Calculating Length of line for Total Seizing Length

dbdriskell

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#1
Please see Greg's entry #17 near bottom of this thread. MY CALCULATIONS FROM HERE ON ARE FLAWED. like I said. It was an experience in math. I removed most all of my entries as unfortunately they are all wrong. I found some other info on the internet about cols and helix's. Worth looking into.
If I have any defense, I got this info off the internet. The coil obviously grows in length of which my formula does not take into consideration. I think higher math will resolve it.

So, when rigging Shrouds at the Mast Tops, it is customary to seize some line that goes around the Mast Top. Usually, the seizing (in the middle of a double Shroud) is about 60mm or so. Usually, I will use two clips spaced apart to hold the Shroud line. Then I will cut off an approximate length of line to seize 60mm worth of Shroud. That usually equals to about 700 mm or so of line.

However, the line that you are going to seize with of course varies. The Solid Core "WIDTH" of the Shroud varies -- so, I decided that there had to be a math formula to calculate first hand how much seizing line to cut in order to seize the line so many mm total. Kinda confusing. Some formulas I came across were outlandish in complexity.
 
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Peglegreg

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#3
G'day Donnie
Your calculation is not what I normally use.
I have always use the following formula to work out what I need:
Using the Pi x D or the 2R xPi
The formula is
(sL ÷ sD) × (shD + sD) x Pi = LRqd
Where
sL is lenght of seizing required
sD is seize diameter
shD is shroud diameter
sD is seize diameter
PI is 22/7 or 3.142
LRqd is total lenght needed

So, lets experiment:
I have a CORE line that is 1.5 mm in total diameter that is a given (this CORE could be a Shroud's DIAMETER)
I have a SEIZING line that is .72 mm in total diameter that is a given

I want to SEIZE that CORE so that the SEIZING will be 60mm in length.
So using my formula, (photo below) l get 582 mm to be cut. This is assuming that you do a very tight seizing. And of course you have to add the run on and off lenghts, say another 450mm (I like to have plenty to play with) so you should allow 1030mm.
20190111_225610.jpg


In a real case scenario, you might have a .80 mm Shroud line and a .20 mm seizing line and you might want to seize that line for only 30mm. (in other words the "wrap or seize" will measure to be 30mm. So, since you know that you want 30 mm "worth" of seize we can calculate:
Using my formula I get: 472mm and again adding tha wattage of 450mm you nee to cut 922mm.

20190111_232113.jpg
I sorry for not agreeing with you, but my mathematics is what I use all the time.
I must admit I do get 'overs' but it's all depends on how tight you do the seizing.
It's always better to have more than not enough!

I would like someone to test yours and mine and see who's is the closest.
You are right when you said, it's fun to play around with this.
Does anyone else has a way to workthis out?
 

Peglegreg

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#4
I want to use a "constant" that I came up with that is about 47% or just .47
This has made me to think about my formula and maybe you have a good point. The diameter of the total shroud + sieve should not be use but up to the radius of the sieve so my formula should be:
sL ÷ sD) × (shD + (sD ÷ 2)) x Pi = LRqd
So my calculation for the first would be:
20190112_025031.jpg
487 mm + waistage.
And my next calculation would be:
20190112_031356.jpg
424 mm + waistage
Looking at both of ours formulas, we now agree on everything except how to work out the amount of 'rotations' that is needed.
My theory is based on how many sieve's thickness would it take to make the seive's length needed. So I divide the lenght by the diameter of the sieve.
Hmmm what are your thoughts about this Donnie.?
Gee I'm having fun here.. :D;)

Ps. I have done some corrections to some words due to the fact I was writing this at 3am this morning and my phone kept on changing words on me and my eyesight at that time was a bit blurry.:rolleyes:
 
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Peglegreg

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#6
Okay, so , instead of the step by step confusion, I created the formula and it must be executed with operational order. In other words, what ever is in parenthesis is calculated on its own. So therefore taking from Greg's nomenclature:

Sd = seizing line diameter in mm
k = .47 mathematical constant a figure of merit which takes into consideration the pitch of the winding's.
Cd = Core diameter of rope or line that you want to seize.
π = pi
SL = The Length in mm of your final seizing length. (or you can say the length of the coil itself)
SLCut = This is the length that you will actually cut your line to length that you will use to seize with.


( Sd x .47 x 2 + Cd ) x ( π ) x (SL) = SLCut

Perform each operation in parenthesis individually

(.15mm x .47 x 2 + 1mm ) x ( π ) x (75mm)
(1.141 ) x ( π ) = 3.584
(3.584) x (75mm) = 268 mm


On a Calculator you can enter in the data just like this:
.15
x
.47
x
2
+
1
{ENTER} or equals


Next step (take that number 1.141)
x
( π )
{ENTER} or equals
Next step (take that number 3.584........)
x
75
{ENTER} or equals 268


Finally, I hope that I do not get π in my face (no pun intended) for all of this being TOTALLY wrong. But a little research on the net and simplifying some very complicated junk, this is what I come up with.
Your turn Greg !!!! or anyone else ?
Using your above figures, I think you're talking about the rope to be sieved is 1 mm and the sieveing rope is 0.15mm and the lenght to be sieved is 75 mm.
so using my calculator and my formula:
20190112_161949.jpg
1689 mm + waistage.


note: by my calculation there will be 500 rotations the the server would have to do to cover 75 mm with a 0.15 mm rope.
20190112_162915.jpg
So the length of the rope will have to be longer than your figure of 268 mm!
Think about this.
I might be wrong but my gut and my mathematical knowledge, says I'm not.
(I was in the advance mathematics class at school and we were doing the highest level possible. I got the highest grade at my HSC which is your leaving exam, I believe)
 

dbdriskell

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#8
This is funny. You make me laugh, of course in a very entertaining way !!!!

Hey Greg, did you actually perform the real excersize and do in reverse?
Seize a line for about 60 mm. Get your caliper or steel measures and measure as you wrap until you get to 60mm of wrap. Right at the 60 mark, cut the line off flush.
THEN UNWRAP the coil and measure hoe much line it took. Write that figure down. THEN perform the cal and see if your cal is the same length of the uncoiled line !!!
Now that is the real test.
 
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Peglegreg

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#9
I'll do the same thing here when I can, the temperature in the garage during the day is about 113°F or 45°C because of a heat wave we are experiencing.
I might get a ball of twine and do a version in the house. I wouldn't like to waist good rope and twine would be easier to unwind to measurement etc.
Good luck to both of us mate, and let's see who will have Poop on their face! :p LOL.
 

dbdriskell

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#10
Ok. I admit. Something is seriously wrong.

I just wrapped or seized 105mm length of core using .23mm for coil and core is .73
It took 700 mm of line to make this.
The only way I can get my calculations to work is to multiply my entire equation by a factor of 2.
Somehow you have to take into consideration of each coil is also adding length as you go.
So I have to wait to see what Greg comes up with. So I knew I would get pie in my face.
 

dbdriskell

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#11
Well. It was worth a try. Overall if you take the final answer and multiply time 2 you will be very close. I will leave it up to Greg to explain why.
 

Peglegreg

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#12
I haven't had a chance yet to do mine. But using your figures into me formula, I get: 563 mm
20190113_093846.jpg
So adjusting my formula to 80% (instead of 50%) of the seiving rope, I get 700mm.
20190113_093814.jpg
Thanks Donnie for that information.

I'm going to try it with different size rope to check my formula.
 

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#14

Peglegreg

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#16
Err why all the maths?
It's simple... the same reason why a mountaineer climbs a mountain....because it's there.

Seriously, I like to do the serving by hand, you get more control over the outcome, and when you want do do the end of a rope etc, it is easier by hand. There are many occasions that I need to know, how much to cut off.
 

Peglegreg

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#17
Well, I've donel my own test and the results are quite good.
The sizes are 15 mm of length of seiving
Rope to be served diameter: 0.74
Serving rope diameter: 0.34 mm
So the calculation using my formula is:
20190114_162815.jpg
So the lenght of serving rope is 140 mm long.
My result was:
I was able to do the full lenght, but with approx 2.5 mm left of my lenght of 140 mm.
So my new formula with the help from Donnie latest figures is:
(sL ÷ sD) × (shD + (sD × 0.8)) x Pi = LRqd
Where
sL is lenght of seizing required
sD is seize diameter
shD is shroud diameter
sD is seize diameter
PI is 22/7 or 3.142
LRqd is total lenght needed

I think that's is a wrap-up (excuse me pun) on this matter.
This thread may be a bit boooooring for some,Sleep but I had fun!:)
I hoped that Donnie and I wasn't too much of a
math's nerd! :p
What do you think @paulv1958?
ROTF
 

dbdriskell

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#20
Correct Uwe,
this overall was just an excersise in playing around with math. I would not (including myself) would use this unless I were seizing the entire length of the first shroud. I have read that sometimes the first pair of shrouds were wrapped so as the sails would not wear down the shroud.

@Peglegreg I think the formula must be based on the calculations of a "Helix"
 
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