Torque , general question

EDIT: here are some useful links for getting the basics of torque

for the readers (from reply #2)

for the watchers (starting at 3:20 he goes into the specifics of a practical torque calculation)
(Jeremy Fielding, highly recommend him, has some really beginner friendly videos)



Hello,

I'm working on a project where I'd like to use a stepper motor.
I've done some research to get an idea of the torque required and I've seen the following in the arduino forums:

Torque is how much you "weight" you can pick up by wrapping around a shaft. (the amount of Horsepower determines how fast you can move it).

Torque is Measured in "distance/weight" -- "inch/lbs" or "cm/kg" or "foot/lbs" or even "mile/tons" (but that is kind unrealistic, but theoretically possible) The metric world has a N/m or Newton/meters, but I've never felt it was practical.

Torque is a pretty straightforward equation: (think of a cable wrapping around a drum on a shaft)

Torque is [how far you are from the center of the shaft] divided by [how much you are trying to lift by rotation]

So if torque = 10 cm/kg, then you can pick up 10kg that 1cm away from the center of the shaft OR>>OR>OR>> 1kg that is 10cm from the center of the shaft OR>> any "math" that makes the equation true:
2kg that is 5cm away
3kg that is 3.33cm away
.01kg that is 1000cm away

since the farther you get away from the center of the shaft, the "linear speed" increases, so (if HP stays constant) then the weight you can move must decrease

and also

A simple method to get a rough measurement of the required torque.

Attach some sort of wheel or drum to the shaft that the stepper motor will be required to turn. Wrap some strong thread or fine string around the drum and suspend a small plastic beaker from it. This will obviously only work if the shaft is horizontal.

Add coins to the beaker until the weight is just sufficient to make the drum rotate. Weigh the beaker with the coins in it.

Measure the diameter of the drum where the thread is wrapped and calculate the radius. Suppose the radius in 2cm and the weight is 100grams. Then the torque is 200gm-cm.

Repeat the measurement several times and take an average. Choose a motor with perhaps twice that amount of torque to provide a good safety margin.

I have a rough understanding that if the lever arm decreases, the weight can increase.
With this in mind it would be possible to select the correct torque required.

My question is, how do I calculate the torque available if I choose not to use a lever arm and just mount the project directly to the motor shaft?

The way I see it:
I'll take the 10cm/kg example.
It states that at 1cm I should be able to lift 10 kg, and torque is measured in distance/weight but this doesn't make sense if I use it as follows:

Known load to calculate torque required.

  • A distance off 0.01 cm as I'm working directly from the shaft
  • The load of let's say 50kg
    Torque = distance/ mass = 0.01cm/50kg = 0.0002 cm/kg?

Known torque to calculate maximum load

  • A distance off 0.01 cm as I'm working directly from the shaft
  • The load of let's say 50kg
    Torque = distance/ mass --> mass = distance / torque = 0.01 cm/ 10cm/kg = 0.001 kg ?

Both of the calculation results don't make sense to me hence why I'd like to call in the forum help.
I've been staring at the examples* and something tells me that it's really simple but I can't seem to grasp it.

2kg that is 5cm away
3kg that is 3.33cm away
.01kg that is 1000cm away

Thank you

rgds

quebeclima:
The way I see it:
I'll take the 10cm/kg example.
It states that at 1cm I should be able to lift 10 kg, and torque is measured in distance/weight but this doesn't make sense if I use it as follows:

The bit I have highlighted is wrong. Torque is expressed as weight (force) times distance (no division involved). So it should be 10kg.cm where the period signifies multiplication.

Now you should be able to see that if the distance is reduced to 0.1 cm (i.e. one tenth) the weight can be increased ten times because 10 * 1 is the same as 100 * 0.1

...R

Good tutorial on torque and force: Pololu - Force and torque

Robin2:
The bit I have highlighted is wrong...
...R

I see, that does make more sense, thank you!

I didn't want to pinpoint it to that as I've also seen ft-lbs used as a torque measurement and if I looked at the example, cm/kg wouldn't be too strange, but then again: following a formula and it's structure seems more sensible...

I'll just stick to NM as most stepper motors give this as a torque indication on their spec sheet.

jremington:
Good tutorial on torque and force: Pololu - Force and torque

thanks

Nothing in this sentence is correct. Typical internet misinformation, in this case U.S. centric.

Torque is Measured in "distance/weight" -- "inch/lbs" or "cm/kg" or "foot/lbs" or even "mile/tons" (but that is kind unrealistic, but theoretically possible) The metric world has a N/m or Newton/meters, but I've never felt it was practical.

Lets get this right. a torque, a.k.a 'moment' or 'couple' is:

the product of force (not weight, weight is just one kind of force), and a distance.

The distance is the minimum (perpendicular) distance between the line of action of the force and
the axis of the rotating shaft or whatever. For a cable drum / belt drive this is the radius of the drum
or sprocket. The line of action of the force is crucial, don't forget this.

If the force isn't at right angles to the shaft you first have to resolve the force into an axial
component and tangential component. Only the tangential component contributes to torque.

The standard units for force and distance are newton and metre respectively, so torque is
measured in newton-metres (Nm). The hyphen in newton-metres is not mathematical subtraction,
don't be fooled by that, its just a linquistic hyphen.

You can also treat torque as a ratio of energy to angle, and measure it in joules/radian (J/rad).
This is exactly analogous to force being a ratio of energy to distance.

Its very common to see people confused by torque - the way to remember it is to imagine a
lever like a spanner - a longer spanner has more leverage, and a stronger hand on the spanner
also has more leverage.

MarkT:
Lets get this right...

That clears things up, I believe there's alot of confusion especially since people tend to use different units of measurement instead of just Nm, understandably because Newton is not really something tangible.
But I feel I would benefit more using Nm since this is literally what the formulla dictates.

As a bit of a follow up to see if I get it right I've attached a picture.
Am I correct in assuming:

A] If I wanted to attach my project to the shaft (red), would I be ok to use the shafts radius in my calculations to get the 'distance' of the load?
B] (The image might not make much sense CG wise, but it's just to get an idea) If I take into account the CG of the project, am I correct in assuming the arm is portrayed as in the picture (green, where the CG is off-centre) , and therefore I'd have to use said arm in the calculations as my distance?

![](http://<a href=)


what's on my screen
">

I wonder if this diagram illustrates the matter more clearly

...R

Robin2:
I wonder if this diagram illustrates the matter more clearly

yep, this confirms using the radius off the shaft would be my solution for the calculation.

Any idea if my assumption off the CG is correct?

quebeclima:
yep, this confirms using the radius off the shaft would be my solution for the calculation.

Any idea if my assumption off the CG is correct?

About the red part:
I'm not 100% sure what you mean by the red shaft. Is the project supposed to be a globe spinning around an axis, powered by a motor?
In that case:
The globe's center of mass would be exactly in the middle of the axis, and not on the radius of the axis (imagine if you would change the diameter of your axis, the center of mass of the globe would not change.)

In this case the gravity on the globe would not result in a torque on the axis. (imagine not powering the axis with a motor, the axis+globe wouldn't start spinning on it's own ->no net torque on the axis)
If it is supposed to be a globe spinning, then the torque you would have to overcome is simply the rotational resistance that spinning the globe has. (imagine spinning the globe by hand, but not powering it by a motor. If you let it now spin on its own, it will keep spinning for a while, but eventually slow down, this is due to the rotational resistance, which results in a torque in the opposite direction of the spinning motion, hence the globe will slow down.)

When the globe first starts spinning, you will also need some torque to get it up to speed, but unless your project is very heavy, this is not too much. If you want to know more about this, you should google on 'rotational inertia'. It basically means the factor between the torque, and the rate of change in angular velocity (spinning speed).

About your green part:
If you Center of gravity is a distance r removed from the axis, and this distance r is represented by the left vertical green line, then yes, the radius your are looking for is that r that you drew in the right vertical green line. (So indeed NOT the diagonal distance from the motor to the CG, as some people think).

But if you are willing to tell, what is the project? Maybe we can help a little more effectively if we know.

quebeclima:
Any idea if my assumption off the CG is correct?

I don't know because I can't figure out what you are trying to do.

...R

chris_abc:
About the red part:
I'm not 100% sure what you mean by the red shaft. Is the project supposed to be a globe spinning around an axis, powered by a motor?
In that case:
The globe's center of mass would be exactly in the middle of the axis, and not on the radius of the axis (imagine if you would change the diameter of your axis, the center of mass of the globe would not change.)

In this case the gravity on the globe would not result in a torque on the axis. (imagine not powering the axis with a motor, the axis+globe wouldn't start spinning on it's own ->no net torque on the axis)
If it is supposed to be a globe spinning, then the torque you would have to overcome is simply the rotational resistance that spinning the globe has. (imagine spinning the globe by hand, but not powering it by a motor. If you let it now spin on its own, it will keep spinning for a while, but eventually slow down, this is due to the rotational resistance, which results in a torque in the opposite direction of the spinning motion, hence the globe will slow down.)

When the globe first starts spinning, you will also need some torque to get it up to speed, but unless your project is very heavy, this is not too much. If you want to know more about this, you should google on 'rotational inertia'. It basically means the factor between the torque, and the rate of change in angular velocity (spinning speed).

About your green part:
If you Center of gravity is a distance r removed from the axis, and this distance r is represented by the left vertical green line, then yes, the radius your are looking for is that r that you drew in the right vertical green line. (So indeed NOT the diagonal distance from the motor to the CG, as some people think).

But if you are willing to tell, what is the project? Maybe we can help a little more effectively if we know.

You've basically answered all my questions and even made me aware of the rotational inertia which I didn't take into account, I'll have to do some research on that.
As mentioned, I want to attach my load directly to the shaft of the motor, but how strong the motor needs to be depends on the torque needed to move the load.
It's now confirmed that how much torque I need to overcome depends on the CG as this might create a lever arm if it's not centred to the shaft of the motor.
The initial image I shared was to get an idea if I understood the concept of torque with relation to CG
You explained and confirmed it perfectly what I was looking for, looks like I remembered my basic vector physics right.
Like I said, you answered my questions, but to give you an idea what I'm trying to realize I took an image of the model I'm working on :
example

It's a motion platform in very early stages but basically I needed to know the influence of CG so I can balance out the seat before starting the actual build so I don't have to overcome the torque by bad balancing, and just the torque of the weight of the project, hope it makes some sense.

quebeclima:
It's a motion platform in very early stages but basically I needed to know the influence of CG so I can balance out the seat before starting the actual build so I don't have to overcome the torque by bad balancing, and just the torque of the weight of the project, hope it makes some sense.

If the centre of gravity of "the weight of the project" lies on the axis of rotation it won't contribute any torque - all you will need to be concerned with is the moment of inertia.

If (as your diagram suggests) the centre of gravity can be moved above or below the axis of rotation then you will need to allow for the torque due to the displacement of the CoG at its maximum distances - as well as the altered moment of inertia.

...R

PS ... a lot of time could have been saved by posting that diagram in your Original Post :slight_smile:

Robin2:
If the centre of gravity of "the weight of the project" lies on the axis of rotation it won't contribute any torque - all you will need to be concerned with is the moment of inertia.

If (as your diagram suggests) the centre of gravity can be moved above or below the axis of rotation then you will need to allow for the torque due to the displacement of the CoG at its maximum distances - as well as the altered moment of inertia.

...R

PS ... a lot of time could have been saved by posting that diagram in your Original Post :slight_smile:

noted, thank you for your help!

Just a quick point I wanted to add so that you don't make a mistake that I see people often make in this type of problem (and have made my self as a mech. engineering student):

The distance between your gravity vector to your motor axis is the actual arm you are looking for. Notice that this distance changes depending on where the person is w.r.t. the motor.


So the distance you are looking for is as stated in the picture: A=sin(θ)*r
NOT A = r

Imagine rotating your system by hand instead of with the motor, I think you can imagine that it gets heavier to do the bigger theta becomes, this is due to your arm becoming larger and larger for a larger theta (up until theta is 90 degrees, after that it becomes smaller again). In fact, if theta is 90 degrees, sin(θ)=1 and the formula does become M=r*F, but ONLY for theta=90 degrees. Let me know if this confuses you because it's quite important to grasp this before you go further I think.

Another note of safety: Make sure the seat of the person is located so that the COM of the person is BELOW the motor axis, such that if there is a motor failure, the person will simply swing back to the position theta = 0. If the COM were above the axis, the person would fall over in case of a motor failure.

chris_abc:
Just a quick point I wanted to add so that you don't make a mistake that I see people often make in this type of problem (and have made my self as a mech. engineering student):

I can't relate your diagram to the OP's diagram in Reply #12

...R

chris_abc:
Just a quick point I wanted to add so that you don't make a mistake that I see people often make in this type of problem (and have made my self as a mech. engineering student):

The distance between your gravity vector to your motor axis is the actual arm you are looking for. Notice that this distance changes depending on where the person is w.r.t. the motor.


So the distance you are looking for is as stated in the picture: A=sin(θ)*r
NOT A = r

Imagine rotating your system by hand instead of with the motor, I think you can imagine that it gets heavier to do the bigger theta becomes, this is due to your arm becoming larger and larger for a larger theta (up until theta is 90 degrees, after that it becomes smaller again). In fact, if theta is 90 degrees, sin(θ)=1 and the formula does become M=r*F, but ONLY for theta=90 degrees. Let me know if this confuses you because it's quite important to grasp this before you go further I think.

Another note of safety: Make sure the seat of the person is located so that the COM of the person is BELOW the motor axis, such that if there is a motor failure, the person will simply swing back to the position theta = 0. If the COM were above the axis, the person would fall over in case of a motor failure.

I think I understand what you're saying, but I believe your example only applies when the COM is not directly in the axis of the motor right?
Ideally I would try to balance out the structure so the COM coincides with the motor axis, basically eliminating the influence gravity has.

quebeclima:
I think I understand what you're saying, but I believe your example only applies when the COM is not directly in the axis of the motor right?
Ideally I would try to balance out the structure so the COM coincides with the motor axis, basically eliminating the influence gravity has.

In theory, if your COM is exactly in line with the motor axis, you are correct in that the influence of gravity will be zero. However I doubt you can get it to be exactly in line, as modelling a person is quite hard (everyone is different), and the thing you are modelling a COM of will also move and alter it's own geometry (i.e. the person will move their head, their arms, or even their legs), which would change your COM.

If I had to choose between the option of designing my COM slightly under the axis and needing a slightly heavier motor for that, or the risk of the person in the device raising their arm and thereby toppling over, I would definitely go for the first one.
Especially considering that you will need quite a strong motor anyway, if you want to move around rotate a person at a reasonable speed. Coming back to moment of inertia: a heavier, larger object will be harder to start rotating then, for example, a lightweight plastic globe.

chris_abc:
In theory, if your COM is exactly in line with the motor axis, you are correct in that the influence of gravity will be zero. However I doubt you can get it to be exactly in line, as modelling a person is quite hard (everyone is different), and the thing you are modelling a COM of will also move and alter it's own geometry (i.e. the person will move their head, their arms, or even their legs), which would change your COM.

If I had to choose between the option of designing my COM slightly under the axis and needing a slightly heavier motor for that, or the risk of the person in the device raising their arm and thereby toppling over, I would definitely go for the first one.
Especially considering that you will need quite a strong motor anyway, if you want to move around rotate a person at a reasonable speed. Coming back to moment of inertia: a heavier, larger object will be harder to start rotating then, for example, a lightweight plastic globe.

Ok then we're on the same page. I agree with the fail-safe of having the CG slightly below the axis. I'm looking at a motor of 8 to 12NM, maybe have to add a gearbox, but I'm trying to gather as much info as possible to make sure calculation wise (as far as I can) it's correct.