Drag Washer Friction Estimates

[Norcal Pescador]

It occurred to me that there must be a way to roughly determine the drag of a given washer size where the inside and outside diameters are known. So I launched on this project to make it easier to find out how much adding another washer or two to a drag stack will increase the expected drag. Remember this is an estimate. The reel mechanics, amount of force used to tighten the drag star, amount of grease applied to each washer, and other, unknown variables may affect outcomes.

What I did was determine the friction area of the drag washers using a formula of:
pi [(o.d. – i.d.)÷2]²  = D    THIS FORMULA IS NOT VALID! See my later post.
          29            

•   o.d. is the outside diameter of the washer in millimeters
•   i.d. is the inside diameter of the washer in millimeters
•   pi r² determines the area of a circle and 3.14 was used for pi
•   29 is the rough result of using existing maximum drag figures to obtain an average coefficient of drag    washer surface area
•   D is the approximate amount of drag per washer
      
Here are the results of my findings of Penn HT100 drag washers:

Part #       OD(mm) x ID(mm)   OD – ID    Approx. pounds of drag
6-155    17.92 x 10.65           7.27      3
6B-965    24.30 x 15.24            9.06      4
6-114       24.25 x 15.16           9.09      4
6-855       21.48 x 11.99           9.49      4
6-60       20.44 x 10.61           9.83      4
6-113      20.45 x 10.50           9.95      4
6-875       25.01 x 13.06           11.95      4
6-895       29.47 x 16.05            13.42      5
6N-525MAG   28.67 x 15.05           13.62      5
6-309       24.59 x 10.76           13.83      5
6-113H    27.02 x 13.06           13.96      5
6-115       29.54 x 14.82            14.72      6
6-116      32.69 x 16.55            16.14      6*
6-320      29.02 x 12.85            16.17      6*
6-965    24.61 x 8.15            16.46      6*
6-49      28.67 x 10.55           18.12      7*
6-117      38.64 x 16.58            22.06      8*
6-975LD   38.25 x 15.12            23.13      9*

*calculated by extrapolation

The values shown are for one greased Penn HT100 drag washer.
Cal's Reel and Drag Grease (tan) was used on all washers.
---------

Sorry about the hard-to-read table, I copied it over from a word.doc
I hope this holds up under the scrutiny of engineers and mathematicians.

[RowdyW]

Rob, even if it is not accurate it still gives a table to compare one size with another to fit into the same reel. I just was working on a Daiwa Sealine 450h that I was replacing a 3 stack with a 5 stack. I was going to use 6-113h CF washers but instead I used 6-49 washers. Your figures show the differance is quite a gain in drag even though I had to open up the ID about .050 on the 6-49. Gotta install that chart permenently on the site. Its plently close enough to very helpful.       Rudy

[BMITCH]

Rob, very,very helpful. I printed it out and hung it on my bench for future ref. thanks for the hard work.
Bob

[Mandelstam]

Excellent initiative!

But I'm not sure if I follow the formulas here, but then it was a few years ago I went to school. To me it seems that the surface area formula is wrong. You can't subtract the smaller diameter from the larger and use that sum as the new diameter. You have to calculate both areas and subtract the hole from the larger circle. Right?

And what washer size did yield the 29lbs of drag? If you have that you get the pounds of drag per mm2 which you then use to calculate the drag for other sizes.

I'd be happy to help with an Excel file that you can input your washer sizes in and get instant result. I've already started actually as I had to make some control calculations..

/Karl

[Norcal Pescador]

Excellent initiative!

But I'm not sure if I follow the formulas here, but then it was a few years ago I went to school. To me it seems that the surface area formula is wrong. You can't subtract the smaller diameter from the larger and use that sum as the new diameter. You have to calculate both areas and subtract the hole from the larger circle. Right?

And what washer size did yield the 29lbs of drag? If you have that you get the pounds of drag per mm2 which you then use to calculate the drag for other sizes.

I'd be happy to help with an Excel file that you can input your washer sizes in and get instant result. I've already started actually as I had to make some control calculations..

/Karl

Well Karl, I came up with the average of 29 when I measured the drag on several reels. Let's say I had a max lockdown drag of 15 pounds on a jigmaster with three stock washers. When the upper part of the formula was applied and divided by the known drag of the washers, that's where the 29 came from. That amounts to 5 pounds per washer. I used the same process for many of my reels. Some came out to 28, some to 29 and some to 30. 29 is the 'constant' factor. Does that answer your question about '29'? It may be stone age math, but it seemed to work.

As far as subtracting i.d. from o.d. then going from there on figuring surface area on a washer, I definitely could be off-base and I'm open to correction. "It's no good if it's already broken."  

Rob

[Mandelstam]

Well, if you say it works that's good enough for me! And as I said, it was a few years ago i used to tackle formulas like this...

All the best!

/Karl

[Norcal Pescador]

Well, if you say it works that's good enough for me! And as I said, it was a few years ago i used to tackle formulas like this...

All the best!

/Karl

If you do come up with an Excel program, that would be great. I'm not an engineer and squeaked by in high school algebra, so I'm not the sharpest pencil in the drawer.   I'm sure the formula needs work, but...

[Jeto]

have to agree with Karl as to the computation on the surface area.  surface area of the whole washer minus the area of the inner diameter.

[Bryan Young]

Based on the fact that drag washer#6-155 produces 3# of drag per washer, calculated by the area, a linear extrapolation is as follows:

CF #   ID (mm)   OD (mm)   Area (mm^2)   #/Washer
6-25   26.07   54.13   1767.47      32.50
6-49   10.55   28.67   558.16      10.26
6-60   10.61   20.44   239.72      4.41
6-113   10.5   20.45   241.87      4.45
6-113H   13.06   27.02   439.44      8.08
6-114   15.16   24.25   281.36      5.17
6-115   14.82   29.54   512.85      9.43
6-116   16.55   32.69   624.18      11.48
6-114   16.58   38.64   956.74      17.59
6-155   10.65   17.92   163.13      3.00
6-309   10.76   24.95   397.98      7.32
6-320   12.85   29.02   531.74      9.78
6-320LD   20.71   45.88   1316.38      24.21
6-525   15.05   28.67   467.68      8.60
6-855   11.99   24.48   357.76      6.58
6-875   13.06   25.01   357.31      6.57
6-895   16.05   29.47   479.78      8.82
6-965   8.15   24.61   423.51      7.79
6B-965   15.24   24.3   281.36      5.17
6-975LD   15.12   38.25   969.53      17.83
6-5600   14.9   41.15   1155.57      21.25
6-7000   5.08   21.54   344.13      6.33
56-440   8.88   30.93   689.43      12.68
56-710   7.92   14.68   119.99      2.21
56-4200   8.44   26.29   486.89      8.95


Area = PI*(OD/2)^2 - PI*(ID/2)^2

I think there is more to it than meets the eye, such as drag washer compression.  This may be true for Carbontex washers that is solid, but HT-100 washers can be a bit spongy, and therefore, I think we have a slight flaw in the calculation that we are missing.

[Norcal Pescador]

My formula is not valid.

Using a corrected formula of [(o.d.÷2)²pi - (i.d.÷2)²pi] ÷ D does not give a constant factor.
         
Subtracting the surface area of the hole from the overall surface area gives me inconsistent results.  I tried using the revised formula on 155, 60, 309, 525MAG, and 115 washer measurements.

Sorry guys.

[Norcal Pescador]

Based on the fact that drag washer#6-115 produces 3# of drag per washer, calculated by the area, a linear extrapolation is as follows:

CF #      ID (mm)   OD (mm)   Area (mm^2)   #/Washer
........
6-115      14.82   29.54           512.85      9.43
........
6-155   10.65   17.92           163.13      3.00

Area = PI*(OD/2)^2 - PI*(ID/2)^2

Bryan, do you mean 6-155?

[Ron Jones]

Gentlemen, please allow me to get a little geeky in order to explain why these tables don't make sense.

In the late 1700s there was a guy named Coulomb. He came up with a law of friction that says the force required to overcome friction is equal to the force holding them in place times the coefficient of friction. Notice that the surface area of the material is not listed in the equation. A better explanation than I can come up with is listed below.

"Although a larger area of contact between two surfaces would create a larger source of frictional forces, it also reduces the pressure between the two surfaces for a given force holding them together. Since pressure equals force divided by the area of contact, it works out that the increase in friction generating area is exactly offset by the reduction in pressure..."

So, the reason some of the charts you all created don't hold water to real life experience is because the size of the drag surface of a single disk is absolutely immaterial. 5 HT-100 155 disks will have the same drag as 5 HT-100 113 disks assuming they are put under the same pressure, and that's where it gets interesting. Pressure is often measured in Pounds per square inch (PSI), I Jigmaster sized star will exert the same force on the main gear of a model 500 or a model 49. However, the model 49s main gear is much larger and therefore will exert more pressure on the drag disks due to its greater surface area. This increased pressure multiplied by the same coefficient of friction means that it will take more pressure to overcome the 49s drag.

Overcoming the drag is also interesting and brings to light something that has been discussed here several times. The pressure required to overcome the friction of the drag stack is created by a force (hopefully a big fish!) pulling on line on the spool. That force is best expressed as torque being applied on the pinion gear. As anyone who has ever turned a wrench knows, the longer the wrench the more torque generated at the fastener. In reels this means the taller the spool the less force required to overcome the drag. All of these forces are basically trying to pull the reel apart, and this is where the 49 fails. The 49 has a BIG main gear sitting ultimately on a skinny post. It can apply a lot of drag pressure because of the size of the main gear. The 49 also has a tall spool which means that extra torque is being applied to the pinion which is trying to overcome the pressure of the drag stack. It doesn't take a rocket scientist to see that that poor little post is getting its butt kicked and will eventually fail. As it happens, the higher gear ratio doesn't help out at all either.

So, bottom line is the more to it than meets Bryan's eye is that the size of the drag disk doesn't matter.

What does matter is coefficient of friction. I don't know what the CF of HT-100 and Carbontex is but I do know that people often find the reel can create more drag with Carbontex washers. This means that Carbontex probably has a greater coefficient of friction compared to HT-100s.

Ron

[Norcal Pescador]

Ron,
Thanks for your explanation. An engineering genius and physicist I am not, but I think I understand what you're saying.

[Bryan Young]

In the late 1700s there was a guy named Coulomb. He came up with a law of friction that says the force required to overcome friction is equal to the force holding them in place times the coefficient of friction. Notice that the surface area of the material is not listed in the equation. A better explanation than I can come up with is listed below.

Ron, I didn't realize you were so geeky.  lol...and I though it was Joule, Ohm,... and not Coulomb...just kidding.

There are so many other factors that is involved, but I believe one of the main difference between Carbontex and HT-100s are the compression factor of the drag washers.  Greased carbon fiber will have similar coeficient of friction due to the grease, but how much pressure you are able to exert on the washer and each pressure plate (metal washers) to have opposing forces will make a big difference.  I'd say for most people, HT-100s are great because they are the most forgiving drag washer and will allow for slight errors in setting of the drags.  Carbontex is not as forgiving.

Bryan

[BMITCH]

WOW Now I'm completely in the dark. So is it better to have more surface? Say more cf washers or is there a tipping point where it becomes pointless.