A tube with an open (1/2) tube would only cause drag and loss of force whichever way you want to look at it when executing the zig-zag. Note that the zig-zag motion involves loss of energy, rather than create it. There are 5 points where the system incurs energy loss.
- Blue input rod; 2. 2 blue sliding rods; 3. 2 blue balls in tubes. 5 points where energy loss is incurred when in motion.
What exactly is this experiment trying to prove ? That the path of the zig-zag tube is longer than the straight tube?
Where are any laws broken, except the false illustration which seems to ignore the most basic physics?
Figure 1. Energy losses in an incandescent light bulb are very large; most of the input energy is lost in the form of heat energy.[1]
Energy loss
Figure 1. Energy losses in an incandescent light bulb are very large; most of the input energy is lost in the form of heat energy.[1]
When energy is transformed from one form to another, or moved from one place to another, or from one system to another there is energy loss. This means that when energy is converted to a different form, some of the input energy is turned into a highly disordered form of energy, like heat. Functionally, turning all of the input energy into the output energy is nigh impossible, unless one is deliberately turning energy into heat (like in a heater). As well, whenever electrical energy is transported through power lines, the energy into the power lines is always more than the energy that comes out at the other end. Energy losses are what prevent processes from ever being 100% efficient.
Types of Energy Losses
Energy undergoes many conversions and takes on many different forms as it moves. Every conversion that it undergoes has some associated "loss" of energy. Although this energy doesn't actually disappear, some amount of the initial energy turns into forms that are not usable or we do not want to use.
[2]
Some examples of these losses include:
Heat energy, potentially as a result of air drag or friction. Heat energy is the most easily dissipated form of energy.
Light energy is frequently energy seen in combustion, and is a type of wave motion.
Sound energy is another type of wave motion caused by the vibration of molecules in the air. Like heat energy, sound is a type of energy that is generally lost.
Overall, the goal is to reduce the amount of energy lost to increase efficiency. As well, collisions that are inelastic refer to collisions where there is some "loss" of energy during the collision.[3]
For more information on inelastic collisions, see HyperPhysics.
What exactly is this experiment trying to prove ?
Yeah, I think I've lost that point too.
To @mrmhead.
=========================
Hi there,
Ok, I am reminding again what we have been talking about.
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- Look again at PART 3 of the link Perpetual motion and reactionless drive - YouTube . Focus on the “upper” zigzag case.
- Ma = 1 kg.
- Mb = 4 kg.
- Va’ = pre-zig-zag velocity of the blue component = 1 m/s = const.; Va’ = V1.
- Vb’ = pre-zig-zag velocity of the black component = 0 m/s; the black component is motionless.
- Va” = during-zig-zag velocity of the blue component = variable and comparatively difficult (but not impossible) to calculate.
- Vb” = during-zig-zag velocity of the black component = variable and comparatively difficult (but not impossible) to calculate.
8/ Vy = during-zig-zag velocity of each couple blue rod-blue ball along the Y-axis = variable and comparatively difficult (but not impossible) to calculate.
- Va”’ = post-zig-zag velocity of the blue component = 0.6 m/s = const.; Va”’ = V2.
- Vb”’ = post-zig-zag velocity of the black component = 0.1 m/s = const.; Vb”’ = V3.
- According to the third Newton’s law and to the related law of conservation of linear momentum we can write down the equalities
((Ma) x (Va’)) + ((Mb) x (Vb’)) = ((Ma) x (Va”’)) + ((Mb) x (Vb”’)) <=>
<=> ((Ma) x (Va’)) + 0 = ((Ma) x (Va”’)) + ((Mb) x (Vb”’)) <=>
<=> (Ma) x (Va’) = ((Ma) x (Va”’)) + ((Mb) x (Vb”’)) <=>
<=> (1 kg) x (1 m/s) = ((1 kg) x (0.6 m/s)) + ((4 kg) x (0.1 m/s)) <=>
<=> 1 kg.m/s = 1 kg.m/s.
- In one word, the values of Va”, Vb” and Vy are actually of no interest to us. Actually only the values of Va’, Va”’ and Vb”’ are of interest to us as these three values determine the validity of the third Newton’s law and the related law of conservation of linear momentum.
- The mass of each couple blue rod-blue ball is much smaller than the mass of the blue T-shaped component. For example if Ma=1kg, then the mass of each couple blue rod-blue ball must be equal to, let’s say, 0.0001 kg (and even smaller).
- In our numerous real experiments we strongly reduce friction and the mean experimental values of Va”’ and Vb”’ are equal to 0.5999992 m/s and to 0.0999997 m/s, respectively, that is, Va”’ = 0.5999992 m/s and Vb”’ = 0.0999997 m/s. The latter clearly shows that the experimental error (due to friction) is much smaller than 1 % and this experimental error is perfectly acceptable.
- Let me remind only again (it is written in the explanatory text of the link Perpetual motion and reactionless drive - YouTube ) that the zigzags generate a mechanical effect (let us call this mechanical effect the “X effect”), (a) which is absolutely identical and equivalent to friction and (b) which does not generate heat. (We take gravity and friction out of equation and consideration.) And really even if the mean experimental value of force of friction inside the zigzag channels is equal to 0.0000001 N (our last experimental result), then the “X effect” still remains and can be clearly observed as in PART 3 of the link above.
- And now we can easily calculate the pre-zig-zag and post-zig-zag kinetic energies of the bodies, taking part in the experiment. That is, we can write down the inequalities
(0.5) x (Ma) x (Va’) x (Va’) > ((0.5) x (Ma) x (Va”’) x (Va”’)) + ((0.5) x (Mb) x (Vb”’) x (Vb”’)) <=>
<=> (0.5) x (1 kg) x (1 m/s) x (1 m/s) > ((0.5) x (1 kg) x (0.6 m/s)x (0.6 m/s)) + ((0.5) x (4 kg) x (0.1 m/s) x (0.1 m/s)) <=> 0.5 J > 0.2 J
The last three inequalities unambiguously show a severe violation of the law of conservation of mechanical energy in this particular case.
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Please pay special attention to last item 16. This is what the above experiment (and the related theoretical consideration) has been already proved.
Looking forward to your answer.
Any comments related to our last post?
So by dialing the variables (speed, mass, length of zig-zag, zig-zag parameters … magnitude, frequency…) you should be able to better define your x-Factor … as in
x-Factor funtion = F(x)
Me = Mechanical Energy
Me = F(Me’)
I’d bet it tightly tied to the properties of the zig zag.
To @mrmhead.
========================
Hi @mrmhead,
Hi dear colleague,
Welcome to our team! It would be a real pleasure for us to work together with you!
Congratulations for your invention of the really excellent term “X-Factor function”! Sounds very good!
Yes, you grasped perfectly and entirely the essence of the zigzag concept! My respect to your qualification and expertise in mechanics!
Yes, you are absolutely right that our X-factor function would depend on many parameters like initial velocity Va’‘’= V1, Ma, Mb, number of the zigzags, shape of the zigzags, force of friction Ffr inside the zigzag channels (where Ffr varies from, let’s say, 1 N to 0.00000001 N (and to even smaller values) for example), etc. (Did I miss some parameter/variable?)
Do you have any idea for how to build this X-Factor function in a simpler and easier manner? And after that to vary with the related parameters as you suggest?
Looking forward to your answer.
@will34ab
You built the first apparatus. It should be easy enough to reproduce with different measurements.
To @mrmhead.
======================
Hi there,
Thank you for your reply.
Yes, there are even more than one experimental models of the zigzag concept. But these are limited to only a few experimental situations. And because of this it would be as if much better if a mechanical-processes-simulation computer program is involved in our research. (Such a program automatically calculates velocities, accelerations/decelerations, forces, etc., and automatically gives the related mathematical functions.) There are such programs, I am sure. Any idea/suggestion how to find such a program?
To @mrmhead.
======================
Hi there,
There are tens (not to say hundreds) of links, that contain mechanical (and any physical) systems simulation programs. Typical workable examples seem to be Simscape (which can be found in the link What Is Simscape Driveline? - Video - MATLAB & Simulink and in the link Simscape - MATLAB & Simulink) and FreeDyn (which can be found in the link http://www.freedyn.at/).
How to simulate our zig-zags mechanical system in the simplest and easiest manner? Need your help.
Looking forward to your answer.
To @mrmhead.
=====================
Hi there,
Where did you disappear, dear colleague? Some progress related to the mechanical systems simulation programs?
Actually we (or the simulation programs) have to express V2 and V3 as a function of V1, Ma and Mb (and simultaneously as a function of the number and of the shape of the zigzags). Something like that?
Looking forward to your answer.
I think we’ve come to the end of this discussion without more data.
I have neither the time nor the means to produce more data.
One static experiment does not disprove a theory.
To @mrmhead.
====================
Ok, I see. Our team’s collective desire was you to become a member of our team. But you have decided something else. Well, we respect your decision. Anyway if you change your mind, then you are always welcome to join our team.
Regards,
To all other members of this forum, who are interested in the topic.
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Well, may be this forum is not the most suitable place for discussing of the zigzag mechanical principle. If however somebody is still interested in the topic and/or has some questions and/or suggestions, then we (our team) are open to a discussion and/or to answering questions.
Regards,
