Poss. Symmetry Break?

[ME]

0° is down, 45° is the cross-section (enhanced in blue), red cross at 49° from vertical, and the "2 times" at 90° from vertical.

Video: https://www.youtube.com/watch?v=-W937NM11o8
Image-time: around 3:01

Furcurequs, I think it should indeed have been 41.4°.
Height h=radius * (1-cos(alpha)), and V=sqrt(2*g*h)
0.5*v --> 0.25*h --> cos(alpha)=1-0.25

[Furcurequs]

Thanks, ME.

I went back and watched most of the episode, and you've confirmed what I thought I saw.

They started the pendulum a little too high, then, for their 1X speed.

(I'm assuming, of course, that even they were trying to do calculations based on a simple pendulum and so weren't making any corrections for the actual shape of their pendulum bob and how it was attached to the swing arm and the actual location of the center of mass of the system. The initial center of mass of the "bob" itself looked like it could have actually been to the right of the swing arm attachment point, so a correction might have gone in the other direction. Anyway... )

(For a simple pendulum) The height should equal L*(1-cos(A)), where A is the angle measured from the vertical and L is the length of the swing arm.

If 90 degrees was the 2X speed at a full height of L, then for 1X speed the height should have been 1/4 * L.

1-cos(A) = 1/4
cos (A) = 3/4 or 0.75
Taking the inverse cosine of 0.75 gives 41.4 degrees. ...again, measured from the vertical.

So, they should have started the 1X test from below your blue line, and at 41 degrees, not 49.

Their test was still in the ballpark, though.

[Furcurequs]

It looks like you added your calculations while I was typing up my post. We must have been doing our calculations at the same time! ...and we did them the same, too!

We're in agreement, then.

[ME]

:-)

I couldn't have guessed the end-result though: what did they actually measure?

From a quick&dirty linear-fit I get, when starting with a 2.6 inch thick clay (at video-time 4:17):
Clay-thickness = 2.6 -1.823*Sqrt(height)

For 90° (2 speed) that's the fitted 0.777 inch
For 49° (experiment) that's 1.53 inch (close enough)
For 41° (1 speed) that's 1.70 inch (that's what they should have found)
For 0° (0 speed) that's 2.6 inch (no change)

And thus slightly more complicated than getting 1.53/0.777 equals about 2....

..or is that "linear-fit" just invented nonsense?

[Furcurequs]

I suspect the squishing of the clay cylinders was to just get some sense as to the difference in the energy of the collisions in a qualitative way. I'm no expert in squishing clay cylinders myself, so I don't know if there are any guidelines or any sort of standard ways to calibrate such things.

Their use of clay does remind me of experiments of Émilie du Châtelet, though, although she didn't use clay cylinders. Here is an excerpt from wikipedia about her experiments and her contribution to science (and note what should be a couple of familiar names to Bessler enthusiasts):

The Châtelet contribution was the hypothesis of the conservation of total energy, as distinct from momentum. Inspired by the theories of Gottfried Leibniz, she repeated and publicized an experiment originally devised by Willem 's Gravesande in which balls were dropped from different heights into a sheet of soft clay. Each ball's kinetic energy - as indicated by the quantity of material displaced - was shown to be proportional to the square of the velocity. The deformation of the clay was found to be directly proportional to the height the balls were dropped from, equal to the initial potential energy. Earlier workers, such as Newton and Voltaire, had all believed that "energy" (so far as they understood the concept at all) was indistinct from momentum and therefore proportional to velocity.

https://en.wikipedia.org/wiki/%C3%89mil ... C3%A2telet

[james.lindgard]

Gottfried Leibniz Leibniz used clay for his experiments. he measured the depth of impact. This is how he realized that Isaac newton over looked momentum. What he might have over looked is that a planet might not have momentum unless it is the product of another gravitational field acting on it. The earth's gravity gives a 1 kg weight 9.8 newtons of force when at rest.
One thing I have realized with weights when they go from moving downward to moving across bottom center is that momentum seems to be lost. I've wondered if a weight's motion changing relative to the flow of gravity has something to do with it. Of course, with inertia being realized as stress, that might help to explain where momentum goes when a weight changes directions.

[ME]

I thought Newton was basically the Momentum-guy, and Leibniz the Energy-guy..

But how does it work?
Clay is used a lot for impact-analysis, but still hard to find the actual mechanism (at least: I couldn't find a reference or guide).

So I'll try it myself:

From impact there will be Pressure on the clay.
From Blaise Pascal (around 1650) we know p·V/T remains constant (pressure * Volume/Temperature), let's say the temperature remains constant before and after (obviously not).
I don't know the pressure before (besides atmospheric), so I just replace the "remains constant" part with a delta.
Pressure is a Force per Area: p = F/A so we can rewrite ∆p·V = (F/∆A)·V.
As Volume equals Area times Length (V = A·L) we rewrite F·V/∆A = F·∆A·∆L/∆A = F·∆L, looks like the equation for Work: F·∆x
Because F=m·a, we rewrite F·∆L = m·a·∆L; We recognize this formula for Potential Energy: E = m·g·∆h Which we also happen to have because we drop that weight to cause that impact.
Thus height relates to the compressed length: m·a·∆L = m·g·∆h --> ∆L = g·∆h/a

The deformation of the clay was found to be directly proportional to the height the balls were dropped from, equal to the initial potential energy.

Because ∆x = ∆v*∆t, we rewrite F·∆L = m·a·∆L = m·a·∆v*∆t = m·v·∆v; integrated over time we get Kinetic energy: ½·m·v²

Each ball's kinetic energy - as indicated by the quantity of material displaced - was shown to be proportional to the square of the velocity.

So I just assume (?) the Stress factor on clay (the deceleration 'a') is constant, so the length (∆L) becomes an indicator for kinetic energy m·a·∆L = ½·m·v² --> ∆L = ½v²/a
Which is weird because they should have measured 1.98 inch *) for the 1xspeed instead of 1.5 inch.
Perhaps that 'a' is not constant.

---add--
*) That's for the 49° situation

Initial length=2.6 inch
Calibrated for the two speed compression 2.6-0.777 = 1.823 inch
When (∆L = g·∆h/a) then (g/a)=1.823, calibrated for ∆h=1
Thus when ∆h=¼ --> ∆L=1.823/4 = 0.45575 inch --> and measure 2.6-0.46= 2.14 inch

When (∆L = ½v²/a) then (½/a)=0.45575, calibrated for v=2 because 0.45575·2²=1.823
Thus when v=1 --> ∆L=0.45575·v² = 0.45575 inch--> and measure 2.6-0.46= 2.14 inch

eeh, makes sense?

[Furcurequs]

I doubt that squishing the clay cylinders is really comparable to dropping objects onto a sheet of clay.

I would assume the clay behaves dynamically like a highly viscous incompressible fluid.

To depress a spot on a sheet of clay, then, would essentially raise all the clay around it (or a significant amount of it), making it something like a piston problem. I could see how that could be close to a linear function of the energy.

I would say that compressing a cylinder of clay is a different sort of problem and would be a bit more complicated - possibly depending upon lots of factors like the actual geometry and the clay properties themselves.

If I were going to use the squishing of clay cylinders to try to measure collision energy, I would actually go in the other direction and first use the pendulum test to make my calibration curve for my (typical) cylinder rather than try to calculate it.

....or, of course, leave out the pendulum altogether and just drop my weights onto my cylinders from different heights.

[ME]

The earlier linear fit was indeed better.

With unknown formulas, I think I would still use a pendulum and try to shoot a heavy ball horizontally by hitting a vertical post to determine the factor of drop-height versus horizontal velocity in the hope horizontal distance determines speed and ignore the parabolic trajectory.
Then try to shoot it into wet clay, that should give the same result as a plain down drop into the same clay.
When shooting the same weight into the previous dent of the clay, one can record a cumulative effect and relate this to a multiple of velocity and/or weight just by equivalence - and not using clay-behavior, as long as it's consistent.
Which is basically how they determined the equivalence of the crash.

# I am beyond belief that there are no old school clay-experts on this forum!!

... I guess time to get back to MrV's topic.

[james.lindgard]

With clay, if a person knows how to calculate 3 pins on threads, they can calculate the displacement of clay in a piston effect. Still, knowing calculus like Newton and Leibniz would be preferred.
ME, with Newton, he was calculating a planet's mass and gravitational potential. Momentum seems to be a derivative of a gravitational field.
With Leibniz, he calculated both inertia and momentum.

[james.lindgard]

With a pendulum, the stress generated from inertia might be able to be converted into work. Like any system or process, it would take both time and work to consider something.

[MrVibrating]

Textual summary:
http://scienceblogs.com/gregladen/2012/ ... ollisions/

Thanks for that, together with the clarifications it's a helpful reference.

I'm still not entirely clear on the situation tho...

Bear with me (Grrr!) a little further..

If one car's stationary and the other has twice the speed, the relative velocity between them is equal to both cars closing at half that speed, so why isn't the energy?

IE. if a 1 kg mass at 20 m/s really does have 200 J (which it does), then upon what basis does nature differentiate between this and two 1 kg masses closing at 10 m/s? How does it 'know' which mass is 'really' moving, and thus how much energy there is, if motion is entirely subjective?

Maybe if we can determine what this deciding factor is, it'll turn out to be something we can manipulate via gravity / rotation..?

[MrVibrating]

...actually on 2nd thoughts, i think i can see a resolution - the static 1kg mass would be accelerated by 10 m/s = 50 J, the moving 1 kg mass would be decelerated by 10 m/s = 50 J, and because the net system velocity has now changed by 10 m/s the other 100 J has effectively disappeared..? Is that it?

[ME]

Perhaps the compression of a car can be replaced by a spring.

1 car going 50 against a wall compresses one spring.
2 cars heads-on at 50 compresses two springs - where their common mid-point remains stationary on collision.
1 car going 100 against a stationary car will compress its own and the others.
The spring doesn't care from which side it's been compressed, it just absorbs the energy (in these cases each one absorbs the same amount); although an actual spring would give it back.

[Mark]

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