Jim_Mich's beef with Ralph

[LustInBlack]

Waffles evg, a recipe for Waffles! 8]

[jim_mich]

The following boxed quote was the original. Please ignore it. I left it here only as a reference.

Variable definitions:
aº is the droop angle.
R is the Rope radius length.
GF is Gravity Force
CF is Centrifugal Force
GL is the Gravity Lever length.
CL is the Centrifugal force Lever length.
TG is the torque produced by Gravity
TC is the torque produced by Centrifugal force.
W is Weight
K is constant 0.000028403397 when using inches, pounds and RPM

Formulas:
R = 10
RPM = 200
W = 1
GL = R•Sin(aº)
CL = R•Cos(aº)
GF = W
CF = K•W•R•RPM^2
TG = GF•GL
TC = CF•CL

So...
TG = W • R•Sin(aº)
TC = K•W•R•RPM^2 • R•Cos(aº)

TG = TC (both torques must equal for them to balance)

So..
W•R•Sin(aº) = K•W•R•RPM^2•R•Cos(aº)

Divide both sides by W and by R gives...
(This is where I went wrong at first. I canceled out two R's on the right side.)
Sin(aº) = K•RPM^2•R•Cos(aº)

moving both angle to the same side...
Sin(aº)/Cos(aº) = K•R•RPM^2

Cadman reminded me that Tan and substitute for Sin/Cos, so...
Tan(aº) = K•R•RPM^2

Atn is the inverse of Tan so...
aº = Atn(K•R•RPM^2)

aº = 41.35 when RPM = 200, R = 10, W = 1

So in answer to Ralph:
If the mass is increased then weight droops down less, so the droop angle is smaller.
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 higher, so the droop down angle is less, and visa versa.

Variable definitions:
aº is the droop angle.
R is the Rope radius length.
GF is Gravity Force
CF is Centrifugal Force
GL is the Gravity Lever length.
CL is the Centrifugal force Lever length.
TG is the torque produced by Gravity
TC is the torque produced by Centrifugal force.
Rc is the Radius of weight from center. This is tha same as GL.
W is Weight
K is constant 0.000028403397 when using inches, pounds and RPM

Formulas:
R = 10
RPM = 200
W = 1
GL = R•Cos(aº)
CL = R•Sin(aº)
Rc = R•Cos(aº)
GF = W
CF = K•W•Rc•RPM^2
TG = GL•GF
TC = CL•CF

So...
TG = R•Cos(aº)•W
TC = R•Sin(aº)•K•W•R•Cos(aº)•RPM^2

TG = TC (both torques must equal for them to balance)

So..
R•Cos(aº)•W = R•Sin(aº)•K•W•R•Cos(aº)•RPM^2

Cancel both sides by dividing with Cos(aº) gives...
R•W = R•Sin(aº)•K•W•R•RPM^2

Cancel both sides by dividing with W gives...
R = R•Sin(aº)•K•R•RPM^2

Cancel both sides by dividing with R gives...
1 = Sin(aº)•K•R•RPM^2

Divide bothe sides by Sin(aº)...
1/Sin(aº) = K•R•RPM^2

Invert both sides...
Sin(aº) = 1/(K•R•RPM^2)

aº = ASin(1/(K•R•RPM^2))
This is not quite the same as Cadman's formula. Which one is right?

aº = 5.05º when RPM = 200, R = 10, W = 1

But when RPM^2•R > 1/K then this formula fails to work because 1/(K•R•RPM^2) becomes bigger than 1 which cause an error when trying to calculate the ASin.

[rlortie]

This post is aimed at all of interest and not to one particular individual!

If the rope length is changed then nothing happens. The weight stays at the same droop down angle.

This is the first of two parts of what I call the "crutch" of the problem! If the rope length has changed thus changing "R" and a given RPM is retained, then IMO velocity must also change.

If the velocity increases then the mass lifts higher, so the droop down angle is less, and visa versa.

To increase or decreased velocity"V" without altering "R" IMO you must forfeit your given RPM.

The fact that "R" will also be effected by droop angle will be addressed in the second part of the "crutch"

Does this make sense and am I right or wrong?

Ralph

[CADMAN3D]

Jim- cool picture... maybe tonight I'll work on figuring out how to do that (and bullets and superscripts for that matter!). We do disagree on the equation, and the answer, and I think I know why. From my quick calcs, I get aº = 5.03 degrees when RPM = 200, R = 10, W = whatever you'd like.

This evening, I'll try to describe why; I think it starts with some things on your diagram of forces that Newton might take issue with. With any luck, my response won't be compared with breakfast foods ;)

And my quick take on Ralphs questions- you have to be careful what you refer to as V, R, and rotational speed omega (RPM if you use different units). When something is rotating, V = R*omega, but only if the units agree (ie v= in/sec, r=in, then omega needs to be in radians/sec) (For reference, 1 RPM = .105 rad/sec) So only if the rope is completely flat (physically impossible- you'd have to spin it infinitely fast!) would R in this equation = Rope length. With real-world drooping, the R in V=R*omega would be the distance from the person spinning, horizontally out to the weight (sorry- no time for a picture right now!) But, I think the rest of what you are saying is correct- they are all related; you cant change one without changing the other. But it's just not always obvious what "R" represents, or V for that matter- what direction?! Sorry for the lame explanation- maybe more time tonight...[/code]

[rlortie]

Cadman,

My real world is based on blue collar layman terms. My objective is based on a given and retained RPM on/at the driving force. My use of the word velocity pertains to the mass at the end of the rope.

But, I think the rest of what you are saying is correct- they are all related; you cant change one without changing the other. But it's just not always obvious what "R" represents, or V for that matter- what direction?! Sorry for the lame explanation- maybe more time tonight...[/code]

We are getting there! :-) "R" represents the dimension of the radial path from center of axis to the gravitational center of the orbital mass. "V" represents the velocity (speed) of said mass . If droop angle changes yet RPM remains constant then "R" is effected thus "V" must change to maintain given rpm. please do not confuse RPM with velocity, RPM must remain constant.

Mass direction is a true circular pattern based on a level horizontal plane.

Keep in mind that this debate did not start with a "fly ball governor" but soon ended up there. I am not using a fly ball, but the concept and properties excluding any springs is exceptable and beneficial . At least this will confirm that the distance between drive point axis and mass are constant. No matter the droop angle, the distance between axis drive point and mass never change, but "R" (distance from axle) and its horizontal plane does.

I do not consider your explanation lame, You need not dazzle me with formulas and equations.

Some people have phobias, some are for spiders, for others it may be hight. It is well known here that my phobia is algebraic equations. A phobia I have been stuck with since my beginning freshman year in high school. Today if I could find that teacher still alive, I would throttle him!

Not unlike a blind person that develops other senses to compensate, I have learned to do likewise. Either by finding the answer in a book or "hands on" trial and error.

Ralph

[jim_mich]

"R" represents the dimension of the radial path from center of axis to the gravitational center of the orbital mass. "V" represents the velocity (speed) of said mass . If droop angle changes yet RPM remains constant then "R" is effected thus "V" must change to maintain given rpm. please do not confuse RPM with velocity, RPM must remain constant.

Just so that we are clear and in order to avoid confusion, can you look at my picture and confirm which dimension label is the one that you are calling 'R'? The reason for this question is that Cadman and I are using 'R' in our equation to mean the length of the rope, which is not the same as the horizontal distance from the radial path to the center. From your above quote it is not exactly clear which dimension you are talking about when you say 'R'.

Thanks in advance.

[CADMAN3D]

Hey Jim,

Before I get too far into deriving your way and my way and comparing, are TC and TG in your picture "flat" on the drawing page (perpendicular to the rope), or in and out of the page?

And you guys seem all right... you can start calling me Kevin. I'll update my profile. It was getting a bit weird not seeing my name, like I have some secret identity...!

[jim_mich]

Kevin,

Torque from Gravity (TG) is the torque force that develops and causes the weight to swing CCW downward and inward.

Likewise Torque from Centrifugal force (TC) is the torque force that develops and causes the weight to swing CW outward and upward.

Since our scenario involves a rope we don't actually get torque around the pivot point like when using a lever but the two forces will/do act against each other. They are the forces that determine how far the weight swings.

[CADMAN3D]

Hey Ralph,

(whew- two conversations at once- this is getting complicated! Good thing I have multiple personalities)

If droop angle changes yet RPM remains constant then "R" is effected thus "V" must change to maintain given rpm. please do not confuse RPM with velocity, RPM must remain constant.

We're on the same page and I agree- if you you have a rigid link connecting the (higher) swivel point on the axis to a weight, and you find a way to keep RPM constant while changing the droop angle, the velocity V of the weight at the end of the link will change. It has to- it's getting closer and farther from the center of rotation while the rotation speed stays constant.

However...You will need to apply additional force or torque somewhere on the rigid link arm to force it to the desired position- it would not balance itself there. It can only self-balance at one point for a given rigid arm length (previously described as rope length) and RPM, which is where all of these equations have erupted from and debate over sin and cos with respect to droop angles, etc. Maybe this is OK... maybe not?

-Kevin

[CADMAN3D]

Hey Jim,

This is where my problem with "Centripetal Force" (or centrifugal, but most people know better that to use that term any more) begins. As I mentioned previously, centripetal forces do not actually exist, but are a simple way of turning F=mA into F=0 by moving the mA term over to the other side of the equation, thus inventing a force term and saying that it is always pushing the object outward from the center of the circle. To continue justifying my blasphemy, there is no force pushing outwards at all; if the rope were to break, the ball would not turn 90 degrees outward and fly away, it would merely continue flying straight ahead. What appears to be a force pointing outwards is merely the fact that you have to pull inwards with that amount of force to keep the object from flying straight ahead!

Anyway, it would be a harmless simplification to create this imaginary force, except for situations like the one we're dealing with. You must never draw this force on a free body diagram- it does not physically exist, and cannot create any torques as a result. The rope has to pull against the value of this force only to satisfy the F=mA equation. However, it does not generate a torque that you can do a torque balance equation like your approach used.

The correct form of the free body diagram is attached. You could do a torque balance if you wanted, but you'd have to sum torques about the point directly flat with the ball, on the pivot axis and sum the torques. Getting the perpendicular distance to T would be challenging but I have no doubt it would balance the weight torque perfectly. A force balance is easier- set the sum of all F in a given direction = m*A:
1) In the vertical direction, A = 0, so summing all the forces in this direction leads to W= T*sin(a)
2) in the radial direction, F=mA, but A = omega^2*r where r = the true radius from rotational axis to weight and omega is the rotational speed. This leads to T*cos(a)=m*(omega^2)*r

And then you can mix and combine terms, etc, set T=T, and get the equations noted previously:
a = atan(g/((omega^2)*r)) = down angle
T = T = (m*(omega^2)*r)/cos(theta) = tension in the rope

Now I'll take it a step farther so that people can use more common units:
if r is the radius from rotation axis to weight, in inches
a is the down angle in degrees
T is the rope tension in lb
W is the ball weight in lb:

a = atan(35235/(RPM*r))
T = (W*r*RPM^2)/(35235*cos(a))
rope length = r / cos (a)
(be sure your calculator is using degrees for these equations!)

Hopefully I haven't messed up any of these numbers... it's getting late and here I am typing again :(

-Kevin

[rlortie]

Kevin,

We're on the same page and I agree- if you you have a rigid link connecting the (higher) swivel point on the axis to a weight, and you find a way to keep RPM constant while changing the droop angle, the velocity V of the weight at the end of the link will change. It has to- it's getting closer and farther from the center of rotation while the rotation speed stays constant.

However...You will need to apply additional force or torque somewhere on the rigid link arm to force it to the desired position- it would not balance itself there. It can only self-balance at one point for a given rigid arm length (previously described as rope length) and RPM, which is where all of these equations have erupted from and debate over sin and cos with respect to droop angles, etc. Maybe this is OK... maybe not?

Thank you! If everyone involved in this debate can accept this then I am going to frame it. Either way you have satisfied my present wildest desires. To me, it is like a "pie" formula, represented not unlike Ohms law. You cannot change one factor without effecting the others. That was and always has been my point.

I am now leaving the thread, and I leave you and Jim to continue without my interruption.

Jim,

It has been fun, I will see you on the next go-around. :-)

Ralph

[jim_mich]

This is where my problem with "Centripetal Force" (or centrifugal, but most people know better that to use that term any more) begins. As I mentioned previously, centripetal forces do not actually exist, but are a simple way of turning F=mA into F=0 by moving the mA term over to the other side of the equation, thus inventing a force term and saying that it is always pushing the object outward from the center of the circle. To continue justifying my blasphemy, there is no force pushing outwards at all; if the rope were to break, the ball would not turn 90 degrees outward and fly away, it would merely continue flying straight ahead. What appears to be a force pointing outwards is merely the fact that you have to pull inwards with that amount of force to keep the object from flying straight ahead!

Anyway, it would be a harmless simplification to create this imaginary force, except...

First of all, many times we are be told that centrifugal force is a fictitious force that does not really exist. Pay close attention to what Professor Julius Sumner Miller says between 2:45 and 3.32 in his YouTube "Physics - Centrifugal Force pt. 2" demonstration:

Now we must never say that there is a force radially outward on this thing pulling on the string.

Oh, wait a minute! I said something wrong. And I’m glad I did because it may lead some of you to catch me up on it. I said, “There is a force radially outward on this thing pulling on the string.� Yes, yes, yes. There is a force which the string exerts on the body. And there is a force which the body exerts on the string. But there is no force radially outward on the body.

So you see ladies and gentlemen, boys and girls, in the heat of the argument I could very well make an utterance, which the good people will quickly detect. And I’m glad I detected it so you will see there are occasions when I say things that are not right.
-- Professor Julius Sumner Miller

Note that the Professor says, "there is no force radially outward on the body." But the weight does cause a force to develop due to the constant rotation. Contrary to what many are taught to believe, this force is very real. But as soon as the weight is allowed to move away from its curved path into a straighter path then this force diminishes. We might say this force is only momentary. It is an inertial force and like all inertial forces if you use/consume the force then it only lasts until the weight is forced to stop moving.

This force can be used in a force diagram as long as we understand that the calculated force is only accurate when the weight maintains its radial distance.

Before we go on, let’s look at CF and other inertial forces from a relativistic point of view. When we look at straight linear motion it doesn't matter which object is considered to be stationary and which object is in motion. If your automobile crashed into another the results are the same as if the other automobile crashed into yours, at least until/before the pieces land on the ground. If we observe a rotating wheel in the same relativistic way then the inertial forces (centrifugal and centripetal) that develop from the rotation of the wheel and objects are just as real as the inertial forces that develop when one automobile crashes into another.

Of course when a wheel rotates then we always get CF. If you ride in an auto at a constant speed then you are not aware of straight line movement. But if you ride on a rotating merry-go-round then you are constantly aware of rotation because of the constant CF. This might seem strange because how do objects know that the wheel is rotating? Somehow the objects always know that they are rotating relative to some unknown background real-space matrix. I believe this is one key factor for making a PM wheel.
---------------------

So, onward to understanding the rest of your post. It's 1:45 in the morning. I pass until tomorrow.

[scott]

Once again Jim refuses to acknowledge the importance of defining a reference frame before attempting a valid force analysis. This is worse than his insistence that speed and velocity are the same thing. I give up. Good luck Kevin.

[ovyyus]

But if you ride on a rotating wheel then you are constantly aware of rotation because of the constant CF.

On a large rotating wheel I doubt you would be aware of the actual rotation. You would simply feel a constant gravity-like force (until one or more of the force dependent variables changes).

An interesting transformation occurs when the wheel rim is made to suddenly disappear. One moment you feel pinned radially to the inside of the wheel rim, the next moment you find yourself free of force and moving in a straight line that does not extend radially back through the rotation point of your wheel ride.

[jim_mich]

Scott,

Tell me, what are MPH, feet/second, knots. Are they velocities or are they speeds. If your car's velocity is 75 MPH then isn't that your speed?

That which you keep calling velocity is actually vector velocity. Plain velocity and vector velocity are similar but different. A vector velocity has two components, one is the rate of movement and the other is the direction of movement. Plain velocity does not have a vector component. It's very common to omit the vector component and simply state the velocity. Then the velocity and the speed are one and the same. In order to give values to vector velocity you must state both a direction and a velocity, both relative to some other object. Most of the time people omit the vector component and it is assumed that the velocity is relative to the observer. In which case velocity and speed become one and the same.

My old HP-48sx scientific programmable calculator lists a single catagory of speed units. There are no velocity units, only speed units.