Tensioning a fixed backstay on a crusier

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Skylark

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I had my boat lifted yesterday for its winter hibernation. The yard guys had to remove the backstay in order to get the crane cradle in the correct position.

At 4.2m, the boat is quite beamy and they used the cradle (can't think of a better word) to broaden its reach, to ensure that the lifting strops did not crush the hull.

My AWB is a cruiser with 15/16th fractional rig. The backstay is non-adjustable and is an inverted "Y". A single wire to the top of the mast with a wire to each quarter.

I have a Loos rig tension gauge and I'd like to measure and reset the backstay to ensure that the mast is correctly supported over winter (I'm not having it removed).

Given the Y form, if I tension the two lower sections to a nominal 10%, what will that give to the single section? The single section is too high to reach so I will not be able to measure it.

Thanks for your advice.
 
davidjackson said:
I had my boat lifted yesterday for its winter hibernation. The yard guys had to remove the backstay in order to get the crane cradle in the correct position.

At 4.2m, the boat is quite beamy and they used the cradle (can't think of a better word) to broaden its reach, to ensure that the lifting strops did not crush the hull.

My AWB is a cruiser with 15/16th fractional rig. The backstay is non-adjustable and is an inverted "Y". A single wire to the top of the mast with a wire to each quarter.

I have a Loos rig tension gauge and I'd like to measure and reset the backstay to ensure that the mast is correctly supported over winter (I'm not having it removed).

Given the Y form, if I tension the two lower sections to a nominal 10%, what will that give to the single section? The single section is too high to reach so I will not be able to measure it.

Thanks for your advice.
Click to expand...

Depends on the angle between the lower sections

A little school boy trigonometry should enable you to work it out.

Over 50 years since my schooldays so "trig" is a bit rusty now but

tension in the upper part will be = 2 t1 cos A/2

where t1 is the tension in the lower sections

and A is the angle between them.


I think!
 
Its a geometry/vector problem and depends on the relative angles of wires to the horizontal. In a silly example; if the angle was 90 degrees then the force would be equally spread down each arm. However typically in practice the angle will around 60 degrees to spread the forces as evenly as possible, ( ie 120 degrees between each of the legs of the Y shape.) In this case the sum of the forces on the arms will be greater than the force down the main backstay wire. If I remember my schoolboy maths correctly, if you measure the angle (X) from the horizontal to one of the lower wires you can plug it into the formula 50/cos(90-X) this will give you the tension required in each leg to get 100N of tension in the main wire:

eg:
if angle X is 60 degrees 50/cos(90-60) = 57.7n per leg to get 100N of tension in main wire.
if angle X is 45 degrees 50/cos(90-45) = 70.71 per leg to get 100N of tension in main wire.

Just multiply up to get the value you need, so if 10% is 300N and your angle is 60degrees, you will need 212n per leg.
 
The backstay does very little to support the mast on that type of rig - it is mainly there to control bend so not critical.
 
Just checked VicS answer and we concur - it just depends which angle you prefer to measure to get the answer.
 
davidjackson said:
I had my boat lifted yesterday for its winter hibernation. The yard guys had to remove the backstay in order to get the crane cradle in the correct position.

At 4.2m, the boat is quite beamy and they used the cradle (can't think of a better word) to broaden its reach, to ensure that the lifting strops did not crush the hull.

My AWB is a cruiser with 15/16th fractional rig. The backstay is non-adjustable and is an inverted "Y". A single wire to the top of the mast with a wire to each quarter.

I have a Loos rig tension gauge and I'd like to measure and reset the backstay to ensure that the mast is correctly supported over winter (I'm not having it removed).

Given the Y form, if I tension the two lower sections to a nominal 10%, what will that give to the single section? The single section is too high to reach so I will not be able to measure it.

Thanks for your advice.
Click to expand...
I adjust mine all the time when sailing.
 
The wire in the split section may be of smaller diameter than the main section, in which case it becomes quite complex.
You should tension the backstay enough to give the correct tension in the forestay.
 
winsbury said:
Just checked VicS answer and we concur - it just depends which angle you prefer to measure to get the answer.
Click to expand...

It may be difficult measure the angle between the lower sections to use the formula I suggested.

Perhaps therefore easier to measure the angles between the lower sections and an imaginary line between their anchor points at deck level.

If those angles are B then
the tension in the upper section will = 2 t1 sin B

where t1 is again the tension in the lower sections
 
Tranona said:
Not sure the backstay on a 15/16 rig does much for forestay tension. Thought that was for the aft swept spreaders and mainstays - but could be wrong1
Click to expand...

You are generally correct for a fractional rig, but with 15/16 the backstay will have a greater influence than a more radical fractional rig; also, this boat may not have swept back spreaders, in which case you can only use the backstay and tension will be limited by acceptable mast bend.

Some info here: http://www.saltyjohn.co.uk/resources/Rig%20tensioning%20using%20the%20Loos.pdf
 
winsbury said:
Its a geometry/vector problem and depends on the relative angles of wires to the horizontal. In a silly example; if the angle was 90 degrees then the force would be equally spread down each arm. However typically in practice the angle will around 60 degrees to spread the forces as evenly as possible, ( ie 120 degrees between each of the legs of the Y shape.) In this case the sum of the forces on the arms will be greater than the force down the main backstay wire. If I remember my schoolboy maths correctly, if you measure the angle (X) from the horizontal to one of the lower wires you can plug it into the formula 50/cos(90-X) this will give you the tension required in each leg to get 100N of tension in the main wire:

eg:
if angle X is 60 degrees 50/cos(90-60) = 57.7n per leg to get 100N of tension in main wire.
if angle X is 45 degrees 50/cos(90-45) = 70.71 per leg to get 100N of tension in main wire.

Just multiply up to get the value you need, so if 10% is 300N and your angle is 60degrees, you will need 212n per leg.
Click to expand...

NOT between the wire and the horizontal. You need the angle between the wire and the base of the triangle.... ie a line between the two anchor points

Sorry about the red pen... just the easiest way to highlight.


You should see the red ink in some of my old maths books .. esp the "s" level ones :eek:
 
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davidjackson said:
Given the Y form, if I tension the two lower sections to a nominal 10%, what will that give to the single section? The single section is too high to reach so I will not be able to measure it.

Thanks for your advice.
Click to expand...

If the length of each Y part is L and the distance between the chain plates is D then I make it..

total tension on the single part =2 X sqrt (L^2-(D/2)^2)

anyone agree?
 
GHA said:
If the length of each Y part is L and the distance between the chain plates is D then I make it..

total tension on the single part =2 X sqrt (L^2-(D/2)^2)

anyone agree?
Click to expand...

Nearly there, you want the vertical part as a fraction of the diagonal, and to include the tension...

Tension in main part = 2T(sqrt(L^2 - (D/2)^2))/L
 
bignick said:
Nearly there, you want the vertical part as a fraction of the diagonal, and to include the tension...

Tension in main part = 2T(sqrt(L^2 - (D/2)^2))/L
Click to expand...

Don't quite get the /L..


(edit. Got it now, you're right I'm wrong :) )

Pythagorous gives the vertical component from the diagonal, then * 2 as there are 2 triangles.

What about with force added...
total tension on the single part =2 * sqrt (L^2-(D/2)^2)/L * tension along L
 
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Many thanks for all of the replies, much appreciated.

The mast has two sets of swept back spreaders. The backstay is not designed to be adjustable while sailing. I'm not sure how I could easily check forestay tension as the furl mechanism hides the shroud.

To keep things simple, I've assumed the length of the two lower wires to be 10m and also 10m for the single section. Assumed distance between chainplates is 4m.

That suggests that putting a 10% load on the lowers will put pretty much all of that, almost 20% on the upper, single wire. That sounds like too much for a cruising boat.

I'm happy to read from Tranona that the backstay doesn't contribute much to mast stability.

Seems like I should try to put a good bit less than 10% stretch on the lower wires, depending upon how accurate / repeatable the Loos gauge is to use.

I'm particularly impressed how you guys can remember school boy trig.
 
davidjackson said:
Many thanks for all of the replies, much appreciated.

The mast has two sets of swept back spreaders. The backstay is not designed to be adjustable while sailing. I'm not sure how I could easily check forestay tension as the furl mechanism hides the shroud.

To keep things simple, I've assumed the length of the two lower wires to be 10m and also 10m for the single section. Assumed distance between chainplates is 4m.

That suggests that putting a 10% load on the lowers will put pretty much all of that, almost 20% on the upper, single wire. That sounds like too much for a cruising boat.

I'm happy to read from Tranona that the backstay doesn't contribute much to mast stability.

Seems like I should try to put a good bit less than 10% stretch on the lower wires, depending upon how accurate / repeatable the Loos gauge is to use.

I'm particularly impressed how you guys can remember school boy trig.
Click to expand...
They have found google :)
 
GHA said:
Don't quite get the /L..

Pythagorous gives the vertical component from the diagonal, then * 2 as there are 2 triangles.

What about with force added...
total tension on the single part =2 * sqrt (L^2-(D/2)^2) * tension along L
Click to expand...

All you're doing with any of the maths is finding the component of the split part in the direction of the main part. You can draw it on graph paper and measure the relative lengths. I'd draw it on paper and scan it, but I'm on holiday at the moment...

The 1/L part is because you want the fraction of the split part that is in line with the main part. The 2 as you rightly say is because there are two split parts, which cancel out the sideways load, but add together to create the load in the direction of the main part. It's just simple pin-jointed structure analysis.

Would a dimensional analysis help, or would that muddy the water further?

// Edit - your formula sqrt(L^2 - (D/2)^2) gives the length of the vertical part of the triangle i.e. the distance from the joint to the point mid-way between the actual chainplates. It's the angles that are important, not the actual lengths.
 
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