Moving center of coordinates in the polar graph

  • #1
Summary:
Please, check if made a mistake. I get bad result but can`t understand where is my mistake
I have a function in polar coordinates:

t (rho, phi) = H^2 / (H^2 + rho^2) (1)

I have moved the center to the right and want to get the new formulae.

I use cartesian coordinates to simplify the transformation (L = 232.5).

rho^2 = (x')^2+(y')^2
x' = x-L (2)
y' = y

Then I substitute expression (2) into (1) and go back to the polar coordinates (using x=rho*cos(phi) and y=rho*sin(phi) ). The result is:

t (rho, phi) = H^2 / (H^2 + rho^2 - 2 * rho * L * cos(phi) )**2 (3)

The first picture is for (1) function.

The second picture is for (3).
 

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Answers and Replies

  • #2
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7,519
Summary:: Please, check if made a mistake. I get bad result but can`t understand where is my mistake

I have a function in polar coordinates:

t (rho, phi) = H^2 / (H^2 + rho^2) (1)
Or as a bit more readable,
$$t(\rho, \phi) = \frac{H^2}{H^2 + \rho^2}$$
If you click what I wrote above, you can see my LaTeX script. There's more information about LaTeX in the LaTeX Guide line at the lower left of the input pane.
ektov_konstantin said:
I have moved the center to the right and want to get the new formulae.

I use cartesian coordinates to simplify the transformation (L = 232.5).

rho^2 = (x')^2+(y')^2
x' = x-L (2)
y' = y

Then I substitute expression (2) into (1) and go back to the polar coordinates (using x=rho*cos(phi) and y=rho*sin(phi) ). The result is:

t (rho, phi) = H^2 / (H^2 + rho^2 - 2 * rho * L * cos(phi) )**2 (3)
It looks to me like you are omitting the ##L^2## term in the denominator. After the substitution -- in equation (2) -- I get this:
$$t(\rho, \phi) = \frac{H^2}{H^2 + (x - L)^2 + y^2} = \frac{H^2}{H^2 + x^2 - 2Lx + L^2 + y^2} $$
$$ = \frac{H^2}{H^2 +\rho^2 - 2L\rho\cos(\phi) + L^2}$$
ektov_konstantin said:
The first picture is for (1) function.

The second picture is for (3).
Does my change make a difference?
 
  • #3
Or as a bit more readable,
$$t(\rho, \phi) = \frac{H^2}{H^2 + \rho^2}$$
If you click what I wrote above, you can see my LaTeX script. There's more information about LaTeX in the LaTeX Guide line at the lower left of the input pane.
It looks to me like you are omitting the ##L^2## term in the denominator. After the substitution -- in equation (2) -- I get this:
$$t(\rho, \phi) = \frac{H^2}{H^2 + (x - L)^2 + y^2} = \frac{H^2}{H^2 + x^2 - 2Lx + L^2 + y^2} $$
$$ = \frac{H^2}{H^2 +\rho^2 - 2L\rho\cos(\phi) + L^2}$$

Does my change make a difference?
Yes. I made such a stupid mistake. Thanks for advise about LaTeX
 

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