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  #81 (permalink)  
Old 02-02-2012, 08:43 AM
vgb vgb is offline
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Default Re: Simple Harmonic Motion

Quote:
Originally Posted by James9118 View Post
It doesn't surprise me that Lasse can do this - the model predicts it . If you look at the main peak of the velocity versus k (spring constant) graph, you will see that for a given velocity there are two intercepts - one a soft spring and the other a stiffer one.

Click the image to open in full size.
James

Are you referring to axial or material stiffness? If you look at the equations we played with on the previous page, K is a function of mass and frequency. While I agree that frequency is dependant upon material stiffness, it is also a function of taper and without knowing the material construction of the 2 rods you cannot say which is the dominant effect.

However, I do not want to throw the baby out with the bath water because your graph is providing a lot of information.

By plugging the mass and K value back into our previous formula, we know the resonant frequency of the 2 rods. Looking at the response on your graph, I can see that neither rod is fully resonant because at this mass the resonant frequency occurs at a K=1 and putting that back into the formula would give me the best rod frequency for this mass. This would tell me which frequency rod I need for maximum casting distance as the rod would be critically damped.

Going back to the 2 intercept points, I can say that the lower K value is over-lined and the higher rod value under-lined. You can see that effect in the video by looking at the relative tip oscillations.

Another valuable bit of information that I can gather from your graph if it is in the frequency domain is Q (or selectivity). The difference between the frequency values at half power points of the harmonic gives bandwidth. The smaller the bandwidth, the more energetic the cast (conservation of energy) but the more difficult it is to cast because the caster has to hit the precise pause length (t=1/f) in the false cast to add in-phase energy.

What you have in the value of spring constant is a value of frequency and mass which gives a measure for dynamic action angle that is independent of rod length which is heady stuff. But even this does not give you a value of feel unless it is a fully flexing rod.

If you look at the video you can see he has selected 2 rods with the same dynamic action angle (look at the relative bends in the acceleration). However IP is measured as a function of rod length which is why you have different IP and this is important in terms of feel because the longer rod length will give a greater moment of inertia which you will feel in the grip.

In the video, the effect of rod length and greater radial velocity is removed because the tips are taped together and this gives the lower value IP rod the same effective IP as the longer rod. It is a superb demonstration of rod qualities.

Going back to the graph, we can also derive some other factors about our rod. If we have a rod that is operating at its resonant frequency, we can see that maximum velocity with the effect of oscillation is about 16 m/s, with oscillatory effect at resonance it is about 23m/s. This tells us that this our K=1 rod obtains about 70% of its radial velocity from the leverage effect and the remainder from oscillation. Which against Franks estimates tells us that this is a slow action rod.

In terms of quantifying feel, I think we have covered most of the factors that we would need to characterise a rod. We have the dynamic action angle (spring constant), fly line mass, bandwidth and we can easily measure the rod length. Are there any missing?

Last edited by vgb; 02-02-2012 at 04:07 PM. Reason: just twigged may be referring to axial stiffness
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  #82 (permalink)  
Old 02-02-2012, 08:55 AM
vgb vgb is offline
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Default Re: Simple Harmonic Motion

I'll stick my next thoughts in a separate post because it strays into the dark side with damping. If you plot the graph in frequency, you can see not only the resonant frequency but also the harmonics. The more energy in the harmonics, the less available energy is transferred to the fly line as explained below:


To get the most efficient cast,we need to be critically damped. When you look at the video you can see that one of the rods is not critically damped, I'll put my hands up and say that mechanical systems are not my bailiwick but I have a recollection that it is a proportion of the external mass/inertial mass of the spring. However, that is a woolly recollection and I would not want to bet my house on it. I also cannot recall, how inertial mass plays out in the value of K and I'll leave it to the bright young things. I did say I did not want to touch the maths

Last edited by vgb; 02-02-2012 at 03:56 PM.
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