Another one for the R/Os

K urgess
18th November 2006, 14:23
Not sure if putting this in the quiz forum is fair because I have no idea of the solution.

For those who have difficulty with the drawing the array is a cube with voltage applied at diagonally opposite corners.

Maybe the answer is very simple or maybe one should apply Maxwell's, Thevenin's or Norton's theorems either seperately or together.:sweat:

Maybe someone has the solution already complete for our enjoyment.

I've never had the patience, since the answer didn't come easily, to try and work it out. I did, once, try to build it and cheat but ran out of resistors.(Sad)

By the way. Has anybody ever suggested a forum for sparkies? (Thumb)
There are severeal threads knocking about that could do to be lumped together.

benjidog
18th November 2006, 15:00
I think I know how you work this out Fubar. You use the formula for resistors in parallel which I can't write here because it needs fractional notation. Basically it's the 1/R(equivalent = 1/R1 + 1/R2 + .....

So you need to work out all the paths from + to - through the three-dimensional network, work out the resistance of each path and apply the formula.

So the paths are:

R1+R6+R9 = 300 ohms
R1+R6+R11+R8+R7+R4+12 = 500 ohms
R1+R6+R11+R8+R7+R4+R10 = 700 ohms
R1+R5+R10 = 300 ohms
R1+R5+R4+R7+R12 = 500 ohms
R1+R5+R4+R7+R12 = 700 ohms
R2+R11+R9 = 300 ohms
R2+R8+R12 = 300 ohms
R2+R11+R6+R5+R10 = 500 ohms
R2+R11+R6+R5+R4+R7+R12 = 700 ohms
R2+R8+R7+R4+R10 = 500 ohms
R2+R8+R7+R4+R5+R6+R9 = 700 ohms
R3+R7+R12 = 300 ohms
R3+R7+R8+R11+R6+R5+R10 = 700 ohms
R3+R7+R8+R11+R6+R5+R10 = 700 ohms
R3+R4+R10 = 300 ohms
R3+R4+R5+R6+R9 = 500 ohms
R3+R4+R5+R6+R11+R8+R12 = 700 ohms

Blimey - my brain hurts after that.

I will leave it to you to do the 1/r1 etc formula on it! Have to take the dog for a walk before it starts raining again.

Brian

K urgess
18th November 2006, 15:23
So your answer would be

0.0399993 ohms?

Eh?(?HUH)

But R1, R2 and R3 are already in parallel with all the other combinations so doesn't that modify their value before you start?

In actual fact each path from positive to negative is made up of only three resistors in each chain so you could say that there 6 x 300 ohm paths in parallel.

They're coming to take me away, haha, heehee, to the funny farm, dada, deedee.......(==D)

benjidog
18th November 2006, 15:32
I must admit I have been having second thoughts about my answer while taking the dog for a walk and getting pelted with hailstones and running back home.

I think maybe the longer pathways may be invalid as the electrons would maybe have to flow "uphill".(Cloud)

I am sure Charles Wheatstone would have had no trouble with this question! Never got anything that complicated to work out as an electrician thank goodness!

Maybe there is a real genius with the answer out there somewhere?

Tell you what - would you like me to buy you 12 100 ohm resistors for Christmas? ;)

Regards,

Brian

K urgess
18th November 2006, 15:44
I have to venture into the "storage" area this afternoon. In other words the loft 'cos I'm trying to make room for plumbers to gain access.

One of my jobs is to dispose of boxes full of 74 series ICs, capacitors, resistors, PLCs, PSUs, etc., etc., that have been there for 20 years gathering dust. I'll see if I can find enough 100 ohm resistors knocking about before I dump it all. It's not a common size unfortunately. 1k DIL resistors I can find easily.

It may be the simplest solution.[=P]

mikeg
18th November 2006, 17:08
There is an input node with 3 resistors connected to it.
The other end of those 3 resistors must all be at the same
potential. There is an output node with 3 resistors
connected to it. The other end of those 3 resistors must
all be at the same potential. There are 6 other resistors
that, each connect between these two internal nodes, so
there must be an equal drop across each of these 6 resistors.

To solve for the total resistance or the current in each of
the resistors, just connect all points that must be at an
equal resistance, together. This will not change any
current. But with this pair of connections, the whole
circuit collapses into 3 resistors in parallel, that are in
series with 6 resistors in parallel, that are in in series
with 3 resistors in parallel.

benjidog
18th November 2006, 17:40
Hmmmm....

Not sure I agree with you there Mike.

If we consider a simpler circuit with three resistors each 100 ohm. Point A is joined to point B by two pathways - the first is one 100 resistor, the second is two 100 ohm resistors in series. The potential at the midpoint of the resistors in series would be midway between the potential at point A and point B.

Does this not contradict the argument in your second sentence?

Fubar - you have a lot to answer for posing this damn question!!!

Brian

Brian

K urgess
18th November 2006, 17:44
Interesting proposition.

Yes the voltages across R1, R2 & R3 must be the same as the voltages across R9, R10 & R12 but what is that voltage supposing the voltage across the network to be 100 volts for simplicity sake.

The modifying factor is the centre network of resistors which cannot be said to be in parallel because each one connects to a different resistor at each end node of three. R1 connects to R5 in parallel with R6. R5 connects to R10 and R6 connects to R9 so what is in parallel with what?

Attached "simplification" (?) shows the dilemma.(==D)

PS curses Brian beat me to it again. Too much thunking time!

mikeg
18th November 2006, 17:49
Hmmmm....

Not sure I agree with you there Mike.

If we consider a simpler circuit with three resistors each 100 ohm. Point A is joined to point B by two pathways - the first is one 100 resistor, the second is two 100 ohm resistors in series. The potential at the midpoint of the resistors in series would be midway between the potential at point A and point B.

Does this not contradict the argument in your second sentence?

Fubar - you have a lot to answer for posing this damn question!!!

Brian

Brian

That should have read:

To solve for the total resistance or the current in each of
the resistors, just connect all points that must be at an
equal *potential*, together.

PS I also got a bit of a head working this out so I went to another forum for a second opinion that about agrees with my thoughts on this.

Mike

mikeg
18th November 2006, 17:55
Another reply say's why not simply first remove R5,R8.
This, with symmetry considerations, will reduce the network to 2
dividers to which you can now reconnect R5//R8.
What think ye?

Mike

This is a bit like that tune that gets into your head and repeats with variations
(==D)

K urgess
18th November 2006, 18:04
Mike

If you do that shouldn't you remove 3 resistors from the centre array? One from each inner end of R1, R2 & R3?

I have a sneaky feeling I know the answer to this without working it out or maybe I've just lost my nerve.

1100100 - Binary is so much easier.(Whaaa)

pete
18th November 2006, 23:39
Fubar.........Keep remembering Napoleon XV and those nice young men in their clean white coats............sigh, days of my youth...........................pete

mikeg
20th November 2006, 20:25
The definitive answer my friend is blowing in the wind:

http://www.radioelectronicschool.net/files/downloads/resistor_cube_problem.pdf

Have fun.

Mike

Gulpers
20th November 2006, 20:33
Bugger me!

What was the question? (Jester)

Mad Landsman
20th November 2006, 20:38
I've been sat on the sidelines watching this one. every so often breaking off with what I have been doing during the day the grab my calculator.
Thank goodness someone has worked it out - The second solution appeals most.
Just a thought (tongue in cheek) - What difference if any would it make if one constucted the cube with 10% tolerance resistors?
I'll go back into my corner now.

K urgess
20th November 2006, 20:45
Well, well.

As simple as that. For voltage read current. It's the approach that counts.(Thumb)

That original diagram of mine has been knocking around in my college notes since 1974 when I took my electronics ticket. I never got around to solving it and now I don't have to. Great stuff.

I'm naturally drawn to the simpler solution. Equations have never been one of my strong points.(POP)

In this array that would make the answer 83.3333(recurring) ohms.

Don't be naughty Clockman!(Jester)

benjidog
21st November 2006, 00:05
Thank God someone provided a definitive answer.

I don't think we will take you up on your offer Ray - kind as it was! ;)

Now does anyone have a five-dimensional cube version of this puzzle? - that one was a bit too easy really.[=P]

Brian

Trevorw
22nd November 2006, 19:38
Never mind Ohms Law! During the late 50's and early 60's, which was the best receiver out of the following?

CR300
Mercury & Electra
Atalanta
R50M (Redifon)

Without doubt, the worst one of all was the stand-by 500Kc/s Rx, "Alert". It had no BFO, and if some idiot was transmitting in CW the whole thing was blocked out!

K urgess
22nd November 2006, 19:50
IMHO definitely Atalanta, but I'm prejudiced.

Wasn't the "Alert" a TRF and excellent homing device for U-boats? The "Monitor" was a great improvement.

The only good thing about the Mercury and Elettra was the ability to monitor 500 with a decent receiver while listening to GRL TFC LST.

Never seen a Redifon R50M.

Tmac1720
22nd November 2006, 20:10
Watt a shocking puzzle, I've been sat at ohm trying to un earth the current solution before some other bright spark got switched on to volt it was. (==D)

Ron Stringer
22nd November 2006, 20:51
definitely Atalanta, but I'm prejudiced.

Never seen a Redifon R50M.

Now weren't you the lucky one, Fubar?

Ron Stringer
22nd November 2006, 20:54
Trevorw
Sailed with the first 4 (Marconi) receivers and had the ill-luck to have to service the R50M and its equally-rubbish successor the R408. What a mess, don't think they had ever heard of selectivity or protection against IF breakthrough.

K urgess
22nd November 2006, 21:30
R408 rings a bell but I don't know where from. It may have been at college when doing electronics. Martec had all sorts of scrap cr*p for us to play with during practical sessions.

Does anyone remember what a "Seacall" was? Was it any relation to a "Pennant"? I feel a faint stirring of the grey cells but I don't think I had occasion to actually use either of 'em.

I did a coastal on the "City of Ripon" in '75 after getting an electronics ticket. That was a blast from the past. 1955 kit with the only concession to the "modern" age being a Raymarc 16 radar.

Does anyone remember what a "Nebula" looked like?

Ron Stringer
22nd November 2006, 23:06
R408 rings a bell.

It was an early attempt at an SSB receiver but used a long film as the bandspread/tuning dial presentation. Wound across the front window of the receiver as you tuned it. About as stable as the Atalanta but 15 years later.

Does anyone remember what a "Seacall" was?

Not a relation to the Pennant, which was a very good SSB receiver that was crippled by the fact that it only worked on the official 'paired' frequency of whatever HF SSB R/T channel to which the transmitter was tuned. That prevented cross-band working which was frowned upon by the regulators (as being in contravention of the approved use of frequencies) but common practice amongst HF coast stations. When used on paired channels, it couldn't be beaten and produced first quality link calls. If propagation conditions or interference forced you to work cross-band, you were dead.

The 'Seacall' was a multi-channel MF/HF selective calling receiver using the Siemens-developed multitone sequential selective calling system (Selcall). Adopted for international use in the early 1970s, the system was never a success internationally and was overtaken by the development (and adoption for the GMDSS) of DSC - digital selective calling. There was a MF-only version called 'Coastcall'.

Does anyone remember what a "Nebula" looked like?


The 'Nebula' was Marconi's badged version of the Eddystone 958 SSB receiver. Eddystone were owned by Marconi but developed the 958 for coast station and point-to-point use. It was cheaper than the Marconi 'Apollo' receiver and was adopted by Marconi and sold to markets where price was more important than performance, mainly to operators of FOC vessels and the like.

K urgess
23rd November 2006, 00:44
Thanks Ron

I recognise the description of the R408 but never had the "pleasure" of sailing with one. Oh yes I did! You can just see one of the handles on my profile pic which was taken on Big Geordie. I don't remember having any problems with it. I had problems with just about everything else on that ship.

The Pennant I sailed with but it seems to have been paired with an Atalanta as a SSB receiver. I usually used to "phone home" quite a lot but it appears I never used it all trip.

The Seacall was fitted between trips on one ship but I can't remember it receiving a call. The same ship had the Nebula but I don't remember a thing about it. I don't think I've got a radio room pic so I shall have to do some digging.

Ron Stringer
23rd November 2006, 12:44
[QUOTE=Marconi Sahib;90581]The Pennant I sailed with but it seems to have been paired with an Atalanta as a SSB receiver. I usually used to "phone home" quite a lot but it appears I never used it all trip.[QUOTE]

I can assure you that you would have remembered using it! It was self-tuning i.e. its frequency selection and everything else were driven by ledex motors from the 'Crusader' transmitter drive unit. The local oscillator was derived from the Crusdaders high stability, ovened, 5MHz crystal oscillator and employed the same frequency mixing principals as used to generate the transmission frequency (but with the appropriate offsets to allow for the channel send/receive frequency separations on each HF band, and for the receiver IF). Always came up absolutely spot-on the frequency, making the clarifier virtually unnecessary.

But, whenever you changed channels or bands on the transmitter it sounded as though the radio room was in the Great War trenches under machine gun attack, with all the ledex assemblies rattling away in the Crusader transmitter and the Pennant. What a maintenance nightmare.

Oh what a relief when the phase-locked loop came along.

mikeg
23rd November 2006, 14:04
Believe me, it's a good game changing a faulty ledex with complete wafer assembly with the pressure of a looming sailing deadline! How many of you had to give the shaft end screwdriver slot a slight shift to get drive I ponder? The wafer contacts were in a poor state obviously been sprayed many times with WD40 ugh!! plus the shaft centres were very worn giving inaccurate contact alignment. Oh happy days.

Mike

K urgess
23rd November 2006, 17:56
I can remember a lot of clicking of Ledex motors but probably just from the transmitters. I think the Crusader had one in the IF section or some such place.

Having just read through my overtime reports for Big Geordie (102 items on 11 pages) the only thing I did to the R408 was remove it from the console, check it, clean it and put it back. Mind you I appear to have almost rebuilt the Crusader main transmitter because of repeated intermittent problems.

Attached is something Ron asked me about a while ago -

"COVER PHOTO
MARCONI MARINE CO.s NEW FLAG
The lower half of the flag is royal blue, and the upper sky
blue, representing the conquest of sea and air by radio.
Superimposed in the centre, encroaching on both blues, is a
white lozenge carrying the letter 'M' in gold."

Ron Stringer
23rd November 2006, 19:34
Fubar,
Very many thanks for that flag. I have searched high and low for it without success. Tomorrow is our monthly MIMCo old farts lunch (liquid variety) in Chelmsford and I was hoping that someone would come up with one - I asked the others to see if they could find a copy about 3 months ago. I chase 'em up each month but so far without success. Mind you, with the combination of the booze and anno domini, by the time we get home afterwards I don't suppose most of us remember a blind thing about what was said there.

The Crusader had many 60V dc ledex motors each driving multi-wafer rotary switch assemblies - in the frequency generator unit and in the drive amplifier unit prior to the output stages. Some of the switches had multiple shafts connected by universal couplings. As Mikeg says, the very devil to fault-find, repair and realign afterwards. Each wafer was selecting from a number of different tuned circuit assemblies, some with solid state devices connected to them, others with 600V plus RF. Care was needed!
Thanks again for the flag - now I can have an avatar.

K urgess
23rd November 2006, 21:43
Ron

Glad you like the flag. Timely posting then.(Thumb)

The only training I did on the Crusader was a couple of days in Glasgow just before joining a ship. They were so tight fisted we never even got a circuit diagram. The excuse was that the manual was onboard with every transmitter.

As usual it was in at the deep end. By time I'd did the course I'd sailed with the beast twice.

Most trouble was on my last trip where most of the switch wafers in the final stage were knakkered. Plus someone had been at quite a bit of the kit onboard adjusting IF cans with a screwdriver. Lovely mess. Thankfully they hadn't got at the ledex switches.

GALTRA
24th November 2006, 22:08
The original !!!!

mikeg
25th November 2006, 01:50
Ron
Most trouble was on my last trip where most of the switch wafers in the final stage were knakkered. Plus someone had been at quite a bit of the kit onboard adjusting IF cans with a screwdriver. Lovely mess. Thankfully they hadn't got at the ledex switches.

Reminds me of a ship that I sailed twice on and it had this niggling IF instability on the transmitter that only happened once in a blue moon. Many R/O's had tried in the past with no success, I didn't manage to find it either but after quite an intensive bit of tracing I really thought I'd tracked it down to this one can, I carefully open it up and inside the can was a tiny piece of paper which said 'the faults not here!'
(Jester)