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. :-\
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. 8) Rudy
Rob, very,very helpful. I printed it out and hung it on my bench for future ref. thanks for the hard work.
Bob
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
Quote from: Mandelstam on June 03, 2013, 05:56:30 PM
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
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
Quote from: Mandelstam on June 03, 2013, 07:10:38 PM
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. :D I'm sure the formula needs work, but... ???
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.
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.
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. :(
Quote from: Bryan Young on June 04, 2013, 12:07:20 AM
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?
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
Ron,
Thanks for your explanation. An engineering genius and physicist I am not, but I think I understand what you're saying.
Quote from: noyb72 on June 04, 2013, 02:17:03 AM
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
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. ??? :'(
Quote from: BMITCH on June 04, 2013, 01:20:02 PM
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. ??? :'(
Bob, all I can say is I get a lot more drag from reels with bigger washers and more of them.... :-\
Rob, enough said. Bigger is better. Now let's talk about the hex washers and their dual(both sides) being used. This would be more drag surface= bigger, right?
Quote from: BMITCH on June 04, 2013, 02:15:28 PM
Rob, enough said. Bigger is better. Now let's talk about the hex washers and their dual(both sides) being used. This would be more drag surface= bigger, right?
In my opinion... ;D ;D but Jedi Sal knows more about "The Force" than I. ::)
I think it's time to lock this thread. If someone else can devise a formula or program to calculate drag then that would be a benefit. It's just beyond me...
http://answers.yahoo.com/question/index?qid=20070603182434AAxTKI2 (http://answers.yahoo.com/question/index?qid=20070603182434AAxTKI2) - How does the surface area influence the sliding force of friction? Lets say I slide something on a smooth?
...This equation applies to both static and kinetic sliding friction. Static friction is the friction before an object starts to slide. Kinetic friction is the friction when the object is actually moving or sliding.
Static friction and kinetic friction have different coefficient of friction values.
An interesting result of this equation is that in the case of sliding friction of hard surfaces, the friction is independent of the area of the surfaces. In other words, it is just as difficult to move a 1 square-cm object as a 1 square-meter object, if they both are pressed to the surface with the same amount of force.
This is not intuitive. You would think that there is more friction when the surfaces are larger, but the friction equation states otherwise. You can verify this fact with experiments.
In situations where the surfaces deform or there is molecular adhesion, the friction is not independent of the areas in contact. In these cases surface area usually comes into play. This is also true for rolling and fluid friction.
When solid surfaces are soft and deform or when one material is a fluid, the shape of the solid object may be a factor.
Although the standard friction equation still holds, the coefficient of friction may have area, shape and other factors included in it."
-----------------
I think this is beyond me to fully apply to our greased drag washers, but interesting read none the less...
/Karl
We discussed this a bit last year.
http://alantani.com/index.php?topic=4593
Norcal Pescador, your thoughts and calculations are not without merit, even if the formulas themselves are a bit out of whack. As someone else mentioned, if seen only for their comparative values, it still makes some sense. So if you stop quantifying things in terms of millimeters and pounds, and instead describe them as say, The NP Factor, their relationship still carries a comparative value. Sort of.
HOWEVER,
and much to the disappointment of many i'm sure, there is a much easier method of thought. In the example of a given reel with say four drag washers known to make sixteen pounds of drag, each washer is worth four pounds, because 16 / 4 = 4.
Adding another washer would make twenty pounds.
That's all.
Sorry.
.
Quote from: Robert Janssen on June 04, 2013, 04:46:22 PM
..... In the example of a given reel with say four drag washers known to make sixteen pounds of drag, each washer is worth four pounds, because 16 / 4 = 4.
Adding another washer would make twenty pounds.
That's all.
Sorry.
That is where all of this started. I have reels that give a certain amount of drag. Reels that have similar-sized drag washers give similar results. Using the example you gave of 16 / 4 = 4, I grouped washers with similar surface area to give similar values and then tried to create a formula to fit my "findings".
As the carnival hawker would say, "Close, but no cigar."
Quote from: Robert Janssen on June 04, 2013, 04:46:22 PM
and much to the disappointment of many i'm sure, there is a much easier method of thought. In the example of a given reel with say four drag washers known to make sixteen pounds of drag, each washer is worth four pounds, because 16 / 4 = 4.
Adding another washer would make twenty pounds.
That's all.
Sorry.
First, drag disks are normally installed in odd numbers 3-5-7. What you are saying is absolutly true. But it does not mean that 5 big washers produce more drag than 5 small washers. That depends on how much force you can put on the washers and how much surface area that force is applied over.
I guess we also need to talk about what we are calling drag. I (and I think everyone here) considers drag the force required to pull line off the spool when the star is set at a specific position. If that is the definition we are going with then the diameter of the spool and the gear ratio has to be taken into consideration. Not as much as the disks maybe but definatly thought about. For instance, if we took a 500 and connected a spring scale to the line and adjusted the drag to where no more line could be pulled out when the scale reads 12 pounds and then took the same drag/gear trane and put it in a 112h at the same setting the spool would pay out line because of the increased mechanical advantage of the larger spool. So the disks are seeing the same force from the star but the effective drag is different. Swapping in a set of 5:1 gears would change things also.
Ron
Yes, we chatted about that and some other stuff earlier this year.
http://alantani.com/index.php?topic=6203.msg52156;topicseen#msg52156
.
Quote from: noyb72 on June 04, 2013, 09:22:45 PM
First, drag disks are normally installed in odd numbers 3-5-7. .......
Unless you're talking keyed washers like the 113HN (4) or hex washers. ;D
I'm puzzled. What, exactly, are you trying to figure out?
For example, if you have four washers and get 10 pounds of drag, adding one more will give you 12.5 pounds, everything else being equal. You find out how much drag comes out with the 4 washers by actual measuring, in order to avoid the "confounding variables" that you have mentioned (e.g., reel mechanics, amount of force used to tighten, etc).
It is possible to know the coefficient of friction of materials (there are charts for that), but knowing those numbers probably won't help you to find out what actually the result will be when pulling on a line from the spool (because of the variables you mentioned). Yes, you could probably theoretically calculate some number, but it would be much faster, easier, and more accurate to just measure with a scale.
QuoteI 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.
The data would be beneficial for shopping and modifying reels. For instance, would it make sence to 5+1 a small reel to make a dynamite little powerhouse? If the surface area of the drags mattered then this chart would help give you some idea as to what to expect between say a 180 and a 146. However, A 180 and a 146 are about the same except for the gears and so the drag pressure would be pretty much identical. If surface area mattered then the 146 would have the advantage.
Ron
Quote from: noyb72 on June 05, 2013, 08:14:17 PM
The data would be beneficial for shopping and modifying reels. For instance, would it make sence to 5+1 a small reel to make a dynamite little powerhouse? If the surface area of the drags mattered then this chart would help give you some idea as to what to expect between say a 180 and a 146. However, A 180 and a 146 are about the same except for the gears and so the drag pressure would be pretty much identical. If surface area mattered then the 146 would have the advantage.
Ron
This is what confuses me: The size and number of drag disks, alone, do not determine final drag (i.e., line coming out of the reel). As has been mentioned before, gear ratio plays an important part. Thus, to say that the 180 and 146 are about the same except for gears, isn't very revealing since the change in gear ratio throws the whole thing out of whack. Surface area does, in fact, have a great bearing on the outcome. It's the other variables that is making all of this confusing. Such things as gear ratio, diameter of spool, psi that can be applied to the disks, etc. all are different with different reels, and all of these factors affect the final drag (significantly). This is why, unless one is talking about the differences between a Senator 114H using CF or not, or a 114H using 4 or 6 disks in a stack, it's not possible to determine what the effects will be (unless one is willing to do a lot of math, and then the comparison is only theoretical). Trying to predict a 3 disk 114H based on a 6 disk 113H, based on diameter of drag disks
alone, is not possible except, as I've said before, unless doing the engineering math.
Surface area does, in fact, have a great bearing on the outcome
Makule, I'll let you argue with Mr. Comumb about that.
Ron