Creating a Custom Ring: Step-by-Step Guide

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The 6 metals in equal amounts. Through the melting method in a foundry. Not the powder method. Or will this end up not mixing and having cracks due to not adhering to the Hume-Rothery rules? And if not an alloy then what will happen once all 6 are put in a foundry and melted together? Or if in 6 separate foundries and then poured all in together in a 7th foundry?
Purpose is to make a ring.
 
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Maybe put each in separate foundries, each foundry heated up to either the metal's melting point or just all at whichever metal has the highest, and then once all metals are melted, pour all 6 together in an empty 7th foundry
 
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Tungsten is the highest of the six, so we'll go with that temperature (Tungsten's melting point). It's going to be higher than Osmium and Silver anyway. I just realized that Osmium, Chromium, and Vanadium may be toxic so I'm replacing it with Titanium, Molybdenum, and Tantalum instead.

I think having electricity run through the Tungsten in a foundry (since it's the one that's going to be the most difficult to melt) until it melt, and then pour in the other 5 melted metals in, might make it work. Because otherwise the silver would boil off before Tungsten even melts if all are put in one foundry. I haven't figured out the exact way to do it but the question is more of, will these 6 metals actually mix and adhere to the Hume-Rothery rules?
 
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Can you melt and mix the two metals with the highest melting points and create an alloy with a lower melting point than either metal, then melt and add the metal with the third highest melting point, and repeat down to the metal with the lowest melting point?

I don't know, so I'm asking. I do know that tin-lead solder melts at a lower temperature than either lead or tin, and was wondering if this was a general phenomenon. Also if it works for more than two elements.
 
Vanadium 50 said:
I don't see how you can be well above the boiling point of one metal and create an alloy that way.

Wait, you're right, Titanium and Silver would boil off. I'll have to look for an alternative metal
 
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Okay so now the line up is:

Tungsten-Iridium-Tantalum-Molybdenum-Rhodium-Niobium

Would melting together equal parts of Tungsten, Iridium, Tantalum, Molybdenum, Rhodium, Niobium create a substitutional solid solution or become intermetallic?

I think intermetallic means, in this case, they won't fully mix so you'll see layers and streaks of some of the metals not actually mixed together with the rest. The solid solution would be them mixed like an actual alloy or something.
 
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metallurgy said:
Okay so now the line up is:

Tungsten-Iridium-Tantalum-Molybdenum-Rhodium-Niobium
Did you mention your end application for this yet? (If so, apologies that I missed it) Or is it some sort of art project?
 
A ring for a wealthy acquaintance
 
metallurgy said:
Would melting together equal parts of Tungsten, Iridium, Tantalum, Molybdenum, Rhodium, Niobium create a substitutional solid solution or become intermetallic?
Senary alloys are certainly complex, and the solubilities of the various elements vary in the others. One may wish to contact a company that specializes in refractory and noble metal alloys, e.g., Johnson-Matthey, H. Cross, Plansee, H. C. Starck, . . . . One can search for various refractory alloys and which offer different combinations.

Is one looking for equal amounts in terms of atomic fraction or mass fraction?

Usually, in binary systems, one will find a base metal, e.g., Ir, and minor element, e.g., Rh, e.g., Ir-10Rh. One can find examples of Ir-Rh (50/50), Ta-W (1 - 10% W), and Mo-Nb. Probably would want to do three binary alloys in powder form, mix then sinter.
 
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Equal amounts in proportion by mass (%), so 16.6% each, ideally.

Would you recommend an arc vacuum melting method for this? Just to get the metals mixed together and hopefully with no insolubility
 
metallurgy said:
Equal amounts in proportion by mass (%), so 16.6% each, ideally.

Would you recommend an arc vacuum melting method for this? Just to get the metals mixed together and hopefully with no insolubility
Assuming the ring has a mass of 12 to 15 grams, based on a comparable size of a gold ring, one would need 2 to 2.5 grams of each metal. One would probably start with 3 binary alloys, of which I gave example above, although I'm rethinking a couple based on melting point, atomic radius and density.

What is the objection to powder metallurgy?

One group has found microwave sintering to be quite effective in forming alloys of refractory elements.
https://www.mri.psu.edu/sites/default/files/file_attach/198.pdf
 
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I'm still looking into this, but it appears that one could arc-cast an alloy. Consider using three binary starting alloys, Ir-Rh, W-Mo and Ta-Nb, each with 0.5-0.5 by mass, so each binary is approximately 2 light atoms to 1 heavy atom. I'm not familiar with casting of rings however.

One could contact H. Cross Company and see if they could provide the alloys and cast the ring.
https://www.hcrosscompany.com/sitemap/
Otherwise one could contact Ames Laboratory, since they also have the capability to manufacture special alloys. https://www.ameslab.gov/mpc/metal-alloy-services
 
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Astronuc said:
I'm still looking into this, but it appears that one could arc-cast an alloy. Consider using three binary starting alloys, Ir-Rh, W-Mo and Ta-Nb, each with 0.5-0.5 by mass, so each binary is approximately 2 light atoms to 1 heavy atom. I'm not familiar with casting of rings however.

One could contact H. Cross Company and see if they could provide the alloys and cast the ring.
https://www.hcrosscompany.com/sitemap/
Otherwise one could contact Ames Laboratory, since they also have the capability to manufacture special alloys. https://www.ameslab.gov/mpc/metal-alloy-services

Sorry for the late reply, but thank you so much for the info! This was very helpful!

Oh, I see, so it would be best to melt those pairs together first ( Ir-Rh, W-Mo and Ta-Nb), before melting all the pairs together in a vacuum arc furnace?
 
metallurgy said:
Oh, I see, so it would be best to melt those pairs together first ( Ir-Rh, W-Mo and Ta-Nb), before melting all the pairs together in a vacuum arc furnace?
I'd recommend purchasing the alloys and having them mixed, and possibly cast. One can enquire with a company that specializes in these alloys, e.g., H. Cross Co. Arc melting refractory alloys is normally done in a vacuum, and the one has to exceed the melting temperature of the element with the highest melting point, e.g., that of tungsten, 3410°C.

I've seen vacuum arc melting (with consumable electrode) of Zr-alloys and stainless steels of metric tons, but not tungsten alloys. Usually one forms an ingot in a water-cooled copper crucible.
 
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Astronuc said:
Arc melting refractory alloys is normally done in a vacuum, and the one has to exceed the melting temperature of the element with the highest melting point, e.g., that of tungsten, 3410°C.
Where several alloy powders are fused together, the maximum temperature required will be the melting point of the highest temperature alloy. That will be lower than the melting point of the highest temperature elemental metal, tungsten in this case.
Lowering the required temperature can be achieved by specifying a eutectic alloy of tungsten with some other included element.
But then, the highest temperature eutectic alloy or element does not actually have to be melted. It can remain as a crystalline powder. While at high temperatures it will dissolve in, and diffuse through, the surrounding melt.

As the mixture cools, different metals and alloys will precipitate out at different temperatures, until the final minimum temperature eutectic alloy freezes. As different metal structures and alloys are going to crystallise out during cooling, it is probably unnecessary to melt it all together at the start of the process.
 
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With such composition, you might end getting a High Entropy Alloy (HEA). There is some published work on refractory HEAs that use some of the elements you mentioned.
 
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