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

In summary, 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. Conservation of energy might be useful to consider to find the torque induced by the steel ball.
  • #36
Here is a view of your helical tube.
X is to the right. Y is away from viewer. Z is up.
Helix axis passes through origin at angle of 45°.
Yellow line is valley. Yellow circle is where ball will rest in dip.
Magenta circle is a crest. Blue is volume that will hold a liquid.
screengrab.png
 
Engineering news on Phys.org
  • #37
Baluncore said:
Here is a view of your helical tube...
You are very professional! Admire!
Yes, I reserved a little space for the ball to be there. I just want to figure out how to calculate the initial torque of the coil when the ball is in that yellow spot.
 
  • #38
@vxiaoyu18 For someone having to translate what you receive and then translate your response, you have done an amazing job of communicating on this forum with me. I would never realized that you were not fluent in English had you not revealed that; and some of your variations in terminology are just as present in those fluent in English.

@Baluncore : I do not understand your "path" of the ball because the rolling ball will a always remain in the lowest point of each wrap of the helix of the rotating helix tube; and, as a result, its path will be a straight line parallel to the axis of rotation from its bottom entry to its top exit of the helix.
 
  • #40
lb.jpg


I couldn't figure out how to calculate the moment arm.
 
  • #41
  • #42
JBA said:
I do not understand your "path" of the ball because the rolling ball will a always remain in the lowest point of each wrap of the helix of the rotating helix tube; and, as a result, its path will be a straight line parallel to the axis of rotation from its bottom entry to its top exit of the helix.
As the helix rotates the ball will roll in a straight line, parallel with the helix axis.

The yellow line is not the path of the ball. It is the valley down which water would flow with the helix held fixed in the illustrated position. It helps to follow that valley minimum in order to find the crest that sets the local liquid level, and the dip in which a ball would rest.
I wrote the code in BASIC to simulate and measure the capacity of a helical tube pump. The approximate grid position of the ball is a bonus. If the position of the local minimum was needed more accurately I could interpolate between the mesh points.
 
  • #43
Baluncore said:
As the helix rotates the ball will roll in a straight line, ……
You are a professional, a rare talent in mechanics. After five years of efforts, I have mastered the manufacturing principle of perpetual motion machine, which is a machine that continuously converts the gravity of machine parts into mechanical power output by using a machine. I have mastered several styles, and now it is the initial promotion stage. If there is an opportunity in the future, would you like to work with me to develop this technology? It's enough to change the world's power supply patterns.
My machine can't be used yet because it hasn't been patented. In order to facilitate promotion and display, I need to design another machine with reduced functions. I came across that the spiral tube might be useful, so I had to figure it out and see if I could use it, and when I knew how to calculate the torque of the ball in the spiral tube, I could figure out whether the machine was feasible or not.
 
  • #44
Perpetual motion is not possible in physics, only in the mind.
The rules of PF do not allow perpetual motion to be discussed.
 
  • #45
This is intended as a philosophical comment only: "And yet the Earth continues to spin and planets to orbit the sun"
 
  • #46
Baluncore said:
Perpetual motion is not possible in physics, only in the mind.
The rules of PF do not allow perpetual motion to be discussed.
I also know that PF BBS is not allowed to discuss the machine that is considered impossible to achieve. All my machines can be physically calculated. The manufacturing parameters of machine parts are complete and can be manufactured and used. I'm working on a prototype part of this more hidden technology, and when I figure out how to calculate the torque of the steel ball in the spiral tube, I verify that if it works, I may show you the complete machine structure in due course, and ask you to do the physical calculation again, which I believe will be a very interesting thing.
 
  • #47
JBA said:
This is intended as a philosophical comment only: "And yet the Earth continues to spin and planets to orbit the sun"
Light can heat, wind energy using wind, hydraulic, water use is a machine that I use is gravity, strictly speaking, it is only a "gravity" engine, is a force is a force difference, a difference value can cause some changes in the mechanical power, should be able to be used, will not violate any of the existing knowledge of physics.
 
  • #48
The dip is located with a Y value of 0.041113 m. That is the horizontal offset from the axis.
So I think the torque will be; T = y * m * g * Cos( 45° )
T = 0.041113 * 3.03 * 9.8 * 0.7071 = 0.8632 Nm
 
  • #49
Keep trying to find the algorithm for the torque of the steel ball, we now know its trajectory, see where it stays, get closer and closer to the core, find the calculation method for its position, and we can complete the calculation process.
 
  • #50
Baluncore said:
The dip is located with a Y value of 0.041113 m. That is the horizontal offset from the axis.
So I think the torque will be; T = y * m * g * Cos( 45° )
T = 0.041113 * 3.03 * 9.8 * 0.7071 = 0.8632 Nm
If the torque caused by the ball is so small, then my machine can achieve it completely. Because to write the patent, I need to find a more accurate physical formula algorithm. I'll go to bed first. It's almost three o 'clock in the morning. Thank you.
 
  • #51
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.
 
  • #52
JBA said:
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.
 
  • #53
JBA said:
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.
JBA said:
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?
 
  • #54
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.
 
  • #55
JBA said:
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.
 
  • #56
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.
 
  • #57
JBA said:
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.
 
  • #58
vxiaoyu18 said:
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.
 
Last edited:
  • #59
JBA said:
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.
 
Last edited:
  • #60
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
 
  • #61
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.
 
  • #63
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
 
  • #64
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.
 
Last edited:
  • #65
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.
 
  • #66
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.
 
  • #67
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.
 
  • #68
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.
 
  • #69
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.
 
Last edited:
  • #70
JBA said:
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.

vxiaoyu18 said:
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.
 

Similar threads

Replies
2
Views
2K
Replies
1
Views
629
  • Mechanical Engineering
Replies
5
Views
3K
  • Mechanical Engineering
Replies
13
Views
5K
Replies
7
Views
5K
  • Mechanical Engineering
Replies
4
Views
8K
  • Mechanical Engineering
Replies
2
Views
27K
  • Introductory Physics Homework Help
2
Replies
43
Views
2K
  • Mechanical Engineering
Replies
11
Views
15K
Replies
1
Views
1K
Back
Top