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

[JBA]

On a minor point since the result for 45° is the same for both,the direction of force parallel to the helix axis is mg*sin (45°), not cos(45°) which in this case is the identical value anyway. The cos() is the radial force on the tube.
On the other hand, while I agree the force along the axis is as calculated, I am not sure that is the force that provides the actual torque on the screw because the angle of the slope due to the helix's pitch is much less.
As an example, looking at the reference animation, freezing it (mentally) and then observing the immediate slope under the ball indicates not all of the force along the helix axis is actually applied to act as a torque on the screw.
Another way of looking at it is if you unwind the helix tube and lift one end to the same elevation as the top of the helix, the angle of its slope = 23.3°, which exactly the same value @vxiaoyu18 calculated for the combined helix pitch angle and 45° helix axis angle, relative to the horizontal in his post #31.
In other words if you want a high torque from a ball drop then you want a combined helix axis and pitch angle as close to vertical as possible, you just won't get many rotations from each ball drop.

 

[vxiaoyu18]

On a minor point since the result for 45° is the same for both……

I can see from the three-dimensional structure of its running state, but can't find the calculation formula for its physical, and I have to find a calculation way to crack its state, or foreign unconvincing, that's what I'm upset, you know, if you don't have enough calculation and the actual machine, few people would agree with this kind of machine, ha ha. The machine is simple to make, but the spiral tube is difficult to calculate.

 

[Baluncore]

On a minor point since the result for 45° is the same for both,the direction of force parallel to the helix axis is mg*sin (45°), not cos(45°) which in this case is the identical value anyway. The cos() is the radial force on the tube.

I measure the tilt of the helix axis as declination from the vertical, which changes the sine to a cosine.

As an example, looking at the reference animation, freezing it (mentally) and then observing the immediate slope under the ball indicates not all of the force along the helix axis is actually applied to act as a torque on the screw.

The slope where the ball sits is zero. The radius of the ball is not important as the weight is a vertical force.
One turn of the helix will advance the ball one pitch parallel to the helix axis. Maybe there should be a two Pi in there?

 

[JBA]

Another way to look at it is to turn the helix axis vertical and observe the resulting tangent rotation force on the screw from the ball's weight. What you see is as the helix pitch reduces so does the slope the ball sitting on and with its tangent force vector becoming progressively smaller relative to its vertical weight force vector to the point that when the helix pitch becomes 0 then there is a 100% of the downforce on the helix but no tangent force to rotate it.

 

[vxiaoyu18]

You just won't get many rotations from each ball drop...

In fact, my machine is to overcome the torque caused by the steel ball, and the smaller the torque, the easier my machine is to build. If the torque of the steel ball on the spiral tube is greater than the power I give it, I can only change the design and try to use the torque of the steel ball as the driving force. No matter which force is large, my machine can be built as long as the driving force and the resistance are not equal.

 

[JBA]

OK, I am writing a bit more at this time but will revise it for a minimum torque requirement.

You have already reduced the torque arm to a minimum by wrapping the tube directly around the shaft.
I believe the best method of determining the torque from the ball weight on the helix is to first determine the torque from the ball wt on the helix vs a given helix pitch angle is with the helix axis vertical.
The shorter the helix pitch, the lower the the torque from the pitch.
The all other factors being equal, the smaller the angle of the helix axis from the horizontal the lower the torque from the ball.
Having offered that, I think you fully understand the issues at hand; but, will be here to assist if needed.

 

[vxiaoyu18]

The shorter the helix pitch, the lower the the torque from the pitch.
The all other factors being equal, the smaller the angle of the helix axis from the horizontal the lower the torque from the ball.

According to the fuzzy calculation and experimental results, the correlation is correct. In my calculation Excel, only the accurate algorithm of steel ball torque is missing.

 

[JBA]

Length of spiral tube L₂= 2.5 m;
Helix length L₃=4.47 m;

I am working on an equation for the torque and noticed that the above dimensions in your Post#31 do not appear to be correct because they indicate that the helix tube length is shorter than the helix length. What am I misreading?
Edit: I used the calculation to determine the tubing length and that resolved the terminology issue I had.

 

[vxiaoyu18]

I am working on an equation for the torque and noticed that the above dimensions in your Post#31 do not appear to be correct because they indicate that the helix tube length is shorter than the ...

I calculate so spiral length: D ₂ for spiral diameter, namely projection diameter, L ₁ for pitch, lap spiral length: L₄=sqrt((πD₂)^2+L₂^2). Spiral tube high for H ₁, spiral length: L₃ = (H₁/L₁)*L₄ = H₁/L₁*sqrt((πD₂)^2+L₂^2) = 4.4697 m

If you think your calculation is correct, you can use your calculation results, which are calculated by using the physical formula, but I can't guarantee that all my calculations are correct.

 

[JBA]

No problem It was strictly a terminology issue.

Below is a formula I have developed based upon my last best theory (i.e. the pitch angle and tangent force determined with the helix axis vertical and then modified by the the helix angle; and, the ball contact radius being from the helix axis to the centerline of the ball). Whether this equation is actually accurate will be for you to determine.

T = M2*g*[sin(asin(H2/L3))*sin(α)]*((D3/2)-s-r)

The torque values are valid for α from 90°to (the point at which the pitch angle is 90° to the horizontal which is now 23.3°) and the helix angle α is measured from horizontal

 

[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.

 

[JBA]

@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]

@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.

 

[Baluncore]

@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]

@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.

 

[Baluncore]

... 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.

 

[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.

 

[Baluncore]

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.

 

[JBA]

@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.