[hamdani yusuf (OP)]

Anyone who think that current SI standar is already perfect doesn't seem to learn from history. Don't they know that SI standard has been revised many times before? What's their justification that this time, among many other previous times, we already get everything right, while ignoring problems identified by many researchers including AAPT senior members?

[paul cotter]

No system is perfect. I voted for the first option because it was the closest to my views on this subject, not because the SI system is perfect- this is a problem with all polls where one has to select the best fit overall. Although not perfect I can see of no consistent improvements. The simple fact that your idea gives torsional stiffness as a quadratic of displacement angle rules it out.

[alancalverd]

Newton.meter per radian is the frictional loss of a bearing or the viscous drag on a propellor. It is a dynamic quantity, not a static one like torque. 

SI standards are modified according to the needs of professionals, not the misunderstandings of teachers. 

[hamdani yusuf (OP)]

No system is perfect. I voted for the first option because it was the closest to my views on this subject, not because the SI system is perfect- this is a problem with all polls where one has to select the best fit overall. Although not perfect I can see of no consistent improvements. The simple fact that your idea gives torsional stiffness as a quadratic of displacement angle rules it out.

Then you could have chosen the second or third option. Perfect has a meaning commonly understood by most people speaking English. Unless stated otherwise, that's the default we should use in communications.

What convinced you that quadratic of displacement angle is impossible?

[hamdani yusuf (OP)]

Newton.meter per radian is the frictional loss of a bearing or the viscous drag on a propellor. It is a dynamic quantity, not a static one like torque. 

SI standards are modified according to the needs of professionals, not the misunderstandings of teachers. 

If we build a system based on a shaky ground, we will inevitably need to make ad hoc patches and adjustments to make it work in some edge cases.
As I mentioned earlier, if you can include a unit or exclude it from the same quantity without changing its numerical value, then your system must have kept its value at unity. In this case, you can even square it or cube it without changing its numerical value.
In this case, specifying the trajectory as a perfect circle guarantees that the ratio between tangential displacement and angular displacement a constant: the rotational radius. Moreover, the tangential displacement must equal rotational radius when the angular displacement is exactly 1 radian.

[hamdani yusuf (OP)]

In more general cases where the trajectory is not necessarily circular, the rotational radius isn't necessarily constant. In other words, there's a non-zero radial displacement. In these cases, current standard is no longer adequate to describe the rotational system.
I'm working on the problem with elliptical trajectory using my proposed standard. Qwen can solve it in a few prompts, even in older version. Newer AI models will be able too, IMO.

The first case of elliptical trajectory is when the tangential speed is constant. Consequentially, its kinetic energy is constant. The acceleration must be purely orthogonal. But the angular speed must vary, inversely proportional to the rotational radius at the moment.
ω = v/r_rot
r_rot = v/ω
Its unit should reflect the equation above, hence (m/s)/(rad/s) = m/rad.


As shown in the diagram, at major axis and minor axis, the direction of orthogonal acceleration coincides with radial line. But at any other positions, they are different. The deviation is shown by Greek letter φ.

[paul cotter]

Torsional stiffness is known to be linear wrt angle of displacement- your system has it as a quadratic relationship and hence is a fail.

[alancalverd]

Wind a string around a cylinder of radius 1 m and hang a 1 N weight on the string. What is the torque τ?

Lift the weight by rotating the cylinder. What is the increase in potential energy ΔE?

For those of us who understand mechanics, τ= 1 Nm and ΔE = 1 Nm per radian.

[hamdani yusuf (OP)]

Torsional stiffness is known to be linear wrt angle of displacement- your system has it as a quadratic relationship and hence is a fail.

What was known for a long time could turn out to be false.

In linear system, the linear force equals energy per linear displacement, F = E/s  (J/m)
Linear stiffness k_l = F/s (J/m^2)

In angular system, the angular force already contains radian, ie. τ = E/θ (J/rad).
Angular stiffness k_a = τ/θ (J/rad^2)

[hamdani yusuf (OP)]

In more general cases where the trajectory is not necessarily circular, the rotational radius isn't necessarily constant. In other words, there's a non-zero radial displacement. In these cases, current standard is no longer adequate to describe the rotational system.
I'm working on the problem with elliptical trajectory using my proposed standard. Qwen can solve it in a few prompts, even in older version. Newer AI models will be able too, IMO.

The first case of elliptical trajectory is when the tangential speed is constant. Consequentially, its kinetic energy is constant. The acceleration must be purely orthogonal. But the angular speed must vary, inversely proportional to the rotational radius at the moment.

This video could help you visualize what I described above. Imagine the track is elliptical instead of circular.

Superconductor at -196?C, Quantum Levitation | Magnetic Games

With the use of liquid nitrogen, the YBCO compound can be cooled until it becomes a superconductor, and a superconductor placed in a magnetic field has amazing behaviors.
Please activate the subtitles to get more info on the experiment.

The next case is when the angular speed is constant. This can be done by modifying the previous setup with a smooth pipe where the puck can slide inside without friction, and a motor/generator unit equipped with battery to control the rotation of the pipe at a constant angular speed. When accelerating, potential energy from battery flow to the motor, and converted to kinetic energy. When decelerating, kinetic energy from the puck is converted by the generator and stored back to the battery as potential energy.

[hamdani yusuf (OP)]

Wind a string around a cylinder of radius 1 m and hang a 1 N weight on the string. What is the torque τ?

Lift the weight by rotating the cylinder. What is the increase in potential energy ΔE?

For those of us who understand mechanics, τ= 1 Nm and ΔE = 1 Nm per radian.

In current SI system, both τ= 1 Nm and ΔE = 1 Nm.
In my proposed system, τ= 1 Nm/rad and ΔE = 1 Nm.

[alancalverd]

In my proposed system, τ= 1 Nm/rad and ΔE = 1 Nm.

......regardless of how many turns the cylinder makes. So you can't use a plumb bob, a watch spring, or a hydroelectric dam to store energy.

[hamdani yusuf (OP)]

In my proposed system, τ= 1 Nm/rad and ΔE = 1 Nm.

......regardless of how many turns the cylinder makes. So you can't use a plumb bob, a watch spring, or a hydroelectric dam to store energy.

What makes you think that the unit for energy is Nm/rad, instead of Joule, or Nm?

[hamdani yusuf (OP)]

In more general cases where the trajectory is not necessarily circular, the rotational radius isn't necessarily constant. In other words, there's a non-zero radial displacement. In these cases, current standard is no longer adequate to describe the rotational system.
I'm working on the problem with elliptical trajectory using my proposed standard. Qwen can solve it in a few prompts, even in older version. Newer AI models will be able too, IMO.

The first case of elliptical trajectory is when the tangential speed is constant. Consequentially, its kinetic energy is constant. The acceleration must be purely orthogonal. But the angular speed must vary, inversely proportional to the rotational radius at the moment.
ω = v/r_rot
r_rot = v/ω
Its unit should reflect the equation above, hence (m/s)/(rad/s) = m/rad.


As shown in the diagram, at major axis and minor axis, the direction of orthogonal acceleration coincides with radial line. But at any other positions, they are different. The deviation is shown by Greek letter φ.

In the case where only orthogonal (or normal) acceleration exists, which makes tangential speed as well as kinetic energy constant, the rotational axis keeps moving, and rotational radius keeps changing. Nevertheless, the equation
r_rot = ds/dθ always applies.

The diagram below shows the rough tracing points on the elliptical trajectory, with corresponding rotational axes.

Points 0 to 5 show sampling points of the puck positions along the elliptical trajectory with constant tangential speed.
Points o1 to o5 show the estimated rotational axes corresponding to sampling points 1 to 5, respectively.
The position of rotational axes is estimated from the intersection between normal lines of two adjacent sampling points. More accurate estimation can be obtained with more sampling points.
At major axis (point 0 in the diagram), the rotational axis is the nearest focus.
The orthogonal acceleration is v.ω = v^2/r_rot.
But |v| is constant in this case. Thus orthogonal acceleration is inversely proportional to r_rot, which is highest when r_rot is minimum, ie. at major axes (point 0). Likewise, at minor axes (point 5) r_rot is at maximum, thus the orthogonal acceleration is lowest there.

[alancalverd]

What makes you think that the unit for energy is Nm/rad, instead of Joule, or Nm?

I didn't. I asked what was the change in potential energy of the weight, or the work done by rotating the cylinder.

[hamdani yusuf (OP)]

What makes you think that the unit for energy is Nm/rad, instead of Joule, or Nm?

I didn't. I asked what was the change in potential energy of the weight, or the work done by rotating the cylinder.

You did.

Wind a string around a cylinder of radius 1 m and hang a 1 N weight on the string. What is the torque τ?

Lift the weight by rotating the cylinder. What is the increase in potential energy ΔE?

For those of us who understand mechanics, τ= 1 Nm and ΔE = 1 Nm per radian.

[alancalverd]

Exactly. The  change in potential energy as you turn the cylinder, dE/dθ, is 1 N.m per radian

[hamdani yusuf (OP)]

Exactly. The  change in potential energy as you turn the cylinder, dE/dθ, is 1 N.m per radian

dE/dθ describes how hard it is to turn the cylinder. And it's called torque.
In my proposed system, it's also called angular force.
The other types of rotational force are: tangential force, orthogonal force, and radial force. They affect the motion of rotational objects in different ways.
The table below shows a more complete picture if the proposed system.
https://www.thenakedscientists.com/forum/index.php?topic=87006.msg746993#msg746993

[alancalverd]

Would you agree that, in my example, turning the cylinder 1 radian  will raise or lower the weight by 1 m? So the change in potential energy is 1 joule. And if I  turn it 2 rad, ΔE = 2 J. So ΔE = 1 J/rad.

Here's what teachers need to learn:

Energy is a scalar: force x distance moved along the line of action of the force. It is measured in joules and can be converted from mechanical to electrical or heat energy.

Torque is a vector, force x distance perpendicular to the line of action. It is measured in newton.meters and cannot be converted into anything else.

[hamdani yusuf (OP)]

Would you agree that, in my example, turning the cylinder 1 radian  will raise or lower the weight by 1 m? So the change in potential energy is 1 joule. And if I  turn it 2 rad, ΔE = 2 J. So ΔE = 1 J/rad.

Here's what teachers need to learn:

Energy is a scalar: force x distance moved along the line of action of the force. It is measured in joules and can be converted from mechanical to electrical or heat energy.

Torque is a vector, force x distance perpendicular to the line of action. It is measured in newton.meters and cannot be converted into anything else.

In general, ΔE ≠ ΔE/Δθ.
This is an example of common mistakes as a logical consequence from normalizing the abandonment of radian.

The formula E = τ. θ implies that standard unit for torque is Joule per radian. You can tweak your system to keep the numerical value of each quantity 1. But then your system will lost some of its generality.
If your system can maintain the value of E=1, then 1 = τ. θ
Likewise, If your system can maintain the value of τ=1, then E= θ
Meanwhile, If your system can maintain the value of θ=1, then E= τ