Nipkow hole spacing

Forum for discussion of narrow-bandwidth mechanical television

Moderators: Steve Anderson, Dave Moll, Andrew Davie

Nipkow hole spacing

Postby Andrew Davie » Thu Apr 12, 2007 4:22 pm

I am considering having a Nipkow disk manufactured locally, only the manufacturer has a minimum hole size capability of 2mm. I was originally going to do the holes by hand, or have a larger hole created, and then attaching 'slap on' hole patches which I could then position as required.

But I got to thinking... what if I were to use the 2mm holes, how could I space the holes to make it work? And then I started thinking "well, a bit of overlap isn't going to hurt"... and then "actually, a bit of overlap is going to help"... "what is the optimal overlap?"

The attached diagram shows what I think should happen. Given three Nipkow holes, conceptually 'side by side' (over time, they are), 'a', 'b', and 'c', it would appear ideal to me if the light passing through the non-overlapping section (B') is equal to the light passing through the overlapping section (A').

But then I realised that small parts of B' overlap the area of the overlapping section A', even though the section itself isn't in the overlap. So I added the red lines to divide the sections more clearly. Now the task is to ensure that the same amount of light is visible in each red horizontal section, and furthermore to have the width of the horizontal sections the same, or as close as possible. I don't know if the light passing through the sections is equivalent when the sections are the same width -- I highly doubt it -- but there must be some optimal 'width' vs 'light' setting that would make each section appear roughly the 'same'.

So I thought I'd ask anyone if they have any thoughts on how to calculate (a) the light that goes through each section, and (b) the width of the sections (ie: spacing of the holes) and (c) the problems with this idea/approach.

Where I'm leading towards is to actually use 2mm holes, but overlap each hole by (say) 1/3 of the diameter of the hole by shifting it towards the next hole (as shown in the bottom right of the image). This would give me equally spaced 'areas', an unknown light value through each, and (say) a final image width of 42+mm.

Using this technique, the extreme left/right pixel will appear slightly different, but that's not an issue for me.
Attachments
circ.png
circ.png (15.33 KiB) Viewed 7650 times
User avatar
Andrew Davie
"Gomez!", "Oh Morticia."
 
Posts: 1511
Joined: Wed Jan 24, 2007 4:42 pm
Location: Hobart, Tasmania, Australia

Postby Klaas Robers » Thu Apr 12, 2007 6:50 pm

But Andrew, have you calculated then the size of the disc that you are going to make? 2 mm is far too large. I remember that Denis needed for his Baird discs (half a meter in diameter, 30 lines) holes of 0,7 mm wide. So you will end up with a disc of one meter!!

Denis solved this by punching the (square) holes himself and have 30 circular holes on an extra rim around the disc, that guided him to punch the holes. I will see if I have still a photograph of his punching action.
Attachments
disc-Denis1.jpg
This is the laser cut disc. Remark the extra rim with guiding holes, which is removed after punching
disc-Denis1.jpg (136.61 KiB) Viewed 7635 times
disc-Denis2.jpg
This is the disc in the punching table. A slightly tapered pin positions the rotational position of the disc. A digital ruler defines the radial position of the punch
disc-Denis2.jpg (74.02 KiB) Viewed 7634 times
disc-Denis3.jpg
And this is Denis while punching
disc-Denis3.jpg (36.49 KiB) Viewed 7633 times
Last edited by Klaas Robers on Thu Apr 12, 2007 7:01 pm, edited 3 times in total.
User avatar
Klaas Robers
Frankenstein was my uncle.
 
Posts: 1465
Joined: Wed Jan 24, 2007 8:42 pm
Location: Valkenswaard, the Netherlands

Postby DrZarkov » Thu Apr 12, 2007 6:54 pm

Why so difficult? On Monday I met a friend of mine, and we talked about my NBTV monitor. He asked why I had not ask him first to make a Nipkow-disc, they have all machines at his company to make them. All he needs is a usable picture of a Nipkow-disc, he can laser-cut it for me, for free! Of course it will not stay free when I need larger numbers for the club, but it will still be very cheap. I've ordered an aluminium-disc with 1 mm holes and 30 cm diameter, today or tomorrow I will get it. He experimented with a kind of vinyl, too, but I think the about 3 mm thick plastic is not suitable.

So maybe we've got a new source for cheap and high quality(?, I will see today or tomorrow) discs.

As a picture I used the Excel-spreadsheet from the Etzold-homepage, and copied the result to Corel Draw. The difference to the club disc is that the synchron holes are on an inside ring.
User avatar
DrZarkov
I think I've had a cranial implosion.
 
Posts: 1037
Joined: Mon Feb 19, 2007 11:28 pm
Location: Kamp-Lintfort, Germany

Postby Andrew Davie » Thu Apr 12, 2007 7:02 pm

Klaas Robers wrote:But Andrew, have you calculated then the size of the disc that you are going to make? 2 mm is far too large. I remember that Denis needed for his Baird discs (half a meter in diameter, 30 lines) holes of 0,7 mm wide. So you will end up with a disc of one meter!!

Denis solved this by punching the (square) holes himself and have 30 circular holes on an extra rim around the disc, that guided him to punch the holes. I will see if I have still a photograph of his punching action.


OK, well I understand that 2mm is too large, but I hadn't really thought too much about the real implications. I have to have a 2:3 aspect ratio, so I'll take the circumference at the outside of the disc and divide by 32 (as 32 'lines' have to fit into the disc). That then multiplied by 3/2 gives me the required width of the scan hole area.

Assume a disc of diameter 40cm, then the circumference is roughly 125cm and the height of the image is therefore roughly 3.9cm. That makes the width about 2.6cm. Divide by 32 scanlines that gives an area for each scanline of .8125mm.. right? And if I allow 1/3 overlap then the holes can be a 1/3 bigger and fit into the same width. That's 1.2mm.

OK, 2mm isn't going to work --- unless I change the aspect ratio -- which I could do, just for the experiment. Given a 2mm hole with a 1/3 overlap, that would be 1.3mm x 32 = 4.2cm wide image, 3.9cm high -- a very roughly square (well, wedge-shaped) image.

I'm more or less interested more in seeing how overlap of nipkow holes work -- and in particular overlapping them such that the area/width of the overlaps and non-overlaps is optimised -- than I am in getting a "correct" club-format image, at this stage.
User avatar
Andrew Davie
"Gomez!", "Oh Morticia."
 
Posts: 1511
Joined: Wed Jan 24, 2007 4:42 pm
Location: Hobart, Tasmania, Australia

Postby Klaas Robers » Thu Apr 12, 2007 7:06 pm

Oh Volker, lucky you!

Nowadays some laser cutting machines can even make square holes of 1mm or less. The newer discs of Denis for the Daily Express monitor are indeed made directly. But for the Baird replicae he also had the problem that no holes of less than 2 mm could be cut by laser. So this is the reason of his punching actions.

Andrew,
when making these calculations you should use the rough horizontal centre of your picture field to calulate the circumference, then calculate the heighth, then the width and then the size of the holes.
Then you can calculate if the edges of the picture are where you wanted it and eventually reposition the horizontal centre of your picture field. Then do everything again.
The aspect ratio should be indeed 3:2. In practice the heigth of the syncpulse (black bar) is making the picture slightly more a square.

Some overlap is needed for circular holes to make the ilumination of a white picture more even. For square holes this is not needed, but then the precise horizontal position is more important to avoid black lines or bright lines.
Last edited by Klaas Robers on Thu Apr 12, 2007 7:19 pm, edited 1 time in total.
User avatar
Klaas Robers
Frankenstein was my uncle.
 
Posts: 1465
Joined: Wed Jan 24, 2007 8:42 pm
Location: Valkenswaard, the Netherlands

Postby DrZarkov » Thu Apr 12, 2007 7:13 pm

Klaas, write what would be best suitable for the club, and I will ask how much it will be to make some more for the club.

BTW: I'm living about 2 km from the dutch border, my wife is dutch and works in Venlo, so shipping to Valkeswaard would be very easy and cheap:-)
User avatar
DrZarkov
I think I've had a cranial implosion.
 
Posts: 1037
Joined: Mon Feb 19, 2007 11:28 pm
Location: Kamp-Lintfort, Germany

Postby Klaas Robers » Thu Apr 12, 2007 7:29 pm

Volker,
please discuss this with Vic Brown. As far as I know he is searching for a new disc supply for the club. When this can be done via friends this is normally much easier done. In the same way I am supplying the PCBs that are also made by a Ham friend of mine here in NL. Venlo - Valkenswaard is about 45 minutes by car.

However, I am not attending the Convention this year. The fast sea link from Hoek van Holland to Harwich (3,5 hour sailing time) is taken out of operation, too expensive in fuel, and the regular shipping takes twice the time. I haven't found yet a convenient way to go, including transportation of my precious equipment for demonstration, so the airplane is not an option.
User avatar
Klaas Robers
Frankenstein was my uncle.
 
Posts: 1465
Joined: Wed Jan 24, 2007 8:42 pm
Location: Valkenswaard, the Netherlands

Postby DrZarkov » Thu Apr 12, 2007 8:42 pm

For me too it is a little bit difficult, and I'm afraid I can't come. Some years ago I was at an exhibition in London, carrying my modified Amiga-computer in a suitcase. We went by car to Calais and left our car there, then with the ferry (P&O) to Dover, and by train to London. That was horror, my 15 kg computer and another suitcase with clothes. So I would prefer to come by car. But that is a question of time and money...
User avatar
DrZarkov
I think I've had a cranial implosion.
 
Posts: 1037
Joined: Mon Feb 19, 2007 11:28 pm
Location: Kamp-Lintfort, Germany

Postby Viewmaster » Thu Apr 12, 2007 10:32 pm

Klaas Robers wrote:Denis solved this by punching the (square) holes himself and have 30 circular holes on an extra rim around the disc, that guided him to punch the holes. I will see if I have still a photograph of his punching action.


There is something to be said for having a very large dia. rough false disc mounted with the actual Nipkow disc whilst making it.
In this way the guide holes will enable even better accuracy for the Nipkow holes.

Say, for example, the Nipkow disc was 12 inch dia. If the rough outer disc was, say 48 inches dia with the guide holes drilled in its outermost edge, then any error in angular spacing would be reduced by a 1/4 for Nipkow disc.
Kinda 4 to1 leverage reduction of errors.

This larger outer false disc need not be accurately formed nor balanced...pre fabricated ali or plastic sheet etc bolted together would suffice as long as the centre spindle fitted tight with the Nipkow disc (to avoid axial mis alignment.)

Albert.
User avatar
Viewmaster
Frankenstein was my uncle.
 
Posts: 1288
Joined: Fri Apr 06, 2007 4:50 am
Location: UK Midlands

Postby Andrew Davie » Fri Apr 13, 2007 12:56 pm

I posted the problem/question about hole spacing to the [sliderule] list on yahoo -- they like solving this sort of problem :wink: -- and received the following answer/solution...

Steve Treadwell wrote:Hi Andrew,

I'm not sure I understand the problem correctly, but I'll state what I
think you want solved, then show how I solved it:

You have several circles of the same radius r with centers all on a
straight horizontal line, and with their centers equidistant from one
another at a distance d. The circles overlap; i.e., the circle
diameter is greater than d, and you want the area of each overlap (or
intersection) to be equal to one fourth of the area of a circle, so
that if you pick one particular circle, the sum of the two areas where
it is overlapped by adjacent circles is equal to the area of the
circle which is not overlapped.

Here's the solution to the above problem:

Imagine a vertical line (perpendicular to the line through the circle
centers) drawn through the center of one of the overlap areas, from
one point of intersection of the two circles to the other point of
intersection. Considering only one of the intersecting circles and
the line we have just drawn, we see that this line cuts off a segment
of the circle. We know the area of this segment is one half of the
intersection area = (½) (pi*r^2)/4. There is a formula I found in the
"Handbook of Chemistry and Physics" relating the area of a segment of
a circle to the distance from the center of the circle to the segment
line, but this leads to a messy equation, so I used a simpler formula
which relates the area of a segment to the angle in radians between
the two radii from the center of the circle to the ends of the segment
line. That equation is:
Aseg = (½)(r^2)(a – sin a) where Aseg is the area of the segment,
which we know, r is the radius of the circle, which we also know, and
a is the angle (in radians) between the radii, which is unknown.
Derivation of this equation is left as an exercise for the reader <g>.
So we now have:
Aseg = (½)(pi*r^2)/4 = (½)(r^2)(a – sin a) which simplifies to:
pi/4 = 0.785 = a – sin a

Here's where the slide rule is useful, especially a vector type with
angles in radians I used a Lafayette 99-7102 Vectorlog which has an
Sr scale on the slide with angles in radians from 0 to 1.57 and with
sines read on the Q scale (also on the slide, and labeled 0 to 10, but
read as 0 to 1 for reading sines). We need to move the cursor along
the slide to find a position where a – sin a is equal to 0.785. It
becomes immediately obvious that there is no such position in the 0 to
90 degree range, so angle a must be greater than pi/2. We need to
modify our equation – for angles greater than pi/2, the actual angle
is pi minus the angle on the Sr scale, but the actual sine is equal to
the value read on Q.
pi/4 = (pi – a) – sin a or
(3*pi)/4 = 2.36 = a + sin a
Using this equation, we find that at Sr = 1.375, sin Sr = 0.981 and Sr
+ sin Sr = 2.356, so
actual a = pi - 1.375 = 1.767 radians = 101.2 degrees

We now need to find the distance from the center of the circle to the
vertical line which cuts off the segment. Consider the right triangle
formed by a line from the center of the circle to the center of the
segment line, the top half of the segment line, and a radius from the
center of the circle to the top of the segment line. We know that the
angle between the radius and the line from the center of the circle to
the center of the segment line is 1.767/2 radians, or 50.6 degrees,
and because it's a right triangle, the angle between the radius and
the segment line must be (90 – 50.6) degrees = 39.4 degrees. Using
the law of sines, we can write:
d/sin (39.4 degrees) = r/sin (90 degrees) where d is the distance
from the center of the circle to the center of the segment line.
Solving this by proportion using the Stheta and Q scales gives d = 0.635 r
The distance between the centers of two adjacent circles is double
this distance, or 1.27 times the radius.

Whether I understood what you wanted or not, it was a nice break from
doing my tax return, so thanks. :D If I didn't understand correctly,
if you can show me what you are looking for, I'll have another crack
at it.

Steve Treadwell
User avatar
Andrew Davie
"Gomez!", "Oh Morticia."
 
Posts: 1511
Joined: Wed Jan 24, 2007 4:42 pm
Location: Hobart, Tasmania, Australia


Return to Mechanical NBTV

Who is online

Users browsing this forum: No registered users and 6 guests

cron