Energy needed to change the shape of a triangle

[dxtx]

Hello,

Inside the triangle (black+red at start) there is a gas under pressure P. I rotate with an external device two black arms around the black dot axis and I move up the red line in the same time to keep constant the area of the triangle. I drawn the triangle in gray color after an angle of $ 0.1rd$. There are gaskets between the black walls and the red: the gas don't escape. It is a theoretical device, I don't take in account the mass and the friction.

At start, $ a=b=√2/2$ m and $ c=1 m$
Gas under pressure $P=1 Pa$
Area of the triangle = $1 . 1/2 = 0.5 m^2$ = constant
Calculation for an angle of $1e-1 rad$
Depth of the device = $1 m$, the depth don't appear in the calculations


Energy from the black arms :
I have the surface constant so I can write: $ab=0.5$ with a and b two sides of the triangle, so $ a=0.5/b$
Pythagore: $ c^2=a^2+b^2$ with a, b, c the sides of the triangle with $ b=0.5/a$ I have $c^2=0.25/b^2+b^2$
I have $b=c \cos(x)$ so I can write $ c^2=0.25/(c^2 \cos^2(x))+c^2 \cos^2(x) $
$ (c^2-c^2 \cos^2(x))(c^2 \cos^2(x))=0.25$
$c^4(1- \cos^2(x))\cos^2(x)=0.25 $
$c=(0.25/(1-\cos^2(x))\cos^2(x))^{0.25}$
I have 2 arms but I need to divide by 2 for the moment so the energy is:
$ \int_{ \pi/4}^{\pi/4+0.1} (0.25/(\cos(x)^2 (1-\cos(x)^2)))^{0.25} dx = 0.100336 J$

Energy needed by the red line:
The area is constant, so the length of the red line is : 2*0.5/(\sqrt(2)/2-x), so the energy is:
$ \int_{0}^{\sqrt{2}/2-\sqrt{0.5 \cos(\pi/4+0.1)/\sin(\pi/4+0.1)}} 1/(\sqrt{2}/2-x) dx = 0.100673 J$

There is a small difference, I don't know if it is my integrals or the calculations.

I drawn the device:





If someone could help me to find my error ?

Thanks

[dxtx]

It is not clear ?

[photon]

Don't up the thread by uninformative posts

[Munin]

First of all, what is $x$? And what do you mean by 'energy'?

[dxtx]

Thanks for your reply Munin. The variable $x$ is the angle in the calculations of the black arms and $x$ is the distance for the red wall. I'm confused, it is not pretty and readable but I had a problem with the variable \alpha in Latex.

I called energy, the work I recover or I need to give to rotate or move a wall. I change the shape of the triangle, so I need to turn clockwise one black arm, counterclockwise the other and move up the red wall.

The volume is always constant, I suppose an external device controls the rotations of the black walls and the translation of the red wall because the device is unstable.

Little error: $a=b=\sqrt{2}/2 $ m

[Munin]

dxtx в сообщении #1155030 писал(а):

The variable $x$ is the angle in the calculations of the black arms and $x$ is the distance for the red wall. I'm confused, it is not pretty and readable but I had a problem with the variable \alpha in Latex.

Okay. Please rewrite formulas with correct letters.

dxtx в сообщении #1155030 писал(а):

I called energy, the work I recover or I need to give to rotate or move a wall. I change the shape of the triangle, so I need to turn clockwise one black arm, counterclockwise the other and move up the red wall.

Then you should not compare these two parts of work (the black arms and the red wall). You should add them together.

And it's nothing wrong with them being different.

[dxtx]

Hello,

Inside the triangle (black+red at start) there is a gas under pressure P. I rotate with an external device two black arms around the black dot axis and I move up the red line in the same time to keep constant the area of the triangle. I drawn the triangle in gray color after an angle of $ 0.1rd$. There are gaskets between the black walls and the red: the gas don't escape. It is a theoretical device, I don't take in account the mass and the friction.

At start, $ a=b=\frac{\sqrt{2}}{2}$ m and $ c=1 m$
Gas under pressure $P=1 Pa$
Area of the triangle = $1 . 1/2 = 0.5 m^2$ = constant
Calculation for an angle of $1e-1 rad$
Depth of the device = $1 m$, the depth don't appear in the calculations


Energy from the black arms :
I have the surface constant so I can write: $ab=0.5$ with a and b two sides of the triangle, so $ a=\frac{0.5}{b}$
Pythagoreans' theorem: $ c^2=a^2+b^2$ with a, b, c the sides of the triangle with $ b=0.5/a$ I have $c^2=0.25/b^2+b^2$
I have $b=c \cos(x)$ so I can write $ c^2=0.25/(c^2 \cos^2(\alpha))+c^2 \cos^2(\alpha) $
$ (c^2-c^2 \cos^2(\alpha))(c^2 \cos^2(\alpha))=0.25$
$c^4(1- \cos^2(\alpha))\cos^2(\alpha)=0.25 $
$c=(0.25/(1-\cos^2(\alpha))\cos^2(\alpha))^{0.25}$
I have two arms but I need to divide by two for the moment so the energy is:
$ \int_{ \pi/4}^{\pi/4+0.1} (0.25/(\cos(\alpha)^2 (1-\cos(\alpha)^2)))^{0.25} dx = 0.100336 J$

Energy needed by the red line:
The area is constant, so the length of the red line is : $2 \cdot 0.5/(\sqrt{2}/2-x)$, so the energy is:
$ \int_{0}^{\sqrt{2}/2-\sqrt{0.5 \cos(\pi/4+0.1)/\sin(\pi/4+0.1)}} \frac{1}{(\frac{\sqrt{2}}{2}-x)} dx = 0.100673 J$

There is a small difference, I don't know if it is my integrals or the calculations.

I drawn the device:

Цитата:

You should add them together.

I thought the energy from the black arms must be the same than the red arm (the same in value), the sum of energy is not 0.100336-0.100673 ?

Цитата:

And it's nothing wrong with them being different

. If the volume is constant the gas don't lost/win any potential energy. I supposed the mass at 0 even for the gas, no friction, the gas keeps the same temperature, so where goes the difference of energy ?

[Munin]

I still don't see the definition for $x$ in your "Energy needed by the red line" part of calculations. There is no "$x$" anywhere in the figure. Please expand that part.

Since the difference is small, the mistake is done somewhere in approximations or calculations.

dxtx в сообщении #1155134 писал(а):

I have the surface constant so I can write: $ab=0.5$

That's wrong, the triangle's area is $\dfrac{ab}{2}=S_\triangle,$ so $ab=2S_\triangle=1.$

dxtx в сообщении #1155134 писал(а):

I have $b=c \cos(x)$ so I can write $ c^2=0.25/(c^2 \cos^2(\alpha))+c^2 \cos^2(\alpha) $
$ (c^2-c^2 \cos^2(\alpha))(c^2 \cos^2(\alpha))=0.25$
$c^4(1- \cos^2(\alpha))\cos^2(\alpha)=0.25 $
$c=(0.25/(1-\cos^2(\alpha))\cos^2(\alpha))^{0.25}$

This can be simplified to $c=\sqrt[4]{\dfrac{(2S_\triangle)^2}{(1-\cos^2\alpha)\cos^2\alpha}}=\dfrac{\sqrt{2S_\triangle}}{\sqrt{\sin\alpha\,\cos\alpha}}.$ (Here I fixed your mistake $S_\triangle\to 2S_\triangle.$)

After that, the integrating could be done symbolically using the Weierstrass substitution.

[dxtx]

Hello,

'x' used in the integral or the red wall is :



it is the direction of the red wall. I don't know if it is clear enough ?

$c=1$
$a=b=\sqrt{2}/2$
so$ ab=0.5$

$ab$ is the area of the triangle (black+red walls). It is not 0.5 ?

I will simplify my expressions and come back, thanks for the link.

[Munin]

I've mistaken $a$ for the full length if the red side. You're right, $ab=S_\triangle=\tfrac{1}{2}.$ I just wanted to keep it in a symbolyc form. Early substitution is bad.

Still need you to expand the 'red wall' calculations. How did you get the formula?

[dxtx]

The length of the red wall is $2a$. The length $a$ is the area divided by $b$. When I integrate I use the direction 'x', I integrate from $0$ to $d$. So the length of the red wall is $2 area/ (\sqrt{2}/2- x)$. At start, $b$ is $\sqrt{2}/2$ and it is decreased by $x$.

[Munin]

And how do you calculate the upper limit?

[dxtx]

'd' is calculated with the same formula

The error was in the first integral, I integrate c/2 but I need to integrate $c^2/2$ like that the first integral is the same than the second.

thanks for your help

[Munin]

Sorry for being not useful.