IS THIS A REACTIONLESS DRIVE OR A PERPETUAL MOTION MACHINE?

[Sam Peppiatt]

Hi PeterAX,

Yes, looks like it's going to work. I did one pendulum but have to do the other two. I love your description! It's not complicated and it's not simple, but it is defiantly sophisticated!

Where do I start? I you sure you are interested in this? I don't think any one else ----------------------------Sam

[PeterAX]

Hi Sam,
Hi dear colleague,
1) Yes, I am deeply interested in Bessler's wheel machine and in its principle of operation!
2) One pendulum is ready? This is good! Another two pendulums have to be also done? Ok. But how these three pendulums would work together? What is the related method/system of syncronization? Seems to be interesting.
------------------------
Looking forward to your answer.
Regards,

[Sam Peppiatt]

Hi PeterAX,

Good! I'll take my best shot at it. It's not as I had imagined it would be. First off I should point out that the pendulums go around and around in a circle but don't rotate. This type of motion is referred to as translating. Its like the peddles on a bicycle. They go around in a circle but don't rotate, this leads to the very hart of the invention.
The pendulums don't swing out on the down side and back in on the up side, as I had previously thought. They remain straight out towards the down side of the wheel all the times, never moving in or out! How can that be, right?

This is how it's done. The pendulums are "L" shaped. The long side of the "L" is nine inches long and more or less horizontal. It sticks straight out from the rim of the wheel, with a one half pound weight at the end of it. The short leg of the "L" hangs down, is six inches long, with a one and one half pound weight at the bottom end of it. (continued)

[Sam Peppiatt]

PeterAX, Had to rest for a minute.

Do to the angle of the dangle, the weight at the outer end of the pend. drops below the Horizontal 15 to 20 degrees. The pends. are free to swing a little bit on there pivot points, if they chose to do so. If you give the wheel a spin, the pendulums are translating, not rotating, all the way around, and wheel is perfectly balanced.

A small cam on each pend. lifts a lever that stretches out a spring. Then about 2 O'clock energy in the spring is released. It rotates the far end of the pend. up above the horizontal about 20 degrees. The pendulum temporarily latches / locks to the wheel. Now, it is rotating with the wheel, not translating, which means it is out of balance, (OOB). And, OOB quite a lot, remember the 1/2 pond weight sticks out nine inches from the rim of the wheel. This weight drives the wheel down to six O'clock, bottom dead center, (BDC). At this point, it automatically releases, goes back to translating or balanced, and returns to the top of the wheel and the cycle repets. (more)

[Sam Peppiatt]

PeterAX,

I have a hard time with this key board, give me a hacksaw any day.
Anyway, as you can see the weights move very little. They simply latch / rotate going down, an unlatch / translate going up.

I have to verify this; if it does work I'll try to get a picture and send it to John Collins and he can post it if he is willing. Please let me if you have any questions--------------------Sam

[PeterAX]

To Sam Peppiatt.
-------------------------------
Hi Sam,
Thank you for your reply.
1) Please give me some time to consider your three posts carefully and thoroughly. As if I understood the basic principle of operation, more or less. But some questions may appear in the process of the further in-depth consideration.
2) Yes, it would be good if some pictures were available.
Regards,

[PeterAX]

To ME and to all other members of this forum, who are interested in the topic.
=========================
Our simulation/animation is given below.
=========================
Firstly, please always keep in mind and please always refer to our first post of Sat Jul 28, 2018 12:41 pm and to the related two links
https://mypicxbg.files.wordpress.com/20 ... _01-12.pdf
https://mypicxbg.files.wordpress.com/20 ... s01-08.pdf
--------------------------------------------
Secondly, our simulation/animation presentation is subdivided into three consecutive separate parts for an easier understanding.
=========================
PART 1. Please look at the link https://youtu.be/9cBGurYSryw
1) The zigzag device and the straight-line device are put together vertically one to another and are fixed motionless to a horizontal motionless plane.
2) The two blue T-shaped components start free falling together and simultaneously. Friction is negligible.
3) In the straight-line case the T-shaped blue component falls freely without any interruptions and obstacles.
4) In the zigzag case however after the blue balls enter the zigzag channel the blue T-shaped component slows down its vertical downward motion and decreases its downward vertical velocity.
5) In one word, the straight-line modification blue T-shaped component covers the distance between the highest position and the lowest position much faster than the zigzag modification blue T-shaped component.
==========================
PART 2. Please look at the link https://youtu.be/aVOfWLDrYwA
1) How to slow down the vertical downward motion (that is, how to decrease the vertical downward velocity) of the straight-line modification blue T-shaped component?
2) The answer is simple. The straight-line segment "s" is made rough (inside the channels) as the related force of friction is chosen in such a manner so that in the lowest position the linear downward velocities of the two blue T-shaped components are one and same and equal one to another.
3) In one word, the zigzags generate mechanical resistance, (a) which is absolutely identical and equivalent to friction and (b) which does not generate heat.
===========================
PART 3. Please look at the link https://youtu.be/pPGPktU_kpo
The last link simply repeats the experiment, described in our first post of Sat Jul 28, 2018 12:41 pm and in the related two links
https://mypicxbg.files.wordpress.com/20 ... _01-12.pdf
https://mypicxbg.files.wordpress.com/20 ... s01-08.pdf
The zigzags generate again mechanical resistance, (a) which is absolutely identical and equivalent to friction and (b) which does not generate heat.
===========================
In one word, the text above and the related links above unambiguouly show that (a) either the law of conservation of linear momentum is not correct or (b) the law of conservation of mechanical energy is not correct or (c) both the law of conservation of linear momentum and the law of conservation of mechanical energy are not correct simultaneously.
===========================
Looking forward to your answer.

[ME]

Thanks you for the imagery!

The zigzags generate again mechanical resistance, (a) which is absolutely identical and equivalent to friction and (b) which does not generate heat.

With this assessment I agree.

In one word, the text above and the related links above unambiguouly show that (a) either the law of conservation of linear momentum is not correct or (b) the law of conservation of mechanical energy is not correct or (c) both the law of conservation of linear momentum and the law of conservation of mechanical energy are not correct simultaneously.

No one is able to "unambiguously" imply your conclusion from your words and videos.
So please spend some time explaining why and how you come to your conclusion before I explain mine.


For example: It is unambiguous that the statement made in Part 1.5 is false.

- Why is that?: The height is the same, the zig-zag distance is longer.
- What's the effect?: it explains Part 1.4
- How?: The zigzag causes a deflection of the vertical acceleration vector sideways which reduces its effect vertically.

Looking forward to your answer.

[PeterAX]

To ME.
-----------------------------------
Thank you for your prompt reply.
Ok, I will consider very carefully your last post and will start explaining the things one after the other in a step-by-step consequtive method. Please give me some time. I will write to you in the nearest future.

[PeterAX]

To ME.
----------------------------------------
Let us repeat again. (Please look again at your last post.)
----------------------------------------
You agree with the assessment that "The zigzags generate mechanical resistance, (a) which is absolutely identical and equivalent to friction and (b) which does not generate heat."
---------------------------------------
Did I understand correctly your words?

[ME]

Please try to avoid repetition.
Because sometimes repeated words only resonates with your own mind and not with the meaning you try to convey.

To do the same, I'll rephrase "The zigzags generate again mechanical resistance, (a) which is absolutely identical and equivalent to friction and (b) which does not generate heat."
Here it goes as I try to dissect it in multiple sentences.

1. The straight vertical is basically a free fall situation.
2. In this straight line situation the T-shape undergoes the full acceleration that's caused by gravity.
3. Compared to this free fall the zig-zag undergoes a variable deflection of gravitational force.
4. With vector analysis you can determine the percentages of gravitational acceleration that T-Shape will accelerate down the zigzag ramp. And thus determine how much it descends vertically down.
5. Per zig-zag there will be an average downwards percentage. It will be less than the 1 g.
6. Hence one could state that the zig zag undergoes mechanical resistance,
7. Add friction to the straight line vertical that matches with that zig-zag average, and they will undergo the same acceleration (on average).
8. In the straight line situation that friction is lost by heat generation, in the zigzag situation there are no (negligible) losses.

Sure, for the purist there are some nuances to make on "absolutely identical" and "equivalent" but for the resulting effect I'm not going to nitpick and thus, again, I agree with your statement.

So how did you came to your conclusion?

In one word, the text above and the related links above unambiguouly show that (a) either the law of conservation of linear momentum is not correct or (b) the law of conservation of mechanical energy is not correct or (c) both the law of conservation of linear momentum and the law of conservation of mechanical energy are not correct simultaneously.

[ME]

Hmm... I think I finally understand the issue you're having.
Your rollers are massless!

When the T-shape has more mass than the rollers then the zig-zag as "friction machine" becomes less effective.
In that case:

• the rollers will simply shoot in and out with great speed without having much effect on vertical motion

It's then like stepping on marbles that shoot-out because of the experienced pressure.
Why? Because the T-shape itself is unhindered by the zig-zag-ramp and is able to go straight down. The force decomposition that applies to the rollers on the ramp does not apply to the T-shape itself.

But when the T has all the mass and the rollers none (and then zigzag is not shallow) then that would make your whole exercise futile from the beginning.
Especially because you base your theory on the assumption that the zig-zag causes a slowdown.

right?

[PeterAX]

Hi ME,
Thank you for your two replies.
----------------------------------------
I am very glad that you agree with the assessment that the zigzags generate mechanical resistance, (a) which is absolutely identical and equivalent to friction and (b) which does not generate heat. This fact makes much easier our further explanations, which are given below. Actually PART 1 and PART 2 were designed especially for Leafy, who is a member of this forum and who as if does not agree with the the assessment that the zigzags generate mechanical resistance, (a) which is absolutely identical and equivalent to friction and (b) which does not generate heat. (Or may be I did not understand very well his words and his point of view, respectively.)
----------------------------------------
So let us start our explanations.
----------------------------------------
1) Firstly, please have a look for a while at PART 2 and at the related link https://youtu.be/aVOfWLDrYwA from 00:00 to 00:03. This is only for getting a notion about the limits of the segment "s", that is, how this segment "s" is situated in relation to (relative to) the zigzag section.
----------------------------------------
2) Now let us focus on PART 3 and on the related link https://youtu.be/pPGPktU_kpo . The experiment is carried out in a space station under weightlessness conditions. Friction is negligible as the only exception is the friction inside the two straight-line channels of the segment "s". (The inside surfaces of the straight-line channels of the segment "s" are made rough thus able to generate friction (and heat, respectively).)
----------------------------------------
3) The mass of each blue component is Ma.
-----------------------------------------
4) The mass of each black component is Mb.
-----------------------------------------
5) There are four couples blue ball/blue rod. Each blue ball is firmly attached to the related blue rod thus forming one united whole.
-----------------------------------------
5A) The mass of each blue ball is negligible (if compared to Ma or to Mb), but not equal to zero.
-----------------------------------------
5B) The mass of each blue rod is negligible (if compared to Ma or to Mb), but not equal to zero.
-----------------------------------------
6) From 00:00 to 00:03 the two blue components move simultaneously and uniformly. Each blue component's linear velocity is V' as V' = const. The two black components are at rest.
------------------------------------------
7) At 00:03 the four blue balls enter simultaneously (a) the "upper" black component's smooth zigzag channels and (b) the "lower" black component's rough straight-line channels of the segment "s", respectively.
------------------------------------------
8) From 00:03 to 00:15 the four blue balls move (a) inside the "upper" black component's smooth zigzag channels and (b) inside the "lower" black component's rough straight-line channels of the segment "s", respectively.
-------------------------------------------
9) At 00:15 the four blue balls exit simultaneously (a) the "upper" black component's smooth zigzag channels and (b) the "lower" black component's rough straight-line channels of the segment "s", respectively.
-------------------------------------------
10) The force of friction inside the two rough channels of the segment "s" is chosen in such a manner (we could use for example a variable roughness and the related variable force of friction, respectively) that:
-------------------------------------------
a) the blue components decelerate in one and same manner, that is, their decelerations are one and same and equal one to another;
-------------------------------------------
b) the black components accelerate in one and same manner, that is, their accelerations are one and same and equal one to another.
-------------------------------------------
11) From 00:15 to 00:17 the two blue components move simultaneously and uniformly. Each blue component's velocity is V" as V" = const.
-------------------------------------------
12) From 00:15 to 00:17 the two black components also move simultaneously and uniformly. Each black component's velocity is V"' as V"' = const.
-------------------------------------------
13) Therefore for the "upper" zigzag modification we can write down that
(Ma) x (V') = ((Ma) x (V'')) + ((Mb) x (V''')) (1)
(1/2) x (Ma) x (V') x (V') = ((1/2) x (Ma) x (V'') x (V'')) + ((1/2) x (Mb) x (V''') x (V''')) (2)
-------------------------------------------
14) And for the "lower" straight-line modification we can write down that
(Ma) x (V') = ((Ma) x (V'')) + ((Mb) x (V''')) (1)
(1/2) x (Ma) x (V') x (V') = ((1/2) x (Ma) x (V'') x (V'')) + ((1/2) x (Mb) x (V''') x (V''')) + Q (3),
where Q is the heat, which is generated while the two blue balls move inside the two rough channels of the segment "s" in the "lower" modification.
------------------------------------------
15) It is evident that (a) the system of equations in item 13 and (b) the system of equations in item 14 cannot be true simultaneously.
------------------------------------------
16) And it directly follows from the previous item 15 that either (a) the law of conservation of linear momentum is not correct or (b) the law of conservation of mechanical energy is not correct or (c) both the law of conservation of linear momentum and the law of conservation of mechanical energy are not correct simultaneously.
------------------------------------------
17) Please refer, if necessary, to our first post of Sat Jul 28, 2018 12:41 and to the two related links
https://mypicxbg.files.wordpress.com/20 ... _01-12.pdf
https://mypicxbg.files.wordpress.com/20 ... s01-08.pdf
------------------------------------------
Note. And please read carefully my posts, and please DO NOT DISTORT my words. I have never mentioned that the blue balls and the blue rods are massless. Besides I have never mentioned rollers and rolling -- we consider solely and only sliding friction.
------------------------------------------
Looking forward to your answer.

[ME]

I agree with item 15.
Neither are correct.

[PeterAX]

Well, there is some advance.:) You agree with 15. Before that you have agreed with the assessment that the zigzags generate mechanical resistance, which is identical to friction and which does not generate heat. There is a hope.:)