I`m making a flywheel and with the math

[wheelslave1]

First off thanks in advance for any help given here.

Hope you don`t mind bullet points. Makes it easy for me.

I`m building a type of flywheel that consist of:

1) A single rod 2 ft across. Axel in the middle.

2) A 2 lb weight on both ends of the arm.

3) Rod weight 1 lb.

4) There will be a constaint 20 rpms applied to the flywheel at the 1 in axel.

5) A speed change from 20 rpms to 80 rpms (engaged gearing system) will occur for 1/2 rpm.

The question is:

How many foot pounds of force is needed to make the change from 20 rpms to 80 rpms?
As a side note all force is applied to the 1 in axel.

 

[Simon Bridge]

How many foot pounds of force is needed to make the change from 20 rpms to 80 rpms?
As a side note all force is applied to the 1 in axel.

force is rate of change of momentum ... so it depends how fast you want to make the change.
Rotationally - this will be an applied torque:
$\tau=dL/dt = I\alpha$.

You are probably better to do this in terms of energy, and ask how much work you need.
This will mean working out the moment of inertia for your flywheel.
$W=\tau \theta$

The, from knowing how fast you want to make the change, you can work out the required torque.
If you know where the force is to be applied, you can work that out by:
$\vec{\tau}=\vec{r}\times \vec{F}$

 

[wheelslave1]

Simon,

The change will be instant. I understand most of your answer but being self taught has me at a disadvantage. So, being that the change is a jump from 20 rpms to 80 rpms what is the answer?

 

[cjl]

If the change is instant, the required torque is infinite, which is clearly not possible. There has to be a period of acceleration - you cannot instantaneously change the angular velocity.

 

[wheelslave1]

Cil,

The constaint applied rpm is 20. A 4:1 gearbox changes the rpm to 80. I`m just looking for the simplest answer. The project is a form of art that will be viewed by a large number of people. The handmade gears need to be able to handle the load without breaking.

The greatest amount of force will be at the time of rpm change.

 

[Simon Bridge]

Simon,

The change will be instant. I understand most of your answer but being self taught has me at a disadvantage.

depends on your teacher :) You should never teach or learn in isolation - odd ideas can get reinforced. As long as you are talking about your ideas, and using critical inquiry, you should be fine.

So, being that the change is a jump from 20 rpms to 80 rpms what is the answer?

The answer is that it will take an infinite amount of energy. It is impossible.
The shorter the time for the shift the more energy and the bigger the force that is needed.

The constant applied rpm is 20.

You don't "apply" a speed. You apply a torque. There will be some losses (i.e. friction) in the system creating a counter-torque. When the applied torque is the same as the losses, the wheel turns at a constant speed (at a set rpm).

A 4:1 gearbox changes the rpm to 80. I`m just looking for the simplest answer. The project is a form of art that will be viewed by a large number of people. The handmade gears need to be able to handle the load without breaking.

That is a fairly complex situation - presumably something is applying a torque to the wheel - then you want to quickly change gears. Will the something also be able to provide the additional torque and energy to produce the acceleration to the new rpm or will it just jam, or go at a slower speed?

Anyway, I think you'll need to do this in terms of "specific impulse" (you'll have to look it up): τΔt = ΔL = IΔω for the rotational case ... you'll need to look up the term and also look carefully at how the gearbox changes gears. The best bet is just to over-engineer it.

I also think you need to add some classes in Newtonian mechanics to your self-education ... Newtons laws of motion, and how the apply to rotation. Then you can deal with how gears affect things.

 

[cjl]

Wait - so you have one axle turning at 20 rpm, and a different axle turning at 80 rpm? I think we've all been assuming that you're trying to accelerate an axle from 20 to 80 rpm, which is a completely different scenario. If you just have a simple gearbox, the torque on the 20rpm shaft will be 4 times the torque on the 80rpm shaft, ignoring any losses in the system

 

[wheelslave1]

Cil,

"Wait - so you have one axle turning at 20 rpm, and a different axle turning at 80 rpm?"

Yes and no. I`ll explain a little more in detail. There is only one axle that acts like two. The gearbox is configured to, if held in place, turn the rod at 80 rpms for 1/2 rotation. All while the axle remains at a constant 20 rpms.

Now if what I believe you are saying is correct. A 4:1 ratio will be expected.
The wieghts 2lb each and a rod at 1lb would equal 20ft lbs max once the gearbox is held in place. The 20ft lbs of force will act like flash collision on the gears and drop off to less than 1ft lb.

There is a 6in gear locked in place on the 1in axle. The gearbox system runs like this: 6in gear (locked to axle) to 3in gear locked to and across from a 6in gear to a 3in gear connected to the 1in axle but not locked to it.

 

[wheelslave1]

Cil,
Am I correct in my thinking here? The motor will connect to a 40in wheel. So, can you tell how many ft lbs will be needed at 40in to run this balanced system.

 

[cjl]

I can't envision what you're trying to describe here - could you draw a diagram or something like that?

 

[wheelslave1]

Cil,

Thank for your help. Its rare to find someone willing to help someone in need. Would you mind going to private mail?

 

[Simon Bridge]

Cil,

Thank for your help. Its rare to find someone willing to help someone in need. Would you mind going to private mail?

... and deprive others of this assistance. I hope cjl does not agree to that!

The price of free/gratis assistance is usually public exposure.
It's probably in the rules somewhere...

 

[russ_watters]

I don't think it is in the rules, but it is clearly a matter of forum etiquette. That's kinda the whole point of a discussion forum.

However, I'm guessing we have some sort of perpetual motion scheme here...

 

[Simon Bridge]

On that note ... interested students should visit:
http://www.lhup.edu/~dsimanek/museum/unwork.htm
... paying special attention to the "physics gallery" and the "reading room".
OP will find this will fill in the gaps evident in his self-education as well as help him talk to scientists.

 

[cjl]

Cil,

Thank for your help. Its rare to find someone willing to help someone in need. Would you mind going to private mail?

I would really prefer to keep it on the public forum, partially for the reasons already mentioned, and partially because I will be the first to admit that there are a number of posters here with far more physics knowledge than myself, and if I say something misleading, I would hate to have it go unchallenged by those people.

 

[Simon Bridge]

I will be the first to admit that there are a number of posters here with far more physics knowledge than myself ... I would hate to have it go unchallenged by those people

Even the most learned of our colleagues can slip up sometimes ... or find themselves repeating a sloppy explanation. That feedback is part of why we do this after all.

@wheelslave1
Sometimes, someone new to science will feel embarrassed or "stupid" when they are unsure of their reasoning. Nobody likes to be wrong in public, and scientists can appear unfriendly or insensitive. We understand and we've all been there. So no judgement is implied in corrections, criticisms or feedback.

Anyone prepared to post here is treated equally with the most experienced of us.
Explain away and we'll figure it out together.

 

[wheelslave1]

The dark blue 6in gear is locked to the axle. It runs at 20 rpms.

The light blue 3in gear connects to the dark blue gear. It runs at 40 rpms

The green 6in gear and the light blue 3in gear are locked together. Both run at 40 rpms

The green 6in gear connects to the green 3in gear. The 3in gear (free floating on the axle) runs at 80 rpms.

There is a brown rod mounted to the 3in green gear. It and its and its two weights will spin at 80 rpms.

The white rod, if held in place, starts the 80 rpm cycle. Let the white rod go and it will travel with the main wheel at 20 rpms.

My question is: How many ft lbs of force will be applied to my 3in green gear?

Think of it like this. A 40in wheel is running at 20rpms. The white rod stops causing the brown rod to suddenly spin at 80 rpms.

 

[mfb]

How does the white rod work?

Let the white rod go and it will travel with the main wheel at 20 rpms.

Why?

A 40in wheel is running at 20rpms. The white rod stops causing the brown rod to suddenly spin at 80 rpms.

That looke like it would give an enormous force on all components which change their rotation frequency, probably breaking something. The force depends on the ability of the material to stretch/compress/break.

 

[CWatters]

If I've understood correctly...

With the white lever free to rotate the whole gearbox spins round so the weights go at 20rpm. Then if you slow/stop the white lever the weights spin up to 80 rpm.

Work out how fast you want the weights to spin up.
From that you can work out the torque you need to apply to the weights.
I think the torque on the white lever will be roughly twice (or is it four times that?) that.

 

[CWatters]

PS Those weight going at 80rpm could apply quite a lot of shear force to a finger trapped in the mechanisim.

 

[wheelslave1]

CWatters

"Work out how fast you want the weights to spin up.
From that you can work out the torque you need to apply to the weights.
I think the torque on the white lever will be roughly twice (or is it four times that?) that."

Let me try a better example. The 40in wheel turns 3/4 of a rotation along with the gears at 20 rpms. The white rod hits a stable object to begin the 80 rpm 1/2 rotation. During this time the 40in wheel travels 1/4 of a rotation. The white rod is released and the process starts again.

From what I understand from you and others is:

With the current configuration 20 ft lbs (4:1) applied to the white rod.

What I don`t know is it a continuous 20ft lbs or once the 80 rpms is acheived does it lesson.

Also, If I changed my 2lb weights to pendulums would it ease off the 4:1 ratio?

 

[wheelslave1]

Russ Watters

"However, I'm guessing we have some sort of perpetual motion scheme here..."

Yes, you are almost correct. It is a pm but its not designed to work. This artwork nothing else.

 

[Simon Bridge]

Ah - it's a machine designed to look like it might be a pmm just enough to intrigue the onlooker? Presumably you'll add a hidden power source?

For the rest - if you want to change a rotation from one rpm to another, you need accelerate it. This acceleration cannot be instant: it must take some finite time - though it can be a very short time. Once you know the time in which the acceleration can take place, you can work out the torque involved.

Once the correct speed is achieved - you stop applying the torque.
Often the mechanism is rigged to have the applied torque tail off as the desired speed is reached so it tends not to be constant.

 

[wheelslave1]

Simon,

It will have a hidden power source. There will also be a written desciption about its inner workings. The torque required needs to be in the discription.

What am I missing in the discription of this machine that everyone needs for the correct answer?

 

[CWatters]

Let me try a better example. The 40in wheel turns 3/4 of a rotation along with the gears at 20 rpms. The white rod hits a stable object to begin the 80 rpm 1/2 rotation.

At that point I cry foul. No moving object that has mass can stop instantly. I'm sure you are familiar with F = m*a.

An instant stop implies infinite deceleration so the forces or torques would be infinite as well. There is always a "stopping distance" even if it's very short. Either your stable object or the white rod etc must deform however slightly.

Unless you know what that "stopping distance" is for the white rod you can't calculate the acceleration that the masses need to achieve. I refer to earlier posts asking you about the acceleration time.

 

[Simon Bridge]

Simon,

It will have a hidden power source. There will also be a written desciption about its inner workings. The torque required needs to be in the discription.

What am I missing in the discription of this machine that everyone needs for the correct answer?

Calculating forces and stresses can be as vague or as exact as you want. The more precise the description of the device the more precise the resulting calculation can be.

However - the minimum requirement is the time period (or displacement) over which the change occurs. That is why people keep telling you about it. You seem reluctant to address this.

It would also help if you would confirm or expand upon other replies since you posted the diagram. eg. Did CWatters (see post #19) understand you correctly? What about mfb's questions (post #18)? Have you taken on board the comments about how mechanics works?

 

[wheelslave1]

The torque will be on the main 6in gear locked to the axle. Please let me know if this is what you need.

Time:
20rpms = 1 rpms every 3 seconds
change occures every 1/2 rpm = 1.5 seconds
Flywheel travels 1/2 rpm =1.5 seconds
Reponse to for gears to fully engage .25 seconds

Answer for time is .25 seconds

Initial Speed in inches per second
25.125

Final Speed in inches per second
100.26072

time
.25

Acceleration in meters per second
7.6334843525lbs weight converts to 2.2679618500000003 kilograms

7.633484352(a) * 2.2679618500000003(m) = 17.31245129290797(f)
F = m*a.

converts to 38.167421759155566 ft lbs
75.384 pi of 2ft rod /3 for inches per second at 20 rpm =25.128
75.384 pi of 2ft rod *1.33 for inches per second at 80 rpm =100.26072

Please let me know if this is correct

 

[Simon Bridge]

The torque will be on the main 6in gear locked to the axle. Please let me know if this is what you need.

Time:
20rpms = 1 rpms every 3 seconds

You see this sort of statement leads me to suspect you don't understand your own figures.

"rpm" means "revolutions per minute"
it is already plural so what does "rpms" mean?

If you have 20rpm, then it rotates 1 revolution in 3 seconds. Not 1rpm in 3 seconds: that would imply an acceleration.

20rpm is 0.33Hz or $2\pi/3$ radiens per second.

change occures every 1/2 rpm = 1.5 seconds
Flywheel travels 1/2 rpm =1.5 seconds
Reponse to for gears to fully engage .25 seconds

Answer for time is .25 seconds

This should be the time it takes for the wheel that changes speed to go from slow to fast speed. Otherwise I don't know what you mean by "Response for gears to fully engage". In your diagram, the gears are always fully engaged.

Anyway - taking $\Delta t=0.25\text{s}$ to accelerate: the applied torque would be $\tau = I\Delta\omega/\Delta t$ just for that one unit that is accelerating.

The tangential force at a distance r from the center of rotation is $F=\tau/r$.
For instance, if you accelerated it by wrapping a string around an axle of diameter d and hanging a weight mass m off the other end, then the applied torque due to the mass would be $\tau = 2mg/d$

Note: if you have responded to posts #19 and #18 already please point it out.
If not, what's the problem?
Thank you.

 

[wheelslave1]

mfb,

Sorry that I have not answered your question yet

How does the white rod work?

why

A 40in wheel spins at 20rpm. Attached to its axle is a 6in gear also spins at 20 rpm. The gearbox is mounted to the white rod. If the white rod is held in place it causes the brown rod to spin a 80 rpm


To spin the white rod. Something I have mentioned before is on one end of the brown rod is a pendulum. 80 rpm will provide enough centrifugal force to raise a 2lb weight to the top.

As everyone can see this is not a pmm and could never work as a pmm. Unless you see the motor it will make you wonder. I have built test system and I`m almost ready to move forward with the full project.

 

[wheelslave1]

If I've understood correctly...

With the white lever free to rotate the whole gearbox spins round so the weights go at 20rpm. Then if you slow/stop the white lever the weights spin up to 80 rpm.

Work out how fast you want the weights to spin up.
From that you can work out the torque you need to apply to the weights.
I think the torque on the white lever will be roughly twice (or is it four times that?) that.

You are correct and it is .25 second
It took me a very long time to figure that. Thank you for asking the question. It led me to the answer.

 

[wheelslave1]

Simon

You have to understand I am new at this and doing the best I can..

You see this sort of statement leads me to suspect you don't understand your own figures

Yes, I do understand my own figures. I may not use the correct terms but I am trying to learn. Thats why I came here.

Anyway - taking Δt=0.25s to accelerate: the applied torque would be τ=IΔω/Δt just for that one unit that is accelerating.

No idea what that means.

So...your saying that my f=m*a was incorrect. :(

 

[Simon Bridge]

Simon

You have to understand I am new at this and doing the best I can..

No worries - this is why it is important to respond to direct questions even if they seem odd to you. The people helping you have been doing this for a long time and know what they need.

Yes, I do understand my own figures. I may not use the correct terms but I am trying to learn. Thats why I came here.

Willingness to learn is the important part.
One of the things I have been trying to teach you is communication ... you have been used to using technical terms in a non-standard way: very common with self-taught people.

So...your saying that my F=ma was incorrect. :(

No - that is quite correct - just not so useful. It's more useful to use the rotational equivalent since things are spinning here.

That would be $\tau = I\alpha$ ... which is to say that the torque is equal to the moment of inertia multiplied by the angular acceleration.

angular speed ($\omega$) would be "rpm" so angular acceleration ($\alpha$) is rpm per second ... but we prefer to define angular speed in terms of angle-units of "radiens-per-second" (rad/s) because it makes the math easier. So 60rpm is $2\pi$rad/s.

You have an angular acceleration of $(80-20)\text{rpm}/(0.25)\text{s}=240\text{rpm/s} = 8\pi\text{rad/s}^2$

The moment of the inertia of the flywheel is the sum of the moments of inertia of the bits.
You can look up the formulas on wikipedia. The two weights on the ends would be $mr^2$ each while a cylindrical rod (length 2r) connecting them would be $Mr^2/3$ so:

$I=(2m+\frac{1}{3}M)r^2$

I'd put the values in but the list is now over the page :)

Like I said before - once you have the torque, you can work out the linear force to applied at a given radius - that should help you figure how strong to make those gears.

 

[wheelslave1]

Can`t you just give me the answer?

 

[Simon Bridge]

Where's the fun in that? Don't you want to understand your project?

You have (from post #1)
1) A single rod 2 ft across. Axel in the middle.

That would make r=1' = 0.3m

2) A 2 lb weight on both ends of the arm.

2lbs is about 1kg ... you only need ballpark figures right?
so m=1kg

3) Rod weight 1 lb.

M=0.5kg

So $\tau = (2m+\frac{1}{3}M)r^2\alpha = (2+1/6)(0.3)^2(6\pi)=3.68\text{N.m}$

Looks like about 0.25ft-lbs. if I did the conversion right... I'm not used to imperial measures.

You need to multiply this by the distance from the center (in feet) to where you are applying the torque ... that would be the outside of the cog-wheel ... for the 6" one that is 0.125 lbs would be the force on the cog's teeth.

Doesn't look too bad.
You are cautioned to check these figures before going into production ... no responsibility etc etc.

 

[CWatters]

Sounds much lower than I expected but I haven't checked your calculations.

If that's the torque on the output gear shaft I believe you need to multiply up by the gear ratio to get the torque on the input gear shaft (because it's a step up).