Harmonic Oscillator

[Hal Jordan]

Like Grunde I would never use the word never :p

But saying the spring back of the rod eliminates most of the angular velocity is definitely not the same as saying it eliminates all of the angular velocity.

[grunde]

And the "angular velocity of the rod butt" part is also important.

[gordonjudd]

That's what I am doing as a first estimate, but I use the relative speed of the spring.

Merlin,

When I said the spring velocity I was just referring to the derivative of the deflection curve. I expect that is what you mean by the "relative speed of the spring".

Gordy

[Hal Jordan]

And the "of the rod butt" part is also important.

Physics is a bit like law. A single word or phrase can make a huge difference.

[Merlin]

Walter,

For sure, there are Physics laws. One difficulty is to interpret what our eyes are seing and translate that into a model that confirms our intuitions.

I recommend looking at energy. I took the "of the butt" part on board and confirm that the kinetic energy of a rod is by far larger than its elestic energy. That means that you cannot stop a rod using that elastic energy. Another law of Physics.

I am ready to swallow my hat if the contrary can be demonstrated (by something else than intuition).

Merlin

[grunde]

Merlin,

I can't see that anyone have claimed that you can stop the rod completely using that elastic energy?

At RSP the angular velocity of the butt can be quite low, but there is still significant kinetic energy in the upper sections of the rod (in fact the tip is at it's peak speed just prior to RSP). This energy manifests itself in the later counterflex of the rod, and it's damped (taken out of the rod) by the caster mainly between RSP and RSP2 where the torque at the butt and the angular velocity of the butt are in opposite directions.

Cheers,
Grunde

[Merlin]

Grunde

I am a bit caricatural, sorry for that, but I refer to the following:

"So the spring back of the rod has such a strong effect that it actually eliminates most of the angular velocity of the rod butt". Now what "most" means is an open question. I would say "some".

For the rest of your post, I agree with you completely, this is just in line with modeling (hopefully for the modeling).

Merlin

[Hal Jordan]

Now what "most" means is an open question. I would say "some".

Merlin:

50% < most < 100%

[gordonjudd]

50% < most < 100%

Walter,
What calculation did you do to come up with that high of an estimate or is that just a gut feel?

How are you computing the total torque that is being supplied by the shoulder, elbow, and wrist joints during the cast to estimate that more than 50% of the torque being applied to stop the rod from MAV to RSP comes from the P.E. in the rod?

You can see from the change in the angular velocity of the shoulder, elbow, and wrist joints in Chris Korich's cast that the slow down in all of those joint angular velocities are contributing to the slow down in the angular velocity of butt of the rod.


I still do not see how how the P.E. in the rod can supply 50% of the total energy required to stop the K.E. in the rod based on Merlin's comment that:

I recommend looking at energy. I took the "of the butt" part on board and confirm that the kinetic energy of a rod is by far larger than its elastic energy. That means that you cannot stop a rod using that elastic energy. Another law of Physics.

Is there a better way than energy to look at this?

Gordy

[grunde]

50% < most < 100%

Walter,
What calculation did you do to come up with that high of an estimate or is that just a gut feel?

most = more than the half

I guess...

[gordonjudd]

most = more than the half

I guess...

Grunde,
Thanks for that clarification, but as you know that is not what I was asking about in that post. I don't know if "most" implies and even larger percentage of the reversing torque than "more" (I would assume it does however), but I wanted to get some feedback on Walter's statement that:

But saying the spring back of the rod eliminates most of the angular velocityy is definitely not the same as saying it eliminates all of the angular velocity
and
I haven't heard Server say that the rod can stop itself but he does say it will unload without us having to physically stop the rod.

I just wanted to see (and understand) the calculations that would back up those conclusions.

From Merlin's statement about the relative amount of kinetic and potential energy in the rod at MAV I do not see how you could expect that the unloading of the rod is going to supply "most" of the torque required to slow down the angular velocity of the butt of the rod we see from MAV to RSP.

I am not familiar with Sever Sadik's calculation that would imply that is the case, and was hoping that Walter (or you) could give some details about his approach since you both say you agree with him. Just from an energy standpoint that conclusion does not make any sense so I would like to learn something about how you see the energy tradeoffs involved to get the rod to unload as fast as it does.

Gordy

[Merlin]

Gordy

The current model I am using, which is base on the energy standpoint, tells that there is an impact of rod unloading on the torque that the caster has to produce during the cast (original statement from Server Sadik). I can also compute what a strain gauge would say and what it tells you is just when RSP is occurs (straight rod). The torque time story is more influenced by unloading. If your rod is stiff, the unloading will occur much sooner than if it is soft. For a good time tuning or frequency fit, the unloading occurs just as you decelerate, so it eases the torque you have to produce at a critical timing during the cast. That may be part of the feel of the rod by the caster. Since it is depending on cast timing and loaded frequency; that can happen for a given range of line length and maybe one likes a rod because this range is rather wide given one’s casting style and rod characteristics.

The energy which is used to ease our cast comes from the elasticity of the rod, but this same energy is also used to create some force on the moving rod tip during the deceleration. For a normal cast, the maximum elastic energy in the rod may be one quarter of the maximum kinetic energy in the rod (they appear at the same time). When carrying a long line, both energies can be equivalent or so. The question is, in both cases: how much from the elastic energy of the rod goes to ease the deceleration, and how much contributes to generate more speed in the line? We can compute how much there is just at the release time (small), but just between the start of the unloading and the time of line launch, what does really happen to that elastic energy?

I am going to try finding a way to estimate these two contributions.

Merlin

[Bobinmich]

Does someone have the data file for the expert 50 casts that Gordie used to make the plots quoted in post 30 for velocity and accel. It would seem that the acceleration plot shows excessive noise which probably doesn't exist in that form in any of the original casts. Not that noise doesn't exist in the individual casts but any relavance has already been lost by taking an average. It would seem more realistic to curve fit the velocity data with a reasonable quadratic prior to differentiating it. Maybe then some sense could be made as to what the casters are really trying to do during the cast.

Bob

[Tom]

Please take a look:
http://www.youtube.com/user/SuperRattus#p/u/5/B0mmsDtEwaw

Tom.

[grunde]

Excellent Tom, clip #3 is very nice :pirate

Grunde