Emory is there a formula to figure out what a fish's force and or the force in pounds the fish produces when you are fighting him (or her). I.E. 6' rod, 10lb. line, 4oz weight and landed weight of fish 2 lbs.
Please email me or post here
PS See you at Lamiglas

Yes there is but it won't have much to do with the weight of the fish, until you lift him from the water.

If you take the amount of load measured at the reel hand, multiply that by the distance of the reel hand from the rod butt, and then divide that by the length of the rod, you'll have the amount of load being generated by the fish at the tip of the rod.

..............

I dont know much about formulas Tim but maybe we can test the theories up in Woodland. Maybe Emory will take us to a secret spot. Havent fished in either state yet so good place to check out the formula. See ya soon.

Doc The springers (Salmon-Steelhead) should be in the local rivers, when you guys get up here.
Good Wraps Bob

To all reading this post we are getting some fish in now at the mouth of the Willamette and up the river near Sellwood.

Tim Hurst - As Tom Kirkman suggested, one approach to describing this system of forces between a fighting fish and the angler thru a fishing rod with formulas is that of a Lever Arm Analysis. Multiply the applied force times the distance from the fulcrum, and the two opposing forces have to be equal. Solve for the unknown variable, and you have your immediate solution. Done! Just be careful, because the fulcrum may not be where you think it is !!!

Something called a FREE-BODY DIAGRAM or "Finite Elemental Analysis" is another possible approach. The foundations for such things were laid out in your high school Physics, Geometry and Algebra-II & Trigonometry classes. The first few steps in the subject of STATICS would suffice to finish the job.

Technically, a FORCE has a MAGNITUDE and a 3-dimensional DIRECTION, and that combination is referred to as a VECTOR. In order to have a focused discussion about a "force" on a fishing rod, you have to specify (and we have to agree on) exactly which part of the fishing rod or reel you are talking about. Sometimes easier said than done, especially without a diagram.

One generalization that is almost certainly always true is that the force from a fish on a fishing rod is never greater than the line test or the reel drag setting, whichever is lower. Other than that, it becomes an exercise in academics. ...

I have had many such go-rounds with myself on this subject, and some with fellow rod-builders, and some here on the RBO Forum. It is easy to make it more complicated than it needs to be. But it is even easier to wind up very confused and arguing about two different things, and frustrate each other half to death, because we are not on the same page. Been there, done that.

The reason there is no single "formula", Tim, is because you have to know how to define & analyze the element in the system around which you wish to calculate the force. That's a matter of insight, not just "plug & chug".

I will give you two simple and practical example problems for which I am fairly confident that the following presentations are valid. I will not present much explanation or show work or diagrams this time. If you understand what I have said so far, it will be just a corollary for you to derive your own corroborating solution. If you are lost at this point, no mini-tutorial is going to produce the epiphany you need to see what the heck I am talking about

Problem 1: Lift a 10 lb fish out of the water.
Answer 1: The angle of the rod relative to the dangling-free fish determines how much force is in the rod tip, but that force is almost certainly 10 lbs. Only if you are on the deck of a high platform like a pier or party boat, and your rod tip is pointed straight down at the beginning of the lift, and the fish is almost under you, will your rod TIP have little to nil force on it. Otherwise, if you are dead-lifting the fish out of the water like a construction crane, the force at the rod tip is 10 lbs downward. Period. … I told Bill Stevens that last summer, and I am sure that he still does not believe. You are free to believe what you wish. You are just not gonna pass the Professional Engineer's Exam if you have trouble with that one.

Problem 2: What is the force on any given LINE GUIDE?
Answer 2: The Force on the Line Guide, F(Guide), that is directed toward the center of the arc of the curved rod; and lies in the plane of the curvature of the bent rod; and is perpendicular to the Rod Blank; and impinges on the Guide Ring as distributed to it by the fishing LINE TENSION (T) is:

F(Guide) = TENSION * COSINE (THETA / 2)

Where THETA = the THRU-ANGLE of the Fishing Line as it passes THRU the Guide Ring and forms an obtuse angle with itself. The Guide Ring is at the VERTEX of the Thru-Angle, and the 2 sides of the Thru-Angle are the fishing line itself. This Thru-Angle typically has a value between 175-150 degrees.
Example: F(Guide) = (10 lbs) * Cosine (150 / 2) = 10 * Cos(75) = 2.58 lbs.

Almost any other practical STATIC force analysis on a fishing rod or any of its individual components can be described by using a balance of TORQUE and / or LEVER ARM Equations. Beyond that, we would be entering the Rocket Scientist Zone and I wouldn't want to drag anybody's grandmother out into the Ozone Layer with me. …

I hope that helps, ...
Good Friday & Happy Easter, -Cliff Hall, FL-USA.



Edited 1 time(s). Last edit at 04/06/2007 05:01PM by Cliff Hall.

Do yourself a favor, and take a look at See Mark Gibson's Personal website [www.cptelecom.net]

Look at Mark Gibson's slide of R. Otto Nielsen's analysis of reel spool rotary dynamics ... [www.cptelecom.net]

This *.JPG Slide includes the Free-Body Diagrams which are the basis for ANY Static Analysis. The stuff that involves the Velocity and the Acceleration of the Spool or Lure (Otto's "Mass") is the DYNAMIC Analysis. It is possible to learn a LOT about rod-loading and about forces on a rod component (like a line guide) without going crazy doing the dynamics, or ever having to integrate F(m,a)dt. In fact, in my opinion, Equations 1-4 are all that are really needed to achieve a useful analysis of the forces on most rod components. THAT'S the kind of STATICS that I'm always referring to at RBO Forum. A simple Leverage or Torque eqtn. DONE!

Good Friday & Happy Easter, ... -Cliff Hall, FL-USA.



Edited 2 time(s). Last edit at 04/06/2007 05:21PM by Cliff Hall.

Tim,
If the fish were a fixed or static load you could calculate the forces but frankly I do not know how you determine how the forces vary as a result of the fish's movements in the water or how the fish fights if that is what you are asking.

A 150 pound yellowfin tuna in a tight circle headed down at about forty five degrees stripping line at 5 fps with a drag set at 32 pounds will screw up your calculations!

Bill Stevens - It will never be more than the drag setting (32#). I don't care how dang fast that freight train is going !!! ... Once the DRAG SLIPS, you have an entirely new set of operating and controlling conditions. I'm not trying to be rude or closed-minded. But I am trying to prevent an erroneous or confusing description of the system from preventing us from seeing the real factors involved.

No other calculations are needed. Only an understanding of the relevant and IRRELEVANT facts. Eliminating the extraneous facts is about the only way to sort out and solve such a static or dynamic force problem. That, and a set of reasonable assumptions to simplify the problem, to isolate the components involved, and to answer one question at a time.

Once the drag slips, the weight of the fish does not matter. Neither does the speed of the fish; nor the bouyancy of the water's salinity relative to the tuna fish's body fat content. ... ETC.

Once you say that the DRAG SLIPS, then you have 32 lbs operating at 45 degrees, then you have to specify WHERE that 45 degree angle is. Chances are, that if the butt of the rod handle is anywhere above the horizontal, like a typical fish-fighting posture, then the force on the rod tip is ~ 32 lbs, or a little less.

If not, and the rod is horizontal while the fish is diving down at 45 degrees below the horizon,
then the Thru-Angle THETA is (180 - 45 degrees):
Theta = 135 degrees.
Force (Guide) = 32 lbs * COSINE (135 / 2) = 32 * COS(67.5)
Force (Rod-Tip) = 32 lbs * (0.3826) = 12.25 lbs. ... [Q.E.D]

Q.E.D. ... -Cliff Hall, FL-USA.

P.S. - If someone has a more realistic solution that is as simple, I'm all ears. I do realize that I am introducing a simplification of the rod bending and curvature angles, and the Thru-Angle, but it is next to impossible to solve such problems unless we all have the same diagram in our collective minds, and I can tell you that that is about as likely as flipping 3 heads in a row.



Edited 2 time(s). Last edit at 04/06/2007 11:20PM by Cliff Hall.

The force on the tip will not be the same as the force on the angler at the other end of the rod.

Shawn Moore - Yes, Shawn for once you and I are very much in agreement. ... Does that mean I am being useful today, ... ?

Depending on the way the rod is being held and the fish is being pumped, ... and if a gimbal & cup are being used, ... there will likely be an angler who is exerting a force with his lifting hand that is about 2-3 times the force that the fish is exerting at the rod tip.

It is only 2-3 times the fish's force because the CUP exerts a force on the gimbal which cooperates with the angler's efforts, and allows him to reduce his exertion. This is also true for a rear grip which leans into one's forearm on a long-handled spinning rod. Or when a salt-water bottom fisherman tucks the rod butt up into his armpit, we have introduced another applied force, and then the fulcrum may not be so obvious anymore. The wrist of the angler's upper hand can introduce a rotary force along with the pulling force and cause an additional pressure on the fish.

Basically, it may not be quite as simple as the Lever Eqtns:
Fish: 32-lbs * 7-ft rod-tip = 224 ft-lbs
Angler: XXX-lbs * 1.4-ft = 224 ft-lbs
Angler: XXX-lbs = (224 ft-lbs) / (1.4-ft) = 160 lbs.

I think that you will agree that few fisherman, even tuna muscule-men, could hold a 160 lb force on the foregrip or at the reel for very long, ... Even 50-80 lbs at the hand would be a tremendous fatigue, ... which is why I say that the location of the true fulcrum is not always so obvious and that the introduction of cooperative forces by the gimbal cup and especially a harness must also be considered if you want to really isolate the applied forces.

To some extent, it can eventually be argued & demonstrated that the fishing rod itself may be considered as an integral part of the angler, and that his body weight, or leg & torso strength, or even the fighting chair or the boat itself can provide some additional forces to aid the angler to combat the fish. With all that going for us, the poor fish is bound to lose most of the time! ... -Cliff Hall.


P.S. - Shawn your comments the other day regarding factory rod finishes were in fact taken out of context, and words were unfortunately being put into your mouth. It was as if you were being chastised for your past, and not being understood in the present. I just thought I'd share that reality check with you. ... Now we can go back to opposite sides of the field. ... Happy Easter. ...



Edited 1 time(s). Last edit at 04/06/2007 11:13PM by Cliff Hall.

Huh? What past?



Edited 1 time(s). Last edit at 04/07/2007 11:52AM by Tom Kirkman.