Hi.

By its definition, the distance from a parabola to its focus and directrix are the same. Well, this is not a directrix as I knew; it is at the opposite side :

 

Question is -- are the lengths of the paths

Q1 - P1 - F

and Q2 - P2 - F

and Q3 - P3 - F

The same ?

 

Works for me, if you can take the distance when Q is directly above F V (ie on the y axis) to be (QV + FV).

That is a doubling back.

 

As an example if the parabola is (KISS) y=x² (focus at 0,0.25) and the line L = 1 then the length is 1.25 or 5/4

Edited October 13, 201510 yr by studiot

Yes. See fig 5 and the text below it

http://www.antenna-theory.com/antennas/reflectors/dish.php

 

They would have to be for this to work as an antenna. If the lengths weren't the same the signal from different parts of the dish would interfere with each other.

Thanks.

OK, then a directrix can be at the convex side or at the concave side as "L" in the graph; not only as typical tutorials with the directrix shown at the convex side.

 

That was the point. To find in a parabolic antenna if wavefronts would interfere or even cancel with an uneven parabola surface and how 'unsmooth' are the limits for a given wavelenght.

 

I guess that a half wavelength of the intended received signal as 'bump' or 'valley' in the parabola surface would contribute to signal cancellation. Is it right ?

 

Edited: Am receiving 12GHz with an 'ironed' now-unobtanium parabola after tree limbs fall damaged some surfaces.

12GHz ---> ~ 2.5cm ----> half wavelength is 1.25 cm.

Keeping surfaces as flat as possible under 1.25cm bumps and valleys is the aim, correct ?

 

----> http://s588.photobucket.com/user/Innernet/media/HFuWave.jpg.html?sort=3&o=92

Edited October 13, 201510 yr by Externet

Basically. The general paradigm is flatness to some small fraction of a wavelength at the highest operating frequency. At a half wavelength you would be getting complete cancellation of signal (assuming equal amplitudes). The value I often see is that the surface has to be smooth at the 0.1 wavelength level.

Thanks again. Some extra head scratching...

 

Being the distance from the parabola to the directrix equal than from the parabola to the focus, the first post line 'L' (concave side) does not qualify as directrix as I said in post #4.

 

P1-Q1 is not equal to P1-F.

 

But the total paths are the same lenght !.

This is a good example of how a practical physicists can have a feel for the question and use physics to 'prove' the maths rather than the more usual way round.

 

 

So consider the line L (forget the directrix) as the position of a plane wavefront.

By definition of a wavefront the light arrives/leaves at all points along a wavefront at the same time.

We know from physical observations that a parabolic reflector brings parallel rays to a common focus F.

So this can only happen if rays leaving any point Q on the wavefront L travel the same distance to reach the focus, since they all leave at the same time and travel at the same speed.

Note that the line L does not have to be parallel to the x axis as you have drawn it.

 

As a practical chap yourself you might like to read Mark Levi's book

 

The Mathematical Mechanic.

 

Mark sets out to prove many common mathematical theorems using the laws of physics

 

"Using Physical Reasoning to solve problems"

 

Edited October 13, 201510 yr by studiot

This is a good example of how a practicval physicists can have a feel for the question and use physics to 'prove' the maths rather than the more usual way round.

 

It's a matter of asking one's self "Why did they choose a parabolic shape for an antenna?" (or, more precisely, already knowing the answer to that question)