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Atmospheric Pressures and Gravity
- Thread starter RiffRaff
- Start date
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.
Now, here's the thing. Now, how about this one:
$$\frac{2^8}{2^8}=?$$
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 $x$.
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.
Now, here's the thing. Now, how about this one:
$$\frac{2^8}{2^8}=?$$
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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Janus58
Valued Senior Member
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
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'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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RiffRaff
Registered Member
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.
https://metro.co.uk/2019/10/04/dad-...-jumped-out-of-bush-to-surprise-him-10862450/
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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RiffRaff
Registered Member
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
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
RiffRaff
Registered Member
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.
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?
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?
RiffRaff
Registered Member
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.
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...
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 there something there that you did not understand? I'm happy to answer questions.
Do you want to have a discussion, or not?
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.
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 ($8=\sqrt{8}$?)?
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?
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?
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.
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?
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.
You asserted that $8^0=0$, which is incorrect.
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.
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 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?
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.
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.
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?
RiffRaff
Registered Member
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.
RiffRaff
Registered Member
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.
Pinball1970
Valued Senior Member
No. 
Pinball1970
Valued Senior Member
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"
There shouldn't be a therefore in the first part it just should be "and"
RiffRaff
Registered Member
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.
√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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RiffRaff
Registered Member
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.
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.
RiffRaff
Registered Member
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.
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?
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.
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.
Do you want to discuss any of this, or not?
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?
Ah. Projects. I see.
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.
Did somebody here say something irrational? What was it?
You think it's easy to verify that $8=\sqrt{8}=2.828\dots$?
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.
I don't understand what you are claiming the math does not support. Can you explain?
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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