Wheel acceleration...

[ovyyus]

...I know of no force that some form of energy cannot be derived from.

Ralph, enery can be derived from the force generated by a compressed spring - but for how long?

...the Fxd definition of energy is lacking when trying to describe the generation and/or maintaining of forces. Welcome to the other side.

Randall, Fxd obviously = W, not energy or force, hense the apparent lacking. The warth and comfort of the other side seems always just out of reach :D

[AB Hammer]

evgwheel

This topic would have been very welcome in “too weird� thread.
Even better in a thread called “Magnets� as it would be a lasting reference point, instead of hidden in between other post.

I agree with you on this but lets make it one better. Natural forces and Bessler.

The reason I look at it as this is Bessler looked in depth at natural/natures forces

[rlortie]

Bill,

Edited for brevity:

Ralph, energy can be derived from the force generated by a compressed spring - but for how long?

I will play along! :-)

Compressed or tension, energy from a spring can only be derived until the spring reaches it natural state of rest. not unlike two magnets attracted both must be reset. The energy required to reset and store the force is considered the same as what energy is gained when released. Which to my thinking is not quite true due to friction loss.

A compressed spring is a storage receptacle of force/energy that can be released as energy/force.

Force is energy, strength or active power aka The exertion of such power. It is a vector quantity that tends to produce and acceleration of a body in the direction of its application.

Energy is the capacity for work or vigorous activity. Exertion of vigor power. It is the physical system to do work. It also is the source of usable heat or power derived from heat.

Gravity and a compressed spring have a lot in common, difference being once the spring is decompressed it is dormant, gravity how ever never seems to run out of compression, or is it tension?

On a lighter note; does a rock fall from a mountain by a downward force or is it pulled down by the larger mass below it. When the wind blows it is air moving from a high pressure area to a low pressure area, Which is doing the pushing or the pulling? Or is it a combination of both seeking equalization of gradient?

Wong Way Walf!

[bluesgtr44]

Alright, shifting things here a little. Now, most of you know I have always been interested in just what type of a weight distribution we may be looking at to drive a wheel of "X" diameter at "X" rpm's from within. This is about the acceleration rate and why I am putting it here.

Not saying this has anything that we can use....but, I cannot deny the graph if WM2D has any kind of validity. I don't doubt that anyone else would acknowledge that with this set up, a gain is going to be obviously there.....just how legitimate it is, is what I cannot validate.

I know most of you are familiar with this set up. I will relate again as to why I like to use it. It seems to be one of the more "honest" set ups that I can use to test things like this with. It is not, however very useful for other types of tests. The offset is pretty clear and why it happens and this is based on a couple of MT drawings that use this type of offset.

The red spheres do not collide with the yellow and vice-versa. so, once every so often....when they do meet right under the axle at the "collision" point....they pass one another and that creates an offset of weight distribution to the system that can actually be graphed. Now, when I first saw this I didn't think much of it because I was looking at something else specifically. Because I have issues with this crap now....and just can't seem to leave it alone....I go back and review these things sometimes. I saw this again and this time I was reminded of what J. Collins had mentioned about the movement may be not more than 8 inches to produce the effect that Bessler achieved. Don't make this complicated and call John out to the mat....if anything I am making more of it than was intended.

If there is any truth to the program, I believe that with just this mundane shift, so close to the axle....could produce this difference in potential....what do we really have? This was done with gear reductions and there was a difference in potential with this application....and it was expected. I just don't know how to figure up whether or not the significance is in line with reality or not....

See, the thing is....without this program...I know I couldn't see it. It would be that fast! But, with modern technology...I did! So, if it's not real....I'd like to know....if it is can we do anything with it?


Steve

[daxwc]

DT....pg. 244....J. Collins..."It then began to rotate of its own accord with such force that within a minute it had rotated 40 and more times, and could only be stopped by applying great effort."

Mersburg

I wish to show you two different examples, one of a mass dropping at constant acceleration of gravity for the height of one minute rotation of the Mersburg wheel from start, also from of the height of the wheel into constant velocity from just the wheel height then compare them to the velocity of the Mersburg wheel after 60 sec.


3rd wheel,133.8" diameter @ 40 RPM, at Mersburg was 7.12 m/sec (a point on the rim), max theoretical is 8.17 m/s (at gravitational 9.81m/s^2 from 3.398m)

A mass dropped the vertical distance of the circumference of the Mersburg wheel 133.8 inch = 3.398 52 meter,
v = 2( 9.81 m/s^2 x 3.99 m) = 8.17m/s T= 2h/g = 2 x 3.398m / 9.81m/s^2 = .83sec

A point on the wheel has traveled 40 rotations x 3.398m = 135.92m
So a mass dropped at that vertical distance T = 2h/g = 2 x 135.92m / 9.81m/s^2 = 5.26 sec... which is well under the 60 sec for the same distance for the wheel.

Theoretical gravity drop for 60 sec would be h=1/2gt^2 = 1/2 x 9.81m/s^2 x 60 sec x 60 sec = 17658m
Rotations the Mersburg wheel would have done if in constant vertical freefall in 1 minute 17658m / 3.398m = 5196 times
v=2gh = 2 x 9.81m/s^2 x 17658m = 588.6m/s

It sounds like a quick start 40 rpm in the first minute, but is well within the range of a gravity only operated device if one side of the wheel always feels it is falling (remember weight of the mass does not matter just the fact some of it feels it is) or every single mass had to start and stop once ( it did not have continual freefall). The fact alone that the wheel was doing a velocity of 7.12m/sec and the theoretical velocity of a mass from the wheel hieght 3.398m at 9.81 m/s^ 2 is 8.17 m/s which is still under the theoretical max value. So in conclusion there might not need to be secondary driver or energy input, the numbers are still within gravity assisted only range and the wheel would have to be at this fast initial velocity if you expected to get any work out of it. Of course this does not excluded extra energy added in, only that it was not going faster the acceleration or velocity of gravity over the height of the wheel or height in rim travel.

40 plus rpm is defiantly fast though and add in the 8 weight thuds per rotation and suddenly you listening to 320 smacks per minute... no wonder there was conflicting reports as to rpm and weight thuds. The Gera wheel going at 70 rpm would have been 560 whacks and sounded like just a lot of clatter, imagine yourself trying to count those. Thinking about it...Its enough to make your head spin and say the wheel was spring driven ;)))

[ovyyus]

3rd wheel,133.8" diameter...

A small correction: Stewart pointed out to me some time ago that the Merseburg wheel was in fact 12 feet diameter, not the 11.15 feet I have listed on orffyre.com. The reason for the discrepancy had to do with confusion over the reported different measurement units and conversions. I agree with Stewart (thanks Stewart), that the Merseburg (3rd) and the Kessel (4th) wheels were both 12 feet diameter, even though I haven't yet had a chance to update the data on www.orffyre.com

Daxwc, I doubt the difference will significantly alter the general gist of your above post, but it might be worth making a note of it.

[FunWithGravity]

CAN the experts help me.

I am new and learning, I want to understand the acceleration aspect that you guys are speaking of but i want to use a working model.


Suppose i use one of the blank discs that i have made in another thread and want to do acceleration tests with different weights.

Whats the best way to do it, i can make very small incremental increses in a weight box attached to the wheel and measure the time it takes to accelerate through 180 degrees. But how best to measure. it seems to me it would happen very quickly and my reaction time with a stop watch would cause a huge error.

Or does their come a point in the wheel where as long as the weight is heavy enough its passes the point where accerlation mattters since its maxes out its acceleration. In which case i just need to find big enough.

Am i making any sense, sorry if not. I will clarify whatever is needed. Also my thoughts have been as long as my acceleration gets me to a speed that would eqaul the rpm's that were listed for besslers wheel i would be OK.

[daxwc]

Oh, Yes I am sorry I forgot the post about the size correction.

I also just noticed that the upload got rid of all my square root symbols! Guess they thought they were attachments. What do us computer geeks use for square root symbol off the keyboard? By not knowing am I dating myself...lol. Going to have to go back and reread one of Jim's posts, I know I have seen it.

[mickegg]

daxwc

Use "character map" to get the symbols you want

Path on my pc is ....Programs > Accessories > System Tools > Character Map

Select the character you want, copy it (puts it on the clipboard) then paste into your
document at the required places

Regards

Mick

[scott]

Using those characters will give indeterminate behavior depending on the viewer's browser and OS. Better to use the html ascii codes when possible.

http://www.ascii.cl/htmlcodes.htm

Note: only the number codes work in phpbb, not the names.

E.g.

Code: Select all

€

gives €

but

Code: Select all

&# 8364; (no space)

gives €

[mickegg]

Getting technical now Scott.......<grin>

Perhaps it may be possible to have a "Symbol" bar available on the "post reply"
page, from which one could choose, that would always show up in the reply text?

Regards

Mick

[jim_mich]

Many symbols can be typed by holding down the alternate key while pressing four number keys then letting up the alternate key. See list for some examples...

Code: Select all

Keyboard:
| = Alt+0124	ƒ = Alt+0131	† = Alt+0134	‡ = Alt+0135
ˆ = Alt+0136	‹ = Alt+0139	‘ = Alt+0145	’ = Alt+0146
“ = Alt+0147	� = Alt+0148	• = Alt+0149	– = Alt+0150
— = Alt+0151	› = Alt+0155	¢ = Alt+0162	£ = Alt+0163
© = Alt+0169	« = Alt+0171	¬ = Alt+0172	­ = Alt+0173
® = Alt+0174	¯ = Alt+0175	° = Alt+0176	± = Alt+0177
² = Alt+0178	³ = Alt+0179	· = Alt+0183	¹ = Alt+0185
º = Alt+0186	» = Alt+0187	¼ = Alt+0188	½ = Alt+0189
¾ = Alt+0190	¿ = Alt+0191	Ç = Alt+0199	× = Alt+0215
Ø = Alt+0216	ß = Alt+0223	ç = Alt+0231	÷ = Alt+0247
ø = Alt+0248


Microsoft Word:
√ = 221A+Alt+x 

The square root symbol is the hardest one to enter. The only way I've found is to use Microsoft Word where you type 221A and then you hold down the alternate key and press 'x' and the program turns the four digit number into a square root symbol.

I never copy & paste text straight from Word to the forum because Word uses a lot of unseen formating data that gets copied along with the text. This unseen data data can cause problems when parted into places such as the forum. What I do is copy & paste the text into notepad to strip out any unseen formating data and then copy again from notepad to the forum.

[bluesgtr44]

Hey Jim....might be a bit of an imposition here on your idea, but....in your design, is the CF pulling or pushing? I can understand if you would rather not say....


Steve

[bluesgtr44]

Jim, I admit it wasn't very clear. I have attached a couple of images that give an impression of what I am asking and I think you will better understand it. I may be wrong all together in any approach you may be taking. Just looking at possible points of application if one were to try and make use of this force. These are not ever going to produce a damn thing, Jim. It's just that when I look at this mechanically...there has to be a point of application....where does it occur? And again, if you just can't say now, I have no problem with that....


Steve

[jim_mich]

Steve, you ask if the CF is pushing or pulling. That is a strange question. If you place a weight on one side of a lever then it pushes. If you place it on the other side of a lever then it pulls. But when the weight is placed in the middle of the lever then does it push or does it pull?

Also you show both weights moving outward at a same time. Bessler states that as one moves outward the other moves inward; they exchange places. The trick is to arrange them such that the lesser CF of the inner weight overpowers the stronger CF of the outer weight.

(I had an answer almost ready for posting this morning and I got interrupted. When I came back to the computer it was off and most of our clocks were blinking. The post just vanished. Such is life when you're almost the last one on the electric line.)