On the nature of torque ....

[xpensive]

It is situations like these that gives away your lack of technical understanding, just like when you concluded that torque was a form of energy, you can't simply waive a thesis from a university somewhere as eternal evidence to your point.

What if you torque up a torsion spring? Don't you think that the torque represents an energy that you add to that mechanical system? Science is a funny thing. Perhaps my scientific understanding gets in the way of what you call the technical understanding. I would much rather discuss the merits of the source that I mentioned than having a debate about my understanding of technical matters.

When you use a force of X Newtons and move an object Y metres you have performed work of X N* Y m = XY Nm = XY J Work = energy.

In the case of a torsion spring the energy stored is dependent upon torque and angle. While torque is analogous to force, angle is analogous to distance. Angle is dimensionless, while Torque has the dimensions [M][L]^2/[T]^2.

Force has the dimensions [M][L]/[T]^2 and distance has the dimension [L[, so force x distance = [M]{L]^2/[T]^2.

=D> In other words, it is a confusion of units, Torque (Nm), is not the same as Energy (Nm).

[WhiteBlue]

I can offer another solution to the opposing views: If you load a rotating system with torque you typically produce an elastic deformation. The the size of the torque you apply represents the size of the energy that is absorbed by the elastic deformation. In that view torque is not energy in itself but represents the potential energy that is added to the system. I would be happy with that. Perhaps it helps to overcome the semantic hurdles we are facing here.

[wuzak]

Torque is not energy in any way, shape or form.

Torque is the rotational equivalent of force.

Would you ever say that force is "potential energy"?

[wuzak]

=D> In other words, it is a confusion of units, Torque (Nm), is not the same as Energy (Nm).

Correct. And that's why the distinction of using the unit Joules for energy insetad of the dimensionally correct Nm.

[wuzak]

I can offer another solution to the opposing views: If you load a rotating system with torque you typically produce an elastic deformation. The the size of the torque you apply represents the size of the energy that is absorbed by the elastic deformation. In that view torque is not energy in itself but represents the potential energy that is added to the system. I would be happy with that. Perhaps it helps to overcome the semantic hurdles we are facing here.

The energy "stored" in the elastic deformation is due to torque and angular deflection.

[Richard]

Locked to protect the innocent

...

pause

...

and relax

...

Now lets see what this is all about ...

.. unlocked.

[WhiteBlue]

In order to avoid more PM traffic in my mailbox I'm copying all technical content of the messages to this thread. Perhaps we can have a coherent follow up on the discussion here.

I had just typed an explanation on torque vs energy when the mod moved those posts and locked the thread into which he put them.
To summarise, if I lift 1kg against gravity 1m I have raised its potential energy 1J.

If I have a 1Nm motor driving a 1m radius drum with a 1000m rope attached to a hanging 1kg mass I can raise the potential energy of that mass by 1000J. So 1Nm (torque) does not equal 1Nm/J (force x distance).

If I have a 500Nm motor driving the same drum, rope and mass, I can still only raise it 1000m and its potential energy by 1000J.

If we attach the 1Nm motor to a 10:1 reduction gearbox we get 10Nm output. According to the theory of conservation of energy, we cannot get more energy out than in, so we need another torque input to balance things out if torque were energy.

Now, if the speed on the input is 10 rad/s, the power = 1Nm x 10 rad/s = 10W. On the output the torque is up to 10Nm and the speed is down to 1 rad/s. 10Nm x 1rad/s = 10W. Power is conserved. Which makes sense as power is the rate of change or transfer of energy.

Think of a gearbox like a lever. If we have the long side of the fulcrum 10 times longer than the short then with 1N we can lift 10N, but we need to move the input side 10 times that of the output side.

I'm afraid you are comparing apples and bananas. Torque does not relate to the energy that is put through a transmission, at least not immediately. It relates to the deformation energy stored in the transmission or released from the transmission when the torque changes, which is a big difference. Please keep in mind that a crank shaft or transmission is an elastic system that performs torsional oscillations. We are talking about the energy contained in those differential torsional movements between both ends of the shaft and not the energy that is transmitted by the overlaying rotation.

If you use your drum drive as a torsion spring and apply a certain torque to the drive shaft while you block the other end of the shaft you get the energy storage effect. If you then release the shaft from the torque the system will use the stored potential energy of the torsion spring and convert it to a higher potential of your 1 kg weight. So we get an energy conversion of potential energy of one kind to a potential energy of another kind. And those two amounts of energy are equal except for the the heat loss of the conversion. If we assume no losses you can build a system that converts one Nm into one Nm.

You are neglecting the angular displacement for your argument.
The energy stored is dependent upon both torque (equivalent to force) and angular displacement (equivalent to linear displacement).

I have already dismissed that theory in the posts that were deleted by Richard.
Force is not equivalent to torque in a scientific view. It may be in a popular view. But the popular view is not based on a valid physical model.
It is sufficient to apply a given amount of torque to a a torsional system to provide a given amount of stored energy. The angular distortion that results is a consequence of the torque applied and the torsional stiffness of the system. You can apply the same torque and have half the angular distortion and you still have the same stored energy if you make the shaft stronger.

On the other hand force is not sufficient to define energy stored in a linear elastic system. You always have to apply a stroke to complete the definition. If you keep the force constant and you vary the stroke your stored energy will vary. It is a significant difference. It is signalled in the physical dimension of the angular displacement. It is dimensionless compared to a stroke which has the dimension of a length.

Hence you cannot simply assume that torque is the equivalent phenomenon in a polar coordinate system that force is in a Carthesian system.

I hope that clarifies my position.

I was enjoying the thread...

as for wuzak's response;
It cannot be the equivalent of a force if it doesn't have the same units. So he is wrong there.

A torque can be represented by a force acting at a point AND a moment.

They say force times distance is energy correct. That is if you are moving an object along a distance. That'd their claim to fame. however...
A torque is the same thing, only that you are moving the force about a point instead of along.
Because you are moving it about, you will need to describe this motion in vector form. But it is fundamentally no different than the simple energy model of moving a mass along a distance.

The distance is instead expressed radially, and is directly related to energy. Nothing can produce a torque if it has no energy content. this is from wikipedia:

the dimensional equivalence of these units, of course, is not simply a coincidence: A torque of 1 N·m applied through a full revolution will require an energy of exactly 2π joules

...Units are usually the best guide for the nature of things. N.m is to be looked at closely. That dot between N and m is called a dot product. It is used in vector math. It is used to multiply vector quantities. The energy unit and the torque unit a both derived from similar principles.

I'm posting these statements from my private in and out boxes because they are not really meant to be private. They were sent because we had no opportunity to exchange this information on the regular public platform. Right now I feel a bit overwhelmed to continue communication with multiple recipients by private messages. I hope posting the technical content here will make it easier to push the discussion ahead.

[wuzak]

I have already dismissed that theory in the posts that were deleted by Richard.
Force is not equivalent to torque in a scientific view. It may be in a popular view. But the popular view is not based on a valid physical model.
It is sufficient to apply a given amount of torque to a a torsional system to provide a given amount of stored energy. The angular distortion that results is a consequence of the torque applied and the torsional stiffness of the system. You can apply the same torque and have half the angular distortion and you still have the same stored energy if you make the shaft stronger.

Use this test WB.

Shaft A has stiffness X. Shaft B has stiffness 2X.

Connected to each shaft we have a 1m radius drum. On that drum we have a rope, from is suspended a 1kg mass.

Now, since they are both acting at 1m they are both exerting a torque of 1Nm on their shafts.

Shaft A twists 1 radian. Shaft B twists 1/2 a radian, since it is twice as stiff.

The mass on shaft A falls 1m, transferring 1J of energy to the shaft. The mass on shaft B falls 1/2m, transferring 1/2J to the shaft.

[WhiteBlue]

I don't understand what your experiment is going to show. Particularly the use of the same weight appears inconsistent. Perhaps you can elaborate and tell us in numbers what the two systems will do according to your view. I will look at your numbers and comment.

Please keep in mind that we want to demonstrate constant torque being released. It means that the system with the high torsional spring rate will lift twice the weight over half the height. The energy exchanged will be the same and the torque as well. At least that would be the way I would build my experiment to demonstrate the nature of torque.

This experiment will demonstrate that the torsional spring rate and the angular deflection are not necessary elements to the definition of torque. A static system according to our design torqued up with the same amount of torque will provide the same amount of work or energy, regardless of the angular deflection that you selected. QED

Or to come back to our original question: A fixed amount of torque represents a fixed amount of potential energy that is stored in the angular elastic deformation of the system.

Or again put more simply: Torque is a kind of energy.

[wuzak]

I don't understand what your experiment is going to show. Perhaps you can elaborate and tell us in numbers what the two systems will do according to your view. I will look at your numbers and comment.

Please keep in mind that we want to demonstrate constant torque being released. It means that the system with the high torsional spring rate will lift twice the weight over half the height. The energy exchanged will be the same and the torque as well. At least that would be the way I would build my experiment to demonstrate the nature of torque.

Let's see.

First let me apologise for my example above. 1kg being pulled by gravity exerts a force of 9.81Nm on the rope.

The drum is 1000mm radius. Wrap a rope around it and hange a 1kg mass off the rope.

The torque applied, therefore, is 9.81m/s x 1kg x 1000mm = 9810Nmm.

Lest say shaft A and Shaft B are both 5mm in diameter. (Really small to make the defelections large with a small load.)

Shaft A is 1000mm long from where it is fixed to where the torque is applied. Shaft B is 500mm long from where it is fixed.

The polar moment of inertia of each shaft is J = π * d^4/32 = 61.4mm^4.

(I am using millimetres because I am an engineer and it is a habit!)

Suppose the two shafts are made of steel. The Modulus of Elasticity G ~ 80GPa = 80,000MPa = 80,000N/mm^2 (hence using millimetres).

The angular deflection is given by :
w = L*T/(J*G)

For shaft 1 we have:
w1 = 1000mm * 9810Nmm/(61.4mm^4 * 80,000N/mm^2)
w1 = 1.9985 radians. (as you can see, the units all cancel out, leaving us with dimensionless radians)
w1 = 114.5°

Similarly
w2 = 500mm * 9810Nmm/(61.4mm^4 * 80,000N/mm^2)
w2 = 0.9992 radians
w2 = 57.25°

That is w1 = 2 * w2.

Now, to our mass. How much has it dropped?

As you are all aware, radian = length of arc/radius

So, length of arc = radius * radian.

Length of drop: la = 1.9985 rad * 1000mm = 1998.5mm
Length of drop: lb = 0.9992 rad * 1000mm = 999.2mm

As we know, Potential energy = m*g*h
So the change in potential energy for mass A: PEa = 1kg * 9.81m/s^2 * 1.9985m = 19.605J.
So the change in potential energy for mass B: PEb = 1kg * 9.81m/s^2 * 0.9992m = 9.803J.

With the same torque applied Mass A has had twice the potential energy change as that of B. And thus the stiffer shaft stores half the energy than the less stiff shaft does.

[wuzak]

I have also been corresponding with Ringo on the nature of Torque. This is my latest reply to him:

Torque is not a force. That is for certain.
Force is not potential energy i agree, but a torque is not a force as i said. You made that claim.

I made no such claim. What I said is that Torque is the rotational equivalent to Force. Not equal to Force, but analogous.

It is a tendency of a force. It is not a force. It cannot be equivalanet to a force if it is a product of a force and something else.

As above, I never said Torque was a Force or equal to Force. You misinterpret or misrepresent what I mean by rotational equivalent. By that I mean that Torque works in rotational systems the same way force works in linear systems. I will hopefully prove that to you below.

It makes sense, becuase pi is a dimensionless number. it is a quantity. No units, so it doesn't affect the physical nature of what a torque is.
And it goes to show you that For a torque to exist energy must be expended. If you want a torque to exist for 10 revolutions you will obviously see that energy required will be 20π Joules.

No, it doesn't make sense. Dimensionally 2π Nm = J, but physically and mathematically it does not.

The 20πJ after 10 revolutions is the amount of work done by 1Nm over 10 revolutions (20π radians).

Torque can, like Force, exist without doing work. Example, I can push as hard as I like against a big brick wall, but I can't move it. I may have expended energy, but I haven't doen any work.

Similarly Torque can exist without rotation. If I bolt my torque device to a similarly solid structure, I can expend energy applying torque without doing work.

It is not a confusion. It is a similarity. I have stated the difference already. Once has a direction, a vector quantity, the other doesn't, a scalar quantity.
It's like the difference between velocity and speed. Both are the same fundamentally and different functionally. One had direction the other doesn't. But at the end of the the day they can be manipulated in the same manner. Like your example below with the drum.

Yes, you are confused.

Energy/work and torque are very different concepts, whose units just happen to match dimesnionally.

Energy transfer right there. You are shooting your own argument in the foot.

What? That 1J work in a rotational system can be translated to 1J work in a linear system. Shocking!

But I can use 2,000,000Nm to get the same 1J of work. It is all abou the displacement.

Power = Energy/time, Power= torque x angular velocity (radians per second)

Power is not force x time. You calling a torque the equivalent of a force is categorically wrong.
You cannot dispute this.

You are being deliberately diingenuous.

Angular velocity is not time.

Power = torque x angular velocity

is equivalent to

Power = force x velocity

Let's break this down further.

Rotational work is W = T x w (where w = angular displacement).

Linear work is W = F x s (where s = linear displacement).

Power = work/time, so:

Power = (T x w)/t = T x w/t = T x angular velocity
Power = (F x s)/t = F x s/t = F x V


The factor that changes between linear and angular systems is radius (r).

T = F x r
w = s/r
Angular velocity = V/r
Angular acceleration = a/r (where a is linear acceleration)

Torque is the rotation equivalent of force in the same way that angle is the rotational equivalent for position, angular velocity for velocity, and angular momentum for momentum. As a consequence of Newton's First Law of Motion, there exists rotational inertia that ensures that all bodies maintain their angular momentum unless acted upon by an unbalanced torque. Likewise, Newton's Second Law of Motion can be used to derive an analogous equation for the instantaneous angular acceleration of the rigid body

http://en.wikipedia.org/wiki/Force#Rotations_and_torque

It's the conservation of energy that describes the relationship between torque increase with speed reduction.
If you have two gears in contact, both must have the same tangential velocity at the contact point.
torque x angular velocity of gear1 = torque x angular velocity of gear2

If one torque is doubled, the other will be halved. It is this vector quantity that you all are ignoring.

Tangential velocity of the two gears will be the same at the point of action. However, their angular velocity will not be.

If R1 does not equal R2 then:
w1 = V (tangential velocity)/R1
w2 = V/R2

and w1 cannot be equal to w2.

That last staement doesn't make sense. The driving gear's torque will remain whatever it is, while the driven gear's torque will be doubled (for a 2:1 reduction). The angular displacement of the driven gear will be half that of the driving gear, 1Nm x 1rad/s (driving gear) = 2Nm x 0.5rad/s (driven gear), and thus energy is conserved.

Torque is not conserved. Power is (as it is energy over time) - power on the input is the same as power on the output (ignoring losses). As you pointed out, power is work/energy/time.

In conclusion, i am not calling torque scalar energy like heat or whatever other forms you have. All i am saying is that it is more related to energy than a force. In fact it can be converted directly to energy as you have demonstrated with the drum. Only energy can turn to energy.

Work is performed by Torque and angular displacement. And work is performed by Force and linear displacement.

Neither Torque or Force can do work without displacement.

A force is a constituent of torque. They are not equal.

I never said they were. I said, and the all-knowing Wiki backs me up, that Torque is the rotational equivalent of Force.

Any torque that exists can be turned to energy. And this is why you have your KERS working on your drive axle in the first place when you brake. Or a wind turbine, The same can't be said of a simple force in isolation.

If you apply a torque without rotation (which you can do) then you aren't converting it into energy. It needs displacement to do work.

[Richard]

If you apply a torque without rotation (which you can do) then you aren't converting it into energy. It needs displacement to do work.

To me, this is the crux of the conversation. Movement has to occur for there to be any energy.

I think we need to differentiate between N.m and Nm. Torque or bending moments just happen to be forces that existin at an eccnetrictiy from your local axis. Move the axis and the torque disappears *puff*

So my thought experiment is…..

1 - Hold a hammer at the head, ie centre of gravity. Your hand feels N, ie a force on your local axis

2 - Hold the hammer at the handle and your hand feels N.m – ie a force at a distance from your local axis, we call it bending moment or torque (they’re the same)

The hammer has not gained energy between 1 and 2. It is in the same space, it has not moved (potential energy) or moving (knietic energy). Your hand simply experiences forces applied at different eccentricities.

Now place the hammer on a spring and let go. The spring will move as the gravitional potential energy is transferred into compressing the spring. The energy added to the spring is:

Movement of the hammer * hammer self weight * gravity

Gravity and the hammer do not know if the spring is straight, circular or spaghetti. What counts is the input energy (ie movement of the hammer). Anything that happens inside the closed system of the spring can’t create or destroy energy. This means we can calculate input energy and forces regardless of the configutaion of the spring.

So in a nutshell…

- Nm is a measure of energy.
- N.m is the unit for torque or bending, not Nm
- Anyhting measured as Nm can be added to anything else measured as Nm
- Anyhting measured as N.m can be added to anything else measured as N.m
- N.m can not be added to Nm

[amc]

The idea that the distinction between the units in energy and torque is made by the dot product as opposed to the scalar product is immediately flawed. Consider a kN torque applied at a distance of lm from an origin:
The torque vector is (1i + 0j + k(0i + 1j)) (consider it as a vertical line passing through (1,0) )
The displacement vector is (0i + 0j + l(1i + 0j)) (consider it a horizontal line passing through the origin)
The dot product is 0. In fact, the dot product of any two perpendicular vectors is zero, and since torque is defined as the product of two perpendicular vectors... you can conclude for yourself that torque cannot be calculated as a dot product.
That's a vital point: torque is not a product of force and displacement.
What we have in fact demonstrated by finding that the dot product equals zero is that torque on its own transfers zero energy. This has been correctly quoted by many members - the expressions above explain why.

Instead, torque is a product of force and radial distance. Whether this next section has true scientific basis or not is up for debate, but it clarifies some of the confusion surrounding units.
I have always considered angular rotation as having units of rad/s. For torque x angular velocity to still equal power (Nm/s) the torque must have units of Nm/rad. It comes as a product of a force (N) and a radial distance (m/rad).

A radial distance is not the same as a displacement; it is meaningless, for example, to add the circumference of a circle (a distance) and its radius (a radial distance). The fact that torque is calculated using a radial distance, and energy using a displacement, is the difference between the two.

[WhiteBlue]

Apologies for making a late edit to my last post. I had it on the iPad and it was unsend. Then I just send it when I came back without checking what had been written. Sorry for that.

So lets repeat the look at the experiment. My view is it will not work as planned by Wuzak.

It will work just fine if the weight is adjusted. For half the angular elastical displacement, which is reached at double stiffness you must use twice the weight. It will travel half the height. I do not have to do the numbers because it is obvious for everybody that the work applied to the weight is the same in both cases.

And this is the proper demonstration that the energy of an elastic deformation in a torsional system is equivalent to the toque applied. The stiffness or the angular displacement is not relevant as we have seen. Only the torque defines the energy. You can use any stiffness or any angular displacement and you always keep the same torque, provided the stiffness and angular displacement are properly matched.

The same is not true for force in a linear system. It is not sufficient to define the energy of a linear elastic deformation because the linear deformation would be required as well to define the energy.

[Richard]

The idea that the distinction between the units in energy and torque is made by the dot product as opposed to the scalar product is immediately flawed.

.....

A radial distance is not the same as a displacement; it is meaningless, for example, to add the circumference of a circle (a distance) and its radius (a radial distance). The fact that torque is calculated using a radial distance, and energy using a displacement, is the difference between the two.

So the curious thing is that it is possible to have two measurement as Nm, one for energy and one for moment/torque that can't be added.

That's counter to the school explanation of SI which infers that numbers with the same units can be added. The answer is that radical distance and displacement are different units even though they use the same nomenclature.

So that take us back to the point of using Watts and Joules.