fissure_man

Hey now, I think it’s a term of affection, only derogatory if you take it that way ? When I’m after smallies, 90% of the time it’s with a fairy wand in hand. As far as I can gather, people use BFS because they like it – it’s that simple. I don’t think it’s a “macho” thing – hard to look tough wielding a 5 ft trout rod. LOL At the end of the day, who cares? Yikes ? Bit rich to read this kind of thing on a fishing website, bass anglers of course being known for athleticism and good looks ... "half" may be an understatement lol.

@roadwarrior get your name in early!

This is coming from the guy with $700 Stellas on his fairy wand spinning rods? ? lol

If the “doink” (scientific term) of the bite in terms of line movement was charted out on a displacement vs. time chart, you could take a couple derivatives and plot out the acceleration. But to determine the resulting force at the rod tip, you still need to take into account physical characteristics of rod, either dynamically or in a static calculation. Intuitively, wouldn't you agree that the same “doink” on an UL/slow rod results in a different force at the tip (FB) than on a HVY/fast rod? It's like hitting a punching bag vs. a brick wall. That's why it doesn’t make sense to consider FB as a constant when comparing rods. In general this is why, IMO, the flexible beam analogy is more valid than the rigid lever. Not really sure what you mean here, but using a static calculation doesn’t mean nothing changes, we're just considering forces at a given point in time. Ultimately we’re comparing one state to another (“bite” vs. “no bite”).

Both. The premise is that the angler detects the change in MA that results from the bite. In a static sense, maximum change in MA occurs at the maximum tip deflection, so we analyze that case. We're looking at the end point but it's a gradual process. To use the same equations for the interim state, an assumption would need to be made that the deflection is happening slow enough that the resulting tip force at any stage can still be estimated based on deflection and the stiffness properties + length of the rod (i.e. ignoring any considerations of dynamic or inertial effects). Depending on the scenario (rate of change, damping properties of the rod, etc), this may or may not be valid for the interim state. However, plugging in F=ma to the static equations doesn’t solve this complication, and, IMO it doesn’t make sense. It might, or might not, depending on the situation. I don’t think it matters to the discussion. Well, it depends on what you mean. The rod can be in a flexed state with no acceleration occurring – that’s the ‘static equilibrium’ these equations solve for. The internal forces don’t go away because acceleration = 0, it just means that the net force on the system is zero. The rod (supported by the angler) exerts an upward force equal to the downward force of the bite, and nothing is accelerating. Same as if you press your hands together – you're applying force, but zero acceleration occurs because you’re pushing equally hard in opposing directions. On the other hand, for the rod to transition from an unflexed to flexed state, some motion (including acceleration) has occurred. Conceptually, when the line starts to move, the tip moves as well until the resulting force from rod flexure is equal to the force applied by the ‘bite.’ At that point, again, forces are in balance and acceleration is zero.

How are you incorporating acceleration into the model? Do we need more equations!? ? lol I don't think that considering a 'bite' as an acceleration instead of a displacement changes the assumption that the resistant force developed in the rod at that transient moment depends on its stiffness and how far it's been flexed. In other words, it still doesn't make the resistant force a constant. If a dynamic analysis is needed, I'm all ears. When it comes to transmitting a wave, I'm guessing that length doesn't tend to help To be fair, everything I'm talking about is sensitivity to tip deflection or line movement (causing tip deflection). Hanging a weight from your rod is a known and constant force; what you feel is independent of the rods characteristics other than length - not what I'm after.

I’m not saying that acceleration doesn’t or hasn’t occurred; I’m saying that the system defined by the beam equations above is in static equilibrium and in that state, no acceleration is occurring. The situation described is closer to Hooke’s law (F=kδ) than to Newton’s 2nd (F=ma).

IMO it doesn’t make sense to insert acceleration into these equations; they’re solving for an assumed equilibrium state where nothing’s moving. The bass pulls on the line, and the rod pulls back. Like a spring where the reaction force is a function of compression or extension (displacement). I’m not sure what you mean by saying “once you state it as an acceleration then the force equation complies.” As for dynamic analysis, vibrations, damping, etc… There’s no doubt that all of this is an oversimplification, and to what extent it is valid at all is up for debate. I like to refer to George Box: “All models are wrong, but some are useful.”

Worth reiterating that assuming ‘I’ to be constant along the length of the rod is a gross simplification. Obviously the tip section of a rod is much, much less stiff than the butt section. I think, unsurprisingly, it still comes down to the question of what is an "equivalent" rod with different length. Based on the example earlier in this thread, what if we added a stiff butt section to a rod and we assumed that this results in no increase in deflection under a given load, or conversely no increase in resistive force for a given deflection? This would mean that all of the deflection occurs in the unaltered tip section, and the butt extension is perfectly rigid. Perhaps that’s very close to the truth? In that case, comparing the original rod (1) vs. extended (2) F = (δ*3*E*I1)/(L1^3) F = (δ*3*E*I2)/(L2^3) Combining the above and cancelling constant terms: I1/I2 = (L1/L2)^3 I1 = I2 *(L1/L2)^3 Let’s say L2 is extended by some factor greater than 1, in comparison to L1: L2 = a*L1 Sub it into the above: I1 = I2 *(L1/(a*L1))^3 I1 = I2 *(L1/(a*L1))^3 I1 = I2 *(1/a)^3 Plug it into the moment equation for M1: M1 = (δ*3*E*I2*(1/a)^3)/(L1^2) M1 = (δ*3*E*I2)/((a^3)*(L1^2)) And for M2: M2 = (δ*3*E*I2)/((a*L1)^2) M2 = (δ*3*E*I2)/((a^2)*(L1^2)) And finally combining M1 and M2 (recall a > 1.0): M2 = a*M1 The extended rod (2), by this measure, should be more sensitive because the extension did not result in a rod that more easily deflects under load, developing the same force with a longer moment arm. However, one could argue that the average stiffness of the rod was increased to accomplish this, and the rod has been made to have a 'faster' action. And yeah, weight... balance... not going there.

@Deephaven good post ? I wouldn’t say I’m ignoring the force on the end of the cantilever, but instead I’m considering it as a function of displacement. The key difference between my take and yours is that I don’t consider the force at ‘B’ associated with a given ‘bite’ to be a constant. IMO, when a ‘bite’ occurs, the bass is sucking the lure in and maybe moving around a bit (or not), and maybe the angler contributes by lifting the rod tip in an effort to sense the bite. These are all displacements, and the force that results at the rod tip depends on the stiffness of the rod. The same ‘bite’ on a soft rod applies less force at the rod tip than it would on a stiffer rod, both being held rigidly by the angler. It works in reverse, too – it takes a very different hookset to generate the same tip-force with an ultralight rod, vs. a heavy power rod. An equal force scenario would be hanging the same jig off the tip of a long rod, vs the tip of a shorter rod. In that case, absolutely the longer rod produces a greater moment at the grip location. However, if you instead hold both rods straight out and tug each line downward by say, 1 inch (δ), IMO you have a scenario that is more similar to an actual ‘bite’, and the moment at A is given by the equations discussed [M = (δ*3*E*I)/(L^2)]. The basic moment equation [M = F*L] still holds true, but F is variable depending on the characteristics of the rods [F = (δ*3*E*I)/(L^3)].

Perhaps a rod and its sensitivity could be better analogized as a cantilever beam than as a rigid lever. The angler’s grip on the rod is the fixed end of the beam. The rest of the rod is unsupported with length “L” ahead of the grip location. The rod/beam is subject to self-weight plus the force resulting from a ‘bite.’ I don’t think it’s correct to think of the force from a ‘bite’ as a constant – it depends on the flexibility of the rod. If the rod offers little resistance, little force will be developed. Let’s say the line is tight and doesn’t stretch. It the bass’s bite pulls the line a certain distance, that will apply a relatively large force to a stiff rod, or a relatively small force to a rod that is softer and deflects easily. For comparisons sake, maybe the ‘bite’ can be better described as a rod tip deflection (δ) than as a resultant force at the rod tip (F). Continuing with the beam analogy, and in line with earlier posts, perhaps ‘sensitivity’ can be quantified as the moment (static torque, M) produced at the grip location by a given ‘bite’ (δ). A sensitive rod should produce a larger moment at the grip location, in comparison to the same bite transmitted through a less sensitive rod. Classic beam theory tells us that the tip deflection can be calculated as: δ = (F*L^3)/(3*E*I) where E is the elastic modulus of the rod material, I is the moment of inertia for the rod’s cross section, F is the force applied at the rod tip, and L is the length of the rod ahead of the grip location. The above can be rearranged to solve for F: F = (δ*3*E*I)/(L^3) Moment at the fixed end of the beam is a familiar formula: M = F*L Combining the two: M = (δ*3*E*I)/(L^2) Let’s say that for two otherwise identical rods with different lengths, E and I are constant (this is a loaded assumption). The above suggests that the moment that can be felt at the grip location is actually inversely proportional to the square of the rod length – the shorter rod is more sensitive. However, by design, moment of inertia is not a constant through the length of the rod. The rod is ‘tapered’ to produce a desired bending action under load – the rod’s cross-section varies along its length. This significantly complicates the analysis, but it’s still a solvable problem. Unfortunately it would take some calculus and it’s too late at night for that ?. Will need to pick it up another day. Suffice to say, there is a difference between a longer/shorter rod with the same stated action, vs a longer/shorter rod where the difference is a result of adding a stiff butt section, as has been discussed above. I think that this approach could reasonably quantify that difference.

I've never used a tungsten bass jig.

Lead is bad for you if you eat it. For the record, you shouldn't be eating tungsten jigs and spinnerbaits, either. ? The only tungsten spinnerbaits I've ever thrown were Eco Pro - two of them. Both ended up failing where the wire met the head, one in the net and one that resulted in a lost fish. All spinnerbaits will break eventually, but these hadn't seen much use. I guessed that maybe the wire was softened/weakened in that spot by a hotter manufacturing process? Wire was slightly discolored at that spot. Never tried them again.

From the link, is your hand the fulcrum, or is your hand the car? Or is it both at once? Neither? What if the lever arm is not stiff, but is flexible instead (let's say it has a uniform modulus and cross section, for simplicity's sake)? What if we thought of the 'bite' as not a fixed force, but as a fixed displacement (i.e. 1" line movement)? Assuming a long rod is more sensitive, is there a point at which its sensitivity is affected by being too long, and if so, why?

Sounds like it wouldn't be a good idea... for you.