The missing factor

[ME]

Hi
I am interested in the 'some maximum velocity'.
It would be good to have an agreed upper rotation limit.

They are just the observations, see this page of ovyyus, http://www.orffyre.com/measurements.html
Velocity [°/sec]=RPM * (360°/rotation) / (60 sec/min) = RPM * 6
So 50 RPM=300°/s

[daanopperman]

Lord I'm glad it was not

" jim_galilee "

[Sam Peppiatt]

bluesgtry44, I've been thinking about your question a lot and trying to learn more about the wheel(s). It's helpful to experiment with the numbers; like so: Given a 12 foot diameter wheel with 8, 4lb. weights and weighs 212 lbs. with the weights. The wheel is constantly side heavy by 3 weights or @ 12 lbs.
Find acceleration: a=f x g/w, a=12lbs. x 32 / 200, a=1.9 ft. /sec./sec.
If the circumference of the wheel is 36 ft. and reaches full speed in 3 revolutions the time will be: Square root of "t"= d / .5x a, d, distance 3 x 36=108 feet. (108ft. in 3 rev.) Sq. r t=108 / .5 x 1.9, Sq. r t= 108/.95,
Sq r t = 113, Sq. r 113= 10.6 sec.
To summarize: The wheel probably weighed around 200 lbs., had 8, 4lb. weights, accelerated up to speed in 3 rev. which took about 10 sec.. You are right, it's too bad they didn't make a note of it. It would take a lot of the guess work out of it. Sam Peppiatt


Live Your days Inspired Anew, LYDIA

[ME]

Sam, if I follow you correctly it reaches 32 RPM, but keeps accelerating.

[Sam Peppiatt]

Hi ME!! No, the large 12 foot diameter wheel turned @ 26 rpm. It reached full speed in 3 revolutions, if I remember right. The question bluesgtr44 asked is; how long would that take, in other words the time to accelerate to full speed. I'm guessing at a number of things but tried to calculate approximately what that time might have been. Just trying to get some sense of it. I get the feeling that it was pretty snappy, if so it would indicate there was very little resistance to rotation, if it could get up to speed so quickly. Which means the reset device must be a cam of some shape or form. (maybe) This assumes that the acceleration was linear but it might vary a little bit do to friction.
ME, are my numbers screwed up? Which they could be. Please let me know. Sam Peppiatt


Live Your Days Inspired Anew, LYDIA

[ME]

Sam, for me to say your numbers are screwed up, does also imply I know exactly what happened to the wheel: but I don't know.
I guess your guess is as good as my guess.

What do you mean by "linear acceleration"; I think you mean linear from being maximum at start and 0 at finish.

I use a gradual diminishing acceleration and assume its dependency on speed, and actually (in perfect circumstances) would never reach top speed (something like charging a capacitor - considered full when tau=5).
So I also have to estimate when the apparent maximum velocity is reached.

My numbers for reaching 26 RPM in 3 rotations:
These values are (besides causal effects) radius- and weight-independent as displacement (3 rotations) has velocity(max RPM) and acceleration(initial) as derivatives.

tau=3, or 95.02%, or 24.71 RPM:-> t=10.6 s, a=31°/s²; for radius=6 ft, a=3.2 ft/s²
tau=4, or 98.17%, or 25.52 RPM:-> t=9.5 s, a=44°/s² ; for radius=6 ft, a=4.6 ft/s²
tau=5, or 99.33%, or 25.83 RPM:-> t=8.9 s, a=59°/s²; for radius=6 ft, a=6.1 ft/s²

The asked time is similar as yours, so a documented number would be something like: "in about 10 seconds".

Only my acceleration differs by a factor of two (to four) times, that's because my acceleration gets smaller a bit faster, so it initially needs more to get to the given amount of speed at the given amount of rotations.
But acceleration was never asked :-)

[jim_mich]

the large 12 foot diameter wheel turned @ 26 rpm. It reached full speed in 3 revolutions, if I remember right. The question bluesgtr44 asked is; how long would that take, in other words the time to accelerate to full speed.

60 Seconds ÷ 26 RPM = 2.3077 Seconds per Rotation at full speed.
3 Rotations × 2.3077 = 6.9231 Seconds per 3 Rotations at full speed.

When acceleration is constant, and the acceleration rotations are known, then acceleration time is twice the full speed time for the same number of rotations. Thus...

6.9231 Seconds × 2 = 13.8462 seconds acceleration time.

I think this is right, but I make no guarantees. Can someone confirm this?

[Sam Peppiatt]

ME, Would the acceleration taper off as the wheel gained speed? You might be right. I'll have to think about that. For the acceleration to stay constant ,(1.9 ft./ sec. / sec.) the force (12 lbs.) would have to remain constant.

I see what you mean, you might be right. ME, I need more time to resolve it in my mind, (my feeble brain), I will try to get back to you-----------Sam


Live Your Days Inspired Anew, LYDIA

[ME]

Amount of rotations: N=0.5*a*t*t = 3 [Rot]
Velocity: v=a*t = 26 [RPM] /60 [Sec/Min] = 0.433.. = 1/2.3077

rewrite: a=v/t = 0.433/t
rewrite: 6=a*t*t = v*t = 0.433*t
t=13.846 [s]
a=0.021296 [rot/s^2] = 11.266.. deg/s^2

for r=6 ft, a=1.18 ft/s^2

Would the acceleration taper off as the wheel gained speed? You might be right.

I wrote about it somewhere (here or elsewhere). I think as some mechanism takes time to react (especially when it needs to "bang" against some side), it will have less time to react at the right moment when the wheel speeds up or centrifugal gets larger....or whatever reason. At least it stopped accelerating, otherwise those RPM's at "2 or 3 rotations" shouldn't be reported - I think a sudden stop in acceleration would also have been noticed.

[ME]

While I'm doing algebra anyway

We could assume acceleration goes linearly towards zero (and perhaps just stops accelerating at time t)
We introduce the third derivative
N[t]=j*t*t*t/6+a[t]*t*t/2
v[t]=j*t*t/2+a[t]*t
a[t]=j*t+a[0]

Let,, N[t]=3, v[0]=0, v[t]=26/60, a[t]=0, a[0]=a
Rewrite:
j= -a[0]/t
v[t]=a*t/2 --> a=2*v[t]/t
N[t]=a*t*t/3=3 --> 9=2*v[t]*t --> t=(9/2)*(60/26)=135/13=10.38 s
a=2*v/t = 2*(26/60)*(13/135) --> 30.04°/s^2

r=6 ft -->a= 3.15 ft/s^2

[Fletcher]

the large 12 foot diameter wheel turned @ 26 rpm. It reached full speed in 3 revolutions, if I remember right. The question bluesgtr44 asked is; how long would that take, in other words the time to accelerate to full speed.

60 Seconds ÷ 26 RPM = 2.3077 Seconds per Rotation at full speed.
3 Rotations × 2.3077 = 6.9231 Seconds per 3 Rotations at full speed.

When acceleration is constant, and the acceleration rotations are known, then acceleration time is twice the full speed time for the same number of rotations. Thus...

6.9231 Seconds × 2 = 13.8462 seconds acceleration time.

I think this is right, but I make no guarantees. Can someone confirm this?

I think that is right for constant acceleration jim_mich.

Then you have to wonder what stops it accelerating to a greater and greater RPM and why is there a limiting RPM reached ?

As we've all discussed before that could be for at least two reasons AFAIK.

If it is not mass imbalance induced unequal torques then it could be the time taken for 'parts' to transition - they need to accelerate and decelerate again - this takes time.

If the unequal torque is produced by mass imbalance then one assumes that at some RPM the mass is falling or rising at a rate equate to the wheel velocity at the radius it is acting at. IOW's it can't catch up with the wheel anymore and unequal torque is now zero at maximum RPM.

We know the acceleration due to gravity but we don't know the net imbalance force (though it appears not much) so we can guess the number of weights acting at any time over a circumference distance to come up with some figures retrospectively I guess.

Perhaps that is what ME is heading towards by running his math ? I'd be interested in what you determine ME !

I seems quite complex and some assumptions have to be made but we could start with just one Net mass weight imbalance and see how it 'fits'.

If it can be reliably worked out then that could be a good clue to how much extra force there was per revolution and that might rule out some possible mechanical options and narrow the field.

[Sam Peppiatt]

ME, You are quit right, The acceleration at the start would be maximum then drop to zero when as the wheel reached top speed. So the time I came up with, 10sec, using the numbers I suggested, must be wrong. Seams like the time would have to be more than that. I'm probably getting out of my realm of understanding. Sam Peppiatt, LYDIA

[ME]

To add to the previous Math-twitch: the found 'jerk' j= -2197/273375 = 0.008037 rotations/s³ = -2.89317°/s³, and would suddenly drop to zero at time T.

Sam, as mentioned before your 10.6 seconds would give a similar reported observation as my calculus showed (when it actually was measured in Bessler's time)

We know the acceleration due to gravity but we don't know the net imbalance force (though it appears not much) so we can guess the number of weights acting at any time over a circumference distance to come up with some figures retrospectively I guess.

Perhaps that is what ME is heading towards by running his math ? I'd be interested in what you determine ME !

I removed all assumption regarding weight, size and gravity. It's just the derivatives of displacement, measured in amount of complete rotations.

The displacement (in the above case: 3 rotations), the end velocity (26 RPM), and unknown initial acceleration, and the acceleration of 0 the time T which is also unknown)

By simply running the kinematics-formula of displacement we can determine both unknowns, depending on the accuracy of those observed values and also assuming the correct math-description. I assumed a slightly different description in my "tau"-post earlier; I guess when using more derivatives it will close in on those "tau"-values.

- - -

When we reapply the wheel-size to the found acceleration of 0.08346 rot/s² (or 30.04°/s²), then we can speculate further.
A diameter of 12 ft is a radius of 1.8288 m (more comfortable with SI), this means at that distance (obvious some effect would have taken place inside the wheel, meaning a lower radius) things accelerate at 2*pi*r*0.08346=0.959 m/s²
That's about 10% of G.
But I think any speculation on the inside of this wheel is useless (or off-scale) as it was a bidirectional wheel, and thus initially balanced, and thus its initial acceleration was clearly zero.


Let's make it 20%, what would that say?
Let's make it 33%
(N=2.5 rotations, v=50 RPM, diameter=9.3 ft) --> for the above kinematics (3rd derivative-variant) I get t=4.5, a=133°/s² = 3.3 m/s² = 1/3 G

We can hypothesize about the 1 lifts four in two ways (?)
A down-force of 1 over all the fractions under gravity 1/(4+1) , and/or a down-force of 1 over the difference of fractions under gravity 1/(4-1)...
[while immediately glancing at the toy-page when seeing those numbers again]
So this could be a clear sign of a Gravitywheel and some leverage.

I found a relation between the final RPM and time and acceleration: a=RPM^2 * factorA, t=factorB/RPM
Note the difference between 'a' and 't' being RPM^3
So this could be a clear sign of some third derivative usage
(you're welcome Grimer)

But centrifugal acceleration = w^2 / r, where w=angular velocity - a factor of RPM
So this could be a clear sign of some Motionwheel

Or it is some effect of a gravitywheel in motion, or a motionwheel still being limited by gravity, or something completely different.
The one thing we can determine for almost certain: it accelerated significantly but sub-G.

Marchello E.

[Fletcher]

Thanks ME .. I agree it was sub G also.

Usually people cross check by attempting to calculate Power (Rate of doing Work).

One wheel could lift 70 lbs in about a minute with a 4 to 1 reduction from memory. Don't quote me on that.

In that Work is force and displacement. We want to know the size of the asymmetric force that caused the torque that caused the Power.

That force could be something as yet unidentified or it could equate to a weight imbalance. It's a force pure and simple.

From Witness Accounts at BW.com

Examination at Merseberg (October 31, 1715)

... The certificate states:

The inventor first put in motion his six ells (~11 feet) in diameter and one foot thick machine which was still resting on the same wooden support upon which it had previously been mounted. It was stopped and restarted, turned left and right as many times as was requested by the commissaries or the spectators. The machine was started by a very light push with just two fingers and accelerated as one of the weights, hidden inside, began to fall. Gradually, within about one revolution, the machine acquired a powerful and even rotation, which continued until it was forcefully brought to a stop again; the machine preserved the same rapid motion when lifting a box filled with six whole bricks weighing together about 70 pounds. The weight was lifted by means of a rope conducted through a window by means of a pulley. The box was lifted as many times as was requested.

From memory the axle was 2 inches thick and the wooden outer axle about 6 inches (3 to 1 and not 4 to 1 for this machine). No mention of the time to lift and height in this report but I seem to remember about 20 feet from somewhere. Anyways says that turned at the same rapid rate (40 RPM).

http://www.orffyre.com/measurements.html

[ME]

Usually people cross check by attempting to calculate Power (Rate of doing Work).

Perhaps I'm not "usual people" :-)
Also, as mentioned earlier, those speed calculations are not related to the weight and size (only causal).

When assuming the wheel mechanisms under load acts equally when just free-wheeling, then the acceleration at 20 RPM is drained away and not used for further wheel acceleration.
With the math from the previous exercise: After 1.005 rotations when the wheel reaches 20 RPM it has an acceleration of 0.04009 rotations per second squared:
For the Kassel-wheel:
r=6 ft --> a=0.4607 m/s^2,
r=4 inch --> a=0.0256 m/s^2
---
For the Merseburg, I don't see where to begin as there's no change in speed, or how much time equals "rapid".
I think I leave those Force-calculations to someone else.