How to calculate this torque? (steel ball in a spiral tube)

[vxiaoyu18]

TL;DR Summary

Spiral tube torque calculation.

Want to let the steel ball move upwards at a uniform speed, how is the torque of the spiral tube calculated?

The parameters can be assumed by themselves.

**The spiral tube is fixed on the central shaft, the central shaft is mounted in the ball bearing, and the thrust F acts perpendicularly on the surface of the spiral tube.

**Parameters that may be needed: pi = π, Earth gravity coefficient = g, steel density = ρ, steel ball radius = r, bearing friction coefficient = u, steel central axis mass = M1, steel ball mass = M2, steel spiral tube mass = M3, spiral tube inclination = 45, steel spiral tube parameters: inner diameter = d, outer diameter = D, wall thickness = s, pitch = L, number of turns = n, height =H, helix diameter = D1, rotation outer diameter = D2, helix length = C.

 

[vxiaoyu18]

If there is no steel ball, then: F=(M1+M3)*(D2/2)*u

 

[mfb]

Is this a homework question? What did you figure out so far?
Conservation of energy might be useful to consider to find the torque induced by the steel ball.

 

[vxiaoyu18]

Is this a homework question? What did you figure out so far?
Conservation of energy might be useful to consider to find the torque induced by the steel ball.

This is a part of my machine, I can make that machine, but I don't know the calculation here.

 

[AZFIREBALL]

Is there no support of the shaft at the top end? (e.g. another bearing?)
Is the lower bearing the only support for the total assembly?

 

[vxiaoyu18]

Is there no support of the shaft at the top end? (e.g. another bearing?)
Is the lower bearing the only support for the total assembly?

To reduce friction, only one linear bearing is used. When the spiral tube is not big, it can be fully supported.

 

[vxiaoyu18]

I think that the steel ball stays at the lowest point of the spiral circle and may not produce the force that causes the rotating spiral tube.

 

[JBA]

Your diagram arrow on the ball is upward, indicating you are trying to lift the ball from the bottom to the top of the spiral; but, your above post implies you want the ball to rotate the shaft by traveling down the spiral. Which one of those scenarios is the one you are trying to accomplish?

For either case, you need to reduce its vertical downward weight into two resulting force components, one parallel to the shaft center line and the other perpendicular to the shaft centerline. The force parallel to the shaft centerline is the amount of force the ball will exert to either: rotate the shaft; or, resist the rotating of the shaft, if you are trying to lift the ball.

Either way, the torque from the ball = M2 x g x sin(L/D) x sin(45°) x (D2 - s - r)
Note:
1. The angle of the helix to the horizontal it is obtained by first applying the helix slope (L/D) to the shaft C.L. with the shaft vertical and then applying the 45° vertical tilt angle of the shaft to the helix slope.
2. The actual distance from the shaft C.L. to the point of contact of the ball O.D. to the tube I.D. may be slightly more than the distance from the shaft C.L. to the ball C.L. so the torque calculated above may be slightly conservative.

 

[mfb]

View attachment 248835

I think that the steel ball stays at the lowest point of the spiral circle and may not produce the force that causes the rotating spiral tube.

That diagram can only be exact if you have a circular tube around the shaft, for a helix it won't work, your system is inherently three-dimensional.
Consider an idealized case where there is no friction and the whole system is free to rotate: The gravitational force on the ball will certainly make the helix rotate such that the ball moves down. There must be a torque from the ball.

 

[vxiaoyu18]

Your diagram arrow on the ball is upward, indicating you are trying to lift the ball from the bottom to the top of the spiral; but, your above post implies...

My goal is to calculate the torque that the spiral tube will allow the ball to rise at a constant rate.
Many thanks to JBA for your answers.
As you explained:
F=(M1+M2+ M3)*(D2 / 2)* u+M2*g*sin(L / D)* sin(45°)*(D2 / 2 - s - r)

 

[vxiaoyu18]

That diagram can only be exact if you have a circular tube around the shaft...

Your understanding is correct. When the spiral tube has an inclined angle, the steel ball has a certain torque to the rotation of the spiral tube. My goal is to figure out how to calculate this torque.

 

[vxiaoyu18]

F =（M1 + M2 + M3） * g*（D2 / 2）* u + M2 * g * sin（L / D）* sin（45°）*（D2 / 2 - s - r）

 

[vxiaoyu18]

sin(L/D)

Are these 2 parameters the one?
pitch = L，outer diameter = D ？
------------------
Whether the torque generated by the steel ball on the spiral tube is equal to = M2 * g * sin (the angle between the spiral and the horizontal line) * sin(45°)*(D2 / 2 - s - r)

 

[JBA]

F =（M1 + M2 + M3） * g*（D2 / 2）* u + M2 * g * sin（L / D）* sin（45°）*（D2 / 2 - s - r）

Just as a note: Since you are turning this with some device to drive the helix and lift the ball, the inertia of M1 & M3 only affect the Torque on startup and not during continuous operation. Once the spiral assembly is rotating they will have no affect on the required operating torque of the unit other than any friction their cantilevered load might cause in the supporting bearing.
In that respect, you do not show any driving motor in your diagram so how do you intend to rotate the assembly?

 

[vxiaoyu18]

Just as a note: Since you are turning this ……

I will use an external force to make the spiral tube rotate at a constant speed to raise the steel ball. I still don't understand what "sin(L / D)* sin(45°)" means in your calculation formula. The horizontal angle of the helix should be: arctan (pitch/circumference of spiral diameter) =arctan(nL/πD2) ) .
That is: F =（M1 + M2 + M3） * g*（D2 / 2）* u + M2 * g *sin(arctan（nL / πD2）)* sin（45°）*（D2 / 2 - s - r）？

 

[Baluncore]

One turn of the helix will lift the ball, h = L * Sin(45°)
Knowing the mass of the ball, m, the energy required for one turn can be calculated.
PE = m·g·h;
Power = rate of energy flow = PE / turn / second.
Torque = power / angular velocity.

 

[vxiaoyu18]

One turn of the helix will lift the ball, h = L * Sin(45°)……

This machine cannot determine its power, so I can't calculate it with power. I look for how much resistance I need to overcome to spin it, which is the starting torque.

 

[vxiaoyu18]

Just as a note: Since you are turning this with some device to drive the helix and …

I will use an external force to rotate the spiral tube at a constant speed to raise the steel ball. I still don't understand what "sin(L / D)* sin(45°)" means in your calculation formula. The horizontal angle of the spiral should be: the inclined height of the spiral tube / the length of the spiral = H1/C.
That is: F = (M1 + M2 + M3) * g * (D2 / 2) * u + M2 * g * sin(H1/C)*(D2 / 2 - s - R)?

 

[vxiaoyu18]

 

[JBA]

Note:
1. The angle of the helix to the horizontal it is obtained by first applying the helix slope (L/D) to the shaft C.L. with the shaft vertical and then applying the 45° vertical tilt angle of the shaft to the helix slope.

I might have been in error using D (the helix O.D.) instead of D1 for a spiral tube, L (pitch)/D (thread O.D.) represents the slope of a thread from the crest of one thread to the crest of the next at the thread; and, therefore establishes the slope of the thread from one turn of the spiral to the next turn of the spiral relative to the shaft centerline; and 45° is the tilt of the combined spiral and shaft centerline to the horizontal.
The less the tilt of the shaft the less torque required to move the ball along the spiral but also the more rotations and longer spiral length required to lift the ball from one height to the next.
The work required to lift a ball from one height to the next is the same for both cases.

 

[vxiaoyu18]

I might have been in error using D (the helix O.D.) instead of...

I think I can understand what you mean, which is to calculate the actual Angle at which the ball is balanced when it rises on the inclined plane. Based on this Angle, the initial rotational resistance can be calculated, which is very close to the result of my experiment.
Thanks again.

 

[Baluncore]

If the helix is not rotating, the ball will settle at the lowest point of one helix turn. That point will be offset horizontally from the helix axis. The mass of the ball at that offset represents a static torque.
Is that the torque you require?

Without friction, that torque will drive helix rotation downwards, until the ball escapes at the lower end of the helix.

 

[vxiaoyu18]

If the helix is not rotating, the ball will settle at the lowest point of one helix turn...

Yeah, whether the ball goes up or down, I want to figure out how to calculate its initial torque.

 

[vxiaoyu18]

Without friction, that torque will drive helix rotation downwards...

I just don't know how to calculate this torque because the force analysis is too complicated for me. Can you show me how to calculate?

 

[Baluncore]

I think your first question must be; exactly where does the ball rest in a static tube?
Do you know that answer yet ?

This problem is related to helical-tube positive-displacement pumps. I see two fundamental pump questions as being of general interest.

1. I want to know the volume of liquid pool or plug that can be held in a part turn of the coiled tube before it flows back over the crest and down the tube.

2. I want to know the x & y position of the deepest point in the liquid, relative to the axis, for you, that is where a static ball would sit. Knowing that position makes it possible to calculate the initial static torque. The additional dynamic torque needed to advance a ball, or water, can then be solved using conservation of energy.

A helical-tube positive-displacement pump can also be configured as a turbine to convert potential energy into rotational energy. Your "ball in the tube" problem is a special case of the general geometrical problem.

At this stage I have been looking into making a general numerical model, rather than for an algebraic solution. Later I will be able to check any algebra against the numbers. My interest is in the water turbine, or the generation of a depression or compression of air from a multi-turn pump. This has analogy with a potential multi-stage tromp.
https://en.wikipedia.org/wiki/Trompe

Sorry, but I was distracted for a while by G.H. Mortimer, 1988; thesis “The Coil Pumps”.
https://repository.lboro.ac.uk/articles/The_coil_pumps/9454172I am now momentarily deliberately ignoring the complexity of differential pressure on each “manometer” plug that allows inlet suction and positive pressure possible in advanced spiral or helical tube pumps.

 

[Baluncore]

I believe I can now identify the position of the dip that a ball will rest in.
I need numbers for these four parameters;

1. Pitch of the helix; = axial advance per turn.
2. The angle of the helix rotational axis, measured from the vertical.
3. Radius of the helix, that the central axis of the tube follows.
4. The internal radius of the tube.

 

[Spinnor]

To me it seems there are only 3 important numbers in this problem to 1st approximation,

vertical distance the ball will drop, distance the ball is from the central axis of the spiral, angular rotation angle of the spiral tube.

Can we equate torque times angular rotation of the spiral tube with energy the ball loses when it drops some distance d, mgd?

 

[Baluncore]

Can we equate torque times angular rotation of the spiral tube with energy the ball loses when it drops some distance d, mgd?

Energy can only solve the dynamic situation when the helix lifts a ball or is driven by a ball.

If the tilted helix of tube is held so it will not rotate, then the ball will rest at a minimum on the tube inside surface. In that static situation there is a torque generated by the mass of the ball offset from the helix axis. Without movement there can be no energy expended.

Where in a turn of the tube will the ball rest? and what will that static torque be?

 

[Spinnor]

I did not read carefully, as normal, and solved a different problem (but maybe related?). If we let the tube rotate freely and allow the ball to drop in the gravitational field then the torque is given as I suggested? But is not the reverse problem raising the ball by rotating the spiral tube just the time reverse of the problem I tried to solve?

Thanks.

 

[JBA]

Ignoring any energy loss due to the shaft supporting bearing friction, the energy to lift the ball with your device once it starting rolling inside the spiral is equal to the PE of the ball at its full lift - PE of the ball at its elevation entering the bottom of the tube.
i.e. M2*g* (h2-h1).

 

[vxiaoyu18]

I believe I can now identify the position of the dip that a ball will rest in.
I need numbers for these four parameters;

1. Pitch of the helix; = axial advance per turn.
2. The angle of the helix rotational axis, measured from the vertical.
3. Radius of the helix, that the central axis of the tube follows.
4. The internal radius of the tube.

Let me just reimagine the problem for the sake of math.

Gravity coefficient g= 9.8n /kg;
π= 3.14;
Density of 304 steel ρ =7930 kg/m³;
Rolling friction coefficient between steel ball and stainless steel material u₁=0.0015;
Friction coefficient of deep groove ball bearing u₂=0.0015.

Steel ball:
Radius r₁= 0.045m;
The quality of the m₁ = 3.03 kg;
Gravity G₁ = 29.65 N;
Radius of rotation of steel ball in tube is r₂=0.064 m.

Center axis:
radius r₃= 0.005m;
The length of the L₁= 2.8 m;
The quality of the m₂ = 1.74 kg;
Gravity G₂= 17.08 N

Spiral pipe (304 stainless steel) :
Inner diameter d= 0.1 m;
Outside diameter D₁ = 0.108 m;
Wall thickness s = 0.004 m;
Pitch L₁ = 0.25 m;
Number of turns n = 10;
H₁ = 2.5 m;
Diameter of helix D₂=0.118 m;
Outside diameter of spiral tube rotation D₃=0.226 m;
Length of spiral tube L₂= 2.5 m;
Helix length L₃=4.47 m;
Mass m₃=46.30 kg;
Gravity of spiral tube G₃= 453.74 n;
Helix Angle α =34°;
Angle of inclination between spiral tube and horizontal plane β =45°;
Angle of inclination between helix and horizontal plane θ=23.30°;
The inclination height of spiral pipe is H₂= 1.77 m.

 

[vxiaoyu18]

These are all the parameters that I calculated here, so let's see what we can use. For the convenience of calculation, only the initial torque of steel ball to spiral tube can be calculated.

 

[Baluncore]

Let me just reimagine the problem for the sake of math.

Unlike an engineering drawing, a numerical model can be very simple.
1. Pitch of the helix; = axial advance per turn.
2. The angle of the helix rotational axis, measured from the vertical.
3. Radius of the helix, that the central axis of the tube follows.
4. The internal radius of the tube.
When I asked for those four specific numbers, you did a snow job that confuses things.

Parts of your reply, for example...

Helix Angle α =34°;
Angle of inclination between spiral tube and horizontal plane β =45°;
Angle of inclination between helix and horizontal plane θ=23.30°;

seem to make no sense together. You may know what you meant, but you need to be more precise in your description.

The length of the helix is irrelevant to the torque. One turn is sufficient.
The simple spherical ball has a mass m, and fits freely inside the tube.
The ball rolls on the inside surface of the tube. That contact is one tube radius from the tube axis.
The tube axis follows a helical path about the helix axis.

A “helix” is like the thread on a bolt or cylinder. It has a pitch independent of radius.
Mathematically, a “spiral” is a line drawn about a point on a plane. It has a changing radius.

Please don't use the informal term "spiral" when referring to a "helix". Not only are there helical tube pumps, but there are also spiral tube pumps. It is important to avoid confusion.

You seem to use the word “quality” to refer to a “quantity” of mass. Why not just use the word mass?

 

[vxiaoyu18]

Unlike an engineering drawing, a numerical model can be very simple.
...

I'm very sorry, because I don't know English, many sentences are translated by translation software. Some inaccuracies can only be understood as best as possible. These parameters are the parameters of the machine model I am actually calculating at present. I have listed all the parameters I know and can calculate, so that you can choose which ones you need. If you find the data too cluttered, I take the time to re-create a simpler model, illustrated graphically, perhaps more directly.

 

[vxiaoyu18]

1. Pitch of the helix; = axial advance per turn.
2. The angle of the helix rotational axis, measured from the vertical.
3. Radius of the helix, that the central axis of the tube follows.
4. The internal radius of the tube.

1. Pitch of the helix; = axial advance per turn.
L₁ = 0.25 m
2. The angle of the helix rotational axis, measured from the vertical.
β=45°
3. Radius of the helix, that the central axis of the tube follows.
r₄=D₂/2=0.059 m
4. The internal radius of the tube.
r₅=d/2=0.05 m

 

[Baluncore]

Here is a view of your helical tube.
X is to the right. Y is away from viewer. Z is up.
Helix axis passes through origin at angle of 45°.
Yellow line is valley. Yellow circle is where ball will rest in dip.
Magenta circle is a crest. Blue is volume that will hold a liquid.

 

[vxiaoyu18]

Here is a view of your helical tube...

You are very professional! Admire!
Yes, I reserved a little space for the ball to be there. I just want to figure out how to calculate the initial torque of the coil when the ball is in that yellow spot.

 

[JBA]

@vxiaoyu18 For someone having to translate what you receive and then translate your response, you have done an amazing job of communicating on this forum with me. I would never realized that you were not fluent in English had you not revealed that; and some of your variations in terminology are just as present in those fluent in English.

@Baluncore : I do not understand your "path" of the ball because the rolling ball will a always remain in the lowest point of each wrap of the helix of the rotating helix tube; and, as a result, its path will be a straight line parallel to the axis of rotation from its bottom entry to its top exit of the helix.

 

[JBA]

@vxiaoyu18 Go to the below website to see an animation of a ball being lifted in a Screw Conveyor which is a accurate representative of the path of a ball in your device.

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

 

[vxiaoyu18]

I couldn't figure out how to calculate the moment arm.

 

[vxiaoyu18]

@vxiaoyu18 Go to the below...
https://en.wikipedia.org/wiki/Screw_conveyor

Thank you. I mainly want to find out how to calculate it. I have searched many places on the Internet, but I haven't been able to find a detailed introduction.

 

[Baluncore]

I do not understand your "path" of the ball because the rolling ball will a always remain in the lowest point of each wrap of the helix of the rotating helix tube; and, as a result, its path will be a straight line parallel to the axis of rotation from its bottom entry to its top exit of the helix.

As the helix rotates the ball will roll in a straight line, parallel with the helix axis.

The yellow line is not the path of the ball. It is the valley down which water would flow with the helix held fixed in the illustrated position. It helps to follow that valley minimum in order to find the crest that sets the local liquid level, and the dip in which a ball would rest.
I wrote the code in BASIC to simulate and measure the capacity of a helical tube pump. The approximate grid position of the ball is a bonus. If the position of the local minimum was needed more accurately I could interpolate between the mesh points.

 

[vxiaoyu18]

As the helix rotates the ball will roll in a straight line, ……

You are a professional, a rare talent in mechanics. After five years of efforts, I have mastered the manufacturing principle of perpetual motion machine, which is a machine that continuously converts the gravity of machine parts into mechanical power output by using a machine. I have mastered several styles, and now it is the initial promotion stage. If there is an opportunity in the future, would you like to work with me to develop this technology? It's enough to change the world's power supply patterns.
My machine can't be used yet because it hasn't been patented. In order to facilitate promotion and display, I need to design another machine with reduced functions. I came across that the spiral tube might be useful, so I had to figure it out and see if I could use it, and when I knew how to calculate the torque of the ball in the spiral tube, I could figure out whether the machine was feasible or not.

 

[Baluncore]

Perpetual motion is not possible in physics, only in the mind.
The rules of PF do not allow perpetual motion to be discussed.

 

[JBA]

This is intended as a philosophical comment only: "And yet the Earth continues to spin and planets to orbit the sun"

 

[vxiaoyu18]

Perpetual motion is not possible in physics, only in the mind.
The rules of PF do not allow perpetual motion to be discussed.

I also know that PF BBS is not allowed to discuss the machine that is considered impossible to achieve. All my machines can be physically calculated. The manufacturing parameters of machine parts are complete and can be manufactured and used. I'm working on a prototype part of this more hidden technology, and when I figure out how to calculate the torque of the steel ball in the spiral tube, I verify that if it works, I may show you the complete machine structure in due course, and ask you to do the physical calculation again, which I believe will be a very interesting thing.

 

[vxiaoyu18]

This is intended as a philosophical comment only: "And yet the Earth continues to spin and planets to orbit the sun"

Light can heat, wind energy using wind, hydraulic, water use is a machine that I use is gravity, strictly speaking, it is only a "gravity" engine, is a force is a force difference, a difference value can cause some changes in the mechanical power, should be able to be used, will not violate any of the existing knowledge of physics.

 

[Baluncore]

The dip is located with a Y value of 0.041113 m. That is the horizontal offset from the axis.
So I think the torque will be; T = y * m * g * Cos( 45° )
T = 0.041113 * 3.03 * 9.8 * 0.7071 = 0.8632 Nm

 

[vxiaoyu18]

Keep trying to find the algorithm for the torque of the steel ball, we now know its trajectory, see where it stays, get closer and closer to the core, find the calculation method for its position, and we can complete the calculation process.

 

[vxiaoyu18]

The dip is located with a Y value of 0.041113 m. That is the horizontal offset from the axis.
So I think the torque will be; T = y * m * g * Cos( 45° )
T = 0.041113 * 3.03 * 9.8 * 0.7071 = 0.8632 Nm

If the torque caused by the ball is so small, then my machine can achieve it completely. Because to write the patent, I need to find a more accurate physical formula algorithm. I'll go to bed first. It's almost three o 'clock in the morning. Thank you.