# How much HP to spin this ring?

1. Jul 16, 2012

### prodriverex

OK I'm super frustrated.. I spent 50min typing the whole text here earlier and click preview only to be told I was logged out. Click back and and all was gone. Damn..

Ok here's the question, AGAIN.

Weight: 1kg
Type: Flywheel, OD is 80mm, ID is 40mm if that matters.
So that's a typical Ring calculation?

Torque, T = Moment of Inertial, I * Angular Acceleration, α
HP = N * T/5252

MOI, I, depends on Ring or Solid
Solid cylinder = (1/2)*m*R2
Hollow cylinder = m*R2

So if this is ring, then I = 1kg * 0.08 * 0.08 = 0.0064

Angular Acceleration, α (rad/s2): dω / dt = αT/ r.

ω is the angular velocity
αT is the linear tangential acceleration.

So if the unit is rad/s2 is essentially how many rad in a squared-second?

If it is true, t being 8000rpm or 133.333 rev/sec. In square sec, divide further by 60.

Torque, T = 0.0064 * 2.222 = 0.0142 (Nm?)

So HP = 8000 * 0.0142 / 5252 = 0.02166 hp? Doesn't look right!

Appreciate any enlightenment please. Thank you..

2. Jul 16, 2012

### tiny-tim

welcome to pf!

hi prodriverex! welcome to pf!
no, it's not a ring, it has an OD and an ID, so it's a solid cylinder with a cylindrical hole …

you'll need to find the moment of inertia of the "whole" cylinder, and subtract the moment of inertia of the "hole" cylinder

(i haven't checked the rest of what you've done, since i'm not sure what the question is asking for … the HP needed to achieve what acceleration? )

3. Jul 16, 2012

### prodriverex

Thanks Tim!
So the simplest formula is I = M*R2
And OD is easy as what was quoted at 0.0064 taking 0.08 as R. This is the whole

As for the hole, say the bearing and spokes weigh 80gm, that will be I = 0.08kg * 0.042 = 0.000128.

Nett I is thus 0.006272!?

But if I put a weight to the hole, then it should be also sum into the equation, not subtract. If I don't put a weight, then anything * zero is meaningless?

4. Jul 16, 2012

### the_emi_guy

prodriverex,
There are a couple issues here:
1 - In your formula for horsepower, torque should be in lb-ft not newton-meters.
2 - One radian is not one revolution. There are 2π radians in one revolution.
3- We do not get from velocity (rad/s) to acceleration (rad/s^2) by dividing by 60.

5. Jul 16, 2012

### tiny-tim

no, the moment of inertia of a solid cylinder is 1/2 MR2 (with R = OD)

from that, you need to subtract the moment of inertia of a solid cylinder with R = ID

6. Jul 16, 2012

### CWatters

Prodriverex: It doesn't take any power to keep it spinning at a constant rpm. It only takes power to accelerate it upto that rpm. That's why the units are per second2

7. Jul 16, 2012

### the_emi_guy

He is actually trying to use the formula for engine horsepower given constant RPM and constant torque (braking device in dynamometer loading engine). But he is calculating torque using unloaded kinematic equations from acceleration. But he is getting acceleration by dividing angular velocity by 60.

8. Jul 16, 2012

### CWatters

There is a brake?

Prodriverex: I think we need more info on what the exact set up is.

9. Jul 16, 2012

### prodriverex

Iwhole = 0.5 * 1 * 0.08 * 0.08 = 0.0032

And what to do with the hole? Subtract by 0.000128?

Ha I was taking ref from https://www.physicsforums.com/showthread.php?t=201254 and got, probably, confused too.

So as in one rev = 2.∏ rad.

Much as i want to get the result, i am also interested to know the process and learn.
So if i have a radius of 0.08m, one radian = 0.08m.
One rev = the circumference = 2∏R = 0.50265544m
∴one rev = 0.50265544/0.08 = 6.283193 rad?

Ok now, how do i get to Angular Acceleration, α, from here?

No brake. spool up and braking relies on the engine's torque.
I had design a generator that shapes like a fly wheel which is the easiest to imagine. But I was afraid that the weight is too heavy on that puny 3W 2 stroke engine. So I like to get a feel of how breathless the engine will go spinning that generator. If it is ok, then what if I add in another pair of plate to churn out double the kW output at double the weight at 2kg, can it cope? I do not want the fuel consumption of the engine to suffer.

10. Jul 16, 2012

### the_emi_guy

prodriverex,
Are you trying to analyze a motor that is running at final rpm, or are you trying to analyze the spinup of the motor.

By the way arc length is not the same as radius.

11. Jul 16, 2012

### AlephZero

So there IS a "brake". It's the electrical power your generator is producing.

Apart from the time it takes to run up to speed, you can forget everything you have done so far. Just compare the eletrical power output with the HP of the engine (1 HP = 746 watts) and add say another 20% to the engine power because the system isn't 100% efficient.

12. Jul 17, 2012

### prodriverex

I think both. Spin up will determine how long it takes to get up to speed to reach desire power.

Running at final rpm will determine the burn rate of the engine.

Arc length is different as radius I know. But 1 rad is the ratio of arc length to radius.. no?

13. Jul 17, 2012

### prodriverex

Not to sure if you are referring to cogging effect or final motor efficiency. Yes I am concern about final motor efficiency as it will determine how useful this gen is. That is why the final product must go for further diet..

3hp engine say 80% efficient, the gen must produce more than 2700W to be classified as a good gen, that's what you mean?

14. Jul 17, 2012

### tiny-tim

hi prodriverex!

(just got up :zzz:)
no, you add the moment of inertia of the hole, which is minus 0.5 * 1 * 0.04 * 0.04 = -0.0008

(because the hole is made of the same material as the solid cylinder, except that its mass is negative! )

15. Jul 17, 2012

### prodriverex

Good morning. I'm from another side of the earth. Ha..
Ok that make sense to me now.

I is then, 0.0032 - 0.0008 = 0.0024

Let me try again on Angular Acceleration (rad/s2) again.

Radian is the ratio between the length of an arc and its radius.

That's what I meant. My radius is 0.04m (not 0.08 as in previous cal).
So 1 rad is ratio of arc length, 0.04m / radius 0.04m
Ok this has not much meaning.

Circular Motion & Acceleration
Uniform circular motion is an example of a body experiencing acceleration resulting in velocity of a constant magnitude but change of direction. In this case, because the direction of the object's motion is constantly changing, being tangential to the circle, the object's velocity also changes, but its speed does not. This acceleration is directed toward the centre of the circle and takes the value:

where v is the object's speed

In a linear speed, we measure in m/s. In circular speed, we measure in rad/s.
So if we are talking about 8000rpm, the circumference being 0.2513m, in equivalent circular speed it will be 133.333rev/s * 0.2513 = 33.51 m/s.
And we need to convert it to rad/s.
Since one rev is 0.2513m/0.04m = 6.2825 rad,

Ok still wrong..

16. Jul 17, 2012

### tiny-tim

yes (and rpm is also a circular speed)
it's a lot easier to convert the "r" in "rpm"

r = revolution = 2π radians,

so 1 rpm = 2π rad/minute

17. Jul 17, 2012

### prodriverex

8000rpm = 8000 * 2∏ = 50,265.48 rad/min

or 837.758 rad/s - and this is angular velocity.

So now to calculate the Angular Acceleration, , I can use v2 / r?

Then T = I * a: 0.0024 * 111,701 = 268 lb-ft2?

18. Jul 17, 2012

### tiny-tim

no!!!

v2/r isn't angular acceleration

-v2/r is the radial component of linear acceleration, and is in m/s2

angular acceleration is simply d/dt of angular speed (in rad/s2)

to calculate angular acceleration (if it's constant), use the usual constant acceleration equations

(your original post was correct …)

19. Jul 17, 2012

### prodriverex

α: dω / dt = 837.758 / dt

And what is dt now? I know it must be a timing of t or something, timing per rev in rad, 2∏? Argh..
My apology for being slow. TIA!

20. Jul 17, 2012

### tiny-tim

well, now we come back to my original question …
… what exactly is the question asking for? accelerating from what speed to what speed in what time???