Cookies-law

Cookies help us to bring you our services at overunity.com . If you use this website and our services you declare yourself okay with using cookies .More Infos here:
http://www.overunity.com/5553/privacy-policy/
If you do not agree with storing cookies, please leave this website now. Many thanks for your understanding.
Amazon Warehouse Deals ! Now even more Deep Discounts ! Check out these great prices on slightly used or just opened once only items.I always buy my gadgets via these great Warehouse deals ! Highly recommended ! Many thanks for supporting OverUnity.com this way.

User Menu

FireMatch

FireMatch

CCKnife

CCKnife

Poplamp

poplamp

CCTool

CCTool

LEDTVforSale

Magpi Magazine

Magpi Magazine Free Rasberry Pi Magazine

Battery Recondition

Battery Recondition

OverUnity Book

overunity principles book

Arduino

Ultracaps

YT Subscribe

Gravity Machines

Tesla-Ebook

Magnet Secrets

Lindemann Video

Navigation

Products

Statistics

  • *Total Members: 81835
  • *Latest: Hydrtec

  • *Total Posts: 488295
  • *Total Topics: 14366
  • *Online Today: 44
  • *Most Online: 103
(December 19, 2006, 11:27:19 PM)
  • *Users: 7
  • *Guests: 232
  • *Total: 239

Facebook

Author Topic: Waldens inertial drive is an optical illusion  (Read 10283 times)

Offline CuriousChris

  • Sr. Member
  • ****
  • Posts: 261
Waldens inertial drive is an optical illusion
« on: September 10, 2010, 11:57:32 AM »
Hi All I am new here.

Sorry my first post is a bit of a downer, I have recently decided I wanted to try and create a reactionless drive or inertial drive as its often called, in fact I prefer the latter.

Of course I did my due diligence and searched over many patents. I came accross a thread here talking about waldens inertial drive
http://www.overunity.com/index.php?topic=6387.5

I hadn't heard of that one so immediately took a look at the patent.
http://www.freepatentsonline.com/20080168862.pdf

At first I thought wow so ingenious, so simple, an offset axis creating an ellipse where the inactive load travels a shorter path in the same time as the active load. therefore it must travel slower than the active load and by that virtue the active load must have greater inertial value.

It wasn't until later that night lying in bed thinking about how to construct my own device that it hit me.

Wait a sec! The path the weights follow is a circle not the ellipse that it appears to follow. the offset axis makes it look like an ellipse, but if you ignore (cover the axis and arms with a white sheet of paper if you cant imagine it) you will see the weights travel an exactly circular path.
Therefore the centrifugal force is equal in all directions.
Therefore the device creates exactly nil thrust.

Sorry to be the bearer of bad news to anyone that was holding out hope that this invention may have been for real.

If the inventor claims that it passes the pendulum test then he must be thinking of another pendulum (it may make a great weight to hang on a pendulum and use to regulate a big clock) but it will not exert a force in only one direction. any claim that it produces thrust or unidirectional force is a total fallacy.

If you then say what about his other 'embodiments' then I am afraid they are all the same just a little more confusing to decipher.

Anyway back to my research...

CC


P.S. inertial thrusters are not antigravity devices any more than a rocket is an antigravity device.

Free Energy | searching for free energy and discussing free energy


Offline FredWalter

  • Full Member
  • ***
  • Posts: 121
Re: Waldens inertial drive is an optical illusion
« Reply #1 on: March 29, 2011, 03:17:18 PM »
Yes, the path that the weights follow is a circle.

However the offset axis results in constantly changing velocity and acceleration vectors, which results in a constantly changing force vector.

Do the math, eh?

The question is, does the force vector cancel out over one full revolution.


Offline CuriousChris

  • Sr. Member
  • ****
  • Posts: 261
Re: Waldens inertial drive is an optical illusion
« Reply #2 on: March 30, 2011, 10:56:19 AM »

The illusion is because when looking at it we automatically cut the circle at the pivot point. It looks like there are two arcs, one longer than the other. Therefore for the masses to travel over these two arcs one must go faster then the other.

This is NOT the case, the curvature of both arcs are of course identical.

It may look like the angular velocity changes but it doesn't. Because both masses are always in line they MUST both be traveling at the same velocity otherwise one would start to catch up to the other.

The masses don't have a changing velocity in respect to each other or the supporting structure.

Don't get me wrong I am convinced Inertial Drives can be done. I am close to building my own which will either prove or disprove my own theories. Hence why I looked at as many patents and ideas as I could find.

I do think it is important that we report when we believe a claim is in fact false and why it is. Otherwise people devote too much time and resources following a line of enquiry which can only fail.


Offline FredWalter

  • Full Member
  • ***
  • Posts: 121
Re: Waldens inertial drive is an optical illusion
« Reply #3 on: March 30, 2011, 04:18:25 PM »
It may look like the angular velocity changes but it doesn't.

Force/velocity/accelleration are based on the center of rotation *not* the center of the circle that is used to constrain the paths of the weights.

The center of rotation is offset from the center of the circle that is used to constrain the paths of the weights.

The angular velocity changes w.r.t. the center of rotation.

Do the math.


Offline CuriousChris

  • Sr. Member
  • ****
  • Posts: 261
Re: Waldens inertial drive is an optical illusion
« Reply #4 on: March 31, 2011, 11:19:16 AM »
I really screwed that up didn't I :(

I answered too quickly and couldn't remember much of my original reasoning.

It wasn't until I lay in bed and thought, "You Idiot!" (about myself)

So I decided to review it again. If it did work then we should see it in use already. We don't and the obvious answer is because it doesn't work.

So why doesn't it?

It didn't take long for me to work it out.

Firstly most inertial drives that use varying speeds of the mass or masses to provide the kinetic energy, use a weight propelled at speed to generate the 'main' force, then return the weight to the start point at a slower speed.

What they fail to realise is that the weight at slower speed may have less kinetic energy, but its acting for (much) longer therefore the total kinetic energy is exactly the same as the weight at higher speed for a shorter duration. the two vector forces thus cancel out.

The good news is, due to the off centre pivot point this is not happening here.

The bad news is, due to the off centre pivot point, during a portion of one full cycle you have two periods where both weights are on the same side of the true center of the circle. This means both weights are exerting force in the opposite direction of what is actually wanted.

My guess is that the summation of these vector forces will result in a net vector force of zero. See the attached image.

Please feel free to "do the math" and prove me wrong.

CC



Free Energy | searching for free energy and discussing free energy

Re: Waldens inertial drive is an optical illusion
« Reply #4 on: March 31, 2011, 11:19:16 AM »
Sponsored links:




Offline FredWalter

  • Full Member
  • ***
  • Posts: 121
Re: Waldens inertial drive is an optical illusion
« Reply #5 on: March 31, 2011, 02:32:03 PM »
Please feel free to "do the math" and prove me wrong.

I've attached a simulation that I did using Maple.

Here is the Maple code:

---attempt01.txt
restart:
with(plots):
setoptions(scaling=constrained):

printf("For the mass track, use a 2-dimensional circle, centered at the origin (0, 0), with radius='RADIUS'.\n");
printf("Let the mass arm be rotating at 'RPM' rotations per minute.\n");
printf("Put the center of rotation of the mass arms at (0, OFFSET).\n");
printf("Put one end of the rotating arm at point P on the circle.\n");
printf("Let 'L'=the length from (0, OFFSET) to P.\n");
printf("Let 'A' be the angle between lines (0, 0) to (0, OFFSET), and (0, OFFSET) to P.\n\n");

printf("Using the Law of Cosines for the triangle formed by the above points, we get the following equation:\n");
printf("\tRADIUS*RADIUS = OFFSET*OFFSET+L*L-2*OFFSET*L*cos(A)\n");
EQN01 := RADIUS*RADIUS = OFFSET*OFFSET+L*L-2*OFFSET*L*cos(A):
ANS_EQN01 := solve(EQN01, L):
L := A -> OFFSET*cos(A) + sqrt(RADIUS*RADIUS+OFFSET*OFFSET*(cos(A)*cos(A)-1)):

printf("Solving '%a' for 'L' gives us:\n\tL(A)=%a\nand\n\tL(A)=%a\n\n", EQN01, ANS_EQN01[1], ANS_EQN01[2]);

printf("Let us use the first solution for L:\n\tL(%a)=%a\n\n", A, L(A));

Px_A := A -> L(A)*sin(A):
Py_A := A -> OFFSET-L(A)*cos(A):

printf("Using the relationship between the sides of our triangle, and angle A, we get the position of point P as a function of angle A:\n\tPx_A(A)=%a\n\tPy_A(A)=%a\n\n", Px_A(A), Py_A(A));

A := t -> t*(2*Pi)/RPM:
Px := t -> simplify(Px_A(A(t))):
Py := t -> simplify(Py_A(A(t))):
Vx := t -> simplify(diff(Px(t), t)):
Vy := t -> simplify(diff(Py(t), t)):
Ax := t -> simplify(diff(Vx(t), t)):
Ay := t -> simplify(diff(Vy(t), t)):
Sx := t -> simplify(diff(Ax(t), t)):
Sy := t -> simplify(diff(Ay(t), t)):

printf("Angle A is a function of time:\n\tA(t)=%a\n\n", A(t));

printf("Position of point P as a function of time:\n\tPx(t)=%a\n\tPy(t)=%a\n\n", Px(t), Py(t));
printf("Velocity vector at of point P as a function of time:\n\tVx(t)=%a\n\tVy(t)=%a\n\n", Vx(t), Vy(t));
printf("Acceleration vector at of point P as a function of time:\n\tAx(t)=%a\n\tAy(t)=%a\n\n", Ax(t), Ay(t));
printf("Surge vector at of point P as a function of time:\n\tSx(t)=%a\n\tSy(t)=%a\n\n", Sx(t), Sy(t));

RADIUS := 5: # 10" diameter mass track
OFFSET := RADIUS/2: # arbitrary
RPM := 1:
Alpha := 0: # do one full pass around the mass track
Omega := 1:
Curve := plot([Px(t), Py(t), t=Alpha..Omega], style=line, color=blue, thickness=2):

NumFrames := 30:
t := i -> Alpha + i*(Omega-Alpha)/NumFrames:
VScale := 0.1: # arbitrary
AScale := 0.01: # arbitrary
SScale := 0.001: # arbitrary

Points := display(seq(pointplot([Px(t(i)), Py(t(i))], style=point, symbol=circle, symbolsize=17, color=black), i=0..NumFrames), insequence=true):
VVectors := display(seq(arrow([Px(t(i)), Py(t(i))], VScale*<eval(Vx(a), a=t(i)), eval(Vy(a), a=t(i))>, width=.02, color=red), i=0..NumFrames), insequence=true):
AVectors := display(seq(arrow([Px(t(i)), Py(t(i))], AScale*<eval(Ax(a), a=t(i)), eval(Ay(a), a=t(i))>, width=.02, color=red), i=0..NumFrames), insequence=true):
SVectors := display(seq(arrow([Px(t(i)), Py(t(i))], SScale*<eval(Sx(a), a=t(i)), eval(Sy(a), a=t(i))>, width=.02, color=red), i=0..NumFrames), insequence=true):

printf("Here are some graphs following the point P for one full revolution of the mass arm, using the following values:\n\tRADIUS=%f\n\rOFFSET=%f\n\tRPM=%f\n\n", RADIUS, OFFSET, RPM);

display(Curve, Points, title="Mass Path");
display(Curve, VVectors, AVectors, title="Velocity+Acceleration");
display(Curve, AVectors, SVectors, title="Acceleration+Surge");





Offline CuriousChris

  • Sr. Member
  • ****
  • Posts: 261
Re: Waldens inertial drive is an optical illusion
« Reply #6 on: March 31, 2011, 03:05:22 PM »
Thats really cool.

I wonder if I can use that tool to test my own theory.

One thing though. As far as I can tell. and this may be just because I have never read a maple script before and may be relying too much on the images but are you modeling just one mass. Is that correct?

The image that shows the mass motion (mass_path.gif) only displays one and the vectors in the other gifs only show one. Waldens patent relies on multiples of two masses. its the summation of these forces I contend equals a vector force of 0.

CC


Free Energy | searching for free energy and discussing free energy

Re: Waldens inertial drive is an optical illusion
« Reply #6 on: March 31, 2011, 03:05:22 PM »
Sponsored links:




Offline FredWalter

  • Full Member
  • ***
  • Posts: 121
Re: Waldens inertial drive is an optical illusion
« Reply #7 on: March 31, 2011, 11:39:16 PM »
I wonder if I can use that tool to test my own theory.

If you can represent everything in your theory using equations then you probably can use Maple to help test your theory.

Quote
As far as I can tell. and this may be just because I have never read a maple script before and may be relying too much on the images but are you modeling just one mass. Is that correct?

Yes. I did this last year sometime, and haven't had time to extend it to cover two masses.

Quote
The image that shows the mass motion (mass_path.gif) only displays one and the vectors in the other gifs only show one. Waldens patent relies on multiples of two masses. its the summation of these forces I contend equals a vector force of 0.

You can integrate the force equation over the circle for one mass.

If F=M*V*V/R (as I've found in some text books) then you get a non-zero result.

If F=MA then you get 0.

When I look at the values of A and V as the mass moves over one revolution of the mass track, A=V*V/R only when V is tangential to the mass track (IE. for my setup, when x = 0).

It looks to me that F=MA is the general equation for force, and F=M*V*V/R is the equation that applies only when the velocity vector is tangential to the curve upon which the mass is moving.

I still haven't found a derivation of F=M*V*V/R, and until I find a derviation that shows it is always valid, I'm going to have to wonder if it valid to apply it to the WIP device.

Here is the maple code that I used to get these results (append it to the maple code in attempt01.txt):

---

Amagnitude := t -> sqrt(eval(Ax(a), a=t)^2+eval(Ay(a), a=t)^2):
Vmagnitude := t -> sqrt(eval(Vx(a), a=t)^2+eval(Vy(a), a=t)^2):
printf("\n");
for i from 0 by 0.05 to 1 do
printf("Time=%f Amagnitude=%f Vmagnitude^2/RADIUS=%f\n",
i, evalf(Amagnitude(i)), evalf(Vmagnitude(i)^2/RADIUS));
end do;
printf("\n");

FAx := t -> simplify(eval(MASS*Ax(a), a=t)):
FAy := t -> simplify(eval(MASS*Ay(a), a=t)):
FAmagnitude := t -> simplify(sqrt(eval(FAx(a), a=t)^2+eval(FAy(a), a=t)^2)):
Total_FAx := simplify(int(FAx(t), t=0..1)):
Total_FAy := simplify(int(FAy(t), t=0..1)):
Total_FAmagnitude := simplify(sqrt(Total_FAx^2+Total_FAy^2)):

FVx := t -> simplify(eval(MASS*Vx(a)^2/RADIUS, a=t)):
FVy := t -> simplify(eval(MASS*Vy(a)^2/RADIUS, a=t)):
FVmagnitude := t -> simplify(sqrt(eval(FVx(a), a=t)^2+eval(FVy(a), a=t)^2)):
Total_FVx := simplify(int(FVx(t), t=0..1)):
Total_FVy := simplify(int(FVy(t), t=0..1)):
Total_FVmagnitude := simplify(sqrt(Total_FVx^2+Total_FVy^2)):

printf("integrate F=MA over one revolution and you get: %a\n",
evalf(Total_FAmagnitude));
printf("integrate F=MV^2/R over one revolution and you get: %a\n",
evalf(Total_FVmagnitude));

---

Here is the output from the above Maple code:

---

Time=0.000000 Amagnitude=444.132198 Vmagnitude^2/RADIUS=444.132198
Time=0.050000 Amagnitude=433.781166 Vmagnitude^2/RADIUS=433.132245
Time=0.100000 Amagnitude=402.908288 Vmagnitude^2/RADIUS=399.815957
Time=0.150000 Amagnitude=353.424767 Vmagnitude^2/RADIUS=344.643860
Time=0.200000 Amagnitude=291.761146 Vmagnitude^2/RADIUS=272.820308
Time=0.250000 Amagnitude=227.928750 Vmagnitude^2/RADIUS=197.392088
Time=0.300000 Amagnitude=169.374817 Vmagnitude^2/RADIUS=134.142398
Time=0.350000 Amagnitude=119.976251 Vmagnitude^2/RADIUS=90.909984
Time=0.400000 Amagnitude=82.432846 Vmagnitude^2/RADIUS=65.672556
Time=0.450000 Amagnitude=58.162465 Vmagnitude^2/RADIUS=53.106629
Time=0.500000 Amagnitude=49.348022 Vmagnitude^2/RADIUS=49.348022
Time=0.550000 Amagnitude=58.162465 Vmagnitude^2/RADIUS=53.106628
Time=0.600000 Amagnitude=82.432846 Vmagnitude^2/RADIUS=65.672556
Time=0.650000 Amagnitude=119.976251 Vmagnitude^2/RADIUS=90.909985
Time=0.700000 Amagnitude=169.374817 Vmagnitude^2/RADIUS=134.142398
Time=0.750000 Amagnitude=227.928750 Vmagnitude^2/RADIUS=197.392088
Time=0.800000 Amagnitude=291.761146 Vmagnitude^2/RADIUS=272.820308
Time=0.850000 Amagnitude=353.424766 Vmagnitude^2/RADIUS=344.643860
Time=0.900000 Amagnitude=402.908288 Vmagnitude^2/RADIUS=399.815957
Time=0.950000 Amagnitude=433.781166 Vmagnitude^2/RADIUS=433.132245
Time=1.000000 Amagnitude=444.132198 Vmagnitude^2/RADIUS=444.132198

integrate F=MA over one revolution and you get: 0.
integrate F=MV^2/R over one revolution and you get: 158.5148465*csgn(MASS)*MASS


 

Share this topic to your favourite Social and Bookmark site

Please SHARE this topic at: