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SHO model trends - what we can get from it

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VGB
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Post by VGB »

Merlin

I am comfortable with your model representation.

However, there is a difference in what we are talking about for damping. I understand that you are using damping co-efficient to describe losses in the spring. I am talking about the damping ratio (Sorry, I made a mistake in my Greek, I should have called it zeta not sigma).

The damping ratio details the frequency response of a 2nd order system and characterises the relationship between the damping co-efficient of the spring and the mass attached to a damped harmonic oscillator. Zeta is shown just before the damping text:

http://en.wikipedia.org/wiki....illator

This is a good resource but you only get a few goes before you are asked to join. If you vary the relationship between the mass and the damping co-efficient, you can zeta and the damped frequency change:

http://www.efunda.com/formulae/vibrations/sdof_cal.cfm#calc

In this model, the damping co-efficient that you are talking about is described by Cv but I take M as being the fly line. If you vary M, the frequency changes. You can also see this effect in the Lovoll/Borger analysis. If you look at the cord length, there is a step change in frequency somewhere near -0.25 seconds as the mass of the fly line is de-coupled from the rod and the frequency returns to the natural frequency.

I think that this information is in your model but you have not looked at it yet.
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Merlin
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Post by Merlin »

Vince

The value I use is zeta = damping ratio (dimensionless) = 0.025; and not the damping value which is linked to the mass. Since you solve the equation by dividing by mass, it disappears from the equation and the remaining parameters are zeta and the pulsation (2 pi * frequency which depends on rod stiffness and mass in motion).

The response of the SHO is linked to zeta (damping ratio) and not to the damping value. Zeta is not variable for the mechanical system. The main parameter is the frequency, zeta is here to take friction losses on board.

The decoupling of frequency as the line is launched is included in my model. It has been tested against pro software and is a valuable tool to understand the underlying mechanics of the cast. Not pefect for sure. In fact I use three SHO models: an explicit one, which uses the exact mathematical solutions (so you need 5 different configurations to model completly the cast (accel, decel, launch, counterflex, in various situations), a combination of four explicit models for testing shoulder, elbow, wrist, translation/thrust which Bernd called the obi wan model recently, and a numerical model allowing to enter the non linear domain.

Merlin
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Post by VGB »

Merlin

I think we are looking at the elephant form opposite ends and I am hoping we will meet in the middle.

Because your model works with a coupled and uncoupled load you should have 2 values of zeta. When the load is coupled, your system is a damped harmonic oscillator, when your load is decoupled it operates as a simple harmonic oscillator. The 2 different states will have different damping ratios

Because you are removing the mass from your equation, it tells me that your value of zeta at 0.025 is correct for after the line is released. This is as you have previously mentioned a severely underdamped system. If you do not remove the value of the mass it will provide a zeta for the coupled system.

My particular interest is in the value of zeta while the fly line mass is still coupled. The reason for this is that I believe that it provides a measure of efficiency of potential to kinetic energy conversion. The mechanism is the same as matching a car radio to its aerial. For us, it will allow us to match a rod and line mass.

Zeta must be much higher than 0.025 for a loaded rod, otherwise if we stop the cast but do not release the line, the rod will continue to oscillate and wrap the caster in line. I can manage this effect when I am casting with a high modulus, fast action rod and no fly line beyond the rod tip

I think that we can also prove this by practical experiment, if we carry out the cast and stop without releasing the line. An underdamped system would drag the line back towards the caster, the overdamped line would drop the line before its cast length was reached (loose at the casters end) and a critically damped system would lay the line flat.

I hope you do not think that I am criticising the model because that is not my intention. I firmly believe that there is more valuable information that can gained from it.

regards

Vince
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Post by VGB »

Merlin

I expect this video has been seen many times here but it does illustrate the point I was trying to make about critical damping:

http://www.gofishn.com/ned/videos/9066-a-fly-cast-in-super-slow-motion

The response in of the rod tip is not typical of a severely underdamped system. I expect that once the line is released then the rod will return to its natural frequency.

regards

Vince
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Post by Merlin »

Hi Vince

Impossible to look at the video, my computer just refuse to do it, I can't tell why, a link issue maybe.

I understand your point about having different damping ratios, but until now the necessity never appeared.

I modeled a number of casts and if the damping ratio for the loaded rod would have been too small (which is possible because of the losses due to the air drag on the line), it never showed up. It would have been easy to correct that situation.

Secondly, my model has been tested with the help of a research center of a supplier for auto industry specialized in car suspensions, so damping is well known by these people and they never pointed your issue, although we had long discussions about how to damp a rod.

I would like to understand your analogy to an electrical problem but I just cannot understand it (sorry, I'm a mechanic).

Merlin
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Post by VGB »

Hi Merlin

The link will not work for me either, I think the gofishn website is down. It shows a cast at 500 frames per second where the rod tip recovers to its rest point after 1 cycle, the line was not released. I hope we can get it back it illustrates my point perfectly.

By training I am an aerosystems engineer and have been a flight test engineer in the past. But for a long time I have been managing, not doing so have forgotten many details.

A large part of my work involved jettisoning equipment from aircraft, so I had to look at the effects of coupled and uncoupled loads, measure reaction loads into the structure and a fair bit of aerodynamics. There is a lot in common with rod casting but I am a generalist rather than having a deep knowledge.

I have absolutely no doubts about the capability of your model. However, in one of your earlier posts you mentioned :

"The value I use is zeta = damping ratio (dimensionless) = 0.025; and not the damping value which is linked to the mass. Since you solve the equation by dividing by mass, it disappears from the equation and the remaining parameters are zeta and the pulsation (2 pi * frequency which depends on rod stiffness and mass in motion)."

I suspect that your model is using the equation twice. Firstly, with mass in the equation which equates to the coupled condition and when you solve the equation by removing the mass it equates to the uncoupled condition. Alternatively, your model may assume that it is critically damped and that the line mass is matched to the rod. Can you vary the fly line mass?

In practice, I think it would be difficult to get exactly a zeta of one for the coupled condition and that may account for the differences between your model and the real world.

The reason that I am so interested is that I think that the critical damping governs the efficiency of the transfer of potential energy between the spring and the mass in a damped harmonic oscillator. I was trying to equate it to impedance matching in electronics:

http://en.wikipedia.org/wiki/Impedance_matching

When I was lurking I seem to recall that you had noticed reflections in your model, this is typical of an unmatched system. In a cast I would expect to see that as an oscillating rod tip. However, impedance matching in electronics is used to ensure that you get maximum power transfer between an aerial and a radio or a music system and speakers.

It is difficult to discuss this subject like this and I suspect face to face it would be much quicker.

Thank you for your time and patience

regards

Vince
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Post by gordonjudd »

The reason for this is that I believe that it provides a measure of efficiency of potential to kinetic energy conversion. The mechanism is the same as matching a car radio to its aerial.

Vince,
Based on the mechanical impedance mismatch between the flexural wave in the rod and the line, I do not think there is much direct transfer of wave energy in the rod to wave energy in line as discussed here. You can imagine the transmission loss between your radio and antenna that would result from an impedance mismatch ratio of 25:1. One table I found calculates that about 15% of the output signal level from the antenna would pass through to the front end of the radio with a 25:1 impedance mismatch.

I have no experience with the concept of mechanical impedance or flexural wave transmission, (I am an EE) but as you can see from the article referenced in the post above mechanical impedance (The characteristic impedance of a material is the product of mass density and wave speed) has nothing to do with the damping factor that slightly impacts the rod/line SHO response from RSP0 to RSP1.
The reason that I am so interested is that I think that the critical damping governs the efficiency of the transfer of potential energy between the spring and the mass in a damped harmonic oscillator.

Contrary to traditional wisdom, the KE in the line does not come from the PE in the rod. Rather it comes from the deflection force from the bent rod being applied over some distance determined by the path the caster moves the tip of the rod during the cast.

Thus as Merlin's model shows the rod/line system is acting like a forced harmonic oscillator, not as a big spring.

As you know critical damping involves considerable energy loss from the damping mechanism, and thus would adversely affect the spring speed produced by the rod at RSP1. You can see the effect that damping has on the velocity of the moving mass by comparing the slope of the blue line to the red line in this plot from Wikipedia.
Image
Consequently even if you could achieve it, applying critical damping to the rod as it unloads from MRF to RSP1 would not be a wise choice in terms of producing maximum line speed.

As noted before Merlin does include a small damping factor in his model to include the damping related to the internal friction losses in the rod, but as Phillips found that is a very small factor of .01 to .03 for graphite rods. Looking at the exponential weighting factor you get when the butt of the rod is hard clamped with a 10g tip load I found the internal friction loss factor to be around .016 as shown below:
Image

That internal friction damping factor gives an exponential weighting factor that is equal to 2*pi*f*c=2*pi*1.69*.0165*t=.175*t as noted in that plot. I think Merlin mentioned he uses a factor of .025 in his model.

When the caster relaxes his grip at RSP1 the damping coefficient increases, but I do not know how you would get an accurate measured value for it.

Gordy
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Merlin
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Post by Merlin »

Thanks a lot Gordy, now I see what Vince was telling

Vince,

A fly rod is a "speed generator", not a"wave generator". Its function is to create speed in the line, and let it go when that speed is maximum. Then it counterlexes and shapes the loop. Wave transmission, if any (pending question), can occur during loop shaping, but we have no evidence of it. The analogy with electricity is possible for longitudinal waves (along the fly rod axis), but unfortunatly not in the case of flexural waves.

A "critical" damping can occur after line launch, here is the example for Mathia's cast (Grunde's publication):

Image

This correspond to the red dotted line. The solid blue line is a model of the rebound, and the dotted blue line illustrates the same model with an "absolute" stop. You can see that the damping corresponding to the red line can be considered as "critical", but that this is not completely the case for the rebound model (solid blue line), even if there is a significant improvement by comparison to the "absolute" stop.

In terms of rotation speed, it corresponds to this picture:

Image

The model consist of assuming the caster is passive and can be considered as a mass associated to a damper. During the rebound, the rod is the source of motion.

Under these conditions, the damping comes from the coupling between the caster and the fly rod.

Merlin
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Post by VGB »

Gordon/Merlin

Thank you for your explanation. I was trying to use impedance matching as an analogy for an efficient transfer of energy in a mechanical system and it seems I missed the target.

I was not considering the transfer of wave energy as being the baseline for a rod model, maybe for some forms of cast it may be possible but I do not think that is the case for an overhead cast. A few years ago, I was managing some of the development trials of composite rotor blades and hubs for a helicopter and found the flexural components a real headache to understand. Fortunately, I had some serious boffins looking after the modelling and analysis of blade performance.

I also was not considering Merlins model as a big spring, but was considering it as a damped harmonic oscillator at the point that there was not translational or rotational hand movement going on. When I used my biro/spring analogy it was for the movement of the tip of the spring to explain the effect on rotational speed due to a fast or slow action rod.

I think from your remark "As you know critical damping involves considerable energy loss from the damping mechanism, and thus would adversely affect the spring speed produced by the rod at RSP1." that I am not explaining myself in enough detail.

As you know in the harmonic oscillator, there is a transfer of energy going on between the spring and the mass to sustain the oscillation. The considerable loss I was considering was by transfer of energy from rod to line as per the conservation of energy. It can only dissipate into the line or the caster.

If you are only damped by the rod such as in the "no line" case, the energy stays with the rod and it continues to oscillate for much longer than if it had additional damping from the line, unless the caster dampens it. This is because it is severely underdamped, the rod is intended to be matched to a line. In his Itinerant Angler podcast, Gary Loomis uses a great diving board analogy when he describes this.

I believe that we are in furious agreement on almost every point but one point of definition which is the mass element of the damping system. If we look at the mass/spring/damper diagram here:

http://en.wikipedia.org/wiki....3damper

I consider that the mass has 2 values. The 1st when the fly line is coupled to the rod ( intrinsic mass of the rod and the line), the 2nd when the line has been released (inertial rod mass only, although there might be some loose coupling with the line).

In the 1st state from the pause until RSP1 the frequency is lower than the 2nd state. I understand the 0.025 value to be based upon B in the diagram.

If I understand Merlins diagrams correctly, they are showing the tip action in the 2nd state, after the line has been released. I am interested in the before as well, to see if the rod and line are matched.


I hope I do not come across as trite because that is not my intention. I am getting a bit frustrated at the internet as a medium to explain complex ideas. I like drawing pictures on boards.

I looked at your discussion with Massimo, I do not think he was talking about flexural waves but about about harmonics induced by the forcing frequency. But best we park that discussion for now. ;)
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Post by Merlin »

Vince

The illustrations in my post correspond to the full cast. The first state goes up to about 0.57s and the second state starts from there.

We agree about the mass issue (first state versus second state).

Incidently I did not catch up Gary Loomis point when I listened to his podacst (I have difficulties to download anything from the Itinerant Angler web site today, do not know why).

Massimo's point was on harmonics, but you know, these phenomenon are just linked to flexural waves...

Merlin
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Post by VGB »

Thank you Merlin, I can see it now :p

I can also see that the 1st peak corresponds to about 1.7Hz (coupled) and 3.1Hz (uncoupled). There is also a significant reduction in the peak to peak amplitudes. Due to transfer of energy from the rod to the line?

My contention has been that when you have a matched rod and line (critically damped), then you are driving the 2nd peak to a minimum. Is your model assuming a zeta of 1 for the coupled point or are you able to vary the mass of the fly line?

I would suggest that you keep trying with the podcast, it is very entertaining :)

Going back to harmonics, here comes a real can of worms.

Resonance will occur when the forcing frequency of the input activity, such as the cast, corresponds to the natural frequency of the rod. The harmonics of the forcing frequency are multiples of the forcing frequency and may also cause a resonance condition if they correspond to a natural frequency of the rod.

Your model will generate harmonics without flexural waves. If you built your model and forced it, you would see harmonics at 6.2 and 9.3Hz on a spectrum analyser. The levels would be dependant upon the bandwidth of the rod. In this instance they would be quite high because the rod has a relatively broad bandwidth. You would probably need some mass damper system to get rid of it.

This is a huge problem on helicopters because you have 2 rotating sets of multiple blades. Early development flights always show harmonics to both the main and tail rotor forcing frequencies that are large enough to make fillings fall out.

I think what Massimo was going for was that the TLT casting style can excite one of the higher harmonics such that the the rod is artificially "faster". Theoretically it is possible and Gordon would recognise a band pass filter. I do not know how you would do that on a fly rod.

regards

Vince
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Post by Merlin »

Vince,

Frequencies are not easy to detect precisely from the graphic; the numbers are respectively 1.2 Hz and 2.5 Hz. The reduction of amplitude is due to both the damping ratio, zeta is set at 0.03 here, and to the rebound model (depending on what curve you are looking at). If you do not vary the mass (you do not launch the line), it increases. I vary the mass of the line of course and zeta remains at 0.03, the damping which looks like a critical one is due to the rebound model (caster seen as a mass and a damper). In that case the damping ratio for the “caster” is very high (17), but this is some kind of artifact to mimic the rebound as it was recorded.

There is no realistic way to get rid of the harmonics (harmonics correspond to a steady state situation for flexural waves), the mechanical solution is much too heavy. Controlling harmonic is key in rod design. The issue with the TLT was that there was a claim stating the third harmonic doubles the line speed. The fact is one can mimic a TLT cast and get a very good estimate of line speed without taking any harmonic into consideration. Use of pro software confirmed the presence of higher harmonics without any impact on tip speed.

Merlin
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Post by VGB »

Merlin

My frequency calculation was done with thumbs :???:

This caught my attention "I vary the mass of the line of course and zeta remains at 0.03". It should not happen.

zeta (damping ratio) = damping co-efficient(spring)/2 sq rt (mk)

I do not know enough about your model to understand why a mass change is not varying zeta.

I agree about mechanical solutions to harmonics. On even quite small helicopters the rotor head mass dampers can be 50-100kg. I can understand why harmonic suppression would be desirable but I must admit I did not read all 50 pages of the thread.
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Post by Merlin »

The damping coefficient = 2 * zeta * m * omega

State 1 : m1 = mline + mo (equivalent rod mass) and omega1 = sqrt (K/m)

State 2: m2 = mo and omega2 = sqrt (K/mo)

The damping coefficient are thus different in between the states, but the (dimensionless) damping ratio can remain the same (although it should be slightly higher for state 1 because of the drag due to the line).

Merlin
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Post by gordonjudd »

(although it should be slightly higher for state 1 because of the drag due to the line).

Merlin,
As you know the drag will vary as the square of the velocity, so including a drag effect for the rod and line would require adding another damping term to the standard SHO differential equation where the viscous damping depends on the velocity.

I expect that is something that would be easy for me do since MATLAB is using a numerical solution to the ODE, but I have no idea of how it would impact your closed form solution.

I don't know how you would arrive at an "effective" coefficient for the that x_dot squared drag term since its impact on the rod is not straightforward (to me at least). However, Phillips did say that the drag on the rod had a significant damping effect, so maybe that is something we should add to your model.

Gordy
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