# Atmospheric Pressures and Gravity

8^0 =
8 x 0 =
8^2 =
8 x 8 =

Did you get 2 different answers for the first 2 problems but not the 2nd two problems? If so, why? Both sets of numbers use the same math function.

$$8^0 \neq 8 \times 0$$
$$8^2 = 8 \times 8$$

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RiffRaff:

I'd like to give you a few examples of calculations. Let me know what you think, and where you think things go wrong.

Let's start with this: $2^5=2\times 2\times 2\times 2\times 2 = 32$.
And this: $2^3=2\times 2\times 2= 8$.

And, bearing the above in mind:
$$\frac{2^5}{2^3}=\frac{32}{8}=4$$
Do you agree?

Now you can notice something interesting:
$$\frac{2^5}{2^3}=2^{5-3}=2^2=4$$

See how that works? More generally, the rule is:
$$\frac{2^a}{2^b}=2^{a-b}$$

for any values of $a$ and $b$.

With this, we can simplify a lot of calculations. For instance:
$$\frac{2^{50}}{2^{46}}=2^{50-46}=2^4=16$$

You can check that this is correct with a calculator.

$$\frac{2^8}{2^8}=?$$

Do you agree that this is equal to 1?

But, using the general rule above, we can also write:
$$\frac{2^8}{2^8}=2^{8-8}=2^0$$

It follows, then, that $2^0 = 1$.

See?

And, more generally, since there's nothing special about the number 2, we have $x^0=1$ for any value of $x$.

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Or another way to look at what James R said:
8^3 = 8 x 8 x 8
8^2 = 8 x 8
8^1 = 8
Note that we remove "x 8" from the end the the right side as we go (we do not replace it with "x 0")

Now we could write the above sequence as:
8^3 = 1 x 8 x 8 x 8
8^2 = 1 x 8 x 8
8^1 = 1 x 8
without changing the equality

So if we go one more step from where we left off, and follow the same rule( removing"x 8"):
8^0 = 1

[...] As has been said, I am the problem, right? I've been stalked online by a Christian for over a decade. He doesn't like my attacking his "credibility",
his friends will tell you how good he is. He even told me that he will not let me use my being a disabled Veteran as a crutch to have a life in the U.S.
Because he kept me from pursuing surgery in a timely manner, to have surgery I will need to move to another country. Doctors and hospitals will
cover up malpractice and I have the imaging to prove it.
With my interest in Atmospheric Chemistry and Astrophysics (I'm including gravity and the Van Allen radiation belts into the equation which is not
a part of atmospheric chemistry at this time) it started when I was stationed on board the USS Kitty Hawk (I am from Dayton, Ohio) when it was
ported at Coronado which is in San Diego Bay across from San Diego.
With the chemical reactions, what helped me to create a flow chart was modifying how to verify trigonometric identities. It's a similar process. And
who'd think going to school for Propulsion Engineering in the U.S. Navy and then working in an engine room would lead to something like this?
Where I went wrong I read a biography about Albert Einstein when I was 13 or 14 years old. That got into his 1915 paper on General Relativity and
light bending more when passing the Sun from a distant star than what Newtonian gravity allows for.

I faced these types of attitudes at work, at church and from the family I was born into. Considering I have to leave the U.S. to have surgery and a life, what does that
say about what kind of country America is? I'm not going to compete with immigrants for approval to have a life here. Punching numbers into a calculator or pasting
an equation in the address bar and hitting enter is easy to do.

[...] With my medical situation, people will find out that the healthcare industry is corrupt...

[...] This is Alan Bauldree's dream and not mine. It's my nightmare. Why I'll leave the U.S. To live here I'll have to say that I am living his dream for him. It's never been what I wanted in my life but is what he wants. After all, it can't let me have a life in America so it doesn't matter to me.

As I mentioned, I made a mistake in ever considering science or math. My woodworking project will allow me to ask another country for both sanctuary and surgery.
My being a disabled Veteran does not mean I deserve a life in the U.S. Kind of why I need to go elsewhere. What is it Christians told me? They'll decide who can breed. [...]

You've got techniques like "stream of consciousness" down pat, RR. Plus a bevy of personal ordeals and experiences as a source to found creative narratives on. When you get this Climate Change Crusader phase out of your system, consider taking up writing (whether fiction or autobiographical accounts like Kerouac). Or even surreal painting. Arts and lit may be your true calling.
_

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Or another way to look at what James R said:
8^3 = 8 x 8 x 8
8^2 = 8 x 8
8^1 = 8
Note that we remove "x 8" from the end the the right side as we go (we do not replace it with "x 0")

Now we could write the above sequence as:
8^3 = 1 x 8 x 8 x 8
8^2 = 1 x 8 x 8
8^1 = 1 x 8
without changing the equality

So if we go one more step from where we left off, and follow the same rule( removing"x 8"):
8^0 = 1

You came up with this f(x) = 93/(1+ 9.65)^x = y, right? You know, showing Venus', Earth's and Mars' atmospheric pressures are all relative, right?
You didn't but I did so why are you acting like you're teaching me what I've already showed I know? Why I don't like forums.

God Jul!!

p.s., Will not post in here anymore but being a "real" American in the U.S. does matter.
Since I learned to snakk Norsk in Norge, couldn't hide my accent when I started school in the U.S. (am from Ohio with an American mother who got lucky my father married her.
Just an FYI, my "pure" Englisher brothers have had easy lives. Christopher Bergen of Norway
was wish his American father-in-law when his father-in-law killed him.

Yep, An American will kill his daughter's husband because he scared her father because he was saying Happy Birthday! in Norske. Just not a problem.

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p.s., No one explained the difference between multiplying 8^0 and 8 x 0. Why I plan on talking with Dr. Tom Crawford about that.
https://www.youtube.com/@TomRocksMaths A group he belongs to has a science forum. I felt like I was a Norwegian in occupied Norway.
Family told me what that was like. And since he has a PhD, why would he care? Why I have projects that I have been pursuing.
BTW, were you guys aware that Isaac Newton claimed that Gottfried Leibniz plagiarized his work in calculus?
https://amsi.org.au/ESA_Senior_Years/SeniorTopic3/3b/3b_4history_2.html

Hmm, this poster’s hovercraft is full of eels, I’m afraid.

With research on my website, this is a part of it; https://photos.app.goo.gl/QZ24smzLjWKXqJLSA
Anvil clouds are stopped by a cold air inversion called the tropopause. What creates that effect? This is also PhD level research and if I'm
right, what kept me from having a life in the U.S.?
Why does the tropopause matter? It's basically considered insignificant in science. And with my experiment that's on my website
https://climate-cycling.com/cold-fusion , it might help me to discover the secrets that the tropopause is hiding.
On a personal note, are you guys Christians or Judes? It's like when Jesus said;
Matthew 7:5; You hypocrite, first take the plank out of your own eye, and then you will see clearly to remove the speck from your brother’s eye.

Yep, I'm the problem because I let you guys know about research that I've been pursuing. And with what I say about the tropopause;
https://climate-cycling.com/earths-greenhouse-the-tropopause that could explain the cold air inversion which I think has a lapse rate of less than
2º C/km in altitude. When cumulonimbus incus (anvil) clouds stay in the troposphere then the heat energy they contain stays in the troposphere.
I like to call that the "Greenhouse" effect. And what I'll be able to say is that I could've been complaining about my medical situation but instead
I chose to work through it.
In a sense this gets into 8^0 = 1. Why? Why does the cold air inversion that is the tropopause is a constant phenomena of the Earth's atmosphere
with seemingly no reason for its continued effect?
And I have 2 woodworking projects that I've been working on. They'll keep me busy. I do need to my own self interest first. It's an American thing.

RiffRaff:

Are you not going to thank Janus58 or myself for helping you to understand why $8^0\ne 0$? That would be the polite thing to do.

Do you agree with what we told you, or not?

Are we in error? If so, what mistakes did we make? Can you point them out to us?

Or are you just going to ignore what other people say to you here?

Did you come here for a conversation, or just to preach a sermon of your own on various topics, to anybody who will listen?

I've been asked to slow it down which I do have some woodworking projects I am working on and I need to pace myself.
With atmospheric pressures and gravity, bringing gravity into the equation would actually expand what influences the Earth's atmosphere.
With p = 93/(1 + 0.02395)^x, why the exponent is a percentage and not a real number does matter. I've told students who
follow Dr. Tom Crawford on his YouTube channel https://www.youtube.com/@TomRocksMaths that one day I would explain why 8^0 = 1 is wrong. This gets into function of which math is based on function just as x, /, + or - is a function.
And I know calculators will say 8^0 = 1. With using Venus as the limit in the same fashion as
f(x) = 93/(1 + 0.02395)^x
lim
x ----> 0 (0 = 1 or 100%)

You can approach 0 from higher or lower but you can never reach it because it's the limit. This allows for 8^-1 as an
example to allow x to be 12.5% which is what 1/8 or .125 is. And this allows for ∫ with upper and lower limits but it isn't necessary for this discussion.

This means when using Venus as the basis for showing a relationship with Earth and Mars and with Mercury it's theoretical then when (1 + 0.02395)^0 = 1 is the same as 8^0 = 1. But I'm not multiplying a number. What is being factored is surface area. An example is 8^0 = 1 is a surface area of 1 x 1. Then 8^1 = 8 is a surface area of 8 squares.
And the values that describe the surface area is 2.828427125^2. That is the square around the 8 squares.
I realized this when trying to understand why I needed to use the exponent differently for the equation to work. And it came back to am I describing a surface area or am I calculating a value? With the example of 8 squares, it's values are
2.828^2. And the surface area is a quantity. And with understanding the relationship between space and atmospheric pressure, the inverse square law applies.
With Newton's g = Gm/r^2, the radius squared is the inverse law increasing the surface area of space which has the inverse affect of decreasing the acceleration of gravity.
It'll probably be a few days before I have anything else to add. This is because 8 can also be multiplied using an exponent kind of like what the log function is. But to understand how atmospheric pressures relate to the Sun's gravity, then it needs to stay within what the values in the equation represent. And for people who haven't considered this before, it might take well into next year to consider that a number to a power is actually describing the surface area of something.
This is a link to the inverse square law, it might help you guys to consider what I posted in relation to what has been made known. And they do show when moving from 1 the numbers are squared and it is surface area.If you notice they are using whole numbers and with my example of 8 squares, then instead of 3r it uses something like 2.8r instead.
http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/isq.html

James R, Is this what you corrected me on? I'm supposed to thank you for this? You're the one preaching because what I posted allows for a discussion but a discussion did not happen.
I took the time to explain the function of an exponent. And this gets into gravity as well because when 3 planets have atmospheric pressures all relative to one another and the Sun's
gravitational field just know that gravity is involved. You probably don't know that 8^1 = 8 = √8 = 2.828^2. This is what I realized when working at mathematically defining a relationship
that has previously been missed. And you guys want me to thank you because you said what I already know? I'll give everyone here a clue, okay?
When it's 8^1 = 8, the surface area of 1 will increase by multiples of 8 just as 9.81m/s adds 9.81m/s of velocity for every second, with the exponent the surface area is multiplied by the factor
which in this case is 8. It supports the inverse square law. What value is 1? It's undefined because we don't know the distance from the vertex. When they use examples of the inverse square
law they're like this; http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/isq.html
My question is, do you guys agree or disagree with me? I tried to start a discussion but then I was given a textbook answer and not an explanation. This post did not start a discussion which
is why I posted about 8^0 = 0, etc.

I took the time to explain the function of an exponent.
But you were wrong, that is not my opinion, it is just how math works.
You probably don't know that 8^1 = 8 = √8 = 2.828^2.
The problem again is that you are wrong. 8 is not equal to the square root of 8! I have no idea how to help you when you cannot see something so obvious.
You wrote 8^1 = 8 = √8, so that means 8 = 2.828. How do you think that makes any sense?
When it's 8^1 = 8, the surface area of 1 will increase by multiples of 8
The surface area of 1!?!

I'm at a loss...

RiffRaff:

Did you read posts #42 and #43, above?
Did you understand them?
Do you agree with them, or do you believe Janus and/or myself made an error?
If there is an error, can you explain where we went wrong?
Why didn't you respond to the questions I asked you in post #49?
Is this what you corrected me on?
I thought I was clear in post #42 about what I was correcting you on.

Is there something there that you did not understand? I'm happy to answer questions.
I'm supposed to thank you for this?
That would be the polite thing to do: for you to acknowledge that somebody tried to help you to clear up a point of misunderstanding or confusion that you have.

Do you want to have a discussion, or not?
You're the one preaching because what I posted allows for a discussion but a discussion did not happen.
Once you have acknowledged what I posted, I am quite happy to move on and discuss some of the other matters you have raised. However, if you are not interested in actually discussing anything, I won't bother engaging with you any further. Let me know how you'd like to proceed.
I took the time to explain the function of an exponent.
You asserted that $8^0=0$, which is incorrect.
You probably don't know that 8^1 = 8 = √8 = 2.828^2.
While it is true that $8^1=8$, it is certainly not true that $8=\sqrt{8}$.

Do you recognise that you just made another error? Are you willing to be corrected on this (or anything)? If you are happy to admit you made some mistakes, and you agree that I am correct, we can move on to other points of dicussion.
And you guys want me to thank you because you said what I already know?
No. We guys want you to thank us because we politely engaged with what you wrote and tried to help you to understand a mistake you made.

Clearly, you did not already know you were wrong. If you already knew that, why would you post something you knew to be wrong? And why follow it up with an even more obvious error ($8=\sqrt{8}$?)?
What value is 1? It's undefined because we don't know the distance from the vertex.
What vertex?

Look, this is just a simple matter of arithmetic.

Do you agree that $8^0=1$, for the reasons we gave to you in posts #42 and #43, or don't you?
My question is, do you guys agree or disagree with me?
You don't need to ask whether we agree with your arithmetic. We would not be trying to correct you if we agreed with you. We replied to you, giving reasons why we disagree with you, in posts #42 and #43. The ball is now back in your court. It is your turn to respond to posts #42 and #43. This is how a conversation works.
I tried to start a discussion but then I was given a textbook answer and not an explanation.
You tried to raise about 8 different topics at once in your early posts. I have tried to start by discussing one I thought we could easily deal with, but so far you have not acknowledged any error on your part. Can you do that? If so, then we can move on to discussing something more interesting - one of your other topics.
This post did not start a discussion which is why I posted about 8^0 = 0, etc.
It is a side issue, so it should be quick to clear up any confusion. Right? If we can't agree on the simplest of things, what hope is there of agreeing about anything that is more complicated?

But you were wrong, that is not my opinion, it is just how math works.

The problem again is that you are wrong. 8 is not equal to the square root of 8! I have no idea how to help you when you cannot see something so obvious.
You wrote 8^1 = 8 = √8, so that means 8 = 2.828. How do you think that makes any sense?

The surface area of 1!?!

I'm at a loss...

I have projects I’m working on right now. As I mentioned, I told people who follow Tom Rocks Maths YouTube channel that one day I’ll explain it. They were rational when wondering about it and they were aware that I know what the textbook answer is. You didn’t understand it but people familiar with calculus might understand it better simply because they’re more involved with math. What I say is easy to verify.

But you were wrong, that is not my opinion, it is just how math works.

The problem again is that you are wrong. 8 is not equal to the square root of 8! I have no idea how to help you when you cannot see something so obvious.
You wrote 8^1 = 8 = √8, so that means 8 = 2.828. How do you think that makes any sense?

The surface area of 1!?!

I'm at a loss...

The inverse square law is surface area. When 8^1 = 8. That is saying the surface area has a value of 8 squares. Why it can be said 8^1 = 1 x 8 = 8, how to factor an
increase in the surface area for each increased value of the exponent. 2.828^2 is the dimension for that surface area because √8 = 2.828^2.
When 8^0 = 1, the surface area is one as they show here; http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/isq.html The radius is not known. Math does support this.
This gets into gravity and is also what I used to find the mathematical relationship between the atmospheric pressures of 3 planets and their distance from the Sun. If
anyone used a calculator or copy and paste in your browser then you'd know the math is correct.
With this equation; f(x) = 93/(1+ 9.65)^x = y for Venus, x = 0. With the Earth and Mars, how much further from the Sun are they than Venus?

Venus is 108,208,930 km. The Earth is 149,000,000, divided by Venus’ orbit that value is 1.377. When squared it’s 1.896.

Mars is 228,000,000 km from the Sun. When divided by Venus’ orbit that value is 2.107. When squared it is 4.44.

Then factoring the Earth’s value, it is f(x) = 93/(1+ 9.65)^1.896 = 1.048 bars (actual is 1.013 bars) while Mars is f(x) = 93/(1+ 9.65)^4.44 = 3.955mb (actual is 6.518mb).

Too lazy to use a calculator to to copy and paste? Then what you have is an opinion.

No.

That is what James and others are trying to explain

There shouldn't be a therefore in the first part it just should be "and"

When the surface area is 1, it's 1 x 1. The center of the surface area is 1/2 of 1 from the point of origin. This makes the radius
√2/2 or a 45º angle as the hyperphysics link shows. This is what exponents as taught are used for. And 1 = n^0, any number to the 0 power.
n represents the rate of change/increase in surface area when it is increased exponentially. Then n^x = y and then the surface area can be graphed.
This represents using right angles to establish dimensional relationships.
Distance from the point of origin would be n^x = √y/2. An example 8^2 - a surface area of 64. √64/2 = 8/2 =4.
With the inverse square law they use the radius while showing a flat surface area.
https://photos.app.goo.gl/jdgta92rPL1xiCDp9

The flat surface area actually represents a section of a sphere.
https://photos.app.goo.gl/MYq6dgatpoMU6gaV7

Since I am using the distance from the Sun, the center of the area of the sphere is the radius as are the edges of the section of a sphere. Basically
when the 4 corners disappear the surface area remains the same.
And this gets into gravity like g = Gm/r^2. And with the gravitational effect that I am considering, ^x is further than Venus from the Sun squared.
With the inverse square law, using squares simplifies use and understanding of that law.

p.s., not sure why the images didn't load.

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I am seeing if using a different source for photos allows for them to be displayed, ie., I'm finding out if it's Google or not.

n^0 = 1

And because the inverse square law is based on radius, what it is actually describing

I'll do the math in the next day or 2 and see how accurate it is.

https://flic.kr/p/2ppczAy
With the inverse square law, hopefully everyone knows that any emitted energy or force will be distributed according to this law (usually).
And since I am considering the atmospheric pressure of different planets a spherical cap that has been squared works better for me. As the
planets orbit the Sun they'll follow closer to a spherical shape than a square. And this allows for both perihelion and aphelion. This might
cause a small change in atmospheric pressure.

I used a calculator. It calculated the surface area of a spherical cap based on a radius of 1. https://photos.app.goo.gl/78gk4AJoevTDF7Du9 The inverse square law says it is 1 x 1 = 1^2. The inverse square
law uses the length of the radius to create the square. If the distance from the vertex were used then it'd be 1.414 x 1.414 = 1.414^2 = 2.0736. 2.0736 is a lot closer than 1. I think I just found another mistake
in math. With this link to the inverse square law http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/isq.html If the distance is 1 then with 2 right triangles, a square is formed. At 45º the sin or cos, √2/2 or
.707. This is actually representative of the inverse square law https://photos.app.goo.gl/vqKpgE7pQ2LpZap76 Everything moves equally from the vertex so to speak in 6 different directions.
I might have to rework my math to agree with the surface area of a spherical cap.

p.s., With this, the sphere I calculated is inside of one face, it would use the 4 midpoints between corners for the base of the spherical cap. A more accurate surface area might be about 1.1 times larger.

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RiffRaff:

Are you going to respond to any of my posts, or have you decided to ignore me?

Is it because you're unwilling to acknowledge that you've made mistakes?
I have projects I’m working on right now.
Ah. Projects. I see.
As I mentioned, I told people who follow Tom Rocks Maths YouTube channel that one day I’ll explain it.
Why not clear up your confusion now? You'll never be able to explain why $8=\sqrt{8}$, because that will never be true. The same goes for attempting to explain why $8^0=0$, because that will never be true either.

The thing about maths is that it is definite. Things are either right or wrong. Once an error is demonstrated, it remains demonstrated. Once a mathematical fact is proved, it stays proved. You can't unprove that $8^0=1$. You'll never do that, no matter how much you try. So, I'd advise you not to waste time better spent on your "projects". Just accept the correction and move on.
They were rational when wondering about it and they were aware that I know what the textbook answer is.
Did somebody here say something irrational? What was it?
You didn’t understand it but people familiar with calculus might understand it better simply because they’re more involved with math. What I say is easy to verify.
You think it's easy to verify that $8=\sqrt{8}=2.828\dots$?
The inverse square law is surface area.
Allow me to help you again, so that next time you have this conversation you can express your meaning more clearly.

The surface area of a sphere of radius $r$ is given by the formula $SA=4\pi r^2$.

It follows that a sphere of radius $r=1$ has surface area $4\pi$, while a sphere of radius $r=2$ has a surface area of $4\pi\times 2^2=4\pi\times 4 = 16\pi$.

The surface area of the sphere of radius 2 is four times as great as the surface area of a sphere of radius 1. By a similar argument, the surface area of a sphere of radius 3 is nine times as great as the surface area of a sphere of radius 1; the surface area of sphere of radius 4 is sixteen times a great, and so on.

The surface area of a sphere is proportional to the square of the radius.

Newton's expression for the acceleration due to gravity at a distance $r$ from a uniform spherical mass $M$ is:
$g=\frac{GM}{r^2}$,
where $g$ is the acceleration due to gravity and $G$ is a constant. This is an inverse square law. It is "inverse" because the $r^2$ factor is in the denominator of the fraction on the right-hand side of the equals sign. This tells us that when we double the distance $r$ from the mass, the acceleration due to gravity will be 1/4 of the original. If we triple the distance, the acceleration drops to 1/9 of the original, and so on. Inverse square.
When 8^0 = 1, the surface area is one as they show here; http://hyperphysics.phy-astr.gsu.edu/hbase/Forces/isq.html The radius is not known. Math does support this.
I don't understand what you are claiming the math does not support. Can you explain?
This gets into gravity and is also what I used to find the mathematical relationship between the atmospheric pressures of 3 planets and their distance from the Sun. If
anyone used a calculator or copy and paste in your browser then you'd know the math is correct.
I asked you some questions about your formula in a previous post. Why did you ignore them? Do you want to discuss your formula for the air pressure, or not? Note that I have not said it is incorrect (yet). I want to know more about it. But only you can give me the information I want, because it's your work. Right?

Do you want to discuss any of this, or not?

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