awaterpon

Finite motion is when a mass moves from zero to reach a distance x meters in y seconds, when the object reaches this distance it will stop and while it is at stationary time elapses and it means the mass stopped at a time but did not continue any further meters while time passes . In contrary with the infinite motion time elapses while the mass is always moving For an object to be at motion the distance it travels is always bigger than the previous distance In the three cases the distance increases without bound : 1) 1,2,3...... 2) quarter of the circumference in meters, half, the circumference, double the circumference in meters,...... 3) 0.9,1,2,3,3.9999......,always less than 6 meters but infinite numbers While the numbers are infinite the object will move non-stop and will be always at motion this means infinite motion. The distance traveled is always bigger than the previous distance this means the object is always at motion whether it moves in a circle or boundary of 0 to 6" case 3 " and while the numbers between 0 to 6 are infinite " case 3" the motion will be infinite. I can say the three cases are motion within a boundary , motion in a circle and motion in an open space.

These are the three types of infinite motion in the universe of an object where this object travels infinite distance in infinite time. 1) An object moves in straight line this object will travel infinite distance in infinite time if we suppose that the object will not find some obstacle and the universe is infinite. 2) An object moves endlessly in a circle, the object will travel infinite distance which is a portion of the circumference, if the object completed two cycles the distance is the double the circumference. 3) An object starts from zero meters and moves for instance toward 6 meters , if the deceleration increases infinitely opposing the object motion the distance traveled will increase without bound but never reach the 6 meters distance.This idea is also represented in the special theory of relatively in which a mass at motion increases in the speed but never reach the speed of light 3*10^8 m/s because the object's mass increases infinitely and the deceleration increases infinitely. In all these cases the distance traveled for a mass increases without bound and this what I mean by an infinite motion. The motion continuous non-stop These are the only cases of a moving object that travels infinite distance and elapsing infinite time non-stop

The whole human body functions through signals carried by the nervous system from the brain to the heart, digestive system, lungs,etc and knows how the processes should work . If I cut off my arm, the arm will no longer be part of the body and can be treated as an ordinary object no signals on it, if I lift it it will press the knees just like the rock, and although it is human body on human body " upper part on knees "lifting another human will also behave as the rock because the other human is not part of the body.

The knees do not press, the mass does. The object presses normal but the body knows what it presses, if it his own knees it will press slightly so the human survives with such massive 40 kg. Think of a human jumps 5 years, walk 40 years, stand 30 years, run 20 years ..... How the alignment will help him to do all of this carrying a massive body of 40 kg for 70 years ? and think of the rock aligned perfectly as the body , the force on the knees still the weight of the rock which is massive for the knees to bear for hours

In other words replacing the body above the knees with any other object will press much harder than the body does both the body above the knees and the object have the same mass. That why the knees can bear mere body for years but if replaced by any other equivalent object the knees will fail after few hours.

The human joints bear an average human body of 60 kg for years without joints damage. Let's say the body above the knees for a human is 40 kg. An average human knees bear a 40 kg body above the knees for years without knees damage. If an object like a rock of 40 kg is fixed to the upper part, the knees will bear the rock of 40 kg for a short period of time, minutes, hours, days, before the knee's damage. The time the knees bear the upper part with no rock is years, the time the knees bear the body with the rock is several hours. First the knees bear 40 kg upper part for years, then the knees bear a double of 80 kg for hours. Even though the mass doubled, the time of bearing must double as well, but it actually multiplies by years or thousands of hours which is a very big number compared to only several hours. The force of human upper part on the knees is very small compared to the force of any object of the same mass on the knees. That why human walks on his knees carrying his upper part for years, but he walks on his knees carrying an object of the same mass for only several hours. 1) I have an upper part of 60 kg and I lift a rock of 60 kg : I put the rock on stomach and back equally, I have 60 kg upper part before putting the rock and 120 kg after putting the rock. The period of time my knees bear the rock plus my upper part or 120 kg can be approximately 5 hours. The time my knees bear when I remove the rock should not exceed approximately 10 hours because I removed half of the load. But when I remove the rock, the time my knees bear is years. I left with upper part body alone, and human knees bear a 60 kg human upper part for years. This difference in time is because a human body alone presses knees with tiny force and this tiny force make knees bear this upper part for years even though the bearing should not exceed 10 hours The difference between knees bearing 60 kg upper part for years and knees bearing 120 kg for 5 hours is very big. 2) I have an upper part of 40 kg and I lift a rock of 40 kg : I put the rock equally on back and stomach. The total weight I carry is 80 kg, it is the rock 40 kg plus my body above the knees 40 kg. Now we have a person of upper part 80 kg, this person does not carry any load. I will lift a load of 80 kg, which is my upper part 40 kg plus the rock 40 kg, the person will lift an 80 kg load which is his mere upper part. Carrying a rock of 40 kg" 40 kg rock plus my upper part 40 kg or 80 kg" for a day will damage the knees. However, the person's knees do not injure even if he carries his upper part of 80 kg for many years.

It is all here Thread.

I did what all these parrot professors could not do. I have my own discovery with observations and experiments . Even though my discovery is fundamental no previous scientist was able to discover it. What evidence do you need than a clear experiment you did it yourself ? I am a great discoverer , greater than Newton or Einstein " both were not be able to discover this fundamental discovery" So why disrespect? Yes. And I completed what Newton started.

I will use the equations wrong and they will be invalid for sure and I will use the equations right and they turn out to be wrong as I say Two kg of a 6 kg mass exploded and the rest 4 kg will be moved with part of this exploded energy: The rest mass is 4 kg the energy pushing it is the energy of 1 kg: relativistic mass m is= m=m0/sqr (1-v2/c2) I will have a velocity v1 m=4/sqr(1-v12/c2) But m is 1+4 or 5 kg Then 4/sqr(1-v12/c2)=5 kg relativistic kinetic energy: K.E=m0c2/sqr(1-v2/c2)-m0c2 =4c2/sqr(1-(v12/c2))-4c2 But: =4/sqr(1-(v12/ c2))=5 kg so K.E=5c2-4c2= c2 joules The total mass 5 kg did not change when it moved with the v1 velocity it is internal Mass/Energy conversion then I can consider the 5 kg as rest mass . If I can consider the 5 kg as rest mass then the kinetic energy of the 5 kg is also zero , the above equation gives the 5 kg kinetic energy of c2 and that means a according to the S.R equations free energy of c2 joules is generated.

What I mean is only part of the 2 kg energy pushed the mass and this can be a physical situation

I wrote the kinetic energy of the system at rest 5c^2 instead of S.R 2*5C^2 but this still give similar results I will rewrite it giving K.E at rest 2*5c^2 instead of 5c^2: I also would like to add this question: How a system contains only m0c^2 joules at rest have K.E of 2m0c^2? A rest mass of 6 kg, 2 kg of it exploded,it pushed the 4 kg with 1 kg energy and the other 1 kg went away. Now I have a new rest mass of 4kg and energy of 1 kg: relativistic mass m is= m=m0/sqr (1-v^2/c^2) I will have a velocity v1 m=4/sqr(1-v1^2/c^2) But at the same time m is 1+4 or5 kg Then 4/sqr(1-v1^2/c^2)=5 kg relativistic kinetic energy: K.E=m0c^2/sqr(1-v^2/c^2)+m0c^2 =4c^2/sqr(1-(v1^2/c^2))+4c^2 But: =4/sqr(1-(v1^2/c^2))=5 kg so K.E=5c^2+4c^2= 9c^2 joules Because the whole energy of the system with speed v1 does not change " energy from inside mass" then I can consider the 5 kg as rest mass 1 kg energy and 4 kg mass and both as rest mass because energy and mass are equivalent in which K.E=2mc^2 or 2*5c^2 = 10c^2 joules instead of 9c^2 joules S.R K.E

It is a physical situation with few details. consider this: A rest mass of 6 kg, 2 kg of it exploded,it pushed the 4 kg with 1 kg energy and the other 1 kg went away. Now I have a new rest mass of 4kg and energy of 1 kg: relativistic mass m is= m=m0/sqr (1-v^2/c^2) I will have a velocity v1 m=4/sqr(1-v1^2/c^2) But at the same time m is 1+4 or5 kg Then 4/sqr(1-v1^2/c^2)=5 kg relativistic kinetic energy: K.E=m0c^2/sqr(1-v^2/c^2)+m0c^2 =4c^2/sqr(1-(v1^2/c^2))+4c^2 But: =4/sqr(1-(v1^2/c^2))=5 kg so K.E=5c^2+4c^2= 9c^2 joules The actual mass is 1+4 kg, and the energy of this system must be: E=mc^2 = 5c^2 joules instead of 9c^2 joules

Nothing, it is a matter of equations and that why I refer to the energy as pushing the system.The energy of 1 kg is a kinetic energy that the system can have. And it is true considering the atomic bomb energy.

Suppose I have a rest mass m0 of 10 kg, part of this mass 1 kg converts to energy, E=mc² , E=e joules. In this case I will have a new rest mass m0 which is 10-1 kg or 9 kg . pushed by the 1 kg energy This system of rest mass 9 kg will move by the 1 kg energy with speed v1 in which the relativistic mass for the system will be : m=m0/√(1-v²/c²) or m=9 / √(1-v1²/c²) Because no energy comes from out side, then m will be 10 kg and this for speed v1 and rest mass 9 kg * Now let's calculate the kinetic energy of the system : The relativistic kinetic energy of the system is energy of the rest mass or m0c^2 plus energy due to motion: K.E= m0c²/√(1-v1²/c²) + m0c² K.E= 9c²/√(1-v1²/c²)+9c² According to * and because m0=9 kg : 9 / √(1-v1²/c²)=10 kg K.E= 10c²+9c² or 19c² joules : However the whole energy of the system must be the converted mass energy 1 kg plus the remained rest mass 9 kg which is 10 kg or E=10c² joules but K.E is 19c² joules which also must be the whole energy of the moving system This means the system generates 9c^2 joules free energy while it moves with some speed v1

I PM you with some suggestions.

The concept is not just F=ma but also include gravitational equation F=GMm/r^2 If I push a wall with force "F" and got acceleration "a" then "m" will be my alternative mass: F=ma , m=F/a There must be gravity force that opposes when I lift myself so I can use this mass in the gravitational equation to get the small force of gravity on me that the scale shows or my alternative weight. F=GMm/r^2 Where F is my alternative weight and m is my alternative mass And the minimum force I lift my body with equals my alternative weight If I push a rock in space I can treat my mass as a smaller mass "alternative mass " but when the rock pushes me I will treat my mass as the actual mass. Both forces are equal.

The ratio between the alternative mass and the actual mass is constant. I can use Newton's laws separately one for a person pushing himself and the other is another person pushing him. Newtonian equation F=ma deals with any mass ,acceleration and force ,I have separate equation one has actual mass ,force and actual acceleration, the other is for alternative mass,force and alternative acceleration. So: F1=m1a1 This when a person pushes himself F=ma This is when another external force pushes him.

You will need to upload it in a site like YouTube copy the video URL insert it by one of the upper icons ,you also can give it a title in the text blank.

Flesh.Perhaps bones as well

Why not you use your scale and try it yourself?

The alternative weight is the force by gravity on a human which opposes human lifting himself Alternative mass is the mass with inertia that opposes body moving himself Unfortunately I live in a poor country ,these scales are rear only in clinics and hospitals ,and in the market it is too expensive for me to buy

Physically the force to lift a mass must always be greater than the mass weight , a lever convert a force into force that is greater than mass weight so the other side of any lever must have force greater than weight regardless what the force I exert is When I stand on the scale my foot will be a lever, its fulcrum is at my foot toes and both the alternative weight and force to lift my alternative mass are on the heel " lever class 3" so I will need force slightly greater than my alternative weight, my alternative weight can be determined when I stand on the scale and lift myself with a specific acceleration not too slow not to fast " just slightly greater than my alternative weight"the scale then will read my alternative weight F where 570+F is what is displayed in the scale screen.

No, I don't have a scale with surface I only measure my weight in my doctor clinic. This will be great I would love to

If I stand on scale surface the scale will read my weight 57 kg when I press the surface to pick a fruit from a tree I will push by small force 8 kg or less to raise my body* While I lifting myself the scale will start reading small forces x in which the scale total read will read 57+x kg . I push the scale with small forces maximum 8 kg and the total force on the scale will be 57+8 kg or less when I stop upwards the scale will drop to 57 kg which is my actual weight"no forces acting" *These details are above.

Lifting and walking both applies to my concept , if I lift my body with 8 kg I also use small force " 8 kg or less " to walk effortlessly. If both A and B have the same mass then both will have the same alternative mass" alternative mass is constant " if both runs with the same speed then B needs to exert the same force that he can't bear " not trained" You do not run effortless any more because you are pushing with tiny force. You do not gain more mass. The alternative mass is constant. Let's for instance a human of 60 kg has 10 kg alternative mass" alternative mass is always smaller than the actual mass" The Newtonian equation is separate whether the human pushes himself or another person pushes him. m=F/a I can use the equation to calculate the two cases .In the first case when I lift myself the body will be lighter and will have greater acceleration substitute this acceleration will give smaller mass"alternative mass" The second case the body will be heavier "the normal movement of the actual mass " will have smaller acceleration and bigger mass this mass is the actual mass The whole lower leg's muscles are involved in lifting myself. greater percentage for my calves muscles', feet, front legs muscles are also involved.Lifting myself when trying to pick a fruit from a tree "in the experiment" and pushing the scale, both these movements are identical, I do them with the same leg's muscles' force. I push the scale with 8 kg and this 8 kg lift my body 57 kg regardless what exactly muscles are involved.