Calculate Torque to lift 7'-9' pole with 5-8 lbs at the end

[Awagenhurst]

I'm trying to figure out what type of motor I need to use that has the ability to lift a 7'-9' pole that has about 5-8 pound weight attached to it at the end. The max angle that the poll will be is 180 degrees and the motor will need to lift it to 90 degrees. As for the weight its not a free swinging weight its attached. The poll will need to be lifted from the bottom end.

 

[anorlunda]

A picture or a sketch will help us to visualize what you are talking about. Use the Upload button.

 

[Awagenhurst]

Anorlunda,
Thank you for the reply, hopefully this diagram helps.

Thank you

 

[jbriggs444]

So you have a pole up to 9 feet in length which masses 20 pounds. On the end of this pole is a mass of up to 8 pounds. You wish to be able to rotate this pole in a controlled fashion up from horizontal to vertical or down from vertical to horizontal. No speed requirements are given.

Since there is no speed requirement, any size motor will do. You just need to gear it down sufficiently to provide the requisite torque.

 

[Awagenhurst]

Jbriggs444,
This much I know. I just need to know what is the minimum amount of torque that is required to perform such a feat . Once I know the torque, I can than get a decent motor and apply the appropriate, gears ration, pulley ratio and or worm gear to complete this task. I just don't know the formula to calculate the torque I need.

Thank you

 

[jbriggs444]

From ordinary experience, it should be clear that the required torque is at its maximum when the pole is horizontal. However, there is a simple approach that can address that concern: A counter-weight. Or springs.

If we are going to determine how much torque you need, we need to know what sort of counter-weight you are able to provision. And how variable the load will be.

Edit:

That said, the formula for torque is easy -- force multiplied by perpendicular distance. If the pole is nine feet long, weighs twenty pounds (force) and is extended horizontally, that's 4.5 feet * 20 pounds = 90 pound-feet of torque about the endpoint. Add to this an eight pound (force) weight nine feet out, that's another 9 feet * 8 pounds = 72 pound-feet of torque about the same endpoint. Total 162 pound-feet.

Minus whatever you can arrange to cancel with suitable counterweights, springs and such.

 

[CWatters]

So you have a pole up to 9 feet in length which masses 20 pounds. On the end of this pole is a mass of up to 8 pounds.

If no counterweight is possible...

The torque required to just hold it horizontal it would be..
(20*4.5) + (8*9) = 162 ft.lbs
In practice you need more torque than that to move it but how much depends on how fast you want to move it.

 

[Awagenhurst]

Jbriggs/CWatters,
Thank you for all the input.

CWatters,
You mentioned about the torque would be greater when I needed the pole to move, how would speed be calculated into the torque. As for the speed I would like it to move, I don't have a set speed in mind, would the formula be the same for any set speed. If so a default speed would be 1 inch per second.

"For my information"
Just wondering as to why the 20 pounds gets multiplied by half the the size of the pole.

Thank you

 

[jbriggs444]

You mentioned about the torque would be greater when I needed the pole to move, how would speed be calculated into the torque.

Speed, as such, does not get calculated into torque (*). Acceleration does. It takes extra torque to get up to speed.

(*) Viscous friction (e.g. air resistance or oil in the bearings) can be proportional to speed. But for the speeds we are considering, air resistance is negligible.

Just wondering as to why the 20 pounds gets multiplied by half the the size of the pole.

The center of mass of the pole is at 4.5 feet from the axis of rotation. The total weight of the pole can be considered as if it acted at that point.

 

[CWatters]

+1

Torque = moment of inertia * angular acceleration.

The pole starts and ends at rest but obviously has some velocity in between. So it has to accelerate from rest up to some angular velocity then decelerate again to a stop.

You can use the equations of motion to work out how fast it must accelerate/decelerate to move from one position to the other in a set time.

The velocity profile/curve may need to be carefully controlled.. For example if the pole comes to a very sudden stop the torque the pole applies to the motor will be large, possibly damaging gears or bending the pole.

 

[Awagenhurst]

CWatters,
Thank you again for all your help. Is it possible to provide me with an example of torque. The pole will not come to a sudden stop, the amount of time to go 90 degrees I would want to take a minute and half to 2 minutes. So if I make it 90 degrees for minute and half it'll be 1 degree every 1 second. Would it be possible to calculate torque with that given information.

Thank you

+1

Torque = moment of inertia * angular acceleration.

The pole starts and ends at rest but obviously has some velocity in between. So it has to accelerate from rest up to some angular velocity then decelerate again to a stop.

You can use the equations of motion to work out how fast it must accelerate/decelerate to move from one position to the other in a set time.

The velocity profile/curve may need to be carefully controlled.. For example if the pole comes to a very sudden stop the torque the pole applies to the motor will be large, possibly damaging gears or bending the pole.

 

[jbriggs444]

CWatters,
Thank you again for all your help. Is it possible to provide me with an example of torque. The pole will not come to a sudden stop, the amount of time to go 90 degrees I would want to take a minute and half to 2 minutes. So if I make it 90 degrees for minute and half it'll be 1 degree every 1 second. Would it be possible to calculate torque with that given information.

That gives the angular velocity, but to calculate the angular acceleration we need to know how suddenly it starts and stops.

 

[CWatters]

Well you could assume that it accelerates with constant acceleration from rest for 45 seconds over 45 degrees, then decelerates at the same rate for 45 seconds. It would cover the 90 degrees in 90 seconds.

You can use the SUVAT equations eg...

S=0.5(u+v)t

to calculate the velocity v in radians/sec at the 45 second/degree point.

s=displacement (eg 45 degrees but in radians)
u=0 as it starts from rest
t=45 seconds

Then plug the value for v into
V²=U²+2as
To calculate the acceleration a in radians/s/s

I'd do the numbers for you but I'm one finger typing this on a useless tablet at the moment.

 

[CWatters]

You also need to calculate the moment of inertia but that can be done using standard formula for a rod rotated about one end and a point mass on a massless rod.

I'll see if I can do the numbers on my pc tomorrow.

 

[CWatters]

Well you could assume that it accelerates with constant acceleration from rest for 45 seconds over 45 degrees, then decelerates at the same rate for 45 seconds. It would cover the 90 degrees in 90 seconds.

You can use the SUVAT equations eg...
S=0.5(u+v)t
to calculate the velocity v in radians/sec at the 45 second/degree point.

u=0 so
v = 2S/t
45 degrees = 0.79 radians so
v = 2*0.79/45 = 0.035 rads/s

Then plug the value for v into
V²=U²+2as
To calculate the acceleration a in radians/s/s

u=0 so..
a = V²/2s
= 0.035²/(2*0.79)
= 7.75 8 10^-4 rads/s/s

More in a bit.

 

[CWatters]

Sorry I prefer working in metric/SI units..

20lbs = 9kg
8lbs = 3.66kg
9ft =2.7m

With reference to http://hyperphysics.phy-astr.gsu.edu/hbase/mi2.html

To calculate the torque we first need to know the moment of inertia. The system comprises two parts:

a) Uniform rod pivoted at one end... I_r = (1/12)m_rL² where m_r is the mass of rod and L the length
b) Point mass on one end of a massless rod I_m = m_mL² where m_m is the mass of the point mass and L the length

Total I = I_r+I_m
= (1/12)m_rL² + m_mL²
= (1/12)*9*(2.7)² + 3.66*(2.7)²
= 5.47 + 26.68
= 32.15 kg.m²

Torque required to accelerate the rod and mass = Moment of inertia * angular acceleration
= 32.15 * 7.75 * 10^-4
= 0.025 Nm

In same units the torque required to hold pole horizontal =

= 9.81(9*1.35 + 3.6*2.7)
= 9.81(12.15 + 9.72)
= 214 Nm

So the torque required to hold it horizontal totally dominates that required to accelerate the rod and mass. It also shows how effective counterbalancing would be in reducing the torque required.

For info 214 Nm is about 157 ft lbs

I'm in a rush so someone best check the sums before relying on them.

 

[jbriggs444]

a) Uniform rod pivoted at one end... Ir = (1/12)mrL2 where mr is the mass of rod and L the length

The 1/12 figure is for a rod pivoted in the middle. It's 1/3 for a rod pivoted at one end. Of course, even with this factor of four, the torque required for the gentle acceleration is still dominated by the torque required to resist gravity.

One could intuit the domination without calculation by comparing how rapidly the rod+mass would "flop" down under gravity to how rapidly the combination is accelerated upward at the desired rate.

 

[jack action]

First, to size a motor, you must define how much power you need. Let's do a simplified analysis.

The power is defined by the torque times the angular velocity. The maximum torque needed will be when the pole is horizontal at 162 lb.ft or 220 N.m. It will decrease at it goes up. The average speed is 1 deg/s or 0.0175 rad/s. Therefore, assuming you've reach the average speed instantly, the maximum power would be:
$$220\ N.m \times 0.0175\ rad/s = 3.85 W$$
So you need a motor than can produce about 4 W. But is it reasonable to assume that the average speed would be reached almost instantaneously?

Let's assume we want to reach the average speed within 1 deg of rotation or 0.0175 rad. So the constant acceleration needed is (as @CWatters showed how):
$$\frac{(0.0175\ rad/s)^2}{2 \times 0.0175\ rad}= 0.00875\ rad/s^2$$
With the corrected mass moment of inertia (1/3 instead of 1/12) as calculated by @CWatters, i.e. 48.6 kg.m², you can find the extra torque needed to accelerate the pole to its average speed within 1 deg of rotation:
$$48.6\ kg.m^2 \times 0.00875\ rad/s^2 = 0.43\ N.m$$
Which is a ridiculous amount compared to the 220 N.m that your motor is already producing to hold the pole. So it is OK to assume that you will reach your average speed almost instantaneously with such a motor.

 

[jbriggs444]

So it is OK to assume that you will reach your average speed almost instantaneously with such a motor.

Given the acceleration profile (accelerate for 45 seconds, decelerate for 45), the average speed is reached 22.5 seconds into the run, not instantaneously. With that profile, the extra power needed to accelerate is negligible.

Given an acceleration profile in which the rod reaches its average speed within 1 degree and an average speed of one degree per second, the acceleration will take 2 seconds. That's not instant -- though the angular acceleration over that 2 second interval would still be low.

 

[sophiecentaur]

The pole will not come to a sudden stop, the amount of time to go 90 degrees I would want to take a minute and half to 2 minutes.

I was just thinking about the 'control' aspect of this problem. How accurate do you want your angle of rotation to be? Can you have an end stop or would you need the motor to slow and then stop in the right position? That could be as big a problem as calculating the basic grunt that the motor can give you.

 

[jack action]

Given the acceleration profile (accelerate for 45 seconds, decelerate for 45), the average speed is reached 22.5 seconds into the run, not instantaneously. With that profile, the extra power needed to accelerate is negligible.

Given an acceleration profile in which the rod reaches its average speed within 1 degree and an average speed of one degree per second, the acceleration will take 2 seconds. That's not instant -- though the angular acceleration over that 2 second interval would still be low.

The first point was that to size a motor, you need to find the power output. To do so, I used the maximum torque (pole horizontal) where obliviously the speed and power are zero. With reaching the 1 deg/s speed within 1 deg of rotation, the total trip takes 88 s (88° @ 1 deg/s) + 2 s (1° under acceleration) + 2 s (1° under deceleration) for a total of 92 s. This is why I'm talking about "almost instantaneously"; The average speed is practically equal to the maximum speed. Therefore, the assumption that the maximum power required is the maximum torque times the maximum speed is valid.

The second point was that with such a small speed, you don't need a slow ramp (45°) for the acceleration.