Author Topic: 2 pound gold "50th Anniversary of the end of WW2"  (Read 8103 times)

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Offline retired44

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2 pound gold "50th Anniversary of the end of WW2"
« on: November 19, 2009, 11:23:37 AM »
Hi there,
I have searched for hours online, google etc for the measurement of the thickness of this gold coin.
I have found plenty of good information on the coin but nowhere can I find any ref to the thickness.
If anyone on this forum has this coin or has a catalogue that lists the thickness can they please reply.
kind regards
Darryl
New Zealand.

Offline retired44

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Re: 2 pound gold "50th Anniversary of the end of WW2"
« Reply #1 on: November 19, 2009, 11:28:51 AM »
I have asked people on ebay selling them what the thickness is but none of them seem to know :(

translateltd

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Re: 2 pound gold "50th Anniversary of the end of WW2"
« Reply #2 on: November 19, 2009, 08:07:41 PM »
I've checked a couple of different pre-1997 nickel-brass £2 commemoratives and they're both something like 3.25 mm thick (measured by ruler and eye, not calipers).  For the gold to weigh the same as the base-metal coin I would assume it would have to be rather thinner than that.

Still not the full solution but I hope this helps a little, at least :-)




Offline retired44

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Re: 2 pound gold "50th Anniversary of the end of WW2"
« Reply #3 on: November 19, 2009, 10:31:39 PM »
thanks translateltd and Phil. 

Yes I emailed the mint two days ago and remember reading at the time about how many emails they are getting.

I am amazed at how a simple specification for a fairly recent modern coin can be so hard to obtain.

I am going to buy the nickel brass version at the end of my search to remind me of my time spend investigating.

cheers
Darryl

Offline Figleaf

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Re: 2 pound gold "50th Anniversary of the end of WW2"
« Reply #4 on: November 20, 2009, 01:27:24 AM »
This may be too simplistic or outright wrong, but let's see how far this solves the question. I used the data above and assumed that the gold (density 19.2) was alloyed with silver (density 10.5)

If the coin weighs 15,98 gram and it is 91.7% gold, it follows that it contains 0.7631927 gram gold and 0.1263238 gram silver (numbers rounded). Dividing these weights by their respective density and adding them up gives a total volume of 0.889535267857. Dividing this number by the result of 100/7 (∏) x 28,4^2 (surface) yields a theoretical average thickness of 0.772 mm. The thickness would be average, because the coins' surface is obviously not flat.

Peter
An unidentified coin is a piece of metal. An identified coin is a piece of history.

translateltd

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Re: 2 pound gold "50th Anniversary of the end of WW2"
« Reply #5 on: November 20, 2009, 02:27:05 AM »
This may be too simplistic or outright wrong, but let's see how far this solves the question. I used the data above and assumed that the gold (density 19.2) was alloyed with silver (density 10.5)

If the coin weighs 15,98 gram and it is 91.7% gold, it follows that it contains 0.7631927 gram gold and 0.1263238 gram silver (numbers rounded). Dividing these weights by their respective density and adding them up gives a total volume of 0.889535267857. Dividing this number by the result of 100/7 (∏) x 28,4^2 (surface) yields a theoretical average thickness of 0.772 mm. The thickness would be average, because the coins' surface is obviously not flat.

Peter

There is a problem with this maths, as pi is approximated to 22/7, not 100/7.  Substituting 22/7 into the equation gives a result that is even smaller, so I think this formula is flawed.  We need to establish what the unit for the volume is (it should be cubic mm for the formula to work) - but does the division of a weight in grams by a value for relative density produce a volume in mm3?

I have to nip out for a meeting so can't spend more time on this at the moment!


translateltd

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Re: 2 pound gold "50th Anniversary of the end of WW2"
« Reply #6 on: November 20, 2009, 02:40:17 AM »
My meeting can wait a moment.  To work out the volume of a cylinder we need the square of the radius, not the diameter.  Since 22/7 is accurate to only two decimal places as an approximation of pi, there's no point in taking any of the other figures to more than two places either:

Peter's formula (with the assumption regarding units as above) thus becomes:

.89 / (22/7 * 14.2^2)

= .89 / (633.73)

= .0014 ... (or, to two significant places, zero!)

Back to our units again: is there any way that this result can be out by three places, giving a value of 1.4mm, or is the logic amiss in the first place?


Offline paddyirish

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Re: 2 pound gold "50th Anniversary of the end of WW2"
« Reply #7 on: November 23, 2009, 02:40:26 PM »
My meeting can wait a moment.  To work out the volume of a cylinder we need the square of the radius, not the diameter.  Since 22/7 is accurate to only two decimal places as an approximation of pi, there's no point in taking any of the other figures to more than two places either:

Peter's formula (with the assumption regarding units as above) thus becomes:

.89 / (22/7 * 14.2^2)

= .89 / (633.73)

= .0014 ... (or, to two significant places, zero!)

Back to our units again: is there any way that this result can be out by three places, giving a value of 1.4mm, or is the logic amiss in the first place?



  I always found it was best to return to SI units (kg for mass and metres cubed for volume) for doing these and this was my result

Density of gold (provided)   19320           kg/m3   
Density of Silver (provided)   10490           kg/m3   
Proportion of Gold (provided)   0.917      
Prportion of silver (provided)   0.083      
Density of coin                   18587.11           kg/m3   (sum of densities * proportions)
Mass of coin (provided)           0.01598           kg   
Volume of coin                    8.59736E-07   m3   (mass/density)
radius of coin (provided)   0.0142           m   
face area of coin                    0.000633726   m2   volume/pi r squared
height of coin                   0.001356637   m   volume/face area
thickness of coin                   1.356636746   mm   

Offline Figleaf

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Re: 2 pound gold "50th Anniversary of the end of WW2"
« Reply #8 on: November 23, 2009, 02:57:12 PM »
Thanks Paddyirish. I think we now have consensus that the correct answer is 1.4 mm.

Peter
An unidentified coin is a piece of metal. An identified coin is a piece of history.

Offline UK Decimal +

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Re: 2 pound gold "50th Anniversary of the end of WW2"
« Reply #9 on: November 23, 2009, 03:20:41 PM »
That's the figure that I kept coming up with, but it still seems pretty thin (although of course it's an average figure and the edge and relief would be thicker).

The base metal version is a thick coin at (about) 3.5mm.

Does anyone have any recent UK coins in rare metals that can be compared with their base metal counterparts?

Bill.
Ilford, Essex, near London, England.

People look for problems and complain.   Engineers find solutions but people still complain.

Offline retired44

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Re: 2 pound gold "50th Anniversary of the end of WW2"
« Reply #10 on: November 25, 2009, 07:52:35 AM »
I have now bought the gold coin.  It measures about the same as our NZ 50cent piece. approx 1.7mm

I am now waiting for one of the brass nickel versions to come in the post.  I bought it on trademe today.

cheers
Darryl

Offline tonyclayton

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Re: 2 pound gold "50th Anniversary of the end of WW2"
« Reply #11 on: December 06, 2009, 05:18:10 PM »
The weights of the gold and nickel-brass versions of the coins are the same, and defined by law.
Thus it is clear that the gold version will be about half the thickness of a nickel-brass version.
(Please note: ABOUT)

In the above calculations it is assumed that density of a gold-silver alloys is a simple arithmetical claculation.  Sadly alloys are not as simple as that.
You cannot assume that the density of a brass (copper-zinc) is in proportion to the copper-zinc content either. This table makes the point

Cu   Zn   Actual   Calculated
100   0   8.96           8.96
70   30   8.52           8.41
65   35   8.45           8.32
63   37   8.40           8.29
60   40   8.38           8.23
0   100   7.14           7.14