The missing factor

[Sam Peppiatt]

Fletcher, I think I know why it stops accelerating to grater and grater RPMs.
When the falling weight approaches the maximum velocity obtainable by the acceleration do to gravity, it ceases to go any faster, which is about 16 feet per sec., ( d=1/2 x a x t Sq.) Or, d=.5 x 32 x 1 Sq. = 16ft. The time that the weight falls is quite short, being interrupted each revolution. Each revolution the acceleration starts over, always starting from zero. in other words the acceleration isn't continuous. The velocity doesn't acuminate as it would for a freely falling body which will continue to accelerate at 32 ft. per sec. per sec. Sam Peppiatt

Live Your days Inspired Anew, LYDIA

[eccentrically1]

They did accelerate significantly, I always thought that is a good clue.

I don't have wm2d, so I don't know the limits it has. Has anyone tried simming a 12 foot wheel and playing around with how much it weighs, how the mass is distributed, and how much force or energy it takes to accelerate different versions up that quickly? Can wm2d do that?

[ME]

Wm2d only shows basic shapes like circles, rectangles or polygons, then one can select some material, weight, and some shape (planar, shell, sphere) and the MoI will be calculated or one can make things up.
It's not fancy, but sufficient.
One still has to assume some internal form and make-up. I think this hollow drum didn't have much weight at the rim and all weight being at its spokes, this would be somewhat similar to a disc

Let's consider two disks (r=1.83m) of oak (730 kg/m3) with a thickness of 1 cm
Its weight 2*(pi*r^2)*(0.01)*730 = 154 kg.
This weight is a bit of a gamble: Perhaps he used less dense wood, so when we add spokes we could still end up around such number.

Let's say 8 weights of 1.8 kg are raised 40 cm two times per rotation (or some equivalent action), the average center of mass would be at (8*0.4/pi=1 meter) on the ascending side (I tried to explain the factor of pi in "Importance of raising weights").
I choose 40cm to fit the acceleration, but basically this is a (8*1.8=14.4 kg) weight located at a 1 meter radius on the ascending side. - note: this is an equivalent weight- and location- factor, and not some actual; and a larger wheel with the same raising factor would have this same equivalence.

According to WM2D that wheel will start to accelerate with 29.75 °/s².

[eccentrically1]

Ok, thanks marchello. I'm not familiar with degrees per second squared for acceleration, so is that a ballpark estimate for the accelerations you and jim and fletcher and sam found earlier?

[ME]

30°/s² is the Link:angular acceleration.
Like any acceleration is a change in velocity, this is basically the change in rotational velocity (or let's say: change in RPM's); Rotation is being used as a measure of (rotational) distance.

That's the result of the Link:jerk-formula - where acceleration goes linearly to 0 at max RPM.
This means a 12 ft diameter wheel will accelerate at the rim:
* a fraction of (30°/360°) for a rotation with length (pi* 12ft) (per s²)
= (12*pi*(30/360))
= 3.14 ft/s²
= 0.958 m/s²
= 10.24% of G
= about 1/10 G

Please remember, it's possible those numbers are way off from what actually happened: time and acceleration are the ones estimated here.
This 30°/s² results from that jerk-formula where maximum velocity of 26 RPM is reached in 3 rotations and 10.38 seconds.

- - -

To make it probably a bit more confusing, but more complete:
I reported with my Link:tau-method: - where acceleration slows down gradually according to the current speed.
Given: tau=4, or 98.17% of max RPM, or 25.52 RPM
Resulting: t=9.5 s, a=44°/s²

for diameter=12 ft -->
a=4.6 ft/s²
= 1.416 m/s²
= 14.44% of G
= about 1/7 G
But this one uses 3 rotations as mark-point.

- - -

I can use this earlier tau-formula differently.
Taking 2.5 rotations as mark (because the reported *about* max RPM was reached in *about* 2 or 3 rotations)
Resulting: t=7.91 s, a= 53.25°/s²
for diameter=12 ft -->
a = 5.58 ft/s²
= 1.70 m/s²
= 17.33% of G
= about 1/6 G

- - -

Soooo... who knows which one is closest to the real event?
Perhaps none, but I hope it gives at least a rough estimate of the range of possible values.

[jim_mich]

My opinion...

Initial rotation of the wheel causes centrifugal forces. These forces cause the weights to oscillate, one inward and one out, then the weights swap. The impulse of the weights hitting the wheel push the wheel rotation.

The speed of the weight oscillations is determined mostly be the strength of their CF's. The strength of CF is relative to rotational velocity squared.

When the wheel is first started then the CF's are weak. Thus the wheel acceleration is initially weak. As the wheel gains speed the CF's became stronger and thus the wheel accelerates very quickly.

Without going into details, when the weights oscillate too fast then the centrifugal force differential causing the oscillations of the weights diminishes.

The acceleration of the wheel then drops down to zero and the wheel maintains its top speed.

The wheel forces are much like a sine curve. They start weak, then peak very strong, then diminish back to just enough force to keep the wheel rotating.

The short quick lifting of a box of brick was not enough for the witnesses to notice the slight decrease of speed. And wheel momentum helped.

But with the sustained load of the water screw, the wheel slowed down to a speed range where it was more efficient.

Just my opinions.

So, assuming a sine wave acceleration curve, how fast would it take to accelerate up to speed in three rotations?

[Trevor Lyn Whatford]

Hi Jim_mich,

there is a too small window of opportunity from a slight push to move the weights on the horizontal, any other position then gravity is pulling down on the weights so the weights are acting like a pendulum and cannot make the horizontal gravity neutral zone again. Gravity remains the dominant force until well over 26 RPM, so CF cannot shift the weights in its own right until CF can be the dominant force, and when that happens it turns back into a pendulum again, not that the wheel could ever reach 26 RPM in the first place.

[ME]

So, assuming a sine wave acceleration curve, how fast would it take to accelerate up to speed in three rotations?

Assuming:
a[t] = f * Sin(pi* v[t]/MaxV )

That factor (f) will be some maximum acceleration to be determined.
Your total time depends obviously on its initial velocity: when 0 it would take forever.

I tried some quick test, and as long as I didn't make a mistake:

v[0] = 0.25 °/sec then f=25.4°/s², T=20.24 s; for (6ft) a=0.811 m/s2; maximum acceleration will be reached in 13.37 s, after 180°

v[0] = 0.5 °/sec then f=25.3°/s², T=18.16 s; for (6ft) 0.807 m/s2 maximum acceleration will be reached in 11.29 s, after 180°

v[0] = 4 °/sec then f=24.8°/s², T=14.95 s; (6ft) 0.792 m/s2; maximum acceleration will be reached in 8.05 s, after 180°

As an exact end-point of reaching Max RPM is hard to determine, I fixed the rotation of reaching maximum acceleration to 180°, as it was for most situations close to this angle.

The velocity over this period follows a sigmoid-curve, and the acceleration a logistics-distribution-curve when plotted over time.

When it follows these curves, then maximum acceleration should (or could) be set at half-time - which I didn't, but almost I guess.

---
Edit for recalculation (I used 26°/s instead of 26 RPM, oops)

[cloud camper]

Hi Jim_mich,

there is a too small window of opportunity from a slight push to move the weights on the horizontal, any other position then gravity is pulling down on the weights so the weights are acting like a pendulum and cannot make the horizontal gravity neutral zone again. Gravity remains the dominant force until well over 26 RPM, so CF cannot shift the weights in its own right until CF can be the dominant force, and when that happens it turns back into a pendulum again, not that the wheel could ever reach 26 RPM in the first place.

Right as usual TLW.

The other factor not addressed here is startup (static) friction. There is way too much static friction with even the best bearings to allow CF to become any factor until the wheel is already spinning at a good clip (and
then running friction plus air friction is very significant).

Any actual builder would of course know this but armchair theorists would
not until they actually complete a build.

Coincidentally my own design uses a falling weight to spin up the wheel,
then "recharging" a spring to help raise the weight once more to repeat the cycle.

[jim_mich]

Obviously cloud camper has no working engineering knowledge.

Cloud camper claims that CF is not strong enough to move weights until the wheel "is already spinning at a good clip" The Kassel (4th) wheel was said to start self-rotating at a speed about 5 to 6 RPM. At such a speed and at a radius of 36 inches inside the wheel, the CF of a four pound weight is about 2 ounces.

This is more than enough force to cause a four pound weight to start moving.

See the attached very simple WM2D example picture.

[ME]

The Kassel (4th) wheel was said to start self-rotating at a speed about 5 to 6 RPM.

I estimate your wheel is up to speed between 9.5 and 10.5 seconds, depending on its acceleration when it's going about 80°/s.

While hanging in my armchair, I theorize that any force the yellow weight applies would offset the blue one.... ok, I'm off to the workshop.

[cloud camper]

the CF of a four pound weight is about 2 ounces

Thanks for doing the math – you have proven my point!

As we see our “inventor� is cherry picking here by showing only the positive side of the ledger.

If we have four weights in the system, only two of them could be producing a driving force at a time assuming for a minute a flawed theory is “workable�.

The other two will be of course producing an opposing force, of course exactly matching the “forward� force but for the sake of discussion let’s assign a value of 1 oz.

So 4 oz of forward force against 2 oz of backward. So we then have 2 oz total forward force to accelerate
16 lbs of weights (16 lbs x 16oz/lb)=256 oz. And this assumes a perfect 100% conversion of CF force to driving force which of course cannot happen.

Anyone beginning to see a problem here?

Now we can determine the number of bearings in the system. Possibly a minimum of 4 if the weights are
cantilevered, but that would be very poor design practice as the crossbars would be subjected to large
bending loads putting high assymetric forces into the bearings, increasing friction.

So let’s figure a bare minimum of 8 bearings to support 4 weights (probably more).

Now we can look up the bearing friction for oiled bronze bearings as we know our inventor would never pony up for proper ball bearings. Coefficient of friction for this type of bearing is .04-.07.
http://www.tribology-abc.com/abc/cof.htm#mu_lagers

So lets give it the benefit of the doubt and call it .05. So .05 x 8 bearings =.4. This means that our forward force of 2 oz is reduced by 40% and then 2 more main axle bearings for a total of 50% reduction.

This leaves only 1 oz of forward force to accelerate 256 oz of weights at 5-6 rpm, assuming otherwise perfect conversion of CF to forward force.

And we have completely ignored the weight of all crossbars and support structures to try and help our inventor.

So we can see that the wheel would have to be otherwise powered by gravity or a spring for the bare minimum of two complete turns or a very hard push (not a gentle push as reported by the witnesses) before CF could drive anything.

These are rough back of the napkin estimations here but clearly illustrate the problem. It appears our “inventor� has some serious explaining to do!

[jim_mich]

Anyone beginning to see a problem here?

Yes. Cloud camper has absolutely no idea of how to make a motion wheel. He makes many erroneous assumptions. I'm not going to point out his errors. Doing so requires disclosure of IP details.

It appears our “inventor� has some serious explaining to do!

It appears our resident troll has some serious explaining as to why he jumps to bad conclusions based upon faulty assumptions.

Cloud camper is making pure guesses, then attempting to calculate results based upon his pure guesses at to how a motion wheel might work. He has no clue. And I'm not about to give him any clues.

Cloud camper will eat crow with the rest of those who doubt me.

[cloud camper]

Typical JM response to attack the messenger when he gets caught with his
pants down and can't think of a response!

Clearly JM has not done his homework on this problem and is trying to deflect.

Very sad he cannot address a simple engineering issue that should have
been put to bed a long time ago!

[jim_mich]

It is very sad that cloud camper keeps seeking the secret of my motion wheel.

Cloud camper makes a bunch of statements as to how he THINKS my wheel works. Then he shows problems with HIS imagined design. And when I clearly state that HE is making erroneous assumptions based upon HIS guessing how my wheel works, then cloud camper writes the I'm the one deflecting. Look in the mirror cloud camper.

Cloud camper keeps repeating a claim that I've not done my homework. But cloud camper is ignorant of facts. Specifically, cloud camper is ignorant of how to make a mechanism such that it gains motion because of its own motions. I assume that cloud camper thinks such can not be done.

Cloud camper writes about some sort of engineering issue. I do not have any engineering issue. If he THINK THAT I DO then IT IS HIM that lacks engineering knowledge.