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I was just reading this article on how Curiosity's aluminum wheels are getting pretty dinged up. Usually titanium alloy is considered an upgrade from aluminum alloys in most applications, from aerospace to bikes to sporks. The "usual" downsides are that it is expensive and difficult to machine. However, those seem like less important factors when you're NASA and the plan is to send an object to another planet and drive over sharp rocks.

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Aluminum is less dense than titanium. For the same mass, the 0.75mm thick aluminum would have to be replaced with 0.45mm thick titanium. Although the titanium sheet would be stronger in tension, it would probably be more susceptible to tearing from a point load like driving over a sharp rock.

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    $\begingroup$ Titanium is brittle. I suggest adding relevant data from MatWeb to your answer. $\endgroup$ – Deer Hunter Feb 5 '14 at 4:03
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    $\begingroup$ Ti alloy is 1.57 more dense than Al, but has 1.75 greater yield strength and 1.66 times stronger in shear (tikore.com/titanium_facts.htm). So wouldn't that mean it would be less susceptible to tearing even considering reduced thickness from its higher density? I don't have numbers to back it up, but I'm suspecting it might be something about brittleness as @DeerHunter suggests -- JPL probably would rather have Al wheels full of holes than a fractured Ti wheel. Plus the cost and inconvenience in working with Ti. $\endgroup$ – joseph_morris Feb 6 '14 at 20:26
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    $\begingroup$ @joseph_morris: puncturing a membrane gets much easier as the membrane gets thinner. I suspect the brittleness may be involved as well. There are many factors that go into material selection. $\endgroup$ – Ross Millikan Feb 6 '14 at 20:41
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    $\begingroup$ Alloys of both metals can be much stronger (or much better wrt. to some other desired property) than the 100% pure material (and still be 99% - thus no concern mass-wise). Which alloys, if any, are we talking about here? In particular, which alloy of aluminium was used for Curiosity's aluminium wheels? 6061-T6? $\endgroup$ – Peter Mortensen Aug 27 '16 at 20:38

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