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

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sms
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Post by sms »

Ok, here we are. I picked a stop where I was able to rotate/move backward so fast that the top of the handle detaches from the cup edge for a split second due to inertia forces until the spring of the wins and it goes back until the rod is so straight that the spring is loose enough for the inertia (and gravity) to win again -> moves to the other cup edge. The rod at halfway up (tip is outside the frame at that point) goes forward, then backward and then forward again. With a tip heavy rod with very soft butt and normal grip would probably ended up with S-curve with the same movement.

I can say that when casting with MPR in the cup and applying backward rotation much before RSP, the line velocity went down compared to normal cast with the same setup. The loop size was much smaller thou. With slower backward rotation than in the video did the same pretty much as in the video except the rod grip did not leave the cup edge before just before RSP.

http://www.youtube.com/watch?v=NESSvMQxklU
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sms
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Post by sms »

In this case of 18' rod from 2 weeks back there is definately backward torque before RSP since I get S-curve in it. The rod butt rotates in the same direction from start of the FC to the bounce so there is no negative rotation needed.

http://www.youtube.com/watch?v=ITNVE6GI2fQ
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Post by Merlin »

Sakke

Thanks for the videos, really interesting. It is not easy to see clearly the S curve but it seems to be large. That may be a consequence of the length of the rod. I cannot tell about the frequencies of such a long rod, especially when it is loaded. Several things may happen, like a low value for the second NF which could be easlily captured by the cast, or a non linear coupling between frequencies (another type of nightmare to understand). What is the line weight?

The towing car decelerates but the bungee is still trying to pull both cars towards each other. Thus, no negative torque.

The question is to define where that negative torque is applied. If the towing car decelerates, it is because there is a negative torque on the wheels. It can be made either by the brakes or more gently by the engine.


Magnus,

The quoted statement refers to a cast without a cup. For a cast with the cup, the stop is provocated by the cup. Then I think that the corresponding "negative torque" exerted by the cup on the rod handle is high and very short (a schock).

Merlin
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Post by grunde »

Thanks sms for posting those very nice casting in a cup videos.

The funny thing is that these videos show exactly the same thing as the video Tom posted and what can be inferred from my rod bend data. These data all show that there is no significant reverting (negative) torque from the caster on the rod during the unload (from max rod bend to rod straight position). In fact it's quite the opposite, in the beginning of the unload there is a significant positive torque from the caster on the rod (the caster is thus "trying hard" to load the rod).

After the unload the rod goes into counterflex and the caster do apply a significant reverting torque on the rod to stop it's rotation.

So no "myth" is debunked, since all this data in fact strengthens Server's statement (but one should of course be careful with the word never). :pirate

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Post by gordonjudd »

After the unload the rod goes into counterflex and the caster do apply a significant reverting torque on the rod to stop it's rotation.

Grunde,
I think Tom and Sakke's examples are much different that the forcing torque profile we normally apply in casting. What you are saying is true when there is no negative torque applied until the the sharp negative torque that comes from the impact with the heel of the hand or edge of the cup.

But I do not think the same thing applies for the more "normal" angular butt velocity profiles that are observed in a cast that result in less deceleration over a longer time to stop the rotation of the rod as it goes from MAV to RSP.

You set up as a convenient time reference point for the plots that I get on the predicted overall forcing torque where t=0 is at RSP. You can see that to get the angular butt deceleration that was measured for the Paradigm cast there was very significant negative torque applied before RSP as shown in the blue curve below.
Image

The torque that comes from the moment of the deflection force on the line is shown in the green curve. That is combined with the torque applied to accelerate and then decelerate the butt rotation of the rod (the red curve) to get the total overall applied forcing torque that is shown in blue.

The torque from the unloading rod reduces the torque required to stop the rod rotation, but some sizable additional negative torque shown in the blue curve is also required to get the rotation of the rod to come to a halt at around RSP.

Tom's cast had a very pointed angular butt rotation velocity followed by a very fast deceleration drop that stopped the rotation of the rod when it impacted the heel of his hand.

The profile in the Paradigm cast was typical of what we generally see, and thus applied the negative torque earlier and less violently than the ones you get from the impact in Tom and Sakke's examples.

Another more direct thing that can be measured is the unloading time of the rod. In another of Tom's video's that shows the bend in the rod it appears that it takes around 81 frames of a 300 fps video (.270s) for the rod to unload with his three-finger grip. The unloading time in the Paradigm cast where a deceleration torque was applied was a much shorter .122 s.

Thus as stated in the first post:
Unloading Time
The loaded frequency of the rod/line for the Paradigm cast was 1.2 Hz. As discussed here if the caster did not put on the brakes to stop the acceleration of the rod rotation then the time for the rod to unload by itself would be .5/1.2=.417 sec.

However, Grunde’s measurements found the actual unloading time was .122 second. Thus to get the fast stopping time we see in casting you would expect that some additional reversing torque must be applied to stop the rod more quickly.


But do not take my word on this. Just set the acceleration in your model to be zero instead of the -138.3 m/sec^2 value that we normally use for the Paradigm cast and see what it does to the unloading time of the rod.

In my version the unloading time goes from around .120 sec with a -138 deg/sec^2 rate to .176 seconds with no deceleration. I expect you will see similar values.

With no deceleration applied after MAV the speeds predicted by your model are also much different than the measured values as shown below.
Image
The torque from the varying acceleration force on the line would provide some deceleration, so this curve does not reflect exactly what you see in Tom and Sakke's examples, but it shows the car speed (which is related to the angular velocity) is still very high at RSP1. The spring in the rod will cause it to unload with no reversing torque applied by the caster, but it will not stop the rotation of the rod as it does so.

That is why there is so still much angular velocity of the rod butt at RSP1 in those videos. The rotation is not stopped until the grip impacts the side of the cup in Sakke's example or the heel of the hand in Tom's. Had a large deceleration torque been applied in addition to the reverse torque produced by the line then the rod rotation would have come to halt as we normally see in most casts at RSP1.

Gordy
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Post by Magnus »

Hi Gordy

I'm probably being dim but I'm having a helluva time trying to visualize this scene. (Maybe beginning to see why physics examples tend to use constant acceleration in explanation .)

Could a free body diagram (or two) help explain?
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Post by gordonjudd »

Could a free body diagram (or two) help explain?

Magnus,
The simple free-body I used to compute the forcing torque for the casting robot was described in post 35. The basic geometry used to get the forcing torque was given in post #3. Only the MOI of the rod and reel about the hub of the stepper motor was considered to get the forcing torque for just the rod, reel, and line.

As Sakke mentioned the actual torque required would be much higher because of the large MOI of the rigid linkages used in the robot.

That should be easier to understand than the rotational model I used that had two massless rods with different masses at the ends of the rod. The rod whose mass gave the same MOI of the rod and reel was driven directly by the angular forcing acceleration. The second having a mass equal to the mass of the line and the mo of the rod was driven by that forcing angular acceleration through a torsional spring.

Grunde's car/spring/brick model has been described in other threads. My version that can use arbitrary acceleration functions is described here.. The rotational model used very similar differential equations to the car/spring/brick model. I have given up trying to describe how Grunde's model works, so maybe Grunde can give you a better idea of how it simulates the rod spring deflections that result from different forcing function..

The mo of the rod is an equivalent mass that accounts for the KE in the portion of the rod that has a velocity component related to the spring velocity of the rod as described in this thread.. We generally use a mo value of .0031 kg for the Paradigm rod.

There is not much difference in the forcing torque values you get with the two models, so take your pick.

Both show negative torques are applied prior to RSP to get the unloading times we see in typical casts. Merlin's model that uses a power/omega approach to get the forcing torque gets similar results.

Gordy
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Post by Magnus »

Thanks Gordy

Likely a dumb question - what's the center of rotation for the recovering rod?
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Post by gordonjudd »

What's the center of rotation for the recovering rod?

Magnus,

It would depend on how the different joints were moving to rotate and translate the butt of the rod while the rod was unloading. If those joints were fixed and the fingers were just holding a pin that the rod could rotate about, then the ICR would be at the pin.

Tom's three-finger grip gets close to that situation when the angular momentum of the rod causes the rod to continue to rotate between the loose grip at this thumb and fingers until the impact with the heel of his hand.

The ICR applies to rigid bodies, so the ICR only shows the point the stiff butt of the rod is rotating about at one instant of time. Thus what the tip of the rod was doing as the rod loads or unloads would not impact the ICR.

In Tom's video it appears the ICR moves from his elbow, up the forearm, and to the point where he is loosely holding the rod between his thumb and fingers as it unloads.

Gordy
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