Spanish experiment

[VGB]

However, a very stiff spring or a very weak spring would not produce opposite results. A very stiff spring would produce the same results as the string driven case. A very weak string would still add to the launch velocity, it just would not be as big as the improvement you would get with an optimum spring.

Gordy

Does your model tell you this is true?

or is James model wrong and Alejandros experiment wrong as well?

[Unregistered]

However, a very stiff spring or a very weak spring would not produce opposite results. A very stiff spring would produce the same results as the string driven case. A very weak string would still add to the launch velocity, it just would not be as big as the improvement you would get with an optimum spring.

Gordy

Does your model tell you this is true?

or is James model wrong and Alejandros experiment wrong as well?

That is not what Merlin states:

Yes, the string can be better than the spring and vice versa...

That is a small contradiction that I mentioned before but didn't get a meaningful reply.

[Unregistered]

In the earlier threads I am familiar with the idea was to have the same angular acceleration of the butt rotation with either the broomstick or the rod. In terms of force (torque in the angular acceleration case) that would require more torque to be applied to the flexible rod than to the broomstick to produce the same angular acceleration function...

I think Aitor just wanted to verify that the spring driven case would produce a higher launch velocity of the brick given the constraint that the input energy available (the PE in the lead mass) was the same. As noted above it takes more input energy to produce the same angular acceleration with a rod so that muddied up what was producing the larger line velocities.

Gordy

This summarizes my doubts perfectly, doubts that I haven't solved yet. I will try again.

If we consider that rotating both rods (flexible and broomstick) with the same acceleration requires more torque to be applied to the flexible one it is clear that the energy input for this rod is bigger than for the broomstick. More energy input equals more energy output (everything else being equal) so the flexible rod will give more line speed.

But in my experiment the force applied by the falling weight is the same in both cases, and, however, the spring combo gives a higher energy output than the string combo.

It seems to me that we have two very different propositions:
- Flexible rod vs. broomstick:
For the same angular acceleration of the butt more energy input with the flexible rod, so it results in higher line speed.

- Falling weight with spring vs. falling weight without spring:
Different linear acceleration of the falling weight, same force input, more energy output for the spring case anyway.

As an aside:
Alejandro has told me that he is working on some prototypes for the trebuchet experiment. If we eventually set up an scenario for that test and use a flexible rod and a broomstick (ideally of the same MOI), with the same falling weights as casters, the flexible rod will have a smaller angular acceleration than the broomstick, right?

[crunch]

It seems to me that we have two very different propositions:
- Flexible rod vs. broomstick:
For the same angular acceleration of the butt more energy input with the flexible rod, so it results in higher line speed.

- Falling weight with spring vs. falling weight without spring:
Different linear acceleration of the falling weight, same force input, more energy output for the spring case anyway.

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

Also a flexible rod butt needs less energy because rod bends.

Simply it looks to me that rod butt is "easier" to accelerate because rod bends so it gains more energy faster than broomstick. Then there is more energy available to keep rod butt moving and turning although rod straightens.

Esa

[gordonjudd]

It seems to me that we have two very different propositions:
- Flexible rod vs. broomstick:
For the same angular acceleration of the butt more energy input with the flexible rod, so it results in higher line speed.

- Falling weight with spring vs. falling weight without spring:
Different linear acceleration of the falling weight, same force input, more energy output for the spring case anyway.

Aitor,
Believe it or not, the reason for what appears to be different causes in the case of casting or your experiment stems from the same basic attribute of an accelerated SHO.

Why do you need to apply more torque to produce the same angular acceleration with a flexible rod as compared to a broomstick? Because the deflection in a flexible rod will overshoot the equilibrium deflection it would have for the applied acceleration. That overshoot produces a larger acceleration force on the line, and thus the torque (force*moment arm) required to achieve the same angular acceleration of the rod is greater. Here is a refresher overview of the torque required to produce a given angular acceleration if you need it.

That overshoot phenomena is just a fundamental property of a SHO. Lets suppose you have a mass that is suspended from a spring. The equilibrium deflection in that case will be such that f_spring= -kx (the spring constant ,k, times the deflection x) spring force will equal the f= ma (acceleration force applied to the mass by gravity where a = 9.81 m/s.^2) ,or weight of the mass. For example lets use a spring with a spring constant of 1 N/m and a 51gram mass whose weight is .5 N. The resulting deflection would then be x=.5N/(1N/m)=-.5m.

Now pull down on the mass until the mass is at -1 m and let it go. It will oscillate about that -.5m equilibrium point as shown below.

With an ideal mass-less spring with no damping that oscillation will go on forever about what limits? It turns out, In order for the sum of the KE and PE in the spring-mass system to have a constant value about its equilibrium value it will overshoot the equilibrium point by an amount that equals the initial deflection. Thus in this ideal case our mass would oscillate back and forth between -1 m and 0 meters and the acceleration force on the mass at x=0 would be twice the acceleration force of gravity. Similarly in a spring with an optimum spring constant value the resulting overshoot of the deflection in the spring will produce a peak acceleration force on the brick that is nearly double the force you would expect for string.

Remember the example given back on page 9 about what would happen if the rod (or spring) had an initial deflection that was equal to that new acceleration-induced equilibrium position prior to applying the acceleration? In that very special case there is no overshoot because there was no need for the deflection to change as the lead mass dropped. In that case the deflection force in the spring caused the brick to accelerate with the same constant value as the lead mass. Not by coincidence, that is the same force as would be produced by the string, and thus the force over distance work energy applied to the brick would be nearly the same and there would be next to no advantage for the spring as shown in that post.

However when there is no deflection in the spring when the mass starts to fall the relative conditions are much different. Now there is an offset from the equilibrium deflection associated with the acceleration of the lead mass. At t=0 there is no deflection in the spring and hence no immediate acceleration of the brick. Therefore it will lag the position it would have compared to being pulled by a string. However as the lead mass continues to fall, the spring stretches more and more and applies larger acceleration forces to the brick.

Does the deflection in the spring suddenly stop changing when it gets to the equilibrium point? No. It overshoots that point and continues deflecting until it reaches an equal offset value on the other side of the equilibrium point just as it did in the SHO example above. Thus in your experiment the spring started out with a deflection of zero that was .277 m less than the equilibrium value as calculated on page 9 of this thread. That means in theory, it will reach a deflection value of .554 m before the deflection starts to turn around and head back to zero.


That means the peak acceleration force it will apply to the brick (.544m *.295 n/m=.163 N) is twice as large as it would be with a string (.082 N). You have to solve the ODE to see how that deflection will vary with time, but when you do you can see that it will produce a force vs distance (or a force vs time) curve that is much different than the one you get with a string as shown below:


The same work-energy advantage applies when using a "matched" rod to a "matched" casting tempo. The peak acceleration force on the line produced by the flexible rod will be nearly twice the force produced by the broomstick and thus you will get much more line speed with a flexible rod as compared to a broomstick as can be inferred from the comparative work energy areas shown for the Paradigm cast below:

Gordy

[Walter]

Say Gordy,

I'm still waiting for you to explain in the other thread that I started why |F1| != |F3|. No need for pretty charts and complex algorithms.

[Unregistered]

It seems to me that we have two very different propositions:
- Flexible rod vs. broomstick:
For the same angular acceleration of the butt more energy input with the flexible rod, so it results in higher line speed.

- Falling weight with spring vs. falling weight without spring:
Different linear acceleration of the falling weight, same force input, more energy output for the spring case anyway.

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

Also a flexible rod butt needs less energy because rod bends.

Simply it looks to me that rod butt is "easier" to accelerate because rod bends so it gains more energy faster than broomstick.

Esa

Esa,

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):

The bigger that force is the more it will reduce the acceleration of the falling mass. You can see from the impulse plot that force is much larger when a spring with the optimum spring constant is pulling on the brick than it is with a non-extendable string. Thus the downward acceleration of the lead mass will be slowed more with the spring than it will with the string.

You also say that a flexible rod needs less energy to accelerate due to its bending. Well, what Gordy stated earlier is:

In terms of force (torque in the angular acceleration case) that would require more torque to be applied to the flexible rod than to the broomstick to produce the same angular acceleration function.

The more I read the more it seems to me that each participant here has a different view of what is going on. You asked some time ago:

Is there someone who does not believe that when spring is "in tune" with weight and brick it isn't more efficient than plain string?

For me it isn't a question of believing but of understanding. In this regard your understanding of why the spring connection gives more speed is exactly the opposite of Gordy's understanding of the issue. Obviously one of the respective points of view is wrong.

[VGB]

Gordy

I am glad to see that you are now accounting for the difference of an accelerated SHO and was with your explanation until this point:

The same work-energy advantage applies when using a "matched" rod to a "matched" casting tempo. The peak acceleration force on the line produced by the flexible rod will be nearly twice the force produced by the broomstick and thus you will get much more line speed with a flexible rod as compared to a broomstick as can be inferred from the comparative work energy areas shown for the Paradigm cast below:

The maximum amplification value of nearly 2 occurs at an "optimum" spring constant as you say but this occurs at the natural frequency of the rod. So if you walk into a shop and waggle a rod sharply you will get almost twice the deflection of a rod compared to where the force is applied comparatively slowly. if you now load that rod with a line such that it is no longer operating at its natural frequency but at a damped frequency, the amplification value is now related to the frequency differential of the natural and damped systems. If you used those values I would expect to see something like the values shown at table 2 of The Rod and the Cast

If you continue to increase the mass the amplification factor will drop below 1 as seen in Alejandros experiment.

Vince

[VGB]

Gordy

If you are using an accelerated SHO with a range of accelerations you should see a gain response something like this:



The top graph has a line, the bottom is the rod only. I have frigged the model for decelerations instead by varying zeta so that a value of zero equates to a dead stop.

Vince

[gordonjudd]

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