Generating more meaningful debate

[john.smith]

@All, simply put, inertia might actually quadruple because the velocity of the weight doubles. This might be what has been over looked.
1/2mv^2 =
.5 kg * 2m/s^2 = .5kg * 4m/s

force is 1kg * 2m/s, when the radius is reduced by half then it's
.5kg * 4m/s
inertia becomes .5kg * 4m/s^2 =
.5kg * 16m/s = 8kgf/m/s

I don't think anyone in here has considered conservation of angular momentum before. And with me, I am tired of all of the crap which is unnecessary. One of the reasons why I'm going to focus on finishing my build. I've tried going over some of this before and unfortunately taking the time to go over the math is necessary. I'm not the one who came up with it and it can be researched online if anyone wants to verify what I just posted. If nobody has the time to do that then no one should have the time for ignorant jokes either. Take the time to prove me wrong.

[eccentrically1]

what E1 doesn't understand about conservation of angular momentum is what everybody else has missed as well, if anything he brought up something that needs more or better explaining. I'll give it a try.
Inertia is mv^2/r. This simply means that if it's raius is reduced by 1/2 then it's inertia increases by a factor of 2. An example of this is a 1kg weight moving at 2 m/s with a radius of 1m. It has 2kgf/m/s. If it's radius is reduced by 1/2 to 0.5m
then it's inertia is 1kg * (2m/s^2 )/ 0.5 = 4kgf/m/s.
Why does the inertial force double when the radius is divided by 0.5 ? This is because the velocity of the weight doubles as well. This allows for a weight to increases it's force f = mv from f = mv to f = m(2v). Since the weight is moving in opposition to gravity it's relativistic mass would decrease. At the same time it's distance from the axis of rotation is decreasing as well allowing for an overbalance to be realized.

. @All, simply put, inertia might actually quadruple because the velocity of the weight doubles. This might be what has been over looked.
1/2mv^2 =
.5 kg * 2m/s^2 = .5kg * 4m/s

force is 1kg * 2m/s, when the radius is reduced by half then it's
.5kg * 4m/s
inertia becomes .5kg * 4m/s^2 =
.5kg * 16m/s = 8kgf/m/s

I don't think anyone in here has considered conservation of angular momentum before. And with me, I am tired of all of the crap which is unnecessary. One of the reasons why I'm going to focus on finishing my build. I've tried going over some of this before and unfortunately taking the time to go over the math is necessary. I'm not the one who came up with it and it can be researched online if anyone wants to verify what I just posted. If nobody has the time to do that then no one should have the time for ignorant jokes either. Take the time to prove me wrong.

I'm going to have to prove you wrong tomorrow. Good night.

[Silvertiger]

Inertia is mv^2/r

That's centripetal force, just fyi.

MoI (what you're looking for) is, for a simple pendulum or point mass (negating the rod), I = mr^2. Attach your weight to one end of a spoke going through an axle and it becomes I = (mr^2)[weight] + ((1/12)mL^2)[rod].

[ME]

Would you understand it better if I said 9.8 n-m's * 2 m/s^2 ?
Inertia is mv^2/r

eeh nope...
[n-m's] ??

9.8 newtons = 1 kg 9.8 n-m's = 1 kg @ 1 meter

"1 kg @ 1 m", could be anything and it's not unambiguous (it needs assumptions to make sense).
Let me demonstrate some possibilities:

• m=1 kg, h=1 m, g=9.8 m/s² --> E[P] = 1 Joule
• m=1 kg, r=1 m, v=? m/s --> F[cp] = (1·v) Newton
• m=1 kg, r=1 m --> I = 1 kg·m²
• m=1 kg, r=1 m, a[cp] = 9.8 m/s² --> v = √(9.8) m/s
• ...

(m·v²/r) is the product of Linear Momentum (m·v) and Angular velocity (ω=v/r), AKA the centripetal force. (for all our information)

(F=m·a), being a Physical force, is the acceleration of a mass in the direction of linear motion: F = m· ∆v/∆t.
Most important: There a kinetic energy factor here because of a change in momentum.

Because you talk about rotation, there's rotational stuff.
The centripetal force is a physical force in any frame of reference and a required tangent to deflect linear momentum into circular motion.
There's no linear acceleration, no change in speed or momentum and thus no exchanges in energy, and only changing the vector.

One could lose this centripetal but that Momentum (m·v) remains, or: it flies off without it with the same speed it had all the time.
In a rotating frame of reference the centripetal force prevents the mass from moving and makes it stationary, as like a normal force which is also a structural reaction force.
Perhaps strange but it's actually the fictitious Centrifugal Force being the inertial force in this rotating frame of reference (Coriolis is another inertial force).

Because that's in magnitude equal to (m·v²/r) you may almost be somehow correct, but not without making assumptions.
As you talk about reducing the radius, you likely mean "Centripetal force" [kg·m/s²] when you mention "Inertia".
Because "inertia" is defined as "tendency to maintain momentum" there's no factor of inertia because both linear as rotational momentum remain the same. Unless the Centripetal Force is cut then the Centrifugal "Force" is the one maintaining linear Motion.
But when you talk about "inertia" and rotation then there's also this Rotational Inertia (m·r²) [ kg·m²].
Hopefully you see "Inertia" as some defining word is not helping to clarify what you mean.

It's entirely possible I too made some minor mistakes here, so I hope I've provided enough context for the sake of "auto-correction". Otherwise return the favor and explain the error.
For a "meaningful debate" I think one needs to lower the assumptions and maximize meaning.

Unfortunately you didn't understand my little gest, perhaps I can elaborate a bit:

I haven't seen anyone take the time to consider why a weight moves inward when nothing is pulling it in.

Besides that shortsightedness being your own problem...

"Shortsightedness": ohw, it was not intended as an insult but a mere teasing reaction to your condescending tone ("being the only one"-rhetoric), irrelevant deflections ("because of situation"-rhetoric) and to your own "observed values" towards fellow members ("those who don't understand or attacks"-rhetoric).
As you say, I fully agree: "I am tired of all of the crap which is unnecessary."

Anyhow, if you had observed carefully there are constantly people considering and trying to move a mass inwards by (two recent examples) focusing on a change in MoI to provoke asymmetries or with build-attempts doing stuff with springs.
Look better and see a lot of accumulated time of considerations by a majority of active members.

[john.smith]

Inertia is mv^2/r

That's centripetal force, just fyi.

MoI (what you're looking for) is, for a simple pendulum or point mass (negating the rod), I = mr^2. Attach your weight to one end of a spoke going through an axle and it becomes I = (mr^2)[weight] + ((1/12)mL^2)[rod].

Since you're closer to the "why it matters". ME does know his science and in this instance we need to stay more towards "classical" science, ie. before Planck and Einstein, etc.
This is one aspect that might be considered more which ME mentioned >
m=1 kg, r=1 m, a[cp] = 9.8 m/s² --> v = √(9.8) m/s

And now for why all of this matters. If the radius a weight is rotating (swinging) is increased by 2x then it's inertia decreases by √.
This is why inertia's v is v^2. Then when it's radius is r/2 it's v increases because it's inertia increases. And with what pequaide has been researching, the one thing he might have overlooked is why inertia increases or decreases as the radius changes.
And as I mentioned, when a radius goes from 1 meter to 0.5 meter inertia increases. As a result velocity also increases. And with Bessler's wheel this does allow for a disc around the axle to be the principle mechanism that allowed for it to work.

to support my assertion there is this;
Answered by: Yasar Safkan, Ph.D. M.I.T., Software Engineer, Istanbul, Turkey

Question
In rotational motion, when the radius is decreased, does velocity increase or decrease?
Asked by: Lisa

Answer
There is no single answer to this question. The answer is, 'it depends on the force'. Let us see why.

First, lets take the case of a rock tied to a string. In this case, the centripetal force is supplied by the string, and is rather arbitrary, as long as the string doesn't snap.

If you change the radius by shortening the string slowly, the speed (velocity is not technically appropriate here, that's a vector) will increase

http://www.physlink.com/education/askexperts/ae154.cfm

pequaide has done energy producing experiments with inertia. Maybe he wouldn't mind finding out if when a weight increases it's velocity if it's amount of force increases as well ? That could be considered independent verification.

[rlortie]

Question from a dummy that uses hands on experience rather than math!

Start with a disk one meter in radius , measure or make a mark at any given point at the circumference then give it a spin.

Now. measure and make a mark 1/2 the radius in line with the axle and outer circumference mark. Give it a spin and note that both marks stay in line and are turning at the same RPM.

The outer mark travels a greater distance than the inner, therefore it must be moving at a greater velocity than the inner mark at 1/2 the radius and less circumference.

So how does everyone come to the conclusion that decreasing the radius and circumference of travel distance, increases the velocity. yet stay at the same RPM? I am not talking about the figure skater who reduces the circumference of total mass!

Ralph

[john.smith]

rlortie asked;

The outer mark travels a greater distance than the inner, therefore it must be moving at a greater velocity than the inner mark at 1/2 the radius and less circumference.

So how does everyone come to the conclusion that decreasing the radius and circumference of travel distance, increases the velocity. yet stay at the same RPM? I am not talking about the figure skater who reduces the circumference of total mass!

This is one reason why I asked if pequaide might be interested in independently verifying this. It would require accelerating a disc rotating perpendicular to gravity. Then when the weight is pulled inward does a scale record a heavier weight striking it ?
An MIT Ph.D. scientist acknowledged the weight does accelerate. I will give my hypothesis as to why this is.

@1m a 1kg weight has 1/2 * 1kg * (2m/s)^2 = inertia
0.5kg * 4m/s = 2kgf/m/s

This example shows 1/2 the radius = 2x the inertia and 2x the inertia might = 2x the velocity which if so then increases the inertia to 4x to be in agreement.
The acceleration could be the result of why the inertial force is increased. And possibly the simplest answer for this is the KE of the weight increases.
After this I'd have to wonder if Einstein's General Theory of Relativity might explain it as his theory allows for the precession of Mercury and of light passing our Sun from a distant star. Precession is a gravitational effect that pushes matter AWAY from the sun.

https://en.wikipedia.org/wiki/Tests_of_ ... relativity

p.s., the reason using a planet as a "slingshot" for satellites works might be for this same reason.

https://en.wikipedia.org/wiki/Gravity_assist

[rlortie]

I am not interested in general relativity or orbital gravity assist. Nor does a simple test need be done on a perpendicular axis to gravity.

If a weight is falling on the descending side of a wheel while pulling inward on an ascending side, then the ascending weight is going to loose velocity the closer it gets to the axis.

Is the velocity on a wall clocks second hand ticking off 60 seconds per 1 RPM turning slower than its shorter end protruding past the axle? Sorry but I think not!

Inertia and kinetic energy may change, but velocity does not speed up.

Check your math against elementary grade school powers of observation.

There is a railroad that makes a perfect circle from station "A" to station "B" located exactly 1/2 the distance. A train leaves station "A" and arrives at station "B" one hour and twenty-six minutes later. It then departs station"B" continuing on and back to station "A" The same distance, It is traveling at the same speed (Velocity) yet it takes eighty-six minutes to make the return trip: Why?

Ralph

[eccentrically1]

Ralph, your marks-on-a-disk example and second hand example aren't examples for proof of conservation of angular momentum. They are examples for proof of the relationship between radius and angular velocity in circular motion. And you are confusing the two different linear velocities of the marks on the disk, and the points on the second hand, with angular velocity. The two different marks have the same angular velocity, but two different linear velocities. The velocity of an object in circular motion is the product of its angular velocity and its radius. As radius increases, velocity must increase to cover the longer distance. I know that sounds obvious, but..
N.B. The moment of inertia (mr^2) for the disk and second hand don't change. This is the part that is different from CoAM. The mass maintains the same relationship with the pivot no matter where you mark the disk, or which points you pick on the second hand.

Conservation of angular momentum involves the ice skater example, and contrary to the clock hand example, as radius increases (arms extending), velocity must decrease.
N.B. The moment of inertia changes. The mass has a changing relationship with the pivot. IOW, the farther the mass gets from the pivot, the more force it takes to rotate it. Absent any external force the mass of the skater has no choice but to slow down.

I hope this is meaningful. And hopefully, it covers some of the same things that need unpacking in john.smith's post.

[rlortie]

eccent,

My marks-on-a-disk example and second hand example are not nor intended to be examples for proof of conservation of angular momentum.
The quantity of rotation of a body, which is the product of its moment of inertia and its angular velocity is not which I wish to convey or deliberate.
For now I wish to leave this for farther discussion dealing with Cf &Cp.

I do not intend to confuse the two different linear velocities of the marks on the disk, and the points on the second hand, with angular velocity. I am attempting to get members to think outside all such mathematical theories and so called laws. look outside the box, look for simplicity

My point is, as circumference reduces it takes less travel to cover the same number of degrees of a circle. So in real world the velocity of a weight pulled in while the wheel speed remains constant must decrease.

My "proof" example lies within a simple lever, imagine a lever eight feet long with fulcrum placed at the two-six foot position. As I push down on the six foot lever the two feet portion must raise at the same velocity otherwise the lever is bending!

As radius increases, velocity must increase to cover the longer distance. I know that sounds obvious,

Then is it not obvious that as a radius decreases velocity must decrease?
Do not confuse velocity with RPM, the latter is a positive no matter the radius,

EDIT:

Conservation of angular momentum involves the ice skater example, and contrary to the clock hand example, as radius increases (arms extending), velocity must decrease.
N.B. The moment of inertia changes. The mass has a changing relationship with the pivot. IOW, the farther the mass gets from the pivot, the more force it takes to rotate it. Absent any external force the mass of the skater has no choice but to slow down.

And here is my problem! Everyone wants to use the ice skater as an example. I am interested in the fact that the wheel does not speed up or slow down but maintains a constant RPM making the ice skater irrelevant. As such something else has got to give! The inward pulling weight looses its AM and slows in velocity!

https://www.khanacademy.org/science/phy ... net-torque

Ralph

[ME]

Take a weight on a string. Make this weight rotate above your head and keep it firm at the string.
Let's use this as a starting situation.
Now pull on the string, with your free hand, to get a smaller radius (say half).
You have provided work.
The radius is (half times) smaller with added centripetal over that distance, which is eight-fold now, a[cp] = v²/r = ω²·r.
The angular momentum (L=I·ω) remains the constant, because things keep rotating.
Because the Moment Of Inertia (I=m·r²) is reduced by one quarter, so does the angular velocity (1 rad/s = 30/Pi RPM) increases four-fold.
The rotational kinetic energy (E=½I·ω²=½m·v²=½L·ω) also increases four-fold as a result of that centripetal work.
The linear velocity (v=ω·r) increases two-fold, and so goes the linear momentum (p=m·v) when you let go of that string.
Demo: https://youtu.be/_eMH07Tghs0

Train: At first that train drove by the clock but then the company decided to make some minute changes to the schedule.

Talking about clocks and speeds (v=ω·r)
You could walk around the clock in one minute where the seconds-hand follows your position (1 RPM)
At 9.5 meters distance one could do a slow walk of about 1 meter per second.
Now try the same on a (semi-)circular running track, luckily the needed speed is still below the 400m-world-record.
That's the same situation as drawing on a rotating wheel.

Hopefully shown here, with math, that a changing radius and its rotation depends on the specific situation as provided or otherwise invites assumptions.
The math should make clear where words are interpretations... and a picture is better (if one could draw)... and a physical device even more (if one could build and touch).[/code]

[john.smith]

Decided to delete comment. Sadly KE is a part of physics and sometimes understanding that precession happens can help to understand KE. This is because a body's motion seems to effect it's KE. And being unwilling to see if a forum member is willing to independently verify what is being discussed does limit to that individual's wants.
I'll stick to my build. I can say it's a mechanical representation of a bygone theorhetical concept I found on paper ;-)

[eccentrically1]

My point is, as circumference reduces it takes less travel to cover the same number of degrees of a circle. So in real world the velocity of a weight pulled in while the wheel speed remains constant must decrease

Trying to unpack your thoughts:

Right, but the wheel speed doesn't remain constant. As the weight is pulled in, the wheel's moment of inertia increases, and the velocity of the wheel and weight also increase.

I am interested in the fact that the wheel does not speed up or slow down but maintains a constant RPM making the ice skater irrelevant. As such something else has got to give! The inward pulling weight looses its AM and slows in velocity!

The inward pulling weight conserves its AM and increases in velocity.

Then is it not obvious that as a radius decreases velocity must decrease?
Do not confuse velocity with RPM, the latter is a positive no matter the radius,

Yes I didn't think I needed to point that out ( for the linear velocity of circular motion. NOT conservation of angular momentum).

Believe me, I am not the one that's confused.

[Art]

Quote Mr V . -"Reactionless accelerations are the only possible explanation for the properties of Bessler's wheels."

-----

or possibly --

-- A 180 Degrees out of phase Reaction (ie a time delay of the Reaction ) to alter the forces direction in relation to its initial vector) !

[john.smith]

I am not interested in general relativity or orbital gravity assist. Nor does a simple test need be done on a perpendicular axis to gravity.

If a weight is falling on the descending side of a wheel while pulling inward on an ascending side, then the ascending weight is going to loose velocity the closer it gets to the axis.


Ralph

I did mention that the ascending weight will not accelerate but that doesn't matter. what you're not understanding is that a decaying orbit increases kinetic energy. Kinetic energy is related to velocity hence the v^2/r in kinetic energy which is 1/2mv^2.

2nd thing, I explained that the velocity of the ascending weight will not increase because of gravity. Instead what happens is the weight's RELATIVE mass decreases.

This is where it helps to have some understanding of physics. Also since a string is pulling the ascending weight in, this allows for the ascending weight to become closer to the axle than the descending weight. This creates an imbalance as well.

@all, when a string pulls a weight in, w = md. And if inertia is 1/2mv^2 then as the string pulls the weight in, the amount of work it does increases as well. Yet this energy does not come from the wheel. This is where "Free Energy" is realized. This is why Bessler called hsi retraction disc his "principle mechanism".
And this is something that can be tested. In over a decade not much if any progress on understanding how Bessler's wheel has been made. If I am successful then it will show that my having taken the time to learn physics and building is what my success will be based on. And I do have every reason to believe my build will work. This is because of my understanding of Conservation of Momentum as well as my understanding of the Conservation of Angular Momentum. And as I have said, basic testing can be done.
Since I am building I have no need to go back and test what I have already done in the past. Someone else who would be independently verifying what I am saying about conservation of angular momentum if they'd like to. Because after all Ralph, without actual testing by a neutral person it's back to word games like it has been for the last 10 years. And with me Ralph, I think this is where finishing my build is important to me. Like you and AB Hammer I have medical problems but then I did build when I had cancer and was being treated for it. It seems neither of you are willing to show any work. This could be why I have Bessler's wheel, I've been about seeing his work realized and then I can have the surgery I need.