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Post by mondo on Sept 3, 2023 22:04:50 GMT
There are several things in this subsection that are difficult to understand:
1. On page 541 in the first paragraph of subsection 2 - "Next, we have arbitrarily chosen a point [not shown] from which to measure circulation and flux, i.e a point at which the potential $\Phi$ and stream function $\Psi$ both vanish". But if both $\Psi$ and $\Phi$ vanish then both $\Psi$ and $\Phi$ are $0$ right? Also, author says that this point is not shown but from the description of it it seems to be one of the two stagnation points $a$ and $b$?
2. First paragraph of page 542: "This is because the geometric signature of a critical point $a$ is the angle $\pi/4$" between the streamline and equipotential.." where is this value $\pi/4$ coming from?
3. At the very bottom of the same page "Since we know that $[\Phi] = 4v$.." how can we know that? Base on what?
4. At the beginning of the second half of page 543, "The complex potential $\Omega(z)$ and $\overline{\Omega(w)}$ map points lying strictly outside $R$ and respectively, $D$ to points in a plane that is slit along the real axis from $0$ to $[\Phi]$" I don't get this idea of slit plane. $\Omega$ should map complex plane $z$ to some other complex plane which represents streamlines and potentials. So how can we "fit" all this information on a single, real line from $0$ to $[\Phi]$?
Thank you.
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Post by Admin on Sept 6, 2023 17:12:56 GMT
Mondo
1. You are not understanding what Needham is saying. Read pages 494-504 again.
2. Since the grid is ultimately square then at the stagnation points where this breaks down the grid lines can only intersect at an angle of $\pi/4$, by symmetry.
3. and 4. I'll come back to you on these. It's such a long time ago that I studied these. I will need some time to read it all again.
Vasco
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Post by Admin on Sept 6, 2023 17:22:20 GMT
Mondo
I have edited your post so that $[\phi]$ is now $[\Phi]$ as it is in the book. It's confusing if you don't use the same notation as in the book.
Vasco
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Post by mondo on Sept 8, 2023 7:24:39 GMT
Mondo 1. You are not understanding what Needham is saying. Read pages 494-504 again. 2. Since the grid is ultimately square then at the stagnation points where this breaks down the grid lines can only intersect at an angle of $\pi/4$, by symmetry. 3. and 4. I'll come back to you on these. It's such a long time ago that I studied these. I will need some time to read it all again. Vasco 1. This is probably the most frequently read part of the book by me  I think I am right with my claim that the points at which both function vanish are critical/stagnation points. Do you agree with that? 2. I don't see it, author says that the "geometric signature" of the point $a$ is an angle $\pi/4$ between a streamline and equipotential through this point - but how can we have a streamline and an equipotential there if this is a critical point? I somehow try to visualize it with:  The red X stands for a critical point, so where is the angle $\pi/4$ there? 3./4. Absolutely, I am impressed how much you remember after reading it years ago anyway. |
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Post by Admin on Sept 8, 2023 7:48:56 GMT
Mondo
1. The velocity is zero at the critical points yes.
2. But if that were a critical point then the grid would not have horizontal and vertical lines near the critical point.
Vasco
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Post by mondo on Sept 8, 2023 8:10:47 GMT
1. The velocity, but what about streamlines and equipotentials at that point, they should be zero too
2. I don't get that. This is somehow connected to point #1 above. How can we talk about streamlines and potentials at a critical point?
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Post by Admin on Sept 8, 2023 8:40:53 GMT
Mondo
1. What do you mean when you say a streamline/equipotential is zero? A streamline/equipotential does not have a value, it is a line of flow.
2. Look at [22] on page 503, you can see the grid lines intersecting at $\pi/4$ at the two critical points.
Vasco
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Post by Admin on Sept 8, 2023 8:45:32 GMT
Mondo
Maybe we should say stagnation point here.
Vasco
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Post by Admin on Sept 8, 2023 9:05:25 GMT
Mondo
You would learn a lot by drawing [22] on page 503. Do some real practical work on streamlines and equipotentials instead of just reading about them.
Vasco
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Post by Admin on Sept 8, 2023 19:23:14 GMT
Mondo
It's easy to see from $\Omega=z+1/z$ that at the stagnation points in [22] the value of $\Phi$, for example, is not zero.
Vasco
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Post by mondo on Sept 9, 2023 19:41:16 GMT
Mondo 1. What do you mean when you say a streamline/equipotential is zero? A streamline/equipotential does not have a value, it is a line of flow. 2. Look at [22] on page 503, you can see the grid lines intersecting at $\pi/4$ at the two critical points. Vasco 1. yes but every lines corresponds with some particular value right? For instance at this line the flow gives this much flux. 2. It looks like a streamline at real axis and the potential (dashed line) at $-1$ meet at at an angle of $45$ degrees but this is judged by visually looking at it, how are we sure it is exactly this angle? Mondo You would learn a lot by drawing [22] on page 503. Do some real practical work on streamlines and equipotentials instead of just reading about them. Vasco Good suggestion, and when I look at figure [22] on p.503 there are things I don't understand i.e let's take the $\Omega(z) = z + \frac{1}{z}$ function on the real axis. $\Phi = \frac{1}{z}$ hence at a point $z = -2$ we should have a streamline going in a negative direction of $X$ but we see it is the other way around. How are we decide what is the general direction of the streamline from $\Psi$? |
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Post by Admin on Sept 9, 2023 22:42:46 GMT
Mondo
To answer your last point: Your expression for $\Phi$ cannot possibly be right since $\Phi$ is a real number and also the values of $\Phi$ and $\Psi$ have nothing to do with the direction of the flow.
Also the solid lines are the level curves of $\Psi$ not $\Phi$.
Vasco
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Post by Admin on Sept 9, 2023 22:55:34 GMT
Mondo
Also on the real axis $\Psi=0$.
Vasco
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Post by Admin on Sept 9, 2023 23:07:36 GMT
Mondo
Also the values of $\Psi$ are positive above the real axis and negative below the real axis, demonstrating that they do not indicate the direction of flow, which is always left to right in [22] outside the unit circle.
Vasco
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Post by mondo on Sept 11, 2023 6:12:14 GMT
Mondo To answer your last point: Your expression for $\Phi$ cannot possibly be right since $\Phi$ is a real number and also the values of $\Phi$ and $\Psi$ have nothing to do with the direction of the flow. Also the solid lines are the level curves of $\Psi$ not $\Phi$. Vasco Yes the mistakes is because of Greek alphabet. I meant to say $\Psi = 1/z$ while $\Phi = z$. Few questions: 1. As for the direction - it is dictated by the sign of $1/z$ right? If it were $-1/z$ then the flow would go in the opposite direction. 2. Also, you say that the value of $\Phi$ is always a real number - here it is a real part of $\Omega$ and that is $z$ so it is a complex number right? But I see it on the plot as a veridical line so from this point of view I agree, it is a real number. 3. I think I am missing something very important in the way we plot complex potentials. Tomorrow I will try to carefully reproduce [22] myself but for now I don't see how we got a unit circle in the flow which boundary is a segment of a streamline $\Psi = 0$, how we got that? 4. The stagnation points at $\pm 1$ on figure [22] p.503 - why? I guess understating above question will also help me here but if I substitute $\pm 1$ into the formula for $\Omega$ I get $\pm 1$ not $0$ at these points. Thank you. |
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I think I am right with my claim that the points at which both function vanish are critical/stagnation points. Do you agree with that?