Rotational KE in the rod - How do we increase angular velocity

Magnus · Post by **Magnus** » Sat Dec 17, 2011 1:10 am

Gordy

Ben has a lot of other clips loaded up in youtube from that tournament - maybe other are from more useful angles.

(Interesting estimate of the angle change at the elbow - I'd put the change of angle at far less than 80 degrees.)

sms · Post by **sms** » Sat Dec 17, 2011 9:22 am

I also do not think there is that much angle change in the elbow.

Here is Gerard's forward cast starting position from August 2010. It seems that his style was pretty much the same in CLA 2011 from where the Ben shot the video.
https://picasaweb.google.com/1171861....1860386

I am going casting today, unfortunately it is quite cloudy and thus dark so the vid quality is not going to be good, but I can try taking video of couple of different distance styles from perpendicular to the casting direction.

Merlin · Post by **Merlin** » Sat Dec 17, 2011 9:56 am

Sakke

It seems that there is a whole bunch of misunderstanding here. If we are looking at the thing from work point of view, the connecting forces on a grip are internal forces, the work is applied via them, the work is not done there

So what I think Gordy is after is that in energy standpoint of view, we can forget the parts that do not create energy and thus do not do work - and could be of one piece. So, I would think that both sides are right, it is just POV difference

Hi Sakke

If I took the example of the “finger only” cast, it was to intentionally have a case for which the separation of contact forces and “motion” forces is difficult. It was not to introduce a new “fingers only” type of cast. So let’s put it aside and use something else for the time being.

I like to take the example of pushing a marble on a table: if I push on the marble with a force, the marble moves with acceleration, and the force I apply is equal to its mass multiplied by the acceleration. This force is moving, and it means I develop energy to move the marble.

What about the contact forces on my finger? Is it something equivalent in intensity to the force which accelerates the marble? No, they are close to zero. It would be the case (contact force = force applied on the marble) if I glue the marble on the table. I use some force intensity, but the marble does not move and sends back to me that force. This is a static problem now, not a dynamic one.

The trouble we have is to identify the intensity of the force we get from the object we are moving, and this is a contact force. Imagine now that I push the marble on a vertical place upwards. What is the intensity of the contact force? It is the weight of the marble. If I move the marble upwards, this contact force will remain the same. I can apply any force or motion to the marble, the contact force will remain the same, and equal to its weight.

Now let’s use a spring in between the marble and my finger. We are back on the table. What is the intensity of the contact force between my finger and the spring now (the marble is at the other end of the spring) as I move it forward? It is no more zero; it is the force corresponding to the squeeze of the spring. The contact force is the effort coming from the spring and changes as the spring shape changes. What is the working force (let’s say the spring has no mass to make things simple, this is a current assumption for exemplifying physics problems)? Hum, not so easy to say: there is a force on the marble (mass*some acceleration), there are forces on the spring (contact forces, one with the finger, one with the marble), and I apply an acceleration to the spring but can’t really tell about the corresponding force. Again if the marble was glued on the table, things would be easy: my force intensity would equal the spring force intensity because nothing moves (back to static).

Then the energy comes to my rescue. If you analyze energies, you do not need to identify forces, it’s all in KE and PE. I know there is kinetic energy for the mass of the marble and elastic energy in the spring. I do produce both. I need to know the speed of the marble and the deflection of the spring to compute their value. OK, I can measure them or use a model to get these data, and then compute the elastic energy and the kinetic energy at any time. Since this total energy comes from my finger, and is the product of a force by the displacement of my finger on the table, here we are: dividing the total energy by the displacement of my finger at any time, I can compute the force intensity history. If I make the comparison with the contact force on my finger, I shall find they are different (and this is where we disagree with some of you in the forum). These forces would be the same only if the marble is glued on the table. Under such conditions KE=0, PE = elastic energy. My work = PE then, and the displacement of my finger equals the deflection of the spring. Bingo.

A bit long, but not too complicate I hope.

Merlin

Stoatstail50 · Post by **Stoatstail50** » Sat Dec 17, 2011 11:21 am

Hi Merlin

Is there any difference between a Contact Force, an Applied Force, a Normal Force, and a Constraint Force ?

Merlin · Post by **Merlin** » Sat Dec 17, 2011 2:02 pm

HI,

I start by the end: I'm no more familiar with Lagrangian mechanics (constraint forces) and I need a serious update in this field.

Normal forces = forces perpendicular to the direction of the motion (unless there is a "Lagrangian" definition)

For sure contact forces and applied forces are different. You can consider the two simplest examples I gave before. I can move a marble on a table with nearly zero contact forces and a significant acceleration. Applied force has no relationship with contact force here. If I consider moving the marble on the vertical plane, I can use the same forces than before (the table example), but the contact force is the weight of the marble under this condition.

I can understand the confusion between applied forces and contact forces, but do these examples bring some light?

Merlin

gordonjudd · Post by **gordonjudd** » Sat Dec 17, 2011 2:31 pm

What about the contact forces on my finger? Is it something equivalent in intensity to the force which accelerates the marble? No, they are close to zero.

Merlin,
I thought about this the same way since when we do free body diagrams and apply a force to a mass to see which way it is going to accelerate the force vector is always drawn with a single-ended arrow.

Walter sent me an example to show that Newton's third law still applies and there is an equal and opposite force pushing back on the finger while it is accelerating the marble.

As explained in Wikipedia:

The Third Law means that all forces are interactions between different bodies,[30][31] and thus that there is no such thing as a unidirectional force or a force that acts on only one body. Whenever a first body exerts a force F on a second body, the second body exerts a force Fprime on the first body. F and Fprime are equal in magnitude and opposite in direction.

Put very simply: a force acts between a pair of objects, and not on a single object. So each and every force has two ends. Each of the two ends is the same except for being opposite in direction. The ends of a force are mirror images of each other, one might say.

We were trained to use a one-ended arrow, because we get the correct answer for the acceleration on the moveable mass thinking about it that way.

As Dr. Lewin says, "Although the forces are equal, the accelerations are not: the less massive skater will have a greater acceleration due to Newton's second law"

Your finger is connected to your body which is connected to the mass of the earth. As a result the acceleration of your finger is a=f/(mass of earth) which would be infinitesimally small. Thus the net effect is the shared force accelerates the marble but has no negative acceleration on your finger or the earth.

The earth does not move, so that force does not do any work in the reverse direction and we get the right answer by considering the force arrow to be going only in one direction. That is the way engineers are trained to look at problems, but the physicists take a world view of what is going one where Newton's third law always applies.

See, Walter by being wrong I learned something. Thanks for your correction.

Normal forces = forces perpendicular to the direction of the motion (unless there is a "Lagrangian" definition)

That is a good way of looking at why constraint forces do no work in Lagrangian dynamics. It is easy to see that for the marble that is constrained to travel in a groove, the forces from the groove are normal to movement of the marble, and thus can do no work.

Then the energy comes to my rescue. If you analyze energies, you do not need to identify forces, it’s all in KE and PE.

Lagrange could not have said it better.

Gordy

gordonjudd · Post by **gordonjudd** » Sat Dec 17, 2011 3:37 pm

I'd put the change of angle at far less than 80 degrees.
and
I also do not think there is that much angle change in the elbow.

Magnus and Sakke,
I would be interested to quantify a bit more what "far less" and "not much angle change" means.

As noted in my post, because of the oblique angle of the camera relative to the casting plane it is difficult to estimate what the angle was between an extension of the line from the upper arm to the forearm, but I would like you to give a value for the angle you would see when the camera was positioned normal to the inclined plane formed by those linkages as shown below.

If the camera was moved to be in line with his shoulders and then elevated on a ladder so it would would be perpendicular to the tilt of the forearm what would the measured angle be? Is it 70 degrees, 30 degrees or some other "not much angle".

Gordy

Magnus · Post by **Magnus** » Sat Dec 17, 2011 3:46 pm

A guess due to the camera position but I'd put it at maybe 50 degrees - pure guesswork though.

Been looking at other clips and its damned difficult to come up with one camera angle which can provide accurate measurements for this - it's definitely 3D - very hard to imagine all that stuff in 2D.

Eugene Moore · Post by **Eugene Moore** » Sat Dec 17, 2011 4:13 pm

A viable dynamic system should only have one constraint.
Input energy is supplied from that constraint and delivered to successive linkages. All other restrictions are contact and are thus capable of recieving and delivering work energy.
Constraints are assumed as terminal and fixed as reflected energy and inefficiency will not be accounted for.
Contacts on the other hand can both recieve and deliver energy in their assigned degrees of freedom.
The grip of the hand is a spring contact with maximum assigned loads. If overcome these loads may allow freedom while storing energy which can be reflected back into the system.
Contacts are assumed to have to have inefficiency loss.
Each contact may have different load and deflection characteristics.
Returning power at it's assigned rate.
If your model is assuming fixed contact with no deflection you may see numbers that are excessive.

sms · Post by **sms** » Sat Dec 17, 2011 4:26 pm

Merlin wrote:I like to take the example of pushing a marble on a table: if I push on the marble with a force, the marble moves with acceleration, and the force I apply is equal to its mass multiplied by the acceleration. This force is moving, and it means I develop energy to move the marble.

What about the contact forces on my finger? Is it something equivalent in intensity to the force which accelerates the marble? No, they are close to zero. It would be the case (contact force = force applied on the marble) if I glue the marble on the table. I use some force intensity, but the marble does not move and sends back to me that force. This is a static problem now, not a dynamic one.

The trouble we have is to identify the intensity of the force we get from the object we are moving, and this is a contact force. Imagine now that I push the marble on a vertical place upwards. What is the intensity of the contact force? It is the weight of the marble. If I move the marble upwards, this contact force will remain the same. I can apply any force or motion to the marble, the contact force will remain the same, and equal to its weight.

The spring does lower the initial force due to inertia of the marble and thus store potential energy into itself. Once the force (acceleration) and the spring compression are in balance, the compression stops. With changing acceleration, the spring compresses or decompresses.

Hi Merlin,

I do not agree. There is always the equal and opposite force. And Newton's laws do not care if the situation is static or dynamic (until we get to velocities where relativity becomes important and I think, I do not know, that basic laws still apply, but the constants, ie mass and time, start to change).

If you push the marble on a table with acceleration of 9,81m/s^2 the table and the finger both exert the same amount of force on the marble, perpendicular to each other. The KE changes, PE does not. Let's say we do this for one meter length and the marble is more like steel ball and weighs 1kg. The KE is 1kg * 9,81m/s^2 * 1m = 9,81kgm^2/s^2 = 9,81J

PE is something, I do not remember how much it is, so just let's consider PE change - it does not change on the table.

In the second case (vertical pushing, going up a wall), the PE change is m * 9,81m/s^2 * distance. Same here, 1m up. The potential energy change is the same 9,81J - regardless of how we we get there as long as velocity (KE) is the same as when starting (if there are volocity changes, then there can be PE to KE and KE to PE changes since and we can go any route, way lower and way over the initial and end point).

sms · Post by **sms** » Sat Dec 17, 2011 4:36 pm

The angle in the elbow is perhaps around 60-65 degrees (looking at still shots from Gerard that I have with 18' rod).

Magnus · Post by **Magnus** » Sat Dec 17, 2011 4:44 pm

In the name of quantification - how many arms does the caster have in that picture?

I see the angle and think it gives a wrong impression Gordy - I'd put that angle at about 30 degrees from straight - rotate the forearm down by 30 degrees and try estimating the angle at his elbow - can't be done - can we leave it there please - much more interesting things going on atm.

If you want a puzzle have a look at Raymond http://www.youtube.com/watch?v=adpPqJgZYo0
Rotates from the shoulder to start the deliver stroke - bends his elbow a little - but I see no straightening. Must see if I can get footage of him casting some time.

Stoatstail50 · Post by **Stoatstail50** » Sat Dec 17, 2011 5:10 pm

For sure contact forces and applied forces are different.

Contact force is a generic term that covers different types of force, like the term "mammals" covers "whales", "wolves" and "wildebeest".

An applied force, for me, is a contact force. A normal force is also a contact force, and a normal force can also be a constraint force.

It is not the case, where I live anyway, that an applied force cannot deliver energy to the rod, on the contrary in fact.

If an applied force is a contact force, which I understand neither you nor Gordy believe to be the case, and a contact force cannot move the rod what sort of force it is that does....?

Edit: Here is Gordys link from post 12 in this thread internal external. to explain why the bead, or marble, moves....

The poke from a stick would be an external force and thus would do work via a force impulse change in the momentum of the bead.

Here is another link from the same site on contact forces. If you look at the two lists you will see that an applied force appears on the external forces list and also appears on the contact forces list.

gordonjudd · Post by **gordonjudd** » Sat Dec 17, 2011 7:20 pm

If you want a puzzle have a look at Raymond http://www.youtube.com/watch?v=adpPqJgZYo0
Rotates from the shoulder to start the deliver stroke - bends his elbow a little - but I see no straightening. Must see if I can get footage of him casting some time.

Magnus,
Of course there are style differences, and just from an energy standpoint some are better than others.

You do understand that the bigger the angle range the torques from our rotating joints are applied means you will produce more rotational work energy don't you?

If two people produce the same amount of torque and one applies it over 45 degrees and the other applies it over 90 degrees then the latter is going to produce twice as much rotational energy.

As Sakke said there is more to a good Spey casting than just producing high rotation speeds, but from an applied energy standpont the person who makes use of larger rotation angles from the shoulder, elbow and wrist is going to produce more 1/2*moi*omega^2 energy than someone who doesn't.

That simple principle from rotational dynamics gives a good hint of what you should aim for in producing the maximum line speed in a cast. Hammer throwers spin to get bigger rotation angles for a reason.

I am sure you could make a cast with no elbow rotation at all. But the fact is that will not produce as much work energy as compared to what could be achieved if you did.

Gordy

Merlin · Post by **Merlin** » Sat Dec 17, 2011 7:36 pm

Contact force is a generic term that covers different types of force,

OK, I shall read the documentation, thanks for the links.

It may be a question of definitions and the way I learned things like that more than 35 years ago.

One simple question: if I push on a marble and the marble applies the same force back to me, how can it move?
I'm sure that if I apply the same force on the wall, it will apply the same force on me in the other direction, because I shall not move the wall.

Merlin

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