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

[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.

 

[JBA]

@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.

Please review this revision and let me know your thoughts.

 

[Baluncore]

I have not yet looked at your revision, but the agreement between our numbers was close to numerically perfect.

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]

 

[JBA]

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

 

[JBA]

@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.

 

[Baluncore]

Sometimes the hardest part of getting an answer to an issue is figuring out what question to ask.

When we know the right question, we also know the answer, and so do not need to ask it.

“The only interesting answers are those which destroy the question”. —Susan Sontag.