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

JBA

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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.
 
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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

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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

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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.
 
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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

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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.
 
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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

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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.
 
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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.
 
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JBA

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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
 
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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.
 

JBA

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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
 
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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.
 
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Baluncore

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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 modelled 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

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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.
 
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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

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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.
 
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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.
 
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Baluncore

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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

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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:

1568393647261.png
 
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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.
 
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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.
 
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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

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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.
 

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