Counterflex, energy loss and absorbation

[hshl]

Hi Daniel and all ,
I‘m thinking about energy loss and the role the counterflex (CF) plays in this context. Starting from Daniels paper „Physics of the overhead cast“ on page 34 the energy loss is highlighted. For a „fair“ fly cast 25 % remains in the rod. 80 % of this amount stays in the tip 25 % * 0.8 = 20 %, so the rest of 25 % * 0.2 = 5 % remains in the butt. Right so far ?

My questions are:
a) Does the 5% of energy loss of the butt section belong to the casters „effort“, the caster has to absorb somehow with his hand / arm respectively ?
b) Is a smaller CF the result of a better absorbation of the caster or a better energy transfer (I assume the latter. Personally I loosen my grip tension at the end of the stroke so that I assume I'm not able to "absorb a lot" of energy with my hand / arm due to the "relaxed" connection between the fly rod and my hand) ?
c) How does the values of page 34 are changing for a „good“ fly cast with a very small CF ?

Thanks in advance,

Tobias

[Merlin]

Hi Tobias

First point: in the butt or more likely in the mid of the rod. That’s small anyway, unless the timing is poor. If the timing is perfect there is only energy left in the tip of the rod. The important loss is the energy remaining in the tip. That one has to be canceled.

All energy is produced by the caster, and as the line is launched, there is kinetic energy left in the rod (tip essentially). This kinetic energy is ½ mo V2, V being the line launch speed and mo being the equivalent mass (at tip) of the rod. To reduce that energy you have either to slow down your motion (V) or to lighten the equivalent mass: shorter rod, lighter rod, tip action rod.

Remember there are 3 equivalent masses (butt, mid, tip) and that one of the objective of the design is to maximize the transfer part of kinetic energy (mid equivalent mass), and minimize the two others.

The amplitude of CF depends of the stiffness of the rod (elastic energy storage equivalent to the remaining kinetic energy in the tip). The stiffer the rod is the smaller the CF is. This energy has to be lost somewhere and the best option is to use a loose grip. If you use a stiff grip, your body acts as an embedment and throws back the energy into the rod (vibrations). A loose grip allows the flexural waves in the rod to travel through your hand and be absorbed by the body which acts as a big damper. Some people have an amazing capability to “kill” the vibrations: the rod counter flexes once and then comes to a standstill, even with an unstable rod. A fast rod helps since it dampens vibrations quicker.

The values I gave are indicative, the energy in the line has several sources (lever, spring, inertia, haul) and the final share depends on the carry you are using. The amplitude of CF is rod dependent, and also cast dependent (line launch speed). There are too many parameters to derive a rule of the thumb.

Merlin

[hshl]

Thanks for your answer, Daniel.

If you use a stiff grip, your body acts as an embedment and throws back the energy into the rod (vibrations). A loose grip allows the flexural waves in the rod to travel through your hand and be absorbed by the body which acts as a big damper.

In the first case you describe above (grip tension is not loosen) I feel a "shock" in my arm and would therefore assume an additional effort for the caster ("braking energy") in contrast to the second case (grip tension is loosen), for which one I don't feel a shock, effort respectively. I expect to feel the absorbation in the second case, but I don't. I understand that the amplitude of the CF depends on several parameters, but even if the same fly rod is casted without a fly line, there are smaller and bigger CF. In this term I'm still not keen with my second question b): the smaller the CF 1.) the more the caster has absorbed (actively) or 2.) the less the caster need to absord (passively) ?!?!

Thanks again,

Tobias

[Merlin]

Hi Tobias

Without a line, the CF mirrors the energy you put into your motion.

I can imagine that both options (1 and 2) can exist. All our motions (reflexes apart maybe) are damped, we are built with damping functions to produce fluid motions. I guess we can damp the counterflex from RSP to MCF and from MCF to the next RSP. The rebound of the rod is damped by hand/wrist/forearm for example, on top of the loose grip which likely takes care of higher modes. Is this counscious or unconscious, I do not know. Both are possible I think.

Merlin

[hshl]

Thanks a lot.

[gordonjudd]

Remember there are 3 equivalent masses (butt, mid, tip) and that one of the objective of the design is to maximize the transfer part of kinetic energy (mid equivalent mass), and minimize the two others.

Merlin,
How do you determine the mo values for the butt and mid sections?

I assume the mo for the tip is determined from the fit you get from the oscillating frequency vs tip mass curve as shown below:


That curve was measured for a high tech rod from China ($100) that compares vary favorably with the natural frequency and mo of a much more expensive ($900) Sage TCR rod.


Gordy

[Merlin]

Hi Gordy

Good question. We can measure the sum of equivalent masses (MOI) and the tip one (mo) just as you show with the frequency curves. Unfortunately I failed to find a measurement process to get an extra equation in between equivalent masses and solve the issue. The current mean I use is a calculation procedure needing a correct description of the rod (materials, dimensions) and a small deflection profile.

I noticed that the best rods always have a high value for the mid part, but that does not mean that all rods showing that characteristic are fine. This only relates to the inertial effect which is modest for a graphite rod (5% to 10% on line launch speed).

I can see that the chinese rod is something like a #9 to #10 rod I think (a bit stiff to be a true #8), so it is naturally fast if I can say so.

Merlin

[gordonjudd]

We can measure the sum of equivalent masses (MOI) and the tip one (mo) just as you show with the frequency curves.

Merlin,
So does the equivalent mass (mo) value we get from the frequency vs tip mass curve represent the sum of the equivalent mass values of the butt, mid, and tip section?

Could the procedure described on the old board http://sexyloops.co.uk/archivedboard/vi ... 20#p248914 to calculate the effective mass in motion for the whole rod be used to get equivalent mass values for the individual sections?

How does the mo of the different sections compare? Does the tip section have the largest mo value since it has the highest velocity even though its mass is smaller?

Gordy

[Merlin]

Gordy

We are not on the same page, let’s go back to basics :
MOI = (mo butt + mo mid + mo tip) L^2
mo butt is linked to the self deceleration mechanism of the butt
mo mid is linked to the mass transfer phenomenon (inertial effect)
mo tip is linked to the loaded frequency curve
The following graphic gives an example of the contribution of mass along the rod to each of those equivalent masses. I added their sum to make the bridge with the MOI.

eqmass.JPG (29.83 KiB) Viewed 6997 times

It is clear that the upper mid of the rod (the tip of a 2 piece rod) dominates the influence on mo tip (green curve). In particular the tip top guide contribution to mo tip represents 100% of his mass. That gives you an idea of the importance of guides and wraps on that section.

The mass transfer effect (mo mid, red curve) is also mostly due to the mass on the tip, but with a different and more balanced distribution. In that case the contribution of the mass of the tip top guide is nil. The maximum contribution lies in the lower part of the tip section of a 4 piece rod, with about 33% of the corresponding mass.

The SDM effect (mo butt, blue curve) depends on a more evenly distributed mass, with emphasis on the upper middle of the rod shaft. Maximum contribution is limited (less than 20%).

The magenta curve represents the contribution of individual masses along the blank to the MOI of the shaft (considered from the middle of the handle approximately). It is the sum of the three mo contributions.

From this graphic you can have an idea of the contribution of individual sections of a multi piece rod to each equivalent masse and to the MOI.
This is the result of a calculation since we miss an extra equation to get mo butt and mo mid. The methodology is the following and is an extension of the one used to calculate mo tip from rod structure. You need to evaluate the kinetic energy of a rod as it rotates and deflects. The original method you refer to is not taking rotation into account. To perform the calculation, you need to describe the linear mass along the shaft, and the velocity of each point of the shaft. The usual simplification is to consider the deflection as a guideline for speed. You do not need to know the exact value of speed, but only its evaluation by comparison to tip speed. So you use the comparison of deflection of each point of the shaft with the tip deflection to describe the speed of each point along the shaft.

The corresponding functions I use for linear density and speed/deflection profile are polynomials (to the fourth power). The integration is painful but it can be performed with a spreadsheet. At the end I can compare the calculated MOI with the one I can directly evaluate from the rod structure for a check.

Merlin

[gordonjudd]

To perform the calculation, you need to describe the linear mass along the shaft, and the velocity of each point of the shaft. The usual simplification is to consider the deflection as a guideline for speed. You do not need to know the exact value of speed, but only its evaluation by comparison to tip speed.

Merlin,
Thank you for going into more detail about how these different effective mass values are determined. I was not even in the same book let alone the same page.

So is the deflection profile you use based on the shape of the first mode's deflection or the more complicated deflection shape that you get in an actual cast? I assume that speed value is related to the the spring velocity of the unloading deflection shape and the velocity due to the rotation of the rod has no impact on your calculations

MOI = (mo butt + mo mid + mo tip) L^2

And does that moment of inertial value have the same nominal value as the one you would get by calculating the second moment of the rod's linear mass density profile about some rotation axis (like the elbow in an actual cast or the grip of the rod as was done in Grunde's calculation)?

I afraid the details of the self deceleration mechanism and the mass transfer effect are a bridge to far for my understanding of how a rod performs in a cast and suspect that I am not alone in that respect.

Gordy