Shorter vs Longer rods sensitivity

[Hammer 4]

I use mid priced rods, i.e. about 150.00. I don't rely on the rod to detect bites, I use my eyes  to watch the line and my finger is always touching the line no matter what bait I'm using.

[Deephaven]

3 hours ago, Team9nine said:

 

So then no fair way to compare sensitivity if you can never compare equally because even identical rod lines are built differently based on length, thereby making the question in a sense, unanswerable?

 

I've got multiple rods from the same line listed as the same action from a couple different manufacturers, just different lengths (6', 6'6", 7'0"), and all I can add is that the shorter rods have the best senstivity in each case.

By default the fish has more leverage on you with the longer rod.  For each unit of force it puts on you will receive more.  Simple Physics.  The part that is unanswerable is the differences between rods.  How much they dampen vibrations, their weight, stiffness, etc.  Of course the simplest way of proving this is to use the same blank but fixture it in a way that changes the length.  

3 hours ago, Team9nine said:

Trying to understaand your mods. So you didn't actually increase blank length, just handle length? So the rod blank now doesn't run through the full handle to the butt, or it does because you added a longer handle/reel seat but shortened the working length of the rod (the part above the reel seat)? Wouldn't that then add weight but create better overall balance? And you rewrapped the guides, which would also change how sensitive the rod is?

 

Asking because I'm not a rod builder, but have studied it a bit.

They aren't mods.  St Croix sells their blanks.  These are both SC5 MH 7' blanks from SC.  The length difference comes in the build.  The butt end of the rod is extended before the cork or guides go on.  Super common practice.  While ideal to get a blank made exactly how you want, that isn't always the case.  By changing the length you can change the action and build something that works exactly how you want it.  

[Team9nine]

19 hours ago, WRB said:

 

It’s all about detecting line movements, vibration don’t travel up the line through water.

Tom

 

 

11 hours ago, Deephaven said:

By default the fish has more leverage on you with the longer rod.  For each unit of force it puts on you will receive more.  Simple Physics.  The part that is unanswerable is the differences between rods.  How much they dampen vibrations, their weight, stiffness, etc.  Of course the simplest way of proving this is to use the same blank but fixture it in a way that changes the length.  

They aren't mods.  St Croix sells their blanks.  These are both SC5 MH 7' blanks from SC.  The length difference comes in the build.  The butt end of the rod is extended before the cork or guides go on.  Super common practice.  While ideal to get a blank made exactly how you want, that isn't always the case.  By changing the length you can change the action and build something that works exactly how you want it.  

 

Can’t say I agree with either of the two above physics concepts, but thanks for the additional build info on your rod.

[Deephaven]

6 minutes ago, Team9nine said:

Can’t say I agree with either of the two above physics concepts, but thanks for the additional build info on your rod.

Why?

 

Some food for thought:  First the subjective response.  Notice how using a 10' panfish rod is more fun than a 5' one when catching 1/3lb panfish?  Fish is able to put more juice on you.  Regardless of whether this is .000001lb of force or .3lbs you feel more.

 

As for an objective, read anywhere on lever/fulcrum math.  When you have questions ask.  

https://www.engineeringtoolbox.com/levers-d_1304.html

 

As for the transmission of vibration, it can...but the line will of course dampen out different frequencies based on the damping and modal parameters of the line in water.  Regardless however force transmittal on a tight line to a rod is easily modeled with fulcrum math.

 

[fissure_man]

2 hours ago, Deephaven said:

When you have questions ask.  

 

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?

 

[Deephaven]

6 hours ago, fissure_man said:

 

From the link, is your hand the fulcrum, or is your hand the car?  Or is it both at once?  Neither?

 

Consider your hand at the fulcrum location the part of the rod after your hand isn't relevant as there is no load acting on it.

6 hours ago, fissure_man said:

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)?

 

Doesn't change the force being applied.  Rod will obviously bend though.

6 hours ago, fissure_man said:

What if we thought of the 'bite' as not a fixed force, but as a fixed displacement (i.e. 1" line movement)?

At what speed?  F=ma.  If there is displacement there is acceleration so there is a force.  Without force there is no motion and therefore no displacement.

6 hours ago, fissure_man said:

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?

 

If sensitivity is your ability to feel a motion on the other end the longer the rod the more it is magnified.  Of course this is assuming exact materials on the blanks.  If you can keep that assumption then no, too long is not a thing.  Since you can't you have to pull in the other factors of weight, stiffness and such to determine this.  Great questions btw!

 

Does any of this mean that a 6'6" SC Avid MF is less or more sensitive than a 7'0 SC Avid MF?  Nope, it doesn't as the blank is completely different.  

[Team9nine]

1 hour ago, Deephaven said:

If sensitivity is your ability to feel a motion on the other end the longer the rod the more it is magnified.  Of course this is assuming exact materials on the blanks.  If you can keep that assumption then no, too long is not a thing.  Since you can't you have to pull in the other factors of weight, stiffness and such to determine this.  Great questions btw!

 

Does any of this mean that a 6'6" SC Avid MF is less or more sensitive than a 7'0 SC Avid MF?  Nope, it doesn't as the blank is completely different.  

 

Yeah, I don't disagree with your longer rod/magnified force point, I'd just argue that's it's not as simple as a single factor. Longer rod has to travel further to move the fulcrum angle the same distance. Your force point might offset that or possibly exceed, but then there is the whole longer rod probably weighs more, likely isn't as stiff, etc. which would all come into play. And again, there's your "completely different blank" argument that makes any exact answer on this whole thing impossible.

 

I will admit longer rods can be more fun, similar to ultralight type actions, but I have friends that are professional crappie pros and they all tout their "super sensitive" long rods (10'-14')...but they can't touch the sensitivity of some of my 6' and 6'6" rods I use for crappie (IMHO). Fun discussion though for a late winter day 

[fissure_man]

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.

[OkobojiEagle]

10 hours ago, fissure_man said:

M = (δ*3*E*I)/(L^2)

Alright, let's go rod shopping!

 

oe

[Fishes in trees]

I'd better not break any rods for a while.   Thinking about rods has suddenly gotten too complicated. I just go by whether it feels good to me or not.  Sometimes it feels good to me at the point of purchase and then later not so much.  Since some of my rods have been purchased as close outs, I wonder how much the close out tag effects my momentary notion of sensitivity.  Obviously it effects it some or I wouldn't have the supply of "spare" rods that I do.  But then I consider some of my extra rods "trading stock".    I've often swapped spare rods for different stuff with fishing buddies..  That is how the carb on my chipper shredder got fixed, i.e. I swapped a couple of fishing rods for a new carb and the knowledge to install it.   Both sides felt like they made out. The young man who installed the carb has 4 little girls between the ages of 9 and 4.   He ain't getting any new fishing rods any other way.

[Deephaven]

12 hours ago, fissure_man said:

Perhaps a rod and its sensitivity could be better analogized as a cantilever beam than as a rigid lever.

I will just quote this portion.  Think your analogy is great in particular when it comes to discussing the differences in rods.  You state:

12 hours ago, fissure_man said:

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.

The thing you aren't stating here is that you are now ignoring the force on the end of the cantilever and only looking at tip deflection.  I don't see how we can ignore the forcing function on the end of the rod.  In a fishing rod scenario, the fish applies this force and that is the answer to the length question and what is more sensitive.  The material and rod design is of course important and the second portion of the topic.  For the rest of the forum I will go into more detail to explain what I mean.  

 

To visualize this better, a fixed cantilever and end points for variable discussion is shown here:

The tip deflection is defined as:

δ = (F_B*L^3)/(3*E*I)

Being we are making the assumption that E, I are the same to make this simpler we can substitute a constant, C.

 δ_B = (F_B*L³_(AB))/C  which you nicely showed we can re-arrange to find which we can re-arrange and solve for the Force at B or F_B

F_B = δ*C/ L³_(AB)

You also shared the smoking gun in the length argument.  The moment at point A is defined 

M_A = F_B*L_(AB)

By default this states that no matter the force applied at point B that the moment will be stronger the longer the rod is.  Your conclusion stated differently when we substitute F_B from above, the question of course is why?  This is more easily seen for those not wanting to do math without the extra E & I.  Sole reason I retyped most of this.

M_A = δ_B*C/ L²_(AB)

Here you can see that the moment at A is indeed inversely proportional to the length squared, but the equation M_A = F_B*L_(AB) shows that it is directly proportional.  The question is how can you have two completely different conclusions from the same set of equations?  The answer is in the stated equation though.  The M_A that was solved only includes the deflection at point B as it's forcing function and not the force.  In other words, if the rods are identical (in E & I) the longer rod will put more force on the fisherman; however, the force felt by the fisherman based on tip deflection is more on a shorter rod and significantly.  Of course, when we go back to the original tip deflection equation:

δ_B = (F_B*L³_(AB))/C

We can see that the deflection is directly proportional to the cubed length of the rod.  Being the moment felt is to the 2nd power based on deflection, but the deflection occuring is to the third power this reinforces the longer rod is more sensitive theory as well....but imagine that, it was conveniently described earlier with a simple equation M_A = F_B*L_(AB)

 

Of course the final portion of the sensitivity discussion in relation to length is also hidden in fissure's response.

δ = (F_B*L^3)/(3*E*I)

 

We have already nailed down tip deflection, δ, Force & Length,F_B*L, but the other key parameters in a sensitivity discussion are E the modulus of elasticity and I the moment of inertia.  These cover the shape and materials in the rod.  Obviously they have a major effect on tip deflection which we saw above has a cubed relation to length.

 

 

[garroyo130]

2 hours ago, Fishes in trees said:

I'd better not break any rods for a while.   Thinking about rods has suddenly gotten too complicated.

 

Its really not that difficult if you can find a rod builder who is good at calculus 

[fissure_man]

@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)].

[Team9nine]

1 hour ago, Deephaven said:

The tip deflection is defined as:

δ = (F_B*L^3)/(3*E*I)

Being we are making the assumption that E, I are the same to make this simpler we can substitute a constant, C.

 δ_B = (F_B*L³_(AB))/C

 

 Wouldn’t it be δ_B = (F_B*L³_(AB))/3C?

[fissure_man]

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]

1 hour ago, Team9nine said:

 

 Wouldn’t it be δ_B = (F_B*L³_(AB))/3C?

I substituted C = 3*E*I, I should have stated that though.  Being C is a constant, 3*C is still a constant.

 

25 minutes ago, fissure_man said:

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.

Too many variables indeed.  Exactly why I tried to simplify the answer.  Based on length alone, the longer is more sensitive.  In practice there are a ton of other things that can screw that up.

1 hour ago, fissure_man said:

@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. 

 

Instead of displacements I would call them acceleration.  Makes the force analogy easier, but they definitely are not done at a constant speed so displacement isn't a perfect representation.  Of course it is a dynamic acceleration.  This dynamic portion of course will play even further in the material properties and damping as it will have a frequency component since it is dynamic and not static.  Once you state it as an acceleration then the force equation complies.  Displacement is too simple, but that displacement of course defines sensitivity once my ridiculous assumption that all things are the same is removed.  The only reason I used it is to make this discussable.  You have clearly demonstrated exactly why my assumption isn't valid in the real world, but for answering only the quesiton of is a longer rod more sensitive it works with that assumption.  Considering no two rods can be identical and different lengths it means that the sensitivity is determined both by length and rod makeup...which amusingly neither of us needed to go so far into the discussion to come to the summary.

 

[garroyo130]

Would all this complex math apply only to visible strikes or also to the tap tap which I most associate with bites. It seems the equations are based on movement of the rod tip when the tap tap may be more of a vibration?

[Derek1]

58 minutes ago, fissure_man said:

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.

That’s what I was gonna say. 

[Teal]

Yall had to break out the calculus and trig...booo

[Skunkmaster-k]

Last weekend I went fishing with a eagle claw telescoping buggy whip .  I do not recommend that. Hope this helps  

[Deephaven]

2 hours ago, garroyo130 said:

Would all this complex math apply only to visible strikes or also to the tap tap which I most associate with bites. It seems the equations are based on movement of the rod tip when the tap tap may be more of a vibration?

Don't think of either as vibration or as displacement but as acceleration.  By default you need acceleration to have any force so it is more direct.  Of course you visible see this acceleration as displacement, but the displacement is not instantaneous.  Vibration is inclusive of frequency content which can occur in either displacement or acceleration.  So yes, the complex math applies to the tap tap tap.

[fissure_man]

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.”

[Deephaven]

44 minutes ago, fissure_man said:

IMO it doesn’t make sense to insert acceleration into these equations; they’re solving for an assumed equilibrium state where nothing’s moving.

How can you have displacement without acceleration?  How can you have weight without force?  If nothing moves then you feel nothing as there is nothing to feel.

46 minutes ago, fissure_man said:

I like to refer to George Box: “All models are wrong, but some are useful.”

Indeed.

[fissure_man]

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 2^nd (F=ma).

[Deephaven]

57 minutes ago, fissure_man said:

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 2^nd (F=ma).

Sensitivity to weight at the end of your rod is just weight.  I am interested in dynamic sensitivity.  I want to feel when a fish breathes on my bait.  And no, no way in hell I can feel that...want to.