Friendly Little Note

[Dwylbtzle]

How much power can you get from only one inch?

that's what SHE said

[Ed]

Sounds like somebody's living in the past. Contemporize, Man!

Shine on you crazy zircon... and while you're at it, you might want to let someone else get a word in. hehehe

[preoccupied]

I stopped reading what Dwylbtzle is writing. Something is making it difficult for me to understand what he is talking about and he sure seems to have a lot to say right now.

I'm thinking of maybe some motion ideas, should I post them in a separate topic?

[smith66]

Sounds like somebody's living in the past. Contemporize, Man!

Shine on you crazy zircon... and while you're at it, you might want to let someone else get a word in. hehehe

Ed,
When you asked about how much force is to be had from an over balanced weight, it's a percentage.
If a weight moves a distance of 10% further out, then 9% of it's mass is over balanced.
Of course, if 8 weights are on the wheel, then the average net force would be 1.125%.

[Dwylbtzle]

Sounds like somebody's living in the past. Contemporize, Man!

Shine on you crazy zircon... and while you're at it, you might want to let someone else get a word in. hehehe

all i'm saying is that the past has had a long time to contain all the tries to explain how to make the wheel spin from internal gizmos
and i'm quite ready and waiting for someone else to go ahead and use as many words as it takes
i think a few of us, here, have been asking that in this very thread
including yourself
specifically about the no-gravity/sheer momentum wrinkle
(no overly specific total-blab details--just a whiff of hope)
gotten any word yet?
hehe

anyway--friendlies
i still think water has something to do with it
maybe not besslers--but it


also--by the way...a zircon has fiery luster
a diamond has brilliant luster

that's how you tell---google it

[smith66]

@dittzle,
you're nothing but a distraction.

@Everybody else;
Please go back and read Jim_Mich's first post.
Bessler states that it's not the arrangement of weignts where some might be further from the center.
And then read wnere Bessler is openly mocking Wagner because Wagner said the same thing that dittzle did, would post it but I'm not on a pc.
Although I will post some math for you guy to consider.
If a lever is 24 inches/60 cm's long and it drops 6 inches/15 cm's, it can use an 4 to 1 ratio. This may allow it to pump twice it's weight.
If the weight moving past top drops ax it should, then it can work with the weight at the bottom by using pulleys.
And this over balance of 100% - 200% takes up less than 2 inches of space for 1 - 2 lbs. or about 450 - 900 grams.

[Dwylbtzle]

distraction from what?
fletcher asked a coupla three times
ed asked a coupla three times
i asked a coupla three times...
here we are on page four
pardon me if i meander while we all wait
(thank christ this note is so friendly and little--lol)

and yer wagner/bessler thing--you threw out something with a question mark
...but it was a quote, i guess?.. or a quote of someone quoting, (or mocking) someone...
in the middle of a lot of stuff you were trying to get in
hard to tell what you meant or who was asking it it to who
lol
sorry
my bad--ok? putchyer shotgun down

and to pre--it's all easily explained: about you, i said i couldn't quite grasp the compound pendulum video--but it was a fascinating idea-...
and maybe it could capture corolis effect...-and then, lets see..i said i liked springs too--that's about it--and then, yeah i said a bunch of cryptic stuff about gordian knots and evil siren mermaid babes
yer right-not necessarily easy to understand my stuff either-my bad

no-one's saying Smith's idea won't work
jim's or pre's either--least of all me
(oh how the frik would any of us here know that anyway?-least of all me)
i WAS showing some interest ya know--a few of us were...no fingers in MY ears, mates
i just think that if ANYONE's does- it'll include a surprise as to why

ditzle--that was good
usually people say dweebtzle
or drill bitzle
that's a cool one

[smith66]

@All,
Am posting the quotes from Bessler that Jim_Mich posted when he started his thread.
Bessler speaking about what and how he learned ;

Bessler, in AP, wrote:

>> Many would-be Mobile-makers think that if they can arrange for some of the weights to be a little more distant from the centre than the others, then the thing will surely revolve. A few years ago I learned all about this the hard way. And then the truth of the old proverb came home to me that one has to learn through bitter experience. <<

And when it comes to mocking Wagner

>> XX Water-power especially is inadequate for perpetual motion

Wagner, red in the face, declares that, just as no arrangement of weights can circle round of its own accord, so too no device using water will work <<

edited to add; why does Wagner even mention water ? end edit
And then ;

>> Oh of course! Its bound to stand still! Wagner says it can't move! Anyone who asks about water is no longer on board the ship. <<

3 exclamation points in a row and then the comment about being on board the ship. Question, why does it matter if someone is onboard the ship or not ?

And an odd statement from Bessler;

>> Even Wagner, wherever he is now, will have heard that one pound can cause the raising of more than one pound. He writes that, to date, no one has ever found a mechanical arrangement sufficient for the required task. He's right! So am I, and does anyone see why? What if I were to teach the proper method of mechanical application? Then people would say: "Now I understand!�

I think this might be as close to a smoking gun as will be found when it comes to Bessler's wheel. After all, what if a weight wasn't a weight ?
The reason I mention this is that fluids are usually measured by volume and not by their mass.

[Dwylbtzle]

well, i certainly always suspected water--(so ya got THAT goin' fer ya)
:{/
but it's a two edged slippery sword (it tends to wanna shlorp both ways--and find the equilibrium)
are there any reports of bessler saying he used water
or anyone hearing water? in his machine?

any reported leakage?
hehe

i do like bessler's line: "he's right and so am I"

anyway--yer idea uses water--my idea uses water-molino's uses water--the cuthbert guy in my "let's test this crap" thread uses water--and he's presented the data of his experiment, (in my thread, when he wrote back, and on nexus), which, (unless he's lying like a dog and faked his own data), seems to have been demonstrated as viably proving the general principle...
(and if you read that thread--at first all the internet blurbs about the guy were saying he was using oil--but i said i thought it MUST be water,
and he writes back and says he's using water)

hell, maybe we're all right
i'm alright/yer alright-right?

[smith66]

Dwylbtzle said;

yes--i see ALL that

but let me tell you something about our friend water
above all things
that stuff tends to FIND its equilibrium

Posted: Sun Jun 02, 2013 8:20 pm

and then today said;
well, i certainly always suspected water--(so ya got THAT goin' fer ya)

Not sure how he could suspect water when as he said, it tends to find it's balance.

Kind of why I think I'm going to step away for a while. have the same thing going on with doctors and that is a bit more important to me.
With this, it's a hobby and would as I told one of my brothers be a cool one to do builds of.
And as Bessler said once again, Anyone who asks about water is no longer on board the ship.

This statement by Bessler means simply that your wrong if you mention water, you're not leaving port, not going any where.

[Dwylbtzle]

Not sure how he could suspect water when as he said, it tends to find it's balance.

because i think you have to use something within the nature of the water
other than just it's weight
to set up a situation where you can take advantage of the quantum level boost--the cheat
that you would need to then be able to use the gravity
which thing (gravity alone) i don't believe works by itself with mere mechanical devices contained exclusively within a wheel
but i think something IS there
and if water weren't involved--i'd be surprised
but don't claim to be sure

i'm trying to get around newton without technically breaking any of his laws--i think that's how you have to do it--because you probably really can't break them

euclid's neither

***************

note: in the tony cuthbert experiment he uses water--but with magnetic ferrofluid mixed with it--and then it's subjected to a magnetic field
that's "cheating" with stuff, (and forces), at the quantum level

my idea uses his same general overall principle to tap the gravity
(but comes from a whole different direction, not a magnetically suspended ferrofluid column)
and (i think ) will, also, depend on a quantum level cheat involving the nature of water

[jim_mich]

It's a pretty simple request Jim .

Two masses have had the same Force [10N] applied to them for the same time - the Force causing the acceleration is stopped & they both coast along at their respective velocities.

Both masses collide with another stationary mass [e.g. 1 kg each] - what FORCE will they apply to the stationary mass ?

If all masses are perfectly elastic the 1 kg mass at v =10 m/s will stop suddenly & transfer all its speed, momentum & energy to the stationary 1 kg mass [like the Newton's cradle] - was the force or impulse it imparted momentum or energy related ?

The 4 kg mas traveling at 2.5 m/s will slow down after collision to 1.5 m/s & have 6 units of momentum while the stationary 1 kg mass will have 4 units of momentum & a velocity of 4 m/s - was the force or impulse it imparted momentum or energy related ?

Definitions:
m1 = mass of object one
m2 = mass of object two
v1 = initial velocity of object one
v2 = initial velocity of object two
u1 = final velocity of object one
u2 = final velocity of object two
Cr = coefficient of restitution
Ref: Cr = 1 for elastic (conservation of energy, parts bounce apart)

Formula:
u1 = ( Cr×m2×(v2-v1) + m1×v1 + m2×v2 ) / (m1+m2) 'final velocity of object one
u2 = ( Cr×m1×(v1-v2) + m1×v1 + m2×v2 ) / (m1+m2) 'final velocity of object two

1st Example Values:
Cr = 1 'want elastic collision (conservation of energy, parts bounce apart)
m1 = 1 'mass (kg) of object one
v1 = 10 'velocity (m/s) of object one (positive direction)
m2 = 1 'mass (kg) of object two
v2 = 0 'velocity (m/s) of object two (negative direction)

Calculations using 1st Example Values:
i1 = m1×v1 'formula initial momentum of object one
i1 = 1×10 'with values entered
i1 = 10 'initial momentum of object one

i2 = m2×v2 'formula initial momentum of object two
i2 = 1×0 'with values entered
i2 = 0 'initial momentum of object two

ii = i1+i2 'initial momentum sum objects one and two
ii = 10+0 'with values entered
ii = 10 'initial momentum sum objects one and two

u1 = ( Cr×m2×(v2-v1) + m1×v1 + m2×v2 ) / (m1+m2) 'formula final velocity of object one
u1 = ( 1×1×(0-10) + 1×10 + 1×0 ) / (1+1) 'with values entered
u1 = 0 'final velocity of object one

u2 = ( Cr×m1×(v1-v2) + m1×v1 + m2×v2 ) / (m1+m2) 'formula final velocity of object two
u2 = ( 1×1×(10-0) + 1×10 + 1×0 ) / (1+1) 'with values entered
u2 = 10 'final velocity of object two

j1 = m1×u1 'formula final momentum of object one
j1 = 10×0 'with values entered
j1 = 0 'final momentum of object one

j2 = m2×u2 'formula final momentum of object two
j2 = 1×10 'with values entered
j2 = 10 'final momentum of object two

jj = j1+j2 'formula final momentum sum objects one and two
jj = 0+10 'with values entered
jj = 10 'final momentum sum objects one and two

-------------------

2nd Example Values:
Cr = 1 'want elastic collision (conservation of energy, parts bounce apart)
m1 = 4 'mass (kg) of object one
v1 = 2.5 'velocity (m/s) of object one (positive direction)
m2 = 1 'mass (kg) of object two
v2 = 0 'velocity (m/s) of object two (negative direction)

Calculations using 2nd Example Values:
i1 = m1×v1 'formula initial momentum of object one
i1 = 4×2.5 'with values entered
i1 = 10 'initial momentum of object one

i2 = m2×v2 'formula initial momentum of object two
i2 = 1×0 'with values entered
i2 = 0 'initial momentum of object two

ii = i1+i2 'initial momentum sum objects one and two
ii = 10+0 'with values entered
ii = 10 'initial momentum sum objects one and two

u1 = ( Cr×m2×(v2-v1) + m1×v1 + m2×v2 ) / (m1+m2) 'formula final velocity of object one
u1 = ( 1×1×(0-2.5) + 4×2.5 + 1×0 ) / (4+1) 'with values entered
u1 = 1.5 'final velocity of object one

u2 = ( Cr×m1×(v1-v2) + m1×v1 + m2×v2 ) / (m1+m2) 'formula final velocity of object two
u2 = ( 1×4×(2.5-0) + 4×2.5 + 1×0 ) / (4+1) 'with values entered
u2 = 4 'final velocity of object two

j1 = m1×u1 'formula final momentum of object one
j1 = 4×1.5 'with values entered
j1 = 6 'final momentum of object one

j2 = m2×u2 'formula final momentum of object two
j2 = 1×4 'with values entered
j2 = 4 'final momentum of object two

jj = j1+j2 'formula final momentum sum objects one and two
jj = 6+4 'with values entered
jj = 10 'final momentum sum objects one and two

My numbers agree with your numbers.

-------------------

3rd (new other) Example Values:
Cr = 0 'want inelastic collision (conservation of KE, parts stick together)
m1 = 4 'mass (kg) of object one
v1 = 2.5 'velocity (m/s) of object one (positive direction)
m2 = 1 'mass (kg) of object two
v2 = 0 'velocity (m/s) of object two (negative direction)

Calculations using 3rd Example Values:
i1 = m1×v1 'formula initial momentum of object one
i1 = 4×2.5 'with values entered
i1 = 10 'initial momentum of object one

i2 = m2×v2 'formula initial momentum of object two
i2 = 1×0 'with values entered
i2 = 0 'initial momentum of object two

ii = i1+i2 'initial momentum sum objects one and two
ii = 10+0 'with values entered
ii = 10 'initial momentum sum objects one and two

u1 = ( Cr×m2×(v2-v1) + m1×v1 + m2×v2 ) / (m1+m2) 'formula final velocity of object one
u1 = ( 0×1×(0-2.5) + 4×2.5 + 1×0 ) / (4+1) 'with values entered
u1 = 2 'final velocity of object one

u2 = ( Cr×m1×(v1-v2) + m1×v1 + m2×v2 ) / (m1+m2) 'formula final velocity of object two
u2 = ( 0×4×(2.5-0) + 4×2.5 + 1×0 ) / (4+1) 'with values entered
u2 = 2 'final velocity of object two

j1 = m1×u1 'formula final momentum of object one
j1 = 4×2 'with values entered
j1 = 8 'final momentum of object one

j2 = m2×u2 'formula final momentum of object two
j2 = 1×2 'with values entered
j2 = 2 'final momentum of object two

jj = j1+j2 'formula final momentum sum objects one and two
jj = 8+2 'with values entered
jj = 10 'final momentum sum objects one and two

Momentum is always conserved.
Kinetic energy is not always conserved, see next post ...

[jim_mich]

Continued...

1st Example Values:
Cr = 1 'want elastic collision (conservation of energy, parts bounce apart)
m1 = 1 'mass (kg) of object one
v1 = 10 'velocity (m/s) of object one (positive direction)
m2 = 1 'mass (kg) of object two
v2 = 0 'velocity (m/s) of object two (negative direction)
u1 = 0 'final velocity of object one
u2 = 10 'final velocity of object two

k1 = 1/2×m1×v1^2 'formula initial kinetic energy of object one
k1 = 1/2×1×10^2 'with values entered
k1 = 50 'initial kinetic energy of object one

k2 = 1/2×m2×v2^2 'formula initial kinetic energy of object two
k2 = 1/2×1×0^2 'with values entered
k2 = 0 'initial kinetic energy of object two

kk = k1+k2 'formula total initial kinetic energy of objects one and two
kk = 50+0 'with values entered
kk = 50 'total initial kinetic energy of objects one and two

e1 = 1/2×m1×u1^2 'formula final kinetic energy of object one
e1 = 1/2×1×0^2 'with values entered
e1 = 0 'final kinetic energy of object one

e2 = 1/2×m2×u2^2 'formula final kinetic energy of object two
e2 = 1/2×1×10^2 'with values entered
e2 = 50 'final kinetic energy of object two

ee = e1+e2 'formula total final kinetic energy of objects one and two
ee = 0+50 'with values entered
ee = 50 'total final kinetic energy of objects one and two

ed = energy difference (lost) due to collision.
ed = kk - ee 'fomula energy difference (lost) due to collision.
ed = 50 - 50
ed = 0 'energy difference (lost) due to collision.

-------------------

2nd Example Values:
Cr = 1 'want elastic collision (conservation of energy, parts bounce apart)
m1 = 4 'mass (kg) of object one
v1 = 2.5 'velocity (m/s) of object one (positive direction)
m2 = 1 'mass (kg) of object two
v2 = 0 'velocity (m/s) of object two (negative direction)
u1 = 1.5 'final velocity of object one
u2 = 4 'final velocity of object two

k1 = 1/2×m1×v1^2 'formula initial kinetic energy of object one
k1 = 1/2×4×2.5^2 'with values entered
k1 = 12.5 'initial kinetic energy of object one

k2 = 1/2×m2×v2^2 'formula initial kinetic energy of object two
k2 = 1/2×1×0^2 'with values entered
k2 = 0 'initial kinetic energy of object two

kk = k1+k2 'formula total initial kinetic energy of objects one and two
kk = 12.5+0 'with values entered
kk = 12.5 'total initial kinetic energy of objects one and two

e1 = 1/2×m1×u1^2 'formula final kinetic energy of object one
e1 = 1/2×4×1.5^2 'with values entered
e1 = 4.5 'final kinetic energy of object one

e2 = 1/2×m2×u2^2 'formula final kinetic energy of object two
e2 = 1/2×1×4^2 'with values entered
e2 = 8 'final kinetic energy of object two

ee = e1+e2 'formula total final kinetic energy of objects one and two
ee = 4.5+8 'with values entered
ee = 12.5 'total final kinetic energy of objects one and two

ed = energy difference (lost) due to collision.
ed = kk - ee 'fomula energy difference (lost) due to collision.
ed = 12.5 - 12.5
ed = 0 'energy difference (lost) due to collision.

-------------------

3rd Example Values:
Cr = 0 'want inelastic collision (conservation of KE, parts stick together)
m1 = 4 'mass (kg) of object one
v1 = 2.5 'velocity (m/s) of object one (positive direction)
m2 = 1 'mass (kg) of object two
v2 = 0 'velocity (m/s) of object two (negative direction)
u1 = 2 'final velocity of object one
u2 = 2 'final velocity of object two

k1 = 1/2×m1×v1^2 'formula initial kinetic energy of object one
k1 = 1/2×4×2.5^2 'with values entered
k1 = 12.5 'initial kinetic energy of object one

k2 = 1/2×m2×v2^2 'formula initial kinetic energy of object two
k2 = 1/2×1×0^2 'with values entered
k2 = 0 'initial kinetic energy of object two

kk = k1+k2 'formula total initial kinetic energy of objects one and two
kk = 12.5+0 'with values entered
kk = 12.5 'total initial kinetic energy of objects one and two

e1 = 1/2×m1×u1^2 'formula final kinetic energy of object one
e1 = 1/2×4×2^2 'with values entered
e1 = 8 'final kinetic energy of object one

e2 = 1/2×m2×u2^2 'formula final kinetic energy of object two
e2 = 1/2×1×2^2 'with values entered
e2 = 2 'final kinetic energy of object two

ee = e1+e2 'formula total final kinetic energy of objects one and two
ee = 8+2 'with values entered
ee = 10 'total final kinetic energy of objects one and two

ed = energy difference (lost) due to collision.
ed = kk - ee 'fomula energy difference (lost) due to collision.
ed = 12.5 - 10
ed = 2.5 'energy difference (lost) due to collision.

[jim_mich]

Notice how the off topic chatter interferes with this discussion. This is why Stewart (in his forum) insists that chatter/comments that are not specific information or specific data be placed in a his comments thread.

Just saying people can always start new threads where they can link or quote to old threads and where they can chatter on and on and on. (That little box just to the left of the word "Posted" will give you a link to the post.)

[Tarsier79]

Jim this is the same stuff Peq has been sayin for years. Have you seen once where in the real world where KE has increased measurably from an impact, or with the use of CF?