Hello out there, I have a reel basic question and somebody can help explain the how to: firstly I want to add more available drag area to a leverdrag reel and it seems physically possible. so how do I calculate the round disk and find the area of a band within a circle. as I also want to see what percentage was gained after computing the original stock drag area in sq. inches..... any easy to explainations would be helpful......thanks gst.
the formula is Pi × radius squared. so you just need the radius of each circle and squared it then times 3.14. take the fifference and you have the difference for the 2 surface areas.
X2
So Surface Area of drag disc = [ pi • ( R(outer radius)^2)] - [ pi •(R(Inner radius)^2)]
I perfectly understand! :P
hmm... i sorta get the feeling that this is leaning towards the more-surface-equals-more-drag-line-of-thinking, is that right?
I don't mean to intrude or dim your light bulb or anything, but if that is the case, then you'll be disappointed. It just simply isn't true.
Or maybe you have something else going on; idunno.
.
Quote from: Robert Janssen on March 05, 2014, 05:19:37 AM
hmm... i sorta get the feeling that this is leaning towards the more-surface-equals-more-drag-line-of-thinking, is that right?
I don't mean to intrude or dim your light bulb or anything, but if that is the case, then you'll be disappointed. It just simply isn't true.
Or maybe you have something else going on; idunno.
.
Robert - I understand that more surface area does not increase the drag but does an increase in the radius increase the amount of drag?
Drag surface area is a component that affects the amount of drag. Other components are pressure, supporting surfaces, pressure plate,...
Quote
Robert - I understand that more surface area does not increase the drag but does an increase in the radius increase the amount of drag?
Yes.
(among other things)
.
It's pretty complicated as Robert Jensen has pointed out and simply increasing the surface area is not enough if the (fixed max) force of the pressure plate is distributed equally over the increased surface area there is no net gain in drag (for a simple but incomplete answer).
Avet uses a dual drag system that has 2 carbon drag discs on the same side in the Raptor series, but as they both experience the force from the pressure plate directly (ie it is not distributed between them) then there is an increase in drag for a given preset, as I understand the system. This is similar to how we increase drag by increasing the number of discs in our star drags.
Accurate puts drag discs on both sides of the spool to accomplish the same.
The simplest way to increase drag in a single drag LD reel is to increase the force that the pressure plate can exert. This either means increase preset or change the drag curve via the cam. Needless to say this stresses other parts of the reel that may not be engineered to withstand the increased loads and it can lead to early failure (ie pinion bearings) or an unfishable reel (high handle turning pressure). Then you have a whole new set of engineering problems to solve :D.
When we "hot rod" a simple design we take it from humble beginnings and slowly add the features that are all currently incorporated in the leading edge designs. A good study of those designs (there is no substitute for taking one apart yourself) is a good way to grasp the ideas.
regards
OK,
I am in the camp of increasing surface area does not increase drag. However, I am hoping that either Robert or John can help explain something to me. In my research on the subject I have come across the term mean diameter. In many equations determining friction of sliding disks mean diameter is included. If you increase the outer diameter of the drag disk without a coresponding increase in inner diameter then you have increased the mean diameter and the surface area and it apears that you have modified the total friction for a specific amount of normal force. Wondering if either of you can explain.
Thanks
Ron
I think I found it. The equation for sliding friction of a lubricated disk is: µsl = φbl µbl + (1 – φbl) µEHL (unlubricated disks is almost the same.) In this equation mean diameter is only utilized in the weighting factor for the sliding friction coefficient equation. With speeds as slow as we are talking about in a conventional reel, the weighting factor can safely be assumed as 1 and therefore has no practical effect. So, I am still in the no effect camp.
That does not mean I don't like big drags. Big drags take longer to heat up which in effect causes them to maintain room temperature drag for a longer period in a situation where a fish takes a lot of drag. Large drags also carry much more lubricant than the smaller drag which is what I theorize produces their dramatic increase in smoothness. But to get higher drag numbers, you need more force or more surfaces being acted on by that force. I believe in a lever drag reel that means changing the ramp of the cam. In a star drag reel that means using pliers on the star or increasing the size of the star.
Quote from: noyb72 on March 05, 2014, 07:13:52 PM
OK,
I am in the camp of increasing surface area does not increase drag. However, I am hoping that either Robert or John can help explain something to me. In my research on the subject I have come across the term mean diameter. In many equations determining friction of sliding disks mean diameter is included. If you increase the outer diameter of the drag disk without a coresponding increase in inner diameter then you have increased the mean diameter and the surface area and it apears that you have modified the total friction for a specific amount of normal force. Wondering if either of you can explain.
Thanks
Ron
It's complicated! But the force that the disc applies is not directly in proportion to its surface area. A huge disc (large diameter) produces way more braking force because of it's ability to use leverage to increase braking force or to negate the leverage of the object it is braking in our case the spool.
Picture a modern brake on a bicycle that acts on the rim. The rim has a tiny surface area and the brake acts on even a smaller portion but because the force is applied far from the axle (the radius distance) it has braking leverage and all you need is a light squeeze of your hand to stop the bike. Conversely, those old bikes with hub brakes took all the force of our legs and weight and we still could barely stop the bike. The wheels had leverage on them even though their braking surface was far larger than the modern caliper brakes. Force at the rim has way more braking power than braking force at the hub.
So, in a nutshell, it is not purely the surface area, it is the amount of force applied, where the force is applied (farther from the axle the better), what mechanical advantages your braking mechanism may use to multiply force plus other variables to do with the rotational speed of the object to be braked and the braking surface's coefficient of friction, how it performs when hot etc etc etc etc.
Hope this helps and I hope I've got it right. Just trying to share my understanding of our toys. :D
ps I'm not sure what you mean by "camps" in this discussion (though you probably use the term very casually). There really is only one right answer to these questions about physics. I may be completely wrong in which case my opinion is of no merit whatsoever it's only *wrong*. Typically in scientific discourse there are "camps" about a legitimate point of dispute where knowledge is incomplete. Here, we may be struggling to understand the physics and explain our understanding to each other, but most of this was figured out 200+ years ago and is completely settled science.
I tend to think about Physics as vectors and levers (which covers most things mechanical) which is why I came up with the analogies I did. :D
best
I completly understand, thanks. I just wan't sure if the difference effected disks as small as drag disks in a reel. I guess theoretically they do, but the effect can practically be called non existant.
Ron
Quote from: noyb72 on March 05, 2014, 08:17:29 PM
I completly understand, thanks. I just wan't sure if the difference effected disks as small as drag disks in a reel. I guess theoretically they do, but the effect can practically be called non existant.
Ron
I believe the reason an 80W has more drag than a 30W is due to the diameter of the disc. etc right on down the line not simply that the disc has more surface area.
Also note that as gears on star drags have become larger and larger Shimano has decreased their thickness to save weight. This makes for shorter drag stacks as well but the effective drag (listed) has gone up if I am not mistaken.
best
so,
(a) if i replace a 1mm c/f disc and ss metal disc
(b) with 2 x .5 c/f and 2 ss disc all the same size
apply same amount of force to (a) and (b)
do i get a smoother drag but no increase
lets make the c/f disc hex shape
thanks
No I think you double up the drag due to more contact surface, and they are operating on parallel planes.
You increase drag, I don't believe it would be absolutly double due to ineficiencies, because you are applying the same force to multiple friction surfaces. The pressure seen by the top washer will be seen by the bottom washer.
Ron
Quote from: Robert Janssen on March 05, 2014, 05:19:37 AM
hmm... i sorta get the feeling that this is leaning towards the more-surface-equals-more-drag-line-of-thinking, is that right?
I don't mean to intrude or dim your light bulb or anything, but if that is the case, then you'll be disappointed. It just simply isn't true.
So, from Doc's and others comments, I get the impression that adding additional drag washers doesn't increase drag. If this is so, then what is the advantage of increasing a drag stack from, say, a 1+3 to a 1+5 or 1+7?
That is not true. Stacked drag washers are subjected to the same force, so each friction surface increases drag. Increasing drag washer size decreases the force per unit square that the washer is exposed to.As it happens, the decrease in force is equivalent to the increase in friction due to surface area.So there is no net increase in drag.
Ronald
Quote from: George4741 on March 06, 2014, 04:03:27 AM
Quote from: Robert Janssen on March 05, 2014, 05:19:37 AM
hmm... i sorta get the feeling that this is leaning towards the more-surface-equals-more-drag-line-of-thinking, is that right?
I don't mean to intrude or dim your light bulb or anything, but if that is the case, then you'll be disappointed. It just simply isn't true.
So, from Doc's and others comments, I get the impression that adding additional drag washers doesn't increase drag. If this is so, then what is the advantage of increasing a drag stack from, say, a 1+3 to a 1+5 or 1+7?
No, they add drag because instead of distributing the limited force of one pressure plate they magnify it when they are place in sequence. When you add surface area under the pressure plate of a lever drag all you are doing is distributing the force over that larger area (and as well that extra area doesn't have as much effect being close to the spindle). You would need a larger drag surface *and* a larger pressure plate to create more drag ie instead of a 30w you have to go to a 50w reel.
Take a look at this picture:
(http://i266.photobucket.com/albums/ii261/haugusnord/9918d02c-b0eb-4203-b07a-a5968cfd450d_zps38e3659c.jpg) (http://s266.photobucket.com/user/haugusnord/media/9918d02c-b0eb-4203-b07a-a5968cfd450d_zps38e3659c.jpg.html)
If you add surface area in "A" it has virtually no effect. Area "B" is doing all the work because it has some distance from the spindle and you would just be spreading out the force from the pressure plate. If you added the surface area in "C" by increasing the diameter you would have a dramatic increase in drag and it would continue to increase the farther you got from the spindle assuming we had a pressure plate to cover the new diameter. This is also why drag increases as the line runs off the spool. The diameter of the spool is decreasing as line gets run off reducing the force on the spindle. The drag disc, conversely gets effectively larger in diameter in relation and the drag force effectively increases. Of course this is due to the effective leverage changing, getting less for the spool and more for the drag itself.
If you add another carbon fiber drag washer under the pressure plate like Avet does in their "dual drag" design then you do increase the drag for the same force from the pressure plate. In the case of the drag "stack" in an Avet or in a star drag we are adding washers in sequence so we are effectively adding another area "B" for every washer we add but we are not diluting the force of the pressure plate. Similarly, Accurate is doing effectively the same thing but they have the extra drag on the other side with another pressure plate.
So, it's not a case of simply increasing surface area it depends on *where* you do it and how you apply pressure.
TL;DR if you can increase the surface area and apply equal force to that new area and it has the same relationship (leverage) to the spindle as the existing drag then you will increase drag. But filling in the center of a lever drag's disc has minimal effect because you have just spread out the force from the pressure plate and the center of the disc has minimal effective braking power.
Hope this makes sense! :-\
:) regards
ps sorry hard to explain in words, very easy if we could rapidly draw a picture to address various points. Think of a bike wheel and how little pressure it takes to brake when the cantilever brakes work on the rim as compared to the old hub brakes right next to the axle that took all our weight and leg strength to stop the bike. This is what increasing the diameter of the disc does for braking power and also why the interior of the disc has so little effective power: different leverage.
Hello again. and I,m sure glad that I ducked down after asking a question like this for their reply. and thanks too for the feedback. Its good for all of us to learn something here. I learn by listening. so here goes my explanation of why I asked the question to start with..... I was thinking that for this popular older lever drag reel if I could increase the pounds of available drag at strike position, that I would get a non dragging freespool at the slightly higher strike position settings......I have added bellville washers and reconfigured them, and at lower drag setting the freespool is good, its when the drag settings are increased that you begin to lose freespool......so it seems like if I had more available drag the settings would not have to be so high........I hope that I explained my intent????
This reel could easily have an additional 1/4inch larger O.D. to the drag washer on the plate, and maybe 3/8inch larger at the inner diameter as well. to me this is a significant change, Bur what would I really achieve???????????.
I welcome all Ideas, and we can do the math soon. and I can be more specific with measurements. to help calculations and comparisons. I think this will be a good place to learn something........thanks to all so far......
Yes, you are correct, if you were to increase the OD of the drag disc and had extra pressure plate to cover it should take less preset force to have the same effective drag at strike. Increasing coverage in the center will have virtually no effect.
This *may* reduce handle pressure (and pressure on the pinion bearing which is the source of the handle pressure and loss of freespool) depending on the design of the pressure plate. As I understand it as the force from the pressure hub has to apply pressure to the pressure plate it does not uniformly apply pressure to the drag across it's entire surface, particularly near the center.
Usually the pressure plate is designed to mate to the characteristics of the diameter/surface of the drag disc. So you may not experience any gains.
Or, the smaller disc is intended for structural reasons by the manufacturer and other components may not be designed to handle the additional pressure if you increase the size.
But heck, if it's not your last reel it may be worth a try!
Quote from: noyb72 on March 06, 2014, 12:31:23 AM
You increase drag, I don't believe it would be absolutly double due to ineficiencies, because you are applying the same force to multiple friction surfaces. The pressure seen by the top washer will be seen by the bottom washer.
Ron
.
Ron, from some of my tests with drags I found that adding additional drag will increase the numbers, more so with the custom hex inserts.
I placed a bunch of washers in a stock 9/0, I believe (11) carbon fiber washers. Thant 9/0 wasn't holding drag at 15lb, but was ok from there :-\.
John, thanks so much for the bicycle caliper brake analogy. Anyone who has adjusted or even taken a close look at the stopping power of these breaks will understand the principal behind it. Great stopping power for such a small contact area. The lightbulb came "on". I wasn't able to wrap my head around the fact that more surface area on a single washer couldn't increase the drag. It just depends on where the surface area comes in contact with the pressure plate/drag washer. Thanks again.
Bob
by increasing the drag washer size,id,will the drag be smoother?
cheers
Good question Wallace. I dunno?
Quote from: BMITCH on March 06, 2014, 11:19:21 AM
John, thanks so much for the bicycle caliper brake analogy. Anyone who has adjusted or even taken a close look at the stopping power of these breaks will understand the principal behind it. Great stopping power for such a small contact area. The lightbulb came "on". I wasn't able to wrap my head around the fact that more surface area on a single washer couldn't increase the drag. It just depends on where the surface area comes in contact with the pressure plate/drag washer. Thanks again.
Bob
That analogy helped me as well!
And I think this is why so many people get confused. The math states that an increase in surface area doesn't effect drag, but because an increase in area can result in greater leverage it clearly can. But then it's not a discussion about the surface area anymore but of the placement of the break in relation to the axle. Two different things.
Karl
I have some miss cut 114H washers that are undersize, one of these days I'll do some testing but I want to make a socket so I can set the drag with a inch pound torque wrench first.
The best way to increase drag is to add washers in the stack.
Quote from: Keta on March 06, 2014, 02:46:17 PM
I have some miss cut 114H washers that are undersize, one of these days I'll do some testing but I want to make a socket so I can set the drag with a inch pound torque wrench first.
The best way to increase drag is to add washers in the stack.
Hey Lee, I miss cut some 114 washers that are oversized (that I sent to you)...too bad we could not magically combine them and separate them and it would be all good ;D
Quote from: BMITCH on March 06, 2014, 11:19:21 AM
John, thanks so much for the bicycle caliper brake analogy. Anyone who has adjusted or even taken a close look at the stopping power of these breaks will understand the principal behind it. Great stopping power for such a small contact area. The lightbulb came "on". I wasn't able to wrap my head around the fact that more surface area on a single washer couldn't increase the drag. It just depends on where the surface area comes in contact with the pressure plate/drag washer. Thanks again.
Bob
Not to worry. I used to tutor Chem and Organic Chem and it helps me to think through how things work by putting it into words.
The analogy I chose is a simplification that ignores certain things but should hold up in our simple example and for our purposes of basic understanding.
Quote from: johndtuttle on March 06, 2014, 06:31:14 PMNot to worry. I used to tutor Chem and Organic Chem
Those are the 2 very courses that kept me from being a pharmacist
I too tutored Chem, and Organic in college...not quantitative, P-chem or inorganic though.
Thanks again for Tutoring us old fisherpersons , on the merits of stayin in school. But seriously, explaining how things work will help us all to understand and make better choices and help others as well. maybe make a better world........
The reel I chose could accept a larger drag washer (O.D.) by almost .75'' and would have to do a minor amount of machine work. as well as make a new (larger dia.) sst drag plate, as the plate has 2 ears it just sits in there, being held by the ears from rotating.
So lets do the math again, what would be gained????? the current dragwasher is 1.70 O.D. and is approx. .370'' wide so its center hole is approx .685'' I could increase the new drag washer to 2.00''dia. with a center hole of .950''dia. then the width of the washer would be .675'', this is what was planned, the reel is overbuilt for its earlier rating with monofiliment, it didnt need to be better as at the time it was considered the best, at the time 1960-70 remember when gas was 39-59cents????????
It seems like what I,m hearing increasing the diameter will be the reason it can achieve higher drag ratings, and it would be a great project for me, as I pondered the outcome. what percentage of area would be gained? how much more drag before loosing freespool? Will I Have pictures for the show and tell??????
gst,
The gain in the actual area will probably be very small but as John has stated earlier, the extra .300" of outer radius you increase will actually help you increase the drag substantially. Probably more than you expected.
I am not a lever drag guy, so correct me if I am wrong, but it seems to me that the cam acting on the larger plate will significantly increase the force on the drag washer and that will increase drag considerably.
Ronald
Gary, you already know your gonna increase it. Then put it on the "modified" drag testing machine where you hang some ridiculous amount of lead ball weight and than post some really cool pictures. ;) ;D can't wait to see what ya got.
Bob