B Filling the spool of a spinning reel with braid & monofilament

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The discussion focuses on how to fill a Daiwa BG3000 spinning reel spool with two different fishing lines: 20 lb monofilament as a filler and 20 lb J-Braid for the remainder. The spool can hold 240 yards of braid, and the goal is to use 150 yards of braid while leaving a margin of 1/8 inch. Calculations reveal that approximately 23.5 yards of monofilament should be used to achieve this balance, based on the volume of the lines. Concerns about the changing diameter of the spool as line is added are addressed, confirming that volume is the critical factor, not the diameter. Understanding these calculations allows for effective line management on the reel.
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TL;DR Summary
How to calculate the amount of mono filler so that the balance will be 150 yds of braid line.
I need help on figuring how to fill a spool on a spinning reel with two different diameter fishing lines.

BACKGROUND INFO:
The Daiwa BG3000 spinning reel’s spool can hold 240 yds of 20 lb J-Braid line.
The Daiwa J-Braid X8 Grand line comes 300 yds on a spool.
J-Braid X8 Grand 20 lb line is considerably more expensive than 20 lb monofilament line.
The diameter of 20 lb mono line is bigger than the diameter of 20 lb braid line.

WHAT I WANT TO DO:
I want to fill the reel’s spool with 20 lb test monofilament line first. (Called “filler” line”) Then fill the rest of the reel’s spool with 150 yds of the 20 lb test J-Braid line. The object is to fill the reel’s spool so that when there is 1/8” margin left on the spool as shown in the picture below, that will be 150 yds of braid line used:

fill to 1-8 inch margin.jpg


NEED HELP ON THE CALCULATION:

If the reel’s spool is rated to hold 240 yds of 20 lb braid line, how much of the outer diameter of the spool takes up 150 yds of line when it just reaches the 1/8” margin?

Put another way, how much of the spool should I fill with the mono line so that the remainder of the spool will use 150 yds of braid line?

Spool depth marked with red line.jpg


The red line indicates the “depth” of the spool with the 1/8" margin. What percentage of the red line should I fill with mono line so that the balance of the red line will use about 150 yds of braid line?

I’ve ordered the reel but have not received it yet so I can’t measure the actual dimensions of the reel’s spool. But the actual dimensions may not be needed for this calculation in percentage of the red line.

Hope you physics gurus can figure this out.
Thanks.
 
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You could probably get a good estimation just calculating the volume of the line in each case. How thick is each line?
 
Dale said:
You could probably get a good estimation just calculating the volume of the line in each case. How thick is each line?
The spool that the J-Braid X8 Grand 20 lb comes on says 0.23mm diameter.
The spool that the cheap monofilament 20 lb that I plan to use does not list the diameter. But 20 lb mono line diameter ranges from 0.41mm to 0.46mm according to a chart on the web. Since I'm using the cheap mono, I suspect the diameter will be on the larger side so call it 0.45mm diameter.

But without knowing the diameter of the mono line, I'm hoping some guru can calculate the percentage of the red line in picture-2 to fill the outer portion with 150 yds of braid knowing that the reel's spool completely filled with the braid line takes 240 yds. Then I can fill with 20lb or 25lb mono line until the the balance of the red line = the calculation, then fill the rest with 150 yds of braid.

Thanks!
 
The volume of a cylinder is ##L\ A=L \ \pi r^2##. So the volume of braided line that the reel can handle is
##240 yd * \pi * (0.23/2)^2 mm^2 = 9.97 yd*mm^2##
(I know it is a weird unit of volume, but it is fine as long as we are consistent).

If you fill it with 150 yards of braid then that takes
##150 yd * \pi * (0.23/2)^2 mm = 6.23 yd*mm^2##
so you have 3.74 yd*mm^2 of volume left for the mono-filament.

So solving
##3.74 yd*mm^2 = L * \pi * (0.45/2)^2 mm^2##
gives L = 23.5 yd
 
Dale said:
The volume of a cylinder is ... snip ...
Hi Dale,

Thanks much for your help with this calculation.

My concern is that the inner diameter of the reel's metal spool before any line is installed, (call it the core), is small compared to the outer diameter.

So let's assume that the entire spool is filled with 20 lb braid line (no mono filler). The braid wound on the spool in the beginning is being wound on a small diameter core. As the spool is gradually filled with line, the diameter of where the line is being wound on is getting larger and larger. Therefore looking at the red line in picture-2, maybe 30% of the outer portion holds an equal amount of line as the inner 70% of the red line. (Just made up percentages as I have no idea of the correct percentage difference.)

Hope I explained my concern clearly enough to understand what I mean.

Thanks!
 
HRG said:
Hope I explained my concern clearly enough to understand what I mean.
Yes, you explained the concern clearly. You don’t need to worry about it. A given amount of line will take the same amount of volume on the spool regardless of if it is the part of the volume next to the spool or on the outside of the spool. What matters is the volume.
 
Dale said:
Yes, you explained the concern clearly. You don’t need to worry about it. A given amount of line will take the same amount of volume on the spool regardless of if it is the part of the volume next to the spool or on the outside of the spool. What matters is the volume.
(light bulb) Now I see what you mean about volume being the key.
Thanks for your calculations!

Out of curiosity's sake, I'm trying to understand how you derived the formula:

"The volume of a cylinder is L A=L πr2. So the volume of braided line that the reel can handle is
240yd∗π∗(0.23/2)2mm2=9.97yd∗mm2"


from the formula for the volume of a clyinder being: V=π * r^2 * h
For example, what is mm^2?
 
HRG said:
For example, what is mm^2?
That is just square-millimeters. Since the radius is in millimeters then the cross sectional area is in square-millimeters or mm^2.
 
Dale said:
That is just square-millimeters. Since the radius is in millimeters then the cross sectional area is in square-millimeters or mm^2.
Thanks Dale.
 
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