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

Gold Member
Well I have a new analysis (much better than my last effort) that seems to work but does not correlate with the Bernoulli or a modification of that equation as discussed above.
The sin(α + β) equation simply converts angle inputs in terms of "β" to be converted into an equivalent rotation of "α" about its horizontal perpendicular axis so that when β = α then α = 0.
Note: Without the requirement to use the "β" angle of the helix axis as an input, the torque values due to the helix pitch rotation can be calculated by simply entering increasing "α" values from 33.995 to 90 degrees.
A screenshot of my Excel worksheet calculation is below: vxiaoyu18
Well I have a new analysis (much better than my last effort) that seems to work but does not ……
The calculation result of your Excel sheet is similar to that of my Excel sheet, except that I don't understand how to calculate the torque of steel ball. If your calculation result is correct, it will be no problem to use this spiral tube to manufacture my machine.Your calculation result is within my experimental value, it may be correct.

vxiaoyu18
Thank you very much for "Baluncore" and "JBA" for helping me so much. I will remember your efforts and may be surprised in the future.

vxiaoyu18
In my experiment, the torque of 3 steel balls (7.59NM) can drive the spiral tube to rotate, making the same 10 steel balls inside the spiral tube rise. If the results do not match the experimental results, the calculation may be wrong.

In my experiment, the torque of 3 steel balls (7.59NM) can drive the spiral tube to rotate, making the same 10 steel balls inside the spiral tube rise.
Don't forget that conservation of energy limits the rise possible and that friction must result in an overall loss of potential energy. Perpetual motion and over-unity efficiency are clearly not possible.

vxiaoyu18
Don't forget that conservation of energy limits the rise ……
There is no conflict between machines and conservation of energy, only a clever use of force. I have mastered several methods of machine manufacturing and can complete the calculation. I am now at the stage of building small engines to improve the running speed. The helical tube is used to make a demonstration prototype, and if it can be accurately calculated, I can share it with you, otherwise, without verification of physical calculation, people won't believe it.

@JBA I have some doubts about your numbers, the reason being that your value for torque falls smoothly to zero as the available minimum, that holds the ball, ceases to exist at alpha = (90 – beta).

You specify alpha as helix axis angle measured from the horizontal, but then tabulate from 90 deg down to beta. I would expect there to be torque results only in the range of alpha = (90 – beta) down to zero. Have you changed to alpha from the vertical?

When I consider a horizontal axis helix I see the dip or minimum directly below the helix axis, so no torque can be induced. The crest or maximum is then directly above the axis, so the tube can be almost completely filled with water.

As the slope of the helix axis reaches the pitch angle away from the vertical, the minimum and the maximum meet to the side of and at the same height as the helix axis. That must produce a non-zero torque.

When one of us finds the right result the others will take some time to realize and be convinced. To understand the problem and recognise a correct solution we rely on progressive independent solutions. Keep up the good work.

Gold Member
@Baluncore, Good catch, I understand your point and you are correct. While first developing my calculation, whenever α rotated below α (34°) I was getting negative tangent force values, so I worked out a method to correct that issue.
However, while now looking at a diagram I drew today, it appears the issue is that while the pitch angle of the helix is 34°, when viewed from the side of the helix each the wrap appears as V with its perpendicular to the helix axis being 1/2 of the wrap pitch angle.
With that in mind, when α reaches 0° the side view of one wrap of the helix becomes a vertical V and the ball sits in the bottom of that V and no torque exists.
Looks like I have more work to do!

Gold Member
Steel ball:

Above you state that the ball "radius" is .045m but the:
helix tubing inside diameter = .5*(D3 - D2) - 2*s = .5*(.226 - .118) - 2*.004 = .046m
So is the above a typo and ball "diameter" = .045m

vxiaoyu18
The spiral tube has a supporting cylinder in the middle. Center axis: radius r₃= 0.005m;

Gold Member
You are right, I got s and .004 switched somehow in the above equation. I think I need to back away from this for a bit to clear my mind.

vxiaoyu18
The view was that the ball's gravity produced no torque on the spiral tube. To rotate the spiral tube, you need to overcome the bearing friction caused by the spiral tube gravity and the steel ball gravity, and the friction caused by the steel ball rolling, right?

Gold Member
The view was that the ball's gravity produced no torque on the spiral tube.

The issue was that when the helix axis becomes horizontal α = 0°, that is the point at which there will be no torque on the helix due to the ball wt. and my current program does not do that. I am now going to work to correct that issue.

To rotate the spiral tube, you need to overcome the bearing friction caused by the spiral tube gravity and the steel ball gravity, and the friction caused by the steel ball rolling, right?

That is correct.

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Here is my minimum solution. The derivation of this is a little longer.

First Specify Helix and Ball Parameters.
d = 45 ' tilt of helix axis above horizonal in degrees
r = 0.059 ' radius of helix cylinder, metre
p = 0.250 ' pitch, helix advance per turn, metre
m = 3.030 ' mass of ball in kg
g = 9.8 ' acceleration due to gravity

' Precalculate
Pi = 4 * Atn( 1 ) ' a value for Pi = 3.14159
a = d * Pi / 180 ' convert tilt of helix in degrees to alpha in radians

' main computation
ka = Cos( a ) * r ' precompute two coefficients of the derivative to find minimum
kb = Sin( a ) * p / ( 2 * Pi )
t = Pi - Asin( kb / ka ) ' theta is the angular position of minimum on each helix turn
q = m * g * r * Sin( t ) * Cos( a ) ' torque on helix axis

' The result for the specified ball in a helical tube is;
Torque = 0.835437 Nm
' and if you need it;
b = Atn( p / ( 2 * Pi * r ) ) * 180 / Pi ' pitch angle in degrees
Pitch angle = 33.995 °

Gold Member
Now, it is time for me to see if I can develop a realistic version of mine for comparison. Obviously my final q equation should be in line with your's.

Obviously my final q equation should be in line with your's.
It should not be so obvious that my result is correct.
Finding the position of the minimum on the helix turn is necessary. I wrote the equation for the height of the helical filament, then took the derivative of that and solved it for zero. The position of the second zero is the minimum.
This graph shows height of first turn for different alpha. The position of the analytic minimum is marked with a small circle. • • Spinnor, vxiaoyu18 and JBA
Gold Member
@Baluncore I now believe you are correct, another quick review of my method now convinces that there is no simple geometric calculation solution to this issue; and, beyond that all I can say is that to me, what you have done is amazing work.

• Baluncore
Here is my derivation of the equation for torque due to a ball in a helical tube.
I believe this follows a similar method to Daniel Bernoulli in 1738.

We start by defining a filamentary helix wound on a cylinder of radius, r.
The advance per turn of the helix is the pitch, p.
The bottom of the helix axis remains at the origin of x, y and z coordinates.
As the helix axis is elevated in the y plane it will sweep from the +x axis to the +z axis.

We now view from +x, the circular section of the helix, on the y - z plane.
We follow the first turn of the helix by angle theta, t, from 0 to 2Pi.
The filament advances towards us, as a right handed screw.
x = p * t / 2Pi;
y = r * Sin( t );
z = r * Cos( t );
That gives us the coordinates of points on the filament at helix axis elevation, a = 0;

We now walk round to view the helix from the -y axis.
To rotate the axis to the slope angle, a, we multiply by a complex unit vector.
That unit vector will be; u = Cos( a ); v = Sin( a );

For beginners, to multiply the two vectors; ( u + i v ) * ( x + i z ) =
= ( u * x ) + i( u * z ) + i( v * x) + ii( v * z ); where ii = -1;
= ( u * x - v * z ) + i( u * z + v * x );
So the new value for; x = u * x - v * z;
and the new value for; z = u * z + v * x;
Note that the y value does not change during this rotation about the y axis.

We can now write the equation for height, z, of the filament as a function of t.
h = Cos( a ) * r * Cos( t ) + Sin( a ) * p * t / 2Pi
We want to find the minimum of that curve, so we look for where the derivative = 0.
Simplify the equation by removing two parameters, ka and kb.
ka = Cos( a ) * r; and kb = Sin( a ) * p / ( 2 * Pi );
So it takes the form; h = ka * Cos( t ) + kb * t;

The derivative is; h' = kb - ka * Sin( t );
0 = kb - ka * Sin( t ); will have a couple of zeros.
Rearrange it to; t = Asin( kb / ka ); which will give t values between -Pi/2 < t < Pi/2;
But the minimum we want lies between Pi/2 and Pi, so we must fold it by Pi - Asin();
Then the minimum is at; t = Pi - Asin( kb / ka );

At the start we wrote; y = r * Sin( t ); which gives the perpendicular radius arm length.
Then we multiply by; Cos( a ); to allow for the slope of the helix axis.
And by the vertical force; m * g; to get the helix axis torque.
Then torque; q = m * g * r * Sin( t ) * Cos( a );

I hope that does not have too many errors or typos.

• vxiaoyu18 and Bystander
vxiaoyu18
Your knowledge is too strong, and this calculation method is beyond my knowledge. Maybe you can arrange this problem and publish a technical paper on how to calculate. That's your credit. I can't use your method to build a machine.

Maybe you can arrange this problem and publish a technical paper on how to calculate.
In post #84 I gave the 4 steps needed to calculate the torque on the helix due to a ball.
' main computation
ka = Cos( a ) * r ' precompute two coefficients of the derivative to find minimum
kb = Sin( a ) * p / ( 2 * Pi )
t = Pi - Asin( kb / ka ) ' theta is the angular position of minimum on each helix turn
q = m * g * r * Sin( t ) * Cos( a ) ' torque on helix axis

Then in post #88 I gave my full derivation of the method. So others may compare their results.
You do not need to understand the full derivation. Do the calculation in post #84.

Gold Member
@Baluncore
I submit the below for you review and comment
( I have no confidence it is fully accurate; but, the 45° correlation is interesting) • Baluncore and vxiaoyu18
@JBA. Well done.
I think our methods give the same results up until the minimum disappears at alpha = 90° – pitch angle.

I believe the energy method should work because applying a fixed helix torque to counter the torque induced by the ball on the helix will hold the ball in position, with no movement there will be no energy transferred. Increasing the torque slightly will cause the ball to rise slowly, reducing the torque slightly will cause the ball to descend slowly. The question becomes, what torque is needed to raise the ball at a rate approaching zero, and that can be solved using the energy method.

vxiaoyu18
Congratulations to the two gods for calculating the same result, with this method, it is very easy to make a magic machine.
Although using spiral tubes is not the best way to do it, it can be used to illustrate what kind of machine can be made, which is why I want to use it.

@JBA.
When I collapse your energy method equations to eliminate common factors and temporary variables, I get, using my variable names;

Torque = m * g * Sin( a ) * p / ( 2 * Pi );

Note that, as I expected, the number of turns and the radius do not appear in the calculation.
That is the simplest equation yet. It agrees with my derivation, with the usual constraint on maximum alpha.

• JBA and vxiaoyu18
Gold Member
@Baluncore , I had no doubt that your solution is correct; but, if my actually would match it results. At the same time, it is nice to have a correspondence of two methods.
This morning I fell into the mode of: Well, if my other method wasn't going to work then it was time to grab sometime off of the wall and see what would happen; and, an energy method was the only thing I could think of that would not require solving the extremely difficult trigonometry that you had conquered, so I decided to give it a try.

... an energy method was the only thing I could think of that would not require solving the extremely difficult trigonometry that you had conquered ...
I am no conquering mathematician, I just crunch numbers. If I was any good at math then perhaps I would have recognised my ratio of kb/ka contained Sin(a)/Cos(a) = the Tan(a) function and that it too might cancel the Asin() and collapse to your simpler solution. Trigonometric identities never were my strong point.

It needed two or more minds to follow two or more paths to reach the same result to get confirmation. @JBA Thank you for diversifying and persevering.

• Bystander
vxiaoyu18
I am in China, you are in the United States, thousands of miles apart, can not toast together to solve this physics problem, hope to have a chance in the future.

Cheers from Australia, where it is 11PM and 1°C.

vxiaoyu18
I'm in guangzhou, China. It's 9:10 PM. The temperature is 28℃, very comfortable night, ha ha, I am glad to meet you in PF.

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Gold Member
@vxiaoyu18 , Hello, from Houston, Texas, USA, the temperature is 23° C and it is raining.
I am happy we managed to find a solution for your issue and learn a bit for ourselves as well. This is what worldwide collaboration can achieve in solving problems . It is a shame that some of our governments can't understand that as well.
Thank you for bringing this challenge to the forum.

Gold Member
@Baluncore , After all of our work I hardly believe I am posting this.

Last night, I began to consider that I had made an error in my above calculation by using D2*π as the distance the ball traveled for each wrap of the helix in my in my F = E/n/(D2*π) calculation; when, I should have used:
F =E/n/(sqrt((D2*p)^2+H2^2) for that formula.

After making that revision this morning, I appear to have verified its accuracy because with that revision my calculation result for F at 90° is exactly the same as the simple F = Ma*sin(33°) = 16.60N result, (unfortunately I failed to check that for my original calculation after seeing my correlation with your result at 45°).
Obviously, the main issue that creates is that at α = 45°, that revision reduces my original T = .835N-m value to
T = .693N-m and reduces the values in my curve but not its profile and still calculates T = 0 at α= 0°; but, no longer correlates with your results.

I have not yet looked at your revision, but the agreement between our numbers was close to numerically perfect.
Code:
Pitch angle 33.995 °
Torque at a = 45°, 0.835437 Nm
alpha Baluncore JBA difference
90.0 1.#NAN00000 1.181486720 % 1.$E+00 85.0 1.#NAN00000 1.176990806 % 1.$E+00
80.0 1.#NAN00000 1.163537282 % 1.$E+00 75.0 1.#NAN00000 1.141228536 % 1.$E+00
70.0 1.#NAN00000 1.110234352 % 1.$E+00 65.0 1.#NAN00000 1.070790615 % 1.$E+00
60.0 1.#NAN00000 1.023197514 % 1.\$E+00
55.0 0.967817262 0.967817262 0.0E+00
50.0 0.905071337 0.905071337 0.0E+00
45.0 0.835437272 0.835437272 3.3E-16
40.0 0.759445025 0.759445025 2.2E-16
35.0 0.677672942 0.677672942 0.0E+00
30.0 0.590743360 0.590743360 1.1E-16
25.0 0.499317864 0.499317864 5.6E-17
20.0 0.404092257 0.404092257 -5.6E-17
15.0 0.305791265 0.305791265 3.3E-16
10.0 0.205163016 0.205163016 4.4E-16
5.0 0.102973353 0.102973353 -1.2E-16
0.0 0.000000000 0.000000000 2.1E-16
[code]

Gold Member
Note: I have done multiple edits of this post so see the below post for my latest input.

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Gold Member
@Baluncore

Thankfully, my issue is now fully resolved and we still have full correlation.

My late night investigation just revealed that: Atan(L1/(D2*PI) = 33.995°, so using D2*PI() is the correct interpretation for my analysis.
Sometimes the hardest part of getting an answer to an issue is figuring out what question to ask.

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