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## Spanish experiment

Moderator: Torsten

Walter
Posts: 1498
Joined: Thu Jan 10, 2013 7:06 pm

### Spanish experiment

It is interesting how two people can read the same sentence and come away with a completely different understanding of what was meant.
Gee Gordy,

Are you trying to tell me that your statement:
tells me you did not understand what was going on with the force applied to the brick by the spring or the net force on the falling lead mass in this "Spanish" experiment.
actually has another meaning than I don't understand the interaction between the forces applied to the string/string/brick? Or are you going to respond to my question in the other thread? Are you trying to find the answer in matlab?
"There can be only one." - The Highlander.

PS. I have a flying tank. Your argument is irrelevant.

PSS. How to generate a climbing loop through control of the casting stroke is left as a (considerable) exercise to the reader.

Unregistered
Posts: 747
Joined: Wed Jan 09, 2013 10:22 pm

### Spanish experiment

gordonjudd wrote:
To me it looks because weight with spring accelerates faster it is going to gain more energy.

For Gordy the weight with the spring doesn't accelerate faster as you state; as written by him on February third (I miss the posts being numbered, like in the old board):
Esa and Aitor,
It is interesting how two people can read the same sentence and come away with a completely different understanding of what was meant.

My reading of Esa's comment was that the mass of the brick was being accelerated more with the spring and thus it ended up with a higher velocity and more KE.

Why does the brick have higher acceleration? Because the overshoot of the deflection in the spring produces a larger peak force on it than does the string.

What does that larger spring force mean to the net accelerating force acting on the lead mass as it falls? A larger spring force means the lead mass will have a larger upward force and thus its downward acceleration is reduced. That means it will have slower velocity when it hits the ground, and thus less energy that is lost to the system. That is where the added energy going into the brick with the spring comes from.

Gordy
Gordy,

Yes, maybe it is a misunderstanding on my part.
But, what about rotating a flexible rod butt asks for less energy due to its bending? I think that I am not misunderstanding Esa's words in this case.

Unregistered
Posts: 747
Joined: Wed Jan 09, 2013 10:22 pm

### Spanish experiment

gordonjudd wrote:
To me it looks because weight with spring accelerates faster it is going to gain more energy.

For Gordy the weight with the spring doesn't accelerate faster as you state; as written by him on February third (I miss the posts being numbered, like in the old board):
Esa and Aitor,
It is interesting how two people can read the same sentence and come away with a completely different understanding of what was meant.

My reading of Esa's comment was that the mass of the brick was being accelerated more with the spring and thus it ended up with a higher velocity and more KE.

Why does the brick have higher acceleration? Because the overshoot of the deflection in the spring produces a larger peak force on it than does the string.

What does that larger spring force mean to the net accelerating force acting on the lead mass as it falls? A larger spring force means the lead mass will have a larger upward force and thus its downward acceleration is reduced. That means it will have slower velocity when it hits the ground, and thus less energy that is lost to the system. That is where the added energy going into the brick with the spring comes from.

Gordy
Gordy,

Yes, maybe it is a misunderstanding on my part due to my use of the term "weight" always applied to the falling mass.
But, what about rotating a flexible rod butt asks for less energy due to its bending? I think that I am not misunderstanding Esa's words in this case.

crunch
Posts: 138
Joined: Sat Feb 02, 2013 6:58 am
Location: Kerava, Finland

### Spanish experiment

Sorry, I wrote weight when I meant falling mass.

This is how I understand rod function just now:

When a rod handle is mounted to an angle somewhere before RSP with 10g line and then "spring" is bent and line is released producing 10m/s line speed it is 0.5J

Then when you cast the rod "the lever" accelerates line to 20m/s which produces 2.0J to line and rod reaches same bend. Then when rod straightens lets assume it still accelerates line 10m/s making line launch speed 30m/s it is 4,5J

So only 0,5J is 2,5J of line energy in the end which is five times more.

Of course in real life it does not happen like that because when rod bends there is line energy loss and because of various other reasons but doesn't it approaches it?

Esa

Walter
Posts: 1498
Joined: Thu Jan 10, 2013 7:06 pm

### Spanish experiment

Or does the extra spring at the end accelerate the line from 20 m/s (2J) to 22.36 m/s (2.5J)?
"There can be only one." - The Highlander.

PS. I have a flying tank. Your argument is irrelevant.

PSS. How to generate a climbing loop through control of the casting stroke is left as a (considerable) exercise to the reader.

Unregistered
Posts: 747
Joined: Wed Jan 09, 2013 10:22 pm

### Spanish experiment

crunch wrote:Sorry, I wrote weight when I meant falling mass.

Esa
So what you meant is that the falling mass with spring accelerates faster than without spring? That was what I understood in the first place... and exactly the opposite of Gordy's explanations. Back to square number one.

As usual the more I read the less I understand.

Merlin
Posts: 1372
Joined: Wed Jan 09, 2013 8:12 pm
Location: France

### Spanish experiment

Hi everybody

Speculating on gut feelings is troublesome, here is the acceleration profiles corresponding to post number one:

You can pick up the point that fits with your conviction, but this is how it looks like for a rather well tuned experiment.

Merlin
Fly rods are like women, they won't play if they're maltreated
Charles Ritz, A Flyfisher's Life

gordonjudd
Posts: 1298
Joined: Sat Jan 19, 2013 11:36 pm
Location: Southern California

### Spanish experiment

When a rod handle is mounted to an angle somewhere before RSP with 10g line and then "spring" is bent and line is released producing 10m/s line speed it is 0.5J

Then when you cast the rod "the lever" accelerates line to 20m/s which produces 2.0J to line and rod reaches same bend. Then when rod straightens lets assume it still accelerates line 10m/s making line launch speed 30m/s it is 4,5J
and
Of course in real life it does not happen like that because when rod bends there is line energy loss and because of various other reasons but doesn't it approaches it?
Esa,
You are correct in thinking that the SHO is a very simplified model of what is going on in casting, but your idea that the maximum PE in the rod is only about 20% of the total KE in the line is certainly in the right ballpark. I think you have the right idea about what is going on, but I might differ with you about how much energy is applied to the line with the bending and unbending of the rod. In casting the caster continues to rotate the rod forward during the deceleration phase of the cast, so you will find the tip travels a bigger distance as the rod unloads.

I like to think of the basic physics of casting in terms of work energy. To get a better understanding of the work energy principle and how it impacts the change in the KE of an object you might want to read (or re-read) this Wikipedia article.

Back so soon? Now you should know that the work applied to an object is just equal to the integral of the force (dot) distanace curve where dot indicates a dot product. Thus the direction the force is applied relative to the path the object is moving is important in a three dimensional world.

The SHO model only has one dimension (either x or phi) so that makes it easier to compute and maybe understand. The basic thing to understand is that force applied to the line is related to the deflection in the rod. In the basic form of the model it is assumed that force is equal to:
f_spring=-k*x. The non-linear characteristics of the rod can be somewhat accommodated by adding in a cubic term to that Hook's law equation

In the real world some 30% of the deflection in the rod is just due to inertial bending. Consequently it gets complicated when you measure the deflection and then compute the actual force being applied to the line. We try to adjust for that by adding in an effective mass value for the rod to the mass of the line in the model, but that is only a first order correction.

The distance that force is applied is determined by how far the caster moves the tip of the rod from RSP0 to RSP1.

In you scenario let's assume in the clamped bow and arrow cast the rod was deflected 1 m before it was let go. Thus as the rod straightened it will apply a varying force to the line over that 1 meter distance. With no losses you would find the work energy from the force over distance integral would equal 1/2*(mo+m)*v.^2. In the loss-less case that energy would also equal the 1/2*k*x.^2 the initial PE in the rod. Thus some of that available energy went into increasing the KE of the line, and some went into increasing the KE associated with the moving mass in the rod.

Now lets assume that from RSP0 to MRF the caster moves the rod tip 3 meters and gets the same maximum deflection of 1m. The shape of the deflection vs distance curve over that path will not be an exact stretched, mirror-image replica of the force vs distance curve for the bow and arrow cast, put for our ball park estimate lets assume that was the case. That would mean the work energy done on the line would be 3 times larger than the value for the bow and arrow cast for that portion of the cast. As the rod unloads from MRF to RSP1 lets assume the caster rotates the rod such that the tip 2 meters (1 meter from rotation + 1 m for the deflection) . That will not have the exact shape of the f vs d curve for the bow and arrow either, but assuming it is close would mean the work energy applied to the rod as the it unloads in the cast would be 2 times the bow and arrow cast value.

That means the work energy applied to the line for that cast would be roughly 5 times the value for the bow and arrow cast for the same maximum deflection in rod.

The force for a given rod deflection is important to the force over distance work energy calculation, but you will notice the PE in the rod never enters directly into computing how much KE you could expect to be induced in the line.

Gordy

Walter
Posts: 1498
Joined: Thu Jan 10, 2013 7:06 pm

### Spanish experiment

Gordy

James, Merlin and I are waiting for you in the other thread to explain to us why we don't understand the SHO...

"There can be only one." - The Highlander.

PS. I have a flying tank. Your argument is irrelevant.

PSS. How to generate a climbing loop through control of the casting stroke is left as a (considerable) exercise to the reader.

Unregistered
Posts: 747
Joined: Wed Jan 09, 2013 10:22 pm

### Spanish experiment

Merlin wrote:Hi everybody

Speculating on gut feelings is troublesome, here is the acceleration profiles corresponding to post number one:

You can pick up the point that fits with your conviction, but this is how it looks like for a rather well tuned experiment.

Merlin
At the start the lead (falling weight) with the spring connection is getting a higher acceleration value, but soon after (as the spring gets stretched) its acceleration decreases to a value smaller than that of the lead without spring.

Consequently, for the same acceleration profile, the butt of a bendy rod asks for more force (torque) than a rigid rod.

Thanks Merlin. Very clarifying.