[guest39538]

Explain why you think this simple piece of physics would alter any travelling faster?

https://en.wikipedia.org/wiki/Special_relativity#Composition_of_velocities

I can understand that, however I stated the relative speeds so I know from my stated speeds of 0.5c and 0.5c that it would be double the force.
I did not state 0.5c and 0.75c did I?

[Kryptid]

I can understand that, however I stated the relative speeds so I know from my stated speeds of 0.5c and 0.5c that is would be double the force.
I did not state 0.5c and 0.75c did I?

Even without taking relativity into consideration, you don't get double the force by colliding two objects travelling at the same speed. See what Mythbusters demonstrated about this:

[guest39538]

I can understand that, however I stated the relative speeds so I know from my stated speeds of 0.5c and 0.5c that is would be double the force.
I did not state 0.5c and 0.75c did I?

Even without taking relativity into consideration, you don't get double the force by colliding two objects travelling at the same speed. See what Mythbusters demonstrated about this:

OK, I have watched your video link, the test is flawed and they calculated incorrectly.   

At the end it is 1x + 1x  = 2x,   They need to only have one piece of clay in the final test on one arm only.

They are not calculating F / 2

If two cars are travelling at 50 mph head on, neither car knows who is moving, either car measure the other car travelling at 100 mph. 

[Kryptid]

OK, I have watched your video link, the test is flawed and they calculated incorrectly.   

At the end it is 1x + 1x  = 2x,   They need to only have one piece of clay in the final test on one arm only.

They are not calculating F / 2

I'm not sure I understand the situation you are describing. You are talking about two particles moving at the same speed (relative to an outside observer) and then colliding head-on, right? If so, the two pendulum approach is exactly the same as that. The one pendulum method would be analogous to a single particle hitting a wall.

[guest39538]

OK, I have watched your video link, the test is flawed and they calculated incorrectly.   

At the end it is 1x + 1x  = 2x,   They need to only have one piece of clay in the final test on one arm only.

They are not calculating F / 2

I'm not sure I understand the situation you are describing. You are talking about two particles moving at the same speed (relative to an outside observer) and then colliding head-on, right? If so, the two pendulum approach is exactly the same as that. The one pendulum method would be analogous to a single particle hitting a wall.

Because they used two pieces of clay at the end, the squash of the clays was shared between the two clays, both of the squashes added together would equal the 1 smash of more swing , they used a single clay for that ,

[Kryptid]

Because they used two pieces of clay at the end, the squash of the clays was shared between the two clays, both of the squashes added together would equal the 1 smash of more swing , they used a single clay for that ,

I suppose then the question is, "what force are you measuring?" Are you talking about the force experienced by a hypothetical target right in the middle of the collision, or the force experienced by each particle? Another reason you can't add force up linearly like that is because doubling the speed of an object doesn't double its kinetic energy, it actually quadruples it (at subrelativistic speeds). The kinetic energy equation is exponential, not linear.

[guest39538]

Because they used two pieces of clay at the end, the squash of the clays was shared between the two clays, both of the squashes added together would equal the 1 smash of more swing , they used a single clay for that ,

I suppose then the question is, "what force are you measuring?" Are you talking about the force experienced by a hypothetical target right in the middle of the collision, or the force experienced by each particle? Another reason you can't add force up linearly like that is because doubling the speed of an object doesn't double its kinetic energy, it actually quadruples it. The kinetic energy equation is exponential, not linear.

I am measuring the force of impact by the observation.

If the clay compresses example 1cm from 1x and 2cm from 2x against the metal block

Then when both arms are dropped with the clay on each arm, from 1x, each clay is compressed 1cm , a total of 2cm

[Kryptid]

I am measuring the force of impact by the observation.

If the clay compresses example 1cm from 1x and 2cm from 2x against the metal block

Then when both arms are dropped with the clay on each arm, from 1x, each clay is compressed 1cm , a total of 2cm

I'm doubtful that the 2x speed pendulum compressed the clay exactly twice as much as the 1x speed pendulum, considering that the pendulum would have had four times as much kinetic energy when moving twice as fast. Clay doesn't compress linearly as force increases, so it's not a great way to get an exact measurement unless you know exactly how it behaves under compression.

[guest39538]

I am measuring the force of impact by the observation.

If the clay compresses example 1cm from 1x and 2cm from 2x against the metal block

Then when both arms are dropped with the clay on each arm, from 1x, each clay is compressed 1cm , a total of 2cm

I'm doubtful that the 2x speed pendulum compressed the clay exactly twice as much as the 1x speed pendulum, considering that the pendulum would have had four times as much kinetic energy when moving twice as fast. Clay doesn't compress linearly as force increases, so it's not a great way to get an exact measurement unless you know exactly how it behaves under compression.

True, it is not an ideal test , the steel block having more density etc.    I would not be sure about kinetic energy, I do not think that is what is meant to be exact.

[Kryptid]

For the sake of continued consideration, this is what happened when the Mythbusters tested it with actual cars:

[guest39538]

For the sake of continued consideration, this is what happened when the Mythbusters tested it with actual cars:

The results I expected to see.    There is no contradiction , there is twice the damage.   

100mph (F)  / 2  = damage

The fans are wrong not right.   

Putting it a different way ,  you see one car hit the wall at 50 mph, you see one car of damage

You see two cars head on at 50 mph, you see twice the damage of the 1 car  because there is two cars with equal damage tot he one car,

2 is not equal to 1

[Kryptid]

For the sake of continued consideration, this is what happened when the Mythbusters tested it with actual cars:

The results I expected to see.    There is no contradiction , there is twice the damage.   

100mph (F)  / 2  = damage

The fans are wrong not right.   

Which is still less than that of a crash into a wall at 100 mph, which would be four times the damage (since it's four times the kinetic energy, not two times).

[guest39538]

For the sake of continued consideration, this is what happened when the Mythbusters tested it with actual cars:

The results I expected to see.    There is no contradiction , there is twice the damage.   

100mph (F)  / 2  = damage

The fans are wrong not right.   

Which is still less than that of a crash into a wall at 100 mph, which would be four times the damage (since it's four times the kinetic energy, not two times).

Lol I dont think you understand, the car crashing into the wall's damage at 100 mph is the same as the two cars crashing at 50 mph added together.

[guest39538]

I will do the math for you using 3 springs

A) under 100 lb pressure

B) under 50 lb pressure

c) under 50 lb pressure

b+c = a

c ≠ a

b ≠ a

[Kryptid]

Lol I dont think you understand, the car crashing into the wall's damage at 100 mph is the same as the two cars crashing at 50 mph added together.

You realize that you can't say, "it's twice as short now therefore it experienced twice the force", right? The more a material is compressed, the more it resists compression and the harder it becomes to compress it further. It's not linear. Doubling the force equals less than double the compression.

I will do the math for you using 3 springs

A) under 100 lb pressure

B) under 50 lb pressure

c) under 50 lb pressure

b+c = a

That's true, but impact force is not linear. The car travelling at 100 miles per hour has four times the impact force as either of the cars travelling at 50 miles per hour. Here is how you do the actual math for a car crash: https://www.engineeringtoolbox.com/impact-force-d_1780.html

[guest39538]

You realize that you can't say, "it's twice as short now therefore it experienced twice the force", right?

I was not been exact, just general chit chat.  I understand density plays a role. In general I was considering an equality of substances where I presume it the compression of material would be equal .

[guest39538]

That's true, but impact force is not linear. The car travelling at 100 miles per hour has four times the impact force as either of the cars travelling at 50 miles per hour. Here is how you do the actual math for a car crash: https://www.engineeringtoolbox.com/impact-force-d_1780.html

Why 4 times the force, that does not make much sense unless the momentum adds force?  second thought I think I see your point but I also think there is a equality in there somewhere. 
Meaning two identical masses would have the same qualities so equally compress ?

[guest39538]

F = (m1 + m2)  * v ²   = 4 lol just playing

F=mp²   looks good lol.

Shouldn't 1/2 mv² be something different than v, v being speed and direction, how do you square speed and direction?

[Colin2B]

The correct answer is yes.
And, in spite of what Colin2B says, there is not scientific theory which lets you get away from that, because it's aan empirically observed fact.

I think you may have misunderstood what i said:

You need to caveat this statement with “in the N Field new theory an electron cannot exist outside the atom.

The reason is that in current physics the electron has been observed outside the atom.

I didn’t say it was a scientific theory

[guest39538]

I didn’t say it was a scientific theory

Shucks......

So let us discuss the independence of an electron,  are you saying this electron has a volume?

added- I have the answer that allows my theory to continue. 

For the purpose of discussion I ask for you to consider the following definition, an individual free electron is an electron shell with a hollow space nucleus. It's inability to complete its annihilation into a n-wave is because of the n-field enclosure it occupies.  The likewise electron half of the n-field sufficiently enclosing the electron within itself allowing the electron to retain form.