Maths challenge…..

[Sea-Fever]

I could probably work this out if I put my mind to it but I’m far too lazy……

If your roller furling drum has length L, and you accept that the number of turns required to furl/unfurl remains constant (let’s say N) what is the relationship between the thickness of the furling line D and the resultant total diameter of the furled reel (let’s say T).

It’s just a bit of fun but with a real world application…..

Thicker furling lines are easier to handle and also provide a greater turning moment at the drum but you can’t exceed the max diameter of the reel….so how do you work out the max size of line for your setup?

…or alternatively maybe I’ll just buy some thicker rope and see how I get on!

 

[Refueler]

Theoretically yes ... you could calculate .....

But in actual real world - it is not unusual for turns to ride up ... then it destroys any theoretical layer heights on the drum.

The simplest solution - is {forgive me for saying it) to use the manufacturers quoted specs of max line length etc.

 

[Aeolus]

Needs to take account of the width (or should that be height?) of the furling drum, its inner diameter and outer diameter. You can calculate a theoretical capacity for different diameter ropes but rope doesn't usually lie perfectly on the drum and the imperfections (overlaps etc.) would be difficult if not impossible to predict. Much simpler to try it out.

 

[Bav32]

The length of rope that will be on the drum would be the the same as the LP (luff perpendicular) measurement of the sail. In practice there would be a few extra turns on the drum too.
So now you have the length.
You know the thickness of the rope and the height / width of the drum. Knowing the diameter of the drum, and usi g Pi D gives you the ideal length to fill the first "full"drums worth.
If it rolls up perfectly with each subsequent turn fitting g neatly against the previous, you could easily calculate the perfect number of turns etc.
For the second and subsequent full perfect drum fills you have to recalculate Pi D for each layer.
You get the point.
It's not a perfect world, so over,appiah g turns will not give an even roll.

 

[Stemar]

I reckon the maths is beyond my pay grade. A 50+ year-old A level suggests that a bit of calculus is involved and I haven't used that since the exam - nor much else from the A level syllabus, though a good bit of O level stuff has come in useful.

I think it should be possible to get round it with a spreadsheet, using a row for each layer. The number of turns per layer (t) is the width of the drum divided by the diameter of the line (d). The number of layers (l) is the height of the drum sides divided by d.
Now you can put in a formula with a row for each layer. If r is the radius of the empty drum, the length of line on the first layer is, near enough, 2πr*t. The second row for the layer will be 2π(r+d)*t. Next layer, R+2d, and so on. Now add the layers.

That come with all the caveats mentioned above, but should give an idea and allow you to play around with different diameters.

 

[boomerangben]

If you know:
1 The length of the existing rope that goes onto the drum

2 The volume of the drum which is Pi/4 x (outside diameter squared - inside diameter squared) x height of the drum

3 The diameter of the new rope

4 You then calculate the volume of new rope needed on the drum which is rope diameter squared x Pi/4 x length of the old rope

Make sure the answer to 2 is larger than the answer to 4 by a suitable margin (I would guess answer 4 should be 80% of answer 2)

 

[ash2020]

I could probably work this out if I put my mind to it but I’m far too lazy……

If your roller furling drum has length L, and you accept that the number of turns required to furl/unfurl remains constant (let’s say N) what is the relationship between the thickness of the furling line D and the resultant total diameter of the furled reel (let’s say T).

It’s just a bit of fun but with a real world application…..

Thicker furling lines are easier to handle and also provide a greater turning moment at the drum but you can’t exceed the max diameter of the reel….so how do you work out the max size of line for your setup?

…or alternatively maybe I’ll just buy some thicker rope and see how I get on!

Or you could do what I did when my furling line didn't quite fit on the drum. Strip the inner core out of half of the line. It works nicely, with still plenty of strength and almost a boat length of full thickness.

 

[Refueler]

The length of rope that will be on the drum would be the the same as the LP (luff perpendicular) measurement of the sail. In practice there would be a few extra turns on the drum too.
So now you have the length.
You know the thickness of the rope and the height / width of the drum. Knowing the diameter of the drum, and usi g Pi D gives you the ideal length to fill the first "full"drums worth.
If it rolls up perfectly with each subsequent turn fitting g neatly against the previous, you could easily calculate the perfect number of turns etc.
For the second and subsequent full perfect drum fills you have to recalculate Pi D for each layer.
You get the point.
It's not a perfect world, so over,appiah g turns will not give an even roll.

I do not agree .... with your LP idea.

Why ?

You are assuming a) the drum OD is same as the foil on the forestay - which it is not, b) that rope turns will lay correctly and to same depth as the sail as it furls.

The lengths will be different as the turns of drum will take more rope than sail on the foil and increase as more rope goes on the drum.

 

[Fr J Hackett]

Volume is the easy answer, easy to calculate the available volume of the drum and easy to work out the volumes of the lines.

 

[earlybird]

If you know:
1 The length of the existing rope that goes onto the drum

2 The volume of the drum which is Pi/4 x (outside diameter squared - inside diameter squared) x height of the drum

3 The diameter of the new rope

4 You then calculate the volume of new rope needed on the drum which is rope diameter squared x Pi/4 x length of the old rope

Make sure the answer to 2 is larger than the answer to 4 by a suitable margin (I would guess answer 4 should be 80% of answer 2)

The best packing density for equal, rigid circles is just over 90%, so rope, which is squidgy, should do better. Also, rope under load will probably reduce below nominal diameter, allowing an even greater length; complicated!!

 

[boomerangben]

The best packing density for equal, rigid circles is just over 90%, so rope, which is squidgy, should do better. Also, rope under load will probably reduce below nominal diameter, allowing an even greater length; complicated!!

Thank you for the more accurate number. My only concern is that the last layer might leave upto 1 x rope diameter of the drum radius unused which is quite a big chunk of volume. It would be interesting to compare Stenmar’s more rigorous calculations to see what factor would be reasonable.

 

[LadyInBed]

I'm not sure what the question is!
'your roller furling drum has length L' - Surely the drum has a circumferance and the line has an incremental circumferance.
Are you asking what length of line will it take to fill the drum using different diameter lines?

 

[Aja]

I always assumed that the rope used on furlers was of a harder consistency than normal rope so that when the sail is furled on a good breeze the rope doesn't compress and get stuck ending with the drum inoperative.

The suggestion that you can use a softer or larger diameter rope may not be the best idea when the going gets tough.

 

[Boathook]

I've done / did the same as @ash2020 . Strip the inner core out of the furling line from the furler to to the cockpit so that that when you use the line you are grabbing the full diameter of the rope. Make quite a difference to the line on the furler its self.

 

[johnalison]

If the drum were of zero diameter, then the width of the coil after one turn would be twice the diameter of the line. As far as I can see, however the line coils it will continue to increase in diameter in proportion to the diameter of the line. I think that this would hold whether line coils improbably by lining itself up with itself, or more likely forming a hexagonal packing, when the rate of increase will be least but the proportionality the same. In practice the coiling will be bulkier than this idealised version, and the degree of irregularity will determine how much additional space is needed. Possibly.

 

[thinwater]

Maths for 11-12 year olds? Really?

Hint. You do not need to calculate a bunch of turns stuff.

• volume of rope per foot.
• volume inner cylinder.
• volume of outer cylinder (subtract the two).
• make an assumption of gaps. Something around than pi/4. And some allowance for not filling the drum (or it will jam for sure), perhaps 80%.

Capacity = 80% x (pi/4)(volume of drum/specfic volume of the rope)

 

[Fr J Hackett]

Maths for 11-12 year olds? Really?

Hint. You do not need to calculate a bunch of turns stuff.

• volume of rope per foot.
• volume inner cylinder.
• volume of outer cylinder (subtract the two).
• make an assumption of gaps. Something around than pi/4. And some allowance for not filling the drum (or it will jam for sure), perhaps 80%.

Capacity = 80% x (pi/4)(volume of drum/specfic volume of the rope)

May I refer the gentleman to my previous statement #9

 

[alahol2]

Agree with boathook and ash2020. Buy the thicker line and strip out the core from the first several metres on the drum. According to Marlow the strength is split more or less 50/50 between core and cover. Can't find the reference at the moment.

 

[thinwater]

... According to Marlow the strength is split more or less 50/50 between core and cover. Can't find the reference at the moment.

I've tested them and been told the same. Standard eye splices depend on this fact.

 

[LadyInBed]

Buy the thicker line and strip out the core from the first several metres on the drum.

I milked the core back, then cut off the exposed core at the cockpit end and made a soft shackel from it.