How is the drag measured on a conventional reel?
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Posted 1/10/2010 5:33:08 PM


Ruby Red Lip

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When the max drag on a reel is measured, is it measured with a full spool? I see max drag ratings on reels. Is this measured with a full spool of line? If so, the drag would be many times more when the spool is emptied.

thanks

Greg

Post #499959
Posted 1/10/2010 7:40:09 PM


Trigger

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The drag on the reel is always the same no matter how full the spool is.  The reason you lower the drag when

you have a lot of line out is that the line going through the water especially in an arc creates resistance for the fish.

 

 

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Post #500022
Posted 1/10/2010 7:45:49 PM


Trigger

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I'm pretty sure that the max drag ratings are with an empty spool, so you are correct that when you have a full spool you won't necessarily get the same amount of drag.

 

 

Post #500023
Posted 1/10/2010 7:47:22 PM


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I think the drag is working a lot harder when there is less line on the reel. If a fish takes as foot of line on a full reel that might mean a single revolution of the spool. If a fish takes a foot of line from a nearly empty spool it might mean 10 revolutions.

A full reel doesn't seem to get hot... a half empty one sure can.

I could be wrong though.

Jim

Post #500025
Posted 1/11/2010 6:35:22 AM


Ruby Red Lip

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Looking at it from a technical viewpoint, the more line you have on the reel, the more leverage you have to turn the spool. Just like using a cheater bar to tighten a bolt, the longer the bar, the more torque you can put on the nut. Same for the reel. The more line on the reel, the longer your "cheater bar" would be. Lets say you have a 6/0. With full line, your cheater bar would be lets say, about 2 inches. When the spool is empty, your cheater bar is about 1/4 inch. Now for the full spool of line, if you pull 20# on the line, you are putting 2"x20#=20 inch-pounds of torque on the spool. When the spool is empty, you are putting .25"x20#=5 inch-pounds of torque on the spool. The highter the drag is set on a reel is just the higher torque required to overcome the resistance and make it rotate.

thanks

Post #500204
Posted 1/11/2010 6:44:43 AM


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1bandit (1/10/2010)
The drag on the reel is always the same no matter how full the spool is.  The reason you lower the drag when

you have a lot of line out is that the line going through the water especially in an arc creates resistance for the fish.

Are you Sure about that?

Let's say when a spool is full there is 14 inches of line in a full revolution of the spool and when it's low there is 4 inches of line in a full revolution of the spool.   That being said and the drag not being adjusted, it would take the same amount of energy to get 4 inches of line on a low spool that it would also take to get 14 inches on the same reel's full spool.  Meaning that a low spool with an unadjusted drag would be heavier drag than when the same spool was full.

 

 

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Post #500207
Posted 1/11/2010 7:14:09 AM


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The drag, (preset when adjusted on a trolling reel), doesn't care how much line is on the spool.  It may take less turns of the spool as dictated by the amount of line, but the same amount of drag is being used. 

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Post #500221
Posted 1/11/2010 7:38:49 AM


Ruby Red Lip

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You are right Garbo. All the drag does on a reel is to keep the spool from rotating. As you tighten the drag, it increases the resistance to rotate. The reel doesn't know how much line is on it.

From an energy standpoint, each revolution of the spool takes x amount of energy. If the spool is full, and you pull one foot of line off, you have rotated the spool ( lets make a guess for illustration) one revolution. Now if the spool is empty, and you pull one foot of line off the reel, the spool rotates (again lets guess for illustration) four revolutions. It takes more energy to pull the one foot off with an empty spool than with a full spool. So, it is harder to pull the line off an empty spool than an full spool.

Post #500235
Posted 1/11/2010 7:56:10 AM


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Downtime2 (1/11/2010)
The drag, (preset when adjusted on a trolling reel), doesn't care how much line is on the spool.  It may take less turns of the spool as dictated by the amount of line, but the same amount of drag is being used. 

Wade, are you saying that the amount of drag a running fish has to overcome is the same on a low spool as a full spool on a trolling reel?    

I would agree that the preset lbs would be the same, but I would have to think the amount of energy the fish has to use to overcome to take line/drag is greater on a low spool than a full spool if the lbs of drag is not adjusted, even on a trolling reel. 

Please correct me if I don't understand something.

 

 

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Post #500244
Posted 1/11/2010 8:17:35 AM


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Post #500258
Posted 1/11/2010 9:00:41 AM


Ruby Red Lip

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Let’s think about our old physics class. (yea, I know, it makes some of us a little sea sick).

Torque is the twisting “force” on a shaft. It is the moment arm times the force. The resistance the spool has against turning is the drag. This resistance to turning is described by the units inch-pounds, or foot-pounds. If you think of torquing a bolt with a wrench, the length of your wrench is the moment arm, L, and the force you are pulling on the wrench is F. So the torque on the bolt is T. Torque =moment arm x F, or

T=LxF.

From the wrench example, if you have a wrench that is 12 inches long, and you put a force of 5 pounds on the wrench, you exert a torque of 12x5=60 inch pounds.

Now let’s look at the fishing reel. See my illustration.

The distance from the center of the spool to where the line leaves the spool is the moment arm, L. The force pulling on the fishing line is F.

So, let’s say you have a 6/0. The distance from the centerline of the spool to where the line leaves the full spool (the moment arm, L) is 2 inches. You pull on the line with a force of 20 pounds( the force F). You are exerting a torque of 2x20=40 inch-pounds on the spool. Now let’s say your 6/0 spool is empty. Now, your moment arm (L) is 1/4 inch. You again pull 20 pounds. This time you have exerted .25x20=5 inch-pounds. To exert the same torque on the empty spool, you would have to pull 80 pounds (.25x80=20 inch-pounds).

So with the drag set the same, it takes more force (or pull) on the line to turn the empty spool than it does for a full spool.

So back to my original question, is the max drag measured at full spool? I emailed shimano but never heard from them.

 

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Posted 1/11/2010 9:32:28 AM


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GregBR549 (1/11/2010)

Let’s think about our old physics class. (yea, I know, it makes some of us a little sea sick).

Torque is the twisting “force” on a shaft. It is the moment arm times the force. The resistance the spool has against turning is the drag. This resistance to turning is described by the units inch-pounds, or foot-pounds. If you think of torquing a bolt with a wrench, the length of your wrench is the moment arm, L, and the force you are pulling on the wrench is F. So the torque on the bolt is T. Torque =moment arm x F, or

T=LxF.

From the wrench example, if you have a wrench that is 12 inches long, and you put a force of 5 pounds on the wrench, you exert a torque of 12x5=60 inch pounds.

Now let’s look at the fishing reel. See my illustration.

The distance from the center of the spool to where the line leaves the spool is the moment arm, L. The force pulling on the fishing line is F.

So, let’s say you have a 6/0. The distance from the centerline of the spool to where the line leaves the full spool (the moment arm, L) is 2 inches. You pull on the line with a force of 20 pounds( the force F). You are exerting a torque of 2x20=40 inch-pounds on the spool. Now let’s say your 6/0 spool is empty. Now, your moment arm (L) is 1/4 inch. You again pull 20 pounds. This time you have exerted .25x20=5 inch-pounds. To exert the same torque on the empty spool, you would have to pull 80 pounds (.25x80=20 inch-pounds).

So with the drag set the same, it takes more force (or pull) on the line to turn the empty spool than it does for a full spool.

So back to my original question, is the max drag measured at full spool? I emailed shimano but never heard from them.

 

There is no "twist" effect on the actual drag.  There drag surface area doesn't change.  How much leverage is used is a whole different matter.  Drag surface doesn't change..

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Posted 1/11/2010 9:38:21 AM


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I would think the rule of setting your drag at 1/3 the breaking strength of the # test is the margin applied to presetting drag at a full spool.  With conventionals this safe setting is a good rule of thumb due to drastic drag increase at a low line levels. 

Modern fixed spool reels (spinning) have helped big time with consistent drag by going with longer shallower spools.

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Posted 1/11/2010 10:58:52 PM


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GregBR549 (1/11/2010)

Let’s think about our old physics class. (yea, I know, it makes some of us a little sea sick).

Torque is the twisting “force” on a shaft. It is the moment arm times the force. The resistance the spool has against turning is the drag. This resistance to turning is described by the units inch-pounds, or foot-pounds. If you think of torquing a bolt with a wrench, the length of your wrench is the moment arm, L, and the force you are pulling on the wrench is F. So the torque on the bolt is T. Torque =moment arm x F, or

T=LxF.

From the wrench example, if you have a wrench that is 12 inches long, and you put a force of 5 pounds on the wrench, you exert a torque of 12x5=60 inch pounds.

Now let’s look at the fishing reel. See my illustration.

The distance from the center of the spool to where the line leaves the spool is the moment arm, L. The force pulling on the fishing line is F.

So, let’s say you have a 6/0. The distance from the centerline of the spool to where the line leaves the full spool (the moment arm, L) is 2 inches. You pull on the line with a force of 20 pounds( the force F). You are exerting a torque of 2x20=40 inch-pounds on the spool. Now let’s say your 6/0 spool is empty. Now, your moment arm (L) is 1/4 inch. You again pull 20 pounds. This time you have exerted .25x20=5 inch-pounds. To exert the same torque on the empty spool, you would have to pull 80 pounds (.25x80=20 inch-pounds).

So with the drag set the same, it takes more force (or pull) on the line to turn the empty spool than it does for a full spool.

So back to my original question, is the max drag measured at full spool? I emailed shimano but never heard from them.

 

A 6/0 reel is a fairly large reel but the diameter of a full spool is only 2 1/2".  Moment arm is always measured from the center of rotation.

L  could never = 1/4" the spool would be past empty.  By the time you have lost 1/2"  depth of line on the spool ( now your spool

diameter is 1 1/2 ) 80% of your line is out. If you rework your numbers with proper moment arm figures and realistic spool measurements

you will find the torque gained like 30% not 400% The amount of torque gained is negligible as compared to the resistance created by the

line going through the water. If you ever carried a bait 150 yds off the beach you would be amazed at the drag created by the water even

though the reel is in free spool. (don't try to hold the line in your teeth). 

The only accurate way to measure your drag is with a drag scale.  Of course this should be done with a full spool (I can't see someone

stripping 80% of the line off a reel and trying to hook up a scale)  Normally the drag is set at 1/3 of the rated line strength. As a fish takes

a large amount line out the drag should be decreased.

 

 

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Posted 1/11/2010 11:32:46 PM


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The amount of drag is set depending on the amount of line is on the spool. Say the reel is full and and the drag is set at 20#, as the spool gets to about half empty the drag setting will be that of approx 30#. Now as we empty the spool to about empty the drag setting will be that of 60#. As the line is being taken the drag setting needs to be decreased.

Also if the line is pulled and measured in a horizontal (no rod bending), the drag set will be less then when the rod is bowed due to friction. 10# drag set at horizontal might be a 12-15# drag set fighting a fish.

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Post #500934
Posted 1/12/2010 6:23:23 AM


Ruby Red Lip

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The numbers are just for an example. I don't have an empty reel around to measure the spool.

In any case, if the moment arm is cut in half, the force required to pull the line off the reel is doubled with the drag unchanged. It is just simple physics.

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Posted 1/12/2010 7:37:29 AM


Ruby Red Lip

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I found some dimensions for the spool. These are not exact, but much closer than my previous guesses.

At full spool, the diameter is about 3.75". So the moment arm is 1/2 that which is 1.875".

So, with 20 pounds of force, the torque is 37.5 inch-pounds.

The empty spool is about 1.25" in diameter (much larger than my first guess). So the moment arm is 1/2 that which is .625".

So with 20 pounds of force, the torque is 12.5 inch-pounds.

Therefore, it takes 3x or 300% (37.5/12.5=3) more force to pull the line off the reel with an empty spool than a full spool. Again, these numbers are not exact, but closer than my first estimate. If someone has some better spool dimensions, we can calculate some better numbers.

that means that if your drag is set for 20# at full spool, it will take about 60# to pull the line off with an empty spool.

The increase in drag is a simple ratio of the diameter of the full spool divided by the empty spool.

So, the bottom line is that you must decrease your drag setting as the line comes off the spool to maintain a steady pull on the fish (not accounting for other varibles like rod angle, etc).

Post #500996
Posted 1/12/2010 8:40:07 AM
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the drag on a conventional reel is measured using a scale such as a manley brass chatillion scale. no matter how much line is on your reel, you hook scale to line and pull off6-10 feet of line at an even pace.  adjust your preset to the amount you want. that will be your strike drag regardless of the amount of line on the reel.  Yes as line is removed from the reel, the force on the fish,or line in the water, is greater but the drag is the same. I might also add that if you wanted to set your drag w/ an empty spool you would have to make a sizeable investment in a dynamometer to measure the pressure.
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Posted 1/12/2010 8:48:13 AM


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Wade, Bandit1

If you take an 80W and preset the strike at say 20lbs and then remove half the spool and put a scale to it you will see a rather large increase in the drag pressure. It is true you should back off on a drag when a big fish has a lot of line out due to the presure being exerted on it through the water BUT also to compensate the large increase in the drag. It takes a lot more force to pull that spool. You are right Wade on the fact that the drag surface isn't experiencing any more stress but it is physically harder to rotate the spool from that close to the axis.

Heres a simple test you can do with a reel with no line on it to see firsthand.

Set the drag at whatever moderate setting you wish. Make sure though when its set you can slip drag by simply applying pressure with one finger to the outside rim of the spool. Now take your finger and move to the inside edge of the spool near the axis and try to slip the drag again. Either it will take much more force to rotate the spool or your finger will slip and the spool will not turn.

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Posted 1/12/2010 7:31:44 PM


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GregBR549 (1/12/2010)
I found some dimensions for the spool. These are not exact, but much closer than my previous guesses.

At full spool, the diameter is about 3.75". So the moment arm is 1/2 that which is 1.875".

So, with 20 pounds of force, the torque is 37.5 inch-pounds.

The empty spool is about 1.25" in diameter (much larger than my first guess). So the moment arm is 1/2 that which is .625".

So with 20 pounds of force, the torque is 12.5 inch-pounds.

Therefore, it takes 3x or 300% (37.5/12.5=3) more force to pull the line off the reel with an empty spool than a full spool. Again, these numbers are not exact, but closer than my first estimate. If someone has some better spool dimensions, we can calculate some better numbers.

that means that if your drag is set for 20# at full spool, it will take about 60# to pull the line off with an empty spool.

The increase in drag is a simple ratio of the diameter of the full spool divided by the empty spool.

So, the bottom line is that you must decrease your drag setting as the line comes off the spool to maintain a steady pull on the fish (not accounting for other varibles like rod angle, etc).

If you have a reel with a 3 3/4" diameter spool it must be a 12/0 or 120 international. My 6/0 measures about 2 3/8" full and the 50w

international measures almost 2 3/4" diameter full.  The moment arm on the spool is a simple linear function but what you are not taking in to account is that the diameter of the spool as compared to the amount of line on the spool is not a linear function. By the time a spool gets down to within a 1/4" of line on the spool you probably only have 2 or 3% of your total line left. Real world dynamic problems are rarely linear so you must keep all the likely forces in mind, including friction form line being out in the water (also not a linear function) and friction from the guides (dependent on rod angle and stiffness of rod.)  And yes I am a mechanical engineer.



 

 

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Posted 1/12/2010 7:40:17 PM


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Regardless of any technical breakdown on spool sizes, more force is required to move a spool half empty than full. Yes, the drag has remained the same but the factor that is now causing the increase is not the drag washers but the spool and axis itself.

Try the example I gave in my above post if there is an empty reel around. It will immediately prove the drag increase. take your finger to the spool rim and push until the drag slips and then move finger to the inside of the spool a little further. You either won't be able to move it and your finger will slip or it will move but requiring far more pressure than when done to the outside spool lip.

I am not a mechanical engineer

Post #501491
Posted 1/12/2010 7:43:49 PM


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I guess the question should be asked this way... Is there a better chance of line breaking while fighting a big fish with less line on the spool if the drag lever or knob has not been moved?  (Let's leave out line in the water drag).

Jim

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Posted 1/12/2010 7:49:02 PM


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jim t (1/12/2010)
I guess the question should be asked this way... Is there a better chance of line breaking while fighting a big fish with less line on the spool if the drag lever or knob has not been moved?  (Let's leave out line in the water drag).

Jim

There's no doubt about it. With more force needed to turn the spool your drag will not react properly

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Posted 1/12/2010 10:39:11 PM


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tunapopper (1/12/2010)
Regardless of any technical breakdown on spool sizes, more force is required to move a spool half empty than full. Yes, the drag has remained the same but the factor that is now causing the increase is not the drag washers but the spool and axis itself.

Try the example I gave in my above post if there is an empty reel around. It will immediately prove the drag increase. take your finger to the spool rim and push until the drag slips and then move finger to the inside of the spool a little further. You either won't be able to move it and your finger will slip or it will move but requiring far more pressure than when done to the outside spool lip.

I am not a mechanical engineer

YES  As moment arm decreases more force is required to turn the spool.  I think thats been established.

Is it the only or most significant factor for having to reducing the drag as line goes out?  No, not anywhere close.

Is there a simple formula or linear function to reduce your drag by as line goes out?  NO,  to many other forces at work.  The amount to lessen your drag is done by feel and years of experience.  The engineers at MIYA EPOCH have been working on a computer controlled drag for years.  They have several new models that have it now, but it is my understanding it has not worked out as well as they hoped and they are still tuning the systems.  Each computer controlled reel has an ideal line diameter it is designed for. Changing lines can reduce the effectiveness of the system of cause it to fail.

Should drag be measured with a full spool?  Yes.  (it would not be practical or necessary to do it differently)  Setting the drag at 1/3 the line strength gives a good safety factor but you still need to loosen it as line goes out.  How much to loosen it can only come from experience and being familiar with your tackle.

Is there a greater chance of breaking off a fish if the drag is not moved?  Yes.  You should not have to start making adjustments until at least 1/3 of you line is gone. One other factor you have to watch for is how large an arch the fish has put in your line. The fish can put much more pressure with an arch in the line as compared to pulling it  straight  from the tip of your pole.



 

 

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Post #501649
Posted 1/12/2010 10:47:29 PM


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Good God almighty thats a beast of a reel.

Its good of you to bring up the point is there a formula to use to compensate and like you stated there isn't.

As far as I know the only reel company that has tried to come up with a full line of reels with drag increase indicators is Everol. They are great reels but a little pricey.

The only thing you can do when a lot of line is out and big blue is greyhounding is to ease off the drag with your best judgement using the rod load as a scale and keep backing down on her

Post #501654
Posted 1/12/2010 11:52:15 PM
Mingo

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OK. So how many of you wait until the reel is 1/2 gone or more to "set the drag"??? You set it when the reel is full, yes???  

Wrap your hand around the line and pull it, until the drag feels right to your hand, and let it go at that. I've never seen any captain or boat owner measure it. Go with what feels right. If it runs, tighten it. If it breaks, loosen it. Have another beer and it'll be all right.

Say what you think, own what you say.

Post #501689
Posted 1/13/2010 7:12:40 AM


Trigger

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WE always set our drags with a scale, after we warm up the drag washers a few times. Even our bottom torque's get the scale check. We also go as far as to put tape under the fiction knob with the drag settings written on the tape  with sharpies. To us it does not matter how much line is being dump on a hot fish untill half a spool is left. This past year we have reduced the drag on only three fish. Line angle is more of a concern to me with half a spool of line left on the reel. At that point is when we will back off on the drag, 25% of the drags full setting as marked on the tape. We have also gone as far as testing the drag setting at half a spool full while changing line. Thats were we come up with a 25% increase in drag measured at half a spool close to 4 pounds on a 50 vsw.  Gene

            

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OLD GUYS RULE !!!!

Post #501743
Posted 1/13/2010 8:05:42 AM


White Marlin

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flappininthebreeze (1/12/2010)
 I've never seen any captain or boat owner measure it.

Most bottom fishermen don't but most serious bluewater fishermen always set their drags with a scale including myself. To say "it'll be alright" is extremely untrue when you have the fish of a lifetime on and you lose it cause you weren't willing to have a properly set drag. It'll really piss you off when it costs you a tournament winning fish.

Post #501772
Posted 1/13/2010 10:09:04 AM


Sailfish

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I wrote Shimano an e-mail asking this question. Here was the response from a Ptroduct Support Specialist:

"

Jim,The drag will actually increase slightly as the spools radius(as a result of line capacity diminishing
) decreases.Of course the radius is the distance from the center of the spool to the edge.As the radius decreases as a 
result of line paying out the amount of leverage decreases as well,creating more resistance.This is true for all reels, 
but so minor on spinning reels and reels with shallow spools you probaby can't measure the effect.
It is more prevalent on large diameter spools like Tiagras and TLD's 50 and up.I hope my explaination makes sense."
 
Jim


Post #501834
Posted 1/13/2010 9:23:20 PM
Trigger

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A lot of the confusion is caused by the terms  "drag", "pressure" "force" and "torque"

Lets call drag as the tension force required in the line to just start a reel spool rotating -

Then the drag force (when measured in good old english units - pounds-force) is inversely proportional to the the spool diameter.

Less line - less diameter - more drag needed to start rotating the spool.

Drag adjustments really only changes the amount of torque, or moment (forces x distance)  it takes to just start the rotating the spool -

(english units for torque - ft-pounds or inch pounds)

This torque in independent of the spool diameter.

There are several factors that can and do change this 'start-up'  torque  - and example is drag pad temperature. The age and wear of the pad also changes finteral friction with time and use - Think of the drag pads like your car's brake pads.  Larger diameter tires need larger, beefer  brake pads to stop the same size car (or truck - ask me about how 20" rims cuased my Dodge to eat brake pads & rotors)  

There can be a significant deference between the torque or moment needed to start the spool rotating, and the torque needed to keep it rotating.  Typically on cheaper reels - the drag "sticks" unit the line force overcomes the torque needed to start the spool rolling and once it breaks free- it takes less drag to keep pulling the the line off the spool. 

If anyone still thinks that drag is the same for a full spool and 1/4 full spool on a good quality - well maintained trolling reel - let me know and I will meet you at the marina and I will bet you a tank of gas it takes more force to pull the drag out on the 1/4 spool than the full spool.

And ...while we all say we put pressure on a fish - we really can't - pressure is force over area (pounds per square foot) - all we can control is the amount of force - and the direction - of the fishing line.

Cabin fever has set -in,  I need to hear that drag sing (units decibels) -

Post #502289
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