My world has a prospecting vehicle—a large robot, actually, that operates in 850°F volcanic regions. To bring it into a hangar for maintenance, I have decided that the material will be damaged by cooling too rapidly since dissimilar metals and alloys are used. The robot is 24 feet tall and proportioned like a portly human, with a pilot occupying a hollow cockpit in the abdominal cavity.

The materials are too varied to detail, so let's assume that alloys in question which are used to construct the hull and appendages have a variance of thermal expansion coefficients α equal to 7x$10^{-6}/{°F}$. I hope those units are correct.

Assume there is some critical component where the largest and the smallest coefficient materials are bound together, so that the one contracting more quickly is in a band around the one compressing more slowly by that variance, in the temperature range noted.

What would be a believable rate of cooling a machine such as this might need to come in for maintenance without permanent deformation or breaking one of the materials? The bands would be around the arms or ankles or shoulders, probably, having diameters proportional to a human equivalent.

The materials are iron, copper, tin, nickel, zinc, or their alloys.


  • The materials operate properly at the full temperature range; either by expansion joints or alloy selection. It is the stress from the rate of linear change that is sought. Expansion design can not accommodate those stresses.
  • FWIW, the plot has an injury due to material failure under quenching. I do not know if the thing could break in 5 minutes or after 5 days, honestly. Just need to put a reasonable number on the build-up to the incident.
  • 1
    $\begingroup$ Traveling musicians have this problem -- instrument can be damaged if warmed/cooled/humidified/dehumidified too rapidly. I know a professional violinist who lets their instrument acclimate over 4-6 hours in each new location before playing. First half with the case closed (temperature), second half with the case open (humidity). Then it needs to be re-tuned, then sit another half hour to check that acclimatization is done and it;s going to stay in tune. That's for delicate glued wood joints with tight tolerances and temperature swings of up to 40C. $\endgroup$
    – user535733
    Aug 8 at 18:36
  • 1
    $\begingroup$ The denominator would be 'K or 'C not 'F and the stress depends on the hole pitch and shear strength with thermal mismatch shearing the weakest link. Then for plastic memory effects the accumulation of displacement and thermal cycles is another fracture interval. To prevent damage the insulation must increase the thermal constant so that both materials cool at the same rate. Thus minimal stresses by insulation. $\endgroup$ Aug 8 at 20:00

4 Answers 4


53 minutes.

At least, that's my ballpark value. Allow me to walk you through the calculation.

Newton's Law of Cooling:

Newton's law of cooling states the temperature as a function of time for a body cooling from an initial temperature $T_i$ to a lower, final temperature $T_f$ is given as: $$T(t) = T_i + (T_f - T_i) e^{-kt}$$

The coefficient $k$ (units $s^{-1}$)controls the rate of cooling. Two objects made of different materials, or different shape, will have different cooling rates. For example, if $k_1 > k_2$, then the graph of temperature vs time, $T_1(t)$ and $T_2(t)$ of two different objects will look like: cooling curves

The difference in temperature is the temperature gap: $T_g(t) = T_1(t) - T_2(t)$. It will have a maximum somewhere between the starting and ending times. This temperature difference will drive the additional stress two different materials will face when they are both constrained, not the expansion coefficients. We therefore must find the time when the temperature difference is a maximum. Since this isn't Math-SE, ill skip the derivation and state that the maximum temperature difference, $t_m$, occurs at:

$$t_m = \frac{\ln{k_1 / k_2}}{k_1 - k_2}$$

Interestingly, it doesn't depend on the starting or ending temperatures, only the relative cooling rates. To calculate $t_m$ we will need to make some choices of materials and at least a little bit of shape. I get 53 minutes by choosing copper for material-1 and alumina (a ceramic, not aluminum) for material 2, since they have vastly different cooling rates. $k$ is calculated as:

$$k = \frac{h}{A C \rho}$$

Where $h$ is the thermal conductivity (W/m/K), $A$ is the surface area ($m^2$) the cooling takes place over, $C$ is the specific heat (J/kg/K), and $\rho$ the density (kg/m^3). Using an arbitrary value of $A = 0.10\text{m}^2$, and the internet for the rest of the values (I won't be citing them -- they are well defined values):

$$\begin{array}{|c|c|c|c|c|} \hline \text{Material} & h & C & \rho & A & k\\ \hline \text{Copper} & 400 & 380 & 8920 & 0.10 & \text{1.18E-3}\\ \hline \text{alumina} & 10 & 850 & 3950 & 0.10 & \text{2.98E-5} \\ \hline \end{array}$$

Using those values, $t_m = 3200\text{s}$ or 53.3 minutes.


  1. Why did you pick $A$ to be 0.10 square metes? It was an arbitrary choice. You can make $t_m$ any number you want by picking the correct materials and surface area. In general, $t_m$ will be directly proportional to surface area. So if 53 minutes is too short, have your story focus on some component of larger surface area of two different materials.
  2. Wait a minute.. these two materials are supposed to be in thermal contact with each other. Doesn't that change the calculation? Yes, yes it does. But im not getting paid to write this, and in general thermal engineering is hard. This is the best bang for your wild-ass-guess buck.

It's not just a matter of cooling rate: if two joined materials have highly mismatched expansion coefficient, once their expansion/contraction start it will induce stress in them and after a certain point the weakest one will break. That's why designing electronic to operate in deep space is tricky: because we realize it at our normal temperatures and then they have to work 200 K colder.

Sure, the cooling rate induces additional stress by adding an uneven distribution of temperature across the material, but too cold is too cold no matter how slowly you reach it.

  • $\begingroup$ Can I assume for the purpose that they have engineered sufficiently close expansion coeffeicients over the temperature range but the linear rate itself is mismatched by seven? That is what I was trying to write. I suppose that unit would be an integral of the coefficient? $\endgroup$
    – Vogon Poet
    Aug 8 at 18:38
  • $\begingroup$ @VogonPoet: The expansion coefficient is the "linear rate". $\endgroup$
    – AlexP
    Aug 8 at 18:58
  • $\begingroup$ OK. I see that the difference is governed by the amount of time each material reaches its final dimension, which is actually a function of rate of change of temperature. That is the variance I wish to be 7 then, for whatever design reason that causes the two materials to cool at a different rate. I suppose the denominator would be /°Fs then. What is the opposite of quenching? Slow cooling? $\endgroup$
    – Vogon Poet
    Aug 8 at 19:12
  • $\begingroup$ Most coefficients of expansion for metals are in the 10-25 um / m / K, and at the other end ceramics are 2-6 um / m / K. Roughly a 5x difference. However, if you look at cooling rates, those are x200 difference (see my post). Therefore, differential cooling is going to drive stress more than the absolute difference in size at the final temperature. $\endgroup$
    – cms
    Aug 8 at 21:52

Global Cooling Rates Won't Save You

It doesn't matter how quickly or slowly you reach a temperature: the mismatch between coefficients of thermal expansion will get you either way! That is, if your material doesn't shatter because you cool it too fast (due to internal stresses).

But What Can I Do?

Assuming there is no design path around this and destruction is going to happen due to reasons*, there are a few options to account for this:

  • accept that this part will break every time maintenance occurs and replace it every time the robot comes in. Could be costly!
  • take into account cooling rates such that you can do your maintenance before one material cools down and breaks another. This is dependent on the difference between specific heats of the materials and presents hazards for anyone placing their arms inside.
  • keep the 'outside' part hot to prevent damage to the inside part. This is not ideal, because it is an active hazard!
  • design the robot's joints to quickly disassemble when coming in for maintenance. Have some heat resistant arms grab and separate the thermally mismatched components. Heat these up as part of the re-assembly process.
  • $\begingroup$ Not my specialty, but does this imply that the robot must have been manufactured in a facility kept at or near its "operating" temperature and somehow maintained at that temperature continuously as it was transported subsequently, deployed etc. Or am I misunderstanding? $\endgroup$ Aug 8 at 22:34
  • $\begingroup$ @KerrAvon2055 Yes, some of these solutions would imply the robot is assembled at an elevated temperature, or at least with at-temperature parts. This is actually not uncommon in certain applications: sometimes manufacturers assemble things hot so they 'clamp down' on each other after cooling. (Cams on drive shafts, metal bands on barrels, and wagon-rims come to mind...) $\endgroup$
    – PipperChip
    Aug 8 at 22:43
  • $\begingroup$ These solutions to prevent damage are insightful but the time to failure was what the story needed, as this robot was brought inside without proper procedure (it was stolen by pirates). It breaks, they get hurt. I was trying to ballpark that time to failure to create my scene. $\endgroup$
    – Vogon Poet
    Aug 9 at 4:32

You want a giant Annealing Oven. [1]

Glassblowers use an Annealing Oven to cool their pieces.

The oven is kept at 960°F while the piece is worked, then the artist picks up their piece with Kevlar gloves and transfers it to the oven. The oven is then cooled down to room temperature over a period of 14 hours.

The mechanics of this can be easily adapted to your needs. You could have an "airlock" that your robot walks into, which then (over whatever time period your story needs) is slowly cooled down from 850°F to the internal temperature of the maintenance hanger.

It would probably be more efficient to have a succession of 'airlocks', each room deeper into the complex and more insulated from the outside temperature. So the robot walks into the first, and waits a couple hours to cool a bit, and then walks into the next chamber, waits a couple more hours, etc.

[1]This will likely require that your reasons for needing to be cooled slowly is adjusted from different rates of cooling of different materials, to something like the cooling system must be be slowly brought offline, or the special alloy used as the outer shell will crack and shatter if cooled too quickly.

  • 1
    $\begingroup$ I do not think this solves the thermal-expansion-mismatch issue. It seems annealing ovens are more to prevent the piece from shattering due to internal stresses, not thermal mismatch. ("Annealing" is a catch all term for removing internal stresses via heat.) $\endgroup$
    – PipperChip
    Aug 8 at 20:35
  • $\begingroup$ This is what prompted the scenario however. There's a trick you do with soda bottles: Heat a spot with a butane torch, put the bottle in a microwave. The hot glass becomes molten while the cooler glass is unaffected. During cooling the bottle breaks. This robot was stolen by pirates, brought inside their enclave, and it suffered this sort of casualty injuring the crew. I assumed different rates of thermal contraction cause the internal stresses. Bottom line: how long until it pops? $\endgroup$
    – Vogon Poet
    Aug 9 at 4:27
  • $\begingroup$ @PipperChip Yeah, with the other answers stating the issues with thermal-mismatch, this answer is a little bit of frame challenge. The overall picture is the same (need a way to slow the cooling of the robot), but the mechanics of why? need to be adjusted. I'll add that to the answer. $\endgroup$ Aug 9 at 14:09

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