Calculating the required motor size to move a load.

In summary: So the force is proportional to the mass and the acceleration?Your set up is similar to a piston and con rod in a car. You can use geometry to work out the peak acceleration (which won't be equal to g) and then apply f=ma.Hmmmm... If there is no friction, then once the mechanism is up to speed, is there any more work to be done?Thank you for that answer. Yes, there is more work to be done as the box has to reach its final destination in 0.06 seconds.
  • #1
Anasazi
18
0
Hello,

From what I've read in the stickies, this question may be more suited to this homework forum, although please note I'm not a student nor is this homework (it's simply for a personal project I'm doing), so please accept my apologies in advance if there are significant gaps in my knowledge that would be expected of a physics student...

I would really appreciate it if somebody could check the following for me. There's not a lot here, however I've tried to go through every step I need in detail so things are clearer hence the size:


Homework Statement



Right, I've got a box that moves along a straight path, then once reaching the end of the path it reverses and moves back along the straight path.

The mass of the box is 5 kg.
The distance the box moves (from one side to the other) is 5 cm.
The time the box must move this distance is 0.06 seconds.

I'm trying to work out the size of the motor required to move this box within the given time constraints. Now obviously the motor is rotary and the box moves in a linear fashion, so the motor would be fixed to a crank and so would work similar to how a steam engine moves a locomotives wheel.


Homework Equations



Just a couple of pointers, I'm ignoring any inefficiencies in the motor, gearing and friction etc. So, for the object to move, I have the following:


Distance to move object: 0.05 metres (5cm)
Mass of object: 2 KG
Time to move object: 0.06 second

Force required to move object: Mass x 9.8
Force required to move object: 2 x 9.8
Force required to move object: 19.6 Newtons

Torque required to move object: Distance (m) x Force (Newtons)
Torque required to move object: 0.05 x 19.6
Torque required to move object: 0.98 Newton-metres

RPM required to complete distance in time: 60 (seconds per minute) / time to move object
RPM required to complete distance in time: 60 / 0.06
RPM required to complete distance in time: 1000 rpm
One RPM actually moves the object twice, so halve the RPM:
RPM required to complete distance in time: 500 rpm


For the motor calculations, I have the following. Unfortunately this is all I know of the motor, so I'm unsure if I even have all the information required!

Given motor rated no-load RPM: 7000
Given motor rated input power: 650 watts

I don't know the motor efficiency, etc. but I know the motor the version down is 82% efficient, so I shall use that - I'm more interested in the process of these calculations rather than the accuracy as if need be I'll find a motor with full specs.

Calculated output power: input power x efficiency
Calculated output power: 650 x 0.82
Calculated output power: 533 watts

Horsepower: power / 745.69
Horsepower: 533 / 745.69
Horsepower: 0.71

(The braking torque formula I got from this website: <a href="http://www.elec-toolbox.com/Formulas/Motor/mtrform.htm">http://www.elec-toolbox.com/Formulas/Motor/mtrform.htm</a>)

Braking torque (lb-ft): (5252 x HP) / rpm
Braking torque (lb-ft): (5252 x 0.71) / 7000
Braking torque (lb-ft): 0.53
Braking torque (NM): lb-ft x 1.35
Braking torque (NM): 0.53 x 1.35
Braking torque (NM): 0.7155

However, this is for the motor running at the given rpm, not the required RPM. So I now need to calculate the gearing required to drop the motor rpm to the required rpm:

Gearing ratio: Motor RPM / Required RPM
Gearing ratio: 14

This means that the final braking torque is:

Braking torque (NM): Braking torque (NM) x gearing ratio
Braking torque (NM): 0.7155 x 14
Braking torque (NM): 10.02

So, as the required torque to move the object initially was 0.98NM, this motor is just over 10 times the required power.


Could somebody verify the above for me please?

Thank you.
 
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  • #2
Force required to move object: Mass x 9.8
Force required to move object: 2 x 9.8
Force required to move object: 19.6 Newtons

In which direction is the box moving? Horizontally or vertically? I ask because elsewhere you say

The distance the box moves (from one side to the other) is 5 cm.

So I don't see where g (9.8) comes into your calculation for the force?

Your set up is similar to a piston and con rod in a car. The piston starts from rest in one position, accelerates upto some speed V in the middle then decelerates again to come to rest at the other position.

You can use geometry to work out the peak acceleration (which won't be equal to g) and then apply f=ma.
 
  • #3
Hmmmm... If there is no friction, then once the mechanism is up to speed, is there any more work to be done?
 
  • #4
Hello there CWatters,

Thanks for taking the time to reply.

In which direction is the box moving? Horizontally or vertically? I ask because elsewhere you say
My apologies, I hadn't realized the direction was needed as I assumed that the difference between the two over such a small distance would be negligible. Although to answer your question, the box is moving horizontally.

So I don't see where g (9.8) comes into your calculation for the force?
Ah okay. So if the box is moving only horizontally then I don't count gravity as a force as I'm never having to oppose it, although if the box was moving say up and to one side I'd have gravity and the acceleration of moving to one side. Correct?

Your set up is similar to a piston and con rod in a car. The piston starts from rest in one position, accelerates upto some speed V in the middle then decelerates again to come to rest at the other position.
Yes, that's a perfect description! I've dug up a few formulas (and a couple of calculators for piston acceleration - oh boy! This may take a while to work out! :D
 
  • #5
Hello BarryJ,

Hmmmm... If there is no friction, then once the mechanism is up to speed, is there any more work to be done?
Ah, so friction has to come into it somehow then? I'll have a look for some sample friction coefficients...

Thanks.
 
  • #6
Think about this. You could have a slot, i.e. the path, where your mass would move. At each end, put a spring. To start the system, pull your mass so it compresses one of the springs and let go. If there were no friction and everything is ideal, the mass would bounce back and forth forever. The only work needed would be to compress the spring initially. I think you will definitely account for friction somewhere.
 
  • #7
Well... After putting it like that, it makes my previous statement seem a little silly! :) Haha, thank you. I shall gather an appropriate friction coefficient. :)
 
  • #8
No, the statement is not silly. It shows you are trying to learn and that is good.

In your proposed problem, there is friction at all moving parts. The motor must be sized to overcome the friction at various parts. CWatters said your problem is sort of like a piston moving in a cylinder. In this case, friction is between the piston and the cylinder walls. If there were no friction, and you removed the cylinder head, you could spin the crank of a motor and it would spin very easily. If the cylinder head were attached, there would be work done on the gas as it is compressed and let to expand, unless there was no heat exchanged between the cylinder head and the environment.
 
  • #9
Hmmmm... If there is no friction, then once the mechanism is up to speed, is there any more work to be done?

Possibly. Don't forget it's actually an oscillating mass. It takes energy to accelerate it upto speed which you might be able to recover when it's decelerated at the other side. Using a heavy flywheel would be one way to do it. Using multiple "cylinders" out of phase might also help. That way you have one mass accelerating while another is decelerating.
 
  • #10
Revised question.

Hello!
Thanks once again for the tips on things I’ve missed such as acceleration and friction. I’ve spent the past couple of days rooting round trying to find things out. I’ve posted below a revised question that should hopefully explain things a little better along with some further calculations. I would really appreciate somebody taking a quick look over how I'm doing these calculations to make sure things are okay.

Problem:

I have an object that is moved horizontally by a given distance and is supported by a table. The object will be moved by a crank system (similar to this Crankshaft image on Wikipedia). I want to be able to source a motor that is capable of moving this object at a given speed. I’ve done a diagram (albeit, not terribly nice) that helps illustrate the problem and the components involved:

https://www.physicsforums.com/attachment.php?attachmentid=58766&stc=1&d=1368715566

Formulas and data:


Object Inputs
I have the following data for the object:
Code:
Distance to move (cm): 8
Mass of object (kg)  : 2.1
Movements per second : 16
Friction coefficient : 0.5

I'm guessing it's obvious, but the movements per second is how many times the load is expected to move per second and the friction coefficient is between the load and the supporting table.


Object Calculations
It seems the only calculation that is useful from the above is the friction calculation. This was previously ignored, although the above posts makes it quite clear that friction must be taken into account otherwise the object would move forever... Quite obvious when said like that! :-)
Code:
Normal Force (Newtons)       : (Mass of object) x 9.8
Normal Force (Newtons)       : 2.1 x 9.8
Normal Force (Newtons)       : 20.58

Code:
Force for movement (Newtons) : (Normal Force) x (Friction coefficient)
Force for movement (Newtons) : 20.58 x 0.5
Force for movement (Newtons) : 10.29


Crank Inputs & Calculations
Previously, I took no notice of the crank only using the acceleration of the object, however after reading the above posts, it is clear that I need to calculate the acceleration of the crank instead, which would let me work out the actual force needed to move the crank (and so the object). Just in case anybody comes across this, I had quite a bit of difficulty working out the acceleration of the crank (partly due to a silly mistake and partly due to a misunderstanding of my part in what I was looking for), a thread I created helped tremendously. These calculations are correct as per the thread just mentioned, so I shall simply post the results rather than the workings:

Code:
Crank Inputs    
Revolutions per minute: 480
Rod length (cm)       : 15
Crank angle (deg)     : 0

The RPM is derived from the desired speed of the object / 2 (one whole revolution moves the object twice).
A Crank Angle of 0 is about the point of maximum acceleration of the crank.


Crank Calculations
Crank angle (rads)           : 0
Crank angular vel. (rad/s)   : 50.26548246
Crank radius                 : 4
COS Theta                    : 1
COS2                         : 1
n                            : 3.75
-rw2                         : -10106.47491
Crank acceleration (cm/s)    : -12801.53488
Crank acceleration (m/s)     : -128.0153488
Crank acceleration ABS (m/s) : 128.0153488


Crank & Object Force Calculations
From the above calculations, I can work out the final force required to move the object and the crank.

Code:
Distance to move (m) : (Distance to move (cm)) / 100
Distance to move (m) : 8 / 100
Distance to move (m) : 0.08

Code:
Force required for crank (Newtons) : (Mass of object x Crank acceleration (ABS))
Force required for crank (Newtons) : 2.1 x 128.0153488
Force required for crank (Newtons):  268.8322325

Code:
Total Force to move crank and object (Newtons) : (Friction force for movement + Force required for crank)
Total Force to move crank and object (Newtons) : 10.29 + 268.8322325
Total Force to move crank and object (Newtons) : 279.1222325

Code:
Total Torque to move crank and object (Newtons) : (Total Force to move crank and object x Distance to move)
Total Torque to move crank and object (Newtons) : 279.1222325 x 0.08
Total Torque to move crank and object (Newtons) : 22.3297786


Motor Inputs
Assuming the above is correct, in summary I need a motor that has:

Code:
Minimum torque (Newtons) : 23

Unfortunately, the motors I've got at hand I only know the following characteristics (I've chosen one motor at random here):

Code:
Rated RPM (RPM)         : 12000
Rated input power (W)   : 650
Rated efficiency (%)    : 80


Motor Calculations
From the above motor inputs I can work out the following:

Code:
Output power (W) : Rated input power x efficiency
Output power (W) : 650 x 0.8
Output power (W) : 520

Code:
Horsepower (HP) : Output power / 745.69
Horsepower (HP) : 520 / 745.69
Horsepower (HP) : 0.697

Code:
The formula for braking torque was found at [URL="http://www.elec-toolbox.com/Formulas/Motor/mtrform.htm"]Elec Toolbox[/URL].
Braking Torque (Lb.ft) : (5252 x Horsepower) / Motor RPM
Braking Torque (Lb.ft) : (5252 x 0.697) / 12000
Braking Torque (Lb.ft) : 0.5232

Code:
Braking Torque (Newtons) : Braking Torque (lb.ft) * 0.737
Braking Torque (Newtons) : 0.5232 * 1.3558
Braking Torque (Newtons) : 0.3856

This is the braking torque of the motor running at the full RPM, however as the crank needs to rotate at 480 RPM, I can reduce the motor RPM with gearing and so increase the torque.

Code:
Gear ratio motor to crank: Motor RPM / Crank RPM
Gear ratio motor to crank: 7000 / 480
Gear ratio motor to crank: 14.58

Code:
Final braking torque of motor (Newtons) : Braking Torque * Gear ratio
Final braking torque of motor (Newtons) : 0.3856 * 14.58
Final braking torque of motor (Newtons) : 5.6233


Results
So, I can now see that the crank and the object required a torque of 22.3297786, however, the motor can only provide a torque of 5.6233, so the motor is only around 25% of the power required.
 
  • #11
You are doing a lot with torques, braking, and etc, but think about this.
Where is the energy consumed that the motor has to overcome? It is in the friction of the mass moving back and forth. I would suggest finding out how much energy is required to move the mass 8 cm, and then continue to find the power required to move the mass at your 16 cycles per sec. Then you have the power needed from the motor. I assume you are ignoring the friction associated with the crank bearings and motor shaft.
 
  • #12
Hello there,

Thanks for the prompt reply.

You are doing a lot with torques, braking, and etc, but think about this.
So, am I going down the wrong route by using braking torques etc. and overcomplicating it? If so, where would these things fit in?

I would suggest finding out how much energy is required to move the mass 8 cm
Well, I know that I need to apply more than 10.29 Newtons of force to get the object to move due to the static friction, so if I applied 11 Newtons:

Code:
Energy (J) = Force x Distance
Energy (J) = 11 x 0.08
Energy (J) = 0.88

I know that the total force needed to move both the crank (running at the required RPM) and object is 279.1222, which gives an energy of:

Code:
Energy (J) = Force x Distance
Energy (J) = 279.1222 x 0.08
Energy (J) = 22.329776

So that is the energy consumed per second, however I'm not 100% sure where to go from here...
 
  • #13
Actually, I've been searching about and I wonder would I be better getting a moderator to move this thread to the "Mechanical Engineering" forum?
 
  • #14
I don't see where you got the 279 number. Your force of 10.29 is correct. The energy to go back and forth one time will be 10.29 X 0.08 X 2 = 1.64 J. At 16 cycles per sec you will use 1.64 X 16 = 26.34 joules/sec or 26.34 watts if you motor is 100% efficient.
 
  • #15
Hello,

The 279 number is in the "Crank & Object Force Calculations" section, but here it is below:

Code:
Total Force to move crank and object (Newtons) : (Friction force for movement + Force required for crank)
Total Force to move crank and object (Newtons) : 10.29 + 268.8322325
Total Force to move crank and object (Newtons) : 279.1222325

Is what I've actually got in the revised post (#10) wrong?
 
  • #16
Continuing, 26 watts is about 26/760 = 0.034 hp. at 80% efficiency, you would need a .043 hp motor. Now you will have to figure out the rpm of the motor and the desired gear ratio to reduce the speed.
 
  • #17
I don't see why the crank force would be so high??
 
  • #18
I don't see why the crank force would be so high??
Well, the crank has a maximum acceleration of around 128.015 metres per second and the weight is 2.1 KG, which gives a force of approx. 268 Newtons as per the first post by CWatters:

You can use geometry to work out the peak acceleration (which won't be equal to g) and then apply f=ma.
 
  • #19
The equation for d(t) = .04sin(2Xpi/(1/16))t = .04sin 32pit
v(t) = 4.02cos(2Xpi/(1/16))t
a(t) = -404sin(2Xpi/(1/16))t

This is just for the moving 2.1 kg mass. The maximum acceleration will be 404m/sec^2
However, the average acceleration is 0.

If you have a motor driving a crank and flywheel, the inertia of the flywheel will average out these what seem to be high forces.

Bottom line, the only consumer of energy is friction. If you use a small hp motor, yes, it might take a time to get up to speed but there is really little work done by the motor.
 
  • #20
Hello Barry,

Thanks very much for that. I have to admit though, I don't understand what's happening with the maths there, it seems you're calculating distance, velocity and acceleration. Nevertheless, I'm a little more interested in the writing there than the maths...

If you have a motor driving a crank and flywheel, the inertia of the flywheel will average out these what seem to be high forces.
My apologies, there isn't a flywheel present - I'm sorry if I gave you that impression...

Bottom line, the only consumer of energy is friction.
This makes perfect sense after your earlier posts - obviously initially I need to overcome static friction, then to keep the velocity constant I need to overcome the kinetic friction... However...

I calculated the force needed to move the crank a given time through a given distance as follows:

Code:
Distance to move (m) : (Distance to move (cm)) / 100
Distance to move (m) : 8 / 100
Distance to move (m) : 0.08

Code:
Force required for crank (Newtons) : (Mass of object x Crank acceleration (ABS))
Force required for crank (Newtons) : 2.1 x 128.0153488
Force required for crank (Newtons):  268.8322325

Code:
Total Force to move crank and object (Newtons) : (Friction force for movement + Force required for crank)
Total Force to move crank and object (Newtons) : 10.29 + 268.8322325
Total Force to move crank and object (Newtons) : 279.1222325

The object will have a velocity of 0 whenever it changes direction and so the acceleration of the object must be made each time the crank reverses direction, so surely this force is always present so to bring the object up to speed - I'm struggling to see how we can skip over it? If so, I'm quite confused why CWatters said I needed to calculate this.

Thanks.
 
  • #21
When the mass changes direction, you are correct in that the velocity will be 0. At this time, the acceleration will be the greatest. Any motor will have some sort of flywheel even if it is just the rotor of the motor. The crank will also act as a flywheel. Even if there were no specific flywheel, there would be one. The example I gave with the mass moving at a sinusoidal rate is not exact but close. The connecting rod between the flywheel, or crank, and the mass will cause a slight deviation. In any case, you do not need a large hp motor.
 
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Related to Calculating the required motor size to move a load.

1. How do I calculate the required motor size?

To calculate the required motor size, you will need to know the weight of the load, the distance it needs to be moved, and the time it needs to be moved in. You will also need to consider factors such as friction and incline.

2. What units of measurement should I use for calculating motor size?

It is best to use consistent units of measurement when calculating motor size. Common units used include kilograms for weight, meters for distance, and seconds for time. Make sure to convert any units that are not consistent before beginning calculations.

3. How do I factor in friction when calculating motor size?

Friction is an important factor to consider when calculating motor size. This resistance force will require the motor to work harder, so it is important to estimate the coefficient of friction between the load and the surface it will be moved on. This can then be factored into the calculation.

4. Can I use a motor with a higher or lower horsepower than the calculated size?

It is generally recommended to use a motor with a higher horsepower than the calculated size. This allows for any unexpected circumstances or additional weight that may need to be moved in the future. Using a motor with lower horsepower may result in the motor being overworked and potentially failing.

5. Are there any online tools available for calculating motor size?

Yes, there are several online calculators and tools available for calculating motor size. However, it is important to double check the calculations and consider any other factors that may not be accounted for in these tools. It is always best to consult with a professional engineer for accurate and customized calculations.

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