“Personal Perfect Rod” Recipe - Taylor make your rods

[Merlin]

Tom,

I made some calculations using Grunde's data about the XP 590. Again we face the difficulty of having data corresponding to the same blank. Grunde's data indicate that the unloaded frequency for the finished rod is about 3 Hz, slightly above yours (2.8 Hz)

If I speculate on the unloaded blank frequency corresponding to Grunde's data, I can compute 1.6 grams for the "equivalent mass" of the rod. Then the fittings contributions would be around 1 gram. Small differences, large impacts.

If I take your own data, I can deduct from them that mo for the rod is 2.1 grams, and that the contribution of guides is thus 0.5 grams, which is well below what I use to find.

Again, we are trapped by small differences which make straight estimations questionnable.

Merlin

[gordonjudd]

mo is a strange characteristic in fact.

Merlin,
I like to look at the dynamics of the mo factor that comes out of a long linear spring which is calculated by including the kinetic energy in the moving portion of the mass in the spring.

This procedure is discussed at about 27:00 minutes into this MIT lecture by Dr. Lewin. He stumps his students about this difference between the measured and expected frequency for an added mass at about 32 minutes into the lecture. (Bernd, I was looking at the wrong set of lectures the other day, and I agree that Dr, Lewin is a great instructor. That is no doubt why he is at MIT.)

I will check out the MIT vibration book today and send you a copy of the derivation of the mass/3 factor for a linear spring.

This would say the mo for a vibrating rod with less moving mass (tip flex) would be smaller than when more mass is moving (full flex).

Gordy

[Merlin]

Excellent course on uncertainties and proper equations, Thanks Gordy.

In our case, the basic equation is valid on a certain range (e.g. from zero to 30 grams), but if we try to apply it to small masses, it is not exact (the mo value in the range from zero to 1 or 2 grams differs from the average over the larger range).

This is what makes the estimations difficult for small masses.

The important fact in my last test on the computer is that although I do not change the mass distribution of the rod, and it remains with the same mass, the changes in modulus change the value of mo in coordination with the flex profile, and I wondered why. Your last comment gives the clue.

Merlin

[Merlin]

Gordy

I had a look at the theory behind and kicked myself not to have investigated the issue before. For a simple spring, mo equals 1/3 of its mass, and for a uniform cantilever beam, mo is nearly equal to 1/4 of its mass.

For a fly rod, it is less than that, of course. Although mo looks fine as a parameter related to rod deflection there is a remaining problem: mo varies with the length of rods: the longer the length, the higher mo. This is because the rod becomes heavier and heavier, and relatively much more that the line it is designed to cast. Missed again! More work to be done, then.

Merlin

[gordonjudd]

Missed again!

Merlin,
And now you are found.

Dr. Robson used a different form for the frequency vs. added tip mass curve as described here.

As with the effective mass we have been using the "effective" MOI Robson gets in his fit is much smaller than the real MOI of the rod.

Gordy

[Bill Hanneman]

Gordy,
It seems our old conversations are again germain, so let us bring the "lurkers" up to date a bit.

Gordy, and Torsten

I'm back from vacation and now I need advice from you engineers.

Basically, I a looking for a graphical solution for "Joe". Consequently, I have five questions. I would appreciate your comments.

1. Am I correct in assuming the plotted curves in Grunde's chart relating frequency to mass on the tip represent differences due to the different values of k?

2. If so, isn't it possible to construct a similar chart consisting of a series of "parallel" curves which intersect the zero mass line at various points corresponding to the fundamental frequency for each curve value of k?

3. That being done, shouldn't it be possible to measure the frequency of any rod having a "known mass added" and determine which curve that point lies on—thereby approximating k?

4. Once the curve for k is identified, shouldn't it be possible to use this graph to determine change in frequency as a result of change in mass?

5. If all of the above are possible, is there any reason one cannot simply recalibrate the odinate (Hz) in units of cpm and the abscissa in units of cents, ELN, or grams—or all three?

Followed by:

Am I correct in assuming the plotted curves in Grunde's chart relating frequency to mass on the tip represent differences due to the different values of k?

As noted by the scales in Grunde’s plot he is showing the predicted and measured oscillating frequency as a function of the added mass attached to the tip of the rods he measured.

The fixed k he used for each rod is also noted in the graph. For the SP rod the k was .91 N/m, the PA#5 used a spring constant of .75 N/m and so on.

Since your initial assumption was incorrect, I don’t think your following procedure questions are germane.

-----
At this point, the conversation ended.
Not being discouraged, I proceeded to create my graph which you so kindly published in your Post 85 in this thread. Perhaps, we could digress a bit and dicuss it. The graph is obviously showing something, and it does have predictive value. Could it be that what I have called K is actually a function of the spring constant and should be called something else? Your comments will be appreciated.
Bill

[gordonjudd]

Could it be that what I have called K is actually a function of the spring constant and should be called something else?

Bill,
K is the "effective" spring constant that comes out of finding a value for k that gives the best fit between the measured frequency vs added tip mass curve, and Grunde's equation that predicts the simple harmonic oscillator frequency as a function of added mass.

If you have not already done it, I would recommend that you watch Dr. Lewin's lecture so you will have a better understanding of how to apply Grunde's equation., and the physical significance of what the "effective mo" is all about.

As note previously the "effective k" is in the ball park of the measured spring constant value, but they are not the same.

Grunde's frequency equation depends on the "effective mass of the rod"(m0) , the effective spring constant of the rod "k" and the added tip mass m. Thus it has three variables that must be taken into account, not just one that is based on the measured spring constant of the rod.

Gordy

[Bill Hanneman]

Gordy,
Thanks for the quick reply. Can I take that to mean the answer to my question is Yes?

[Merlin]

Gordy

Dr Robson wrote the equations for a rotation instead of a one dimensional motion, so he used MOI instead of mass. The Equivalent MOI was 1/3 of the real one, I guess.

I think the one dimension model is easier to understand, like the spring & marble or car and brick. Incidentally, the name "spring & marble" was given to me by Noel Perkins, my model had no name before. Those illustrations (marble, car) are easy to figure out by comparison to a lever rotating around some point.

When I say "missed again", I mean that the remaining difficulty is to define a real "action parameter" which is valid for a 6 feet and a 15 feet rod. mo has the potential to do that.

The basic equation used in a previous post is an approximation, with a limited range for its validity, but it is sufficient for our needs. With my computer, I can mimic the frequency test with a finished rod or with a blank, so it is quite easy to get an mo value for the fittings. I do not know of a comparable experience in real (same blank, same rod). I never did it myself, if someone is tempted, we can compare results.

Merlin

[gordonjudd]

Could it be that what I have called K is actually a function of the spring constant and should be called something else?
and
Can I take that to mean the answer to my question is Yes?

Bill,
You are free to take:

As noted previously the "effective k" is in the ball park of the measured spring constant value, but they are not the same.

anyway you want.

I would call it the "effective spring constant value" since it does not have the same value as the measured spring constant of the rod. In that respect it is much the same as the "effective mass, m0" value that comes out of the fit to get the values used in Grunde's SHO equation. m0 is different than the measured mass of the rod, and k is different than the measured spring constant of the rod.

If you can determine the function that relates the "effective k" value to the measured spring constant curve, then you will have advanced our understanding of how Grunde's SHO equation relates to the physical qualities of the rod. That would be a very useful endeavor.

Determining how to determine the divisor to get the "effective mo" value from the measured mass of the rod would also be useful. As Merlin noted the divisor is 4 for a uniform cantilevered beam but will be higher for a tapered rod. In the TCR example the measured mass (that I assumed included the mass of the grip and guides) was equal to .1074kg and the m0 was equal to .0027 kg. That gives a divisor of around 40. A big difference indeed!

For my full flex saltwater rod blank the measured mass for just the blank was 76.1 g and the effective m0 was 6.45 grams. That gives a divisor of 11.8.

Tom might have some measured values for the mass of the reel seat, cork grip, and guides for one of his TCR rods. That could then be used to get better mass estimate for just the blank in Grunde's mass measurement of his TCR. That could then be used to get a better divisor (it would be much less than 40) for the TCR blank. Such a comparison would give an apples to apples relative divisor difference for a tip flex or full flex rod.

Coming up with that number will give you a rough idea of the rod's action as tip flex rods will have a bigger divisor than full flex ones.

Gordy

[Merlin]

I do agree Gordy. K, mo and MOI are the best candidates to characterise a fly rod. I wonder if casters have a favourite mo. I mean that since mo tells you about the way a tackle slows down with load, one might have preferences in that field. Here again, the caster himself is a parameter.

Merlin

[gordonjudd]

The Equivalent MOI was 1/3 of the real one, I guess.

Merlin,
His effective length goes down as well (2.7 to 2.0 m) in my TCR example so his MOI is about 15% of the real one based on numbers for a TCR as described in the other thread.

I don’t think there is any particular significance to the effective bending length (2.01 m) and effective moment of inertia (.01098 kg-m.^2) that is derived from the fit to Dr. Robson’s equation.

The values Grunde and Magnus got for their MOI measurement of the overall TCR rod was Rod_mass=.1074kg,rod_length=2.722m,rod_moi=.0722 kg-m.^2.

Thus the “effective mass” (.0027kg) you get from the fit to the oscillation data is only 2.5% of the total rod mass, and the “effective MOI(.01098kg-m.^2) is only 15% of the MOI for the rod. The bending length fit of 2.01m was 74% of the total rod length, so there appears to be no simple function that maps the physical mass, length, MOI, and spring constant of a rod to the corresponding “effective” values.

Coming up with those relationships is something that Dr. Bill can chew on while we try to digest his "fundamental equation of flyfishing".

Dr. Robson must have been looking at the MOI as being the MOI of a mass at the end of a thin, "massless" rod. Consequently the mo in Grunde's equation is equal to:
m0=I/c.^2
Where I is the effective MOI value and c is the effective length.
Here is an example of the fit you get for a set of frequency vs added tip mass values using Robson's equation:

That gives an equivalent mo value of
m0=MOI/c.^2=(.01098 kg-m.^2)/2.01.^2 m.^2=.0027 kg.

His d value (.93033 m/N is just the inverse of the k value so k=1/.93033=1.07 N/m.

Thus if you determine the MOI,c, and d values in Robson's fit you can get the equivalent effective m0 and k values in Grunde's equation.

Gordy

[Merlin]

Gordy

Thanks, I shall have a look at it.

Today I put Dr. Lewin’s lesson about uncertainties into practice for testing precision of measurements and consecutive calculations. I use two “computer test benches”, one about the frequency with load, and another for static deflection. This is done through calculation, but the programs have been checked against powerful pro software, so we take the assumption that my “computed measures” are correct. I then can find the equivalent mass and stiffness for a blank and for a finished rod, so let’s see what it tells. Same for the deflection for a given load (I can derive the relative deflection like in the case of CCS, or Graig’s methodology, “angles” to compare to Theo’s methodology, or anything else I can invent).

We start by mo for a 9 feet rod for a #5 line (one of my model rods, some kind of average):

mo blank: 2.33 +/- 0.05
mo rod: 3.60 +/- 0.05

So the estimation for the (very light) guides is mo gw = 1.27 +/- 0.07

The MOI of the finished rod is 48.5 (blank) + 13.2 (guides and wraps) = 61.8 gm2 (I do not take the handle contribution in this evaluation).

From Fo, supposed to be exact, I can estimate Kd (d for dynamic) the equivalent stiffness of the rod with an uncertainty: 1.054 N/m +/- 0.022.

Now let’s see what my static bench finds for Ks (s for static). My problem is that I can find various values of Ks depending on how I measure it, because my rod is a non linear spring. So I use a complete set of measurements with masses up to 200 grams, so I can be able to give a value as a function of weight, and the asymptotic one (zero load) is 0.908 N/m. So there is some 15% difference in between the static and the dynamic stiffness values under these conditions.

To satisfy my intellectual frustration coming from the fact that the two stiffness values are different, I look after the conditions which would allow me to find 1.054 N/m with the static bench, and I also estimate the consequences of the uncertainties in evaluating mo. What load (mass)? What relative deflection (%), what angle (degrees). Here are the results:

Estimated loads: nominal 27.5 grams (between 25.9 and 29.1), so approximately one ounce.

The relative deflections: all estimates are close to 13% to 14% of rod length: this is smaller than what Graig and Bill have chosen.

The angle, as in Theo’s methodology: 7 degrees to take a rounded number.

Interesting, isn’t it? I did not check for other rods, but I shall do it sometime.

Merlin

[Bill Hanneman]

Gordy,

I would call it the "effective spring constant value" ...

A great idea, may I use it? Of course I’ll give you full credit for originating the name.
I think ESC will fit in nicely with ERN and ELN.

Coming up with that number will give you a rough idea of the rod's action as tip flex rods will have a bigger divisor than full flex ones.

I find AA to be sufficient for that task. Besides, the desired action is determined by one’s personal choice.

Bill
---------------------
Merlin,

I do agree Gordy. K, mo and MOI are the best candidates to characterize a fly rod.

While I won’t argue, I prefer CCS in explaining it to Joe. Nevertheless, I am taking the route of frequency to arrive at my “Perfect Personal Rod".

I wonder if casters have a favourite mo. I mean that since mo tells you about the way a tackle slows down with load, one might have preferences in that field. Here again, the caster himself is a parameter.

At last, you seem to be coming to my point of view. Simply replace your term mo with frequency and you have discovered the secret ingredient in the perfect rod (PPF).
In a nut shell, The Perfect Personal Rod has the same frequency (mo, if you prefer, or is it effective mo and K?) as that desired by the client. Of course, the rod builder has to determine that value, but that is a trivial task.

[gordonjudd]

At last, you seem to be coming to my point of view. Simply replace your term mo with frequency and you have discovered the secret ingredient in the perfect rod (PPF).

Bill,
Stop digging!

These kind of statements show you do not understand the constraints given by the effective mo, k, and added tip mass on the loaded frequency used in the car/spring/brick model. You cannot just replace mo with PPF. M0 enters into to the equation, but you also need k and the tip mass to tell the whole story.

Merlin's approach is much more complicated than some mythical fixed PPF value as the loaded frequency is going to change with the length of line being cast on a given rod.

Dr. Robert Grober has has used accelerometers to measure the rotational velocities in the golf swing, and finds that in putting at least:

It is shown that the putting stroke of world class golfers can be described as the motion of a pendulum driven at twice its natural resonance frequency.

His sonic golf approach would be interesting to apply to fly casting so you could get some auditory feedback on how to make a good fly cast. I would not be surprised that we use a casting tempo that is related to the loaded frequency of the rod and line system just as he found in golf.

It will come as no surprise that that a competent caster is going to adjust his tempo when casting different line lengths. Consequently he will have a wide range of casting tempos rather than some fixed value of 83 CPS that supposedly matches up with the tempo use to cast 30 feet of line.

Merlin's computer modelling is trying to come up with ways to get expected values for the effective mo and k values for a given blank design that can be used in his model and is finding it to be more complicated than one might think. That is a worthwhile pursuit, and I will try to do some more studying to better understand what he is sharing.

I do not always casts 30 feet, so I do not see the usefulness of using CCS to give me an idea of what rod I might want to use to cast from 20 feet to a hundred feet. One single PPF does not apply to my casting experience, so I do not see any utility in applying your graphs to my needs, since I for one do not have a single PPF.

Your mileage might differ.

Gordy