so what about medium format bellowsChrisR wrote:Lothar - 52mm bellows? Often bellows + adapters is just too ungainly, or the diameter of the bayonet is too small.

http://www.ebay.de/itm/P-Six-Pentacon-T ... Swax5YrrZg

**Moderators:** rjlittlefield, ChrisR, Chris S., Pau

so what about medium format bellowsChrisR wrote:Lothar - 52mm bellows? Often bellows + adapters is just too ungainly, or the diameter of the bayonet is too small.

http://www.ebay.de/itm/P-Six-Pentacon-T ... Swax5YrrZg

It's not the only best way to go , Lou.

*never *better. It is better in the cases I explained. In the particular case of a number of sets of the *same *tubes, only, yes you are correct *which is why some series quoted in the original post were irregular.* If you're worried about having zero redundancy in multiple sets then yes have a set of primes, or, sets with different lengths in each.

Particularly before you buy your second set, you have arithmetically uneven steps, which some folk would describe as*not better*.

It depends what you care about.

I have about 10 of the long tube (~20mm) of a Nikon K set, with fewer shorter ones. It's quite handy having them like that. I cerrtainly wouldn't want them all different lengths.

No, notChris, it is never better to make regular series.

Particularly before you buy your second set, you have arithmetically uneven steps, which some folk would describe as

Not always, only if you have multiples of the same tubes. 5,10,15 wasn't suggested.Never 5, 10, 15. You'll always have more choices with non-integer multiples.

It depends what you care about.

I have about 10 of the long tube (~20mm) of a Nikon K set, with fewer shorter ones. It's quite handy having them like that. I cerrtainly wouldn't want them all different lengths.

Chris R

If you have 40, 80 and 160,

then 5, 10, 20,

where are the unnecessary gaps?

You have N rings, spacing S, here S=5.

the maximum number of permutations is 2^N

You can have lengths of every integer multiple of S, with no duplications, which gives from zero to

((2^N)-1) * S

if you want closer steps, then you need a slimmer skinniest tube, but the total length you can make is then shorter. You can't get any more different steps.

You're picking the tubes as a binary selection

000000 0.... 0

000001 1.... 5

000010 2 ....10

000011 3.... 15

..

111111 63.. 315

then 5, 10, 20,

where are the unnecessary gaps?

You have N rings, spacing S, here S=5.

the maximum number of permutations is 2^N

You can have lengths of every integer multiple of S, with no duplications, which gives from zero to

((2^N)-1) * S

if you want closer steps, then you need a slimmer skinniest tube, but the total length you can make is then shorter. You can't get any more different steps.

You're picking the tubes as a binary selection

000000 0.... 0

000001 1.... 5

000010 2 ....10

000011 3.... 15

..

111111 63.. 315

Chris R

My 5, 9, 29 gives 21 lengths from 5 to 40mm. Your 5, 10, 20 just gives seven choices. And all seven are included in my series. You lose nothing and gain many options.

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Lou is optimizing under different constraints, by allowing to purchase multiple sets of a small number of different lengths.

So, ChrisR has 6 tubes all of different lengths; Lou has 12 tubes = 4 identical sets of 3.

The idea of only a few different rings with carefully selected lengths, purchased in multiple quantities, is an interesting concept to explore.

So, here's a start...

Allowing 5 sets drops the large gaps to size 2 mm. At 6 sets you have enough to get all values beyond 41 mm, but in the early going you're still stuck with a couple of gaps of size 4 mm, plus several 3's and 2's.

As an alternative, we might consider 5, 7, 29, where it takes only 3 sets (not 4) to get max gaps of size 3 mm.

Or better yet, use 5, 7.5, 35, which with 3 sets will cover all multiples of 2.5 out to 137.5. If the criterion is smallest gaps with 3 sets of 3 rings, then this sequence beats the others (despite that 35 is a integer multiple of 5).

If we only allow 2 sets of 3 different, then I'll submit 5, 7.5, and 22.5, which covers all multiples of 2.5 from 5 to 65.

BTW, it's no coincidence that 22.5 = 2*5+2*7.5-2.5 and 35 = 3*5+3*7.5-2.5 . The construction method for these last two sequences is that the second ring is 1.5 times the first ring, and the third ring is all the others, in all sets, minus half the first ring.

The above sets of 6 rings (2 sets of 3) are pretty good, but they're not as good as what we could do with 2.5, 3.75, 5, 10, 20, 40. That's a 1:2:4:8:16 sequence extended with an intermediate ring instead of an impossibly thin one to start. That sequence would cover every 1.25 mm from 2.5 to 78.75 . Add in an 80 to extend full coverage out to 158.75.

At the moment, if I were going to request custom lengths I'd start with something like this last set, 1:2:4:8:16 augmented with a 1.5.

Then if the seller balked at too many different rings, I'd suggest the other approach.

--Rik

So, ChrisR has 6 tubes all of different lengths; Lou has 12 tubes = 4 identical sets of 3.

The idea of only a few different rings with carefully selected lengths, purchased in multiple quantities, is an interesting concept to explore.

So, here's a start...

If I've counted correctly, it takes at least 4 sets to do that well.Lou Jost wrote:Maybe 5, 9, 29 if limited to three. With multiple sets you could then cover 5, 9, 10, 14, 15, 18, 19, 20, 23, 25, 27, 28, 29, 30, 33, 34, 35, 36, 38, 39, 40, etc.

That's a tempting intuition, but I disagree. My tabulation shows a gap of 3 mm between 53 and 56, and between 58 and 61, and repeating regularly after those every 29 mm.Coverage would be virtually complete from this point onward.

Allowing 5 sets drops the large gaps to size 2 mm. At 6 sets you have enough to get all values beyond 41 mm, but in the early going you're still stuck with a couple of gaps of size 4 mm, plus several 3's and 2's.

As an alternative, we might consider 5, 7, 29, where it takes only 3 sets (not 4) to get max gaps of size 3 mm.

Or better yet, use 5, 7.5, 35, which with 3 sets will cover all multiples of 2.5 out to 137.5. If the criterion is smallest gaps with 3 sets of 3 rings, then this sequence beats the others (despite that 35 is a integer multiple of 5).

If we only allow 2 sets of 3 different, then I'll submit 5, 7.5, and 22.5, which covers all multiples of 2.5 from 5 to 65.

BTW, it's no coincidence that 22.5 = 2*5+2*7.5-2.5 and 35 = 3*5+3*7.5-2.5 . The construction method for these last two sequences is that the second ring is 1.5 times the first ring, and the third ring is all the others, in all sets, minus half the first ring.

The above sets of 6 rings (2 sets of 3) are pretty good, but they're not as good as what we could do with 2.5, 3.75, 5, 10, 20, 40. That's a 1:2:4:8:16 sequence extended with an intermediate ring instead of an impossibly thin one to start. That sequence would cover every 1.25 mm from 2.5 to 78.75 . Add in an 80 to extend full coverage out to 158.75.

At the moment, if I were going to request custom lengths I'd start with something like this last set, 1:2:4:8:16 augmented with a 1.5.

Then if the seller balked at too many different rings, I'd suggest the other approach.

--Rik

Edit to add: This is exactly what we need if we want to use Raynox and other short-mount lenses as tube lenses focused at infinity.

You add an additional and very practical constraint: find a set that balances gap sizes against the number of replica sets one needs to buy.

I think soelf wrote:Haven't you guys heard of bellows?

Tubes have some advantages over bellows: simpler, more rigid and in principle less expensive and also can have other advantages to put optics inside. Of course bellows advantage is the easy length regulation

In optical benches tubes are much more usual, did you see the Bratcam setup of ChrisS? That kind of tubes are expensive, so having inexpensive and flexible options seems desirable

Pau

It doesn't say 5mm, those are ratios not mm.Look at my example above. The 5mm gaps are big, and unnecessary;

1,2,4,... are ratios.

You can have 3,6,12, or any multiple you like which suits. That will maximize length and number of even steps, - if that's what you're concerned about.

Once again I will not defend something I didn't write.

Lou's confined aim is defined by the results the set gives, therefore it's 100% successful!

I'm not advocating any particular progression. It seems perfectly clear there is no single best solution, as considered since post #1 - and before. I don't find it worthwhile to consider that there is, confronted with errors of logic, fact and comprehension.

That's why I asked what people would prefer.

One can optimise

- maximum number of lengths

overall length

minimum steps

arithmetic progression

geometric progression

Fibonacci progression

- minimum number of different length tubes

minimum number of sets of tubes

minimum cost

. . (and surely more)

Chris R

"You have N rings, spacing S, here S=5....

You can have lengths of every integer multiple of S...

if you want closer steps, then you need a slimmer skinniest tube"

I was just pointing out that you can avoid many of the gaps in that arrangement by using rings that are not integer multiples of each other (especially at the longer end of the series, where there will be NO gaps greater than 0.5mm). If you use tubes that aren't integer multiples of each other, then you do not need a slimmer skinniest tube to reduce the size of the gaps at the longer end of the series.

The problem I'm trying to solve is not an arbitrary one, Chris. Many of us need to construct tubes that can reach a particular given length (for example, to mount Raynoxes as tube lenses).

Yes, I realise that you misread. Ratios are not millimetres. I am repeating myself, again.

You can optimise one aspect to the detriment of others, sure. What you're saying was understood and covered a few posts back.

The suggestion was*not arbitrary* so - use 4-5 sets of 5, 9 and 29 to focus a 208mm lens at infinity??* No thankyou.*

--

The "Long screw" above would be rigid, adjustable and probably cheaper than a normal helicoid. (I don't have any to compare with).

I have scratched a few 52mm filters - with glass removed they make skinny 52mm rings. Four would be about 10mm.

Close enough for a NA 0.8.

I can't remember the length required for the slightly telephoto Raynox 150, after a camera. Tubes of say 10mm, 20mm, 40mm and 80mm, plus the empty filters, would be easy to work with and give 2.5mm steps to 160mm.

You can optimise one aspect to the detriment of others, sure. What you're saying was understood and covered a few posts back.

Then you need an adjustable element to get it exact - or state a number you'll accept as close enough, and in what range. The goalposts are moved.The problem I'm trying to solve is not an arbitrary one, Chris. Many of us need to construct tubes that can reach a particular given length (for example, to mount Raynoxes as tube lenses).

The suggestion was

--

The "Long screw" above would be rigid, adjustable and probably cheaper than a normal helicoid. (I don't have any to compare with).

I have scratched a few 52mm filters - with glass removed they make skinny 52mm rings. Four would be about 10mm.

Close enough for a NA 0.8.

I can't remember the length required for the slightly telephoto Raynox 150, after a camera. Tubes of say 10mm, 20mm, 40mm and 80mm, plus the empty filters, would be easy to work with and give 2.5mm steps to 160mm.

Chris R

For any desired length greater than 40mm, you can use a big tube for most of it and leave the last 40mm to be covered by different combinations of 5, 9, 29mm, exact to within 0.5mm.

Yes, a filter ring can get you even closer, and I use this too. The nice thing about them is that there is no standard width, so filters from different manufacturers are not integer multiples of each other. So you can get as close as you want to a given length, by combining rings of different manufacturers. Same principle as the one I suggest for rings.

So, if Rik was right, that's 4 sets (Edit : or some other multiple ) of tubes plus an unspecified "big tube"?!!For any desired length greater than 40mm, you can use a big tube for most of it and leave the last 40mm to be covered by different combinations of 5, 9, 29mm, exact to within 0.5mm.

Last edited by ChrisR on Wed Mar 08, 2017 10:26 am, edited 2 times in total.

Chris R