helpin do the math on square area calculations!

[gstours]

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.

[maxpowers]

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.

[Tightlines667]

X2

So Surface Area of drag disc = [ pi • ( R(outer radius)^2)] - [ pi •(R(Inner radius)^2)]

[Shark Hunter]

I perfectly understand!

[Robert Janssen]

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.

.

[maxpowers]

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?

[Bryan Young]

Drag surface area is a component that affects the amount of drag.  Other components are pressure, supporting surfaces, pressure plate,...

[Robert Janssen]

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)

.

[johndtuttle]

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 .

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

[Ron Jones]

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.

[johndtuttle]

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.


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.

best

[Ron Jones]

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

[johndtuttle]

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

[wallacewt]

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

[maxpowers]

No I think you double up the drag due to more contact surface, and they are operating on  parallel planes.