Traveler and the Strange Tale of the Confounded Experimental Design

[D H]

Huge, because it's not the product, it's the difference divided by the small distance.

 

A=(L-S)/R2

 

Lets say the proton is 1

Lets say the electron is .0000001

Lets say the distance is .000000001

 

(1-.0000001)/.000000001=999,999,900

 

Units, please. One of the signs of a crackpot is someone who doesn't know a m/s^2 from a kg/m.

 

 

The reason you lose is that science has hundreds of years of observation that confirms their models of gravitation. You, on the other hand, have nothing to offer.

 

You claim to have performed some experiments. Please explain your experimental setup, how you made measurements, how your experimental setup and procedures addressed bias and error, how you analyzed your data, and the results of your experiment. Am I being mean? No. The scientists who have performed real experiments have had to endure much harder questioning.

 

You claim to have a model of gravitation, A=(L-S)/R2. This equation doesn't even have the right units. As it stands it is invalid. The units problem can of course be solved by adding a proportionality constant. Suppose you do just that: A=k*(L-S)/R. Some questions need to be addressed at this point: What is the basis for this model? Is there anything behind it, or did you just make it up out of thin air? What tests can be be made that distinguishes your "theory" from others?

 

With this new, improved equation you should be able to estimate the masses of Jupiter and Saturn from the orbits of their relatively tiny natural satellites; the masses of Mercury, Venus, Earth, Mars from the incredibly tiny artificial satellites in orbit around them; and the mass of the Sun from the planets. Your new, improved theory should agree with things outside the solar system as well, things such as 61 Cygnii. Finally, your estimated values must be consistent with one another.

 

The problem with your "theory" is that that won't happen.

[traveler]

Mass of proton: 1.67x10^-27

Mass of electron: 9.11x10^-31

Radius of hydrogen atom (separation distance): 52.9177x10⁻¹²

 

They're all in si so your answer should be in m/s/s as it's an acceleration, and then we can find a force from that using F=ma, or would you disagree with that?

 

Can you do the math for me to find an "A value?" My calculator won't hold that many digits. I'm only using the windows calculator and it barely made it using my other example.

[D H]

Your calculator does have scientific notation, doesn't it?

[Klaynos]

so...

 

[math] A = \frac {1.67 \times 10^{-27} - 9.11 \times 10^{-31}} { 52.9 \times 10^{-12} }[/math]

 

A=3.155 x 10^-17 [some unit]

 

So now I think we should work out the acceleration due to their charge... And compare...

Edited October 5, 2008 by Klaynos

[traveler]

Your calculator does have scientific notation, doesn't it?

 

I have poor math skills. It took me a long time to just figure out how to get to where I am. I already explained that earlier, that I am putting my idea out there so that maybe someone with good math skills and a good understanding of science can help me.

 

If you don't want to help me then that is as far as I can go. Sorry.

[Klaynos]

I have poor math skills

 

How did you derive your equation then?

 

Charge of proton: 1.6 * 10 ^{-19 C}

^{Charge of electron: -1.6 * 10 -19 C}

^{Radius as above...}

 

^{Using Coulomb's law we find:}

 

^{F=–8.2x10-8}

 

^{So what force answer does yours give? As we really need to compare the forces here....}

[traveler]

How did you derive your equation then?

 

I found the net mass with L-S and divided it by the distance. That gave me an A value.

 

The A value is then multiplied by the constant derived from a known test result of an actual distance and time. That gave me an acceleration according to the A value.

 

I then used that acceleration and found the time of impact from a known distance.

[Klaynos]

I found the net mass with L-S and divided it by the distance. That gave me an A value.

 

The A value is then multiplied by the constant derived from a known test result of an actual distance and time. That gave me an acceleration according to the A value.

 

I then used that acceleration and found the time of impact from a known distance.

 

OK, so above I calcuated an A value, how do I make that into a proper force?

[traveler]

OK, so above I calcuated an A value, how do I make that into a proper force?

 

Well, as in my initial example, if an object is dropped from 16.087 feet, and it hits the ground in EXACTLY one second, the acceleration is 32.174 ft/sec^2.

 

The constant for that example of the Earth would then be:

 

.000000000000000000000086636392822469954136118658002141 ft/sec^2

[D H]

I found the net mass with L-S and divided it by the distance. That gave me an A value.

In other words, it is ad hoc. It also disagrees with some very important observations:

1. We live, as far as we can tell, in a three dimensional universe. Newton's law of gravitation has gravitational force between two objects inversely proportional to the square of the distance between the objects. This squares very well with the three dimensional nature of space. The surface of a sphere is proportional to the square of the radius of the sphere. Naively, the gravitational gravitation force that emanates from some object should decrease as the surface area of spheres centered on the object increases.
    
    
   
2. The planets have nearly elliptic orbits about the Sun. With the exception of Mercury, Newton's law of gravitation accounts for the deviation from true elliptical orbits by virtue of inter-planetary gravitational interactions. Suppose gravitation somehow had something other than an inverse square relationship with respect to distance: [math]F=kr^{\alpha}[/math]. Here, [math]k[/math] is a proportionality constant and [math]\alpha[/math] is the power that relates distance to force. There are only two powers that yield closed elliptical orbits: [math]\alpha=1[/math] (e.g., a spring) and [math]\alpha=-2[/math] (e.g., gravitation). Your 1/r relationship will not yield closed elliptical orbits.
    
    
   
3. Newton's third law. Newton's law of gravitation, in which the relative acceleration between two objects is proportional to the sum of the masses of the objects, is consistent with Newton's third law. Yours is not.

[Klaynos]

So your idea currently makes no predictions? Investigate how the value of G is found and you should be able to do the same for your idea...

[Bignose]

I already explained that earlier, that I am putting my idea out there so that maybe someone with good math skills and a good understanding of science can help me.

 

Then please accept the help that is being offered in this thread. People are trying to help you. They are trying to tell you that your equation doesn't even have the correct units. They are trying to help you understand that having the correct units is essential in any kind of calculation. You know this. When you cash your paycheck, you expect to be paid in units of money -- not bananas or peacocks or telephone poles. When the cop pulls you over for speeding he doesn't tell you you were going 42 pounds per pizza in a 35 vortexes per milliliter zone. Units matter. That is the help that you appear to be asking for in the quote above, but you are refusing.

 

You are asking for help, and yet you are refusing it so far. Accept the help. Unless this is just an empty request, in which case you are just a troll. Sorry to be so blunt, but that's the long and the short of it. There is lots of help in this thread, if you'd just accept it.

[insane_alien]

i had a brain wave about this last night,

 

does traveller think its the difference in mass because one object is exherting a force in one direction and the other object a force in the oposite direction and that these cancel out at least partially?

 

if you do, then the froces cancel out exactly ALWAYS which means the system as a whole does not accelerate but both the objects will as they have unbalanced forces acting upon them untill they collide.

[John Cuthber]

OK, bear with me on this for a sentence or two.

At one level Traveler is right.

There is no way that we could drop a feather and a hammer and time their fall accurately enough to distinguish between his "theory" and the conventional one. You would need to measure the time of flight to about twenty something significant figures and the best determinations of time "only" give about 15 significant figures.

 

All that tells us is that droping rocks is a dumb way to test the "theory".

 

 

I could say that I have a theory that says the boiling points of alcohol and water (at 1 atmosphere pressure) are the same.

I could then say I have tried to find their boiling points by an ab initio molecular orbital calculation based on things like the electron charge and plancks constamt etc.

Unfortunately it's an exceptionally difficult calculation and I can only get the answer to an accuracy of the nearest factor of 2 (in thermodynamic temperature)

Accordingly I say that there is no evidence that my "theory" is wrong- nobody else can do the calculation sufficiently accurately to show that the boiling points are different.

Since it really is a difficult problem, it's possible that I'm right- nobody can do the calculation.

In the same way, nobody has a stopwatch that could test Traveler;'s "theory"

 

But who cares?

Nobody in their right mind would try to compare two boiling points this way- they would just measure them with a thermometer.

Similarly, nobody would try to see if Traveler's theory is correct by comparing the tiny differences he talked about. They would look at something that's a better test- a couple were mentioned- the torsion balance would work but it's a difficult experiment to do to that sensitivity. An easier one is to look at things that are big (so the force is measureable) and similar in mass to one another (so the effect is clearly distinct from the established theory.)

An example of this is the solar system. Jupiter's mass is roughly comparable with that of the sun (it isn't very close, but it's a lot closer than the mass of a feather and the mass of the earth).

So we can look at the orbit of Jupiter and see how it orbits. Then we can look at the moons of Jupiter. They orbit the sun with the same period as Jupiter does (gosh).

But we know that the moons are less massive than jupiter (because they would pull it about more if they weren't).

So we know that relatively light thing (the moons of Jupiter) and really quite heavy things (like Jupiter) orbit in the same way.

The only way you can do that is with conventional gravity.

If Traveler's "theory" were right then the orbits of the planets would all be screwed up. They are not; so the "theory" is wrong.

 

Face it Traveler; if your theory doesn't agree with reallity it's not reallity that needs changing.

[D H]

At one level Traveler is right.

There is no way that we could drop a feather and a hammer and time their fall accurately enough to distinguish between his "theory" and the conventional one. You would need to measure the time of flight to about twenty something significant figures and the best determinations of time "only" give about 15 significant figures.

 

All that tells us is that droping rocks is a dumb way to test the "theory".

Nobody here has claimed that the feather and hammer experiment had any scientific merit. This demonstration was targeted at a lay audience, many (or even most) of whom have a pre-Archimedean view of the world. A Cavendish-style experiment would not have the visual impact that the feather and hammer did.

 

An example of this is the solar system. Jupiter's mass is roughly comparable with that of the sun (it isn't very close, but it's a lot closer than the mass of a feather and the mass of the earth).

So we can look at the orbit of Jupiter and see how it orbits. Then we can look at the moons of Jupiter. They orbit the sun with the same period as Jupiter does (gosh).

But we know that the moons are less massive than jupiter (because they would pull it about more if they weren't).

So we know that relatively light thing (the moons of Jupiter) and really quite heavy things (like Jupiter) orbit in the same way.

That is not correct. If Jupiter was a lot smaller (for example, the size of one of its moons), the mini-Jupiter would take longer to orbit the Sun than does our big honkin' Jupiter. The orbital period of two bodies about each other is, ignoring all other influences (i.e., other gravitational bodies),

 

[math]T = 2\,\pi\,\sqrt{\frac{a^3}{G(m_1+m_2)}}[/math]

 

Ignoring the mass of Jupiter in the above equation (or making it very small) yields a period of 4,335.4 days at Jupiter's distance from the Sun. Accounting for the mass of Jupiter but still using the above simple two-body problem equation yields a period of 4,333.3 days. The actual value: 4,332.589 days. Accounting for Jupiter's mass removes 75% of the error.

[traveler]

DH,

 

Let's suppose for a moment that the acceleration of an object due to gravity is proportional to the difference in mass, and inversely proportional to the distance between them. If that were the case, and you knew it to be true (still supposing), how would you write the equation? Will you rewrite my formula to properly express my idea?

 

In my original example, you will see that I multiplied the "A value" by a constant to arrive at the proper units. I understand you that the constant needs to be in the formula. As I said, my math (or lack thereof) sucks.

Edited October 6, 2008 by traveler

[Sayonara]

I think people might be a bit concerned that you don't really know what your "A value" represents. This would be a valid concern, since it has various implications for your interpretations.

[traveler]

I think people might be a bit concerned that you don't really know what your "A value" represents. This would be a valid concern, since it has various implications for your interpretations.

 

A few years ago I had this idea. I was "arguing" with a guy that had tremendous math skills. During the discussion, I concluded that an "A value" would be needed, because as the distance decreases, the acceleration increases. So basically, the "A value" is an "instantaneous" value of acceleration at any specific distance. How could I give the formula a constant, and as soon as the object is released, and travels any distance, the acceleration increases? In other words, yes, acceleration is the rate of change of velocity, but as the distance decreases, not only is the velocity increasing, but so is the acceleration. The acceleration is accelerating as the object gets closer.

[Klaynos]

I think people might be a bit concerned that you don't really know what your "A value" represents. This would be a valid concern, since it has various implications for your interpretations.

 

Agreed, in particular this comment worries me:

 

Well, as in my initial example, if an object is dropped from 16.087 feet, and it hits the ground in EXACTLY one second, the acceleration is 32.174 ft/sec^2.

 

The constant for that example of the Earth would then be:

 

.000000000000000000000086636392822469954136118658002141 ft/sec^2

 

 

Is a new value required for every situation, how can you present an idea without knowing this, how can you possibly have tested it to your own satisfaction let alone be able to convince others!

[traveler]

Is a new value required for every situation, how can you present an idea without knowing this, how can you possibly have tested it to your own satisfaction let alone be able to convince others!

 

I didn't test it or measure it, 32.174 ft/sec^2 is accepted to be the acceleration of gravity (although it varies according to where on Earth you are) as is 9.8 m/s^2.

 

If an object was dropped from 16.087 feet, and the acceleration was 32.174 ft/sec^2, it would take exactly 1 second to impact.

 

So in my example, I was comparing impact times of a 1kg object to a 10 kg object when dropped from an exact height of 16.087 feet, so that I could show the difference between an object taking exactly 1 second, and the other object taking longer, and I could show how small the difference was, as compared to exactly one second for the 1kg object.

 

The "A value" of the 1kg object was divided into 32.174 to arrive at a constant for a "A value" of 1. I then used that constant and multiplied it by the "A value" of the 10 kg object, and did the math to figure the impact time.

 

That is how I arrived at the constant for that example. There is no way I would be capable of measuring such a small number, I calculated it from the "accepted" 32.174 ft/sec^2 and used the constant for both objects.

[Edtharan]

Traveller, I have a few questions:

 

If I were to drop a Hammer, Feather and a Piece of String, would they all fall at different rates to each other?

 

If I then tie the feather tightly to the hammer with the piece of string, how does this effect the fall rate of the objects?

 

As the combined masses of the Hammer, Feather and String mean that the mass of this new object, being closer to that of the planet Earth than any of them are by themselves, falls at a faster rate or a slower rate?

 

Does the fall rate average out between the three objects (is slows down compared to the hammer by its self, but falls faster than the feather does by its self)?

 

How would this fall rate compare to an object that was the same mass, but not actually separate objects tied together?

 

This last question is important. As the The String is Heavier than the Feather, the Hammer is heavier than the String, and the Combined mass is heavier than the Hammer by its self.

 

If your formula is correct then the trend in variation of fall rates at the mass increases should continue, so the combined object should continue the trend of increasing of mass.

 

But.

 

Why then would they fall at a different rate, just because there were or were not tied together? Tying an object to another does not actually change any physical properties of the object like mass. In fact, when you look at the objects on the atomic scale, even if they are tied together, they are not actually physically touching. Even the atoms of the objects themselves are not physically touching, but bound together by electrical attractions, and can be considered separate (like when they are not tied together)

 

So in that sense, how could tying something together change its rate of falling?

 

If the rate of falling does not change when they are tied together, then it proves your formula wrong (as that is what you and your formula say). But the mass of an object can't effect the rate of falling, because if you increased the mass by tying them together, at the atomic level they are not really the same object.

 

It used to be thought that different objects fell faster because they were heavy. But this same argument (about tying object together) showed that that idea did not make any sense, as the act of tying together objects should make them fall faster than either of them alone, but if the objects were only loosely tied together, then each would fall at their own rate and the object as a whole should simultaneously fall faster than either of them.

 

How can an object fall simultaneously at two different rates?

 

It is the same problem that your idea has. Tying two objects together should make them fall at a different rate than they would by themselves, but if you loosely ties them, then they should fall at the original rates but as they re tied together they should also be falling at the rate dictated by their combined masses.

 

How can an object fall at two different rates simultaneously?

[traveler]

I don't mind answering questions to the best of my ability, and I have and will continue to, but it's aggravating when I do all the answering, and my questions go unanswered.

 

My last three posts summed up my idea well. I asked for help from DH, and responded to Klaynos and Sayonara (cubed), and yet none of you three responded back. I would really appreciate some type of response to my last 3 posts.

 

Ed,

 

If each object was tested separately, the object with the least amount of mass would accelerate at the greatest rate, then the next more massive object would have a slightly lower acceleration rate, and the most massive object would have the lowest acceleration rate.

 

Combining the objects makes one object of greater mass, therefore it will take a longer time than any of the three separate objects.

 

Adding mass adds inertia, which resists the acceleration greater than a lower inertia, so a more massive object has more inertia, and a lower acceleration rate, and takes longer to travel the same distance.

 

You tying the objects together with a strong bond means that ALL of that mass has to be accelerated as a whole, where separate objects have less mass, less inertia, and a greater acceleration rate, because the NET FORCE is greater, and that means a greater acceleration and lower elapsed time.

 

Does that answer your question properly?

Edited October 7, 2008 by traveler

[Klaynos]

That is how I arrived at the constant for that example. There is no way I would be capable of measuring such a small number, I calculated it from the "accepted" 32.174 ft/sec^2 and used the constant for both objects.

 

Again I suggest you investigate how the value for G is calculated, you can probably find these people original data as well without having to conduct the experiments.

 

I also asked you which questions I ignored?

 

You are presenting a new theory here, one of the precursors to presenting a new theory is being an expert on the old... this doens't mean an expert in the traditional sense but you should understand the theory fully and be able to answer questions for both theories and show how yours is better.

[Bignose]

I don't mind answering questions to the best of my ability, and I have and will continue to, but it's aggravating when I do all the answering, and my questions go unanswered.

 

It is equally aggravating to post a reply and to be completely ignored.

 

Was my post, #87, unclear?

 

You asked for help, and I explained how people are trying to help you. You said nothing. You certainly haven't given a serious look at the help members are trying to provide.

 

You didn't address any of the issues that were brought up by me or anyone else. Such as the incorrect dimensions.

 

As I said in my previous post, this is awfully a lot like troll behavior. I want to give you the benefit of the doubt, but you have to look at the issues being raised and address them. If your theory cannot take a few questions from random people on an Internet forum, how is it ever going to survive some serious scrutiny by something like a peer-review process to get a paper published?

[Edtharan]

You tying the objects together with a strong bond means that ALL of that mass has to be accelerated as a whole, where separate objects have less mass, less inertia, and a greater acceleration rate, because the NET FORCE is greater, and that means a greater acceleration and lower elapsed time.

So, what happens if you have two objects (A Feather and a Hammer) tied together by a Rope so that initially, the Feather and Hammer are initially allowed to move freely (ie the rope is slack).

 

As the Feather falls faster, it will eventually move away from the Hammer, and tighten the Rope.

 

At the point, just before this, say the Feather was moving at 2m/s and the Hammer was moving at 1m/s

 

When the Rope goes tight, the Inertia of the Feather should transfer some of its motion to the Hammer along the Rope as the faster motion of the Feather pulls the Hammer. Also, the Hammer should slow down the Feather as it is moving Slower (essentially the Feather looses some of its kinetic energy and it is transferred to the Hammer).

 

So the Hammer should speed up and the Feather should slow down. A balance should then be reached where the Hammer and Feather both travel at the same speed (because if they weren't then one would pull on the other and either speed it up or slow it down).

 

But, this means that the combined object of the Hammer and Feather (and rope), has to be faster than the Hammer as the Hammer has been sped up by the motion of the Feather as it pulled on it. It also means that it has to be slower than the Feather as the Hammer has pulled on it.

 

BUT, you state that the system should be SLOWER than the Hammer because the combined mass is greater than the Hammer.

 

This is where your theory fails.

 

However, if we assume that all objects fall at the same rate (when started from the same height), then we do not run into this problem at all.

 

Parachutists know the exact phenomena I am talking about. A Parachute will fall at a slower rate than a person without a Parachute and that if they are joind, then they will fall at a rate faster than the slowest speed and slower than the fastest speed (yes I know it is due to air resistance not gravity, but what I am using this analogy for is to point out that if objects falling at different rates are joined, then the resulting final speed will be slower than the fastest and faster than the slower speed).

 

This occurs because the Faster moving object pulls on the Slower moving object making it go faster, and the Slower moving object pulls on the Faster moving object making it go slower until they both reach the same speed.

 

Your theory can not work because of this effect. As this effect is well know and really does occur (the reality check of science), then your theory can not possible be correct.