Another one for the R/Os - Ships Nostalgia
19:19

Welcome
Welcome!Welcome to Ships Nostalgia, the world's greatest online community for people worldwide with an interest in ships and shipping. Whether you are crew, ex-crew, ship enthusiasts or cruisers, this is the forum for you. And what's more, it's completely FREE.

Click here to go to the forums home page and find out more.
Click here to join.
Log in
User Name Password

Another one for the R/Os

User Tag List

Reply
 
Thread Tools
  #1  
Old 18th November 2006, 13:23
K urgess K urgess is offline
user
 
Join Date: Aug 2006
My location
Posts: 83
Another one for the R/Os

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.

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.

By the way. Has anybody ever suggested a forum for sparkies?
There are severeal threads knocking about that could do to be lumped together.
Attached Images
File Type: jpg Resistor - 001s.jpg (43.7 KB, 73 views)
Reply With Quote
  #2  
Old 18th November 2006, 14:00
benjidog benjidog is offline
member
 
Join Date: Oct 2005
My location
Posts: 17
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
Reply With Quote
  #3  
Old 18th November 2006, 14:23
K urgess K urgess is offline
user
 
Join Date: Aug 2006
My location
Posts: 83
So your answer would be

0.0399993 ohms?

Eh?

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.......
Reply With Quote
  #4  
Old 18th November 2006, 14:32
benjidog benjidog is offline
member
 
Join Date: Oct 2005
My location
Posts: 17
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".

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
Reply With Quote
  #5  
Old 18th November 2006, 14:44
K urgess K urgess is offline
user
 
Join Date: Aug 2006
My location
Posts: 83
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.
Reply With Quote
  #6  
Old 18th November 2006, 16:08
mikeg's Avatar
mikeg mikeg is offline  
Senior Member
Organisation: Merchant Navy
Department: Radio Officer
Active: 1966 - 1987
 
Join Date: Aug 2006
Posts: 1,264
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.
Reply With Quote
  #7  
Old 18th November 2006, 16:40
benjidog benjidog is offline
member
 
Join Date: Oct 2005
My location
Posts: 17
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
Reply With Quote
  #8  
Old 18th November 2006, 16:44
K urgess K urgess is offline
user
 
Join Date: Aug 2006
My location
Posts: 83
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.

PS curses Brian beat me to it again. Too much thunking time!
Attached Images
File Type: jpg Resistor - 002s.jpg (57.4 KB, 41 views)
Reply With Quote
  #9  
Old 18th November 2006, 16:49
mikeg's Avatar
mikeg mikeg is offline  
Senior Member
Organisation: Merchant Navy
Department: Radio Officer
Active: 1966 - 1987
 
Join Date: Aug 2006
Posts: 1,264
Quote:
Originally Posted by benjidog View Post
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
Reply With Quote
  #10  
Old 18th November 2006, 16:55
mikeg's Avatar
mikeg mikeg is offline  
Senior Member
Organisation: Merchant Navy
Department: Radio Officer
Active: 1966 - 1987
 
Join Date: Aug 2006
Posts: 1,264
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
Reply With Quote
  #11  
Old 18th November 2006, 17:04
K urgess K urgess is offline
user
 
Join Date: Aug 2006
My location
Posts: 83
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.
Reply With Quote
  #12  
Old 18th November 2006, 22:39
pete's Avatar
pete pete is offline  
Senior Member
 
Join Date: Mar 2005
My location
Posts: 935
Fubar.........Keep remembering Napoleon XV and those nice young men in their clean white coats............sigh, days of my youth...........................pete
Reply With Quote
  #13  
Old 20th November 2006, 19:25
mikeg's Avatar
mikeg mikeg is offline  
Senior Member
Organisation: Merchant Navy
Department: Radio Officer
Active: 1966 - 1987
 
Join Date: Aug 2006
Posts: 1,264
The definitive answer my friend is blowing in the wind:

http://www.radioelectronicschool.net...be_problem.pdf

Have fun.

Mike
Reply With Quote
  #14  
Old 20th November 2006, 19:33
Gulpers's Avatar
Gulpers Gulpers is offline   SN Supporter
ex-Denholm Moderator
Organisation: Merchant Navy
Department: Navigation
Active: 1972 - 1981
 
Join Date: Sep 2005
My location
Posts: 11,591
Bugger me!

What was the question?
__________________
Ray
. . . . A closed mouth gathers no feet!
Reply With Quote
  #15  
Old 20th November 2006, 19:38
Mad Landsman's Avatar
Mad Landsman Mad Landsman is online now
Senior Member
 
Join Date: Dec 2005
Posts: 6,831
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.
__________________
Malcolm.
Reply With Quote
  #16  
Old 20th November 2006, 19:45
K urgess K urgess is offline
user
 
Join Date: Aug 2006
My location
Posts: 83
Well, well.

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

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.

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

Don't be naughty Clockman!
Reply With Quote
  #17  
Old 20th November 2006, 23:05
benjidog benjidog is offline
member
 
Join Date: Oct 2005
My location
Posts: 17
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.

Brian
Reply With Quote
  #18  
Old 22nd November 2006, 18:38
Trevorw's Avatar
Trevorw Trevorw is offline  
Senior Member
 
Join Date: Jun 2006
Posts: 269
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!
Reply With Quote
  #19  
Old 22nd November 2006, 18:50
K urgess K urgess is offline
user
 
Join Date: Aug 2006
My location
Posts: 83
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.
Reply With Quote
  #20  
Old 22nd November 2006, 19:10
Tmac1720's Avatar
Tmac1720 Tmac1720 is offline  
Senior Member
Organisation: Merchant Navy
Department: Engineering
Active: 1964 - 1997
 
Join Date: Jun 2005
My location
Posts: 7,014
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.
__________________
[SIGPIC][/SIGPIC]Oul hand
Money can't buy happiness but it's more comfortable to cry in a Porsche than a Skoda
Reply With Quote
  #21  
Old 22nd November 2006, 19:51
Ron Stringer's Avatar
Ron Stringer Ron Stringer is online now
Senior Member
Organisation: Merchant Navy
Department: Radio Officer
Active: 1960 - 1966
 
Join Date: Mar 2005
My location
Posts: 6,508
Redifon Receiver

Quote:
Originally Posted by Marconi Sahib View Post
definitely Atalanta, but I'm prejudiced.

Never seen a Redifon R50M.
Now weren't you the lucky one, Fubar?
__________________
Ron
_____________________________________________

Never regret growing older. It is a privilege denied to many. Don't worry about old age - it doesn't last.
Reply With Quote
  #22  
Old 22nd November 2006, 19:54
Ron Stringer's Avatar
Ron Stringer Ron Stringer is online now
Senior Member
Organisation: Merchant Navy
Department: Radio Officer
Active: 1960 - 1966
 
Join Date: Mar 2005
My location
Posts: 6,508
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.
__________________
Ron
_____________________________________________

Never regret growing older. It is a privilege denied to many. Don't worry about old age - it doesn't last.
Reply With Quote
  #23  
Old 22nd November 2006, 20:30
K urgess K urgess is offline
user
 
Join Date: Aug 2006
My location
Posts: 83
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?
Reply With Quote
  #24  
Old 22nd November 2006, 22:06
Ron Stringer's Avatar
Ron Stringer Ron Stringer is online now
Senior Member
Organisation: Merchant Navy
Department: Radio Officer
Active: 1960 - 1966
 
Join Date: Mar 2005
My location
Posts: 6,508
Quote:
Originally Posted by Marconi Sahib View Post
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.
__________________
Ron
_____________________________________________

Never regret growing older. It is a privilege denied to many. Don't worry about old age - it doesn't last.
Reply With Quote
  #25  
Old 22nd November 2006, 23:44
K urgess K urgess is offline
user
 
Join Date: Aug 2006
My location
Posts: 83
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.
Reply With Quote
Reply


Currently Active Users Viewing This Thread: 1 (0 members and 1 guests)
 
Thread Tools

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is Off
HTML code is Off


Similar Threads
Thread Thread Starter Forum Replies Last Post
Looking for British Sparks captainrodaway Looking for Old Shipmates 39 17th May 2012 12:57
Sparks and pics and braid Doug H The Radio Room 60 24th April 2012 00:03
Hi to all RO's tonycooper Say Hello 17 22nd August 2005 23:34



Support SN


Powered by vBulletin® Version 3.8.8
Copyright ©2000 - 2020, vBulletin Solutions, Inc.
User Alert System provided by Advanced User Tagging v3.1.0 (Pro) - vBulletin Mods & Addons Copyright © 2020 DragonByte Technologies Ltd.
vBulletin Security provided by vBSecurity v2.2.2 (Pro) - vBulletin Mods & Addons Copyright © 2020 DragonByte Technologies Ltd.