Jim_Mich's beef with Ralph

[CADMAN3D]

LOL... now I've stepped in it!

I did misunderstand the technical issue being squabbled about. And with my graceful entrance here, I've made quite the *ss out of myself. But I'm plenty enough of a man to say when I'm wrong, and this it not likely to be the last time I'll have to say it. So let me start with some statements that I believe are true:

1) If the technical question related to a rope spinning overhead at a given "down" angle and rope radius, any weight, spinning at a given speed and a given rope length, will always have the same angle down from horizontal. (this assumes that aerodynamic drag is ignored, something that few would argue with!) The forces in the rope are significantly different, since the rope is at an angle down from horizontal, but it appears that this was not the question being debated. I started doing a diagram and equations to show this, but I'm betting that lots of equations with trigonometry probably would have a hard time convincing people who don't do a lot of trig!

2) The whole Jim/Ralph thing has been discussed ad nauseum at this point, and I think it's clear to everyone that both are excellent resources here, each in their own way, but they just don't get along! Perhaps now everyone can complain about me and start getting along better ;)

Lemee just climb back under this rock now...

[rlortie]

LIB,

For one, Ralph seem to give up on things quite fast without looking or thinking about the small and subtle details..

I experienced this once.

Thank you for sharing your opinion, I often wondered why our private communication dried up.

Calculating a design and scrutinizing it with a physical approach does not always require the same amount of time. One may be able to draw a conclusion by either approach through awareness of previous similar designs. Awareness that is gained through experience. I have always made a point to clarify my findings to a submitter. If I give up I say so and recommend another reliable source. Or I ask permission to consult with a third member that I feel has the skills that could prove of value.

When I tell a submitter that his or her design will not work and I include the reasoning behind my conclusion. I also make it clear that it is "my opinion" ... It is with my blessing if they should seek another opinion.

In my opinion, you have both your skills, and everybody here have their own skills.. Put that together and we are a strong team ..

Thank you for the recognition! And I agree; there is a wide spread of diverse skills represented on this forum. Putting them together is what this forum stands for, getting them to work together is another story.

The forum brings together those that respect and trust one another, those with ideas but without the skills or resources. Some only seek consultation or opinions on a design. What ever the need or what ever they lack is represented here in one form or another. That is; everything but a working proto-type :-)

Others join and never post, do you feel it would be fare to include them in sharing what the active members may someday produce? I do not. With that in mind I do not believe you will see more than small groups (usually two to three) joining hands in a joint effort. Three representing the submitter, the calculator, and the builder.

I have a partner-builder making up a group, we call upon on a third member if a simulation or calculations are required which is rare but has been known to happen. There are times when an idea is submitted ending up with a total of four involved, all approved and agreed upon by the submitter. I do not think you will ever see it get any better than this.

Ralph

[scott]

Now wait a minute cadman, you aren't saying you think Jim understands kinetics and the importance of defining your frame of reference are you?

Not taking sides, but this is still my biggest beef with Jim.

See:
http://www.besslerwheel.com/forum/viewt ... 0591#40591
http://www.besslerwheel.com/forum/viewt ... 0594#40594
http://www.besslerwheel.com/forum/viewt ... 0617#40617
http://www.besslerwheel.com/forum/viewt ... 0640#40640

Jim apparently doesn't believe that frame of reference matters when computing centrifugal vs centripetal forces.

I thought you were calling him on it like I tried to do a while back.

No?

Thanks,
Scott

[DrWhat]

We should all be praised for our persistance.

So many would have given up by now, but we continue on this road.

Each of us has our own strengths. I just build and build and don't post lengthy discussions as I am not an expert on the most technical aspects although I have a generally broad background in various areas.

I brag that I am clever in that I can look at things from many angles, different perspectives. I also work on the theory "try every possibility and hopefully you'll hit the jackpot".

There are many clever people on this site and you should all be praised.

Please keep the discussions and polite disagreements going. We need to keep throwing ideas around. If we dont then this site becomes just a Perpetual Motion Support Group. As a minimum I certainly need to know that I am not alone in this quest.

I agree that we tend to keep ideas to ourselves. I also agree that perhaps if we were a closed (non public) group and had some sort of binding agreement to share the spoils that perhaps much could be accomplished. As for 'who did most of the work', and who should be included, well that becomes the issue.

Otherwise we should simply make the Bessler phenomenon public and our aim should be to hound the media, history and science shows, in the hope that if enough people know the truth about Bessler that someone somehow will hit the jackpot. I'm not a big fan of this idea as we end up receiving no credit.

Any of us who miss out on 'the final glory' will be sorely disappointed. And it is clear that most of us will be, especially those who have worked so hard for so long. Better for all of us to share rather than most of us be left out.

So it's my same old idea. Lets find a way to unite. None of us has the answer so for now we have nothing to lose.

[jim_mich]

Cadman, you are correct that any weight spinning at at a given speed and a given rope length will (almost) always have the same angle down from horizontal.

I added "almost" because there are a few other criteria that affect the angle but are seldom mentioned. The first is the strength of gravity; the angle will be significantly different on the moon. It will also be slightly different depending on its location on the Earth and depending on the phase of the moon. Also the shape of the weight will affect the angle somewhat due to the radius of gyration being different with a large shaped mass verses a small shaped mass. Most of the time we ignore these other criteria as being more or less insignificant.

[rlortie]

Cadman,

First off: right or wrong I do not believe you earned the right to consider yourself an *ss... And I thank you for coming on board.

1) If the technical question related to a rope spinning overhead at a given "down" angle and rope radius, any weight, spinning at a given speed and a given rope length, will always have the same angle down from horizontal. (this assumes that aerodynamic drag is ignored, something that few would argue with!) The forces in the rope are significantly different, since the rope is at an angle down from horizontal, but it appears that this was not the question being debated.

In layman terms we have R=radius, M=mass, and V=velocity to reach or = a given down angle. What happens to the given down angle if any one of the other three factors is changed? Or define the results of changing one at a time. What must the other two do to keep the given down angle.

As I recall without going back and attempting to re-a simulate the problem I believe it was about one of the given above factors changing and the results of that change. For obvious reasons I am not digging up any bones, I wish to approach this from here!

I remain un-opinionated.

Ralph

[rlortie]

Not aimed at any particular person: Pardon me but, is this what you call an astute observation?

that any weight spinning at at a given speed and a given rope length will (almost) always have the same angle down from horizontal.

I cannot refute, argue or find reason for this statement. Sounds almost like tying an apple of given weight on a given length of string and it is going to stay at a given angle providing a given speed is maintained. Why should it not? Help me find what is considered pertinent here, please!

I am interested in learning what happens if one of the "givens" is changed.

Ralph

[ovyyus]

What is pertinent is the relationship between mass, inertia and gravity.

[jim_mich]

Ralph, what are you trying to say?

[rlortie]

Thank you ovyyus for the quick reply!

[CADMAN3D]

Well, thanks everyone for not tearing me a new one :)

First,
The equation the covers the motion of a spinning weight on a rope with a given down angle Theta is:
Theta = atan(g/((omega^2)*r))

where
theta is the down angle
g is the acceleration due to gravity (Jim- I did consider this constant... interesting point!)
omega is the rotational speed
r is the radius from the rotational center to the weight (shorter than the rope length!)

Or, if you're interested in the tension of the rope,
T = (m*(omega^2)*r)/cos(theta)
where m is mass

The hard part about using these is that the units have to agree- that's messy! But just looking at them lets you draw conclusions such as the previous one (angle is not dependent on tension).

THEN... to jump in with a position on centrifugal force/centripetal acceleration and further annoy everyone, I think the only way to tackle really tough problems that involve rotations is to consider that there are no centrifugal forces, centripetal forces, etc, acting on objects in motion. Let me explain...

If an object is not moving, you can sum all the forces acting on it in any given direction and they must equal zero. This is obviously called Statics. This is a subset of Dynamics, the study of moving bodies. The key equation in Dynamics is that F=mA for any given direction of moving objects (assuming mass is not changing!). Clearly, if A = 0, then the dynamics reverts to statics (sum of all F = 0). I apologize for the basic discussion but I'd like to come back to this in a second.

So, if you're standing in the middle with a weight on a rope, one direction would be straight outward. In this direction, a fraction of the rope tension is the only force acting on the weight in that direction. There are no other forces on that weight in that direction! So if you do F=mA, the F is merely T*cos(theta). And mass is obviously constant. However, there is an acceleration in that direction. In order to travel on a curved path, the weight must always accelerate inward, towards you. Turns out, this acceleration is (omega^2)*r (rotational speed^2 * radius); again- units are important! Again- if I'm Captain Obvious here.

So what happened is that at some point, somebody decided to simplify dynamics problems into statics problems by adding a "virtual" force (centripetal, centrifugal, whatever you want to call it) to the force side of the equation and bringing back the sum of F = 0 in any given direction.

That simplifies things a bit, but confuses other things. It can get messy when considering accelerations- particularly with rotation + translation, and it can be easy to overlook forces or accelerations and lead to faulty conclusions. But you can't go wrong if you draw in only the physical forces (including gravity, contact, ropes, etc, but no CF!) and make them = mA. Figuring out A for any given direction can be tough, but if you can get that right, you'll never make a mistake.

[ovyyus]

...But you can't go wrong if you draw in only the physical forces (including gravity, contact, ropes, etc, but no CF!) and make them = mA.

This will be interesting :)

Welcome to the fray cadman!

[jim_mich]

I usually end up writing a computer program to tackle these messy conditions because their is no simple formula that will give the exact angle the weight will droop down to when spun at a certain speed and rope radius. The simple formula for CF is CF = K•W•R•RPM^2 where the constant 'K' is 0.000028403397, W is weight as pounds, 'R' is radius as inches. When we figure in the droop angle 'a' and calculate the CF on one side of an equation and gravity force on the other side and allow for the droop angle then the formula becomes K•W•R•Sin(a)•RPM^2=W•R•Cos(a). The weight 'W' can be canceled from both sides. This is why the weight does not figure into the equation. So we end up with K•R•Sin(a)•RPM^2=R•Cos(a). I then use my HP calculator's solver routine to find an angle that causes both sides of the equation to be equal. For instance at a string radius of 10 inches and a speed of 200 RPM the angle 'a' would be 41.35º down from straight out.

In layman terms we have R=radius, M=mass, and V=velocity to reach or = a given down angle. What happens to the given down angle if any one of the other three factors is changed? Or define the results of changing one at a time. What must the other two do to keep the given down angle.

In this scenario the CF lifts the weight outward and thus upward while balancing against gravity pushing the weight downward and thus inward. (Note that this is slightly different from a fly-ball governor that includes spring tension.)

If the mass is changed then nothing happens. The weight stays at the same droop down angle. (Not Correct)
If the mass is increased then weight droops down less, so the droop angle is smaller. (Added)
If the rope length is changed then nothing happens. The weight stays at the same droop down angle.
If the velocity increases then the mass lifts to a higher less droop down angle, and visa versa. (Not Correct)
If the velocity increases then the mass lifts higher, so the droop down angle is less, and visa versa. (Added)

[CADMAN3D]

the formula becomes K•W•R•Sin(a)•RPM^2=W•R•Cos(a)

Now we're getting somewhere! I haven't checked the units to get K, but I'd bet they are right- they let you use speed in RPM (instead of radians/min), weight in lb (instead of mass in lbm or slugs), etc. But there is something funny with that formula... R should not be on both sides; it would cancel out and become as irrelevant as W, which I think we all agree must not be true.

If we pretend the CF is a real force (temporarily ignoring my previous assertion that F=mA, not F=0!), and it's balanced by the radial component of rope tension, we could sum forces in the horizontal plane and get:
Cos(a)=(K•W•R•RPM^2)/T

Now to sum forces in the vertical plane, you could write:
Sin(a) =W/T

Rearranging both equations to get T on the side by itself, you get:
T=(K•W•R•RPM^2)/Cos(a)
and
T=W/Sin(a)

And since T=T, you can make one equal the other:
W/Sin(a) =(K•W•R•RPM^2)/Cos(a)
(moving the (a) units to the same side of the equation)
Sin(a)/Cos(a)=W/(K•W•R•RPM^2)
And since Tan(a) = Sin(a)/Cos(a),
Tan(a) = W/(K•W•R•RPM^2)
And now canceling the Ws,
Tan(a) = 1/(K•R•RPM^2)

Now you can use your ATAN function, or Tan^-1 as some calculators call it, to solve directly for a:
a = Arctan(1/(K•R•RPM^2)

So the angle clearly depends on R and RPM, but nothing else. (except K, which is constant - it just makes all the units work out). Again, I havent verified that K = 0.000028403397, but I'd bet it does...

[evgwheel]

For those who can’t understand the above post, I think it is a recipe to make your own Kellogg’s Special “K� breakfast cereal.