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## Swing, Spring, Whip

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gordonjudd
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Location: Southern California

### Swing, Spring, Whip

Tobias,
As you know angular momentum is proportional to the moment of inertia and the angular velocity of a rotating object.

As given by the parallel axis theorem the moment of inertial of a rod section about its center of gravity is equal to

plus its mass * the distance from the center of mass to its rotation axis squared. Thus the distance to the center of rotation can have a big effect on its angular momentum of an object that is not rotating about its center of mass.

Because we translate the rod while we rotate it the instant center of rotation is greatly impacted by ratio of the translation velocity to the angular rotation velocity. Because of the translation effect on the velocity center of rotation I don't think the rotation of the rod stays fixed at the wrist as you imply in your diagram.
https://imageshack.com/a/img922/9162/zrhvEM.jpg

Rather if you calculate the actual velocity center of rotation of a rod section near the tip you find it is a long way from the wrist at MAV because the ratio of its translation velocity to its rotation velocity is quite large. The same effect puts the instant center of rotation for the butt to be near the elbow as shown below.
https://imageshack.com/a/img921/2749/1ozhJ5.jpg

If the mass of a 10 cm section of the rod near the tip was 1\16 the mass near the butt then using the above values I calculate the angular momentum of the tip section would be about 2.5 times the angular momentum of 10 cm section of the butt because the distance to the rotation center of the tip section (4.6m) is so much larger than the distance to the rotation center for the butt (.5m).

You should check my values, but this says the fact that the MOI of the tip section is about 5 times the MOI of the butt section that difference in MOI is enough to offset the difference in the angular velocity of the two sections.

Thus if you take the actual center of rotation of the different sections into account I don't see that there is a flow of angular momentum from the butt to the tip during the cast.

Gordy

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### Swing, Spring, Whip

Daniel/Tobias

Thank you both for your considered responses, I will reread your relevant pieces.

Gordy

We’ve been around this buoy of ICR quite recently, did you manage to solve how to apply a rigid body concept to a flexible object?

http://www.sexyloops.co.uk/theboard/vie ... 614#p45614

Tobias provided an answer to your original question

http://www.sexyloops.co.uk/theboard/vie ... 618#p45618

Regards

Vince
Bright but shite

IANACI - There’s no such thing as absolutes

Free the Mark One

hshl
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Location: close Potsdam
Contact:

### Swing, Spring, Whip

Gordy,

in annex 2 of my investigations (page 51) I calculated the moment of inertia exactly the way you wrote. Didn't you read the chart ? It seems you still try to explain a complex issue with simple equations applicable to rigid bodies ?!
Gordy wrote:Because of the translation effect on the velocity center of rotation I don't think the rotation of the rod stays fixed at the wrist as you imply in your diagram.
I think this was the question of Graeme I already answered before.
Graeme H wrote: I've always tried to teach my students that they need to be moving the whole rod - also known as translation - in addition to the rotational phase of a cast. That avoids a "windscreen wiper action". How does such instruction fit with Figure XIII?

hshl wrote:good point and you pointed out an assumption I had to make to keep my investigations simple. Figure XIII is reducing the casting stroke to the rotary motion only and in fact my investigated casting stroke includes a significant translatory motion at the beginning.
VGB wrote:Gordy

We’ve been around this buoy of ICR quite recently, did you manage to solve how to apply a rigid body concept to a flexible object?

viewtopic.php?p=45614#p45614

Tobias provided an answer to your original question

viewtopic.php?p=45618#p45618
Yes Vince, I don't have any other glue the explain the redistribution effect (whip, inertia ...).

Regards, Tobias
http://www.passion-fliegenfischen.de/_en
All in its proper time ...

Graeme H
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Location: Perth, Western Australia

### Swing, Spring, Whip

hshl wrote:Hi Graeme,

....

So how I recomment to cast is shown on the attachment. With this video I also tried to work out some motion properties which should benefit the described energy transfer. However, this video is only a kind of approximation. Fell free to continue.

Cheers, Tobias
Thanks Tobias.

Cheers,
Graeme
FFi CCI

hshl
Posts: 62
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Location: close Potsdam
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### Swing, Spring, Whip

You’re welcome Greame,

since for the sake of simplicity that figure XIII is focusing on the rotary motion I added section F2 and figure XIV.

I just noticed that annex 3 includes a quite good visualization about the whip effect how Franz- Josef is describing it.
hshl wrote: Simply spoken the whip effect is visualized by the additional / further shortening of the chord length between the tip of the fly rod and the upward moving "center of the rotating mass".
Regards, Tobias
Attachments
http://www.passion-fliegenfischen.de/_en
All in its proper time ...

gordonjudd
Posts: 1265
Joined: Sat Jan 19, 2013 11:36 pm
Location: Southern California

### Swing, Spring, Whip

We’ve been around this buoy of ICR quite recently, did you manage to solve how to apply a rigid body concept to a flexible object
`
Vince,
It is quite simple actually. You just isolate a small section of the flexible fly rod (lets say 1 cm) that remains quite straight throughout the cast and use its varying rotation angle and position to come up with the I.C. for that short section.

Obviously for a flexible body different short sections of the body will have different rotation centers. You can see there is a big difference in the I.C. computed for the short section of rod near the tip compared to the I.C. for a short section near the butt in that cast recorded by Grunde years ago.
https://imageshack.com/a/img921/2749/1ozhJ5.jpg

Gordy

gordonjudd
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### Swing, Spring, Whip

in annex 2 of my investigations (page 51) I calculated the moment of inertia exactly the way you wrote.
Tobias,
From you diagram, I assumed that when you did the calculation for the MOI of tip and the butt sections you used a common rotation center that was fixed at the wrist rather than the different rotation centers you get for the different sections of a flexible rod.

If that was not the case and you used a rotation center for the tip section that was meters away from the I.C. for the butt to compute the m*r.^2 component of the MOI for the different sections then we are in agreement of how the MOI should be calculated.

Gordy

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### Swing, Spring, Whip

Hi Gordy

So if we take each piece of 1cm mass and assign an instantaneous centre of rotation to each section, a bending rod will have many instantaneous centres of rotation. This is why we cannot simply apply it to a fly rod analysis as you suggested.

Vince
Bright but shite

IANACI - There’s no such thing as absolutes

Free the Mark One

hshl
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### Swing, Spring, Whip

gordonjudd wrote:From you diagram, I assumed that when you did the calculation for the MOI of tip and the butt sections you used a common rotation center that was fixed at the wrist rather than the different rotation centers you get for the different sections of a flexible rod.
By calculating the MOI in annex 2 I just estimate the influence of the mass of the fly rod. This center of rotation stays at the grip since it is hold there by a constrain as the caster rotates the fly rod "near the grip". The "center of the rotating mass" is a different thing, and I already answered.

Tobias
http://www.passion-fliegenfischen.de/_en
All in its proper time ...

gordonjudd
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Joined: Sat Jan 19, 2013 11:36 pm
Location: Southern California

### Swing, Spring, Whip

This center of rotation stays at the grip since it is hold there by a constrain as the caster rotates the fly rod "near the grip"
Tobias,
Try comparing the K.E. you get at for a point mass on the rod based on K.E.=1/2mv.^2 with the K.E.= 1/2*I*omega.^2 value given by rotational dynamics. I think you will find that the I (MOI) you get assuming the rotational center is at the grip will produce an energy value that is much smaller than the 1/2mv.^2 value.

If you use the velocity rotation center (and its much larger r value) to compute the MOI of that mass then the 1/2*I*omega.^2 value will equal the 1/2mv.^2 value.

It does not make sense to me to use one I value for angular momentum calculations but use a much larger I value when you calculate the K.E. of a point on the rod. But that is just me.

Gordy