Penn long beach 60 frame

[Robert Janssen]

With the same force being applied the more contact area the more friction thus more "breaking".

Actually, no.

That's correct Lee, as I have said all along. Some people think they have the answer to any problem. They just don't sit back and listen once in a while.

Again: actually, no. This subject has been discussed several times in the past at this forum, as well as any number of physics and engineering texts for a couple of hundred years. If people would just sit back and listen or read once in a while, or use the search feature here or on their browser...

To be honest, I am no longer interested in explaining it. But you can start here for example
http://alantani.com/index.php?topic=4593.msg36161#msg36161

PS: Please do bear in mind the subtle differences between mean diameter and surface area, as well as number of sliding surfaces.

...Who makes a reel with hydraulic pressure to apply drag pressure? Maybe you should invent one.

Garcia Corporation invented one in the '70s. I guess the idea didn't pan out.

Larger area with equal pressure means more stopping power.

No. It may seem strange at first, but no. Given the example of two drag washers of equal diameter but where one has a very small hole and the other a very large hole (= different surface area) under equal pressure, the resulting drag will be the same.

If said drag washers are coupled or paired in tandem, the resulting drag will be doubled. (=greater number of sliding surfaces)

In the case of differing drag washers where one washer is of large diameter, and the other of small diameter, the one of greater diameter will provide greater drag. This is so even if one were very thin and the other very broad, so as to provide frictional surfaces of equal area.

:

[Ron Jones]

Thank you Robert, I figured you would be along eventually. I would apreciate some aplication on your last point, I don't quite follow what you are saying.

Ron

[Mandelstam]

For us that went to school a long time ago...



This next video demonstrates how surface area doesn't affect the frictional force. I actually felt like a kid again watching this...

[Ron Jones]

The second video is almost exactly my 8th grade advanced science project. I used flat 4 oz sinkers and my question was why do Surfmasters and Jigmasters have the same drag with different sized washers. My Grandfather laughed his butt off when I thought I was doing something wrong (he recommended the experiment.) Thanks for that Mandelstam, haven't thought of that story for quite a few years. 

Ron

[RowdyW]

Larger area with equal pressure means more stopping power.

In the case of differing drag washers where one washer is of large diameter, and the other of small diameter, the one of greater diameter will provide greater drag. This is so even if one were very thin and the other very broad, so as to provide frictional surfaces of equal area.

:

[/quote]

[Ron Jones]

In the case of differing drag washers where one washer is of large diameter, and the other of small diameter, the one of greater diameter will provide greater drag. This is so even if one were very thin and the other very broad, so as to provide frictional surfaces of equal area.

:

[/quote]

Ah, got it.
You were referring to mean diameter. So basically, when Sal bored out the 49 drags to accommodate the 113H gear sleeve he didn't reduce drag pressure for a given star setting because the mean diameter is the same, even though he reduced the surface area. I understand that mean diameter is effectively a measurement of torque, but I'm not sure how much it comes into play when the difference in diameter is so small. Robert, would you happen to have an equation? My mind works better with numbers.

Thanks
Ron

[Alto Mare]

Let me start by saying that unfortunately, I'm not as smart as you guys, I didn't have the opportunity to proceed with school .
I'm just a mason/carpenter. One good thing though, after 37+ years,  I'm able to build a house on my own .
Here is my take on the test above with the gentleman using the (4) 2x4 pieces.
To me, the test is inaccurate and it would never be accurate.
He is showing that pulling four block stacked on top of each other is the same as pulling four blocks next to each other.
He never mentioned the vertical weight, the single block had 4x the weight as the 4 spread out blacks directed at the same area. 
To me, the only way that test it accurate is to pull the block twice using the same stack
He could have probably use 16 block and divide the total by 4, but that wouldn't have been accurate as well, other things to take in consideration.
....just my two cents....
Sal

[Mandelstam]

Let me start by saying that unfortunately, I'm not as smart as you guys, I didn't have the opportunity to proceed with school .
I'm just a mason/carpenter. One good thing though, after 37+ years,  I'm able to build a house on my own .
Here is my take on the test above with the gentleman using the (4) 2x4 pieces.
To me, the test is inaccurate and it would never be accurate.
He is showing that pulling four block stacked on top of each other is the same as pulling four blocks next to each other.
He never mentioned the vertical weight, the single block had 4x the weight as the 4 spread out blacks directed at the same area. 
To me, the only way that test it accurate is to pull the block twice using the same stack
He could have probably use 16 block and divide the total by 4, but that wouldn't have been accurate as well, other things to take in consideration.
....just my two cents....
Sal

Not entirely sure about your arguments here Sal, but what he's doing is dragging the same amount of mass (total of four blocks), therefore the same amount of force against the table, but with different surface areas depending on how they are stacked. Don't think of them as four individual blocks, think of them as one big block.

[Robert Janssen]

Let me start by saying that unfortunately, I'm not as smart as you guys, I didn't have the opportunity to proceed with school  I'm just a mason/carpenter...

Not true, man.... I have greater respect for that than much else

Ah, got it.
You were referring to mean diameter.

Yup

So basically, when Sal bored out the 49 drags to accommodate the 113H gear sleeve he didn't reduce drag pressure for a given star setting because the mean diameter is the same, even though he reduced the surface area.

Precisely.

I understand that mean diameter is effectively a measurement of torque, but I'm not sure how much it comes into play when the difference in diameter is so small. Robert, would you happen to have an equation? My mind works better with numbers.

Thanks
Ron

Mmm, maybe. I'm not particularly up-to-date with such things, often falling back on the it-is-what-it-is method.

Here, try this, borrowed from a car clutch website:




Torque and force calculations



Torque and Power Transmitted by a clutch

With reference to the figure, let



W= total spring force (N)

r1 = ro = external radius of friction (m)

r2 = ri = internal radius of friction (m)

n = number of pairs of friction surfaces

     in contact

μ = coefficient of friction between disc

      and driving surfaces.



Now,

Mean or effective radius, R = ½ (r1 + r2) = ½ (ro + ri)

Tangential force acting at distance R from centre of rotation,

F = μ W

\ Friction torque transmitted,

TF = F ´ R

    = ½ μ W (r1 + r2)

Since there are n pars of friction surfaces in contact (for a single-plate clutch, n = 2), then the torque transmitted by a clutch is given by:



TF = ½ μ W n (r1 + r2)      (N m)



If N is the rotational speed of the clutch in rev/min, then,



Power transmitted = TF ´ (2 π N/60)    (W)





Area of clutch friction lining (A)



Allowable surface pressure for the lining material 0.05 N/mm2 to 0.20 N/mm2 (p)



Normal force (F)



F = A p = (π/4) (D2-d2) p



Ff = n μ F



Torque transmitted



(Constant wear)

(Constant pressure)

μ = 0.2 : 0.3

p = 0.02 N/mm2

Tc = 50% : 100% Te max = 1.5 Te max : 2.0 Te max



From  http://www.thecartech.com/subjects/design/Automobile_clutchs.htm

.

[Keta]

When I say equal force I'm referring to equal force per square inch, with equal amount of force PER SQUARE INCH more contact surface = more friction. 

[George4741]

The second video is almost exactly my 8th grade advanced science project. I used flat 4 oz sinkers and my question was why do Surfmasters and Jigmasters have the same drag with different sized washers. My Grandfather laughed his butt off when I thought I was doing something wrong (he recommended the experiment.) Thanks for that Mandelstam, haven't thought of that story for quite a few years.  

Ron

Ron, I wonder if the surfmaster and jigmaster are an ideal comparison.  Despite the fact that the surfmaster has smaller drags, it has a mechanical advantage over the jigmaster because of the gear ratio.  Doc explains this in the following link:

http://alantani.com/index.php?topic=6203.msg52156#msg52156

Here is the specific paragraph that illustrates what I mean:  

(quote)
About spool radius and gear ratio:

Yes, that matters too. Not only as a result of the greater lever arm and inherent moment of torque of a larger diameter spool vs a smaller one, but it should also be considered that the reel's gear ratio offers a mechanical advantage when operated in reverse- as is the case on an outgoing line.
(unquote)


I noticed this gear ratio thing when when comparing drag numbers on a 349 (1:2.3 gear ratio) and 349H (1:3.25). Even though the 349H drag washers are larger in diameter than the 349, I installed the same diameter drag washers in both reels.  The 349, it's lower gear ratio having the mechanical advantage when operating in reverse with the outgoing line, had the higher drag numbers.  Installing the correct larger diameter drag washers in the 349H would have increased it's drag.  

Until we are able to accurately apply a specific pressure to the drags, as opposed to just tightening the star by feel, ACCURATELY comparing drag figures between two reels is difficult.  

George

[George4741]

Mentonemoose,
We've really digressed from your original question.  
 George  

[Alto Mare]

Not entirely sure about your arguments here Sal, but what he's doing is dragging the same amount of mass (total of four blocks), therefore the same amount of force against the table, but with different surface areas depending on how they are stacked. Don't think of them as four individual blocks, think of them as one big block.

I understand what the professor was showing us on the video about surface area, but here is what got me puzzled:
let's say that the block of wood was 8mm x 16mm, that would equal 128 sq. mm.  Let's also assume that the same block of wood was 1lb in weight.  He was dragging 4  blacks of wood stacked on top of each other, that means 4lb on an area of 128 sq. mm.
When he was dragging 4 block next to each other, the 128sq. mm area didn't have the same weight it only had 1lb, the force of 4lb was spread out to 512 sq. mm.

[Mandelstam]

Not entirely sure about your arguments here Sal, but what he's doing is dragging the same amount of mass (total of four blocks), therefore the same amount of force against the table, but with different surface areas depending on how they are stacked. Don't think of them as four individual blocks, think of them as one big block.

I understand what the professor was showing us on the video about surface area, but here is what got me puzzled:
let's say that the block of wood was 8mm x 16mm, that would equal 128 sq. mm.  Let's also assume that the same block of wood was 1lb in weight.  He was dragging 4  blacks of wood stacked on top of each other, that means 4lb on an area of 128 sq. mm.
When he was dragging 4 block next to each other, the 128sq. mm area didn't have the same weight it only had 1lb, the force of 4lb was spread out to 512 sq. mm.

It sounds as you maybe mix up pressure with force. Force is in this case the gravitational force from the blocks on the table. Doesn't have anything to do with surface area. Pressure is then how much force per square unit is done to the table.

The force is the same because the mass of the four blocks is the same (4lb= 1.8kg x 9.8= 17.64 Newton). That is independent on the surface area.

The pressure on the table is dependent on surface area though and in the stacked example will be 17.64 (N)/128 (mm2)=0.13 N/mm2. In the side by side example it will be 17.64/512=0.03 N/mm2. (N/mm2 isn't a correct unit I know...)

[Ron Jones]

THANK YOU SAL!!

You are describing exactly what we are saying. Because the star is putting the same force on a larger disk their is less force per square inch (thanks Lee.) That reduction in force reduces friction by exactly the same amount that the increased surface area would iincrease friction. So surface area makes no difference. As you pointed out if you used more blocks (increased total force) stacked on top of each other you would increase friction. That would be demonstrated on a reel by tightening the star drag.

Ron