 3d model all connected, a space truss, you would move the boom in a radial plane, doing so for a angle range and z height, get a squillion answers repeat again for a change in tang position do again another squillion answers. Your after strain on one component correct ? Put a constant load in block plane.

An oldy spacetruss. A new one do you know a formula one team they could work it out.

What about something like solidworks or inventor to do 3d model. Analysis no idea. I will ask guy I was talking to a few minutes ago he competes at elite sailing level.

[leeminardi]

I am not a sailor so do not fully understand the rigging but this seems to me a basic 3D statics problem.  Here's my understanding of the configuration.  Please clarify or better yet post a file with the 3D geometry.  I've labeled your drawing to help in the analysis.  Please confirm/correct my statements below.

1.  Points B1 and B2 are on the deck and can be assumed to have a fixed z = 0.0. 
2. P2's location is fixed in x,y,z.
3. Distance P1 P2 is fixed.
4. Distance P1 B3 is fixed
5.  Are the lines at B3 fixed there or is there a pulley?  If there is a pulley then the force in B1B3 is equal to the force B2B3 and ang1 = ang2.  It the connection at B3 is fixed then ang1 may not equal ang2.
6. Points B1, B2, B3 and P1 all lie on the same plane 

[john.uhden]

Thank you very much, Alan, but I have started my code in AutoLisp.  I think my approach is similar to what @leeminardi posted.  It can't solve for a direct solution, but will iterate to converge on a close-enough answer for a range of boom angles.  I won't know if my idea is valid until I complete and test the code, which won't be very soon since my honeydo list takes precedence.  Have you ever tried to fix a broken spring on a dishwasher door?

[john.uhden]

Your mind is working just like mine.  See my latest response to @Sea-Haven 

[leeminardi]

@john.uhden What is the answer to my statement #5?

     Are the lines at B3 fixed there or is there a pulley? 

I'd like to take a stab at a VLISP program to solve this.  What is the typical input variable(s)?   

If B3 is not a pulley I would assume the length of B1 B3 and B2 B3.

If B3 is a pulley then it would be the total line length B1 B3 B2.

Do you have an AutoCAD model?

[john.uhden]

The image I posted is of just a 2D model. I am visualizing it in 3D.
Yes, B3 is a block (pulley) that rolls along the vang rope which is
continuous. My assumption is that (equal ang1 ang2 fuzz) since each leg is
taking half the force.
My intent is to solve for the Z value of P1 over a range of boom angles. I
am not getting into actual forces. The presumption is that the lower the Z
value at P1, then the greater the downward pull on the boom. Remember that
of equal importance to the downward pull is the forward pull achieved by
the vector passing outside of the gooseneck/mast, which is contrary to the
conventional design described by @Anonymous.

[leeminardi]

"My intent is to solve for the Z value of P1 over a range of boom angles."

If that is all you want then the z value of P1 is:

P1z = P2 + P1P2 * sin(90 - alpha)

where P1P2 is the distance from P1 to P2

    alpha is the angle between the mast and the boom.  If the boom is perpendicular to the mast then alpha = 90.

It's interesting to note that the path of B3 is an ellipse with foci at B1 and B2.  You can locate B3 by finding the intersection of the line from P2 to the midpoint between B1 and B2 with the ellipse that lies on the plane defined by B1, B2, and P1.  This should be the same as:

B3 = P1 + uP1M * P1B3

where

uP1M is a unit vector from P1 to the midpoint between B1 and B2

  =  (((B1 + B2)/2) - P1)/ (magnitude of (((B1 + B2)/2) - P1))

P1B3  is the distance from P1 to B3

I realize that the boom could swing to one side or the other but this would change the height of P1 for a given angle alpha although it would change the force in the lines connecting B1 and B2 to the boom. To calculate the forces you would need to know the forces the sail places on the boom.

I haven't checked this out with AutoCAD yet.  It's too hot here right now to use my CAD computer.

Lee

[john.uhden]

I really don't know vector math.
When it cools off, can you translate those equations into AutoLisp?
Where are you that it is so hot... Baghdad or Needles, CA or Damascus?
What, no air conditioning?

[leeminardi]

@john.uhden , In my post #17 I said "I realize that the boom could swing to one side or the other but this would change the height of P1 for a given angle alpha..."  I meant  to write "I realize that the boom could swing to one side or the other but this would not change the height of P1 for a given angle alpha..."

I also wrote that:

P1z = P2 + P1P2 * sin(90 - alpha)

This should have been:

P1z = P2z + P1P2 * sin(90 - alpha)

That is, the z coordinate of P1 is equal to the z coordinate of P2 plus the distance from P1 to P2 times the sine of (90 degrees minus alpha).

I was watching some sailboats this afternoon and it didn't look like the angle alpha changed much. I guess it could decrease a bit as the wind put more strain on the sail causing the boom to raise a bit.  Perhaps I am not understanding how alpha is measured.  Could you clarify?

You asked "can you translate those equations into AutoLisp"?  Is the following what you want?

1. User is asked to specify P2 to which he can click a point on the screen or type in an x,y,z value.

2. User is prompted for the angle alpha.

The program outputs the value of the z coordinate of P1.

BTW, since you asked, I live near Boston, Massachusetts and my home office does not have A/C.  It's a long story why. We have had about 2 weeks of hot and humid weather.  My CAD computer has 2 screens and one of them was continually blacking out.  When I googled  potential reasons I learned that it could be due to heat.  My office temp was in the 90's (F).  I am writing this from my cheap Chromebook (which I love but it doesn't support CAD) in an air conditioned room.

[john.uhden]

Thank you, Lee.

In terms other than what I said about Z, it may be better to think in terms of alpha, though they are mathematically interchangeable.

Anyway, I am now thinking of simplifying the design and the math by attaching the block at P3 directly to the boom at P4.  I think that will also reduce the tension on the vang and make it easier to adjust.  I often overcomplicate things.  You can see it in my code.

Boston, eh?  Whether it's Maine, where my oldest daughter is, or here at the Jersey Shore, the weather has been unbearable.  Even just watering the tomato plants, I'm dripping in 2 minutes.  I have to water my feet too.  Thank goodness it's well water at 55°.  Luckily we have both air conditioning and a pool, but even the pool temperature has been 88°.