421
New Theories / T-V not constant with that device ?
« on: 24/05/2016 17:07:50 »
Look the device at the last message post please.
This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.
I am trying to understand you but sorry your wording is not helping,
I think you are saying , imagine a vacuum packed full of positive ion's, the positive ion's repel positive ions's in the ''chamber''. In the centre of this chamber is an odd shape that is negative and the particles apply a torque on the odd shape making it move?
If that is the case, what do you suggest a negative only mass is? surely the negative mass absorbs the energy of the positive ion,s creating equilibrium of the entirety of the volume?
Positive ion's repelling positive ion's causes expansion not centripetal acceleration/rotation. Where is your rotational velocity coming from?
I think the problem here is that you are setting up a thought experiment with parameters that can not exist in any natural system and attempting to apply the laws of nature to it.
What you are describing can not exist in this Universe so can not be analyzed by this Universe's laws of nature.
Either that or you have to explain it better.
In a disk,Quote
do you mean on a disk? if not and you do mean in a disk , do you mean a flat circle formation?
Yes, a flat circle (2 dimensions), it's easier than a disk. Outside the circle there is no particle. Inside the circle there are particles. Inside the object there is no particle.
QuoteI place N theoretical particles.I presume you mean a random number of particles?
Yes, a random number, for example I can place 1000 in a circle with a radius of 10 cmQuoteThese particles have the same sign,
Public toilets? This way up? or do you mean they all have equal polarity signs of positive?
Equal polarity signs (positive for example)Quoteso they repulse themselves with the law 1/d² with 'd' the distance.
They would repel themselves according to magnitude of the positive charge, I do not know where you are getting 1/d² from.
I use the law of electrostatic for example 1/d² or something like k/d²QuoteThere is no particle outside the disk.
But they may repel each other out of the disk, so they may be particles outside of the disk.
No, there is no particle outside, it's a theoretical problemQuoteIf I place in the disk an asymmetrical object in rotation in the center of the disk, why the object don't accelerate more and more in one direction and the particles accelerate in the other direction ? The object has no particle inside. For me the asymmetric shape will give a torque in one direction to the object and the particles must receive the torque in the other direction.
I got lost here in understanding you .
If the object is asymmetrical and the friction on the walls are not the same, why the torque on the object would be at 0 ? The walls of the object don't have an electrostatic charge so a lot of particles will be in contact with the walls of the object and push the object in a lot of directions. With the friction and the shape the torque is changed so for me the object would have a net torque in one direction and the charges will have the same torque in the other direction. But maybe someone had done the calculations and prove there is no torque ?
from math import *
N=2000
c1=0 #number of points of the torus1 inside the container1 or the surface if divised by N²
c2=0 #number of points of the torus2 inside the container2 or the surface if divised by N²
c3=0 #number of points of the hatching area or the surface if divised by N²
couple1=0.0 #clockwise torque on the torus1 (from springs) IF all part is attracted by the green line
couple2=0.0 #clockwise torque on the torus2 (from springs) IF all part is attracted by the red line
couple3=0.0 #counterclockwise torque on the torus1 or the torus2 due to the presence of the hatching area
Ra=7.0 #inner radius of the tore
Rb=8.72 #outer radius of the tore
dx=Ra+Rb #the center of the torus1 is (0,0), the center of the torus2 is (dx,dy)
dy=4.92
for i in range (0,N) :
for j in range (0,N):
x=i*(Ra+Rb)/N
y=j*dy/N
if x**2+y**2>Ra**2 and x**2+y**2<Rb**2 and x>Ra and x<Rb and y>0 and y<dy:
c1+=1
couple1+=y*x
if (x-dx)**2+(y-dy)**2>Ra**2 and (x-dx)**2+(y-dy)**2<Rb**2 and (x-dx)<-Ra and (x-dx)>-Rb and (y-dy)<0 and (y-dy)>-dy:
c2+=1
couple2+=abs(y-dy)*abs(x-dx)
if x**2+y**2>Ra**2 and x**2+y**2<Rb**2 and x>Ra and x<Rb and y>0 and y<dy and (x-dx)**2+(y-dy)**2>Ra**2 and (x-dx)**2+(y-dy)**2<Rb**2 and (x-dx)<-Ra and (x-dx)>-Rb and (y-dy)<0 and (y-dy)>-dy:
c3+=1
couple3+=y*x
print c1
print c2
print c3
print couple1/N**2
print couple2/N**2
print couple3/N**2
Frankly I'd have to say that you're lying about being an electrical engineer. No such engineer is as ignorant about electrodynamics as you are.
Because angular momentum is defined as rxg where g is momentum density which is integrated over the field. If the field is too far from the center then its moving faster than the speed of light and it then becomes infinite. You should learn relativistic electrodynamics before attempting to do the things you're doing.
It's meaningless to say that a particle rotates because by definition a particle is a point object and a point object has nothing which can be thought of as rotating. Even in the case of non-point particles like hadrons they can't be thought of as rotating because of the quantum mechanical nature of such entities doesn't permit such a description.
If a field that extends to infinity is rotating then it will have an infinite angular velocity.Why, could you explain please ?