Torque applied to stop rod rotation - What are the primary sources?

[gordonjudd]

To be precise I stated that's where I believe the uncoupling to begin.

Eugene,
And to be precise what to you mean by "uncoupling"?

Uncoupled to me means the rod tip is no longer applying a force on the line that can change its momentum.

If that is the case when to you see that the rod tip is no longer impacting the momentum of the line?

Straight rod equals zero tension.

And I think we can agree that tension is an example of an external force.

It is obvious from Grunde's video that the rod tip is pulling the line around as the loop is being formed, and unless Newton was wrong about how momentum is changed that means the rod tip is still applying a force (that creates tension in the line) after RSP1 when the deflection force from the spring in the rod goes to zero.

Do you see something different in that video?

Gordy

[Eugene Moore]

Gordon,
To me uncoupling is the removal of mechanical deformations and lash prior to a dynamic system undergoing load reversal.
Tension imposed stretch on the line and rod deformation caused by wind drag must be removed. The system is in contact but no work is being accomplished. Lash must then be taken up and pre-tension of rod and line performed before work may continue in the opposite direction. This takes place after the system reaches max velocity, prior to the incept of deceleration loads.

[gordonjudd]

Tension imposed stretch on the line and rod deformation caused by wind drag must be removed.

Eugene,
Your description is too technical for me to grasp the point you are trying to make.

How do you see the tension in line as it relates to its change in momentum?

Gordy

[Eugene Moore]

Gordon,
All mechanical systems I'm aware of experience loss of motion due to drag. friction and mechanical deflection of imposed loads.
In the case of a line moving thru air there is drag imposed by air resistance and stretch imposed by tension. If you take 10 meters of line into an 20 meter per second air stream a force can be measured in tension. This is drag related and increases as wind velocity increases and as line length increases.
If you place your rod in that same airstream deflection and load can also be measured. This is the minimum amount of rod deflection required to account for drag. More deflection means the rod can apply load to the line increasing it's velocity. Less deflection than this is lost from work to air drag.
At max line velocity the line has max stored energy and becomes the system driver. If the fly line were nipped at this point in time the cast would continue at it's motion vector slowed only by drag and gravity. The rod however is still attached and though it isn't capable of adding energy it can deflect the line's direction vector. The rod is now taking energy from the line because it's incapable of continuing upward at the line's direction vector and it's deflection is now created by it's own air resistance. Some of this may be recovered by thrusting the rod upward. No more rod power but the caster may still contribute. The line is giving energy to the rod through the line. The deflection of the line also increases frontal area and drag on the line. Rod tip velocity is dropping reducing the rod air drag deflection allowing the rod to straighten while the line continues to deflect. Once the rod straigtens the line is now applying tension to the tip from the opposite side creating the loop and once again adding tension to the lower leg of the fly line.
This is all lost motion from the line's potential energy.

[gordonjudd]

This is all lost motion from the line's potential energy.

Eugene,
Do you mean KE here or are you referring to its PE due to gravity?

As noted in this discussion doesn't the fact that the rod is bent as it goes through counter flex indicate that it is still applying a force on the line (at least up to the point where the line starts going ahead of the tip) that is changing the line's momentum, not the other way around?

Gordy

[Eugene Moore]

Gordon,
The energy is kinetic.
The tension you observe is caused by a direction vector difference between the rod and the line. The line wants to contiue on the max velicity vector that the rod is not capable of following.
Rather than the rod adding energy it's actually pulling it out since the rod deflection is less than that necessary to overcome air drag.
The goal, I presume, is to achieve max launch velocity in a spring driven system. For this you apply max torque as the load is moving the opposite direction from the previuos backcast. The load is increasing as the loop is straightening. The upper leg adds mass to the lower leg as the loop progresses. If timing is correct rod is accelerated from a stop with maximum acceleration occuring as the line mass increases to max. Straight line in backcast direction. The line mass has dictated the proper timing point. This effectively couples the two forces, in opposite directions adding rod deflection.
The rod and caster continue to accelerate the line (with the rod heavily deflected. As the caster reaches max stroke the additional rod deflection continues to drive line velocity higher. When max velocity is obtained air drag is also at it's greatest and the rod lost motion is still present.

[sms]

An MPR study:
http://www.youtube.com/watch?v=5MrQSpMy6RQ

[xyl]

Cool video and great idea Sakke!

[gordonjudd]

Sake,
Thanks for posting a clever idea.

What do you see as the sources of the negative torque being applied to stop the rod rotation?

I see it as very similar to Tom's example. The reversing torque is applied by stopping (or slowing down) the rotation from the elbow and in your case the extra rotational acceleration you added prior to the stop with your wrist. As shown in frame grab (#325) below the butt continues to rotate with roughly the angular momentum it had at MAV.

There is some torque coming from the decreasing acceleration force on the line that reduces that angular momentum, but it is not sufficient to slow down the angular velocity very much. Thus the rod continues to rotate and leaves the back of the cup. The rod starts to unload at that point since it is no longer being accelerated.

That reduces the rod deflection force even more so the torque associated with that force (which also the acceleration force on the line) continues to get smaller and would go to zero at RSP1.

It is tough to see the bend in the rod, but going through the blurred images of the rod (there are two of them since this is an interleaved video) it appears the maximum bend was probably at a time in between the 1/50 second rate in this video. The rod butt was against the back of the cup in the odd frame and is leaving the back of the cup in the even frame.

After the deceleration from the wrist and elbow is applied to end the acceleration phase, the rod continues forward with a slightly reducing angular momentum until it impacts the front of the cup. The force impact with the cup (and its resulting torque) then stops the the rotating rod very quickly.

Do you see it the same way?

Also since you brought it up (and it is beginning to change the way I see the forces that impact the rod rotation) what is your view that the contact forces holding the rod to the rotating linkage in the robot are internal forces? As such they would simple oppose each other and not impact the work energy that is causing an increase in the rotational velocity of the rod.

That energy is coming from the torque over angle work energy done by the stepper motor as you can see from the sketch of the "true" free body in the robot. That is the true external force.

The same concept would hold for the stiff-wrist casting style exhibited by Chris Korich. In that case the external rotation force is coming from the elbow. The forces on the grip are just internal forces (such as they are in the robot), and could be ignored.

Thanks for taking the time to take some data on this. I am beginning to think our focus on the grip of the rod is missing the dominate torque forces that are mainly responsible for increasing the rotational velocity of the rod.

You would look at forces from the wrist rotation (and the grip) if the forearm was strapped so that the torque from the shoulder and elbow would not come into play, but (aside from a Tim Rajeff demonstration) I have never seen anyone cast that way. You certainly use a lot of shoulder and elbow rotation in your casting style.

Thanks again for having the interest to take your video. I think It shows the reversing torque does not necessary only come from the wrist as is generally thought.

Gordy

[sms]

Hi Gordy,

Just tried casting with the same setup and using way less elbow rotation and much more wrist, but from physical point of view, nothing changes.

I am not sure what you mean by internal and external forces. I would put the dividing interface on the grip since the rod is what we are trying to look at and understand it's behaviour and see how our input impacts it.

The retaining (=deceleration) and acceleration forces of course are applied via the grip, but they also go through the elbow, shoulder etc. So, in my opinion, the border between internal and external forces is where we decide to put them. But I if we are looking at the rod or rod and line system, we need to consider anything outside the rod or the rod and line system as external forces regardless where the rotation (or any other movement) occurs.

If one is to use a casting robot, one needs to run it without the rod to calibrate it for zero level and then use it with a rod and line.

I am not sure if I answered your question as I am not totally understanding your point or goal about the internal/external force divide. And I might have missed something too as I haven't been keeping my eye on this thread since my previous post. I did read your post above thou.

Cheers,
Sakke

[Paul Arden]

Would it make any difference if the cup was full of beer?

[Lasse Karlsson]

Would it make any difference if the cup was full of beer?

Good beer or bad beer?

[sms]

No, Paul since I am drinking red wine at the moment. Palinka thou, might be different.

Tomorrow is a national holiday here so Palinka really could be it. :p

Lasse, is there such a thing as bad beer?

(Yes there is)

[Lasse Karlsson]

Lasse, is there such a thing as bad beer?

(Yes there is)

Good thing you answered that one yourself Sakke :p

Otherwise I would know what to offer you next time

Cheers
Lasse

[gordonjudd]

Would it make any difference if the cup was full of beer?

Paul,
That would be an effective way to add some viscous damping to the rotation of the rod. Molasses in January would be even better.

It would also provide some refreshment if you did not mind filtering out some cork flakes through your teeth as long as it did not all slosh out when you stopped the acceleration of the cup.

Gordy