Thread starter
Anasazi
Start date
May 14, 2013Tags
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.
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:
CWatters
Science Advisor
Homework Helper
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In which direction is the box moving? Horizontally or vertically? I ask because elsewhere you saySo 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.
barryj
Hmmmm... If there is no friction, then once the mechanism is up to speed, is there any more work to be done?
Anasazi
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.
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?
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
Hello there CWatters,Thanks for taking the time to reply.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.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?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
Anasazi
Ah, so friction has to come into it somehow then? I'll have a look for some sample friction coefficients...
Hello BarryJ,Ah, so friction has to come into it somehow then? I'll have a look for some sample friction coefficients...Thanks.
barryj
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.
Anasazi
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. :)
barryj
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.
CWatters
Science Advisor
Homework Helper
Gold Member
Possibly. Don't forget it's actually an oscillating mass. It takes energy to accelerate it upto speed which yoube 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.
Anasazi
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
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
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),
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
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
Assuming the above is correct, in summary I need a motor that has:
Code:
Minimum torque (Newtons) : 23
Code:
Further reading:
Rated RPM (RPM) : 12000
Rated input power (W) : 650
Rated efficiency (%) : 80
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
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
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.
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.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=1368715566I have the following data for the object: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.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! :-)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:From the above calculations, I can work out the final force required to move the object and the crank.Assuming the above is correct, in summary I need a motor that has:Unfortunately, the motors I've got at hand I only know the following characteristics (I've chosen one motor at random here):From the above motor inputs I can work out the following: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.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.
barryj
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.
Anasazi
So, am I going down the wrong route by using braking torques etc. and overcomplicating it? If so, where would these things fit in?
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
Code:
Energy (J) = Force x Distance
Energy (J) = 279.1222 x 0.08
Energy (J) = 22.329776
Hello there,Thanks for the prompt reply.So, am I going down the wrong route by using braking torques etc. and overcomplicating it? If so, where would these things fit in?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: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:So that is the energy consumed per second, however I'm not 100% sure where to go from here...
Anasazi
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?
barryj
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.
Anasazi
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
Hello,The 279 number is in the "Crank & Object Force Calculations" section, but here it is below:Is what I've actually got in the revised post (#10) wrong?
barryj
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.
barryj
I don't see why the crank force would be so high??
Anasazi
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:
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:
barryj
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.
Anasazi
My apologies, there isn't a flywheel present - I'm sorry if I gave you that impression...
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...
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
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...My apologies, there isn't a flywheel present - I'm sorry if I gave you that impression...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:, 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.
barryj
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.
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.
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.
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.
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.
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.
It's there an minimum HP rating on the pump itself?
I'm no expert but it seems gas motors are rated by their peak power output at whichever RPM. An electric motor is rated at its peak continuous power. Most electric motors can be run up to 2.5 times their rated HP before blowing a fuse or letting the smoke out. Which is why most CNC equipment says something like 30HP continuous/40HP 30 minutes.
So maybe the 5hp rated electric motor will get the same job done as the 8HP gas engine was. But is it only drawing 5hp's worth of electricity? You can certainly do the test and let us know. All you need is an amp and volt meter. As far as I know you're still better of going with a slightly larger motor than needed.
I will say that they do usualy put slightly bigger gas engines on such equipment. Not because 1HP from a gas engine is not the same as 1HP from an electric motor. But the gas engine generaly needs a bit longer to respond when there's a sudden change in load and wouldn't be able to instantly draw way more than its rated output in order to stay up to speed, as an electric motor does. Generators are probably the best example of this.
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