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.

11 hours ago, swansont said:

Provide evidence of this claim. Not assertion. Evidence.

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.

1 hour ago, swansont said:

How would the knee know the nature of the mass providing this weight?

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.

1 hour ago, exchemist said:

No. Read the post by @TheVat. The difference is that a 40kg rock is not aligned with the skeleton to load it in the same way as body weight would. Furthermore it is dead weight, unlike body weight, which is a mass of bone, fat and muscle that moves dynamically in response to the action of walking. 

Both things make a huge difference to how the knees and other joints are loaded, during the action of walking or other movements.  

(It is noteworthy that people used to carrying heavy loads in many parts of the world choose to place them on their heads. This will be because having the extra load taken by the spine is the best way to avoid off-centre loading of the joints.)  

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

2 hours ago, swansont said:

Beyond that, I have no idea what you are trying to say. Clarification is needed.

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. 

3 minutes ago, exchemist said:

That’s really interesting. What experiments did you do and what did you discover? Perhaps you should start a new thread, though, since it would be a new topic.

It is all here Thread.

2 hours ago, Ghideon said:

A friendly reminder: You are posting on a forum where plenty of members are experts; scientists, teachers, engineers and more.
Imagine what you would be able to learn if you posted questions instead of incorrect claims. You would get advanced answers, good references, reading advices, links to free online courses etc. You seem interested in physics and capable of attempting basic mathematics. The predictive power of the current models of mainstream physics is quite good, why not try to find out how those models work before attempting to change them? 

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?

2 hours ago, StringJunky said:

He's probably read about Galileo...

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.

1 hour ago, swansont said:

What posseses this 1 kg’s worth of energy that “went away”? Energy doesn’t go away. It is conserved.

Energy is a property, not a substance. 

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

27 minutes ago, awaterpon said:

 

2 hours ago, exchemist said:

You are still using a wrong sign in the formula for relativistic k.e.

In your formula, if you set v to zero, you get a kinetic energy of 2mc², which is an obvious nonsense. This has already been pointed out to you. If you fix that, you have a chance of making sense, at least. 

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

35 minutes ago, Ghideon said:

If you use the equations to an unphysical situation, for instance a system that breaks conservation of momentum is not conserved*, the equations does not predict the behaviour of the system. The equations loose their predictive power; conclusions you draw from using the equations outside of their scope of applicability, will be incorrect. 

 

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 

5 hours ago, swansont said:

Energy doesn’t push things.

What is happening that results in this motion, and conserves both energy and momentum?

 

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

On 4/14/2021 at 9:23 PM, studiot said:

+1

Thanks for the response I was hoping someone would try it as I have never put anything on youtube before.
It did say something about private and then something about waiting a day whilst it 'processed' when I asked to make it public. That dfay has now passed.

But I really don't know what I am doing so welcome all the help I can get.

I PM you with some suggestions.

On 4/13/2021 at 12:54 PM, swansont said:

"Using Newton's Laws" and "using Newton's laws correctly" are two different claims.  There is no "alternative mass" and "alternative acceleration" This is just an excuse to do the analysis incorrectly.

 

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
 

On 4/13/2021 at 1:32 PM, Ghideon said:

Apply* Newton's laws F=ma (2nd law) and when an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A (3rd law). You will find that the mass is always m, or in other words, that the ratio m/m1=1. Your definition will result in F1=F, m1=m and a1=a or an unphysical situation according to Newton.

I may get some time to post an example later if you wish.

*) correctly applying them that is.

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.

On 3/26/2021 at 7:53 PM, Prof Reza Sanaye said:

 

As an almost direct result , intelligent people like awaterpon come to fall in doubt as about the feasibility of : 

 

 

According to classical mechanics for a force to  lift a mass it should be slightly greater than its weight .

My hypothesis  is that a human body can lift itself  by a force far less than its weight .

It is obvious phenomenon that when lifting an object  of 60 kg up , it would be extremely hard than lifting one's body " 60 kg" .while standing.

This applied to many phenomenon  .A body will seem to have inertia far less than its actual mass inertia , moving and walking effortlessly , standing effortlessly , lifting one's body parts easily.

In this special case the Newtonian equations doesn't apply , however we could measure the ratio between the force lifting a body and the force lifting an object both body and the object have the same mass.

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.

23 hours ago, studiot said:

Ok I have enlisted the aid of a small person with a mobile phone and obtained a short video of what happens on my bathroom scale.

It is 6.2MB and a couple of seconds long, in MP4 format.

So I am seeking some help posting it her for all to see as I have never posted a video here before.

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.

6 minutes ago, Ghideon said:

Let's try something else. What objects can replace the word "human" in your definition and still give exactly the same results as you claim? Can "human" be replaced by "dog", "humanoid robot" or something else?  

Flesh.Perhaps bones as well

1 hour ago, Bufofrog said:

Sorry to hear that.

Why not you use your scale and try it yourself?

On 3/22/2021 at 11:23 AM, Ghideon said:

I would like to see a rigorous definition of "alternative weight" "alternative mass". 

 

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

On 3/22/2021 at 2:07 PM, studiot said:

You did not respond.

Have you repeated the experiment for yourself ?

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

On 3/19/2021 at 4:13 PM, studiot said:

When you jump or rise slowly you are configuring your body as a see-saw or medieval catapault so that you can get necessary leverage to multiply the force on one side of the fulcrum.

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.

1 minute ago, studiot said:

Have you ever tried this ?

No, I don't have a scale with surface I only measure my weight in my doctor clinic.

8 minutes ago, studiot said:

Would you like me to post some photographs of what happens when I stand on my domestic bathroom scales and press up to stand on tiptoe ?

 

This will be great I would love to

1 hour ago, swansont said:

I'm not asking for your explanation. I'm asking how you are actually doing this measurement; i.e. I want an explanation as if I were going to try and replicate it myself. You stand on the scale and it reads 57 kg. When you reach up like you're picking fruit it drops to 8 kg?

 

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.

11 hours ago, Bufofrog said:

You screwed up your experiment, if you could only lift 8 kg with your calves you wouldn't be able to walk.  I can easily calf press 120 kg.

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.
 

12 hours ago, Ghideon said:

17 hours ago, awaterpon said:

Human body effortless walking, running  jumping, standing, dancing and other movements by human force on its own mass.

 

12 hours ago, Ghideon said:

Let's try another point: "effortless" walking is a psychological effect, not physics, in this case. An experience of "effortless" walking is not an indicator that mass magically is reduced. An example of your flawed logic: Assume a well trained human A runs "effortlessly" at the same speed as a not so well trained human B. A and B has the same mass. The fact that B struggles to keep up with A does not mean that B have some unspecified "alternative mass", A and B have the same mass. B's struggle is more likely due to being less fit than A. 

 

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"
 

12 hours ago, Ghideon said:

Another example: After 10km of running I do not run effortless anymore. How much "alternative weight" have I gained according to your idea?

 

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

11 hours ago, swansont said:

How do you measure just your calves? 

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.