Realistic Relativy (Corrected)

[arkain101]

Let me ask you a question: If instead of doubling the initial velocity you _tripled_ it, how many times greater would the stopping/braking distance be? For example, if at the velocity 20 km/h the distance was 30 meters, how much would it be at 60 km/h?

 

^ Lets call the object 100kg and 20m/s = 30m.

 

so 20m/s = 30m to stop. which is relavent to 10m/s = 7.5m

 

40m/s = 120m (when you are driving twice as fast)

 

so tripple. 60m/s = 180m (when you are driving 3 times as fast)

 

This is not a collision incident as we know.

The object is only increasing its velocity by Distance/Time not velocity of potential impact energy. The potential impact velocity is different. If this object hits a heat treated titanium object weighing 1million kg's. It will reflect backwards likely the same speed it hit the object. Taking away all its kenetic for a brieft moment then giving it all back again in the opposite direction.

 

I am seeing it as, coservation of momentum is due to the relativity that each object hits eachother, if one has more mass it can look like it sends it back where it came from. but if all that was visible was two objects (nothing else in the universe) , you wouldnt be able to tell who hit who and who sent who flying.

 

Lets look at this from this point of view.

 

When you drop a 1kg ball from a height to give it 1m/s velocity when it hits a clay pad it sinks 12mm into the clay lets say.

Now lets drop the ball from a height that gives it half the velocity, 0.5m/s, it will give the object 4times less energy causing it sink 3mm into the clay.

Next, drop it from a height giving it 0.25m/s it should sink 0.75mm into the clay

drop it from a height giving the ball 0.0625m/s it will sink 0.1875mm into the clay.

Even if you dropped it from: a height to give it 0.0009766m/s it should sink 0.000061mm into the clay.

Eventually you reach a point where it should have an unmeasureable velocity because it is so slow. We should call this point 0m/s which = 0mm of clay penitration. This is the beginning, of which you have 0 kenetic energy.

eventually there is a point where you have a beginning. so we have an impossible paradox here twice the speed of zero is whatever you want to call it. But the energy is not 4times 0 for the moving object. No matter what you do it will always contain 4times the energy if you give twice the velocity. This is because they both hit eachother.

[arkain101]

Maybe someone with stronger math skills could work this out in a more comprehendible way for me.

:)

[ixa]

^ Lets call the object 100kg and 20m/s = 30m.

 

so 20m/s = 30m to stop. which is relavent to 10m/s = 7.5m

 

40m/s = 120m (when you are driving twice as fast)

 

so tripple. 60m/s = 180m (when you are driving 3 times as fast)

 

 

To put it shortly, that is not correct. The correct answer would be 270 meters.

 

However, I am not very eager elaborate or continue discussing this topic after having actually read all the previous posts in this thread. Earlier I thought the mistakes in the car/space ship example, to which I originally replied, were an accident on your part. But having read through the previous discussion I see that is not the case.

 

As many others have already pointed out, physics is both a theoretical and experimental science. The theory part relies heavily on mathematics (and not only basic calculus, but also on some finer branches of mathematics), and physics cannot be fully understood without it and its logic. On the other hand any physical theory that cannot be tested experimentally is not a real theory at all, and has no significance outside the lunch break discussions.

 

Your "theory of realistic relativity", as you call it, has some major shortcomings. The logic in it is flawed, elements of actual physics have been taken out of context and used erroneously. These have already been pointed out by other persons in this forum, and it is needless for me to go into details. Your theory can, however, be tested experimentally, as you have pointed out. But the truth is that any test you could perform would only prove your theory wrong.

 

My advice would be to gather more skills and knowledge in basic mathematics and classical mechanics and after that think things through once more.

[arkain101]

yah. this post was a mess.. and I appologize for it.

 

The one titled "New Relativity" is much clearer. forget this post for now.

[CraigD]

Several questions came up in this thread concerning the distance required to stop a car traveling a different speed. Here’re the basic mechanical equations needed to answer them:

 

Force = Mass * Acceleration. This is a fundamental mechanical definition, simply defining what the terms mean. Simple algebraic rearrangement of it gives A = M / F. With cars accelerated by the force of tires against pavement, we can reasonably ignore M and F, and simply assume A is a constant, so we won’t worry further about it, just writing the constant A in the following equations. This is not very realistic for increases in velocity, since most cars are unable to sustain a constant acceleration, nor is it realistic for velocities (> 25 m/s), where air drag begins to dominate, but in this example, these can be ignored. It is realistic for decreases in velocity, because car brakes can easily sustain a constant acceleration as long as they do not overheat.

 

V = A * Time. Another fundamental definition. Often given as V = A * T + V0, where V0 is initial velocity. In this example, we can assume V0 = 0, and ignore it.

 

D = Vconstant * T. A fundamental definition. Not useful in this example, thought, because it applies only for motion with constant velocity. This one also has a form with an initial term D = V*T+ D0.

 

D = (A/2) * T^2. This is the first equation so far that’s not a fundamental definition. Deriving it actually requires a very simple use of the Calculus, thought one can confirm it approximately by repeatedly calculating D from the equations V=A*T+V0 and D=V*T+D0, and comparing results.

 

Now we have the mechanical equations to answer questions like

if at the velocity 20 km/h the distance was 30 meters, how much would it be at 60 km/h?

We’ve been given a specific V and D, 20 km/h (20 km/h = 20 km/h*(1000 m/1 km)*(3600 s/1 hr) = about 5.56 m/s) and 30 m, allowing us to calculate A, as follows:

V=A*T

D= (A/2)*T^2 = (A/2)*(V/A)^2 = V^2/(2*A)

A=V^2/(2*D) = about ((5.56 m/s)^2/(2*30) = about 0.514 m/s^2. This can be use to calculate D for a velocity of 60 m/s (about 16.7 m/s):

D=16.7^2/1.02 = about 270 m.

 

We can avoid the approximations by picking our units to match the given values 20 km/h = 1 speed unit, 30 m = 1 distance unit.

Then A= 1^2/(2*1) = (1/2), and

D= 3^2/(2*(1/2)) = 9 distance units. Converting back into meters,9 distance units *(30 m/1 distance unit) = 270 m.

 

Put simply, the equation D=V^2/(2*A) states that a car’s stopping distance is proportional to the square of its velocity. 2 times speed give 4 times distance, 3 times, 9, 4 time, 16, etc.

 

This sort of example is found in nearly all introductory Physics textbooks, and well known by most people here at scienceforums (ixa, for example, gave the correct answer to his question shortly after posting it).

[ixa]

Thank you for the excellent post, CraigD. I realize I should've done the same in my previous reply to this topic, but I suppose I was too shocked and disheartened by the previous discussions to do that. :) But it's good that someone is willing put that much effort into explaining things.

[arkain101]

^^

Thanks for the math.. didnt notice it there for awhile. I am not the best at the math but I still understand how it works.

 

 

 

I will try to sum up my theorization in one example, and regardless of what you know of math or what math says, this is a fact of reality.

 

If you drop a ball into clay from height A and it sinks 10mm of its height into the clay. Then raise the ball to hieght B which gives it double the velocity on impact (so double the hieght from impact) the ball will fall and sink 4 times further, 40mm, than it did when dropped from height A.

 

Regardless of knowing math this proves that doubling a velocity of an object causes an impact energy of 2 times (the double speed).

 

In effect,

the mass either increases of the moving object

or

the law of "Every action has an equal and opposite reaction" can really say that objects hit back just as hard as they are hit, saying, they both travel together at the same speed (but only one can viewed as having the total velocity), and the observing reference frame chooses which is to be the one with energy.

 

It might help to use different tems. But something is going on here regardless of what math equations say. Reality says that every interacting set of objects has every object in the system combined as the energy and the total energy can only be applied to one object (hence the 1/2 of Ke total). If you had a ball with a machine inside it that spun a wheel when it impacted, the total mass would still be put to use on the impact but more energy should be in the system than just the moving ball.