Looking forward to your two answers for the 1st time.Go play with your perpetual motion train.
To @write4u and to @citizenschallengev3.
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Your masters will not pay you your month’s salary because you fail to manipulate successfully the audience. Your masters may even beat you! You have to run quickly!
V2 = ? V3 = ? Answer these two simple questions.
Looking forward to your two answers for the 2nd time.
Bye, bye Will,
you have just lost all respect I might have had for you. As of now you are on ignore. Click!
To @write4u.
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You obviously failed to manipulate successfully the audience. Your masters will beat you for sure! You really have to run quickly!
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V2 = ? V3 = ? Answer these two simple questions.
Looking forward to your two answers for the 3rd time.
V2 = ? V3 = ? Answer these two simple questions.
Looking forward to your two answers for the 4th time.
V2 = ? V3 = ? Answer these two simple questions.
Looking forward to your two answers for the 5th time.
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We do not have for an object to conflict with anybody. On the contrary, we are simply trying to search honestly for the truth.
Inadequate information.
The forces on moving body of fluid mass are: 1. Inertial due to mass (ρ density) 2. Gravitational associated with specific gravity 3. Viscous related to viscosity (μ) 4. Pressure to water/fluidSteady flow vs. transient flow
Uniform vs. varied flow
Pipeflow velocity changes with CS area
Open channel velocity “V” changes with slope (s), CS area (A) and roughness (n)Shear stress:velocity relationship
τ = μ dv/dy
Basic Hydraulics - Rabi H. Mohtar ABE 325
Now, will, what does that have to do with your mind experiment and that drawing of yours?
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Your last post text has absolutely nothing to do with our zigzag concept. You are trying again to manipulate the audience in a clumsy and primitive manner. Your masters will not pay you your month’s salary because you fail to manipulate successfully the audience. Your masters may even beat you! You (as well as the other clumsy manipulator @write4u) have to run quickly! You and write4u both are absolutely unworthy persons! Shame on you!
To the clumsy primitive manipulators @citizenschallengev3 and @write4u.
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V2 = ? V3 = ? Answer these two simple questions.
Looking forward to your two answers for the 6th time.
42
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We might as well try to learn something while we’re here,
https://www.pipeflow.com/pipe-pressure-drop-calculations/pipe-friction-lossPipe Friction Loss Calculations
Flow of fluid through a pipe is resisted by viscous shear stresses within the fluid and the turbulence that occurs along the internal pipe wall, which is dependent on the roughness of the pipe material.
This resistance is termed pipe friction and is usually measured in feet or metres head of the fluid, which is why it is also refered to as the head loss due to pipe friction.
I graduated with a BA in “Aeronautics and Mathematics” - two subjects, one degree (no use ? )
So I’ve had some fluid dynamics studies. But about all I use it for now is to put my thumb on the end of the hose to squirt the dog.
Oddly enough, I now work for a “Gas and Fluid Solutions” company (to keep it generic enough) … in the IT department.
To @mrmhead and to @citizenschallengev3.
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We are talking about solid body mechanics. We are not talking about fluid mechanics. Solid body mechanics and fluid mechanics are practically two different fields of knowledge.
Please consider carefully and thoroughly again the link Perpetual motion and reactionless drive - YouTube
V2 = ? V3 = ? Answer these two simple questions.
Looking forward to your two answers for the 7th time.
AND LET ME REMIND YOU AGAIN: WE ARE TALKING SOLELY AND ONLY ABOUT SOLID BODY MECHANICS! WE ARE NOT TALKING ABOUT FLUID MECHANICS! PLEASE ALWAYS KEEP IN MIND THIS SIMPLE FACT!
Well, I was impressed with a commenter and noticed you use the same tactics with him, not listening and being paranoid:
DarkLight 2.1 3 weeks agoOops. You’ve made a basic mistake. Your “zigzags” do not generate mechanical resistance as you chose to eliminate friction .
In part 2 you had to increase the force acting on the zigzag side in order for the balls to arrive at the bottom at the same time.
F=MA
A=V/T
V=D/T
Remember those?
You changed D, (the distance the balls traveled) but kept T equal , therefore V has to change. If V changes and T still remains the same, A has to change. If A changes and the mass remains the same, then F must change, so F for the zigzag side cannot be equivalent to the straight side in that example.
So you started with 2 unequal systems, added friction to one side and then claim to have overturned a law of physics when heat is generated.
Sorry, you are wrong.
Peter Axe Responds:
To @DarkLight 2.1
You are the same clumsy and unskillful manipulator as @Malachi Wiens! Stop imitating pathological ignorance and pathological lack of understanding thus trying to manipulate the audience in a clumsy and primitive manner! You are a payed agent of the official science mafia and this fact is beyond any doubt! How much do they pay you? Shame on you!
----------------------------------------- Here are our last REAL experimental results. 1) Look again at our video above. 2) From 3:45 to 3:48 we have Ma = 1 kg, Mb = 4 kg and V1 = 0.1 m/s. (Consider only the “upper” zigzag device.) 3) From 3:59 to 4:01 we have Ma = 1 kg, Mb = 4 kg, V2 = 0.06 m/s and V3 = 0.01 m/s. (Consider only the “upper” zigzag device.) 4) (1 kg) x (0.1 m/s) = ((1kg) x (0.06 m/s)) + ((4kg) x (0.01 m/s)). The last equality unambiguously shows the validity of the law of conservation of linear momentum in this particular case. 5) (0.5) x (1 kg) x (0.1 m/s) x (0.1 m/s) > ((0.5) x (1 kg) x (0.06 m/s) x (0.06 m/s)) + ((0.5) x (4 kg) x (0.01 m/s) x (0.01 m/s)). The last inequality unambiguously shows the invalidity of the law of conservation of mechanical energy in this particular case. 6) How to reduce friction inside the zigzag channels? The answer is simple – by using permanent magnet slides. (There are literally hundreds of permanent magnet slide designs in YouTube and in Google.) Look at the links below for example: https://www.youtube.com/watch?v=KQH2UhHss6c https://www.youtube.com/watch?v=HXQqfIb-NXc https://www.youtube.com/watch?v=BQ4VGJCZUYE 7) The permanent magnet slide design: a) reduces friction (and the related generated heat) practically to zero; b) reduces the experimental error (due to friction and to the related generated heat) practically to zero. 8) And if the above mentioned experimental error is practically equal to zero, then this experimental error can be neglected (as it is much smaller than 1 %). 9) Alternatively you can use electrostatic levitation, rolling friction instead of sliding friction, etc. Besides modern tribology (this is the science, which focuses on sliding/friction phenomena) suggests a great variety of high-tech materials and/or lubricants’ which are also able to reduce sliding friction (and the related generated heat) practically to zero. 10)
In one word, you can carry out easily the above mentioned experiments in your garage or in any standard school laboratory (or by using any other simple DIY (DoItYourself) methods).
Firstly carry out the experiments and then try to discuss anything!
DarkLight 2.1 responds:@Peter Axe Congratulations-you managed to hurl insults while remaining completely confused about basic physics. -
You go on and on about reducing friction…who cares?
It’s a thought experiment-you can set friction to whatever value you want or eliminate it entirely AS LONG AS YOU ARE CONSISTANT ACROSS THE EQUATION.
You aren’t, and that’s: FAIL #1. -
You are still claiming that your zigzag configuration creates mechanical resistance but no heat. However, you removed friction from that side of the equation, remember? With no friction there is NO MECHANICAL RESISTANCE, and of course, NO HEAT. Literally the only difference between the sides of the equation is that you increased the distance with the zigzag path, meaning that you hade to increase the force required to push the balls through in the same amount of time. This is incredibly simple physics that you ignore and that’s: FAIL #2
Your math is utterly confused. You seem to be mixing up units across the board while making basic mathematical errors and your equations are not only meaningless, but you even get those garbled and meaningless results wrong (which hardly seems possible, but there it is…): FAIL #3
Your bizarre and paranoid accusations seem to indicate some form of mental aberration and have me wondering if you are simply a troll or perhaps on some form of medication. You invite critique, then hurl invective when people criticize your claims. Pathetic behavior for an adult, troll or not, and that’s: FAIL #4
There. You have been thoroughly debunked again and for the last time by myself as you are getting boring. One last question-if this experiment is so easy to recreate, why don’t you demonstrate it yourself and show the results?
Of course you won’t, you’ll just respond with more nonsense. Good luck and, as they say, don’t quit your day job.
Well at least Will/Pete doesn’t waste any energy trying to come up with new insults for different platforms.
Maybe his masters pay him for every time he uses a form of the word Manipulate.
But it does get a little redundant and repetitious. Not to mention he repeats himself.
And if I may re-Pete the answer: 42
But it does get a little redundant and repetitious. Not to mention he repeats himself. -- mrmThat redundant repetitious repeating is the worst kind.
Some people here keep constantly avoiding to answer clearly our two simple quesions.
Consider carefully and thoroughly again the link Perpetual motion and reactionless drive - YouTube
Ma = 1 kg
Mb = 4 kg
V1 = 1 m/s
V2 = ? (What would be the value of V2? How many meters per second would V2 equal to?)
V3 = ? (What would be the value of V3? How many meters per second would V3 equal to?)
Looking forward to your two answers for the 8th time.
Ma = 1 kg
Mb = 4 kg
V1 = 1 m/s
V2 = ? (What would be the value of V2? How many meters per second would V2 be equal to?)
V3 = ? (What would be the value of V3? How many meters per second would V3 be equal to?)
Looking forward to your two answers for the 9th time.
Ma = 1 kg
Mb = 4 kg
V1 = 1 m/s
V2 = ? (What would be the value of V2? How many meters per second would V2 be equal to?)
V3 = ? (What would be the value of V3? How many meters per second would V3 be equal to?)
Looking forward to your two answers for the 10th time.