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

[vxiaoyu18]

TL;DR

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).