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

[vxiaoyu18]

You are all better at physics than I am. My physical knowledge level is not enough to judge whether your calculation is correct or not. I can only see whether your analysis is reasonable and whether the calculation result is close to the experimental result based on my experiment.

 

[Baluncore]

There is a solution here by Daniel Bernoulli; from page 189.
http://www.pumppower.com.au/wp-cont...median-Screw-Pump-a-note-on-its-invention.pdf
Torque = weight of ball * Tan( angle of axis from the horizontal ) * Cos( helix pitch angle )

 

[JBA]

I have calculated the torque based upon the Bernoulli's method and there is a problem for your application and there is an issue for it relative to your application because as α approaches 90° its calculated torques for your ball M*g increase unrealistically, i.e. for α = 80° its calculated torque:
T = 3.03*9.8*5.67*.918 = 154.6 N-m

 

[vxiaoyu18]

According to my experiment, When the diameter of the axis of rotation is equal to the diameter of the spiral tube, the gravity of 3 balls can easily drive 10 balls to rise in the spiral tube, and the torque of 3 balls is equal to 5.69 Nm, so if the calculation result is contrary to this experiment result, it should be wrong. I'll take the time to do another rigorous experiment and see how it goes.

 

[Baluncore]

I did notice the suspicious use of the Tan( alpha ).
I expect the multiple translation has corrupted the original terminology and summary.
I looked for a pattern and found that swapping the pitch and helix angle produced a more realistic result when alpha was measured from the vertical. It needs more investigation.
Pitch angle 33.995° alpha torque Nm 0.0 20.025199 5.0 19.948997 10.0 19.720971 15.0 19.342857 20.0 18.817531 25.0 18.148993 30.0 17.342331 35.0 16.403682 40.0 15.340192 45.0 14.159954 50.0 12.871950 55.0 11.485982 60.0 10.012599 65.0 8.463015 70.0 6.849021 75.0 5.182903 80.0 3.477339 85.0 1.745311 90.0 0.000000
It is interesting that Bernoulli's method also modeled the dip and crest.
The ball and tube diameter are not important so long as they are a close and free fit.
The helix radius is not important to torque as the effect of arm length and slope cancel.

 

[JBA]

Good work, I also see that you revised the helix pitch angle to = asin(L1/.44 (one wrap of the helix coil)) = 33.99°, which seems to make makes sense to me as well.

 

[vxiaoyu18]

The ball rises along the Helix Angle with the horizontal plane, which I calculated in #31 and gave the result: Helix Angle α =34°, which is useful for estimating the result.

 

[JBA]

I fully agree.

On the issue of the cancellation of the radius, I observed that in the Bernoulli formula discussion but was unable to really comprehend it; so, I need to go back and take another look at that issue.

 

[vxiaoyu18]

I think it is necessary to have a moment arm, otherwise the moment =0. In this way, the spiral tube can rotate against the friction resistance of the bearing with a small force, and lift one or more steel balls high enough to obtain a large gravitational potential energy.

 

[Baluncore]

On the issue of the cancellation of the radius, I observed that in the Bernoulli formula discussion but was unable to really comprehend it; so, I need to go back and take another look at that issue.

The helix pitch_angle = Atan( pitch / ( TwoPi * radius ) ).
The gradient or slope of a vertical axis helix = pitch / circumference.
If we double the radius of the helix, we double the circumference which will halve the rise over run that separates the vector forces, but it compensates by doubling the length of the radius arm, so those two effects cancel with regard to torque.

I think it is necessary to have a moment arm, otherwise the moment =0.

Yes, the model and the mathematics collapses with zero radius arm.
The pitch angle becomes 90°, so there can be no minimum dip to hold a ball.

 

[JBA]

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.

 

[Baluncore]

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.

 

[Baluncore]

@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.

 

[JBA]

@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!

 

[JBA]

Steel ball:
Radius r₁= 0.045m;

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;

 

[JBA]

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?

 

[JBA]

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.

 

[Baluncore]

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 °

 

[JBA]

@Baluncore I am glad to see your calculation. Well done.
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.

 

[Baluncore]

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.

 

[JBA]

@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]

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.

 

[Baluncore]

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.