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[Sam Peppiatt]

To all, Just to simplify it; Wheel rpm =16/C x 60. Sam Peppiatt

Live Your Days Inspired anew, LYDIA

[ME]

: How fast would a 12 foot gravity wheel turn?? (no load). C=36ft., acceleration 32ft./sec. d =distance d=1/2at squared. d=.5x32x1 squared=16, d=16ft. in one sec. 16/36=.444rps
.444x60=26.6rpm! Minus a little bit for friction. Doesn't this prove Bessler's Wheels ran on gravity? If I'm right that is------------maybe someone can verify it.. My math doesn't always come out quite right. Sam Peppiatt

What math?

a 12 ft wheel... what's that? radius or diameter?
C=36 ft why and how 3 times the previous, what's C?
Acceleration 32 ft/sec how come? Acceleration should be ft/sec^2
distance: from where to where? up, down, circumference... is that C: 12*pi ?--> so "12" is the diameter
Is that a kinetics formula d=0.5*a*t^2 ?

ok...I'll try

radius, r=6 ft
acceleration, a=32 ft/sec^2 <-- but why?
circumference, c=2*pi*r = 37.7 ft

Why do you calculate the speed after 1 second, and not 2 seconds? (it still accelerates right?)

I give up. You try again.

[Sam Peppiatt]

ME, Thanks for writing. The wheel was 12 ft. in diameter. C is the circumference of the wheel, @ pi, 3 x 12 = 36 feet.. d = distance. The formula to calculate the distance a weight will fall, with an acceleration of @ 32ft. per sec. in one sec. or any number of sec. is: d = 1/2 a x t Squared. So: d = .5 x 32 x 1 squared = 16 feet. I.E, 16 feet along the circumference of the wheel in one sec. or .444 revolutions in one sec. or x 60 = 26.6 rev. per minute. Supposedly the biggest wheel did turn @ 26rpm.

You might ask why 16? I think it's because it starts from zero. The formula is a basic formula associated with gravity. Not only does it prove the wheel ran on gravity but, it also proves Bessler's wheels must be valid. If I'm right that is. Does it make sense? Please let me know, Sam Peppiatt.

Live Your Days Inspired Anew, LYDIA

[Sam Peppiatt]

ME, Why not 2 sec.? That's a very good question. I don't know the answer.
Maybe someone smarter than me can figure it out. I think it must be right other wise it would keep accelerating until it blew up. My feeble brain can't give you a good answer. Maybe its because the weight is repeatedly falling, always being reset, always starting from zero, so to speak? Sam Peppiatt

LYDIA

[ME]

The formula is a basic formula associated with gravity

Not specifically, it's associated with any acceleration in general: derivatives of displacement over time

0. There's a position "x"
1. There's a velocity "v", which is a change in position over time;
2. There's an acceleration "a", which is a change in velocity over time;

v[t] = a*t + v[0]
x[t] = 0.5*a*t^2 + v[0]*t + x[0]

We could add a third derivative, so you can see how it's extended (and saves us some dull paragraphs of text, but get some dull formula's instead :-)

3. There's a Jerk "j"(that's how it's called), which is a change in acceleration of time: this means the acceleration is not a constant.

now that list becomes:
a[t] = j*t + a[0]
v[t] = (1/2)*j*t^2 + a[0]*t + v [0]
x[t] = (1/6)*j*t^3 + (1/2)*a[0]*t^2 + v[0]*t + x[0]

*For fun and education*
Old Dutch: Ter leering ende vermaeck

--I know it doesn't read like a book, but perhaps you can follow what I try to do--

Perhaps interesting, it works with any displacement, so also arcs when rewritten;

Your acceleration (for a yet unknown reason) is 32 ft/s². At a 6 ft radius this equals to (32/6) rad/s²; so working with this number results in a velocity in [rad/s] and a position in radials; (basically this fraction is my reason I use the radius instead of diameter)
We can just invent some new metric: Rotations/minute², we could call this RPMM

a = (32/6) rad/s² = (32/6)/(2*pi) rotations/s²
...multiply by (60/60)*(60/60)... (which actually equals to 1)
= 60*60*(32/6)/(2*pi) rotations/(60 seconds * 60 seconds)
= 3600*(32/6)/(2*pi) rotations/(minute²)
= 3055.775 RPMM

Now we can use the same formula
x=0.5*a*t^2 +v[0]*t+x[0]
v=a*t + v[0]
true, v[0] and x[0] are zero, so we can just ignore those
Let t be (1/60) minute
x=0.5*3055.775*(1/3600) = 0.4244 rotations (so Yes, finally confirmed :-)
v=3055.775*(1/60) = 50.93 RPM (thus, I do not agree)

Why is this so
This formula is a parabola.
It starts with 0 RPM, and then it accelerates;
I can show the numerical derivative, Velocity as a change in position v=dx/dt
(now it's annoying I used RPMM)

Let's take dt= 2*0.01 = 0.02 seconds, or actually (0.02/60) minutes, and calculate around that time
x[before] = 0.5*3055.775*((1-0.01)/60)^2 =0.41597 rotations
x[after] = 0.5*3055.775*((1+0.01)/60)^2=0.43294 rotations

The numerical velocity dx/dt = (0.43294-0.41597)/(0.02/60) = 60*0.01697/0.02 = 50.91 RPM, which is about the same as found before.

---
Let's use this annoying formula again for 2 seconds

x=0.5*a*t^2
v=a*t
Let t be (2/60) minute
x=0.5*3055.775*(2*2/3600) = 1.697 rotations
v=3055.775*(2/60) = 101.86 RPM
----

If you want 25 RPM
then we use v=a*t, or t=v/a
so t=25/3055.775 = 0.0081812 minutes = 0.49 seconds
Reached within 0.102 rotations, or 36.82°

---
Another note:

Your "pi" need a bit more significant numbers
pi=3 --> 3*12=36
pi=3.1 --> 3.1*12=37.2
pi=3.14 --> 3.14*12=37.68
pi=3.14159 --> 3.14159*12=37.69908

----
With all this one can calculate velocity and displacement when acceleration is known, but unfortunately it's unknown how this acceleration came to be.
---

[jim_mich]

pi=3.14159 --> 3.14159*12=36.69908

That should be: pi=3.14159 --> 3.14159*12=37.69908

Note that I was crucified a few years back for making a math error. And every since a certain troll keeps claiming that I don't have any math skills.

[ME]

Basically that's the reason I write it all out... any mistake should thus (hopefully) be a correctable one, unless I made some error in logic or method.

Thanks Jim, but I can't help what others think and do.
I often make mistakes - and then edit my posts after some proof reading as I just use it as scrap-paper -

[John doe]

The formula is a basic formula associated with gravity

Not specifically, it's associated with any acceleration in general: derivatives of displacement over time

0. There's a position "x"
1. There's a velocity "v", which is a change in position over time;
2. There's an acceleration "a", which is a change in velocity over time;

v[t] = a*t + v[0]
x[t] = 0.5*a*t^2 + v[0]*t + x[0]

We could add a third derivative, so you can see how it's extended (and saves us some dull paragraphs of text, but get some dull formula's instead :-)

3. There's a Jerk "j"(that's how it's called), which is a change in acceleration of time: this means the acceleration is not a constant.

now that list becomes:
a[t] = j*t + a[0]
v[t] = (1/2)*j*t^2 + a[0]*t + v [0]
x[t] = (1/6)*j*t^3 + (1/2)*a[0]*t^2 + v[0]*t + x[0]

*For fun and education*
Old Dutch: Ter leering ende vermaeck

--I know it doesn't read like a book, but perhaps you can follow what I try to do--

Perhaps interesting, it works with any displacement, so also arcs when rewritten;

Your acceleration (for a yet unknown reason) is 32 ft/s². At a 6 ft radius this equals to (32/6) rad/s²; so working with this number results in a velocity in [rad/s] and a position in radials; (basically this fraction is my reason I use the radius instead of diameter)
We can just invent some new metric: Rotations/minute², we could call this RPMM

a = (32/6) rad/s² = (32/6)/(2*pi) rotations/s²
...multiply by (60/60)*(60/60)... (which actually equals to 1)
= 60*60*(32/6)/(2*pi) rotations/(60 seconds * 60 seconds)
= 3600*(32/6)/(2*pi) rotations/(minute²)
= 3055.775 RPMM

Now we can use the same formula
x=0.5*a*t^2 +v[0]*t+x[0]
v=a*t + v[0]
true, v[0] and x[0] are zero, so we can just ignore those
Let t be (1/60) minute
x=0.5*3055.775*(1/3600) = 0.4244 rotations (so Yes, finally confirmed :-)
v=3055.775*(1/60) = 50.93 RPM (thus, I do not agree)

Why is this so
This formula is a parabola.
It starts with 0 RPM, and then it accelerates;
I can show the numerical derivative, Velocity as a change in position v=dx/dt
(now it's annoying I used RPMM)

Let's take dt= 2*0.01 = 0.02 seconds, or actually (0.02/60) minutes, and calculate around that time
x[before] = 0.5*3055.775*((1-0.01)/60)^2 =0.41597 rotations
x[after] = 0.5*3055.775*((1+0.01)/60)^2=0.43294 rotations

The numerical velocity dx/dt = (0.43294-0.41597)/(0.02/60) = 60*0.01697/0.02 = 50.91 RPM, which is about the same as found before.

---
Let's use this annoying formula again for 2 seconds

x=0.5*a*t^2
v=a*t
Let t be (2/60) minute
x=0.5*3055.775*(2*2/3600) = 1.697 rotations
v=3055.775*(2/60) = 101.86 RPM
----

If you want 25 RPM
then we use v=a*t, or t=v/a
so t=25/3055.775 = 0.0081812 minutes = 0.49 seconds
Reached within 0.102 rotations, or 36.82°

---
Another note:

Your "pi" need a bit more significant numbers
pi=3 --> 3*12=36
pi=3.1 --> 3.1*12=37.2
pi=3.14 --> 3.14*12=37.68
pi=3.14159 --> 3.14159*12=37.69908

----
With all this one can calculate velocity and displacement when acceleration is known, but unfortunately it's unknown how this acceleration came to be.


---

I thought the universal measurement for acceleration was in G's?

[ME]

I think it must be right other wise it would keep accelerating until it blew up. My feeble brain can't give you a good answer. Maybe its because the weight is repeatedly falling, always being reset, always starting from zero, so to speak? Sam Peppiatt
LYDIA

Then your time shouldn't be t=1 second, but you should set the rotation to x=1

In another recent post I use some similar formula as posted here, but with the premise the velocity will become constant and acceleration stops;
The reason it stops accelerating is because some (unknown imaginary) mechanism can't react fast enough, so I used (for simple reasons) a cosine transition: a keeling overbalance.
The same 'scrap-paper'-excuse applies there and realized during write-up that weights and radius are no factor (doh!) when only considering the end-velocity within so-much rotations.

http://www.besslerwheel.com/forum/viewt ... 671#141671

I thought the universal measurement for acceleration was in G's?

You thought wrong.
Some use even plain seconds

[Sam Peppiatt]

ME, You lost be, I'm afraid my level of understanding is pretty low. But the wheel didn't turn 50 rpm, right? It turned 26 rpm, so how can 50 be right?
Sam Peppiatt

LYDIA

[ME]

Oh sorry for that, I had the impression it could freshen up some old and rusty knowledge.
Otherwise I don't see the reason of picking some random formula and conclude gravity being the source.

The 50 RPM are related to the earlier unidirectional wheels.
Showed it somewhere earlier:, but here it is again: a nice summary by Ovyyus: http://www.orffyre.com/measurements.html.

---
add:
If you want further explanation about that math-stuff, feel free to ask: I'll see if I can type a bit slower :-)

[John doe]

I think it must be right other wise it would keep accelerating until it blew up. My feeble brain can't give you a good answer. Maybe its because the weight is repeatedly falling, always being reset, always starting from zero, so to speak? Sam Peppiatt
LYDIA

Then your time shouldn't be t=1 second, but you should set the rotation to x=1

In another recent post I use some similar formula as posted here, but with the premise the velocity will become constant and acceleration stops;
The reason it stops accelerating is because some (unknown imaginary) mechanism can't react fast enough, so I used (for simple reasons) a cosine transition: a keeling overbalance.
The same 'scrap-paper'-excuse applies there and realized during write-up that weights and radius are no factor (doh!) when only considering the end-velocity within so-much rotations.

http://www.besslerwheel.com/forum/viewt ... 671#141671

I thought the universal measurement for acceleration was in G's?

You thought wrong.
Some use even plain seconds

Automobiles do that because consumers are more interested in the performance characteristics of the car. Many additional factors come into play when dealing with acceleration of an automobile. Gearing, tires. traction etc etc. however we are only concerned with the motor itself. As far as we know there was no transmission or differential gears and traction was not an issue, all of this would only confuse the issue. We are only concerned with the performance characteristics of the engine. I'm not saying G's are the only measurement but since we are looking at if it was gravity powered it seems that there would be a relationship to 1g or possibly there was an acceleration of 1g which is 32.2 fps. Unless it's just a coincidence...

[ME]

OK, I can "translate" FPS: 32.174 ft/s²
But NUMEROLOGY, seriously ?!!
Please proof such wheel could accelerate that fast (be creative), I say it never can besides by tossing it out of the window.

[John doe]

Sadly I cannot...(yet)

[ME]

Not even a single argument?