Dimensions for a reel to hold kedge warp

[Poignard]

I'm thinking of making a reel on which to store my rope kedge warp.

Assuming I know the length of the rope and its diameter, how would I calculate the dimensions of a reel large enough to hold all the rope and no more?

Obviously there are several variables to be juggled with, ie: rope diameter, length, core diameter, width of reel and diameter of the outer discs.

What I need is a formula combining these.

 

[Shanty]

Did this a few years ago. The trick that worked for me was to assume the rope occupies a square with side equal to its diameter. Then work out the volumes involved:

If the reel has an outer diameter of D, a core diameter of d and an internal width of w, then the volume available for the rope is:

(pi*(D2-d2)/4)*w

If the rope has diameter R and length L, the volume occupied is:

L*R2

When these are equal, the reel is full, and the length of rope that can be stored is given by:

L = ((pi*(D2-d2)/4)*w)/R2

Note: haven't bothered to work out how to do any fancy characters on here yet - I've used D2 etc to mean D squared, * to mean multiply

This method assumes the rope will be coiled neatly on the reel, and that the width w is an exact multiple of the rope diameter R. Given these, it worked for me. You might want to add a rope diameter or two to the reel outer diameter D to allow for less than perfect stowage of the rope.

 

[Poignard]

Thanks very much for that.

 

[Bajansailor]

I am sure that there must be a standard formula incorporating all of the variables in one equation, but that is a bit beyond me.
However it should be possible to calculate by iteration, and you shouldn't need too many iterations.

Edit - Shanty has it above!

No worries, here is how I calculated it.

Lets say you have 100 m. of 20 mm diameter rope.
The cross section area of this rope is pi x r x r (pies are squared) = 314 sq mm.

The effective volume taken up by this rope on a reel is square shaped though, ie a 20 mm square, rather than a circle of 10 mm diameter.
The ratio of the area of the square to the area of the circle is 1.27
So use this as a multiplying factor - or perhaps go up to 1.3 to add a bit of a fudge factor.

So, the effective square cross section area of the rope is 314 x 1.3 = 408 sq mm.

The total volume required for 100 m of this rope is 100m x 408 sq mm = 0.0408 (say 0.041) cubic metres.

Lets say that the length (along the spindle) of the reel is 0.5 m.

Then cheek side area x length of reel = volume of rope on reel = 0.041

Hence cheek side area = 0.082 sq m. for a reel length of 0.5 m.

And the formula for the cheek side area is = 0.7854 (D squared - d squared)
Where D is the overall cheek diameter, d is the diameter of the spindle, and 0.7854 is pie divided by 4.

If the spindle diameter d is say 2" (50 mm, or 0.050 m) diameter, then d squared is 0.0025 sq. m.

So cheek side area = 0.082 sq m = (0.7854 x D x D) - 0.0025

0.7855 x D x D = 0.0845

D x D = 0.1076

Hence D = 0.328 m (say 0.33 m., or 13") for a reel length of 0.5 m (20")

Can anybody see any flaws in the maths?

 

[onesea]

Did this a few years ago. The trick that worked for me was to assume the rope occupies a square with side equal to its diameter. Then work out the volumes involved:

If the reel has an outer diameter of D, a core diameter of d and an internal width of w, then the volume available for the rope is:

(pi*(D2-d2)/4)*w

If the rope has diameter R and length L, the volume occupied is:

L*R2

When these are equal, the reel is full, and the length of rope that can be stored is given by:

L = ((pi*(D2-d2)/4)*w)/R2

Note: haven't bothered to work out how to do any fancy characters on here yet - I've used D2 etc to mean D squared, * to mean multiply

This method assumes the rope will be coiled neatly on the reel, and that the width w is an exact multiple of the rope diameter R. Given these, it worked for me. You might want to add a rope diameter or two to the reel outer diameter D to allow for less than perfect stowage of the rope.

YUp basically do it by volume, volume of rope (described as a square)+ 5%? = Volume needed on drum.

Have done it that way for wires and worked well rope should compress a little more than wire...

 

[rogerthebodger]

Have a look here

http://www.tecni-cable.co.uk/s.nl/ctype.KB/it.I/id.132/KB.109/.f

for a downloadable calculator

 

[guernseyman]

Or you could try this:

D^2 = d^2 + (4LR^2/pi.W)

where D = diameter of reel required,
d = diameter of core of reel,
L = length of rope,
R = diameter of rope,
W = width of reel.

 

[davidej]

Or buy one of these neat gismos

http://www.compass24.com/web/catalog/shop/technics_chandlery_anchor/8397239

 

[Wansworth]

I used a garden hose reelwith 12mm rope.