[guest39538]

Kryptid ; 

m³ + c³ = m³

V³ + V³ = V³

?

Added- rethought

V³ = m³

m³c³ = V³



 

u = m³

E³ = c³

[guest39538]

I've got it , I've invented  rhetorical maths for beginners .   


''A rhetorical question is a figure of speech in the form of a question that is asked to make a point rather than to elicit an answer. Though a rhetorical question does not require a direct answer, in many cases it may be intended to start a discussion or at least draw an acknowledgement that the listener understands the intended message.''

Why does a snow man melt ?

Δ

What is a change in temperature ?

Δ

How big is the universe ?



How does metal expand ?

Δ

See , it's proper good .

Added- more demonstration of it's simplicity brilliance

How do things age ?

Δ

How does time change ?

Δ

What is the speed of time ?

Δ

What is mass ?

Δ


etc, etc , 

[The Spoon]

The equation tells you that

Δu = ΔE³ = ΔkE = Δ T / t


u / t is a constant and the idol measure , u always returns to u , nothing lost and nothing gained .

Natural temperature 0 .

u = m

m+m = m2

m*m=m³

m/V= <m

I won't add a big bang equation .

What is an idol measure? Is it related to some reality TV that has obviously rotted your mind?
Why not add a big bang equation. It is bound to have comedy value.

[guest39538]

The equation tells you that

Δu = ΔE³ = ΔkE = Δ T / t


u / t is a constant and the idol measure , u always returns to u , nothing lost and nothing gained .

Natural temperature 0 .

u = m

m+m = m2

m*m=m³

m/V= <m

I won't add a big bang equation .

What is an idol measure? Is it related to some reality TV that has obviously rotted your mind?
Why not add a big bang equation. It is bound to have comedy value.

Which  big bang ? The beginning of time or the end of time ?

Additionally here is the model for the cool maths .


super model.jpg (49.75 kB . 744x478 - viewed 1064 times)

[The Spoon]

The equation tells you that

Δu = ΔE³ = ΔkE = Δ T / t


u / t is a constant and the idol measure , u always returns to u , nothing lost and nothing gained .

Natural temperature 0 .

u = m

m+m = m2

m*m=m³

m/V= <m

I won't add a big bang equation .

What is an idol measure? Is it related to some reality TV that has obviously rotted your mind?
Why not add a big bang equation. It is bound to have comedy value.

Which  big bang ? The beginning of time or the end of time ?

Additionally here is the model for the cool maths .


super model.jpg (49.75 kB . 744x478 - viewed 1064 times)

Lines and random letters. I've seen better Airfix models.

[guest39538]

Lines and random letters. I've seen better Airfix models.

Of course you have ! 

You couldn't do no better put it that way .

[Bored chemist]

If you already have the value and equation you suppose to use that equation, my equation just generalises the process's and is meant as an education aid  in simplicity .  In other words you don't have to work out an answer , it tells the ''story'' of the process's .

So can it calculate the energy released when hydrogen combusts or not?

Well in theory yes if you're good at maths and can work out the values .

You have to work out u then you have to work out ignition energy added to the system and the kinetic response energy .

Alternatively we could work backwards , 1ltr of hydrogen  volumes energy combusted,  divided by a volume of water , temp measure etc and reverse engineer the results to get an exact measure .

I have a simpler suggestion for an equation that is just as much use as yours.
anything =  1-x

Obviously, to calculate it you need to know x
But, as you have suggested, that's easy- you just work backwards from the answer you want.

For example, if you want to know the  heat of combustion of hydrogen you start off by finding out what the heat of combustion of hydrogen is- you can measure it or google it, or whatever.

It's 286 JK/mol
Now, from that we can calculate x
x = 1- 286 So x (for this particular question) is -285

And then, using my formula we can calculate the heat of combustion for hydrogen.
we know that x = -285 so we calculate 1- (-285)
And that gives the answer 286.

Do you see any problems with this method which do not apply to yours?

[The Spoon]

Lines and random letters. I've seen better Airfix models.

Of course you have ! 

You couldn't do no better put it that way .

Depends what it was trying to represent. As yours does not represent anything it is hard to judge.

[guest39538]

Lines and random letters. I've seen better Airfix models.

Of course you have ! 

You couldn't do no better put it that way .

Depends what it was trying to represent. As yours does not represent anything it is hard to judge.

That's strange it represents almost everything , thermodynamic process for a start , the process of time and aging, how the universe works , big bang singularity etc. 

Demonstrate it doesn't ?

[guest39538]

If you already have the value and equation you suppose to use that equation, my equation just generalises the process's and is meant as an education aid  in simplicity .  In other words you don't have to work out an answer , it tells the ''story'' of the process's .

So can it calculate the energy released when hydrogen combusts or not?

Well in theory yes if you're good at maths and can work out the values .

You have to work out u then you have to work out ignition energy added to the system and the kinetic response energy .

Alternatively we could work backwards , 1ltr of hydrogen  volumes energy combusted,  divided by a volume of water , temp measure etc and reverse engineer the results to get an exact measure .

I have a simpler suggestion for an equation that is just as much use as yours.
anything =  1-x

Obviously, to calculate it you need to know x
But, as you have suggested, that's easy- you just work backwards from the answer you want.

For example, if you want to know the  heat of combustion of hydrogen you start off by finding out what the heat of combustion of hydrogen is- you can measure it or google it, or whatever.

It's 286 JK/mol
Now, from that we can calculate x
x = 1- 286 So x (for this particular question) is -285

And then, using my formula we can calculate the heat of combustion for hydrogen.
we know that x = -285 so we calculate 1- (-285)
And that gives the answer 286.

Do you see any problems with this method which do not apply to yours?

You're a genius like me after all 

Yes exactly , I always work backwards too . 

Not hydrogen ;

x-y = (<x)

x+y=(>x)

Where V is volume and y is internal energy .

[The Spoon]

Lines and random letters. I've seen better Airfix models.

Of course you have ! 

You couldn't do no better put it that way .

Depends what it was trying to represent. As yours does not represent anything it is hard to judge.

That's strange it represents almost everything , thermodynamic process for a start , the process of time and aging, how the universe works , big bang singularity etc. 

Demonstrate it doesn't ?

You are right there. It doesn't.

[Bored chemist]

That's strange it represents almost everything , thermodynamic process for a start , the process of time and aging, how the universe works , big bang singularity etc. 

Demonstrate it doesn't ?

It gave an answer that was wrong by about 50 orders of magnitude.
That's a pretty convincing demonstration of failure

[guest39538]

That's strange it represents almost everything , thermodynamic process for a start , the process of time and aging, how the universe works , big bang singularity etc. 

Demonstrate it doesn't ?

It gave an answer that was wrong by about 50 orders of magnitude.
That's a pretty convincing demonstration of failure

No, I've stated it is a new form of maths that is a generalised simplified equation that explains most processes .  I've recently give the math a name of rhetorical math .

I ask again demonstrate it's failure to explain ?

[Bored chemist]

I ask again demonstrate it's failure to explain ?

It failed to explain why you got the answer a million billion billion billion billion fold wrong.

It fails to do anything.
It fails to explain anything.
And you have the burden of proof the wrong way round anyway.
It is your job to prove that it does explain things.

[guest39538]

It is your job to prove that it does explain things.

Ok, I've now altered it slightly

Lets ask it a question

Why does metal expand when heated ?

Δ

Where U³ is internal energy volume 1 =

And where u is external energy propagating through volume Vⁿ

[Kryptid]

Perhaps the answer shouldn't be in joules

Joules is the standard metric unit for energy, so it's quite appropriate.

how do they derive their equation ?

It's basic arithmetic. Two pieces of information are required: a balanced chemical equation and the enthalpy of formation of each chemical involves. The enthalpy of formation of a chemical element in its standard state is set to be zero, since they are the basic building blocks of other compounds (the exception being allotropes of those same elements, which will either absorb energy or release it when they form from the standard state of those elements. The difference between diatomic oxygen (O₂) and ozone (O₃) is one example). The enthalpy of formation of a chemical compound is usually determined experimentally.

I'll show an example. The combustion of gaseous hydrogen with gaseous oxygen to produce liquid water. The balanced equation is as follows:

2H₂ (g) + O₂ (g) → 2H₂O (l)

Now we multiply the coefficients by the known enthalpies of formation to get the following equations:

(2 x 0 kJ/mol) + (1 x 0 kJ/mol) and (2 x -285.83 kJ/mol)

The equation on the left is the total enthalpy of formations of the reactants (the hydrogen and oxygen) while the equation on the right is the total enthalpy of formations of the products (the water). Since the enthalpies of formation of the elements are zero, the answer to that equation is also zero. The equation on the right works out to -571.66 kilojoules per mole (a negative number indicates that the reaction releases energy instead of absorbing it). Now we subtract the answer on the right from the answer on the left, (-571.66 - 0 = -571.66). So the total energy released is 571.66 kilojoules per mole. That is if two moles of hydrogen are burned. If we want to know the energy released when one mole is burned, then you simply divide the result by two, which is 285.83 kilojoules per mole.

Are they doing it correctly  ?

Algebra is one of the easiest forms of math to do, so I'm going to say yes.

nX + mO2  →  xCO2 (g) + yH2O (l) + zZ + heat of combustion

HUh what's that suppose to mean ?

It's a generalized equation for combustion of a carbon and hydrogen-containing compound. The coefficients "n", "m", "x", "y" and "z" are the number of moles of each chemical in the equation. The "X" is the formula for whatever chemical you are burning (such as, for example, C₆H₁₂O₆ for glucose). The O₂ is oxygen, which is the usual chemical used as an oxidizer. The CO₂ is carbon dioxide while H₂O is water, both of which are common chemical compounds formed as a result of combustion of organic compounds. The "Z" is for any product other than carbon dioxide and water (such as if the burned compound contained phosphorus, sulfur, nitrogen or some other element).

Would be correct , the energy released is kinetics .

Antoine - nothing is lost and nothing is gained .
The hydrogen atom loses no energy in the process of combustion .

The stated purpose of the calculation was to determine the amount of energy released when hydrogen burns. Regardless of where the energy actually comes from, energy is released in the process. So either your equation is wrong or you gave me the wrong information to use in the calculation.

[guest39538]

The stated purpose of the calculation was to determine the amount of energy released when hydrogen burns. Regardless of where the energy actually comes from, energy is released in the process. So either your equation is wrong or you gave me the wrong information to use in the calculation.

The problem is no matter what value to derive , it is relative .  I could express the combustion of hydrogen releases 1 in energy , 1 would be equal to a more derived value .  I may of give you the wrong values to my equation .  It does work for the purpose it is designed for in expressing many different processes using one equation .
The purpose of my equation is for people like myself who take one look at equations like you've just provided and think stuff that it looks complex .

I think it would make a good teaching aid and help people understand why math is important , it shows how the maths works for physical process without anything complex added  and also will help people understand how things work in simplicity .

It's weird to be honest and seems kinda magical because I can ask a question and it gives the answer .

What is space-time in example ?



What is a Black hole ?



etc,

It's really like magic and ''it'' knows the answer .

Added- How about we do the speed of light ?

   

Added- shall we try gravity ?

F(G) =   +    

Added-  What is the physical process of filling a bath tub ?



Added- Sorry I love it ,  How does my shoe size change ?



How does my hard drive fill up ?

[Kryptid]

The problem is no matter what value to derive , it is relative .  I could express the combustion of hydrogen releases 1 in energy , 1 would be equal to a more derived value .

So then the equation is useless for answering questions that we don't already have the answer to.

[guest39538]

The problem is no matter what value to derive , it is relative .  I could express the combustion of hydrogen releases 1 in energy , 1 would be equal to a more derived value .

So then the equation is useless for answering questions that we don't already have the answer to.

What don't you have an answer too ?  I'll have a go at working it out .

[Kryptid]

What don't you have an answer too ?

There are a lot of things that we don't know yet, but I'll give you something that could potentially be addressed with math: what is the melting point of the yet-to-be-synthesized element 119?